Thinking Geometrically: A Survey of Geometries 1939512085, 9781939512086, 9781614446194, 1614446199

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Thinking Geometrically A Survey of Geometries

c 2015 by ⃝ The Mathematical Association of America (Incorporated) Library of Congress Control Number: 2015936100 Print ISBN: 978-1-93951-208-6 Electronic ISBN: 978-1-61444-619-4 Printed in the United States of America Current Printing (last digit): 10 9 8 7 6 5 4 3 2 1

Thinking Geometrically A Survey of Geometries

Thomas Q. Sibley St. John’s University

Published and distributed by The Mathematical Association of America

Council on Publications and Communications Jennifer J. Quinn, Chair Committee on Books Fernando Gouvˆea, Chair MAA Textbooks Editorial Board Stanley E. Seltzer, Editor Matthias Beck Richard E. Bedient Otto Bretscher Heather Ann Dye Charles R. Hampton Suzanne Lynne Larson John Lorch Susan F. Pustejovsky

MAA TEXTBOOKS Bridge to Abstract Mathematics, Ralph W. Oberste-Vorth, Aristides Mouzakitis, and Bonita A. Lawrence Calculus Deconstructed: A Second Course in First-Year Calculus, Zbigniew H. Nitecki Calculus for the Life Sciences: A Modeling Approach, James L. Cornette and Ralph A. Ackerman Combinatorics: A Guided Tour, David R. Mazur Combinatorics: A Problem Oriented Approach, Daniel A. Marcus Complex Numbers and Geometry, Liang-shin Hahn A Course in Mathematical Modeling, Douglas Mooney and Randall Swift Cryptological Mathematics, Robert Edward Lewand Differential Geometry and its Applications, John Oprea Distilling Ideas: An Introduction to Mathematical Thinking, Brian P. Katz and Michael Starbird Elementary Cryptanalysis, Abraham Sinkov Elementary Mathematical Models, Dan Kalman An Episodic History of Mathematics: Mathematical Culture Through Problem Solving, Steven G. Krantz Essentials of Mathematics, Margie Hale Field Theory and its Classical Problems, Charles Hadlock Fourier Series, Rajendra Bhatia Game Theory and Strategy, Philip D. Straffin Geometry Illuminated: An Illustrated Introduction to Euclidean and Hyperbolic Plane Geometry, Matthew Harvey Geometry Revisited, H. S. M. Coxeter and S. L. Greitzer Graph Theory: A Problem Oriented Approach, Daniel Marcus An Invitation to Real Analysis, Luis F. Moreno Knot Theory, Charles Livingston Learning Modern Algebra: From Early Attempts to Prove Fermat’s Last Theorem, Al Cuoco and Joseph J. Rotman The Lebesgue Integral for Undergraduates, William Johnston Lie Groups: A Problem-Oriented Introduction via Matrix Groups, Harriet Pollatsek Mathematical Connections: A Companion for Teachers and Others, Al Cuoco Mathematical Interest Theory, Second Edition, Leslie Jane Federer Vaaler and James W. Daniel Mathematical Modeling in the Environment, Charles Hadlock Mathematics for Business Decisions Part 1: Probability and Simulation (electronic textbook), Richard B. Thompson and Christopher G. Lamoureux Mathematics for Business Decisions Part 2: Calculus and Optimization (electronic textbook), Richard B. Thompson and Christopher G. Lamoureux Mathematics for Secondary School Teachers, Elizabeth G. Bremigan, Ralph J. Bremigan, and John D. Lorch The Mathematics of Choice, Ivan Niven The Mathematics of Games and Gambling, Edward Packel Math Through the Ages, William Berlinghoff and Fernando Gouvea Noncommutative Rings, I. N. Herstein

Non-Euclidean Geometry, H. S. M. Coxeter Number Theory Through Inquiry, David C. Marshall, Edward Odell, and Michael Starbird Ordinary Differential Equations: from Calculus to Dynamical Systems, V. W. Noonburg A Primer of Real Functions, Ralph P. Boas A Radical Approach to Lebesgue’s Theory of Integration, David M. Bressoud A Radical Approach to Real Analysis, 2nd edition, David M. Bressoud Real Infinite Series, Daniel D. Bonar and Michael Khoury, Jr. Thinking Geometrically: A Survey of Geometries, Thomas Q. Sibley Topology Now!, Robert Messer and Philip Straffin Understanding our Quantitative World, Janet Andersen and Todd Swanson

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Contents Preface

xv

1 Euclidean Geometry 1.1 Overview and History........................................................................... 1.1.1 The Pythagoreans and Zeno .......................................................... 1.1.2 Plato and Aristotle ...................................................................... 1.1.3 Exercises for Section 1.1 .............................................................. 1.2 Euclid’s Approach to Geometry I: Congruence and Constructions................... 1.2.1 Congruence............................................................................... 1.2.2 Constructions ............................................................................ 1.2.3 Equality of Measure .................................................................... 1.2.4 The Greek Legacy....................................................................... 1.2.5 Exercises for Section 1.2 .............................................................. 1.2.6 Archimedes............................................................................... 1.3 Euclid’s Approach II: Parallel Lines ......................................................... 1.3.1 Exercises for Section 1.3 .............................................................. 1.4 Similar Figures.................................................................................... 1.4.1 Exercises for Section 1.4 .............................................................. 1.5 Three-Dimensional Geometry................................................................. 1.5.1 Polyhedra.................................................................................. 1.5.2 Geodesic Domes ........................................................................ 1.5.3 The Geometry of the Sphere.......................................................... 1.5.4 Exercises for Section 1.5 .............................................................. 1.5.5 Buckminster Fuller ..................................................................... 1.5.6 Projects for Chapter 1.................................................................. 1.5.7 Suggested Readings ....................................................................

1 1 2 4 5 10 11 12 15 16 17 25 26 30 33 37 42 42 46 48 51 59 59 65

2 Axiomatic Systems 2.1 From Euclid to Modern Axiomatics ......................................................... 2.1.1 Overview and History.................................................................. 2.1.2 Axiomatic Systems ..................................................................... 2.1.3 A Simplified Axiomatic System ..................................................... 2.1.4 Exercises for Section 2.1 ..............................................................

67 67 67 68 71 73

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2.2 Axiomatic Systems for Euclidean Geometry .............................................. 2.2.1 SMSG Postulates........................................................................ 2.2.2 Hilbert’s Axioms ........................................................................ 2.2.3 Exercises for Section 2.2 .............................................................. 2.2.4 David Hilbert............................................................................. 2.3 Models and Metamathematics................................................................. 2.3.1 Exercises for Section 2.3 .............................................................. 2.3.2 Kurt G¨odel................................................................................ 2.3.3 Projects for Chapter 2.................................................................. 2.3.4 Suggested Readings ....................................................................

75 76 77 78 81 82 88 94 94 96

3 Analytic Geometry 3.1 Overview and History........................................................................... 3.1.1 The Analytic Model .................................................................... 3.1.2 Exercises for Section 3.1 .............................................................. 3.1.3 Ren´e Descartes .......................................................................... 3.2 Conics and Locus Problems ................................................................... 3.2.1 Exercises for Section 3.2 .............................................................. 3.2.2 Pierre de Fermat......................................................................... 3.3 Further Topics in Analytic Geometry........................................................ 3.3.1 Parametric Equations................................................................... 3.3.2 Polar Coordinates ....................................................................... 3.3.3 Barycentric Coordinates............................................................... 3.3.4 Other Analytic Geometries............................................................ 3.3.5 Exercises for Section 3.3 .............................................................. 3.4 Curves in Computer-Aided Design .......................................................... 3.4.1 Exercises for Section 3.4 .............................................................. 3.5 Higher Dimensional Analytic Geometry.................................................... 3.5.1 Analytic Geometry in Rn .............................................................. 3.5.2 Regular Polytopes....................................................................... 3.5.3 Exercises for Section 3.5 .............................................................. 3.5.4 Gaspard Monge.......................................................................... 3.5.5 Projects for Chapter 3.................................................................. 3.5.6 Suggested Readings ....................................................................

97 98 98 100 104 104 110 114 114 114 116 118 120 121 126 131 133 133 137 140 145 145 149

4 Non-Euclidean Geometries 4.1 Overview and History........................................................................... 4.1.1 The Advent of Hyperbolic Geometry............................................... 4.1.2 Models of Hyperbolic Geometry .................................................... 4.1.3 Exercises for Section 4.1 .............................................................. 4.1.4 Carl Friedrich Gauss.................................................................... 4.2 Properties of Lines and Omega Triangles .................................................. 4.2.1 Omega Triangles ........................................................................ 4.2.2 Exercises for Section 4.2 .............................................................. 4.2.3 Nikolai Lobachevsky and J´anos Bolyai ............................................

151 152 153 155 158 160 161 164 167 169

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4.3 Saccheri Quadrilaterals and Triangles....................................................... 4.3.1 Exercises for Section 4.3 .............................................................. 4.3.2 Omar Khayyam.......................................................................... 4.3.3 Giovanni Girolamo Saccheri.......................................................... 4.4 Area, Distance, and Designs................................................................... 4.4.1 Exercises for Section 4.4 .............................................................. 4.5 Spherical and Single Elliptic Geometries................................................... 4.5.1 Exercises for Section 4.5 .............................................................. 4.5.2 Georg Friedrich Bernhard Riemann................................................. 4.5.3 Projects for Chapter 4.................................................................. 4.5.4 Suggested Readings ....................................................................

169 173 174 175 176 183 185 189 189 190 192

5 Transformational Geometry 5.1 Overview and History........................................................................... 5.1.1 Exercises for Section 5.1 .............................................................. 5.2 Isometries.......................................................................................... 5.2.1 Classifying Isometries ................................................................. 5.2.2 Congruence and Isometries........................................................... 5.2.3 Klein’s Definition of Geometry ...................................................... 5.2.4 Exercises for Section 5.2 .............................................................. 5.2.5 Felix Klein................................................................................ 5.3 Algebraic Representation of Transformations............................................. 5.3.1 Isometries................................................................................. 5.3.2 Exercises for Section 5.3 .............................................................. 5.4 Similarities and Affine Transformations .................................................... 5.4.1 Similarities................................................................................ 5.4.2 Affine Transformations ................................................................ 5.4.3 Iterated Function Systems............................................................. 5.4.4 Exercises for Section 5.4 .............................................................. 5.4.5 Sophus Lie................................................................................ 5.5 Transformations in Higher Dimensions; Computer-Aided Design ................... 5.5.1 Isometries of the Sphere............................................................... 5.5.2 Transformations in Three and More Dimensions................................ 5.5.3 Computer-Aided Design and Transformations ................................... 5.5.4 Exercises for Section 5.5 .............................................................. 5.6 Inversions and the Complex Plane ........................................................... 5.6.1 Exercises for Section 5.6 .............................................................. 5.6.2 Augustus M¨obius........................................................................ 5.6.3 Projects for Chapter 5.................................................................. 5.6.4 Suggested Readings ....................................................................

195 195 200 201 203 208 209 209 212 213 217 220 223 223 226 228 231 234 235 235 239 241 243 245 252 254 254 259

261 6 Symmetry 6.1 Overview and History........................................................................... 262 6.1.1 Exercises for Section 6.1 .............................................................. 264

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6.2 Finite Plane Symmetry Groups ............................................................... 6.2.1 Exercises for Section 6.2 .............................................................. 6.3 Symmetry in the Plane.......................................................................... 6.3.1 Frieze Patterns ........................................................................... 6.3.2 Wallpaper Patterns...................................................................... 6.3.3 Exercises for Section 6.3 .............................................................. 6.3.4 M. C. Escher ............................................................................. 6.4 Symmetries in Higher Dimensions........................................................... 6.4.1 Finite Three-Dimensional Symmetry Groups .................................... 6.4.2 The Crystallographic Groups......................................................... 6.4.3 General Finite Symmetry Groups ................................................... 6.4.4 Exercises for Section 6.4 .............................................................. 6.4.5 H. S. M. Coxeter......................................................................... 6.5 Symmetry in Science............................................................................ 6.5.1 Chemical Structure ..................................................................... 6.5.2 Quasicrystals............................................................................. 6.5.3 Symmetry and Relativity.............................................................. 6.5.4 Exercises for Section 6.5 .............................................................. 6.5.5 Marjorie Senechal....................................................................... 6.6 Fractals ............................................................................................. 6.6.1 Exercises for Section 6.6 .............................................................. 6.6.2 Benoit Mandelbrot...................................................................... 6.6.3 Projects for Chapter 6.................................................................. 6.6.4 Suggested Readings ....................................................................

267 270 272 273 276 282 288 288 288 290 291 291 293 294 294 296 297 299 304 304 310 312 312 314

7 Projective Geometry 7.1 Overview and History........................................................................... 7.1.1 Projective Intuitions .................................................................... 7.1.2 Exercises for Section 7.1 .............................................................. 7.2 Axiomatic Projective Geometry .............................................................. 7.2.1 Duality..................................................................................... 7.2.2 Perspectivities and Projectivities..................................................... 7.2.3 Exercises for Section 7.2 .............................................................. 7.2.4 Jean Victor Poncelet.................................................................... 7.3 Analytic Projective Geometry................................................................. 7.3.1 Cross Ratios.............................................................................. 7.3.2 Conics ..................................................................................... 7.3.3 Exercises for Section 7.3 .............................................................. 7.3.4 Julius Pl¨ucker ............................................................................ 7.4 Projective Transformations .................................................................... 7.4.1 Exercises for Section 7.4 .............................................................. 7.5 Subgeometries .................................................................................... 7.5.1 Hyperbolic Geometry as a Subgeometry .......................................... 7.5.2 Single Elliptic Geometry as a Subgeometry ...................................... 7.5.3 Affine and Euclidean Geometries as Subgeometries............................

317 318 319 322 327 331 332 333 336 337 339 340 342 346 346 351 354 355 358 358

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7.5.4 Exercises for Section 7.5 .............................................................. 7.5.5 Arthur Cayley............................................................................ 7.6 Projective Space.................................................................................. 7.6.1 Perspective and Computer-Aided Design.......................................... 7.6.2 Subgeometries of Projective Space ................................................. 7.6.3 Exercises for Section 7.6 .............................................................. 7.6.4 Projects for Chapter 7.................................................................. 7.6.5 Suggested Readings ....................................................................

359 361 361 362 367 368 370 372

8 Finite Geometries 8.1 Overview and History........................................................................... 8.1.1 Exercises for Section 8.1 .............................................................. 8.1.2 Leonhard Euler .......................................................................... 8.1.3 Rev. Thomas Kirkman ................................................................. 8.2 Affine and Projective Planes................................................................... 8.2.1 Affine Planes............................................................................. 8.2.2 Projective Planes. ....................................................................... 8.2.3 Exercises for Section 8.2 .............................................................. 8.3 Design Theory .................................................................................... 8.3.1 Error-correcting Codes ................................................................ 8.3.2 Exercises for Section 8.3 .............................................................. 8.3.3 Sir Ronald A. Fisher.................................................................... 8.4 Finite Analytic Geometry ...................................................................... 8.4.1 Finite Analytic Planes.................................................................. 8.4.2 Ovals in Finite Projective Planes .................................................... 8.4.3 Finite Analytic Spaces ................................................................. 8.4.4 Exercises for Section 8.4 .............................................................. 8.4.5 Projects for Chapter 8.................................................................. 8.4.6 Suggested Readings ....................................................................

373 373 376 377 378 378 378 382 384 385 389 391 393 394 395 398 399 401 404 407

9 Differential Geometry 9.1 Overview and History........................................................................... 9.1.1 Exercises for Section 9.1 .............................................................. 9.2 Curves and Curvature ........................................................................... 9.2.1 Exercise for Section 9.2................................................................ 9.2.2 Sir Isaac Newton ........................................................................ 9.3 Surfaces and Curvature ......................................................................... 9.3.1 Surfaces ................................................................................... 9.3.2 Curvature.................................................................................. 9.3.3 Surfaces of Revolution................................................................. 9.3.4 Exercises for Section 9.3 .............................................................. 9.4 Geodesics and the Geometry of Surfaces................................................... 9.4.1 Geodesics ................................................................................. 9.4.2 Geodesics on Surfaces of Revolution............................................... 9.4.3 Arc Length on Surfaces................................................................

409 409 411 413 420 423 424 424 426 428 429 432 432 435 436

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9.4.4 9.4.5 9.4.6 9.4.7 9.4.8 9.4.9

The Gauss-Bonnet Theorem......................................................... Higher Dimensions.................................................................... Exercises for Section 9.4............................................................. Albert Einstein ......................................................................... Projects Chapter 9 ..................................................................... Suggested Readings ...................................................................

439 439 441 444 445 446

10 Discrete Geometry 10.1 Overview and Explorations .................................................................... 10.1.1 Distances between Points ............................................................ 10.1.2 Triangulations .......................................................................... 10.1.3 The Art Gallery Problem............................................................. 10.1.4 Tilings .................................................................................... 10.1.5 Voronoi Diagrams ..................................................................... 10.1.6 Exercises for Section 10.1 ........................................................... 10.1.7 Paul Erd os............................................................................... ˝ 10.2 Points and Polygons ............................................................................. 10.2.1 Distances between Points ............................................................ 10.2.2 Triangulations .......................................................................... 10.2.3 The Art Gallery Problem............................................................. 10.2.4 Exercises for Section 10.2 ........................................................... 10.3 Tilings .............................................................................................. 10.3.1 Exercises for Section 10.3 ........................................................... 10.3.2 Branko Gr¨unbaum ..................................................................... 10.4 Voronoi Diagrams................................................................................ 10.4.1 Exercises for Section 10.4 ........................................................... 10.4.2 Projects for Chapter 10 ............................................................... 10.4.3 References...............................................................................

447 448 448 449 449 450 451 453 457 457 457 459 462 465 469 474 477 477 481 483 485

11 Epilogue 487 11.1 Topology ........................................................................................... 488 11.2 Henri Poincar´e .................................................................................... 489 A Definitions, Postulates, Common Notions, and Propositions from Book I of Euclid’s Elements A.1 Definitions ......................................................................................... A.2 The Postulates (Axioms) ....................................................................... A.3 Common Notions ................................................................................ A.4 The Propositions of Book I ....................................................................

491 491 492 492 492

B SMSG Axioms for Euclidean Geometry

497

C Hilbert’s Axioms for Euclidean Plane Geometry

499

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D Linear Algebra Summary D.1 Vectors.............................................................................................. D.2 Matrices ............................................................................................ D.3 Determinants...................................................................................... D.4 Properties of Matrices........................................................................... D.5 Eigenvalues and Eigenvectors.................................................................

503 503 504 505 506 506

E Multivariable Calculus Summary E.1 Vector Functions ................................................................................. E.2 Surfaces ............................................................................................ E.3 Partial Derivatives................................................................................

509 509 509 510

F Elements of Proofs F.1 Direct Proofs ...................................................................................... F.2 Proofs by Contradiction......................................................................... F.3 Induction Proofs.................................................................................. F.4 Other Remarks on Proofs.......................................................................

511 511 512 512 513

Answers to Selected Exercises

515

Acknowledgements

549

Index

551

Preface I begin to understand that while logic is a most excellent guide in governing our reason, it does not, as regards stimulation to discovery, compare with the power of sharp distinction which belongs to geometry. — Galileo Galilei (1564–1642)

Geometry combines visual delights and powerful abstractions, concrete intuitions and general theories, historical perspective and contemporary applications, and surprising insights and satisfying certainty. In this textbook, I try to weave together these facets of geometry. I also want to convey the multiple connections that topics in geometry have with each other and that geometry has with other areas of mathematics. The connections link chapters together without sacrificing the survey nature of the whole text. Geometric thinking fuses reasoning and intuition in a characteristic fashion. The enduring appeal and importance of geometry stem from this synthesis. Mathematical insight is as hard for mathematics students to develop as is the skill of proving theorems. Geometry is an ideal subject for developing both, leading to deeper understanding. However, geometry texts for mathematics majors often emphasize proofs over visualization, whereas some texts for mathematics education majors focus on intuition instead. This book strives to build both and so geometrical thinking throughout the text, the exercises, and the projects. The dynamic software now available provides one valuable way for students to build their intuition and prepare them for proofs. Thus exercises benefiting from technology join hands-on explorations, proofs, and other types of problems. This book builds on the momentum of the NCTM Standards, the calculus reform movement, the Common Core State Standards (CCSS), and the ongoing discussion of how to help students internalize mathematical concepts and thinking. (We abbreviate the National Council of Teachers of Mathematics throughout as NCTM.) There are two natural audiences for a geometry course at the college level—mathematics majors and future secondary mathematics teachers. Of course, the two audiences overlap considerably. In many states, however, requirements for secondary mathematics education majors can differ substantially from those for traditional mathematics majors. This book seeks to serve both audiences, partially by using a survey format so that instructors can choose among topics. In addition, those sections matching CCSS expectations and NCTM recommendations for teacher preparation assume less mathematical sophistication, although they have plenty of problems

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and projects to challenge more advanced undergraduates. In particular, in those sections vital for teacher preparation, earlier exercises require less sophistication than later ones. A later subsection of this preface discusses possible course choices. It has been a treat to revisit the material in this text, re-envision the problems and projects, and add new ones. I have also enjoyed adding the material in Chapters 9 and 10. In the process I realized again how geometric my thinking is and how much I enjoy sharing the elegance and excitement I experience in geometry. I hope that some of my enthusiasm shows through.

Geometric Intuition Everyday speech equates intuitive with easy and obvious. However, psychology research confirms what mathematicians have always understood: people build their own intuitions through reflection on their experiences. My students often describe this process as learning to think in a new geometry. As Galileo’s quote introducing this preface suggests, geometry has for centuries been an ideal place for developing mathematical intuition. Since the advent of analytic geometry (at the end of Galileo’s life) mathematicians have repeatedly turned geometric insights into algebraic formulations. The applicability, efficiency, simplicity, and power of algebra have reasonably led educators from middle school through graduate school to emphasize algebraic representations. In my view, the success of algebra has so focused the curriculum that students’ geometrical thinking often lags far behind. My text tries to correct that imbalance without neglecting the power of algebra. The NCTM calls for high school students to develop geometric intuition and understanding. Similarly the CCSS expects students to build geometrical thinking in a variety of ways. Throughout this book I seek to help students develop their geometrical intuition. Visualization is an important part of this effort, and the hundreds of figures in the book provide an obvious means to this end. Many of the more than 750 exercises ask students to draw or create their own figures and models. In addition, I have included many exercises and projects requesting students to explore and conjecture new ideas, as well as explain or prove unusual properties. I advocate having students use dynamic geometry software, such as Geometer’s Sketchpad or Geogebra, to explore geometric ideas. While some texts explicitly incorporate such software, I don’t want to tie my book to one program. However, in my teaching experience students gain different and often more insight working with physical models than from manipulating computer models. So I strongly encourage instructors to give students physical models to use for as many topics as the class time, budget, and their creativity allows. I provide several suggestions in the text and in the projects. I hope the text’s explanations are clear and provide new insights, but I know that students need to reflect on the text and the exercises. I also hope that students find the many non-routine problems and projects challenging, but solvable with effort; such challenges enrich intuition.

The Role of Proofs Since Euclid, over two thousand years ago, proofs have had a central place in mathematical thought. Non-mathematicians often think the value of mathematics is restricted to its amazing ability to calculate “answers.” Certainly, many applications of mathematics rely on

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computational power. However, people applying mathematics want to know more than that there is an answer—after all, an astrologer gives answers. We want to know that the answers are valid. While confidence in much of science depends on experiments, it also often depends on mathematical models. The models make explicit assumptions about how some aspect of the world behaves and recasts them in mathematical terms. So applications of mathematics require that someone—a mathematician—actually prove the results that others use. However, the need for someone to have proved any given result doesn’t lead to a need for every mathematics student to prove every result. Most people acknowledge the value of honing students’ ability to reason critically, and mathematical proofs certainly contribute greatly to that skill. Educators actively debate how much students need to prove and at what level of rigor, although all agree that the answers depend on the level of the student. The amount of proof in high school courses now varies greatly across the United States. Still, the NCTM and CCSS call for high school students to do a certain amount of work with proofs. It follows that high school mathematics teachers need substantial background and facility in proof. I have written this text for mathematics majors and future high school mathematics teachers, and I think there is a range of proof experiences both audiences need. All these students need facility in making good arguments in a mathematical context, something I ask for repeatedly throughout the book. At a minimum, that means they need to make their assumptions explicit and use clear reasoning leading from them to their conclusions. I think it also means that they should be introduced to more formal proofs and axiomatic systems, although I don’t think that should be the primary focus of an entire course at this level. In Chapter I, I employ a fairly informal focus on proof to fit that chapter’s goals. One goal is to include enough coverage of the content of Euclidean geometry for students who have not had a solid year-long high school geometry course (and provide a review for others, as needed). Another goal is to develop students’ ability to prove substantive results in a context where they already have a comfortable intuition. Hence in that chapter I haven’t made explicit the many subtleties and assumptions discussed in Chapter 2. Instead, I ask students to build on Euclid’s theorems so that they can prove results that aren’t instantly apparent, although they should be plausible. Chapter 2, which looks more carefully at axiomatics, makes axiomatic systems explicit and builds up theorems carefully from the axioms. The level of proof in later chapters lies in between the informality of Chapter 1 and the axiomatically based proofs in Chapter 2, depending on the chapter. In most chapters the content is less familiar and the mathematics more sophisticated than in Chapter 1. Thus the value of proofs in them also connects with the goal of increasing geometric intuition. Proofs at the college level are written in paragraph form, unlike in high school geometry, where two-column proofs appear frequently. The range of argumentation appropriate at the college level can make the two-column format artificial and overly restrictive. A good proof ought to help the reader understand why the theorem is correct, as well as make the correctness of the reasoning clear, and two-column proofs can sacrifice understanding for clarity of reasoning. Appendix F gives an introduction to proof techniques used in this text.

General Notation Proofs end with the symbol !. Examples end with the symbol ♦. Exercises or parts of exercises with answers or partial answers in the back of the book are marked with an asterisk (*).

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Definitions italicize the word being defined. We use the abbreviation B.C.E. (before the common era) for dates predating the common era, which started somewhat more than 2000 years ago. Dates in the common era will not have the abbreviation C.E. added to them.

Prerequisites In general, students need the maturity of Calculus I and II, although only Chapter 9 and Section 2.4 use calculus extensively. Chapter 9 needs some content from Calculus III as well. Appendix E summarizes the material from Calculus III used in Chapter 9. Of course, additional mathematical maturity and familiarity with proofs will help throughout the book. Sections 3.3 and 3.5 and Chapter 9 require an understanding of vectors. Sections 5.3, 5.4, 5.5, 7.3, 7.4, 7.5, 7.6, and 8.4 depend on a more extensive understanding of linear algebra. Appendix D summarizes the linear algebra material needed in the text. Although Chapter 6 builds on concepts from Chapter 5, it doesn’t require linear algebra. Section 5.6 makes use of complex numbers and their arithmetic. Chapters 5, 6, and 7 discuss groups and Section 8.4 discusses finite fields, but don’t assume any prior familiarity with the concepts.

Exercises Learning mathematics centers on doing mathematics, so problems are the heart of any mathematics textbook. A number of exercises appear in the text and are meant to be done while reading that material. Far more appear at the end of each section. I hope that both students and instructors enjoy spending time pondering, solving, discussing, and even extending the problems. They should make lots of diagrams and, when relevant, physical and computer models. The problems include routine and non-routine ones, traditional proofs and computations, hands-on experimentation, conjecturing, and more. Exercises or parts of exercises with answers or partial answers in the back of the book are marked with an asterisk (*).

Projects Too often textbooks and courses shift to a new topic just when students are ready to make their own connections. And geometry is a particularly fertile area for such connections. Projects encourage extending ideas discussed in the text and appear at the end of each chapter. They include essay questions, paper topics, and more extended and open-ended problems. Many of the projects benefit from group efforts. The most succinct projects, of the form “investigate . . . ,” are leads for paper topics.

History Geometry reveals the rich influences over the centuries between areas of mathematics and between mathematics and other fields. Students in geometry, even more than other areas of mathematics, benefit from historical background. The introductory sections of each chapter seek to link the material of the chapter to a broader context and to students’ general knowledge. The biographies give additional historical perspective and add a personal flavor to some of the work discussed in the text. One common thread I found in reading about these geometers

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was the importance of intuition and visual thinking. As a student I sometimes questioned my mathematical ability because I needed to visualize and construct my own intuitive understanding, instead of grasping abstract ideas directly from a text or a lecture. Now I realize that far greater mathematicians than I built on intuition and visualization for their abstract insights, proofs, and theories. Perhaps this understanding will help the next generation as well.

Chapter Content Each chapter starts with an overview, including a discussion of the relevant history, and ends with projects and a list of suggested readings. Geometry is blessed, more than other areas of mathematics, with many wonderful and accessible expository writings as well as texts. (The vast number of web sites, software, and other media devoted to geometry surpass my ability to view, let alone recommend a helpful selection. Further, any printed list would quickly be outdated. So, while I do not make suggestions, I encourage instructors to find media that support their courses.) 1. Euclidean Geometry. Most of this chapter considers plane geometry and follows the lead of the ancient Greeks’ approach, especially Euclid’s synthesis. (Appendix A gives the definitions, axioms, and propositions of Book I of Euclid’s Elements.) Since high school geometry courses include much of Euclid’s emphases, this approach adds context to a teacher preparation course. Students’ preparation varies greatly, so instructors should adjust their pace and coverage according to how much this material is review for the students. All the exercises in Section 1.1 and many of the others have Greek or pre-Greek roots. The threedimensional material has a more modern focus, considering polyhedra, including geodesic domes, and the sphere. The material on the sphere is a useful transition into a study of non-Euclidean geometry, although it is presented as part of Euclidean geometry. 2. Axiomatic Systems and Models. The first section introduces axiomatic systems and investigates simple ones. The next section considers a high school axiomatic system for Euclidean geometry and Hilbert’s axiomatization. (Appendices B and C give the axiomatic systems.) I chose to use the SMSG axioms, one of the “ancestors” of all high school axioms systems, rather than try to choose among contemporary ones. (SMSG is an abbreviation for the School Mathematics Study Group.) The final section explores models and metamathematics. Instructors wishing to include more experience with axiomatic systems and models can include material from Chapter 8. 3. Analytic Geometry. While high school students use analytic geometry, they often don’t understand it and often don’t see many of the traditional topics. And, although calculus texts include topics such as parametric equations and polar coordinates, instructors often leave them out for lack of time. In addition to these topics, later sections discuss B´ezier curves in computer aided design and geometry in three and more dimensions. 4. Non-Euclidean Geometry. The bulk of the chapter develops hyperbolic geometry axiomatically. In addition to typical axioms, we assume the first twenty-eight of Euclid’s theorems, which also hold in hyperbolic geometry. By assuming them we can use familiar approaches to focus on how this geometry differs from Euclidean geometry. Models help illustrate the concepts and theorems. The final section considers spherical and single elliptic geometry.

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5. Transformational Geometry. The first two sections develop the key ideas of transformations and plane isometries without linear algebra. The CCSS strongly emphasize transformations in high school geometry, so at least these first two sections are vital for teacher preparation. Students who have already studied Chapter 4 can consider the corresponding theorems in hyperbolic and spherical geometries. (See Project 22.) The next three sections use linear algebra to delve into isometries more deeply, and into similarities, affine transformations, and transformations in higher dimensions. The final section investigates inversions using complex numbers and relates to the Poincar´e disk model of hyperbolic geometry and is not used elsewhere in the text. Appendix D covers the linear algebra needed for this and subsequent chapters. 6. Symmetry. While this material uses concepts from Chapter 5, it doesn’t depend on linear algebra. Students find this material accessible, compared with some of Chapter 5, and gain insight into the power of the transformational approach, including for applications. 7. Projective Geometry. Projective geometry historically and pedagogically provides a capstone unifying Chapters 1, 4, and 5. The first two sections briefly develop it intuitively and axiomatically. Later sections use linear algebra extensively and provide connections with computer graphics and the special theory of relativity. 8. Finite Geometry. Since the late nineteenth century geometers have drawn important insights about traditional geometry from the study of simplified finite systems. Section 8.2 discusses the most important of these, finite affine and projective planes, axiomatically. Section 8.3 generalizes the material to balanced incomplete block designs. The final section explores analytic models of finite affine and projective planes and spaces over the fields Z p , the integers (mod p), where p is prime. 9. Differential Geometry. Differential geometry deserves an entire undergraduate semester course, but many schools can’t offer it. I think students in a survey course benefit from an introduction to this vital area of geometry. I try to convey here its geometric insight and introduce some key geometric ideas—curvature and geodesics, and I endeavor to minimize the machinery of multivariable calculus. The chapter connects differential geometry to Euclidean, spherical, and hyperbolic geometries. Students need little more from multivariable calculus than a familiarity with parametric equations, partial derivatives, and cross products. Appendix E covers the needed multivariable material. 10. Discrete Geometry. This relatively new and rapidly growing area focuses on problems, especially ones that remain unsolved. Therefore I organized the chapter around problems. Students are encouraged to explore them in the first section. Subsequent sections develop them more fully, with relevant theorems and more in-depth problems.

Course Suggestions This text supports a variety of approaches to geometry and different levels of coverage of the material. Many of the sections benefit from more than one class period, especially to enable students to present problems or projects. The entire book would require a full year geometry course to cover, a luxury few mathematics department can offer. A. Teacher Preparation. To meet the goals of the CCSS and of the NCTM for teacher preparation, a course should include at least Euclidean geometry (Chapter 1), axiomatic systems

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B.

C.

D. E. F.

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and models (Chapter 2), transformational geometry (Chapter 5, except Section 5.6), and some of non-Euclidean geometry (Chapter 4). As time, interest and student background indicate, topics from analytic geometry (Chapters 3) and symmetry (Chapter 6) are valuable supplements for future teachers. Historical Survey. Chapters 1, 2, 4, 5, and 7 provide an understanding of the important historical sweep of geometry through the nineteenth and early twentieth centuries. In 1800 there was just Euclidean geometry (Chapter 1). Geometrical thinking expanded enormously, including non-Euclidean geometry (Chapter 4), transformational geometry (Chapter 5), and projective geometry (Chapter 7), among others. The transformations of projective geometry provided a vital unification of geometric thought, both historically and pedagogically. Because of these advances, mathematicians realized the need for a careful investigation of proofs, theories, and models (Chapter 2). Euclidean Geometry. Chapters 1, 2, 3, 5, and, as time permits, topics from Chapters 6, 9, and 10. If the class needs little review of Euclidean geometry, instructors could interleave Chapters 1 and 10 together at the start of the class. Transformational Geometry. Chapters 5, 6, and 7 and as much of Chapters 1 and 2 as needed. Axiomatic Systems and Models. Chapters 1, 2, 4, 8, and Sections 3.1, 7.1, 7.2, 7.3. Topics. Instructors of courses for mathematics majors have fewer constraints than those teaching mathematics education majors and so can choose topics more freely. Students’ background and interest will suggest different options. Students with a weaker background will benefit from Chapters 1, 2, 3, 5, and 6. Chapters 4, 7, 8, 9, and 10 can stretch better prepared students in different ways.

Dependence and Links Between Chapters I have tried to keep chapters as independent as reasonable. Students with a decent geometry understanding from high school will have adequate Euclidean and analytic geometry background for all chapters except Chapter 4, which depends explicitly on Chapter 1. The basic concepts of axiomatic systems and models from Chapter 2 appear in Chapters 4, 7, and 8. Chapter 5 is a prerequisite for Chapters 6 and 7. Sections 5.4 and 6.6 consider aspects of fractals. Sections 6.5 and 7.6 briefly consider aspects of the special theory of relativity, and Section 9.4 touches on the general theory of relativity. Section 9.3 refers to Chapter 4. A number of exercises (denoted with #) connect with other material. Section 1.1 See Example 1 of Section 10.3 for another proof of Theorem 1.1.2 (the Pythagorean theorem). # 1.2.9 anticipates Theorem 4.3.1. # 1.2.10 asks students to prove the converse of the Pythagorean theorem. # 1.2.14 anticipates Section 4.4. # 1.2.15 Compare this approach with #3.1.8. Section 10.4 uses this result. # 1.2.17 is used in a number of later sections. # 1.2.23, the law of cosines is used in # 3.3.12, # 3.3.21, #3.5.17, #3.5.18, and #10.3.14. # 1.2.25 is referred to in Section 3.3. Playfair’s axiom appears in Sections 1.3 and 2.2. # 1.3.8 (d) Compare this approach with # 3.1.4 (a).

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Section 1.5 Euler’s formula (Theorem 1.5.1) is used in Section 10.4. Project 14 in Chapter 3 asks students to investigate this formula in higher dimensions. Section 1.5 Shortest paths on a sphere are explored more in # 9.1.6 and in Section 9.4. Section 1.5 Theorem 1.5.3 relates to Theorem 4.1.1 and their generalization Theorem 9.4.3. # 1.5.35 anticipates concepts in Sections 2.1 and 2.3. Chapter 1, Project 6 is used in Project 11 of Chapter 10. Chapter 1, Project 7 relates to the art gallery theorem in Chapter 10. Many axiomatic systems in Section 2.1 are developed further in Section 2.3. Here are the pairings: Subsection 2.1.3 and # 2.3.6, # 2.1.7 and # 2.3.8, #2.1.8 and #2.3.9, #2.1.9 and #2.3.11, # 2.1.10 and # 2.3.12, # 2.1.12 and # 2.3.13. # 2.1.12 and # 2.3.13 anticipate projective planes, discussed in more detail in Section 8.2. Section 2.3 Example 3 (taxicab geometry) is used in # 3.3.21 , in Project 23 of Chapter 5, and in Sections 10.1 and 10.4. #3.1.6 and #3.1.7 relate to Section 2.3. # 3.1.9 and # 3.1.10 relate to the first fundamental form in Chapter 9. # 3.1.14 and # 3.1.15 introduce complex numbers and their arithmetic, used in Section 5.6. Section 3.3 Parametric equations are used extensively in Chapter 9. #3.3.17 to #3.3.22 relate to Section 2.3. # 3.5.14 connects with Project 2 in Chapter 9. # 3.5.17 and # 3.5.18 relate to #10.3.14. Section 4.1 The pseudosphere is discussed in #9.3.16. Section 4.1 Theorem 4.1.1 connects with Theorem 1.5.3 and the generalization Theorem 9.4.3. Chapter 4, Project 1—Compare with Chapter 5, Project 1 and Chapter 6, Project 3. # 5.610 (b)—Compare with Section 7.1 # 14. # 5.6.14 relates the Poincar´e model and the half plane model of Section 4.1. Section 6.3 relates to tilings in Sections 10.1 and 10.3. # 7.5.7 connects with relativistic velocities in Section 6.5 and Lorentz transformations in Section 7.6. Chapter 7, Project 8 connects Euclidean isometries in Section 5.2 with hyperbolic isometries. # 8.1.8 investigates an axiomatic system and its models.

Acknowledgments This text is based on my book The Geometric Viewpoint: A Survey of Geometries, published by Addison Wesley in 1998. First, the acknowledgments from the earlier text: I appreciate the support and helpful suggestions that many people made to improve this book, starting with those students who studied from the rough versions. My previous editors, Marianne Lepp and Jennifer Albanese, and my production supervisor, Kim Ellwood, provided much needed direction, encouragement and suggestions for improvement. I am grateful to the reviewers, whose insightful and careful critiques helped me greatly. They are: Bradford Findell, University of New Hampshire; Yvonne Greenleaf, Rivier College; Daisy McCoy, Lyndon State University; Jeanette Palmiter, Portland State University; and Diana Venters, University of North Carolina,

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Charlotte. They pointed out many mistakes and unclear passages; any remaining faults are, naturally, my full responsibility. I particularly want to thank two people involved in the earlier text. Paul Krueger not only painstakingly made many of the fine illustrations that are an integral part of this book. He also pondered with me over the years the nature of geometric thinking and the connections of geometry with other fields. Connie Gerads Fournelle helped greatly with her valuable perspective on the text, both as a student using an early version and as my assistant writing answers to selected problems. I want to thank a number of people who have helped in my rewriting this book. First of all, Prof. Peiyi Zhao at St. Cloud State University and Prof. Sue Hagen at Virginia Tech and their students were kind enough to use an early version of this rewrite in their classes and provide me feedback. I thank my students who also used an early version. Prof. Peter Haskell and Prof. Nick Robbins of Virginia Tech kindly provided feedback on Chapter 9. Prof. Ezra Brown, also of Virginia Tech, provided advice for Chapter 8. Prof. Frank Farris of Santa Clara University, pushed me to rewrite this text and provided enthusiastic support over the years. I am grateful to St. John’s University for funding the sabbatical making this rewrite possible. Thanks go to the Mathematics Department at Virginia Tech for welcoming me, encouraging me, and providing the materials and space needed in the long rewriting process. I appreciate the extensive and cheerful secretarial support Suzette Ehlinger and Gail Schneider have provided. I am grateful to the editorial staff at the Mathematical Association of America for their hard work: Beverly Ruedi, Carol Baxter, the copy editor, and editors Stan Selzer and Zaven Karien provided the technical support and positive encouragement needed to make this book a reality. Finally I thank my wife, Jennifer Galovich, for her unswerving support and love throughout the years of the writing and now rewriting. If you have comments or suggestions for improvement, please contact me by e-mail at [email protected]. Thomas Q. Sibley Mathematics Department St. John’s University Collegeville, MN 56321-3000

1

Euclidean Geometry

Figure 1.0 Geodesic domes, such as the U.S. Pavilion at Expo ’67 pictured here, satisfy pragmatic engineering demands with an elegant geometric shape. Buckminster Fuller used elementary two- and three-dimensional geometry, familiar to mathematicians for more than 2000 years, to fashion the first geodesic domes in the 1950s. A!E"METPHTO# MH$EI# EI#IT" “Let no one enter who doesn’t know geometry.” (the inscription Plato is said to have placed over the entrance to his Academy)

1.1 Overview and History Geometry has a rich heritage as well as contemporary importance. Euclidean geometry combines both: Its roots go back to the oldest mathematics texts from 4000 years ago while modern

1

2

Euclidean Geometry

mathematicians find new results and applications regularly. This chapter supplements a historical examination of the fundamental contributions of the ancient Greeks with more modern aspects. Much of high school geometry covers the same territory well known to the Greeks so future secondary mathematics teachers need to deepen their understanding of it. Geometric knowledge developed in all ancient cultures, consisting largely of solving problems involving geometric patterns and finding methods for determining areas and volumes. The best preserved and most developed pre-Greek mathematics came from Egypt and Babylonia (which is a shorthand for land in modern day Iraq occupied by many ancient peoples). An Egyptian papyrus dated 1850 B.C.E. gave an exact procedure for finding the volume of a truncated square pyramid. However, the Egyptians were quite likely not aware that it was exact or which of their other methods, such as approximating the area of a circle, were not exact. A century earlier (by 1950 B.C.E.) the Babylonians possessed a sophisticated number system and methods to solve problems that we would describe as first- and second-degree equations in one and two variables. The Egyptians and Babylonians, along with the Chinese and others, had numerical examples of what we call the Pythagorean theorem. Later Babylonian astronomers developed mathematics for the sphere. Although Egyptian and Babylonian mathematics dealt with specific numbers rather than general formulas, the variety of examples that survived convinces scholars that these peoples understood something of the generality of their methods. (Exercises 1.1.3 to 1.1.9 consider problems from non-Greek cultures.)

1.1.1 The Pythagoreans and Zeno The heritage of deductive mathematics started in ancient Greece and built on the insights of the Babylonians and Egyptians. The Greeks found and proved many mathematical properties, including the familiar ones of high school geometry. They also organized this knowledge into a systematic axiomatic system, now known as Euclidean geometry, which honors the Greek mathematician Euclid. We will discuss Euclid’s contributions in the following sections, while in this section we focus on the developments leading to Euclid’s synthesis. The Pythagoreans, followers of Pythagoras (580 to 500 B.C.E.), were among the first groups to focus on theoretical mathematics. The relationship we now call the Pythagorean theorem (Theorem 1.1.2) had been known independently at least in numerical form in many cultures. The Pythagoreans, however, are credited with giving a geometrical argument for the truth of this key link between geometry and numbers. (See Exercises 1.1.17, 1.1.18, and 1.1.19 for several ways people have proved this theorem.) The Pythagoreans at least initially founded their mathematics and mystical musings on positive whole numbers and their ratios and properties. The Pythagoreans developed the theory of positive whole numbers, studying prime numbers, square numbers, and triangular numbers among others. They also investigated geometry and developed proofs. Heath [11, vol. I, 317–320] cites evidence that the Pythagoreans found the argument given in proof of Theorem 1.1.1, a theorem as fundamental as the Pythagorean theorem (Theorem 1.1.2). While the Pythagoreans weren’t the first to note the relationship of the Pythagorean theorem, they are credited with its first proof, as well as of its converse. (See Exercise 1.2.10 for the converse.) However, the systematic organization of theorems and proofs, such as the reference in the proof to one of Euclid’s theorems, didn’t happen until later. We refer to particular Euclidean theorems using Roman numerals to describe the book and ordinary numerals for the particular proposition. For instance, Theorem 1.1.1 is a more modern rendering of I-32, the thirty-second proposition in Euclid’s first book.

3

1.1 Overview and History

Notation. The line segment between two points A and B is denoted AB, the length of AB is ← → denoted AB, and the line through A and B is denoted AB. The triangle with vertices A, B, and C is denoted △ABC, and the angle of that triangle with vertex at B is denoted ∠ABC. Congruent is denoted ∼ =. We abbreviate “sum of the measures of the angles” as angle sum. Theorem 1.1.1. (Euclid I-32) In Euclidean geometry the angle sum of a triangle is 180◦ . ← → Proof. Let △ABC be a triangle and construct the line D E parallel to BC through A (Figure 1.1). Then (by Euclid’s proposition I-29) ∠D AB ∼ = ∠C B A and ∠C AE ∼ = ∠AC B. Thus the angles of the triangle are congruent to the angles ∠D AB, ∠B AC, and ∠C AE that comprise a straight angle. Hence the angle sum of the triangle equals the measure of a straight angle, 180◦ . ! C B

E D

A

Figure 1.1 The angle sum of a triangle.

Theorem 1.1.2. (Euclid I-47) In a Euclidean right triangle the square on the hypotenuse has the same area as the squares on the sides: that is, a 2 + b2 = c2 . Proof. (See Exercises 1.1.17 and 1.1.18.) ! The Pythagorean attempt to ground mathematics on numbers ran into an irreconcilable conflict with the discovery of incommensurables (irrational numbers, in modern terms). Commensurable lengths have a common measure—a unit length such that the given lengths are 1 as a common measure. The integer multiples of the unit. For example, 23 and 45 have 15 Pythagoreans proved that the diagonal and side of a square were incommensurable. We say that √ √ the diagonal is 2 times as long as the side and that 2 is not a rational number. Recall that a rational number can be written as a fraction. Theorem 1.1.3. No rational number equals rable with its side.)

√

2. (The diagonal of a square is incommensu-

√ Proof. Suppose, for a contradiction, that there were two integers p and q such that p/q = 2. Without loss of generality, assume that p and q are not both even. Otherwise we could factor out any common factors of 2. Then ( p/q)2 = 2, or p 2 = 2q 2 . Thus p 2 must be even, which in turn, forces p to be even. (To illustrate, suppose that p were odd, say, p = 2k + 1. Then p 2 = 4k 2 + 4k + 1, an odd number.) Now rewrite p as 2r , for some integer r . Then (2r )2 = 2q 2 , p and q are not both even, giving or more simply 2r 2 = q 2 . As before, q must be even. However,√ us a contradiction. Thus our initial supposition is invalid, and 2 is not a rational number. !

4

Euclidean Geometry

√ Exercise 1.1.1.∗ Modify the proof of Theorem 1.1.3 to prove that √ 3 is not rational. Explain where the corresponding argument fails when you try to show that 4 is not rational. Theorem 1.1.3 ruined the Pythagoreans’ philosophical commitment to explain everything in terms of whole numbers and their ratios. This and other philosophical problems led later Greek mathematicians to base their mathematics on geometry. For example, they no longer thought of lengths, areas, and volumes as numbers because the values could be incommensurable (irrational). The lack of numbers for measurement ruled out geometric formulas, although they proved theorems that we now readily convert to formulas. Nevertheless, the Greeks made impressive advances in geometry and developed careful, well-founded proofs. The theoretical, abstract nature of Greek mathematics separated it from practical and computational mathematics. However, modern scientists and mathematicians have found important applications of Greek discoveries and the theoretical approach. √ Zeno’s paradoxes—and the irrationality of 2—spurred a careful study of the foundations of geometry. Zeno (circa 450 B.C.E.) proposed four paradoxes purporting to disprove obvious facts about motion. Zeno’s most famous paradox, Achilles and the Tortoise, continues to puzzle people. (See Salmon [23].) Achilles and the Tortoise. Achilles, the swiftest human gives a tortoise a head start in a race. Zeno argued that Achilles can never pass the tortoise. For Achilles to catch the tortoise, he must first run to where the tortoise started, but by then the tortoise will have crawled a bit farther. Achilles must continue running to this new place, but the tortoise will then be a tiny bit farther along. No matter how often this process is repeated and no matter how small the tortoise’s lead, Achilles always remains behind. Exercise 1.1.2.∗ Zeno’s argument is paradoxical because each step seems reasonable, yet we know that faster things regularly pass slower ones. Discuss this paradox, trying especially to find an error in Zeno’s argument.

1.1.2 Plato and Aristotle The school of philosophy founded by Plato (429–348 B.C.E.) next took the lead in the study of geometry. One of Plato’s pupils, Eudoxus, developed a theory of proportion and a way to give careful proofs of sophisticated results that applied equally to commensurables and incommensurables. Eudoxus’s work, like limits in calculus, avoided Zeno’s paradoxes. Another pupil, Theaetetus, developed a classification of incommensurable lengths and gave the first proof that there are five regular polyhedra. Plato viewed geometry as vital training for philosophy. He thought that only those who understood the truths of geometry could grasp philosophical truths. In his view, mathematics was certain because it was about ideal, eternal truths, and mathematics was applicable because the physical world was an imperfect reflection of the ideal truth. Now mathematics is often viewed as part of science with its emphasis on physical reality rather than Plato’s ideal view. But mathematics, with its astounding certainty that surpasses any other subject’s reliability, seems to have a different content than any science. After all, no one can physically measure π to one hundred decimal place accuracy, let alone the more than one trillion decimal places that have been found with the aid of computers. Obviously, though, mathematics is not isolated from the real world, as its sophisticated and varied applications reveal.

5

1.1 Overview and History

Aristotle (384–322 B.C.E.), Plato’s most famous student, established his own school of philosophy. Aristotle considered mathematics to be an abstraction of concrete experience. Thus for him the applicability of mathematics derived from its origin in the world. Aristotle thought that mathematics owed its certainty to its careful proofs. He emphasized the necessity of starting with simple, unquestionable truths (axioms or postulates) and carefully proving all other truths from them. His work on logic set the standards of reasoning for two thousand years just as his contributions to many other areas—science, law, ethics, and esthetics—profoundly influenced Western culture. Mathematics has been an important intellectual activity for at least as long as humans have had writing. The Greeks’ theoretical and proof-oriented emphasis took mathematics to a amazingly fruitful level and has come to characterize mathematics. (See Kline [14, Chapters 1–3] for more information on ancient mathematics, including the Greeks.)

1.1.3 Exercises for Section 1.1 In exercises requesting proofs, you may assume any common geometric properties you know, as long as you make your assumptions explicit. *1.1.3. For the area of a circle the Egyptians used the area of a square with side 89 of the circle’s diameter. In measuring a cylindrical granary with a height of 20 units and a radius of 10 units, what error in cubic units and as a percentage would the Egyptians have made? Find the value of k in (8d/9)2 = kr 2 . (The Egyptian method in effect approximates π by k, but it is historically misleading to talk about an “Egyptian value of π .”) 1.1.4. A Babylonian problem: One side of a right triangle is 50 units long. A segment parallel to the other side and 20 units from it cuts off a trapezoid with area 320 square units. Find the length of the other side and the segment. What does this problem suggest about Babylonian mathematics? *1.1.5. A diagram like that shown in Figure 1.2 appears on a Babylonian tablet, but written in their base 60 notation. Convert the fractions to decimals and discuss what the numbers tell about Babylonian mathematics.

30 1+

24 60

+

42 +

51 3600 25 60

+

10

+

216000 35

3600

Figure 1.2

6

Euclidean Geometry

1.1.6. A Chinese problem, 200 or earlier: A reed sticks out one unit above the water at the center of a pond ten units across.When the top of the reed is pulled over to the edge of the pond, its top is even with the water level. How deep is the pond? What does this problem suggest about Chinese mathematics? *1.1.7. A Hindu problem, 630: A man and a wizard live at the top of a cliff 100 units high, whose base is 200 units from a village. The man descends the cliff and walks to the village. The wizard flies up some number of units and then flies directly to the village, traveling the same total distance as the man. How far did the wizard fly up? What does this problem suggest about Hindu mathematics? 1.1.8. Figure 1.3 shows a truncated square pyramid. The following is the Egyptian recipe for the volume of a truncated square pyramid with a height of 6 and lengths of 4 at the base and 2 at the top: “You are to square this 4; result 16. You are to double 4; result 8. You are to square 2; result 4. You are to add the 16, the 8, and the 4; result 28. You are to take one third of 6; result 2. You are to take 28 twice; result 56. See, it is 56. You will find it right.” (a) Verify that this recipe corresponds to the modern formula V = 13 h(a 2 + ab + b2 ). (b) Derive this formula from V = 13 H A for the volume of a pyramid with height H and base area A. Hint: A truncated pyramid is the difference between an entire pyramid and a smaller pyramid removed from the top. Use proportions to relate their bases and heights. Also, a 3 − b3 = (a 2 + ab + b2 )(a − b).

Figure 1.3 A truncated pyramid. 1.1.9. An (abbreviated) Babylonian problem reads “. . . I added the length and the width and [the result is] 6.5 [units] . . . . [The area is] 7.5 [square units] . . . . What are the length and width?” *(a) Solve this problem with modern methods. Explain your √ approach. (b) Verify that when a = 1 the quadratic formula (−b ± b2 − 4ac)/2a reduces to ! −b/2 ± (b/2)2 − c. (c) Use part (b) to discuss the scribe’s recipe for solving this problem: “Halve the length and width which I added together, and you will get 3.25. Square 3.25 and you will get 10.5625. Subtract 7.5 from 10.5625, and you will get 3.2625. Take its square root, and you will get 1.75. Add it [1.75] to the one [3.25], subtract it from the other [3.25 again], and you will get the length and the width. 5 is the length; 1.5 is the width . . . . Such is the procedure.”

7

1.1 Overview and History

1.1.10. Assume that any diagonal of a convex polygon is inside it. Recall that a set S of points is convex if and only if for any two points A and B in S, AB is a subset of S. (a) Find the angle sum of a convex quadrilateral. Repeat for a pentagon and a hexagon. (b) Find a formula for the angle sum of a convex n-gon. (c) Use induction to prove the formula in part (b). (Proclus (410–485) gave an argument without induction for this result.) (d) Does the formula in part (b) hold when the polygon is not convex? Does the proof in part (c) hold? *1.1.11. The Pythagoreans investigated square numbers, oblong numbers, triangular numbers, and other numbers that come from geometrical arrangement of points. (See Figure 1.4.) (a) An oblong number is the number of points in a rectangular array of points where there is one more column of points than there is rows of points. Give a formula for oblong numbers. (b) A triangular number is the number of points in a triangular array of points where the top row has one point and each successive row has one more point in it. Explain why an oblong number is twice the corresponding triangular number. Use this result to give a formula for the sum of the first n positive integers. (c) Arrange points to form pentagonal numbers and hexagonal numbers. Find the first few pentagonal and hexagonal numbers. (d) Find a formula for pentagonal and hexagonal numbers.

Figure 1.4 Oblong number, Triangular number. 1.1.12. (a) Find all rectangles with sides of integer lengths whose perimeters equal their areas. (The Pythagoreans considered this problem, but later Greeks did not because they didn’t consider areas and lengths as numbers.) (b) Find a formula for all rectangles whose perimeters equal their areas. Use it to explain your answer in part (a). (c) Search for all rectangular boxes with sides of integer lengths whose surface areas equal their volumes. *1.1.13. The Pythagoreans investigated different kinds of averages or means. The arithmetic mean of √ the numbers a and b is a+b . The geometric mean of the positive numbers a 2 and b is ab. relates to them. Use (a) Consider a and b as lengths. Explain how the length a+b 2 geometry and particularly area to explain how the length of the geometric mean √ ab relates to the lengths a and b.

8

Euclidean Geometry

(b) The Pythagoreans felt a particular ratio had particular significance. They would split a segment into segments of length a and b with a > b, where a is the geometric mean of b and the whole segment. Find the exact and approximate value of the ratio of a to b. Verify that the ratio of the whole segment to a is the same. This ratio, now called the golden ratio, appears in many natural settings and applications. (See Livio [16].) 1.1.14. Eratosthenes (circa 284–192 B.C.E.) made the most famous and accurate of the Greeks’ estimates of the circumference of the earth. He found that at noon on the summer solstice, the sun was directly overhead at Syene, Egypt; at the same time 5000 stadia 1 (approximately 500 miles) north, in Alexandria, Egypt, the sun was 50 of a circle off vertical. Make a diagram, compute the circumference of the earth based on his data, and explain your procedure. Hint: Eratosthenes assumed that the sun’s rays at Syene and Alexandria were parallel. *1.1.15. The Pythagoreans thought that the pentagram, a pentagon with its diagonals, as shown in Figure 1.5, had mystical qualities. (a) Find the measures of the following angles. ∠AO B, ∠O B A, ∠ABC, ∠B AC, ∠AG B, ∠C G B, ∠AB H , and ∠C AD. (b) Verify that △ABC and △AG B are isosceles and similar. List other similar triangles. (c) If the side has length AB = 1 and AG = x, explain why the diagonal has length AC = 1 + x. Hint: Use △BC G. (d) Explain why in part (c) x satisfies 1 + x = 1/x. Find the exact value of 1 + x, the diagonal of the pentagon. Verify that the ratio of AG to G H is also 1 + x. Verify that 1 + x is the golden ratio, as defined in Exercise 1.1.13. D J

E

I

F

O x

A

B

C H

G 1

B

Figure 1.5 A regular pentagon.

A

O

Figure 1.6 A lune.

1.1.16. In the only surviving Greek mathematical text from before Euclid, Hippocrates (circa 440 B.C.E.) investigated the areas of lunes, regions bounded by two circles. Explain his result that the lightly shaded lune shown in Figure 1.6 has the same area as △AB O. What properties of circles and triangles did he need to know to obtain his result? Hint: Use the unshaded area between A and B.

9

1.1 Overview and History

*1.1.17. In Figure 1.7 ∠AC B is a right angle and C D is perpendicular to AB. Why are △ABC, △AC D, and △C B D all similar? Show that cy = a 2 and cx = b2 and develop a proof of the Pythagorean theorem.

C b

a x

A

y D

B

c

Figure 1.7 The Pythagorean Theorem.

Figure 1.8 BEHOLD!.

1.1.18. Give a proof of the Pythagorean theorem based on Figure 1.8. The Indian mathematician Bhaskara (1114–1185) devised such a proof but provided only this diagram with the word “BEHOLD!” written below it. 1.1.19. Study Euclid’s proof of the Pythagorean theorem (proposition I-47 in Heath [11, 349]) and compare it with the proofs in Exercises 1.1.17 and 1.1.18. Discuss each proof ’s assumptions, how convincing it is, and how easy it is to follow. 1.1.20. The Greeks proved that a cylinder has three times the volume of a corresponding cone. (Euclid gives a proof in XII-20.) Use each of the following methods to verify this fact and discuss their advantages, disadvantages, and assumptions. (a) (Empirical) Find or make a cylinder and a cone with the same height and radius (Figure 1.9). Use the cone three times to fill the cylinder with sand. How nearly does the volume of sand in the cylinder come to filling it exactly?

r

2πr

h

r h2+r2

Figure 1.9 A cylinder, a cone, and a pattern for making the cone.

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Euclidean Geometry

"b (b) (Calculus) Recall that a π ( f (x))2 d x gives the volume of revolution generated by rotating around the x-axis the curve y = f (x) from x = a to x = b. Use calculus to compare the volumes of a cone and a cylinder. (c) (Stacked disks) You can approximate the volume of a cone with the volume of a stack of disks (Figure 1.10). If h is the height of the cone and n is the number of disks, the height of each disk is h/n. Find a formula for the radius and the volume of the ith disk from the top in terms of the radius. Find the volume of the stack of disks and simplify it to get the approximation formula n π hr 2 # 2 i . n 3 i=1 $n 2 i = (2n 3 + 3n 2 + n)/6. Assume (or, better, prove by induction) that i=1 What happens to the approximation above as n approaches infinity?

Volume ≈

Figure 1.10 Approximating a cone with disks. 1.1.21. Heron (circa 75) found and proved a formula for the area of a triangle given the . Then Heron showed that the area is lengths a, b, and c of its sides. Let s = a+b+c 2 ! s(s − a)(s − b)(s − c). (a) Verify Heron’s formula for equilateral triangles. (b) Verify Heron’s formula for isosceles triangles.

1.2 Euclid’s Approach to Geometry I: Congruence and Constructions We know little about Euclid (circa 300 B.C.E.) except his mathematics, including the most influential mathematics book of all time: the Elements [11]. In it he organized virtually all the elementary mathematics known at the time into a coherent whole. The Elements contains definitions, axioms, and 465 theorems and their proofs—but no explanations or applications. For centuries the format represented the ideal for mathematicians and influenced many other areas

1.2 Euclid’s Approach to Geometry I: Congruence and Constructions

11

of knowledge. This masterpiece exemplifies Greek mathematics, showing the influence of both Plato and Aristotle: It investigates ideal geometric forms, following Plato, and strives to prove all results from self-evident truths, following Aristotle. Euclid’s approach and developments based on it have greatly influenced modern mathematics and thus provide ample reason to study Euclid. High school geometry courses are based on Euclid and provide another reason to look at the Elements in some detail. In this section we consider congruence and constructions, familiar topics of high school geometry found in the Elements. Sections 1.3 and 1.4 consider parallels and similarity, other high school topics based closely on Euclid’s work. Euclid built his work on that of his predecessors. However, he didn’t indicate which of the theorems and proofs were his contribution, and his text was so successful that no prior geometry text was preserved. Scholars credit Euclid with the organization, the choice of axioms (his postulates and common notions), and some of the theorems and proofs. The Greeks called an axiomatic approach synthetic because it synthesizes (proves) new results from statements already known. The Greeks often used a process they called analysis to find new results that they then proved. They analyzed a problem by assuming the desired solution and worked backward to something known. We mimic this procedure in what we call analytic geometry and algebra by assuming that there is an answer, the unknown x, and solving for it. In modern times synthetic geometry has come to mean geometry without coordinates because coordinates are central to analytic geometry. The modern understanding of axiomatic systems (Chapter 2) and non-Euclidean geometries (Chapter 4) arose from careful reflection on Euclid’s work. For this chapter we will use a semiformal approach, intermediate between the presentation found in high school and a modern axiomatic approach, which will be discussed in Chapter 2. In particular, we will make free use of many of the results Euclid proved and listed in Appendix A.

1.2.1 Congruence The familiar triangle congruence results, abbreviated SAS, SSS, ASA, and AAS, are fundamental tools for Euclid as well as for high school geometry. Intuitively, geometric shapes are congruent if they coincide when one is placed on the other. The idea of moving figures leads naturally to transformational geometry, which we discuss in Chapter 5. Euclid’s fourth common notion tries to capture it. However, he implicitly works with the formal idea that when all corresponding segments and angle of two shapes match, then the shapes must match. With two triangles, there are three sides and three angles of each to match up with the other. Euclid’s wellknown triangle congruence theorems side-angle-side or SAS (I-4), side-side-side or SSS (I-8), angle–side-angle or ASA (I-26), and angle-angle-side or AAS (also I-26) save us considerable work by reducing the need of showing all six congruences (“SASASA,” so to speak) to showing only three. The abbreviation SAS indicates, for example, with triangles △BC D and △P Q R that if BC ∼ = P Q (the first “S”), ∠BC D ∼ = ∠P Q R (the “A”), and C D ∼ = Q R (the final “S”), then the two triangles are congruent. Even more importantly, often we don’t have any means to get more than three of the congruences, so the theorems are essential. It is worth noting, as Euclid well knew, that not every three congruences of the six guarantee the triangles are congruent. (See Exercise 1.2.4.) To show congruences between figures with more than three vertices it is easiest to split the figures into collections of triangles and show the triangles congruent. (See Exercise 1.2.23.)

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1.2.2 Constructions Euclid’s first three postulates (or axioms) and many of his propositions reflect the growth of Greek geometry from constructions—figures built from line segments and circles. Euclid assumed the construction of a line segment given the end points (postulate 1), the extension of a line segment (postulate 2), and the construction of a circle given the center and the radius (postulate 3). We now tend to think of a line as a number line with distances embedded already. However, starting before Euclid, there has been a long tradition of using only an unmarked straightedge and a compass in building a construction. The Greeks thought that the two tools mimicked ideal lines and circles, which had no marks on them. We saw that the Greeks avoided basing geometry on numbers. (Instead, Euclid bases number theory on geometry starting in book seven of the Elements.) It can definitely be a challenge to construct a specified geometric shape from some simple givens using only the two tools. Often the given is just an arbitrary line segment. In modern terms, it gives the unit distance. (There is also another tradition just as old that allows other tools in constructions. Exercises 1.2.26 and 1.2.27 consider such constructions, but unless stated otherwise, all constructions should use only straightedge and compass.) In addition to giving the steps of a construction, we also need to justify that the object constructed actually satisfies the desired properties. To justify our conclusions we’ll use Euclid’s postulates and propositions, especially the triangle congruence theorems. Also, we can use the axioms in Appendix B. (See Appendix A for the postulates, definitions, and propositions of Book I of Euclid’s Elements; proposition n of book I is denoted I-n. Appendix B lists the SMSG axioms for Euclidean geometry, which are similar to axioms used in high school geometry texts.) There are three basic constructions from which all others are built: constructing the perpendicular bisector of a line segment (Euclid I-10), constructing the angle bisector of an angle (Euclid I-9), and reproducing an angle on a ray (Euclid I-23). Recall the definition of these terms. Definition. Two lines m and k are perpendicular, written m ⊥ k, if and only if the four angles they create are all congruent. We call the angles right angles. A bisector is a line that splits either a line segment or an angle into two congruent parts. Recall that a straight angle, whose rays make a line, has measure 180◦ , and so a right angle has measure 90◦ . Example 1. Given segment AB, construct its perpendicular bisector. Solution. By postulate 3, we may draw circle C1 with center A and radius AB and circle C2 with center B and radius AB. (See Figure 1.11.) Let D and E be the intersections of the circles. Use postulate 1 to draw D E. We claim that D E is the perpendicular bisector of AB. Since the radii of the circles are the same, we see that AD ∼ = B D and AE ∼ = B E. Of course, D E ∼ = D E. Hence the triangles △AD E and △B D E are congruent by SSS (I-8). Corresponding angles are therefore congruent, so ∠AD E ∼ = ∠B D E. Let F be the intersection of AB and D E. In triangles △AD F and △B D F we have AD ∼ = B D, ∠AD F ∼ = ∠B D F, and D F ∼ = D F. By SAS (I-4), △AD F ∼ = B F, showing that D E bisects AB. Also, ∠AF D ∼ = ∠B F D. = △B D F and so AF ∼

1.2 Euclid’s Approach to Geometry I: Congruence and Constructions

13

D C2 C1

F

A

B

E

Figure 1.11 Constructing a perpendicular bisector. Further, the two angles together make a straight angle, so the sum of their measures is 180◦ . By congruence, they must each be right angles and so D E is perpendicular to AB, completing the argument. ♦ (The solution only used Euclidian propositions that come before I-10, the proposition corresponding to this construction. We can readily show that △AB D is an equilateral triangle, which is Euclid’s proposition I-1.) Exercise 1.2.1.∗ Use Euclidean propositions before I-9 to justify the following construction of −→ B F in Figure 1.12 as the angle bisector of ∠AB D. Circle C1 is the circle centered at B with −→ radius AB. E is the intersection of C1 with B D. The circles C2 and C3 both have radius B A with centers A and E, respectively. Their intersections are at B and F. We claim that the ray −→ B F bisects ∠AB D.

D

C3 E

F C2

C1

B

A

Figure 1.12 Bisecting an angle. Exercise 1.2.2.∗ Use Euclidean propositions before I-23 to justify the following construction −→ of congruent angle ∠G E H to the angle ∠AB D on the ray E F. (See Figure 1.13.) Circle C1 −→ has center E and radius AB. It intersects E F at G. Circle C2 has center G and radius AD and circle C3 has center E and radius B D. Circles C2 and C3 intersect in two points H and I . We claim that both ∠G E H and ∠G E I are congruent to ∠AB D.

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Euclidean Geometry

C2

H

D

F

C3

A B

G E

C1

I

Figure 1.13 Copying an angle. Example 2. Construct a square if you know one side of it, say, AB (Euclid I-46). Solution. Construct circle C1 with center B and radius AB. Use postulate 2 to extend AB to Z on C1 . (See Figure 1.14.) Note that B is the midpoint of AZ . Use Example 1 to construct the perpendicular B D, where D is on circle C1 . Note that AB ∼ = B D. So we have two sides of the square. Construct the two remaining sides similarly. ♦ E E

D

D 5 2

A

B

Z F A

Figure 1.14 Constructing a square.

1

G B

Figure 1.15 Constructing the diagonal of a pentagon.

Examples 1 and 2 show how to construct the first two regular polygons: an equilateral triangle and a square. Exercise 1.2.12(b) asks you to construct a regular hexagon and justify your construction. A regular pentagon presents a bigger challenge. We need the ratio between the side of a regular pentagon and one of its diagonals, given in Exercise 1.1.15. Figure 1.15 uses a square to show one way to construct a segment AG congruent to the diagonal of a pentagon whose side is congruent to AB. In Figure 1.15 F is the midpoint of AB. The remarks after Exercise 1.2.12 discuss which regular polygons are constructible with straightedge and compass. Exercise 1.2.3.∗ If in Figure 1.15 AB has length 1, verify that D F has length has length

√ 1+ 5 , 2

√ 5 2

and so AG

which is the golden ratio.

Example 3. Construct a regular pentagon given one side of it, P Q. Solution. Given the segment P Q, we use Example 2 to construct a square ABC D with√ AB ∼ = P Q and Figure 1.15 to construct AG. By Exercise 1.2.3 the length AG equals P Q 1+2 5 . The Pythagorean theorem (I-47) shows that AG is as long as claimed. Fig 1.16 shows the construction of the pentagon. The circle with center P and radius P Q intersects the circle with center at Q

1.2 Euclid’s Approach to Geometry I: Congruence and Constructions

15

S

T

R

Q

P

Figure 1.16 Constructing a regular pentagon. and radius AG at T . Similarly, R is the intersection of circles centered at A and B of radii AG and P Q, respectively, and S is the intersection of circles centered at P and Q and both of radius AG. Construct the sides to get the pentagon P Q RST . Exercise 1.1.15 of Section 1.1 shows that P Q RST is a regular pentagon. ♦ Example 4. Construct the tangents to a circle from an outside point. Solution. Let P be outside the circle with center C (Fig. 1.17). Construct the midpoint M of C P and the circle with center M and radius MC. The circle intersects the original circle in two points, A and B. We claim that P A and P B are tangent to the circle through P. The very useful Exercise 1.2.16 provides a way here to justify that ∠P AC and ∠P BC are right angles. ♦ A

C

M

P

B

Figure 1.17 Constructing tangents to a circle.

1.2.3 Equality of Measure Euclid uses the same Greek word for “congruent” and for “equal,” meaning “equal in measure.” The measure is a length, area, volume or angle measure, as appropriate. Thus Euclid’s fourth postulate, “All right angles are equal,” could mean either that they are congruent or that their angle measures are equal (and 90◦ ). For angles, congruence matches with having the same measure. However, two triangles with the same area do not need to be congruent, so we need to distinguish between the concepts. Euclid didn’t think of lengths, areas, volumes, and angle measures as numbers, so he studied them without formulas. He started from the intuitive idea that congruent figures have the same area. In general, two regions built from the same congruent shapes must have the same area. He could then, for example, instead of giving the formula 12 bh for the area of a triangle, show

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Euclidean Geometry

in I-41 that a parallelogram with the same base and height as a triangle has twice the area. (Exercise 1.3.5 of Section 1.3 follows Euclid’s approach for this familiar fact.) To compare the areas of two polygons, he constructed two parallelograms with the same areas as these polygons (I-45) and compared the parallelograms. The proposition corresponding to the area formula A = πr 2 for a circle goes beyond Euclid’s many accomplishments. Over fifty years later Archimedes proved it using a deeper understanding of geometry. Actually, he proved that a circle has the same area as a triangle with height equal to the radius of the circle and base equal to the circumference of the circle. (See Figure 1.18.) His proof required the sophisticated “method of exhaustion” developed by Eudoxus, which preceded the idea of a limit by over two thousand years.

r

r 2πr

Figure 1.18 The area of a circle.

1.2.4 The Greek Legacy Three unsuccessful Greek constructions also inspired the development of mathematics. The first, called doubling the cube, was to construct (the side of) a cube with twice the volume of a given cube. The second was trisecting an angle. The third, squaring the circle, was to construct a square with the same area as a given circle. The Greeks did find exact constructions for these problems using methods beyond a straightedge and compass. (See Exercise 1.2.26.) In fact, the development of conics (ellipses, parabolas, and hyperbolas) was connected with doubling the cube and trisecting an angle. The third problem, as Archimedes showed, depends on the quantity we now call π . Not until the nineteenth century could mathematicians prove that the three constructions were impossible with only an unmarked straightedge and compass. Many people misinterpret this impossibility, thinking that mathematicians just haven’t been clever enough to find constructions. Thus people still propose solutions. The impossibility proofs require abstract algebra, and are beyond the scope of this text. (See Durbin [5, 229–235].) In brief, mathematicians converted the geometric problem of what lengths were constructible with straightedge and compass into an algebraic problem about what irrational numbers could be written in a particular form by using rational numbers, repeated square roots, and arithmetic operations. The idea for this conversion from geometry to algebra comes from Descartes in 1637. (See Exercises 1.2.10, 1.2.11, and 1.2.24.) Both doubling the cube and trisecting an angle involve cubic equations whose roots cannot in general be written in that form. Pierre Wantzel proved the impossibility of these constructions in 1837. (See Exercise 1.2.25.) To square the circle requires the construction of π, which also cannot be written in the form just described. Ferdinand Lindemann proved this in

1.2 Euclid’s Approach to Geometry I: Congruence and Constructions

17

1882 by proving that π was transcendental; that is, it is not the root of any polynomial whose coefficients are rational numbers. The decomposition of a figure, something Euclid used to relate areas of polygons, led to other modern investigations. Euclid proved that certain shapes had the same area by decomposing one shape into smaller pieces that could be reassembled to form the other shape. Decomposition puzzles have been popular for centuries, especially the Chinese Tangram puzzle. W. Bolyai in 1832 and P. Gerwien in 1833 independently showed that two polygons in the plane with the same area could be decomposed into one another by using finitely many smaller polygons. (See Boltyanskii [3].) In Chapter 4, we use decomposition to examine area in hyperbolic geometry. David Hilbert posed the corresponding problem for three dimensions in a famous talk in 1900, when he presented a list of 23 important, unsolved problems. The same year Max Dehn proved that a regular tetrahedron (triangular pyramid) could not be cut into finitely many polyhedra and then reassembled to form a cube. This and other results showed that a theory of volumes of polyhedra needed limit arguments for rigorous proofs. (See Boltyanskii [3].) Euclid’s Elements included results now considered to be part of number theory, algebra, and irrational numbers. Mathematicians learned from Euclid’s text for two thousand years, and many important developments in mathematics stem from it. The Elements richly deserves its reputation as the most important mathematics book ever written. (See Kline [14, Chapters 4 and 5] for more historical information.)

1.2.5 Exercises for Section 1.2 For proofs and justifications in these problems you may use any of Euclid’s first 48 propositions (found in Appendix A), unless the problem states otherwise. You may also use the axioms in Appendix B. Explanations can cite other properties, such as similar triangles. *1.2.4. (a) Identify which of Euclid’s first 48 propositions (Appendix A) concern constructions. (b) Repeat part (a), replacing constructions with congruence or equality of measure. (c) In addition to the congruence theorems SAS, SSS, AAS, and ASA, there are four other combinations of three congruences, such as AAA. Write out the other possibilities. (d) For each possibility in part (c), determine whether it qualifies as a triangle congruence theorem. If it does, is it just a variation of one of the known ones? If it does not qualify as a theorem, provide a counterexample: that is, two non-congruent triangles that nevertheless share the three pairs of congruences. 1.2.5. Given △ABC and a point D, explain how to construct points E and F so that △D E F is congruent to △ABC. Illustrate your construction and justify it.

*1.2.6. Suppose in △ABC that AB ∼ = AC. (Thus △ABC is isosceles since it has two congruent sides.) Prove that ∠ABC ∼ = ∠AC B. (This corresponds to Euclid I-5.) 1.2.7. Suppose in △ABC that ∠ABC ∼ = ∠AC B. Prove that AB ∼ = AC. (This corresponds to Euclid I-6.)

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Euclidean Geometry

1.2.8. In Figure 1.19 suppose that AC ⊥ B E, AE ⊥ C F, and AD bisects ∠E AC. Use Theorem 1.1.1, triangle congruence theorems, and previous parts to prove the congruence of pairs of triangles in Figure 1.19: *(a) △AD F and △AD B, *(b) △AFC and △AB E, (d) △F D E and △B DC.

(c) △AD E and △ADC,

E

F D

A

B

C

Figure 1.19 1.2.9. In Figure 1.20 assume that AD ∼ = BC and ∠ADC ∼ = ∠BC D. Use only Euclid’s first 28 propositions in your proofs. ∼ ∠C B A. Hint: Find two pairs of congruent triangles formed (a) Prove that ∠D AB = with the given sides and the two diagonals. (b) If E is the midpoint of C D and F is the midpoint of AB, show that E F is perpendicular to both AB and C D. Hint: Draw D F and C F. Recall: Y is the midpoint of X Z if and only if X Y ∼ = Y Z. A

D

F

E

B

C

Figure 1.20 1.2.10. Prove the converse of the Pythagorean Theorem (I-48) using SSS (I-8). The converse states: In △ABC if AB 2 equals AC 2 + BC 2 , then ∠AC B is a right angle. 1.2.11. *(a) For segments with lengths a and b, where a > b, construct segments with lengths a + b and a − b. ← → ← → (b) In Figure 1.21, assume that Q S∥ RT . Explain how the lengths P Q, P R, P S, and P T are related.

1.2 Euclid’s Approach to Geometry I: Congruence and Constructions

19

(c) Suppose in Figure 1.21 that P Q = 1. Which two of the other lengths in part (b) can be chosen to have lengths a and b so that the fourth length is a · b? Which two of the other lengths in part (b) can be chosen to have lengths a and b so that the fourth length is a/b? Explain your answers. R Q P

S

T

Figure 1.21 1.2.12. (a) In Figure 1.22, let AD = 1 and B D = x. Assume that C D is perpendicular to AB √ and AC is perpendicular to BC. Explain why C D = x. (b) Let M be the midpoint of AB and use algebra to show that C M = AM = B M and so A, B, and C are on a circle centered at M. (c) Given segments of length 1 and x, use parts (a) and (b) to explain how to construct √ a segment of length x. √ √ (d) Construct segments of lengths !4 x and 8 x. Generalize. √ (e) Construct a segment of length 1 + 2. C

x 1 A

x M

D

B

Figure 1.22 1.2.13. Begin with a circle and construct the following regular inscribed polygons. You may use earlier constructions in later ones. (An n-sided polygon, or n-gon, is a set of distinct vertices P1 , P2 , . . . , Pn in a plane and the edges (line segments) P1 P2 , P2 P3 , . . . , Pn P1 , with the condition that two edges intersect only at their common endpoint or don’t intersect at all. A polygon is inscribed in a figure if and only if the vertices of the polygon are on the figure and the rest of the polygon is in its interior. A polygon is regular if and only if all the edges and angles are congruent.) *(a) An equilateral triangle *(b) A regular hexagon *(c) A square (d) An regular octagon (e) A regular dodecagon (12-gon) (f) Explain how to inscribe a regular 2n-gon, from one with n sides. Remarks. Carl F. Gauss determined in 1801 which regular polygons could be constructed with straightedge and compass. He showed that, if the number of sides is a k product of a power of 2 and distinct primes of the form (22 + 1), the regular polygon is

20

Euclidean Geometry

constructible. The only known primes of this form, called Fermat primes, are 3, 5, 17, 257, and 65,536. Gauss actually constructed a regular 17-gon in 1796. He conjectured and Pierre Wantzel proved in 1837 that no other regular polygons are constructible by straightedge and compass alone. 1.2.14. Assume that congruent figures have the same area. (a) Given △ABC construct, as in Figure 1.23, midpoints D and E of sides AB and ← → AC, respectively, and perpendiculars AF, BG, and C H to D E. Show that △ABC has the same area as the quadrilateral BC H G. (You may assume that the order of ← → the points on D E is as shown in Figure 1.23.) (b) What orderings of the points D, E, F, G, and H are possible, other than the order in Figure 1.23? Draw some of them. It is a fact that in all cases the areas of △ABC and of BC H G are equal. A

F

G

E

H

D

B

C

Figure 1.23 *1.2.15. Let △ABC be a triangle and G be the intersection of the perpendicular bisector G D of AB and the perpendicular bisector G E of BC. (See Figure 1.24.) (a) Prove that AG, BG, and C G are congruent. (Thus G is called the circumcenter, the center of the circumscribed circle of the triangle.) (b) Let F be the midpoint of AC. Prove that G F is the perpendicular bisector of AC. (Thus the perpendicular bisectors of a triangle are concurrent, that is, they meet in a common point.) G

A

F D B

E

C

Figure 1.24 G is the circumcenter of △ABC.

21

1.2 Euclid’s Approach to Geometry I: Congruence and Constructions

1.2.16. Let △ABC be a triangle and G be the intersection of the angle bisector AG of ∠C AB and the angle bisector BG of ∠ABC. (See Figure 1.25.) Let DG be perpendicular to AB at D and similarly, E G ⊥ BC and F G ⊥ AC. (a) Prove that DG, E G, and F G are congruent. (Thus G is called the incenter, the center of the inscribed circle of the triangle.) (b) Prove that C G is the angle bisector of ∠AC B. (Thus the angle bisectors are concurrent.) A D G

F C E

B

Figure 1.25 G is the incenter of △ABC. 1.2.17. Prove that in a circle the central angle is twice the inscribed angle. That is, in Figure 1.26, m∠AC D, the measure of ∠AC D, is twice m∠AB D, where C is the center of the circle. Hint: Draw the diameter through B and use the isosceles triangles △BC A and △BC D. Also, consider different placements of the points. For example, if m∠AB D > 90◦ , we need to use the angle at C that is more than half of the circle. A B

C

D

Figure 1.26 m∠AB D is twice m∠AC D. *1.2.18. In Figure 1.27 find the measure of ∠ABC assuming that AC, C D, D E, and B E are all the same unknown length x and △ABC is isosceles. Explain your reasoning. D B

x

x

A x

x

E C

Figure 1.27 Find m∠ABC. 1.2.19. In Figure 1.17, show that ∠P AC and ∠P BC are right angles. Hint: consider the circle with center M and use Exercise 1.2.17.

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Euclidean Geometry

*1.2.20. Suppose that P Q RS is a rhombus—that is, all four sides are congruent—and QT and ST are tangent to the circle with center P and radius P Q. (See Figure 1.28.) If ∠Q P S has measure x, find the measures of ∠QT S and ∠R QT . Justify your answers. S x

P

T

R

Q

Figure 1.28 1.2.21. Fill in the missing steps and justifications in the proof below that proves ASA (I-26) follows from SAS (I-4). That is, suppose that the triangle congruence theorem SAS holds and that for △ABC and △D E F we are given that ∠B AC ∼ = ∠E D F, AB ∼ = D E, ∼ ∼ and ∠ABC = ∠D E F. Then, assuming SAS, we claim that △ABC = △D E F. Use only propositions before I-26. Proof. Suppose that SAS (I-4) holds and that for △ABC and △D E F we are given that ∠B AC ∼ = ∠E D F, AB ∼ = D E, and ∠ABC ∼ = ∠D E F. For a contradiction, suppose that the triangles are not congruent. Then the sides BC and E F are not congruent. (Explain why otherwise △ABC ∼ = △D E F.) Without loss of generality, assume that E F is longer than BC and G is between E and F so that E G ∼ = BC. (See Figure 1.29.) ! (a) Prove that △ABC ∼ = △D E G. (b) Show why ∠E D F and ∠E DG must both be congruent and not congruent, giving a contradiction. A D F G B

C

E

Figure 1.29 1.2.22. Fill in the missing steps and justifications in the argument below that shows SSS (I-8) follows from SAS (I-4). That is, suppose that the triangle congruence theorem SAS holds and that for △ABC and △D E F we are given that AB ∼ = D E, AC ∼ = D F, and ∼ ∼ BC = E F. Then, assuming SAS, we claim that △ABC = △D E F. Proof. Suppose that SAS (I-4) holds and that for △ABC and △D E F we are given that AB ∼ = D E, AC ∼ = D F, and BC ∼ = E F. For a contradiction, suppose that the

23

1.2 Euclid’s Approach to Geometry I: Congruence and Constructions

triangles are not congruent. Then some pair of angles must not be congruent. (Explain why otherwise △ABC ∼ = △D E F.) Without loss of generality, assume that ∠B AC is smaller than ∠E D F. We can construct ∠E DG ∼ = ∠B AC with the ray −→ DG inside ∠D E F since ∠E DG is smaller. We may assume, as in Figure 1.30, that DG ∼ = AC. !

(a) Prove that △ABC ∼ = △D E G. (b) Construct F G. Prove that △D F G and △E F G are isosceles. (c) Explain why Euclid I-21 shows that G is not in the interior of △D E F and similarly F is not in the interior of △D E G. (d) Which is bigger, ∠E G F or ∠DG F? Which is bigger, ∠D F G or ∠E F G? (e) Use parts (b), (c), and (d) give a contradiction. D

A

F B

E

C

G

Figure 1.30 *1.2.23. (a) Use Figure 1.31 to rewrite Euclid II-13, given below, and explain why it is equivalent to the law of cosines for an acute angle, also given below. Hint: In Figure 1.31 let AD be perpendicular to BC and x be the length of DC. Proposition II-13: In a triangle, the square on the side subtending an acute angle is less than the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle, namely, that on which the perpendicular falls, and the straight line cut off within by the perpendicular toward the acute angle. The Law of Cosines: In a triangle with sides of length a, b, and c, c2 = 2 a + b2 − 2ab cos(C), where C is the angle opposite side c. (b) Prove your reformulation of Euclid II-13 by using the Pythagorean theorem. (c) Does your proof in part (b) hold if the angle is obtuse? If so, explain why; if not, modify it to hold. (Euclid II-12 handles the obtuse case.) B D a x A

C

Figure 1.31 1.2.24. Two quadrilaterals P Q RS and P ′ Q ′ R ′ S ′ are by definition congruent if all corresponding sides, angles, and diagonals are congruent, a total of ten possible congruences. In

24

Euclidean Geometry

this problem we look for smaller sets of correspondences that are sufficient to show convex quadrilaterals are congruent. (a) Give an example to show that the congruence of the four pairs of sides (SSSS) is not sufficient. It illustrates a basic engineering property: triangles are rigid (SSS); other polygons need triangular bracing to be rigid. (b) One diagonal brace (SSSSD) ensures congruence for convex quadrilaterals. State SSSSD clearly and completely; prove it. (c) State AAAS clearly and completely. Either prove that AAAS is a congruence theorem for convex quadrilaterals or find a counterexample. (d) Repeat part (c) for SASAA, SAASA, and SASSS for convex quadrilaterals. (e) In Figure 1.32, ABC D and E F G H satisfy AB ∼ = E F, BC ∼ = F G, C D ∼ = GH, ∼ ∼ D A = H E, and B D = F H, yet the quadrilaterals are not congruent. Explain why this does not contradict part (b). If they exist, find similar examples for SASAA and SASSS. (f) Look for quadrilateral congruence theorems that hold without the extra condition of convexity. B

F C G

A

D

E

H

Figure 1.32 1.2.25. Begin with a unit length and investigate what other lengths are constructible with a straightedge and compass. (a) Use exercise 1.2.11 to describe how to construct integer and rational lengths. (b) Use Exercises 1.2.11 and 1.2.12 to describe how to construct lengths corresponding to positive numbers built from rational numbers by using square roots and the four arithmetic operations (+, −,×, ÷). Call such numbers constructible. (c) Recall that a straight line has a first-degree equation ax + by + c = 0 and that a circle has a second-degree equation x 2 + y 2 + d x + ey + f = 0. Suppose that the coefficients of two lines, two circles, or a line and a circle are all constructible numbers. Explain why the coordinates of their intersections must also be constructible numbers. Hint: Remember the quadratic formula. (d) Let (s, t) and (u, v) be the coordinates of two points in the plane. Find the equation of the line through them. The coefficients of the equation can be written in terms of s, t, u, v, and the four arithmetic operations. (e) Repeat part (d) for the circle with center (s, t) and passing through (u, v). (f) Explain why parts (c), (d), and (e) ensure that, if the coordinates of the points are constructible numbers, the coordinates of any points that you can construct with lines and circles through them are constructible numbers.

1.2 Euclid’s Approach to Geometry I: Congruence and Constructions

25

1.2.26. Trisecting an angle generally involves a cubic equation. An angle is constructible only if given a length you can construct two other lengths such that the three form a triangle with the desired angle. (a) Show that an angle is constructible if and only if its cosine is constructible. Draw a diagram. (b) Use trigonometry to show that cos(3z) = 4 cos 3 (z) − 3 cos(z). (c) If the cosine of a constructible angle is b, show that you can trisect it if and only if you can construct a!segment whose length x satisfies the equation 4x 3 − 3x = b. ! √ √ 3 3 (Incidentally, x = ( b + b2 − 1 + b − b2 − 1)/2.)

1.2.27. A well-known method for trisecting an angle requires marking of a length on the straightedge, a slightly stronger condition than an unmarked straightedge. Explain why the following method trisects an angle. Hint: Draw D M, use Exercise 1.2.12(b), and look for isosceles triangles. In Figure 1.33 ∠ABC is the angle to trisect, D is a point on AB, D E is perpendicular to BC, and D F is parallel to BC. On the straightedge mark a length Y Z twice B D. Place the straightedge so that it goes through B and so Y is on the segment D E. Slide the straightedge around, keeping the previous two conditions satisfied, until −→ point Z is on the ray D F. In Figure 1.33, points Y ′ and Z ′ indicate the correct positions of Y and Z . M is the midpoint of Y ′ Z ′ . We claim that m∠C B Z ′ is one-third of m∠ABC. A Z'

D Y' B

F M

E

C

Figure 1.33 Trisecting an angle. 1.2.28. (a) Follow the instructions on the website https://www.math.lsu.edu/∼verrill/origami/ trisect/ to trisect an acute angle using paper folding. (b) Use Euclidean propositions (not rotations and reflections) to justify that the procedure does, indeed, give a trisection. (c) Explain how to modify the procedure to trisect an obtuse angle.

1.2.6 Archimedes Archimedes (287–212 B.C.E.) was the greatest mathematician of the ancient world and its outstanding engineer and physicist. Many legends attest to his engineering feats, as well as to his absentmindedness. Upon understanding his law of the lever, as illustrated in Figure 1.34, Archimedes is supposed to have claimed, “Give me a place to stand and I will move the world.” The story continues with the King of Syracuse asking for a practical demonstration of mechanical advantage. So, by himself, Archimedes pulled a fully loaded ship up a beach using

26

Euclidean Geometry

Figure 1.34 Archimedes’ law of the lever: The mass of one weight times its distance from the fulcrum equals the mass of the other weight times its distance from the fulcrum.

a sophisticated arrangement of pulleys. Another time the King asked Archimedes to determine, without harming it, whether his new crown was made entirely of gold. In a bath Archimedes grasped the principle of buoyancy of water and so found a solution to the King’s problem. In the excitement of discovery, Archimedes ran naked from the bath shouting “Eureka!”, meaning “I have found it!” He apparently forgot the world about him when he worked on a problem— drawing figures in ashes or on the oil rubbed on him after a bath. During the extended siege of Syracuse, Archimedes designed machines that greatly aided in its defense and intimidated the Roman soldiers. He was killed by a soldier when the Romans finally won. Archimedes brought great imagination and supreme mathematical expertise to his mathematical investigations. He found and proved many theorems, including the exact area of a circle and of a parabolic section. His most famous results give the surface area and volume of a sphere. He also proved theorems concerning the tangents and centers of gravity of various shapes. A number of problems in this chapter concern other results of Archimedes. He brilliantly used the geometric properties of each shape he considered. Archimedes’ mathematics went beyond his elegant geometric proofs. He gave upper and lower estimates for π . In the Sand Reckoner he devised a number system capable of handling huge numbers, including his estimate for the number of grains of sand in the earth. In his Method, a treatise rediscovered in 1906 after being lost for more than a thousand years, he explained how he used his law of the lever to discover new mathematical results that he later proved rigorously. (The method anticipated Cavalieri’s principle discussed in Exercise 1.5.32.) Archimedes’ geometry epitomized Greek mathematical thought. His flawless proofs of difficult theorems were unsurpassed for more than 1500 years. His work also showed the limitations of Greek mathematics. Each result required its own ingenious approach for its proof, unlike calculus and other modern mathematics. Archimedes’ writings in physics and mathematics inspired scholars for centuries, especially during the Renaissance. Even today the beauty of Archimedes’ mathematics reminds us of what is best in mathematics. (See Stein [25] for more on Archimedes.)

1.3 Euclid’s Approach II: Parallel Lines In addition to congruence and constructions, notions of parallel lines are fundamental to Euclid’s approach to geometry and to high school geometry. Euclid’s postulates allow construction of parallel lines, as Example 1 illustrates. Euclid also proved the familiar theorems of high school geometry about parallel lines cut by transversals, that is, lines intersecting both lines. Definition. Two lines k and m in the Euclidean plane are parallel (k∥m) if and only if they have no points in common or they are equal.

27

1.3 Euclid’s Approach II: Parallel Lines

Remark. This definition differs from Euclid’s definition because here we define a line to be parallel to itself. This sometimes puzzles students and is often overlooked in proofs—both by students and Euclid. Mathematicians realized that it is necessary in order to have a familiar property, I-30, of parallel lines to be universally true. This property, called transitivity, says that if k is parallel to m and m is parallel to j, then k is parallel to j. Our definition allows the possibility of k = j. Example 1. Using a line k and a point P not on k, construct a line m parallel to k with P on m.

S

C3 P

C4

Q

C2 C3

C4

C1 R

T

Figure 1.35 Constructing a parallel line. Solution. Let k be the line through points Q and R and P be a point not on k. (See Figure 1.35.) Construct the circles C1 with center Q and radius Q R, C2 with center P and radius P R, C3 with center P and radius Q R, and C4 with center Q and radius P R. Since C1 and C2 intersect at R, the circles C3 and C4 will intersect in two points at S and T . Let S be the point on the opposite side of P Q from R. Then we have P R ∼ = Q S, Q R ∼ = P S, and P Q ∼ = P Q. By SSS (I-8) △P Q R ∼ = △Q P S and so ∠P Q R ∼ = Q P S. The angles are “alternate interior angles.” Then m, the line through P and S, is parallel to k by I-27. (It is conceptually easier to use perpendiculars to get parallel lines: the perpendicular l to k on P, and m, the perpendicular to l on P. However, this approach is harder to construct using just straightedge and compass.) ♦ Euclid showed great insight in explicitly assuming his fifth postulate, stated below, which was essential to proving the most important of his theorems. It dissatisfied many people because it seemed far from self-evident and too complicated to qualify as a postulate (axiom). Many mathematicians from 200 B.C.E. until 1800 tried unsuccessfully to prove Euclid’s fifth postulate from his other assumptions. Historians suspect that Euclid wasn’t completely comfortable with his fifth postulate, also called the parallel postulate, for he postponed using it until proposition I-29. (See Kline [14, 60, 863–867] for more historical information.) Playfair’s axiom, a more familiar statement about parallel lines, as Theorem 1.3.1 shows, is logically equivalent to Euclid’s postulate if we accept Euclid’s first 28 propositions. (Two statements A and B are logically equivalent provided that we can prove both “If A, then B” and its converse “If B, then A”.) Once we have shown the equivalence, we can use either property in place of the other.

28

Euclidean Geometry

Euclid’s Fifth Postulate. That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on the side on which are the angles less than the two right angles. (See Figure 1.36.) B A

C

D ← →

← →

Figure 1.36 ∠ABC and ∠DC B are less than two right angles. AB and C D intersect on the side of A and D.

Playfair’s Axiom. If a point P is not on a line k, then there is on P at most one line m that does not intersect k. (See Figure 1.37.) R P

m S

Q ← →

k

T ← →

← →

Figure 1.37 If m ∥ k, m ̸= RS, and P is on both m and RS, then RS is not parallel to k. Theorem 1.3.1. Euclid’s fifth postulate is equivalent to Playfair’s axiom, assuming that Euclid’s first 28 propositions hold. Proof. (Euclid ⇒ Playfair) Suppose that the fifth postulate holds and that we are given a point P not on a line k. Example 4 gives us one parallel, say, m, where P Q is perpendicular to both m and k, as shown in Figure 1.37. For Playfair’s axiom we need to show that m is the only parallel ← → ← → to k through P. We let RS be any other line on P with P between R and S. As RS is not m, either ∠R P Q is acute or ∠S P Q is acute. Without loss of generality we assume ∠S P Q to be acute. We now have fulfilled the hypothesis of the fifth postulate: ∠S P Q and ∠P QT together ← → measure less than 180◦ . Hence RS must meet k, showing there is at most one (and only one) parallel to k on P. ← → ← → (Playfair ⇒ Euclid) Suppose that Playfair’s axiom holds and BC falls on AB and C D ← → ← → so that m∠ABC + m∠BC D < 180◦ . We must show that AB and C D meet on the side of A ← → and D. We construct the line C E, as shown in Figure 1.38, such that ∠BC E ∼ = ∠C B F. By ← → ← → ← → Example 4, C E∥ AB. Playfair’s axiom states that there is only one parallel, meaning that C D ← → must intersect AB, say, at G. However, Playfair’s axiom does not tell us directly on which side of B this point G lies. Note that we have a triangle △BC G. Euclid I-17 guarantees that the measures of any two angles of △BC G add to less than 180◦ . Our original assumption about

29

1.3 Euclid’s Approach II: Parallel Lines

angles ∠ABC and ∠BC D fits I-17 perfectly, but that isn’t enough. We must show that the angles on the other side, ∠F BC and ∠H C B, do not satisfy I-17. The measures of all four of these angles must add to 360◦ , and we assumed that the first two add to less than 180◦ . Hence the last two must add to more than 180◦ , implying that they are not part of the triangle △BC G. Hence G is indeed on the same side of B as A is. ! H

C

E D

F

B

G? -->

A ← →

Figure 1.38 On which side of BC is G? It is easier to use Playfair’s axiom to prove theorems about parallels, as the following proof indicates. Recall that we used Theorem 1.3.2 in proving Theorem 1.1.1, which gives 180◦ as the angle sum of a Euclidean triangle. ← → ← → Theorem 1.3.2. (part of I-29) Suppose AB ∥ C D and A and D are on opposite sides of ← → BC. Then the opposite interior angles are congruent: ∠ABC ∼ = ∠BC D. ← → ← → ← → Proof. Suppose AB ∥ C D and A and D are on opposite sides of BC. Following Example 1, ← → ← → we can construct C E with E on the same side of BC as D and with ∠BC E ∼ = ∠ABC. As in ← → ← → ← → Example 1, C E ∥ AB. By Playfair’s axiom, there is at most one parallel to AB through C, so ← → E is on C D and ∠BC D ∼ = ∠ABC. ! The well known properties of parallel lines lead us to familiar geometry shapes, such as parallelograms, which have many nice geometric properties: For example, opposite sides are parallel, opposite sides are congruent, opposite angles are congruent and the diagonals bisect each other. It is worth noting that we do not want all the properties to be part of the definition of a parallelogram since that would raise the issue of whether one property could exist without the others or even whether they can all exist simultaneously. Instead, we pick one of them for the definition and prove the others from it. In a strong sense, our choice is arbitrary, since, as Theorem 1.3.3 and Exercise 1.3.8 show, any one of the four properties listed above implies all of the others. However, the name “parallelogram” makes our choice natural. Definitions. Three or more points on the same line are called collinear. A quadrilateral ABC D consists of four vertices in a plane A, B, C, and D, no three of which are collinear, together with the four segments AB, BC, C D, and D A, where the intersection of any two of them is at ← → ← → most their common vertex. A quadrilateral ABC D is a parallelogram if and only if AB ∥ C D ← → ← → and AD ∥ BC. Theorem 1.3.3. If ABC D is a parallelogram, then opposite sides are congruent: AB ∼ = CD and AD ∼ = BC.

30

Euclidean Geometry

Proof. Let ABC D be a parallelogram and construct AC. (See Figure 1.39.) By the definition of ← → ← → a parallelogram, AB ∥ C D. Also, ∠C AB and ∠AC D are alternate interior angles, so by I-29, they are congruent. Similarly, ∠C AD ∼ = C A, by ASA (I-26), △ABC ∼ = = ∠AC B. Since AC ∼ ∼ △C D A. Hence AB = C D. A similar argument shows that AD ∼ = BC. ! C D

B A

Figure 1.39 Euclid realized the power of parallel lines in developing geometry. However, he could not have realized the profound impact his sophisticated and controversial parallel postulate would have 2000 years later. Chapter 4 investigates the development of non-Euclidean geometries that contradict Euclid’s parallel postulate along with the many familiar theorems that depend on it.

1.3.1 Exercises for Section 1.3 *1.3.1. (a) In Figure 1.40 assume that lines k1 and k2 are parallel, as are lines m 1 and m 2 . If j and k1 are perpendicular and m∠ABC = 35◦ , determine m∠E F G. Explain your reasoning. (b) In the following situations, we are given a relation between the lines k and l and between l and m. Determine what relation, if any, exists between k and m. Justify your answer. (i) k ∥ l and l ∥ m (ii) k ⊥ l and l ⊥ m (iii) k ∥ l and l⊥m (c) Use the definition of a parallelogram to define a rectangle, a square, and a rhombus. k1 m1

F

m2

k2 E

C A

B

G

D j

Figure 1.40 1.3.2. (a) If AB∥C D and the diagonals are congruent (AC ∼ = B D), must ABC D be a parallelogram? Explain. (b) Suppose we define a quadrilateral ABC D to be a whim if and only if AB ∥ C D and BC ∼ = AD. Must a whim be a parallelogram? If so, explain why; if not, provide an example of a whim that is not a parallelogram.

31

1.3 Euclid’s Approach II: Parallel Lines

(c) Suppose we define a quadrilateral ABC D to be a quirk if and only if it has two right angles. Must a quirk be a rectangle? If so, explain why; if not, provide an example of a quirk that is not a rectangle. (d) If a quadrilateral is both a whim and a quirk, what more can you say about it? Explain your answer. *1.3.3. Let P Q RS be a quadrilateral with P Q ∥ RS, P Q ∼ = RS, and such that Q and S are on opposite sides of P R. Construct the segment P R. (a) Use Euclidean theorems to show that △P Q R ∼ = △RS P. (b) Use Euclidean theorems to show that P Q RS is a parallelogram. 1.3.4. Define a hexaparallelogram to be a convex hexagon ABC D E F whose opposite sides are parallel: AB ∥ D E, BC ∥ E F, and C D ∥ F A. (a) Give an example to show that, unlike ordinary parallelograms, the opposite sides of a hexaparallelogram do not need to be congruent. (b) Suppose ABC D E F is a hexaparallelogram and one pair of opposite sides are congruent: AB ∼ = D E. Draw diagonals AE and B D. Use Exercise 1.3.3, △BC D and △E F A to show that BC ∼ = E F and C D ∼ = F A. 1.3.5. Assume that congruent figures have the same area. *(a) Use Euclidean propositions before I-34 to prove in Figure 1.39 that the diagonal AC cuts the area of parallelogram ABC D in half, which is part of I-34. (b) Use Euclidean propositions before I-35 to prove in Figure 1.41 that the parallelograms ABC E and AB D F, which have the same base and the same height, have the same area, as I-35 states. Hint: Use △AF E, △B DC, and the quadrilateral AB D E. (c) Use parts (a) and (b) to prove in Figure 1.42 that the area of △ABC is half the area of parallelogram AB D E, as I-41 states for a triangle with the same base and height as a parallelogram. C E D F

D

C

E B A

B A

Figure 1.41

Figure 1.42

← → ← → 1.3.6. Assume the points in Figure 1.43 have the relationships shown and that AB and D E are parallel. Use Playfair’s axiom and Euclidean propositions before I-29 to prove each of the following, which are parts of I-29. *(a) ∠AB E ∼ = ∠D E H . (These are corresponding angles.) (b) ∠ABG ∼ = ∠H E F. (These are alternate exterior angles.) (c) m∠AB E + m∠D E B = 180◦ . (These are interior angles on the same side.)

32

Euclidean Geometry

H D

F

E

A

C

B G

Figure 1.43 *1.3.7. Prove that a parallelogram with one right angle has four right angles. 1.3.8. Use only Euclid’s first 29 propositions in your proofs. Let ABC D be a quadrilateral. (a) Prove that opposite angles of ABC D are congruent if and only if ABC D is a parallelogram. (b) If the opposite sides of ABC D are congruent, prove it is a parallelogram. (c) If AB∥C D and AB ∼ = C D, prove that ABC D is a parallelogram. (d) Prove that the diagonals of ABC D bisect each another if and only if ABC D is a parallelogram. *1.3.9. Define a kite as a quadrilateral ABC D so that AB ∼ = BC and C D ∼ = AD. (a) Draw a picture of a kite. (b) Prove that ∠B AD ∼ = ∠BC D. ← → (c) Prove that the line B D is the perpendicular bisector of the other diagonal AC. Must B D bisect AC? Explain. 1.3.10. A quadrilateral is a rhombus if and only if all four sides are congruent. (a) Prove that a quadrilateral is a rhombus if and only if its diagonals are the perpendicular bisectors of each other. (b) If ABC D is a rhombus, prove that the diagonal AC bisects ∠B AD. *1.3.11. Give examples of figures that show why we need the following parts of the definition of a quadrilateral. (a) The vertices are in the same plane. (b) No three of the vertices are collinear. (c) The intersection of two of the segments is at most a common vertex. 1.3.12. Suppose in Figure 1.44 that G H I J K is a regular pentagon; that is, its sides are congruent and its angles are congruent. I H

J

K

G

Figure 1.44

33

1.4 Similar Figures

(a) Prove that the diagonals are all congruent. (b) Prove that a diagonal is parallel to the side that doesn’t intersect it. Hint: Find the measures of the angles ∠G H I and ∠H G I and prove your answers are correct. 1.3.13. In Figure 1.45 we are given that Q is the midpoint of P R and that S and T are on the same side of P R with P S ∼ = QT and Q S ∼ = RT . *(a) Prove that P S and QT are parallel. *(b) Prove that ST and P Q are congruent and parallel. (c) Construct U so that it is on the opposite side of ST from Q, SU ∼ = P S, and TU ∼ = RT . Prove that P, S, and U are collinear and R, T, and U are collinear. U T S R Q P

Figure 1.45 1.3.14. Let ABC D be a parallelogram. Show that AB 2 + BC 2 + C D 2 + AD 2 = AC 2 + B D 2 . That is, the sum of the squares of the sides equals the sum of the squares of the diagonals. Hint: Let h be the height of the parallelogram for the base AB and use the Pythagorean theorem. 1.3.15. Prove that the following are logically equivalent to Playfair’s axiom, using Euclid’s first 28 propositions and Theorem 1.3.1. Draw diagrams. (a) If a straight line cuts one of two parallel lines, it cuts the other. (b) Given two parallel lines and a transversal, the alternate interior angles are congruent (Euclid I-29). (c) If k∥l and l∥m, then k∥m (Euclid I-30).

1.4 Similar Figures The mathematical aspects of scale changes have been studied for thousands of years but continue to yield insights in various fields. The famed Greek mathematician Thales 300 years before Euclid was reported to have determined the height of a pyramid in Egypt by forming a proportion with the lengths of its shadow and his own shadow, together with his height. The Euclidean geometric tradition focuses on similar triangles and polygons, which we follow in this section. In Chapter 5 we use transformations to provide a deeper treatment of similar figures. Definition. The nonzero numbers a1 , a2 , . . . , an are proportional to the nonzero numbers b1 , b2 , . . . , bn , respectively, if and only if there is a nonzero number k such that for each i, kai = bi . The number k is the ratio of proportionality.

34

Euclidean Geometry

Exercise 1.4.1.∗ For nonzero numbers show that a1 , a2 are proportional to b1 , b2 if and only if a1 b2 = a2 b1 if and only if a1 /a2 = b1 /b2 if and only if a1 /b1 = a2 /b2 . Definition. Triangles △ ABC and △A′ B ′ C ′ are similar if and only if corresponding angles are congruent and the lengths of corresponding sides are proportional. We write △ABC ∼ △A′ B ′ C ′ . Example 1. Show that the three triangles in Figure 1.46 are similar. C 8 A

6.4

4.8

6 3.6

10 D

B

Figure 1.46 Solution. Each triangle is a right triangle by the converse of the Pythagorean theorem (Euclid I-48). Elementary trigonometry confirms that the other angles are congruent. Furthermore, the sides of the largest triangle (10, 6, and 8) and the smallest triangle (6, 3.6, and 4.8) are proportional, using k = 0.6. Similarly, the ratio k = 0.8 converts the sides of the largest triangle to the sides of the middle triangle (8, 4.8, and 6.4). ♦ To show two triangles to be similar, we must in principle prove the congruence or proportionality for all their corresponding parts. Similarity theorems provide smaller sufficient sets of corresponding parts, analogous to the congruence theorems SAS, AAS, and SSS. Euclid provides the theorems in the sixth book of his Elements. Given how familiar similar triangles are, it may seem surprising that Euclid postpones their consideration so long after the congruence theorems of book one. Recall, however, that the Greeks didn’t consider lengths, angles, and areas as numbers. So he needed to develop a careful theory of proportion, something he did in book five, using the work of Eudoxus. We follow Euclid’s approach, but simplify it with our modern understanding of real numbers. Euclid laid out the key relationship between parallel lines and similar triangles, as in Theorem 1.4.1. Later geometers realized that the theorem showed that the existence of similar but non-congruent triangles was equivalent to Euclid’s fifth postulate. −→ −→ Theorem 1.4.1. (Euclid VI-2) Let AB and AC be two rays with D between A and B and E between A and C. Then BC∥D E if and only if △ABC ∼ △AD E. Proof. Construct segments B E and C D, as in Figure 1.47. (⇒) First suppose that BC∥D E. Then the corresponding angles are equal by I-29. Triangles △B D E and △C E D have the same area because they have the same base (D E) and the same height (the distance between the parallel lines). Thus the areas of the triangles have the same ratio to the area of △AD E, which also has base D E. Now reconsider △AD E and △D B E, using AD and D B as the bases so

35

1.4 Similar Figures

that the two triangles have the same height. Hence the lengths of their bases and their areas are proportional because of the formula area = 12 (base)(height). That is, 12 (height) is the ratio of proportionality. The same can be said about the triangles △AD E and △DC E. Thus the bases AD and D B are proportional to AE and EC, respectively. By the addition properties of proportions, this result shows AD and AB are proportional to AE and AC. B D

A

E

C

Figure 1.47 Exercise 1.4.14 asks you to prove that the sides D E and BC have the same proportion as the other two pairs of sides. (⇐) Exercise 1.4.15 asks you to show the other direction. ! Theorem 1.4.2. (Euclid VI-4) If two triangles have two pairs of corresponding angles congruent, then the triangles are similar. Proof. Let △ABC and △A′ B ′ C ′ have ∠A ∼ = ∠ A′ and ∠B ∼ = ∠B ′ . By Theorem 1.1.1, ′ ∠C ∼ = A′ B ′ , = ∠C . We must show that the three pairs of sides are proportional. Case 1. If AB ∼ then by AAS the triangles are congruent and so similar. Case 2. So, without loss of generality, suppose that AB is longer than A′ B ′ . Construct D on AB such that AD ∼ = A′ B ′ and construct ′ ′ ′ ∼ D E∥BC with E on AC, as in Figure 1.48. Then △AD E = △A B C by AAS. Furthermore, △ABC ∼ △AD E by Theorem 1.4.1. Then, by the definitions of congruent and similar triangles, △ABC ∼ △A′ B ′ C ′ . ! B A'

D

C' A

E C

B'

Figure 1.48

Theorem 1.4.3. (Euclid VI-5) If the three pairs of corresponding sides of two triangles are in proportion, then the triangles are similar. Proof. Exercise 14.16. !

36

Euclidean Geometry

Theorem 1.4.4. (Euclid VI-6) If two pairs of corresponding sides of two triangles are proportional and their included angles are congruent, then the triangles are similar. Proof. Exercise 1.4.17. ! Exercise 1.4.2.∗ Find two triangles △ABC and △D E F so that the sides AB and BC are proportional to D E and E F, respectively, and that ∠A ∼ = ∠D, but the triangles are not similar. We can intuitively extend the idea of similar figures from triangles to polygons and more general shapes, as in Figure 1.49 and Exercises 1.4.10 and 1.4.20. Exercise 1.4.23 attempts a rigorous definition of similar figures. In Chapter 5 transformations provide an easier way of defining similar figures.

Figure 1.49 Definition. The area of a rectangle is the product of the lengths of two of its adjacent sides. Exercise 1.4.3.∗ Use Figure 1.50 to derive the formula for the area of a triangle. C

E

F D

B

A

Figure 1.50

Theorem 1.4.5. If the lengths of the sides of △ABC are k times the lengths of the corresponding sides of △A′ B ′ C ′ , then the area of △ABC is k 2 times the area of △A′ B ′ C ′ . Proof. Exercise 14.18. ! The generalization of Theorem 1.4.5 for areas and volumes of similar figures has important consequences for animals of different sizes, as Galileo noted in 1638. (See Galileo [9, 131.]) To state the generalization let’s suppose that there is a scaling ratio of k from figure A to B.

37

1.4 Similar Figures

Then the area (or surface area) of B is k 2 times the area of A, and the volume of B is k 3 times the volume of A. Consider increasing the size of an animal. The strength of bones and muscles grows roughly with the areas of their cross sections, a factor of k 2 . However, the weight they must handle grows roughly with the volume of the animal, a factor of k 3 . As animals get larger, the bones and muscles must get disproportionately thicker to support and move the greatly increased weight. (Figure 1.51 shows Galileo’s sketch of this effect.) Similarly, the lift of a bird’s wings is roughly proportional to their surface area. The bird’s weight is roughly proportional to its volume. To compensate for their greater weight, large birds have evolved greater wingspreads compared with smaller birds. (McMahon and Bonner [18] provide more information on area and volume questions in nature.)

Figure 1.51 Galileo’s sketch of bones with proportionate strength.

1.4.1 Exercises for Section 1.4 In these problems, you may use Euclidean propositions from Book I (listed in Appendix A) and properties already proven in this section. *1.4.4. Construct a parallelogram ABC D with E midway between C and D. Let F be the intersection of the segment AC with the segment B E. (See Figure 1.52.) Determine the following ratios of lengths or areas and explain how you found them. (a) (b) (c) (d)

The ratio of the length AF to that of FC. The ratio of the area of △AF B to that of △C F E. The ratio of the area of △AD E to that of AD E B. The ratio of the area of AD E B to that of ABC D. C

E D F B A

Figure 1.52

38

Euclidean Geometry

1.4.5. Two vertical poles have wires connecting the top of one pole to the bottom of the other pole. (See Figure 1.53.) Find the height where the wires cross in terms of the heights of the poles and the distance between them. Explain your answer.

?

Figure 1.53 *1.4.6. Figure 1.54 shows an asymmetric version of the “yin yang” symbol with radii of a and b of the two half circles. Find the ratio of the areas of the regions. Explain your answer.

a

b

Figure 1.54 1.4.7. Consider a rectangle with length x and width y, where x > y. (a) Find the ratio x/y so that you can halve the long side of the rectangle to get two smaller rectangles, both similar to the original rectangle. Explain. (b) Redo part (a), but divide the rectangle into n smaller similar rectangles. (c) Suppose that you remove a square with side y to leave a smaller rectangle. Find the ratio x/y so that the smaller rectangle is similar to the original rectangle. Explain. *1.4.8. Given △ABC and segment D E, explain how to construct a point F so that △ABC and △D E F are similar. Justify and illustrate your construction. −→ 1.4.9. Explain why in a triangle △ABC there is a unique point D on AC such that △ABC ∼ △AD B. Describe AD in terms of the sides of △ABC. *1.4.10. Which of the following functions have graphs that are similar to the graph of y = sin(x)? Explain your answer by using their graphs and your intuitive understanding of similarity. Generalize this problem. (a) y = 2 sin(x)

(b) y = sin(2x)

(c) y = 2 sin(2x)

(d) y =

1 2

sin(2x)

39

1.4 Similar Figures

1.4.11. Dynamic software may help you think about this problem. (a) Let ABC D be a quadrilateral and P, Q, R, and S be the midpoints of the four sides, as in Figure 1.55. What can you say about the quadrilateral P Q RS? Justify your answer. Hint: Use the diagonals AC and B D. (b) A quadrilateral ABC D is a trapezoid if and only if it has exactly one pair of parallel sides. (Some books say at least one pair.) Suppose in ABC D that AB ∥ C D and let M be the midpoint of AD and N the midpoint of BC. What can you say about M N compared with AB and C D? Justify your answer. (c) What happens in part (b) if ABC D is a parallelogram? Justify your answer. B Q P C A

R

S

D

Figure 1.55 *1.4.12. In △P Q R let S, T , and U be the midpoints of the sides P Q, Q R, and P R, respectively. Prove that △T U S is similar to △P Q R. See Figure 1.56. R T

U

P

S

Q

Figure 1.56 1.4.13. *(a) Let P, Q, R, and S be four points on a circle with center C and P R and Q S intersect at T , a point inside the circle. (See Figure 1.57.) Prove that the product P T · RT equals QT · ST . Hint: Use Exercise 1.2.17 to find similar triangles. (b) Prove that the products remain equal when the intersection T is outside of the circle. (See Figure 1.58. The products in both parts are called the power of a point T with respect to the circle.)

P

R

P

Q T C

Figure 1.57

R

T

C S

S

Q

Figure 1.58

40

Euclidean Geometry

F D

R r

A

B

C

E G

Figure 1.59 (c) Two circles are externally tangent if and only if they have exactly one point in common and neither center is inside the other circle. For now assume, as in Figure 1.59, that two externally tangent circles with unequal radii have two common tangents that intersect. If the radii of the circles are r and R, with r < R, find the lengths AB and AC in terms of r and R. Explain your answer. (d) Prove that the assumption in part (c) is valid. Hint: See Example 4 of Section 1.2. *1.4.14. Fill in the proof of Theorem 1.4.1 by showing that the sides D E and BC are in the same proportion as the other two pairs of sides. Hint: Construct E F∥AB with F on BC. 1.4.15. Prove the remaining direction of Theorem 1.4.1. 1.4.16. Prove Theorem 1.4.3. 1.4.17. Prove Theorem 1.4.4. 1.4.18. Prove Theorem 1.4.5. *1.4.19. Generalize Theorem 1.4.5 to polygons. Assume that similar polygons can be divided into similar triangles. *1.4.20. "(Calculus) Recall that the area bounded by y = f (x), the x-axis, x = a, and x = b is b a f (x)d x (Figure 1.60). (a) Explain why the region bounded by y = k f (x/k), the x-axis, x = ka, and x = kb is similar to the region shown in Figure 1.60. In particular, explain why you need to use f (x/k) instead of f (x).

y = f(x)

a

Figure 1.60

b

41

1.4 Similar Figures

(b) Use calculus to verify that the area of the region in part (a) is k 2 times the area of the region shown in Figure 1.60. (c) Generalize the problem to volumes of similar regions in three dimensions. 1.4.21. (a) A median of a triangle is the line segment between the midpoint of one side of the triangle and the opposite vertex. Use similar triangles to prove that the medians of a triangle are concurrent. Hint: In Figure 1.61 the medians B N and C M intersect at Q. Find and prove similar triangles. Now repeat using the medians C M and A P without assuming their intersection is also Q—instead prove Q is that intersection. (b) Explain why the intersection of the medians is called the center of gravity (or centroid) of the triangle. C N

P Q B

M

A

Figure 1.61 1.4.22. An altitude of a triangle is the perpendicular line segment from a vertex of the triangle to the line through its other two vertices. Use Figure 1.62 and the following suggestions to prove that the altitudes of a triangle are concurrent. (The common point is called ← → the orthocenter.) Given △ABC, let AF, B E, and C D be its altitudes. Construct RC ← → ← → parallel to AB, R A parallel to BC and P B parallel to AC. Show ABC R, AB QC, and AC B P are parallelograms and △ABC is similar to △P Q R. Then show the altitudes of △ABC are the perpendicular bisectors of △P Q R. Now use Exercise 1.2.15. C

Q

R E A

P

D B

F

S

Figure 1.62 1.4.23. Let { Ai : i ∈ I } and {Bi : i ∈ I } be the vertices of two polygons A and B, respectively. Then define polygons A and B to be similar ( A ∼ B) if and only if (i) there is a positive r such that for all i, j ∈ I , Ai A j = r Bi B j and (ii) for all i, j, k ∈ I , ∠Ai A j Ak ∼ = ∠Bi B j Bk . *(a) How does this definition compare with your intuition of similar polygons? *(b) Prove that any two squares are similar.

42

Euclidean Geometry

*(c) Provide an example of two quadrilaterals whose sides are proportional but are not similar. *(d) Provide an example of two quadrilaterals whose angles are congruent but are not similar. *(e) Explain why two regular polygons with the same number of sides are similar. (f) Suppose that A and B have at least three vertices. Prove that A ∼ B provided that for all i, j, k ∈ I , △Ai A j Ak ∼ △Bi B j Bk . (g) If you allow the index set I to be infinite, does the definition of similar still make sense? If so, give an example; if not, explain why not. 1.4.24. (a) Given a triangle △ABC and a square, explain why there are three points P, Q, and R on the square such that △P Q R ∼ △ABC. (b) What shapes can you substitute for the square in part (a) and still have the result hold? Explain. (c) Given a quadrilateral ABC D and an equilateral triangle, can you find four points P, Q, R, and S on the triangle such that P Q RS is similar to ABC D? If this can work in general, explain why; if not, give a counterexample.

1.5 Three-Dimensional Geometry Although we live in a three-dimensional world, visualizing three-dimensional geometric figures is harder than visualizing two-dimensional figures. Euclid devoted the final three books of his Elements to three-dimensional geometry. We explore non-axiomatically the geometry of polyhedra (the plural of polyhedron) and the sphere to help you deepen your visual intuition. Working with physical models aids understanding beyond what textbook figures can do.

1.5.1 Polyhedra Polyhedra continue to fascinate people, much as they did the ancient Greeks. We assume an intuitive understanding of a polyhedron because an exact definition is more complicated than our treatment warrants. (See Lakatos [15].) In brief, a polyhedron is composed of vertices, edges, and faces. The faces are polygons. At each vertex (or corner) at least three faces meet, and at each edge (line segment) two faces meet. Further, no two faces intersect in more than their common vertex or edge, if they have something in common. Pyramids are easy to visualize and the simplest polyhedra. A pyramid has a polygon for its base, say, with n edges and n vertices, one more vertex not in the plane of the base called the apex, and n triangular faces that are determined by the apex and the n edges of the base. The polyhedron with the fewest vertices, edges, and faces is a triangular pyramid or tetrahedron, shown in Figure 1.63.

Figure 1.63 Pyramid

43

1.5 Three-Dimensional Geometry

Exercise 1.5.1. Verify that a pyramid with an n-gon for a base has V = n + 1 vertices, E = 2n edges, and F = n + 1 faces. The familiar formula of the volume of a rectangular box as its length times its width times its height generalizes readily to prisms, as defined below. However, as Example 1 indicates, general volumes are more complicated. The ancient Egyptians knew how to find the volume of pyramids and the ancient Greeks proved that volume was correct. Definitions. A prism is a polyhedron with two congruent and parallel polygons for bases connected by parallelograms between corresponding edges of the bases. (See Figure 1.64.) The volume of a prism is Bh, where B is the area of one of the bases and h is the perpendicular distance between them.

Figure 1.64 Prism Example 1. Use calculus to find a formula for the volume of a pyramid. Base = B

Apex

x

h

Figure 1.65

Solution. Without loss of generality select a coordinate axis so that the origin is at the apex of the pyramid and the x-axis is perpendicular to the base (Figure 1.65). Suppose that the area of the base is B and that the x-coordinate of points on the base is h > 0, the height of the pyramid. A cross section of the pyramid in a plane parallel to the base is a polygon similar to the base. If a cross section’s x-coordinate is x, then Exercise 1.4.19 shows its area to be (x/ h)2 B. Thus the

44

Euclidean Geometry

volume of the pyramid is & 3 'h % h & '2 % B h 2 x 1 x Bd x = 2 x dx = B = h B. ♦ 2 h h 0 3h 3 0 0 Exercise 1.5.2.∗ Modify Example 1 to find the formula for the volume of a cone. Example 2. Figure 1.66 illustrates how we can dissect a cube into three pyramids. A cube with a side of s is a prism with a base of area s 2 and so a volume of s 3 . Each of the square pyramids has a base with area s 2 and a height of s. So the sum of the volumes of the pyramids is 3( 13 s · s 2 ) = s 3 , as expected. In general all polyhedra can be dissected into pyramids, much as all polygons can be dissected into triangles. ♦

Figure 1.66 Three pyramids can form a cube. Exercise 1.5.3.∗ Explain how, in principle, you can use pyramids to find the volume of the polyhedra in Figure 1.68. The five polyhedra shown in Figure 1.67 possess regular properties and a high degree of symmetry. They are often called the Platonic solids because the Greek philosopher Plato was fascinated by them and his discussion of them is the oldest that survives. A convex polyhedron is regular provided that all its faces are the same regular polygon and the same number of polygons meet at each vertex. (See Coxeter [4, Chapter 10] for more information.) The noted Swiss mathematician Leonhard Euler (1707–1783) developed a formula relating the number of vertices, edges, and faces for a large collection of polyhedra, including all convex ones. Because a careful proof, even for convex polyhedra, would require an overly long and technical development, we don’t present one here. (See Beck et al [2] and Lakatos [15].) Euler’s formula has a wide variety of applications in geometry, graph theory, and topology. Exercise 1.5.4.∗ Find the number of vertices, V , of edges, E, and of faces, F, for the regular polyhedra in Figure 1.67. Theorem 1.5.1. Euler’s Formula If a convex polyhedron has V vertices, E edges, and F faces, then V − E + F = 2.

45

1.5 Three-Dimensional Geometry

Figure 1.67 The regular polyhedra: tetrahedron, cube, octahedron, dodecahedron, and icosahedron. Outline of Reasoning. Given a convex polyhedron (Figure 1.68) we can stretch the net of its vertices and edges to lay it out on a plane (Figure 1.69). The number of vertices and edges remains the same, but the number of faces is one less than the original polyhedron. If a face of the net is not a triangle, we divide it into triangles by adding edges, as the dashed line DG indicates in Figure 1.69. A technical argument shows that the division can always be G

B

C

E

G

B A

F

E

C

D

Figure 1.68 A convex polyhedron.

A

F

D

Figure 1.69 A planar representation of the polyhedron in Figure 1.68

made and that doing so increases the number of faces and edges by the same amount. Another argument shows that we can carefully eliminate edges one at a time, each time eliminating either a face or a vertex. Thus we preserve the value V − E + F. (Try this process with Figure 1.69.) In the end, we are left with a triangle, for which we have V − E + F = 3 − 3 + 1 = 1. Thus the original polyhedron must satisfy Euler’s formula. ♦

46

Euclidean Geometry

Example 3. The angle sum of the angles of a tetrahedron is 4 × 180◦ = 720◦ because a tetrahedron has four triangles for faces. Similarly the six square faces of a cube give an angle sum of 6 × 360◦ = 2160◦ . A simple formula for the angle sum of all convex polyhedron depends on shifting our focus from the faces to the vertices, as Descartes discovered over one hundred years before Euler found his formula. ♦ Theorem 1.5.2. Descartes’ Formula If a convex polyhedron has V vertices, then the angle sum of all of the angles around all the vertices is 360◦ (V − 2). Proof. We derive this result from Euler’s formula. We rewrite V − E + F = 2 as V − 2 = E − F. The left side multiplied by 360 is Descartes’ formula: 360◦ (V − 2) = 360◦ (E − F) = 180◦ (2E − 2F). Next we relate the right side to the angle sum. Let Fi be the number of faces with i edges on them; for example, F3 is the number of tri$∞ Fi . (Of course, the sum isn’t angles. The number of faces is F = F3 + F4 + F5 + · · · = i=3 infinite.) Figure 1.70 illustrates that a convex face with i edges has i − 3 diagonals from a vertex, dividing the face into i − 2 triangles. By Exercise 1.1.10, a face with i edges has an angle sum $∞ Fi 180◦ (i − 2). We can also count the number of 180◦ (i − 2). Thus the total angle sum is i=3 $∞ i F is of edges in terms of the Fi . There are Fi faces with i edges on each. However, i=3 $∞ i i Fi . too big, as each edge is on two faces and so is counted twice. Therefore 2E = i=3 $∞ $∞ ◦ ◦ ◦ F 180 (i − 2) = 180 (i F − 2F ) = 180 (2E − 2F) = Thus the angle sum is i i i i=3 i=3 360◦ (V − 2). !

Figure 1.70 Exercise 1.5.5. Verify Descartes’ formula for the five regular polyhedra shown in Figure 1.67. The angle sum around each vertex of a convex polyhedron must be less than 360◦ . We call the difference between 360◦ and the angle sum at a vertex the angle defect of that vertex. Descartes’ formula tells us that the total angle defect of all vertices is 720◦ .

1.5.2 Geodesic Domes Buckminster Fuller (1895–1983) related the concept of angle defect to the strength of geodesic domes, which he invented. He was a prolific inventor who devoted years to designing economical, efficient buildings. Traditional buildings, designed as modified rectangular boxes, require the use of a lot of material to enclose a volume and support a roof. In contrast, a sphere, the shape that minimizes the surface area for a given volume, is expensive to build because of its curved surface. Fuller avoided the drawbacks of the box and the sphere by starting with an icosahedron. The domed effect of the icosahedron distributes the weight of the structure evenly,

47

1.5 Three-Dimensional Geometry

as spherical domes do, without needing a curved surface. Furthermore, the triangular faces of the icosahedron are structurally stronger than the rectangular faces of traditional buildings. Fuller was able to increase building size by dividing each of the icosahedron’s twenty faces into smaller triangles. To maximize the strength of the building, he found that he needed to arrange the smaller triangles so that all the vertices were on the surface of a sphere. He coined the name geodesic dome for a convex polyhedron whose faces are all triangles. We consider only domes based on an icosahedron. The frequency of a dome is n, where each of the original triangles is divided into n 2 smaller triangles (Figure 1.71). To determine a geodesic dome we need to know the measures of the edges and angles. As Example 4 illustrates, such three-dimensional calculations rely extensively on two-dimensional geometry.

Figure 1.71 Example 4. Find the edge lengths and angle measures for the two-frequency dome shown in Figure 1.72, if the radius O A of the sphere is 1. C E′

C F′

E

D′ A

D B

O

F

A

B

D′ O

C

C

E A

A

F D

B

B

Figure 1.72 Each face (△ABC) of the icosahedron becomes four faces on the two-frequency dome.

−−→ −−→ −−→ Points D, E, and F are on O D ′ , O E ′ , and O F ′ , respectively, and the same distance from O as A, B, and C are.

48

Euclidean Geometry

Solution. From Exercise 1.5.15 we know that the length AB ≈ 1.05146. Because D ′ and E ′ are the midpoints of AB and AC, AD ′ = D ′ E ′ ≈ 0.52573. By the Pythagorean theorem, −−→ O D ′ ≈ 0.85065. The ray O D ′ intersects the sphere at D and, again by the Pythagorean theorem, AD ≈ 0.54653. By Theorem 1.4.4 the triangles △O D ′ E ′ and △O D E are similar. Hence D E ≈ 0.61803. Next we can use the law of cosines, c2 = a 2 + b2 − 2ab cos(C), to determine that m∠D AE ≈ 68.86◦ . Theorem 1.1.1 gives us 55.57◦ for m∠AD E and m∠AE D. The angles of the other corner triangles are the same, and the angles of the center triangle △D E F are all 60◦ because it is equilateral. ♦ The strength of an icosahedron relies on two facts. First, triangular faces are structurally stable on their own, a consequence of the congruence theorem SSS. Second, the triangular faces distribute forces well because of the angles where they meet. Further, a smaller angle defect of a vertex indicates that the forces are better distributed. There are three regular polyhedra with triangular faces: the tetrahedron, the octahedron, and the icosahedron. Their angle defects at each vertex are 180◦ , 120◦ , and 60◦ , respectively. Clearly the icosahedron is the best for distributing force and it is the most dome-like. Buckminster Fuller’s higher frequency domes extend this idea. As the number of vertices increases with the frequency of the dome, Descartes’ formula necessitates a decrease in the angle defects at the vertices. Furthermore, the proper design of the dome minimizes the largest of the angle defects. Such a design will also make the triangles roughly equilateral. Together the small angle defect and approximately equilateral shapes of the triangles distribute the weight of the dome better and so strengthen the dome. High-precision technology and modern materials enable domes to be structurally stable with ◦ angle defects as small as 12 . Exercise 1.5.6.∗ For the two-frequency icosahedron of Example 4, show that the angle defects of the two kinds of vertices are 15.7◦ and 17.72◦ . The strength of geodesic domes is extraordinary. The U.S. Air Force tested the strength of a 55 foot diameter fiber glass geodesic dome in 1955 before choosing domes to house the radar antenna of the Distant Early Warning system. They linked the vertices of the dome to a winch connected to a 17-ton concrete slab buried under the dome. They had intended to tighten the winch until the dome collapsed under the strain and then measure the breaking point. However, instead of collapsing, the dome withstood the stress and actually lifted the concrete slab. The great strength of geodesic domes enables them to enclose large spaces without interior supports. The largest geodesic dome, over 400 feet in diameter, far surpasses the largest space without interior support of any conventional building. (See Edmondson [6] and Kenner [13] for more on Fuller and geodesic domes.)

1.5.3 The Geometry of the Sphere The needs of astronomy and global navigation have prompted the study of the geometry of the sphere for centuries. Astronomers use the inside surface of a sphere to describe the positions of the stars, planets, and other objects. For navigational purposes, the earth is essentially a sphere. Ship captains and airplane pilots seek the most direct route, following the curving surface of the earth rather than an Euclidean straight line. The shortest trip from Tokyo to San Francisco goes considerably north of either city (Figure 1.73). The shortest path on the surface of a sphere connecting two points is a great circle—a circle with the same radius as the sphere. A great

49

1.5 Three-Dimensional Geometry

circle is the intersection of the sphere and a plane that passes through the center of the sphere (Figure 1.74). Circles of longitude and the equator are great circles. The circles of latitude, except for the equator, are not great circles. (The circles of latitude are sometimes called “parallels of latitude,” a misleading name.)

Figure 1.73 Shortest path from Tokyo to San Francisco.

Figure 1.74 A great circle on a sphere.

Great circles play much the same role on the sphere that straight lines do on the plane. In fact, the great circles on a sphere satisfy the first, second, fourth, and fifth of Euclid’s postulates in Appendix A. (We can also define circles to satisfy the third postulate.) However, the geometric properties of lines and great circles differ in important ways. Two lines intersect in at most one point, whereas two distinct great circles always intersect in two diametrically opposed points, called antipodal points. We use the same letter to denote antipodal points, placing a prime on one of them. Thus there are no parallel great circles. Instead, two great circles divide the sphere into four regions called lunes. We define the angles two great circles make to be the angles at which the two planes meet (Figure 1.75). These angles are also the angles of the corresponding

Figure 1.75 The angles between two great circles.

50

Euclidean Geometry

lunes. Three great circles with no common point of intersection form eight spherical triangles (Figure 1.76). We will measure angles in radians to simplify Theorem 1.5.3 and its proof. Exercise 1.5.7.∗ Use Figure 1.75 to explain why the area of a lune with an angle of α radians is 2αr 2 . (The surface area of a sphere of radius r is 4πr 2 .) Exercise 1.5.8.∗ Suppose that a spherical triangle △ABC has angles measuring π /2, π /3, and π/3. Verify that the angle sums for the seven related spherical triangles, such as △ABC ′ , also are more than π = 180◦ . In Theorem 1.5.3, we show not only that the angle sum of a spherical triangle is more than π, but also that it is related to the triangle’s area. The spherical triangle △ABC shown in Figure 1.76 determines three overlapping lunes B AC A′ , ABC B ′ , and AC BC ′ . The opposite spherical triangle △A′ B ′ C ′ determines three other lunes that do not intersect those for △ABC. Together the six lunes cover the entire sphere. A

C′

B′

C B A′

Figure 1.76 Three great circles form eight spherical triangles.

Theorem 1.5.3. The area of a spherical triangle is proportional to the difference between its angle sum and π . More precisely, on a sphere with radius r , the area of a spherical triangle with angle measures of α, β, and γ is (α + β + γ − π)r 2 . Proof. The area covered by the lunes B AC A′ , ABC B ′ , and AC BC ′ is 2πr 2 , or half the sphere because the opposite lunes cover a symmetric region. The three lunes each cover △ABC. Then 2πr 2 = Area(B AC A′ ) + Area(ABC B ′ ) + Area(AC BC ′ ) − 2 · Area(△ABC) = 2αr 2 + 2βr 2 + 2γ r 2 − 2 · Area(△ABC). Solving for Area(△ABC) gives (α + β + γ − π )r 2 . !

51

1.5 Three-Dimensional Geometry

Definition 1.5.4. The spherical excess of a spherical triangle with angle measures α,β , and γ is α + β + γ − π. Exercise 1.5.9.∗ Convert the formula in Theorem 1.5.3 to degrees. Recall that 180◦ = π radians. Exercise 1.5.10.∗ Explain why, in spherical geometry, if two triangles on a sphere are similar, they are congruent. (For more information on spherical geometry, see McCleary [17, 3ff].)

1.5.4 Exercises for Section 1.5 Physical models greatly aid in visualizing and solving some of these problems. *1.5.11. (a) Redo Example 2 for a rectangular box with length l, width w, and height h. (A rectangular box is a prism all of whose six faces are rectangles.) (b) For the rectangular box of part (a), explain why the long, interior diagonal d satisfies the “three-dimensional Pythagorean theorem” l 2 + w 2 + h 2 = d 2 . (c) Can we redo Example 2 for a triangular prism? Explain your answer with a figure. 1.5.12. Find the volume of the prism whose base is a regular n-gon with side s and whose height is h. Hint: Use trigonometry to find the area of the base. 1.5.13. *(a) Find formulas for the number of vertices, edges, and faces of a prism in terms of n, the number of vertices in one of the bases. (See Figure 1.64.) (b) Verify Descartes’ formula for a general prism. 1.5.14. An antiprism is derived from the corresponding prism. As in Figure 1.77, we rotate one of the bases relative to the other base so that edges of one base align with vertices of the other. An antiprism is a polyhedron with two congruent, parallel, and rotated polygons for bases connected by triangles between each edge of each base and the corresponding vertex of the other base. (a) Find formulas for the number of vertices, edges, and faces of an antiprism in terms of n, the number of vertices in one of the bases. (b) Verify Descartes’ formula for a general antiprism.

Figure 1.77 An antiprism. *1.5.15. Suppose that the edge of a cube is 1 unit long. (a) Find the distances from the center of the cube to the center of a face, to the midpoint of an edge, and to a vertex. (See Figure 1.78.) (b) What percentage of the volume of a circumscribed sphere does the cube occupy?

52

Euclidean Geometry

Figure 1.78 (c) What percentage of the cube’s volume does the inscribed sphere occupy? (d) Describe the polyhedron obtained by connecting four vertices of the cube, no two of which are adjacent. What is its volume? Hint: Consider cutting pyramids from the cube to make it. 1.5.16. Suppose a rectangular box has length l, width w, and height h, where l > w > h. (a) Find the distances from the center of the box to the center of each face, to the midpoint of each type of edge, and to a vertex. (b) Find the ratio of the volume of a circumscribed sphere to the volume of the box. (c) Find the ratio of the volume of an inscribed sphere to the volume of the box. 1.5.17. A cuboctahedron is a polyhedron built from a cube by cutting off a triangular pyramid at each vertex so that adjacent pyramids intersect at the midpoints of the edges of the cube. (See Figure 1.79, where the original cube is shown with dashed segments.) (a) Verify Euler’s formula for a cuboctahedron. (b) Verify Descartes’ formula for a cuboctahedron. *(c) Find the volume of a cuboctahedron whose edges all have length l. Hint: How long is the side of the cube? *(d) Suppose each edge has length l. Find the distance from the center of the cuboctahedron to each vertex, to the center of a square face, and to the center of a triangular face.

Figure 1.79 A cuboctahedron.

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1.5 Three-Dimensional Geometry

1.5.18. (a) Construct models of the five regular polyhedra. (See Figure 1.67 or Wenninger [28].) (b) Prove that there are only five regular polyhedra, using angle defect. Hints: Find the largest and smallest number of equilateral triangles that can fit around a vertex with a positive angle defect. Repeat for squares and regular pentagons. Why do regular polygons with more than five sides not need to be considered? *1.5.19. We can build the dual of certain convex polyhedra by replacing each face of the original polyhedron with a vertex for its dual at the center of gravity of the face. Two such vertices are connected by an edge in the dual if and only if the original faces were connected by an edge. Unfortunately, the edges don’t always define faces, which are polygons in a plane. We restrict our consideration to polyhedra for which we do obtain faces. That is, consider the faces of the original polyhedron that have a particular vertex in common. We require that the centers of gravity for them are in the same plane and so form a polygon with the edges of the dual connecting them. (The concept of a dual of a polyhedron can be defined more generally. See [27].) For example, the dual of a cube is an octahedron. (See Figure 1.67.) (a) Describe the dual of a prism, which is called a bipyramid. Give formulas for the number of vertices, edges, and faces of a bipyramid in terms of n, the number of vertices in a base of the corresponding prism. (b) Draw or build the dual of triangular prism. (c) Describe the duals of each of the regular polyhedra. (d) Describe how the values of the vertices, edges, and faces of a polyhedron and its dual relate to each other. 1.5.20. A polyhedron is self-dual if and only if its dual is similar to the original polyhedron. (a) Explain why a pyramid can be self-dual. (b) Consider the polyhedron formed by gluing together a pyramid and a prism with the same base. Such a polyhedron is called an extended pyramid. Give formulas for the number of vertices, edges, and faces of an extended pyramid in terms of n, the number of vertices in a base of the corresponding prism and pyramid. Can an extended pyramid be self-dual? (c) Look for other self-dual polyhedra. 1.5.21. Repeat Exercise 1.5.15 (a), (b), and (c) for a regular octahedron with edges 1 unit long. (See Figure 1.80.)

Figure 1.80 A regular octahedron.

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1.5.22. Repeat Exercise 1.5.15 (a), (b), and (c) for a regular tetrahedron with edges 1 unit long. (See Figure 1.81.)

Figure 1.81 A regular tetrahedron.

1.5.23.

(a) Construct the shape shown in the left of Figure 1.82 with three 3 × 5 inch note cards. Two of the cards need 3 inch slits in their centers, as in the middle figure. The third card needs the slit extended to one of the 3 inch sides. The twelve corners of the cards approximate the vertices of a regular icosahedron. *(b) Modify part (a) to find the exact coordinates of the vertices of a regular icosahedron as follows. Let the slits be on the x-, y-, and z-axes and let the dimensions of the modified cards be 2 × 2b, as in the right side of Figure 1.82. Why are the twelve vertices at (±1, ±b, 0), (0, ±1, ±b), and (±b, 0, ±1)? Find b. z (0, –1, b)

(0, –1, b)

y

slit x

Figure 1.82

1.5.24. Suppose that the edge AB of a regular icosahedron is 1 unit long (Figure 1.83). (a) Find the lengths of the short diagonal AC and the long diagonal BC. (See Exercise 1.5.23.) (b) Find the length of the edge of a regular icosahedron inscribed in a sphere of radius 1. (c) Repeat Exercise 1.5.15 (a) for a regular icosahedron.

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C

A

B

Figure 1.83 A regular icosahedron.

*1.5.25. Find eight vertices of a regular dodecahedron that are the vertices of a cube. (See Figure 1.67.) 1.5.26. A first stellation of a polyhedron, if it exists, replaces each of its faces with a pyramid whose base is the face and whose triangles are in the planes of adjacent faces, assuming that they meet in a common point on the correct side of the base. (See Figure 1.84, which shows a regular octahedron and its first (and only) stellation.) Johannes Kepler (1571–1630) and later mathematicians investigated stellations. (a) Find the number of vertices, edges, and faces for the first stellation of the regular octahedron, which Kepler called the stella octangula. Explain your values. (b) Find the number of vertices, edges, and faces for the first stellation of the regular dodecahedron, which Kepler called the small stellated dodecahedron. (Kepler found the other two stellations of the dodecahedron as well.) (c) Find the number of vertices, edges, and faces for the first stellation of the regular icosahedron. (d) Explain why the regular tetrahedron and the cube do not have first stellations.

Figure 1.84 A regular octahedron and its first stellation.

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(e) Find the number of vertices, edges, and faces for the first stellation of the cuboctahedron. (See Exercise 1.5.17.) (f) Find formulas for the number of vertices V ′ , edges E ′ , and faces F ′ for the first stellation of a polyhedron in terms of V, E, and F of the original polyhedron and the number F3 of its triangular faces, the number F4 of its quadrilateral faces, etc., assuming that it has a first stellation. Explain your formulas. *1.5.27. On a sphere let N be the north pole and A and B be two points on the equator with ( m∠AN B = 90◦ . Let C, D, and E be the midpoints of the arcs ( AB, ( AN , and B N, respectively. ( ( (a) Explain and illustrate why C N, D B, and ( AE intersect in a common point, F. (b) Find the angle sum of the spherical triangle △AC F and its spherical excess. (c) Verify that the area of △AC F agrees with Theorem 1.5.3.

1.5.28. What is the maximum spherical excess a spherical triangle can have? Explain. *1.5.29. For the spherical triangle △N AB of Exercise 1.5.27, suppose the radius of the sphere is 1. ( (a) Find the length of the arc ( AN . (The arcs ( AB and B N have the same length.) ( (b) Let P be a point on the arc AN . Find the length of the arc ( B P. (c) Relate the length of the arc ( A P to the measure of ∠AB P.

1.5.30. Without using Theorem 1.5.3, show that spherical excesses are additive: If D is on the short arc of the great circle through B and C, prove that the excess of △ABC is the sum of the excesses of △AB D and △ADC. Illustrate the situation. 1.5.31. (Calculus) The derivative d A/dr of πr 2 , the area of a circle of radius r , is the circumference of the circle. Explain geometrically why the area and circumference are related in this way. Explain why the same relationship holds for the volume and surface area of a sphere. Hint: Consider adding a thin strip around a circle of radius r . Approximately how much area is in it? 1.5.32. Cavalieri’s principle is often an axiom in high school texts because it provides an elementary way to prove results about volumes of curved shapes. Bonaventura Cavalieri (1598–1647), a pupil of Galileo, used his principle to find volumes (and areas) before the advent of calculus. Archimedes used a similar idea to find volumes, but he thought this idea wasn’t sufficiently rigorous to count as a proof. Cavalieri’s Principle: Let A and B be two solids included between two parallel planes. If every plane P parallel to the given planes intersects A and B in sections with the same area, then A and B have the same volume. (a) Use Cavalieri’s principle and Figure 1.85 to show the following result of Archimedes. The volume of a cylinder whose height is twice its radius equals the volume of a sphere of the same radius plus twice the volume of a cone with the same radius and a height equal to the radius. (b) Find two solids A and B with different volumes and a family of nonparallel planes so that A and B intersect each plane in sections of equal area.

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Figure 1.85 Compare the areas of the cross sections of a cylinder, a sphere, and a double cone at the same height.

1.5.33. *(a) Find the number of vertices on a two-frequency geodesic dome. Explain your answer. (See Figure 1.72.) Find the average angle defect of the vertices. Verify that this corresponds with the values in Exercise 1.5.6. *(b) Find the number of vertices on a three-frequency dome. Explain your answer. (See Figure 1.86.) (c) Show that the number of vertices on an n-frequency dome based on an icosahedron is 10n 2 + 2. Hint: Find formulas for the number of new vertices on one edge of the icosahedron and the number of new vertices on one face of the icosahedron. Then use the numbers of vertices, edges, and faces of an icosahedron. (d) Use part (c) to find the average angle defect at each vertex of an n-frequency dome. What is the largest practical value of n if the average angle defect must be 1◦ for a dome? 2 (e) Find the number of edges and faces on an n-frequency dome. Verify that Euler’s formula holds for geodesic domes.

C

C

J′

I′ A

B

D′ M′ E′

A

J

I A

D

E

O

O

C

C B

I J A DE

Figure 1.86 A three-frequency dome.

B

B

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1.5.34. Figure 1.86 shows a design for a three-frequency dome. The nine triangles of the dome are “lifted” from the nine equilateral triangles dividing △ABC so that the vertices are all on a sphere of radius 1. Find the lengths and angle measures as follows, assuming that D E = D I . (a) Explain why there are just three different lengths, AD, D E, and D J . Explain how to find all the angle measures from m∠D AI and m∠D J E. (b) From Example 4, AB ≈ 1.05146. Find D ′ E ′ , D ′ M ′ , O M ′ , and O D ′ . (c) Use the law of cosines to find m∠D ′ O E ′ , m∠AO D ′ , AD, D E, O J ′ , D ′ J ′ , D J , m∠D AI , and m∠D J E. (d) Find the angle defect at A, D, and J . (e) Verify that the average angle defect of a three-frequency dome from Exercise 1.5.33 (d) agrees with your answers in part d). 1.5.35. (a) On a sphere reproduce Euclid’s construction of an equilateral triangle (Section 1.3). What happens when the length is more than one-third the circumference? (b) Give an example on a sphere to show that Euclid I-16 is independent of Euclid’s first four postulates. 1.5.36. A spherical cap is the set of points on a sphere of radius R whose distance (measured along the sphere) is at most r from a point A on the sphere. (See Figure 1.87.) (a) Find the circumference of the spherical cap in terms of r and R. (b) (Calculus) Verify that the area of the spherical cap is 2π R 2 − 2π R 2 cos(r/R). Hint: Use radians. Let x be as shown in Figure 1.87. Recall that the area of the surface obtained y = f (x) between x = a and x = b about the x-axis ! " b by revolving is given by a 2π f (x) 1 + ( f ′ (x))2 d x. Calculate the integral for a spherical cap for general a and b. Then determine a and b in terms of r and R. (c) Archimedes proved that the surface area of the spherical cap is equal to the area of a circle whose radius equals the (straight-line) distance from A to a point on the circumference of the cap. Verify Archimedes’ theorem.

r x R

Figure 1.87 A spherical cap.

1.5 Three-Dimensional Geometry

59

1.5.37. Curiously, the Euclidean plane is “triangle complete,” which Euclid showed in I-22. However, Euclidean space is not “tetrahedron complete.” A geometric space is triangle complete if and only if for three lengths a, b, and c satisfying the triangle inequality (a + b ≥ c) in any order there is a triangle with those sides in that space. Find six lengths so that any three satisfy the triangle inequality, but no tetrahedron in Euclidean space has its six edges with those lengths. (For more on the idea of tetrahedron completeness, see [24].)

1.5.5 Buckminster Fuller R. Buckminster (“Bucky”) Fuller (1895–1985) transcended traditional career classifications, including inventor, engineer, and philosopher. Throughout his life he sought to develop ideas and technology that would help all people. His most famous creation is the geodesic dome; there are more than 300,000 such domes in the world. He developed “synergetic geometry,” which he used to analyze the theoretical strength of domes, where forces are distributed in unusual ways. The structural analysis applied to another of his inventions, tensegrity figures, which integrated tension and compression in a novel way. (See Project 14.) In general his many ideas and inventions aimed at efficiency: “to do more with less.” Already in 1927, early in the age of airplanes, he argued for the advantage of flying polar routes between North America and Europe (and other destinations) to save on fuel and time. Some of his many inventions include an automobile with just three wheels, modular apartments and a flat world map based on an icosahedron to minimize distortion. In 1985 chemists honored Fuller by naming a newly discovered form of pure carbon “buckminsterfullerene” because the geometry of the 60 carbon atoms reminded them of a geodesic dome. (See Figure 6.45.)

1.5.6 Projects for Chapter 1 1. Use Geometer’s Sketchpad, Geogebra, or other dynamical geometric software to explore geometric ideas. Here are some suggestions, in addition to replicating figures in this chapter. (a) For a triangle △ABC construct three equilateral triangles △AB D, △AC E, and △BC F so that they are exterior to △ABC. Let G, H , and I be their centers. What can you say about △G H I ? What happens when the equilateral triangles overlap △ABC? (b) For a triangle △ABC that is not equilateral, construct its circumcenter, its centroid, and its orthocenter. (See Exercises 1.2.15, 1.4.21, and 1.4.22.) Euler found interesting properties about the three points. Why can’t the triangle be equilateral? (c) For any triangle investigate the nine-point circle of a triangle. (See [8, 62–63].) (d) Let A, B, C, D, E, and F be six distinct points on a circle (in any order). ← → ← → ← → ← → Find the intersection P of AB and D E, the intersection Q of BC and E F, and ← → ← → the intersection R of C D and AF . Pascal showed that the points are collinear. Investigate this situation. Can two of the six points be the same point? Repeat with A, C, and E on one line and B, D, and F on another line.

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2. To find the area of an irregular quadrilateral the Egyptians used a recipe equivalent to the formula 14 (a + c)(b + d), where a and c are the lengths of two nonadjacent sides of the quadrilateral and b and d are the lengths of the other two sides. (a) For which types of quadrilaterals is the formula exact? (b) Use Geometer’s Sketchpad to verify that the formula always appears to give a result at least as large as the actual area. (The area was used for tax purposes, so an overestimate would be to the advantage of the government.) (c) Use trigonometry to show that the formula does give at least as large a value as the area for convex quadrilaterals. (d) Redo part (c) for nonconvex quadrilaterals. 3. Archimedes and others estimated the ratio of the circumference of a circle to its diameter (π) by using regular polygons—the greater the number of sides, the more accurate the estimate of π. (a) Consider polygons inscribed in a circle of radius 1 (Figure 1.88). Find a formula for the length y of the side of a regular 2n-gon in terms of the length x of the side of a regular n-gon. (b) For a regular hexagon inscribed in a circle of radius 1, the side has a length of 1 and the hexagon has a perimeter of 6, an approximation of 2π . Use the formula from part (a) to find the perimeters of regular 12-gons, 24-gons, etc. to give better lower estimates of 2π. (c) Write a computer program that will print out the approximations of π found by the formula in part (a) for polygons with 3 × 2i sides, where 1 ≤ i ≤ 30. When do round-off errors stop succeeding approximations from being improvements? (d) Find upper estimates for π by using circumscribed regular polygons in parts (a) , (b), and (c). Hint: The formula for tan(2θ ) will help. Remark. The Arab mathematician Nasir-Eddin (1201–1274) used essentially this method (without modern notation or computers!) with a regular polygon of 3 × 228 sides. He wanted to ensure that the error in his estimate of the circumference of the universe, based on then current ideas of its radius, was no more than the width of a horse’s hair. y

1

1

Figure 1.88

x

1.5 Three-Dimensional Geometry

61

4. Tangram is a Chinese puzzle made with the seven shapes shown in Figure 1.89. They can, if properly arranged, form a square, as well as many other interesting shapes. Any figure that can be made with these seven shapes is called a tangram. Construct all convex tangrams. Here are some suggestions to help you: Let the short sides of the smallest triangle have a length of 1. Find the lengths of the sides, the angles, and the area for each of the seven shapes. What angles can appear at the corners of a convex tangram? Why is there only one triangular tangram? Why can a convex tangram not have more than eight sides? (See Read [21] for the convex tangrams, some background, and an open mathematical question on tangrams.)

Figure 1.89 The seven pieces of a tangram puzzle can be arranged to form a square. 5. Figure 1.90 shows a pantograph, a device formerly used for enlarging designs. (a) Explain why the design traced by the pencil is similar to the original design. Hint: Why do the bars need to form a parallelogram? (b) Determine where the holes should be drilled so that the pencil will trace a figure whose dimensions are twice those of the original design. Repeat for k times the original design. (c) Make a pantograph and use it to enlarge several designs. (d) How could you alter a pantograph to reduce a design? 6. Given the vertices of a regular n-gon, we can connect every other vertex or every third vertex, etc., to create a figure with intersecting edges. The figure obtained by connecting every k th vertex of a regular n-gon is denoted {n, k}. If the path following along successive edges goes to every vertex, we call the figure a regular star polygon.

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Figure 1.90 Fix the lower left base of the pantograph. As the stylus in the middle of the bottom traces a design, the pencil in the lower right draws an enlargement. Adjusting where the strips cross changes the scale of the enlargement. (See Figure 1.91.) If the edges form disjoint cycles, we call the figure a regular star figure. (See Figure 1.92.) (a) For which values of n and k do we get a regular star polygon? (b) Some regular star figures are made of unions of regular polygons, as in Figure 1.92. Some are made of unions of regular star polygons. For which values of n and k is the regular star figure a union of regular polygons? In terms of n and k, how many polygons? If a regular star figure is a union of smaller regular star figures, determine how many, in terms of n and k.

Figure 1.91 The star polygon {5,2}.

Figure 1.92 The star figure {6,2}.

7. We can try to measure how close a nonconvex set is to being convex. Call a point P in a set a guard point if and only if for every other point Q in the set, P Q is in the set. (A guard at a guard point can see every spot in the set.) The ratio of the area of the set of guard points to the area of the set is called the inside convexity of a set. (a) Find the inside convexity for each shape depicted in Figure 1.93. Find the worst place to put a guard, that is the point from which the guard would see the smallest

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(b) (c) (d) (e) (f) (g)

percentage of the total area. From the worst place, what percentage of the entire area does a guard see? Explain your answers. Another way to measure the convexity of a set is to measure how much must be added to the set to obtain a convex set. Define the outside convexity of a set S to be the ratio of the area of S to the area of the smallest convex set containing S. Find the outside convexity of the shapes shown in Figure 1.93. Explain why the inside (or outside) convexity of a convex set is 1. Find a set with positive area whose inside convexity is 0. Find a set with positive area whose outside convexity is 0. Find a nonconvex set whose inside and outside convexity is 1. Explain. Compare the definitions of inside and outside convexity with your intuition of how close a set is to convex. Look for a better definition. a b a b

Figure 1.93

8. Investigate statements equivalent to Euclid’s fifth postulate, besides those in Exercise 1.3.15, where we assume Euclid’s first 28 propositions. For example, Theorem 1.1.1 (the angle sum of a triangle), Theorem 1.1.2 (the Pythagorean theorem), and the existence of similar, non-congruent triangles are equivalent. Also, the property that parallel lines are everywhere the same distance apart is equivalent to them. 9. A ribbon wrapped around a box can be removed without cutting, stretching, or untying it (Figure 1.94). Try to do so with an actual ribbon on a box. Then model the situation geometrically and explain why it works. It may be easier to explain first with a rectangle, as at the right of Figure 1.94.

Figure 1.94 10. (a) Construct the thirteen Archimedean solids (Wenninger [28]). The Archimedean solids have two or more kinds of regular polygons for faces and the same number, types, and arrangement of faces around each vertex. There are also an infinite family

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of prisms and an infinite family of antiprisms satisfying the conditions. (See Exercise 1.5.14.) (b) Find the angle defect at a vertex for each Archimedean solids. (c) Explain why there can be no other Archimedean solids. (d) Build duals of some of the Archimedean solids. (See Exercise 1.5.19.) 11. Define a deltahedron to be a strictly convex polyhedron all of whose faces are equilateral triangles. (A convex polyhedron is strictly convex if no two faces lie in the same plane. We use the name deltahedron because the capital Greek letter delta $ looks like a triangle.) (a) Find an equation relating E and F in deltahedra. Justify your equation. (b) Explain why there are at most five triangles at any vertex of a deltahedron. Use this to find an inequality relating V and E in a deltahedron. (c) Use parts (a) and (b) to show that in a deltahedron E ≤ 30 and E must be a multiple of 3. List all the possible candidates for values of V , E, and F of deltahedra. (d) Build a complete set of deltahedra. Hint: There is no deltahedron with E = 27. (See Beck et al [2].) 12. Euler’s formula applies to many, but not all, polyhedra that are not convex. Imagine polyhedra as made of rubber and inflate them. Then the ones that can be inflated to look like a sphere will satisfy Euler’s formula. There are shapes with polygonal faces that have holes. (See Figure 1.95.) We will include these shapes as polyhedra, although some sources exclude them. (a) Draw or build other examples of polyhedra with holes. (b) The generalized Euler’s formula applies to polyhedra with holes. Let H be the number of holes in the polyhedron. Find an equality relating V , E, F, and H . (The value H = 0 should give you Euler’s formula.) You may find that you need to make the definition of a polyhedron explicit. Beck et al [2] gives applications of this generalized formula. (c) Generalize Descartes’ formula to polyhedra with holes.

Figure 1.95

13. Build a geodesic dome. (See Kenner [13].)

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14. Build some tensegrity figures, such as that shown in Figure 1.96. (See Kenner [13] and Pugh [20].)

Figure 1.96 A tensegrity figure. 15. Given a triangle and a point P in the interior of the triangle, let d(P) be the sum of the distance of P to each side of the triangle. Investigate which points P have the minimum and maximum values for d(P). Find a relationship of the points with the same sum of distances. Extend your investigations to quadrilaterals or to pyramids. (See Polster [19].) 16. Investigate the golden ratio and phyllotaxis. (See Coxeter [4] and Livio [16].) 17. Investigate the geometry of four and more dimensions. (See Coxeter [4] and Rucker [22].) 18. Investigate the history of geometry. (See Aaboe [1], Eves [7], Kline [14], Stein [25], and Struik [26].) 19. Investigate the geometry of the sphere. (See Henderson [12] and McCleary [17].) 20. Write an essay giving your understanding of why mathematics is certain and why it is applicable. Compare your ideas with those of Plato and Aristotle. (See also Grabiner [10].)

1.5.7 Suggested Readings 1. Aaboe, A., Episodes from the Early History of Mathematics, Washington, D. C.: Mathematical Association of America, 1964. 2. Beck, A., M. Bleicher, and D. Crowe, Excursions into Mathematics, Natick, Mass.: A K Peters, 2000. 3. Boltyanskii, V., Equivalent and Equidecomposable Figures, Lexington, Mass.: D. C. Heath, 1963.

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4. 5. 6. 7.

Coxeter, H., Introduction to Geometry, New York: John Wiley & Sons, 1969. Durbin, J. R., Modern Algebra, New York: John Wiley & Sons, 2009. Edmondson, A. A Fuller Explanation, New York: Birkh¨auser, 1987. Eves, H., An Introduction to the History of Mathematics, New York: Holt, Rinehart and Winston, 1976. Eves, H., Fundamentals of Modern Elementary Geometry, Boston: Jones and Bartlett, 1992. Galileo (Galilei), Dialogues Concerning Two New Sciences, New York: Dover, 1954. Grabiner, J., The centrality of mathematics in the history of Western thought, Mathematics Magazine, 1988, 61(4):220–230. Heath, T., The Thirteen Books of Euclid’s Elements, 3 vol. Mineola, N.Y.: Dover, 1956. (Also available on line.) Henderson, D., Experiencing Geometry in Euclidean, Spherical, and Hyperbolic Spaces, 2nd ed. Upper Saddle River, N. J.: Prentice-Hall, 2001. Kenner, H., Geodesic Math and How to Use It, Berkeley: University of California Press, 1976. Kline, M., Mathematical Thought from Ancient to Modern Times, New York: Oxford University Press, 1972. Lakatos, I., Proofs and Refutations, New York: Cambridge University Press, 1976. Livio, M., The Golden Ratio, New York: Broadway Books, 2002. McCleary, J., Geometry from a Differentiable Viewpoint, Port Chester, N.Y.: Cambridge University Press, 1994. McMahon, T., and J. Bonner, On Size and Life, New York: Scientific American Library, Scientific American Books, 1983. Polster, B., Viviani a` la Kawasaki: take two, Mathematics Magazine, 2014, 87(4): 280–283. Pugh, A., An Introduction to Tensegrity, Berkeley: University of California Press, 1976. Read, R., Tangrams, Mineola, N.Y.: Dover, 1965. Rucker, R., Geometry, Relativity and the Fourth Dimension, Mineola, N.Y.: Dover, 1977. Salmon, W., Zeno’s Paradoxes, Indianapolis: Bobbs-Merrill, 1970. Sibley, T. Q., The possibility of impossible pyramids, Mathematics Magazine, 2000, 73(3): 185–193. Stein, S., Archimedes, Washington, D. C.: Mathematical Association of America, 1999. Struik, D., A Concise History of Mathematics, Mineola, N.Y.: Dover, 1967. Wenninger, M., Dual Models, New York: Cambridge University Press, 1983. Wenninger, M., Polyhedron Models for the Classroom, Reston, Va.: National Council of Teachers of Mathematics, 1966. (Also available online.)

8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28.

2

Axiomatic Systems A

B

P

D

C

R

Q

E

F

Figure 2.0 Pappus configuration Geometers have used axioms and figures as the basis for proving a variety of Euclidean theorems for centuries. More recently, they have developed new axiomatic systems to investigate geometric properties from different starting points. Figure 2.0 gives a highly symmetric illustration of Pappus’ theorem, proven over 1,600 years ago for Euclidean geometry and generalized to other geometries much more recently. In a separate recent development the Pappus configuration inspired its own finite geometry, discussed in Exercise 2.3.14, for which it is a model. Pappus supposed that A, B, and C are on one line and D, E, and F are on another line. His theorem states that the points P, Q, and R have to be on a line, where P is the intersection of AE and B D, Q is the intersection of AF and C D, and R is the intersection of B F and C E.

2.1 From Euclid to Modern Axiomatics Geometry is the science of correct reasoning on incorrect figures. —George P´olya (1887–1985)

2.1.1 Overview and History At the end of the nineteenth century, Euclid’s work received scrutiny far surpassing the preceding 2000 years’ efforts. Thes examination revealed how much more precise and explicit mathematics has become, without calling into question any of Euclid’s results. For example, consider Euclid’s proof of his first proposition, given in shortened form (from Heath [6, volume I, 241]. Can you find the logical gap in his proof, given his postulates and common notions? (See Figure 2.1.)

67

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D

C

A

E

B

Figure 2.1 Proposition I-1 On a given finite straight line to construct an equilateral triangle. Proof. Let AB be the given finite straight line. . . . With center A and radius AB let the circle BC D be described; again, with center B and radius B A let the circle AC E be described (postulate 3). From point C, in which the circles cut each other, to points A and B, join the straight lines C A and C B (postulate 1). Now, point A is the center of the circle C D B, so AC is equal to AB (definition 15). Again, . . . B A is equal to BC. . . . And things equal to the same thing are also equal to one another . . . (common notion 1). Therefore . . . AC, AB, BC are equal to one another. Therefore △ABC is equilateral. ! Euclid’s construction is straightforward, and he followed it with a proof that the three sides are congruent. However, Euclid never showed, nor could he show from his assumptions, that the circles must intersect. This logical gap indicates one of the many implicit assumptions Euclid made that were hard to detect because they were “obvious.” (We did the same construction and made the same assumption in Example 1 of Section 1.2.) In modern terms Euclid assumed that lines, circles, and other figures are continuous. In the fifth postulate and numerous other places Euclid assumes an order to points on lines with no justification or discussion. (In Example 4 of Section 1.2 we assumed such ordering to talk about a point “outside” a circle and we often used the concept of betweenness in Chapter 1.) A number of mathematicians recognized that rigorous proofs require explicit axioms about continuity, order, and other properties. David Hilbert (1862–1943) developed the most quoted axiomatic system for Euclidean geometry. He, along with others, also developed separate axioms for the real numbers. More recently, high school geometry textbooks give more accessible axioms for Euclidean geometry that provide a basis for proving theorems without requiring the sophistication of systems developed for mathematicians. The high school axiom systems rely on students’ already developed understanding of the real number system for continuity and order. The familiar numerical and algebraic properties simplify geometric proofs. In this section we investigate axiomatic systems in general. Section 2.2 investigates two axiomatic systems for Euclidean geometry—one high school system and Hilbert’s system. In Section 2.3 we study the connection between an axiomatic system and its models. While models have been around informally as long as the idea of an example, the modern concept arose only with the development of new geometries in the nineteenth century.

2.1.2 Axiomatic Systems An axiomatic system provides an explicit foundation for a mathematical subject. Axiomatic systems include seven parts: the logical language, rules of proof, undefined terms, axioms, definitions, theorems, and proofs of theorems. We discuss the last five parts—the ones with geometric content.

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Undefined terms. Euclid attempts to define a point as “that which has no part.” The definition is more a philosophical statement about the nature of a point than a way to prove statements. Euclid’s definition of a straight line, “a line which lies evenly with the points on itself,” is unclear as well as not useful. In essence, points and lines are so basic to Euclid’s work that there is no good way to define them. Other common undefined geometric terms include “on,” “between,” and “congruent.” Mathematicians realized centuries ago the need for undefined terms in order to establish an unambiguous beginning. Otherwise, each term would have to be defined with other terms, leading either to a cycle or an infinite sequence of terms. Neither is acceptable for carefully reasoned mathematics. We then define all other terms from the initial, undefined terms. However, undefined terms are, by their nature, unrestricted. Axioms. How can we be sure that two people mean the same thing when they use undefined terms and the terms defined from them? In short, we can’t. The axioms of a mathematical system provide the key: They tell us how the undefined terms behave. Axioms are sometimes called operational definitions because they describe how to use terms and how they relate to one another, rather than telling us what terms “really mean.” Indeed, mathematicians permit any interpretation of undefined terms, as long as all the axioms hold in that interpretation. In Section 2.3 we explore the interplay between axioms and their interpretations in models. Unlike the Greek understanding of axioms as self-evident truths, we do not claim the truth of axioms. This does not mean that we consider axioms to be false. We freely choose axioms to formulate the fundamental relationships we want to investigate. From a logical point of view, the choice of axioms is arbitrary; in actuality, though, mathematicians carefully pick axioms to focus on particular features. For example, in perspective drawing parallel lines intersect at a point on the horizon. Projective geometry is an axiomatic system in which any two lines intersect in a point. This system, discussed in Chapter 7, enables us to understand many consequences stemming from perspective. However, we don’t need to decide whether “in truth” there are parallel lines or if all lines intersect. In the concrete world of atoms and energy, there are no mathematical lines at all. Nevertheless, these axiomatic systems and many others have given us a profound understanding of the world. Axiomatic systems allow us to formulate and logically explore abstract relationships, freed from the specificity and imprecision of real situations. Mathematicians build two basic types of axiomatic systems. One completely characterizes a particular mathematical system. For example, Hilbert’s axioms characterize Euclidean geometry completely. (This statement may appear to contradict G¨odel’s incompleteness theorem. However, axiom V-2 is a “second-order axiom,” a concept beyond the level of this text. See Delong [2] for information on second-order logic.) The second type focuses on the common features of a family of structures. For example, although infinitely many different vector spaces exist, all satisfy certain essential properties. The general study of vector spaces greatly aids the development of theories in economics, physics, mathematics, and other fields. The axiomatic systems unite a wide variety of examples within one powerful theoretical framework. Indeed, part of the power comes from the fact that they are not tied to any particular application. Mathematical Definitions. Mathematical definitions differ from everyday definitions. We don’t try to give the meaning of the term. Instead a mathematical definition must be suitable for use in proofs: it selects some essential property or properties of the thing defined. This becomes an “if and only if ” characterization. For example, in Section 1.3 we defined a quadrilateral to be a

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parallelogram if and only if non-adjacent sides are parallel. We proved other familiar properties about parallelograms from that. Conversely, we could show that many of the properties forced the quadrilateral to have non-adjacent sides parallel and so were, indeed, parallelograms. Another type of definition deals with notation, such as defining ∥ as a shorthand for “is parallel to.” A mathematical definition shouldn’t assume the existence of the thing defined. Existence needs to be proven or assumed in the axioms. For example, we can define a right pentagon as a five sided polygon all of whose angles measure 90◦ . In Euclidean geometry there are no right pentagons, even though we could prove rather useless theorems about them. (By Euclid I-28, for example, if ABC D E is a right pentagon, then AB ∥ C D.) There are geometries with right pentagons—in particular, hyperbolic geometry, discussed in Chapter 4. Mathematical definitions are built from undefined terms and previously defined terms, even if a definition doesn’t capture our full intuition of the term. Consider, for example, the familiar concept of an angle, which has a primarily visual meaning for most people. We formally define −→ −→ −→ the angle ∠ABC as a point B and two rays B A and BC. The ray B A is the set of points X on ← → the line AB such that X is between A and B, X is B, or B is between A and X . Thus we can reduce the notion of an angle to the undefined terms point, line, on, and between. The formal definition works in proofs, even if it seems more abstract than Euclid’s definition “A plane angle is the inclination to one another of two lines in a plane which meet one another and do not lie in a straight line.” However, it is hard to think of using something as vague as an “inclination” in a proof. The same undefined terms are sufficient to define convex, an important concept in modern mathematics. The Greeks never distinguished between nonconvex sets and convex sets, as in Figure 2.2, perhaps because they never considered betweenness. Intuitively, a convex set has no dents. (See Exercise 1.1.10 for the formal definition.)

Nonconvex Sets

Convex Sets

Figure 2.2 Exercise 2.1.1.∗ Write a definition of the segment P Q using only the undefined terms point, line, on, and between. Exercise 2.1.2.∗ Which of the terms in Euclid’s definitions (Appendix A) do you think should be undefined terms? Which of the remaining definitions fit our modern understanding of definitions? Theorems and Proofs. Theorems and their proofs are the most distinctive parts of mathematics, whether in an axiomatic system or a less formal system. In an axiomatic system, a theorem is a statement whose proof depends only on the rules of logic using the axioms, previously proven theorems, and the definitions. This condition ensures that the entire edifice of theorems rests securely on the explicit axioms of the system. Euclid followed the spirit of this idea. Proofs of theorems in an axiomatic system cannot depend on diagrams, even though diagrams have been part of geometry since the ancient Greeks drew figures in the sand. We need

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the powerful insight and understanding that diagrams provide. However, a corresponding risk comes with the use of pictures: We are liable to accept as intuitive a step that does not follow from the given conditions. Euclid’s first proof, discussed previously, shows how easy it is to include implicit assumptions. Euclid’s minor sins of omission never led him to an erroneous result, but the potential remains for a diagram to mislead us, as in Example 1, used with permission from Dubnov [3, 15]. Even though diagrams are not permissible in a proof in an axiomatic system, they can and should be included to help us understand the ideas. They must be studied critically to ensure that the illustrated relationships are proved or legitimately assumed. Example 1. Claim: A rectangle inscribed in a square is a square. Draw a figure that contradicts the claim. Then try to find the error in the “proof ” below. “Proof.” Let the rectangle M N P Q be inscribed in the square ABC D (Figure 2.3). Drop perpendiculars from P to AB and from Q to BC at R and S, respectively. Clearly, P R ∼ = QS because the segments match the sides of the square ABC D. Furthermore, the rectangle’s diagonals are congruent: P M ∼ = Q N . So △P M R ∼ = △Q N S, and hence ∠P M R ∼ = ∠Q N S. Consider the quadrilateral M B N O, where O is the intersection of Q N and P M. Its exterior angle at N is congruent to the interior angle at M, so the two interior angles at N and M are supplementary. Thus the interior angles at B and O must be supplementary. But ∠ABC is a right angle and hence ∠N O M is a right angle. Therefore the diagonals of the rectangle M N P Q are perpendicular. Hence M N P Q is a square. D

P

Q

S O

A

C

M

N R

B

Figure 2.3 The preceding argument is correct up to the conclusion ∠P M R ∼ = ∠Q N S. Then the diagram shown in Figure 2.3 misleads us to think that ∠O N B is supplementary to ∠O M R. However, these angles can be congruent if we switch N and S in the diagram and correspondingly move Q down. Does this second case fit your earlier figure? ♦

2.1.3 A Simplified Axiomatic System Undefined terms: point between Notation: P(Q)R indicates that Q is between P and R. Axioms: (i) (ii) (iii) (iv)

If P(Q)R, then R(Q)P. If P(Q)R, then not P(R)Q and P ̸= R. There are at least two distinct points. For any two points P and Q there is a point R so that P(Q)R.

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The axioms fit with at least some of our understanding of between. The first axiom allows us to reverse order: If Q is between P and R, it is between R and P. Axiom (ii) restricts the meaning of between. With three points there are six possible betweenness relations. The first two axioms talk about two of them based on an initial one. Our first theorem considers the other three. To compensate for the simplicity of the axioms we will need to use proofs by contradiction frequently. (For more on proofs and proof formats see Appendix F or Sibley [13].) Theorem 2.1.1. For three points A, B, and C, if A(B)C, then not C(A)B, not B(C)A, and not B( A)C. Proof. Let A, B, and C be any points and suppose A(B)C. By axiom (i), C(B) A and so by axiom (ii), not C(A)B. Next suppose for a contradiction that B(C) A was correct. By axiom (i) we would have A(C)B, in violation of axiom (ii). Thus we can’t have B(C) A. Exercise 2.1.3(a) asks you to finish the proof. ! Theorem 2.1.2. For three points A, B, and C, if A(B)C, then A, B, and C are distinct points. Proof. Let A, B, and C be any points and suppose A(B)C. By axiom (ii), A ̸= C. Next suppose for a contradiction that B = C. From this equality and the given A(B)C, we could switch B and C to get A(C)B. But that contradicts axiom (ii). So B ̸= C. Exercise 2.1.3(b) asks you to finish the proof. ! Only axiom (iii) assures us that there any points at all and axiom (iv) gives us a way to generate some more. Together they suggest a natural question: “How many points are there?” Theorem 3 takes a first step in that direction. Exercise 2.1.4 continues the process. In the proof of Theorem 3 we can’t just name points; we need to prove that they are all different. Theorem 2.1.3. There are at least four distinct points. Proof. By axiom (iii) there are at least two distinct points, say, D and E. By axiom (iv) there is a point F so that D(E)F. Again by axiom (iv) there is a point G so that E(F)G. From Theorem 2 we know that D ̸= E, D ̸= F, E ̸= F, E ̸= G, and F ̸= G. Further, if D = G, we could replace G in E(F)G to get E(F)D, contradicting Theorem 1. So D ̸= G. ! Definition. A set S is convex if and only if for all P, Q, and R if P and R are in S and P(Q)R, then Q is in S. Theorem 2.1.4. If S and T are convex sets, then their intersection is convex. Proof. See Exercise 2.1.5. ! The axiomatic system models some of our intuitions of betweenness. Its simplicity allows us to reliably check that our proofs don’t have any hidden assumptions. But our intuitions

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suggest many further properties we expect to be true. In a complicated system how can we possibly make explicit all the assumptions we need and eliminate all risk of incorrect proofs? Mathematical logic, developed by Frege and many others, involves the use of a formal language so austere that a proof can be checked in a purely mechanical manner, free from human intuition. In principle, a computer could check a proof to decide its validity. Consider, for example, the statement, “Two distinct points have a unique line on them.” We can make the statement more explicit as follows: “For all points A and B if A ̸= B, then there is a unique line k such that A is on k and B is on k.” Finally we could turn this more explicit statement into pure symbols: ∀A∀B(¬(A = B) ⇒ (∃!k : (R( A, k) ∧ R(B, k)))). Clearly, we need to decode the symbols so that people not familiar with them can read the statement. (We don’t use these logic symbols, but for your information, ∀, ¬, ⇒, ∃, !, and ∧ represent for all, not, implies, there exists, a unique, and and, respectively. The variable R represents the relation is on.) If we encode the entire axiomatic system, including the rules of proof, into a formal language, we can mechanically determine whether a string of symbols encodes a logical proof of the string of symbols encoding the theorem. There will be no risk of an inappropriate inference, but there will be an incredible barrier to human understanding. Finding a proof in such a language is a daunting task. Axiomatic systems are a workable compromise between the austere formal languages of mathematical logic and Euclid’s work, with its many implicit assumptions. Mathematicians are people: We need both the careful reasoning of proofs and the intuitive understanding of the content. Axiomatic systems provide more than a way to give careful proofs. They enable us to understand the relationship of particular concepts, to explore the consequences of assumptions, to contrast different systems, and to unify seemingly disparate situations under one framework. In short, axiomatic systems are one important way in which mathematicians obtain insight. (Heath’s edition of Euclid’s Elements [6] provides detailed commentary on its logical shortcomings. Delong [2], Sibley [13], and Wilder [15] explore logic and axiomatic systems in more detail.)

2.1.4 Exercises for Section 2.1 *2.1.3. (a) Finish the proof of Theorem 2.1.1. (b) Finish the proof of Theorem 2.1.2. 2.1.4. To the axiomatic system of subsection 2.1.3 add this axiom: (v) If P(Q)R and Q(R)S, then P(R)S. (a) If P(Q)R and Q(R)S, prove that P(Q)S. (b) Prove that there exist at least five points for this augmented axiom system. (c) Use induction to prove that for all natural numbers n there are at least n points. 2.1.5.*(a) Prove Theorem 2.1.4. (b) Extend Theorem 2.1.4 as follows: Suppose each set Si is convex, for i in a finite ) ) or infinite index set I . Prove that i∈I Si is convex, where i∈I Si = {P : for all i ∈ I , P ∈ Si }. 2.1.6. Extend Theorem 2.1.3 to show that there are at least five distinct points. (Do not use the extra axiom of Exercise 2.1.4.)

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2.1.7. Consider the axiomatic system with the undefined terms team and beats and the following axioms. Use t → v to denote “t beats v.” (i) There exists at least one team. (ii) For every team t there are exactly two teams s and r so that t → s and t → r . (iii) For all teams t and s, if t → s, then not s → t. *(a) Prove that no team beats itself. *(b) Prove that there are at least five teams. (c) Suppose there are exactly five teams. Prove that for any team t there are two teams v and w so that v → t and w → t. Hint: Count the total number of →. 2.1.8. Consider the axiomatic system with the undefined terms point and adjacent and the following axioms. Use ◃▹ to denote “is adjacent to.” (i) (ii) (iii) (iv)

There is at least one point. If P ◃▹ Q, then Q ◃▹ P and P ̸= Q. Every point has exactly three distinct points adjacent to it. If P ◃▹ Q and P ◃▹ R, then not Q ◃▹ R. (a) Prove that there are at least six points. (b) Suppose that you omit axiom (iv). Can you still prove that there are at least six points? If so, prove it; otherwise find the largest number of points that must exist and prove your answer. (c) Suppose that we change axiom (iii) to require exactly four distinct points adjacent to any point but leave the other axioms unchanged. How many points must exist? Prove your answer. Generalize.

*2.1.9. Consider the axiomatic system with the undefined terms apples, oranges, and likes and the following axioms. (i) (ii) (iii) (iv) (v)

There is an apple. No apple likes another apple. No orange likes another orange. Everything likes at least two other things. Everything is an apple or an orange. (a) Prove that nothing is both an apple and an orange. (b) Determine the minimum number of apples and oranges and prove your answer. (c) Add the following axiom to this system and redo part (b): If x likes y, then y does not like x.

2.1.10. Consider the axiomatic system with the undefined terms point, circle, and on and the following axioms. (i) There are a circle and a point not on that circle. (ii) Every three distinct points have a unique circle on them all. (iii) Every circle has at least three points on it. (a) Prove that there exist at least four points. (b) Prove that there exist at least four circles.

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(c) Prove that two distinct circles have at most two distinct points on them both. (d) Prove that every point is on at least three circles. Hint: Consider the cases of whether or not the given point is on the circle of axiom (i). 2.1.11. Develop axioms to go with the undefined terms line, ∥, and ⊥. (You may include other undefined terms.) Prove at least one theorem that is not just a reformulated axiom. 2.1.12. Consider the axiomatic system with the undefined terms point, line, and on and the following axioms. (i) (ii) (iii) (iv)

There are a line and a point not on the line. Every two distinct points have a unique line on them both. Every two distinct lines have at least one point on them both. Every line has at least three points on it. (a) Given two distinct lines prove that they have exactly one point on them. (b) Prove that there are at least seven points. (c) Given a point prove that it has at least three lines on it. Hint: First consider a point not on the line of axiom (i). (d) Prove that there are at least seven lines.

2.1.13. Develop axioms to go with the undefined terms point and midpoint. (You may include other undefined terms.) Prove at least one theorem that is not a reformulated axiom. 2.1.14. Figure 2.4 represents a knot by a sequence of strands going over and under each other. It suggests the following axioms with undefined terms strand and under. Use t < u to denote that “t is under u.” (i) For all strands s there are exactly two strands t and u with s < t and s < u. (ii) For all strands s there is at least one strand w with w < s. Develop this axiomatic system, possibly adding axioms, and prove at least one theorem that is not just a reformulated axiom.

Figure 2.4 A knot

2.2 Axiomatic Systems for Euclidean Geometry In this section we consider two axiomatic systems for Euclidean geometry: one from a high school text and the other Hilbert’s axiomatic system. The high school system builds on students’ understanding of the real numbers to make the proofs and the geometry ideas more accessible. Hilbert, in contrast, like Euclid, avoids any dependence on numbers. Proofs in the axiomatic systems must explicitly follow from the axioms. This requires care since it is easy to assume familiar Euclidean properties that aren’t among our axioms or already proven theorems.

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2.2.1 SMSG Postulates High school geometry courses spend less time on developing an axiomatic system now than in previous decades. Still, proofs remain a significant part of geometry at this level. So texts seek to offer an understanding to develop a body of knowledge through proofs from axioms. Axioms in high school texts assume the real number system. The choice, so different from Euclid’s approach, is a testimony to the power of analytic geometry and the emphasis of our mathematics education on algebraic thinking. Also, the texts incorporate the language of elementary set theory, such as union and intersection. Appendix B gives the SMSG postulates (another term for axiom), one of the earliest and most influential high school axiomatic systems for Euclidean geometry. (See [14].) Other high school axiom systems are quite similar. The SMSG system has only three undefined terms, point, line, and plane. The first postulate uses what appears to be an undefined term since it asserts that there is exactly one line that “contains” two points. The authors use elementary set theory and consider a line to be a set of points, each of which it contains. A number of other terms appear in the postulates. Some can be defined in terms of numbers, especially distances and angle measures. For example, the SMSG system defines a point B different from the points A and C to be between A and C if and only if all three are on the same line and we can add the distances: AC = AB + BC. The authors define congruent angles (segments) as angles (segments) with the same measure. They then define congruent triangles as ones where each corresponding pair of angles is congruent and each corresponding pair of segments is congruent. Exercise 2.2.2 asks you to give some other definitions, based on the undefined terms and the context of the postulates. The SMSG approach differs from Euclid’s approach in more than its use of real numbers. It doesn’t pay much attention to constructions. Indeed, the postulates don’t even mention circles, which are essential to many constructions. The parts of high school texts developing the axiomatic system concentrate at least initially on the geometry of lines and triangles. Their axiom systems enable them to approach the topics quickly. In Section 1.3 we saw Playfair’s axiom, which is postulate 16 in the SMSG system. The axiom doesn’t even guarantee that there are parallel lines—only that there are not too many of them. In Section 1.3 we constructed parallel lines by assuming some of Euclid’s propositions. However, in the SMSG system, we have to start from the postulates to show parallel lines exist, as done in Exercise 2.2.7. Most of the SMSG postulates are easy to understand, assuming a student is comfortable with the real numbers. Postulates 9 and 10 on the separation of the plane and of space, however, may become clearer through an example. The x-axis divides the familiar coordinate plane of points (x, y) into two halves, called half planes: One, call it H+ , has the points with positive y-coordinates and the other, H− , has the points with negative y-coordinates. The axis has the points whose y-coordinates are 0. Then H+ and H− are the two convex sets postulate 9 asserts exist. And, indeed, the line segment between any point in H+ and any point in H− must have a point on the x-axis—that is, the axis separates these sets. The same situation holds with any line. Why do we need this postulate? Recall in Section 2.1 that Euclid couldn’t prove that the two circles used to construct an equilateral triangle had to intersect. Without postulate 9, it would be conceivable for a line to slip from one side of another line to the other side without actually crossing it. In any case, we won’t be able to prove there was an intersection. There are in fact geometries, discussed in Section 4.5 and in Chapter 7, where just this situation occurs. Postulate 10 tells us that planes separate three-dimensional space the same way lines separate the plane.

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2.2.2 Hilbert’s Axioms Hilbert first published his axiomatic system in 1898 [8], more than 60 years before the SMSG postulates. His axioms were among the first to correct the logical deficiencies found in Euclid’s work. Hilbert, like Euclid avoided basing geometry on numbers. The avoidance of numbers meant that the axioms don’t address many of the familiar ideas of geometry, such as area. (Hilbert also developed one of the first axiomatic systems for the real numbers, but wanted to keep the subjects from depending on the other.) Hilbert’s purpose wasn’t to write a text like a high school text or Euclid’s, with proofs of all of the familiar theorems. Rather he wanted to show certain properties that we now call metamathematical, discussed in the next section. Hilbert wanted to assume as little as possible while still being able to prove all Euclidean theorems. A consequence of this approach is that many proofs are long and complicated, and so unsuitable for an introduction to geometry. It takes Hilbert eleven axioms to specify the properties of a line, something the SMSG system does in the first four postulates. Without the power of the real numbers, Hilbert needed to state explicitly key properties of betweenness and congruence. The first three groups of Hilbert’s axioms codify elementary properties, many previously overlooked. Axiom I-3 guarantees that there are some points in the geometry, something that seemed totally unnecessary to say before the 1800s. The first three axioms of order give the most elementary properties of betweenness on a line. Pasch’s axiom (II-4), which informally says that a line that enters a triangle must exit as well, deserves more explanation. It is possible for a line to intersect a triangle in just one vertex, but this axiom limits the other options to the line being one side of the triangle or intersecting the triangle in exactly two points. Hilbert notes that the axiom is equivalent to the separation axiom (which is SMSG postulate 9). Theorem 2.2.1 gives a proof of the equivalence to give a deeper feeling for both properties and to give some practice using some of Hilbert’s other axioms. We stay within Hilbert’s axiomatic system so that it is clear what properties we can use. Theorem 2.2.1. The separation axiom (II-4′ ) and Pasch’s axiom (II-4) are logically equivalent.

Exercise 2.2.1. Illustrate both directions of the proof of Theorem 2.2.1. Proof. (⇒) Suppose that the separation axiom holds: A line m separates the points that are not on m into two sets such that if X and Y are in the same set, X Y does not intersect m, and if X and Y are in different sets, X Y does intersect m. To show Pasch’s axiom, suppose that a line l enters △ABC at D on side AB but that A, B, and C are not on l. By the separation axiom, A and B are on opposite sides of l; that is, they are in different sets of points. Now C is in just one of the two sets, which means exactly one of AC and BC have a point on l, where l exits from △ABC, showing Pasch’s axiom. (⇐) Now suppose Pasch’s axiom holds. Let m be a line in the plane. We need to separate the rest of the points of the plane into two disconnected sets. By axiom I-3, we may assume that P and Q are on m and S is not on m. Let k be the line on Q and S, as guaranteed by Axiom I-1. By axiom II-2 there is a point T on k with Q between S and T . We use S and T to define the two desired sets separated by m. Let H1 = {U : SU ∩ m = ∅} and H2 = {U : T U ∩ m = ∅}. We need to show several things about H1 and H2 .

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First we show that the line segment between two points in one of the sets doesn’t intersect m. Let U, W ∈ H1 . For a contradiction, suppose that m did intersect U W , say, at Z . Then for △SU W , we have the conditions of Pasch’s axiom: m enters the triangle at Z and doesn’t contain any of its vertices. However, by the definition of H1 , m doesn’t contain any point on either SU or SW , so it doesn’t exit, a contradiction. The same holds for the set H2 . Next, for U ∈ H1 consider △SU T . Since m intersects ST , by Pasch’s axiom m intersects SU or T U . But m doesn’t intersect SU , so it intersects T U . Hence no point is in both H1 and in H2 . Similarly, for U ∈ H1 and W ∈ H2 , the triangle △T U W has to have two intersections with m and they can’t involve side T W . So m intersects U W . The sets H1 and H2 behave as we’d like. We need to show that every point is in one of them or is on m. To do so we show first that a line can’t intersect all three sides of a triangle, assuming it is not on any of the vertices. For a contradiction, suppose that m intersects △ABC at D on AB, at E on AC, and at F on BC. By Hilbert Axiom III-3 there is some order for these points. Without loss of generality suppose that E is between D and F. Since AB and BC ← → are on different lines, B, D, and F are not collinear. Now AC intersects △B D F on the side ← → ← → D F at E. So it must exit either on B D, which is part of AB, or on B F, which is part of BC. ← → ← → ← → But AC already intersects both AB and BC so it can’t intersect either a second time, giving a contradiction. Hence a line can’t intersect all three sides of a triangle. Let X be any point of the plane. We show that X is in one of H1 , H2 , or m. If X is in H1 or m, we’re done. So suppose it isn’t. Line m intersects △ST X on ST and since X isn’t in H1 , m intersects S X . Hence m doesn’t intersect T X and X is in H2 , as required, proving the separation axiom. ! The congruence axioms III-2 and III-3 make precise Euclid’s first two common notions. Axioms III-1 and III-4 guarantee the existence and uniqueness of congruent segments and angles. The axioms replace Euclid’s use of circles and constructions. Axiom III-5 is Euclid’s Proposition I-4, the familiar SAS property. Euclid’s proof has a logical gap because he used motion without providing any axioms about it. Hilbert’s fourth group contains only Playfair’s axiom, which by Theorem 1.2.1 is logically equivalent to Euclid’s fifth postulate. The axiom distinguishes Euclidean geometry from hyperbolic geometry, which we introduce in Chapter 4. The axioms of group V imply that the points on a line correspond to the real numbers without mentioning numbers. Axiom V-1, the Archimedean axiom, eliminates infinitely large and infinitely small line segments. (Archimedes and a few other Greeks used this idea in some proofs.) Hilbert’s complicated axiom of linear completeness, V-2, ensures that lines have no gaps while avoiding concepts from the real numbers and analysis. Its awkwardness indicates how fundamental our understanding of number has become.

2.2.3 Exercises for Section 2.2 Proofs for problems referring to the SMSG postulates should use only those postulates. Similarly, proofs for problems referring to Hilbert’s axioms should use only those axioms. 2.2.2. Define the following terms within the SMSG axiomatic system. You may use point, line, plane, between, distance, and definitions from previous parts and terms appearing in the SMSG postulates before the number in parentheses. (For example, the term

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“convex” appears in postulate 9, so we can’t use any information in that postulate 9 or later to define convex.) *(a) *(b) *(c) (d) *(e) (f) (g) (h) (i) (j) (k)

segment (9). convex (9) ray (12) angle (11) half plane (12) interior of an angle (13) linear pair (14) supplementary angles (14). parallel lines (16) perpendicular lines vertical angles (In Figure 2.5, ∠AE B and ∠C E D are vertical angles). B

A

E D

C

Figure 2.5 Vertical angles. 2.2.3. *(a) Euclid’s fourth postulate reads “All right angles are equal.” Prove this in the SMSG system, assuming by “equal” we mean they have the same measure as in postulate 11. (b) Use postulates 13 and 14 with the definitions in Exercise 2.2.2 to prove that vertical angles are equal. (See Exercise 2.2.2(k).) 2.2.4. The SMSG postulates do not mention circles at all. (a) Use the terms of the SMSG system to define a circle. (b) Explain how the postulates guarantee that every line through the center of a circle intersects the circle in two points. (c) Define a diameter and a radius of a circle. (d) Define the interior and exterior of a circle. 2.2.5. This problem considers perpendicular lines, defined in Exercise 2.2.2(j). *(a) Given a point P on a line l, explain why the postulates guarantee that in a plane containing l there is exactly one line m containing P and perpendicular to l. (b) Use the SMSG postulates to turn the following argument into a proof that there is at least one line m perpendicular to a line l where m is on a point A that is not on l. Make a figure for the proof. Let A be a point not on a line l and B and C two points on l. If m∠ABC = 90◦ , we’re −→ done. Otherwise, show that there is a ray B D and a point E on the ray so that D is on

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the other side of l from A and △ABC ∼ = △E BC. Prove that the line through A and E intersects l at, say, F. Prove that △AB F ∼ = △E B F. Prove that m∠AF B = 90◦ . (c) Use the SMSG postulates to turn the following argument into a proof that there is at most one line m perpendicular to a line l where m is on a point A that is not on l. Make an appropriate figure for this proof. Suppose, for a contradiction, that there were two distinct lines m and k through A and perpendicular to l. Why must m and k intersect l in different points? Let Q be the intersection of m and l and R be the intersection of k and l. Prove that there is a point S on m with AQ = S Q and Q between A and S. Prove two triangles congruent. Prove that S must be on k. Why is this a contradiction? (d) Use part (c) to prove that no triangle can have two right angles. *2.2.6. Explain how we can use Exercise 2.2.5 and the SMSG postulates to assign coordinates to the points of a plane. Illustrate the process with a figure. 2.2.7. (a) Given two points A and B on a line l, use the SMSG postulates to prove that the perpendiculars to l through A and B must be parallel. Hint: See Exercise 2.2.5. (b) Given a point C not on a line l, use part (a) and Exercise 2.2.5 to prove that there is a line parallel to l through C. (c) Prove that the parallel to l through C in part (b) must intersect the perpendiculars to l through A and B in part (a). 2.2.8. Whereas Euclid builds an equilateral triangle in his first proposition, it is a lengthier process in the SMSG system. This problem gives one way to show that there are equilateral triangles in the SMSG system. −→ (a) Given two points A and B, prove that there is a ray AC so that m∠B AC = 60◦ . −→ (b) Prove that there is a point D on AC so that AD and AB are congruent. So △AB D is isosceles. (c) Prove that ∠AB D ∼ = ∠AD B. (d) Assume Theorem 1.1.1 and prove that △AB D has three congruent angles. (The SMSG textbook takes until page 258 to prove Theorem 1.1.1. It uses the same argument used in Section 1.1, but builds to it carefully from the postulates.) (e) Prove that the three sides are congruent and so △AB D is, by definition, equilateral. *2.2.9. (a) Use your definition of the interior of an angle in exercise 2.2.2(e) to define the interior of a triangle. How does the interior of a triangle relate to the interiors of its angles? Illustrate your definitions. (b) Are the interior of an angle and of a triangle, as you defined them, convex, using your definition from Exercise 2.2.2(b)? Explain your answer. *2.2.10. (a) From Group I of Hilbert’s axioms, how many points and lines must exist in a plane? Prove your answer. (b) Given two distinct lines prove from Group I of Hilbert’s axioms that at most one point lies on both lines. 2.2.11. Compare the axioms and Theorems 2.1.1 and 2.1.2 of the simplified axiomatic system in subsection 2.1.3 with Hilbert’s axioms of order II-1, II-2, and II-3.

2.2 Axiomatic Systems for Euclidean Geometry

81

2.2.12. *(a) List which of Hilbert’s axioms refer to points on just one line. *(b) Use the axioms in part (a) from groups I and II to prove that there are at least five distinct points on any line. *(c) Let P0 and P1 be distinct points on a line m. Use the axioms in part (a) from groups I, II, and III to prove there are points P2 and P3 on m distinct from each other and from P0 and P1 such that P1 P2 ∼ = P0 P1 and P2 P3 ∼ = P0 P1 . (d) In the spirit and notation of part (c), what properties should P−1 and P−2 satisfy? Prove that there are points that satisfy them. (e) Repeat part (d) for P1/2 and P1/4 . (f) Do the axioms in groups I, II, and III suffice to prove there is a point that we would label P1/3 ? Explain. 2.2.13. Verify the claim in Hilbert’s axiom III-2 that the actual axioms do imply the properties of an equivalence relation, as stated there. 2.2.14. Hilbert’s statement of the SAS axiom, unlike SMSG postulate 15, intentionally does not assert that the third pair of sides are congruent. After doing the following steps, write a proof from his axioms that the third pair of sides must be congruent. (a) Given △ABC and △A′ B ′ C ′ state what we are given, where ∠B AC and ∠B ′ A′ C ′ are the given congruent angles. (b) State what Hilbert’s axiom III-5 (SAS) gives us from part (a). (c) For a contradiction, assume that one of the pair of third sides is longer, say BC > B ′ C ′ . (d) Why must there be a point D between B and C so that D B ∼ = C ′ B′? ∼ *(e) Why must ∠B AC = ∠B AD? Why is this a contradiction? 2.2.15. (a) Compare SMSG postulate 12 and Hilbert’s axiom III-4. (b) List the properties that do not involve measurements that the SMSG postulates guarantee but that Hilbert’s axioms do not directly guarantee. (c) Prove Hilbert’s axiom III-3 using the SMSG postulates.

2.2.4 David Hilbert But in the present century, thanks in good part to the influence of Hilbert, we have come to see that the unproved postulates with which we start are purely arbitrary. They MUST be consistent; they HAD BETTER lead to something interesting. —Julian Coolidge

The work of David Hilbert (1862–1943) symbolizes the modern abstract, axiomatic approach to mathematics. He contributed significant results in many fields, including algebraic invariants, number theory, partial and ordinary differential equations, integral equations, geometry, and the foundations of mathematics. He won international renown at age 26 when he published a result in algebraic invariants, a precursor of modern abstract algebra, that had seemed beyond possibility. His proof avoided the laborious and limited constructive methods of previous mathematicians. One of them, Paul Gordan, at first disdained Hilbert’s radical nonconstructive proof saying, “This is not mathematics; it is theology.” He later amended his criticism with “I have convinced

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myself that theology also has its advantages.” Hilbert’s work in integral equations led to Hilbert space, an infinite dimensional analog of Euclidean space important in the study of quantum mechanics in physics. Hilbert was influential in the foundations of mathematics. Developments in nineteenth century analysis and geometry made clear the need for a careful scrutiny of axiomatics. To make hidden assumptions explicit, Hilbert realized that mathematicians must isolate formal axioms from their meaning. He devised ground-breaking axiom systems for both the real numbers and Euclidean geometry. He proved the relative consistency of geometry, assuming the consistency of the real numbers. Hilbert’s program, which he hoped would prove the absolute consistency of all mathematics, led others to a penetrating analysis of logic and the foundations of mathematics. G¨odel’s famous incompleteness theorems came out of that program and proved, among other things, the impossibility of Hilbert’s goal. However, G¨odel’s theorems showed how deeply Hilbert’s axiomatic approach enabled mathematicians to probe the foundations of mathematics. Hilbert’s formal axiomatics devoid of meaning was never intended to replace mathematics. Throughout his career, Hilbert stressed the dynamic interplay of concrete problems and general, abstract theories. His famous list of 23 problems in 1900 illustrates well this link. Each problem has led mathematicians to a more profound understanding of an area of mathematics. His own research revealed the power of particular problems to inspire deep mathematics and of abstract mathematics to elucidate particular problems.

2.3 Models and Metamathematics Mathematics is the art of giving the same name to different things. — Henri Poincar´e

In Section 2.1 we discussed the logical need for undefined terms in an axiomatic system, in which we must ignore what the terms are intended to mean. However, people depend on meaning and intuition to create and understand mathematics. A string stretched taut between its ends provides a strong intuition of a line, but the image is too imprecise for mathematics. Mathematical models provide an explicit link between intuitions and undefined terms. The usual analytic model of Euclidean plane geometry is the set R2 = {(x, y) : x, y ∈ R}, where a point is interpreted as an ordered pair of real numbers (x, y) and a line is interpreted as the points that satisfy an appropriate first degree equation, such as y = mx + b or, to allow vertical lines, more generally ax + by + c = 0. High school students spend considerable time learning how the algebraic model matches geometric intuition. In making a model, we are free to interpret the undefined terms in any way we want, provided that all the axioms hold under our interpretation. We don’t say that the axioms are by themselves true; we need a context to give meaning to the axioms in order for them to be true or false in the model. Poincar´e’s quote addresses the potential for different interpretations and models. Definition. A model of an axiomatic system is a set of objects together with interpretations of all the undefined terms of the axiomatic system such that all the axioms are true in the set using the interpretations.

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Example 1. Figures 2.6 and 2.7 give different models of the axiomatic system with the undefined terms point, line, and on and the axioms (i) Every point is on exactly two lines. (ii) Every line is on exactly three points. In these and later models on has the obvious interpretation. While Figure 2.6 interprets the undefined terms in conventional ways, the choice of undefined terms in Figure 2.7 may seem disconcerting, but our only requirement is that the axioms are true in the model with the interpretation we give. And indeed, each edge is on two faces and each face is on three edges. ♦ A

B d a

b e

D

C

Figure 2.6 Interpret point by a dot, line by a line segment.

f c

Figure 2.7 Interpret point by an edge, line by a face of the pyramid.

Exercise 2.3.1.∗ In the model given by Figure 2.7 reinterpret lines as the vertices of the pyramid and points as its edges. Is the result still a model of the axioms in Example 1? Models do much more than provide concrete examples of axiomatic systems; they lead to important understandings about axiomatic systems. Metamathematics considers axiomatic systems, their models, and relations between them. The prefix meta, Greek for “beyond,” distinguishes metamathematics from the mathematical properties proved within a system. For example, Metatheorem 2.3.1 implies that the axioms of a system are all that are needed to determine the models of the system. The proofs of metatheorems are beyond the level of this book. (See DeLong [2, Chapter 4], Sibley [13, Chapter 9], and Wilder [15, Chapter 2] for more on metamathematics.) Metatheorem 2.3.1 If all the axioms of an axiomatic system are true in a model, then all the theorems of that system are also true in the model. The most important property of an axiomatic system is consistency, which says that we cannot prove two statements that contradict each other. For example, a proof of both 2 + 2 = 4 and 2 + 2 ̸= 4 would render both arithmetic and proofs useless. However, consistency is difficult and often impossible to show directly. Fortunately Metatheorem 2.3.2 provides an easier method with models. As a consequence the models for Example 1 show that the axiomatic system is consistent. Definition. An axiomatic system is consistent if and only if no contradictions can be proven from the axioms. An axiomatic system is relatively consistent if and only if its consistency can be proved assuming the consistency of another axiomatic system. Metatheorem 2.3.2 (G¨odel, 1930) An axiomatic system is consistent if and only if it has a model.

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The analytic model of geometry shows the relative consistency of Euclidean geometry because analytic geometry depends on the real number system. Are the axioms for the real numbers consistent? The model of the real numbers we are likely to choose is a geometric one: a line in Euclidean geometry. We risk entering a vicious circle: The crucial property of a mathematical system, consistency, is beyond our reach for two central mathematical systems. This problem and others like it inspired David Hilbert and others to investigate mathematical logic. Mathematical logic has given us profound insights into the nature of mathematics and proof, including their limitations. In particular, one of G¨odel’s incompleteness theorems, published in 1931, implies that we cannot prove the absolute consistency of elementary arithmetic, and by extension, Euclidean geometry and the real number system. Thus relative consistency is the best we can do for such sophisticated systems. It deserves emphasis that no one seriously doubts their consistency. Other metamathematical properties, though not as essential as consistency, give important insight into axiomatic systems and models. Definition. A statement is independent of a set of axioms if and only if neither the statement nor its negation can be proved from the axioms. If, in an axiomatic system, each axiom is independent of the others, the set of axioms is said to be independent. We can use Metatheorem 2.3.1 to prove independence with models. Suppose that we have two models for a set of axioms and that a statement is true in the first model but not in the second model. Then Metatheorem 2.3.1 applied to the second model shows that the statement cannot be a theorem. Similarly, the first model shows the negation of the statement cannot be a theorem. Hence the statement is independent of the axioms. Example 2. The models in Figures 2.6 and 2.7 show the statement “there are exactly six points” to be independent of the axioms of Example 1. Because the statement is independent of the axioms, the axioms are not complete. We could form a new, consistent axiomatic system with the statement as an additional axiom. ♦ Example 3. Taxicab geometry alters the usual analytic model of Euclidean geometry by giving a different interpretation to distance, while retaining the usual interpretations of other terms, especially lines and angle measure. Rather than use the Pythagorean theorem to measure distance, we use the formula dT ((x1 , y1 ), (x2 , y2 )) = |x2 − x1 | + |y2 − y1 | . This formula gives the distance a taxi goes if it travels only along north–south and east–west streets (Figure 2.8). We still interpret points as ordered pairs (x, y) and lines as the set of points

Figure 2.8 The taxicab distance between (1, 3.5) and (4, 1) is |4 − 1| + |1 − 3.5| = 5.5.

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satisfying a linear equation ax + by + c = 0. We measure angles as usual. The model satisfies Euclid’s postulates, although circles look odd. However, Euclid I-4 (SAS) is no longer valid in this geometry (Figure 2.9). The theorem holds in the usual analytic model, so it is independent of Euclid’s postulates. That is, not only did Euclid’s proof of I-4 have a flaw, but he could never actually prove the proposition from his axioms. (In fairness, Euclid’s work shouldn’t be faulted because of insights occurring 2000 years after his work.) See Exercises 2.3.16, 2.3.17, and 2.3.19 and Krause [10] for more on taxicab geometry. ♦ A

B

C

A' B' C'

Figure 2.9 In taxicab geometry, AC is longer than A′ C ′ , even though AB ∼ = A′ B ′ , BC ∼ = B ′ C ′ , and

∠ ABC ∼ = ∠ A′ B ′ C ′ .

Example 4. We can stretch a string taut on the surface of a sphere to construct a spherical line segment. Similarly, we can construct a spherical circle by fixing one end of a taut string and swinging the other end around it. Experiment informally on a ball or sphere. Confirm that, within reason, Euclid’s five postulates hold in this model. Now draw a spherical line segment with a length of between one-third and one-half the circumference of the sphere (AB in Figure 2.10). Use the line segment to imitate Euclid’s construction in Proposition I-1. (See Section 1.2.) The two spherical circles with centers at A and B and radii length AB do not intersect. The construction does not hold in this model but does in the usual model, therefore it is independent of Euclid’s postulates. For more on spherical geometry, see Section 1.5, Henderson [7], and McCleary [12]. ♦

A

B

Figure 2.10 Example 5. One of Hilbert’s two main goals in developing his axiomatic system was to show the axioms were independent. The entire proof, in [8], is beyond the level of this text, but we

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can investigate independence a bit. Consider axioms I-1, I-2, II-1, II-2, III-1, III-2, III-3, and the first sentence of I-3 that refer only to points on a line. Exercise 2.2.12 asked you to prove that there were at least four points on a line. You might have expected that problem to ask you to continue and prove that there were infinitely many points. But in fact we can’t force more than seven points on a line from the axioms, as the following model shows. By point we mean any of the seven vertices of a regular heptagon, which is what we interpret as a line. (See Figure 2.11.) Interpret a point Q to be between points P and R if and only if Q is on the perpendicular bisector of the Euclidean segment P R. For example, in Figure 2.11, B is between A and C. Similarly, C is between A and E. Interpret two segments P Q and RS as being congruent if and only if the corresponding edges of the heptagon are congruent as Euclidean segments. Verify that the eight axioms listed above are true in this model. The group V axioms force a line to have points matching the real numbers. Since the usual model of a line satisfies all the axioms, the group V axioms are independent of these eight axioms. This seven point model is also a model of the axiomatic system from Subsection 2.1.3. However, it is not a model for the augmented system of Exercise 2.1.4. As a consequence, the additional axiom in that problem is independent of the other four axioms. ♦ A G

B

F

C

E

D

Figure 2.11 Hilbert’s second main goal was to completely characterize Euclidean geometry, a property called completeness. Completeness, which is beyond the level of this text, states that every proposition one can state using the undefined terms is either provable or disprovable from the axioms. Hilbert wanted to ensure that his axioms sufficed to prove every Euclidean theorem. Since we can’t list every theorem, we need to enlist models again through another metamathematical idea. Hilbert showed that any two models of his axiomatic system were structurally identical and so had the same theorems. There were no independent statements one could add as potentially missing axioms. In mathematical terminology, all models of Euclidean geometry are isomorphic, a powerful concept throughout mathematics. Example 6. Two more models of the axiomatic system from Example 1 illustrate isomorphic and non-isomorphic models. The models in Figures 2.12 and 2.13 both have six points and four lines, as the model in Figure 2.7 did. They all look different on the surface, but the models in Figures 2.7 and 2.12 are structurally the same (isomorphic), whereas the one in Figure 2.13 is structurally different, not just cosmetically different. We can match the points and lines from Figure 2.7 with those from Figure 2.12 and have corresponding points on corresponding lines. The labeling of the figures provides a matching. However, no matter how we match the points

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in Figure 2.13 with those of the two models, the relation on will never work the same as in the other models. For example, P and Q share two lines in common, something that never happens in the other models. The model shown in Figure 2.6 has no chance of being isomorphic with any of the models because it has more points and lines. ♦ P

Q

A U

R

E S

B F C

D

Figure 2.12 Interpret point by a dot and line by a line segment.

T

Figure 2.13 Interpret point by a dot and line by an arc.

Definition. Two models are isomorphic if and only if there is a one-to-one onto function that matches all the elements of one model with those of the other model so that all the relations using the undefined terms that hold in one model also hold in the other model. Isomorphism and non-isomorphism can be tricky to distinguish, unless one model has more things than the other. We show two models are not isomorphic by finding a property where they essentially differ and where that property can be described in terms of the undefined terms, not their interpretations or the objects in each model. With regard to the axiomatic system of Examples 1, 6, and 7, we can’t use our interpretations of point, line, or on to show non-isomorphism. Instead we have to show the relation on has different properties in the models. Example 7. Figures 2.14 and 2.15 give two more models of the axiomatic system of Example 1, both with twelve points and eight lines, the same numbers as in the model in Figure 2.6. For determining isomorphism it is not relevant that one of these models uses triangles for lines while the others use line segments. To show isomorphism, we need first to find a potential match of the points. As a first attempt, let’s try matching a and b in Figure 2.14 with A and C, respectively, in Figure 2.6. We quickly run into trouble: a and b share a third dot on the same triangle, but A and C don’t share a segment. This is a structural difference. But that doesn’t mean the two models are non-isomorphic. It might just be our poor start of a matching. For the next attempt, try matching a and b in Figure 2.14 with A and B in Figure 2.6. Do the choices restrict the remaining matches? Yes: only one other dot is on the triangle on a and b and only one other dot on the segment for A and B. So they must be matched. Further, the foursome abcd in Figure 2.14 has each succeeding pair on a segment and it is the only one including a and b that does so. The same thing holds for the foursome ABC D in Figure 2.6. So we need to match c with C and d with D. The same reasoning extends to a complete matching pairing the dots so that the triangles match with the segments. That is, the corresponding points in the two models are on the corresponding lines and we have isomorphic models.

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a

Y

b

X Z

d

Figure 2.14 Interpret point by a dot and line by a colored triangle.

c

Figure 2.15 Interpret point by a dot and line by a line segment.

We could try the same approach with the model in Figure 2.15 with either of the previous two models. But nothing works. Still to prove non-isomorphism this way would require trying all 12! = 479, 001, 600 possible matchings, a very poor strategy. In Figure 2.15 the three dots X , Y , and Z have the property that each two of them share a segment. No three dots in either of the other two models satisfy this property. So no matter what matching we try, it will fail. The model is not isomorphic with the other two. ♦ To completely characterize Euclidean geometry, Hilbert needed an axiomatic system for which all models were isomorphic. However, other mathematical investigations benefit from axiomatic systems that have many models. The quote of Poincar´e at the start of this section emphasizes this aspect of axiomatic mathematics. Starting in the 1840s geometers have investigated geometry in more than three dimensions. Once one goes past the familiar space of three dimensions, there is no compelling reason to concentrate on a particular dimension. So it is advantageous to have an axiomatic system that encompasses all of them at once. Linear algebra provides just such a system. And its generality enabled mathematicians to use it to investigate many things, such as economic systems, that don’t look particularly geometrical at the outset. Many other areas of mathematics, including applied subjects, use the axiomatic approach with systems having many non-isomorphic models.

2.3.1 Exercises for Section 2.3 2.3.2.*(a) In the seven point model of Figure 2.11 in Example 5 determine what points are in −→ −→ −→ the ray AB. Repeat for AG and AC. In the definition of a ray, Hilbert says, “A point A, on a given line m, divides m into two rays such that two points are on the same ray if and only if A is not between them.” Does this statement hold in this model? (b) Modify the model in Example 5 by using the vertices of a regular pentagon. Verify that the sentence quoted in part (a) fails for this model. *(c) For the five point model of a line in part (b), which of the axioms in groups I and II hold? 2.3.3. This problem investigates taxicab geometry, as described in Example 3. Use graph paper to help you with it. (a) Verify in the SMSG system (Appendix B) that postulates 1 to 14 hold in taxicab geometry, where we interpret B is between A and C if and only if A, B, and C lie on a line and the distances add: AB + BC = AC.

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∼ C D in taxicab geometry to mean dT ( A, B) = dT (C, D). Which of *(b) Interpret AB = Hilbert’s congruence axioms hold in taxicab geometry? *(c) Define and describe “taxicab circles,” using the distance dT . How can two taxicab circles intersect? (d) Determine whether SSS, ASA, and AAS hold in taxicab geometry. Give examples for those that fail. 2.3.4 Use a sphere or ball that you can draw on for this problem. (See Example 4.) (a) Construct triangles of different sizes on a sphere and approximate their angle sum by using a copy of the circle in Figure 2.16, which can be used as a protractor. (Place the center of the circle approximately over the vertex of an angle and the mark for 0◦ on one of the sides of the angle.) Does Theorem 1.1.1 hold in this model? (b) Which of Hilbert’s congruence axioms hold for the sphere? (c) Determine whether SAS, SSS, ASA, and AAS hold for a sphere. (d) Determine whether the Pythagorean theorem holds for a sphere. 0 30

30

60

60

90

90

120

120

150

150 180

Figure 2.16 A spherical protractor. 2.3.5. In the SMSG system, interpret plane as the sphere, point as a point on the sphere, and line as a great circle of the sphere. Measure distances along the surface of the sphere, as would an insect crawling along a shortest path between the points. *(a) List the postulates that refer to space. We ignore them for the rest of this problem. *(b) Of the remaining postulates, list those that fail in the spherical model. (c) For each postulate (or part of a postulate) that fails illustrate or explain why.

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2.3.6. This problem refers to the axiomatic system discussed in Subsection 2.1.3. *(a) Find a model with infinitely many points. (b) Verify that the five vertices of a regular pentagon form a model for the axiomatic system of Section 2.1.3 where P(Q)R means that Q is on the perpendicular bisector of P and R. *(c) Prove that each axiom is independent of the others. 2.3.7. Use the axiomatic system of Example 1. (a) What is the smallest positive number of points of a model of the system? Repeat for lines and explain your answers. Find a model with the minimum numbers of points and lines. (b) Find a model with six points not isomorphic to any of the models presented in the text. Explain why it is not isomorphic to any of them. (c) Find two nonisomorphic models with nine points. Explain why they are nonisomorphic. (d) Find and prove a theorem of this system. 2.3.8. Use the axiomatic system of Exercise 2.1.7. *(a) Find a model with exactly five points. *(b) Find two nonisomorphic models with exactly six points. Explain why the models are not isomorphic. (c) Show that each axiom is independent of the others. (d) For each integer n with n > 4, describe a model with exactly n points. *(e) Show that the following sentence is not a theorem: For any team t there are two teams v and w so that v → t and w → t. 2.3.9. Use the axiomatic system of Exercise 2.1.8. (a) (b) (c) (d) (e) (f)

Find several models of this system. Show that axiom (iii) is independent of the other axioms. Repeat part (b) for axiom (iv). Find two nonisomorphic models of this system, each with eight points. Prove that any two models of this system with exactly six points are isomorphic. Find two nonisomorphic models of the modified axiomatic system in Exercise 2.1.8(c).

2.3.10. Consider the axiomatic system with the undefined terms corner, square, and on and the following axioms. (i) There is a square. (ii) Each square is on four distinct corners. (iii) For each square there are four distinct squares with exactly two corners on the given square. (iv) Each corner is on four distinct squares. *(a) Show that the axioms are relatively consistent by finding an infinite model in the Euclidean plane.

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*(b) Find a finite model for the first three axioms. (c) Find a finite model for all four axioms. 2.3.11. Use the axiomatic system of Exercise 2.1.9. (a) Find several models of the system with different numbers of apples or oranges. (b) Find two non-isomorphic models of the system with the same number of apples and the same number of oranges. Explain why the models are nonisomorphic. (c) Prove that each axiom is independent of the others. (d) Repeat part (a) with the augmented system that includes the axiom “If x likes y, then y does not like x.” (e) Repeat part (b) with the augmented system of part (d). (f) Repeat part (c) with the augmented system of part (d). 2.3.12. Use the axiomatic system of Exercise 2.1.10. *(a) (b) (c) *(d)

Find several models with different numbers of points. Prove that each axiom is independent of the others. Find two non-isomorphic models with the same number of points. Show that two models with four points are isomorphic.

2.3.13. Use the axiomatic system of Exercise 2.1.12. (a) Find a model with seven points. (b) Find models to show that all the axioms are independent. 2.3.14. (a) Verify the Pappus configuration of Figure 2.0 with nine points and nine segments is a model of the axiomatic system with undefined terms point, line, and on and the four following axioms. (i) There is a line and a point not on that line. (ii) A line has exactly three points on it. (iii) For two distinct points there is at most one line on both. Definition. Two lines are parallel if and only if no point is on both lines or they are equal. Two points are parallel if and only if no line is on both or they are equal. Denote parallel by ∥. (iv) For a point P and a line k if P is not on k, there is a unique line m on P such that k ∥ m and there is a unique point Q on k such that P ∥ Q. (b) Verify the statement of Pappus’ Theorem with other figures, including figures ← → ← → where the lines AB and D E intersect and the points on the lines are not equally spaced. Do the figures also satisfy the axioms above? *(c) Prove that for two distinct lines there is at most one point on both. (d) Prove that no line contains all the points and no point is on all the lines. *(e) Prove that a point has exactly three lines on it. (f) Prove that for three lines if k ∥ l and l ∥ m , then k ∥ m. *(g) Prove that a line has exactly two other lines parallel to it. (h) Prove that there exist exactly nine points and nine lines. (j) Prove that the axioms are independent.

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The plane dual of a sentence is the sentence obtained by switching the words point and line in the original sentence. (k) Explain why a sentence is provable in the axiomatic system if and only if its dual is also provable. 2.3.15. Develop an axiomatic system for the Desargues configuration in Figure 2.17. This figure illustrates Desargues theorem, which states that if two triangles are perspective from a point, then they are perspective from a line. For example, in Figure 2.17 △ABC and △D E F are perspective from the point P since corresponding vertices, such as A and D, are collinear with P. Similarly △ABC and △D E F are perspective from the line k ← → ← → since corresponding sides, such as BC and E F, meet at U on the line of perspective. (See Chapter 7 for more on Desargues’s Theorem.) P

A C

U

B

T F E

S k

D

Figure 2.17 The Desargues Configuration. 2.3.16. (a) Graph the taxicab circle with center (0, 0) and radius 1. (By definition of a circle, (x, y) is on this circle if and only if dT ((0, 0), (x, y)) = 1.) Which points on the circle are a taxicab distance of 2 away from (1, 0) (a point also on the circle)? (b) Show that the SSS congruence theorem (Euclid I-8) fails in taxicab geometry. (c) Show that the AAS and ASA congruence theorems (Euclid I-26) fail in taxicab geometry. (d) Find the equation for the taxicab circle with center (a, b) and radius r . (e) Graph the taxicab circle with center (3, 2) and radius 4. Find the intersection of the taxicab circle with the taxicab circle of part (a). What unusual property does the intersection have? (f) When does the property of part (e) hold for the intersection of two taxicab circles? (g) Repeat parts (e) and (f) for the taxicab circle with center (3, 0) and radius 4. 2.3.17. We investigate area in taxicab geometry (Example 3) using postulates 17 to 20 of the SMSG system (Appendix B). (a) Suppose a right triangle has a horizontal side of length w and a vertical side of length v. Construct an appropriate rectangle and use it to determine the area of the triangle. Justify your answer using the postulates.

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(b) Determine the area of rectangle ABC D of Figure 2.18 using taxicab distance and postulate 20. Use that postulate and part (a) to determine the areas of the other rectangles and the triangles in Figure 2.18. Compare the areas with the area of the entire square. What can you conclude? (0,4)

A

(4,4)

D

B (0,0)

C

(4,0)

Figure 2.18 2.3.18. *(a) Consider a model for Hilbert’s axioms including for “points” only those points where both coordinates are integers—that is, the positive and negative whole numbers, together with 0. “Lines” are made of “points” that lie on a usual line with at least two “points.” . Interpret “congruence” as having the same Euclidean measure. List those of Hilbert’s axioms that fail in this model. (b) Repeat part (a) using points where both coordinates are rational numbers—that is, numbers of the form a/b, where a and b are integers (and b ̸= 0). (c) Repeat part (a) using points where both coordinates are constructible numbers. (See Exercise 1.2.25.) 2.3.19. Interpret “Q is between P and R” by d(P, Q) + d(Q, R) = d(P, R), where d(X, Y ) is the distance between X and Y . (That is, leave out the requirement that P, Q, and R must be on the same line.) (a) Are all of Hilbert’s axioms satisfied in the analytic model of Euclidean geometry ! with this interpretation and d((a, b), (c, d)) = (a − c)2 + (b − d)2 , as usual ? (b) On a graph, color the set of points in taxicab geometry between (0, 0) and (1, 2) using the given interpretation but with the taxicab distance. Which of Hilbert’s axioms of order are satisfied in taxicab geometry with this interpretation? (c) In Euclidean geometry, if Q is between P and R and R is between Q and S, then Q is between P and S. Does this property hold in taxicab geometry with this interpretation of between? Explain why or give an example where the property fails. (d) If P and R are opposite points on a sphere, which points Q on the sphere would be between P and R under this interpretation and where we measure distance along the surface of the sphere? (e) Repeat part (d), but use the straight line, Euclidean distance between points.

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2.3.2 Kurt G¨odel Either mathematics is too big for the human mind or the human mind is more than a machine. —Kurt G¨odel

The logician Kurt G¨odel (1906–1978) radically changed how mathematicians view mathematics. For his Ph.D. dissertation, he proved Metatheorem 2.3.2, called G¨odel’s completeness theorem, showing the vital connection between axiomatic systems and models. In the process he showed that mathematical logic was sufficient for mathematical reasoning. A year later he stunned the mathematical world with what are now called G¨odel’s incompleteness theorems. They show an inherent limitation to any axiomatic system sophisticated enough to develop number theory—ordinary arithmetic on the positive integers. In informal terms, he showed that we can’t prove its consistency without assuming the consistency of another system at least as sophisticated. For example, the relative consistency of Euclidean geometry and of the real numbers is the best we can do. Another result needs some technical clarification. Even a quick reading of Hilbert’s axioms reveals that axiom V-2 is worded differently from the axioms in the first four groups. It is an example of a second order axiom, one that isn’t stated using only the undefined terms. One of G¨odel’s incompleteness theorems shows, in effect, that no consistent first order axiomatic system of number theory can prove all the theorems of number theory. That is, there are always independent statements that from metamathematical reasoning we can know have to be true about arithmetic. (There are complete second order axiomatizations of number theory.) The same limitation holds for geometry and other systems. The theorems ended Hilbert’s program to establish all of mathematics on the basis of first order axiomatic systems that were provably consistent and complete. However, they showed the amazing power of mathematics to investigate itself as well as the many systems within mathematics. The ability and necessity of mathematicians to think both within and outside of axiomatic systems may have led G¨odel to the quote cited above. He, like Plato, thought of mathematics as concerning itself with ideal things that were true, not just true in a model.

2.3.3 Projects for Chapter 2 1. Develop an axiomatic system describing the geometry of pixels (points) and lines on a computer screen. Find and prove some theorems. 2. Develop an axiomatic system for the geometry of a sphere representing the relationships of great circles. (See Section 1.5.) 3. Investigate the axiomatic system with undefined terms point, line, on, and ⊥ and the following axioms. (i) There exists a point and a line not on that point. (ii) If lines k and m are distinct and both on a point, then k ⊥ l. (iii) For every point P and every line k there is a unique line l on P so that k ⊥ l. (iv) Every line has n points on it. (v) For any two distinct lines there is at most one point on both. (vi) For all lines k, not k ⊥ k.

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(a) Make several models of the axioms. (b) For a given value of n, determine the minimum number of points and lines. Prove your answer. (c) Prove that the axioms are independent. (d) Find two non-isomorphic models with the same number of points on each line. For two lines m and k, define m ∥ k if and only if there is some line l so that m ⊥ l and l ⊥ k. (e) Find and prove some statements about the relation ∥. (f) Develop this system further, adding an axiom or axioms and proving theorems. 4. (a) Redo project 3, but replace axiom (iii) with (iii′ ) For every point P and every line k there are exactly two distinct lines l and m on P so that k ⊥ l and k ⊥ m. Hint: Think in three dimensions. (b) Generalize part (a) using different numbers of lines in axiom (iii′ ). 5. Investigate strengthening the axiomatic system of Subsection 2.1.3 with the axiom (*) “There are exactly five distinct points.” Is this new system completely characterized? That is, are all models isomorphic? Give a proof or provide two non-isomorphic models. In the latter case, provide an additional axiom that completely characterizes the system. 6. Investigate strengthening axiomatic systems from the exercises of Section 2.1 to force all models to be isomorphic. 7. Investigate Pappus’s Theorem, stated below, and its importance in projective geometry. Pappus’ Theorem. Let A, B, and C be collinear and D, E, and F be collinear. Suppose that ← → ← → ← → ← → ← → ← → AD and C F intersect in H , AE and B F intersect in I and B D and C E intersect in G. Then G, H , and I are collinear. (See Figure 2.0.) 8. Investigate taxicab geometry. (See Krause [10].) 9. Investigate an axiomatic system for origami, Japanese paper folding. (See Geretschl¨ager [5].) 10. Investigate what axioms need to be added to the SMSG system for four-dimensional Euclidean geometry. 11. Investigate other axiomatic systems. (See Bryant [1] and Fishback [4].) 12. Investigate metamathematics. (See Delong [2], Sibley [13], and Wilder [15].) 13. Investigate the historical development of axiomatic systems and metamathematics. (See Kline [9].) 14. Write an essay on the roles of intuition and proof in geometry. 15. Write an essay on the different levels of proof demanded in mathematics, in a civil court case (“the preponderance of the evidence”) and a criminal court case (“beyond a reasonable doubt”). What phrase would you offer to describe mathematical proofs? Compare the advantages and disadvantages of the differing levels of proof. 16. Write an essay explaining axiomatic systems and models and their importance to a high school geometry student. Use examples, preferably everyday ones. 17. Write an essay on the understanding of definition in mathematics, in other disciplines, and in everyday language. Discuss the advantages and disadvantages of these different notions of definition.

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2.3.4 Suggested Readings 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

Bryant, V., A fresh look at geometry, Mathematics Magazine, 1971, 44(1): 180–182. DeLong, H., A Profile of Mathematical Logic, Reading, MA: Addison-Wesley, 1971. Dubnov, I., Mistakes in Geometric Proofs, Lexington, MA: D. C. Heath, 1963. Fishback, W., Projective and Euclidean Geometry, New York: John Wiley & Sons, 1962. Geretschl¨ager, R., Euclidean constructions and the geometry of origami, Mathematics Magazine, 1995, 68(5): 357–371. Heath, T., The Thirteen Books of Euclid’s Elements, 3 vol. Mineola, NY: Dover, 1956. (Also available on line.) Henderson, D. and D. Taimina, Experiencing Geometry, 3rd ed., Upper Saddle River, NJ: Prentice-Hall, 2001. Hilbert, D., The Foundations of Geometry, 2nd ed. (transl. E. Townsend). Peru, IL: Open Court, 1921. Kline, M., Mathematical Thought from Ancient to Modern Times, New York: Oxford University Press, 1972. Krause, E., Taxicab Geometry, New York: Dover, 1986. Lakatos, I., Proofs and Refutations, New York: Cambridge University Press, 1976. McCleary, J., Geometry from a Differentiable Viewpoint, Port Chester, NY: Cambridge University Press, 1994. Sibley, T., Foundations of Mathematics, New York: John Wiley & Sons, 2009. SMSG, Geometry, New Haven: Yale University Press, 1960. Wilder, R., The Foundations of Mathematics, New York: John Wiley & Sons, 1965.

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Analytic Geometry

Figure 3.0 Architects no longer need to make small scale physical models of planned buildings. Instead they use computer aided design (CAD) programs to display proposed buildings with interior and exterior views. Analytic geometry provides the structure for these and all computer graphics, as well as numerous other applications. Though the idea behind it all is childishly simple, yet the method of analytic geometry is so powerful that very ordinary [youth] can use it to prove results which would have baffled the greatest of the Greek geometers—Euclid, Archimedes, and Apollonius. —E. T. Bell How can it be that mathematics, being after all a product of human thought independent of experience, is so admirably adapted to the objects of reality? —Albert Einstein

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3.1 Overview and History The fruitful union of algebra and geometry called analytic geometry has become an indispensable tool for mathematicians, scientists, and those in many other fields. Although Ren´e Descartes (1596–1650) and Pierre de Fermat (1601–1665) deserve credit for creating analytic geometry, many others before and after shared in its development. (See Boyer [1].) After 150 B.C.E., Greek astronomers used coordinates to describe the positions of stars, and Greek and Roman geographers used coordinates to describe places on earth. In the fourteenth century Nicole Oresme (1323–1382) used some examples of graphic representation. Arab and Renaissance mathematicians developed algebra into a powerful language. Franc¸ois Vi`ete (1540–1603) used letters to represent unknown values and general situations. Vi`ete called his approach analysis, following the ancient Greeks’ meaning of the word. He started by assuming that the problem had been solved and used a letter to represent the answer, which he then found algebraically in the modern sense. However, Vi`ete retained the Greek limitation of adding only like quantities. Thus, in modern notation, Vi`ete would be willing to add x 3 + a 2 x, but not x 3 + ax, because the latter expression would represent a volume added to an area. Pierre de Fermat united the notational advances of Vi`ete’s algebra with traditional geometry. He realized that first- and second-degree equations correspond to lines and conics and investigated some curves defined by higher degree equations. He solved some questions now considered part of calculus, such as finding maxima and minima of certain functions. Although Fermat developed analytic geometry first, Ren´e Descartes published sooner and was far more influential. Descartes freed algebraic notation from Vi`ete’s restrictions of homogeneous dimensions. His famous book Geometry, published in 1637, showed the power of this new field, solving problems the Greeks could not answer. Mathematicians began investigating the tremendous variety of curves described by algebraic equations. However, Descartes’s book doesn’t look like analytic geometry to us, for he didn’t use coordinates and axes. Rather, he described the length of a line segment in terms of relationships of the lengths of other line segments and translated them into an algebraic equation. Nevertheless, we often call coordinates Cartesian in honor of Descartes (and to distinguish them from other coordinate systems, such as polar coordinates). During the first century of analytic geometry—and the early years of calculus— mathematicians didn’t realize the power and simplicity of functions. Curves such as the folium of Descartes, x 3 − 2x y + y 3 = 0 shown in Figure 3.1, were tackled as were more familiar curves such as y = 13 x 3 + 4x, shown in Figure 3.2. Leonhard Euler (1707–1783), the most prolific mathematician of all time, emphasized functions and recast analytic geometry and calculus in nearly modern form in his influential textbooks. Mathematicians have continued to develop analytic geometry and extend it into new branches of mathematics. In the nineteenth century mathematicians used analytic geometry to overcome visual limitations and investigate four and more dimensions. Transformational geometry and differential geometry grew out of analytic geometry, as did areas no longer thought of as geometry, such as linear algebra and calculus. The advent of computer graphics has renewed interest in analytic geometry. (See Boyer [1] for more on the history of analytic geometry and Eves [7] for more on the model.)

3.1.1 The Analytic Model We make the connection between geometric concepts and their algebraic counterparts explicit by building a model of geometry in algebra. The familiar graphs of analytic geometry are not

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1.5 1.0 0.5 0.0 –0.5 –1.0 –1.5 –1.5

–1.0

–0.5

0.0

0.5

1.0

1.5

Figure 3.1 Folium of Descartes. 4 2 –4

–2

2

4

–2 –4

Figure 3.2 y = 13 x 3 + 4x. actually part of the model, but they make it and its many applications understandable. Geometric axioms and theorems become algebraic facts to be verified. Algebraic equations and relations can be visualized. Interpretation. We interpret the Cartesian product R2 as the Euclidean plane. By point we mean an ordered pair of real numbers (x, y). By line we mean a set of points of the form {(x, y) : ax + by + c = 0}, for a, b, c ∈ R with a and b not both 0. A point (u, v) is on the line {(x, y) : ax + by + c = 0} if and only if ! au + bv + c = 0. The distance between two points P = (x, y) and Q = (u, v) is d(P, Q) = (x − u)2 + (y − v)2 .

Remarks. As usual, we identify a line by its equation. Two lines ax + by + c = 0 and mx + ny + p = 0 are the same provided that there is a nonzero real number k such that ak = m, bk = n, and ck = p. The equation for the distance between two points is in essence the Pythagorean theorem—now no longer a theorem, but a definition. Exercise 3.1.1.∗ Find the slope and y-intercept of the line ax + by + c = 0, if b ̸= 0. (When b = 0, the line is vertical and has no slope. If a and b are both 0, either no points or all points satisfy the equation ax + by + c = 0; either way the equation does not represent a line.)

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Exercise 3.1.2.∗ Find a simple expression for the x and y coefficients of a line perpendicular to ax + by + c = 0. (The usual description that their slopes are negative reciprocals only works when they both have nonzero slopes.) Example 1. Verify Hilbert’s axioms I-1 and I-2 for this model: For any two distinct points, there is exactly one line on both points. Verification. Let (x1 , y1 ) and (x2 , y2 ) be two distinct points. Case 1 x1 = x2 . The line x − x1 = 0 is on both points. Let ax + by + c = 0 be a line through the two points. Then ax1 + by1 + c = 0 and ax1 + by2 + c = 0. The equations reduce to b(y1 − y2 ) = 0. For the points to be distinct, y1 − y2 ̸= 0. So b = 0, which forces c = −ax1 . Thus ax + by + c = 0 is a multiple of x − x1 = 0, showing that only one line passes through the two points. Case 2 x1 ̸= x2 . Verify (y2 − y1 )/(x2 − x1 )x − y + y1 − (y2 − y1 )/(x2 − x1 )x1 = 0 is a line on both points. As in case 1, verify that a line ax + by + c = 0 on both points is a multiple of the line given in this case. ♦ Part of analytic geometry’s power comes from our ability to solve geometric problems with algebra and to understand algebraic expressions geometrically. Example 2 revisits exercise 1.4.21, which asked you to prove that the medians of a triangle meet in a common point. Example 2. Show that the medians of a triangle intersect in the point two-thirds of the way from a vertex to the opposite midpoint. Solution. Without loss of generality, pick the axes so that the vertices of the triangle shown in Figure 3.3 are (0, 0), (a, 0), and (b, c). Verify that the midpoints of the sides are ( a2 , 0), ( b2 , 2c ), , 2c ). The medians connect midpoints to opposite vertices. Verify that the medians and ( a+b 2 are y = (cx/(a + b), y = cx/(b − 2a) − ac/(b − 2a), and y = 2cx/(2b − a) − ac/(2b − a). , 3c ) is on all the lines and two-thirds the distance from each vertex to the opposite Verify that ( a+b 3 side. ♦ (b,c)

(b/2,c/2) (0,0)

((a+b)/2,c/2)

(a/2,0)

(a,0)

Figure 3.3

3.1.2 Exercises for Section 3.1 In the problems use familiar properties of geometry and analytic geometry, such as non-vertical parallel lines have the same slope. 3.1.3. *(a) Imagine the top of a ladder slipping down a wall while the bottom of the ladder slides away from it. Draw a diagram and guess what curve the midpoint of the ladder makes.

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(b) Model the situation in part (a) with a ruler and two adjacent sides of a sheet of paper, marking various midpoints of the ruler on the paper. *(c) Use analytic geometry to find the set of all midpoints of segments of length 1 whose endpoints are on the x- and y-axes. (d) Use a physical model and analytic geometry to explore what happens in part (b) if the corner of the paper doesn’t form a right angle or if you pick a point on the ruler other than the midpoint. 3.1.4. (a) Use analytic geometry to verify that the diagonals of a parallelogram bisect each other. (b) Use analytic geometry to show that the four midpoints of a quadrilateral always form a parallelogram. 3.1.5.

(a) It is easy to find the area of the right triangle with vertices A = (0, 0), B = (3, 0), and C = (0, 4) using the lengths of its two sides. Verify that we get the same area using the hypotenuse BC and the altitude from A. *(b) Use analytic geometry to verify the law of cosines for a triangle: c2 = a 2 + b2 − 2ab cos(C) (See Figure 3.4). B a

C

c

b

A

Figure 3.4 3.1.6. Verify Hilbert’s axiom IV-1 (Appendix C) in the analytic geometry model: Through a given point P not on a given line k there passes at most one line that does not intersect k. 3.1.7. *(a) Define a circle in analytic geometry. (b) Verify Euclid’s postulate 3 (Appendix A): To describe a circle with center (a, b) and radius r . *(c) Give an algebraic expression representing points in the interior of a circle with center (a, b) and radius r . 3.1.8. Use analytic geometry to verify that the perpendicular bisectors of each side of a triangle intersect in a common point that is equidistant from its vertices. *3.1.9. In applications, the x- and y-coordinates may represent quantities with different sizes, so for convenience we use different scales on the axes, as in Figure 3.5. Suppose that on the x-axis a unit represents a distance of k and on the y-axis a unit represents a distance of j. Explain why the equations of lines in this model have the same form as in the usual model. Develop a formula for the distance between two points, P = (x1 , y1 ) and Q = (x2 , y2 ). Give the equation of a circle of radius r and center (a, b).

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5

(2,5)

1 –1 –1

1

2

Figure 3.5 3.1.10. Occasionally axes that don’t meet at a right angle are advantageous, as in Figure 3.6. Repeat exercise 3.1.9, assuming that the angle between the positive x-axis and the positive y-axis is 120◦ . Assume that the units on the axes represent a distance of 1. Hint: See exercise 3.1.5 (b). (2,3)

3 1 –1

–1

1

2

Figure 3.6 3.1.11. (Calculus) Before the advent of calculus, Fermat developed a method for finding the maximum or minimum of certain functions, such as bx 2 − x 3 . He reasoned first that if two values of x, say u and v, give the same height, he would get bu 2 − u 3 = bv 2 − v 3 . (a) Verify that the equality reduces to b(u + v) = u 2 + uv + v 2 , provided u ̸= v. (b) Fermat reasoned that at a maximum (or minimum) for the function the two xvalues are equal. Replace v in part (a) with u and simplify to get u = 23 b. Verify, using calculus, that the value does indeed give a (relative) maximum for the function y = bx 2 − x 3 . *(c) Use Fermat’s algebraic approach to find the (relative) maximum and minima for x 4 − 2b2 x 2 + b4 . Use calculus to verify the values are correct. (Fermat didn’t consider negative numbers or zero as potential answers.) (d) Explain any logical shortcomings of Fermat’s approach. (e) Explain the practical shortcomings of Fermat’s method for expressions that aren’t polynomials. 3.1.12. (Calculus) Most lines that intersect a curve do so in two (or more) places, say a and b. When you solve the system of equations for the line and the curve with two intersections, they reduce to the form (x − a)(x − b)k = 0. Descartes noted that tangents have a double root at a point of tangency a: the system of a curve and its tangent line at a reduces to (x − a)(x − a)k = 0. We can use this idea to find some tangents without calculus.

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(a) Verify that the equation of a line through (2, 2) is y = mx − 2m + 2. (b) Note that (2, 2) is a point on y = x 2 − x. Substitute mx − 2m + 2 for y in y = x 2 − x and find x in terms of m with the aid of the quadratic formula. (c) Which value of m in part (b) gives one value of x (a double root)? Hint: Consider √ what is under .) Use this m to find the tangent’s equation. Why must the line in part (a) with this slope be tangent to y = x 2 − x? (d) Use calculus to verify your answer in part (c). (e) Repeat parts (a) to (d) for the tangent to the hyperbola y 2 − x 2 = 3 at the point (1, 2) with the line y = mx − m + 2. (f) Vary the preceding approach to find the two tangents to y = x 2 − 2x + 4 through the point (0, 0). Graph the parabola and the tangents. (g) Vary this second approach to find the tangent to y = 4x − x 2 parallel to y = 6x. (h) Discuss any logical and practical shortcomings of this method. (Descartes used a method related to this to find tangents.) *3.1.13. (a) Graph the following functions and decide which enclose a convex region of the plane: y = x 2 , y = x − x 2 , y = x 3 , y = x 4 , y = e x , y = sin(x), and y = ln(x). (b) Functions f that satisfy f (a + b) < f (a) + f (b) for all a and b are called convex functions. Which of the functions in part (a) are convex functions? How do convex functions compare with functions that enclose a convex region of the plane? To explain the difference between these uses of convex for functions, define concave functions. Explain how concave functions relate to functions enclosing a convex region. (c) (Calculus) Find the second derivative of the functions in part (a). What is special about the second derivative of the convex functions? Use a graph to explain how the definition of a convex function fits with what you found out about the second derivatives of convex functions. What can you say about the second derivative of the concave functions you defined in part (c)? *3.1.14. The arithmetic of complex numbers (C) has a well-known geometric interpretation in the analytic plane R2 . The complex number a + bi can be represented as the point, or vector, (a, b) in the plane. Addition of complex numbers corresponds to vector addition: (a + bi) + (c + di) = (a + c) + (b + d)i. (a) Explain and illustrate on Cartesian axes why addition satisfies the parallelogram law, which says that the complex numbers 0, a + bi, c + di, and (a + bi) + (c + di) lie on the vertices of a parallelogram. (b) The complex conjugate of a + bi is the number a − bi. Illustrate on Cartesian axes how the numbers are related geometrically. √ (c) The modulus of a + bi is the real number a 2 + b2 . What does the modulus tell you geometrically? 3.1.15. The following formula for complex multiplication doesn’t reveal the geometry. (a + bi)(c + di) = (ac − bd) + (ad + bc)i

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(a) How does the product of a complex number and its conjugate relate to the modulus? (See Exercise 3.1.14.) (b) Illustrate with several examples on Cartesian axes the result of multiplying a + bi by a real number r + 0i. What corresponds geometrically to multiplying by a real number? (c) Illustrate on Cartesian axes the result of multiplying a + bi by i, a complex number on the unit circle. Also illustrate the result of multiplying a + bi by 0.6 + 0.8i and by −0.96 + 0.28i, two other points on the unit circle. What do you think multiplication by a point on the unit circle does geometrically to a + bi? (d) Explain why a complex number c + di can be written as the product of its modulus and a complex number x + yi on the unit circle (for which x 2 + y 2 = 1). Use parts (b) and (c) to describe what multiplication by a complex number c + di does geometrically to a + bi.

3.1.3 Ren´e Descartes Ren´e Descartes (1596–1650) won acclaim in philosophy, as well as in mathematics. In his most famous work, the Discourse on the Method, he tried to show how his rational approach could lead to new knowledge in any domain. He started by doubting everything that he did not know evidently to be true. His famous statement, “I think, therefore I am,” was his first principle of philosophy, the idea that withstood all of his doubting. Next, he sought to reduce each difficulty he studied into many simpler parts and approach them systematically and exhaustively. The method may not seem revolutionary now, but Descartes was challenging traditional learning handed down for centuries. Descartes illustrated his method in three essays—Optics, Geometry, and Meteorology—that followed the Method. Of these, Geometry was by far the most influential. In the first part of Geometry, Descartes showed how to solve a number of problems, including one that the Greeks couldn’t. He translated the geometric descriptions of curves into algebraic equations that he could then solve. In the second part, he provided a method of finding tangents to curves and attempted to classify curves. In the third part, he investigated the theory of equations, including Descartes’ rule of signs, giving a bound on the number of positive roots and what we call negative roots of a polynomial equation. (Descartes would have said that the equation x + 5 = 0 has 5 as a false root, rather than a root of −5.) (See Boyer [1] and Grabiner [8] for more on Geometry.) Descartes intentionally wrote obscurely, making it difficult for others to extend his work. However, later editions of Geometry contained commentaries by others that explained his work, revealing the power of his method. The success of analytic geometry in posing and solving important problems has made it indispensable in mathematics. Of course, calculus depends crucially on the language of analytic geometry. The great insights of Newton and Leibniz establishing calculus in the 1660s and 1670s would have been impossible without Descartes’ pioneering work forty years earlier.

3.2 Conics and Locus Problems The Greeks identified the three types of conics: ellipses, parabolas, and hyperbolas. The books Apollonius of Perga (circa 260–190 B.C.E.) wrote about conics, with complete proofs, were as

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3.2 Conics and Locus Problems

central to the ongoing study of conics as Euclid’s Elements was for the study of elementary geometry. However, nearly two thousand years passed before Johannes Kepler (1571–1630) found one of the first of many applications of conics outside of mathematics. Kepler figured out how to represent the orbits of the planets using ellipses, a major step leading towards Newton’s great advances in physics. We call the curves conics because they are the intersections of a (double-napped) cone with planes at various angles (Figure 3.7). To find their familiar equations we use an easier characterization based on distance. The process of finding a set of points or its equation from a geometric characterization is called a locus problem. See Eves [7] for further information.

(c) (a)

(b)

Figure 3.7 The intersection of a plane and a cone can be (a) an ellipse, (b) a parabola, or (c) a hyperbola.

Example 1. Find the set (locus) of points P such that P is the center of a circle tangent to two given lines. Solution. If the two lines are parallel, then a circle tangent to both must have its center on the line midway between them. In this case the midline is the locus. If the two lines intersect at a point Q, then the centers of circles tangent to both lines must be on one of their two angle bisectors. Illustrate the two cases. ♦ Definition. Given two points F and F ′ , an ellipse is the set of points P in the plane such that the sum of the distances of P to F and F ′ is constant. Given two points F and F ′ , a hyperbola is the set of points P in the plane such that the absolute value of the difference of the distances

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from P to F and F ′ is constant. Given a line m and a point F not on m, a parabola is the set of points P in the plane such that the distance of P from F equals the distance of P from m. The points F and F ′ are called foci, the plural of focus. The line m is called the directrix. (Figures 3.8, 3.9, and 3.10.)

P P

F′ F′

F

F

Figure 3.9 The foci of a hyperbola.

Figure 3.8 The foci of an ellipse.

P F

M

Figure 3.10 The focus and directrix of a parabola.

Exercise 3.2.1. Construct an ellipse as follows. Fix the two ends of a piece of string to different points on a piece of paper. Place a pencil on the paper so that it holds the string taut. Explain why the pencil sweeps out an ellipse as it moves. Example 2. Find an equation for an ellipse. Solution. Without loss of generality, let the foci F and F ′ have coordinates ( f, 0) and (− f, 0). Let P = (x, y) be a point on the ellipse. The definition of an ellipse gives the following equation. !

(x − f )2 + y 2 +

!

(x + f )2 + y 2 = k.

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3.2 Conics and Locus Problems

Here k, the sum of the distances, must be greater than 2 f . We use algebra, including squaring twice, to simplify the equation. ! ! (x − f )2 + y 2 = k − (x + f )2 + y 2 . ! x 2 − 2 f x + f 2 + y 2 = k 2 − 2k (x + f )2 + y 2 + x 2 + 2 f x + f 2 + y 2 . ! −4 f x − k 2 = −2k (x + f )2 + y 2 . 16 f 2 x 2 + 8 f xk 2 + k 4 = 4k 2 (x 2 + 2 f x + f 2 + y 2 ).

x 2 (16 f 2 − 4k 2 ) − 4k 2 y 2 = k 2 (4 f 2 − k 2 ). We can rewrite the equation as

y2 x2 + 2 =1 2 a b because k 2 − 4 f 2 > 0. The diameters of the ellipse along the x- and y-axes have lengths 2a and 2b, respectively. ♦ Exercise 3.2.2. Verify that a circle satisfies the definition of an ellipse and the lengths of the diameters of an ellipse become equal in this case. Example 3. An equation of a hyperbola is x 2 /a 2 − y 2 /b2 = 1, and an equation of a parabola is y = ax 2 . Solution. See Exercise 3.2.7. ♦

Exercise 3.2.3. Paper folding can outline a parabola. In origami we can construct the perpendicular bisector between a point and a point on a line by folding the paper so that the two points match. Use the edge of a piece of paper as the line, which will be the directrix of the parabola. On the piece of paper mark a point, which will be the focus of the parabola. (See Figure 3.11.) Make a number of folds of this type and see that they look like tangents to an unseen parabola. Explain how this process fits the definition of a parabola. (See Broman and Broman [2] for other constructions and an explanation.)

F

m

Figure 3.11 Folds outlining a parabola.

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Example 4. Use analytic geometry and calculus to verify that the method of Exercise 3.2.3 gives tangents to a parabola. Solution. For computational ease and to match the equation y = ax 2 for a parabola from Example 3, let the directrix m have equation y = −1 and the focus F be (0, 1), as in Figure 3.12. For a point W = (w, −1) on m, the midpoint of F W is (w/2, 0) and the slope of F W is −2/w. Hence the perpendicular bisector has equation y = wx/2 − w 2 /4. We need to show that the lines are tangents to y = ax 2 for an appropriate value of a. The point of tangency is not (w/2, 0); it is at some unknown point T = (t, at 2 ). The tangent there has slope 2at and equation y = 2at x − at 2 . We need 2at = w/2 and at 2 = w 2 /4 to match the tangents with the perpendicular bisectors. Then a = 1/4 and t = w. ♦ T = (t, at2) F = (0,1) (w/2,0)

W = (w,-1)

Figure 3.12 Examples 2, 3, and 4 give equations of conics when we pick the easiest starting conditions. Less convenient conditions make the algebra more difficult and the final equations more complicated. The general equation of a conic is ax 2 + 2bx y + cy 2 + 2d x + 2ey + f = 0. The conic is an ellipse, parabola, or hyperbola, depending on whether ac − b2 is positive, zero, or negative, respectively. (Circles are special ellipses with a = c and b = 0.) However, some of the general second-degree equations are degenerate: that is, they are equations for one or two lines or a point or even the empty set. For example, x 2 − y 2 = 0 factors to (x − y)(x + y) = 0, and the points on it are on the lines y = x and y = −x. The locus is a non-degenerate conic provided that ac f + 2bde − ae2 − b2 f − cd 2 ̸= 0. This expression is the determinant of ⎡

a ⎣b d

b c e

⎤ d e ⎦. f

Example 5. The hyperbola x 2 − y 2 = 1 is closely related to the degenerate conic x 2 − y 2 = 0, as illustrated in Figure 3.13. We can factor x 2 − y 2 = 0 into (x − y)(x + y) = 0 or the lines y = x and y = −x. They are called the asymptotes of the hyperbola x 2 − y 2 = 1. As a point moves out on the hyperbola it approaches one of them, although the lines never intersect the hyperbola. √ We can use calculus √ to make this notion more precise. Write the hyperbola as y = ± x 2 − 1. Then the expression x 2 − 1 − x compares the difference between the y-coordinates of a point on the upper right√branch of the hyperbola and the on y = x with the same x-coordinate. Because limx→∞ x 2 − 1 − x = 0, verified using calculus, the hyperbola approaches the asymptote. ♦

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3.2 Conics and Locus Problems

2

1

–2

–1

1

2

–1

–2

Figure 3.13 A hyperbola and its asymptotes. Example 6. If we pick .√ F=

√ / 2 − 2 , , 2 2

. √ √ / 2 − 2 F = , , 2 2 ′

and

√ k = 2 2,

as shown in Figure 3.14, the ellipse has the equation 34 x 2 + 12 x y + 34 y 2 = 1. Note that ac − b2 = (3/4)(3/4) − (1/4)(1/4) = 1/2 > 0 and that the determinant is −1/2. ♦

1

–1

1 –1

Figure 3.14 The ellipse 34 x 2 + 12 x y + 34 y 2 = 1. Example 7. Use calculus and trigonometry to show the reflection property of the parabola: All rays from the focus are reflected by the parabola in rays parallel to its axis of symmetry. Note: Satellite dishes reverse this process, taking essentially parallel signals from a distant satellite and focusing them at one point. Solution. Turn the parabola horizontally, √ √ as shown in Figure 3.15, and use calculus and trigonometry. Verify that the parabola y = ± 4kx has focus F = (k, 0). For a point P =√(x, 4kx) on the upper half of the parabola, verify that the of the line from F to P is 4kx/(x − k) √ slope √ and that the slope of the tangent line at P is k/ x. The reflection property says that the angle between the lines equals the angle between the tangent line and a horizontal line. The slopes are tangents of the various angles. Recall that tan(α − β) =

tan α − tan β . 1 + (tan α tan β)

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Analytic Geometry

√ √ √ If tan α = √ 4kx/(x − k) and tan(β) = k/ x, the right side of the formula for tan(α − β) √ simplifies to k/ x. It is the tangent for the angle between the tangent line and the horizontal line. Explain why the lower half is similar. The shape of reflectors in flashlights make use of the property. The design of satellite disks makes use of it in reverse: The disks take incoming parallel rays and focus them on one point. ♦

Figure 3.15 Reflective property of a parabola. The names of the conics come from properties the Greeks found about areas of conics, as illustrated in Figure 3.16. For a height y, a point (x, y) on the parabola satisfies x 2 = cy. The Greek root for “parabola” means “application,” by which they meant that applying the process of making a square with sides x gave the same area as a rectangle with sides c and y. However, for the same height the corresponding point on the ellipse has x 2 < cy and “ellipse” means “to fall short.” Similarly, the corresponding point on the hyperbola has x 2 > cy and the prefix “hyper” means “too much.” The word “asymptote” means “not falling together” in Greek, meaning that the asymptote and the hyperbola don’t intersect. 4 3 2 1 –2

–1

1

2

Figure 3.16 An ellipse, a parabola and a hyperbola.

3.2.1 Exercises for Section 3.2 *3.2.4. (a) Describe the locus of points in the plane equidistant from two points. (b) Describe the locus of points on a sphere equidistant from two points.

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3.2 Conics and Locus Problems

(c) Describe the locus of points in space equidistant from two points. Relate your answers in parts (a) and (b) to this. (d) Repeat parts (a), (b), and (c) for three noncollinear points. 3.2.5. Derive the equation of a circle with center (a, b) and radius r from the definition of a circle. 3.2.6.

(a) Attach one end of a string to a ruler and fix the other end to a sheet of paper. Hold the ruler and a pencil as indicated in Figure 3.17. As you slide the ruler back and forth while keeping the string taut with the pencil, the pencil will trace out a curve. Identify which conic is created and explain why. (b) On wax paper or tracing paper draw a large circle and mark its center C and any other point A inside it. For various points P on the circle, fold and crease the paper so that points A and P coincide. The set of creases will outline an ellipse. Use the definition of an ellipse together with Figure 3.18 to explain the result. Note that Q is a point on the ellipse, not R. (Figure 3.18 also includes folds for the points S, T , U , V , and W .)

Figure 3.17 Drawing a parabola. *(c) Repeat part (b) with point A outside the circle. Modify Figure 3.18 and explain the result. 3.2.7. (a) Derive the equation of a hyperbola. (b) Derive the equation of a parabola. Hint: Use y = −k for the directrix and (0, k) for the focus. *3.2.8. For the following equations, identify the type of conic it is, determine whether it is degenerate, and sketch its graph.

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Analytic Geometry

S P

T Q

R W

C

A

V

U

Figure 3.18 Folding the outline of an ellipse. (a) (c)

1 2 y 2 1 2 x 4

+ x − 2y = 0

(b) 14 x 2 + y 2 − x − 6y + 9 = 0

+ y 2 − x − 6y + 10 = 0 (d) 2x 2 − 3x y + y 2 − 4 = 0 2 Hint: For graphing, factor 2x − 3x y + y 2 = 0 and then relate the degenerate conic to the given one. (e) x 2 − x y + y 2 − 1 = 0. Hint: Pick x-values and find the y-values.

3.2.9. Solve the following locus of points problems. (a) The set of points equidistant from two intersecting lines. Hint: Why can we let the lines be y = 0 and y = mx or y = 0 and x = 0? The distance from a point to a line is given by the distance between the point and the intersection of the line and the perpendicular to it through the point. (b) The set of points whose distance from one line is half the distance from another line. (c) The set of points whose distance from one line is k times the distance from another line. (d) The set of points P so that the distance of P to y = 0 times the distance from P to x = b equals the square of the distance of P from y = x. (This problem is a special case of the three line locus problem worked on by the Greeks and completely solved by Descartes.) *3.2.10. Apollonius solved the following locus problem without analytic geometry. Use analytic geometry to find the locus of points P whose distance from a point F is k times their distance from another point F ′ . Describe the locus when k > 1 and when 0 < k < 1. Hint: Let F = (k, 0) and F ′ = (−1, 0). Consider the case k = 1 separately from other k > 0. 3.2.11

(a) Use several points on the ellipse in Example 6 to verify that the sum of the distances from a point on the ellipse to the two foci is a constant. (b) Graph y = 2x1 , a familiar form for a hyperbola.

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3.2 Conics and Locus Problems

*(c) Rewrite the equation in part (b) in the general form of a conic, verify that ac − b2 < 0 and that the determinant is nonzero. *(d) Use the definition of a hyperbola and several points on the curve y = 2x1 to verify that its foci are at (1, 1) and (−1, −1). *(e) (Calculus) Find the asymptotes to y = 2x1 and use calculus to verify that they are the asymptotes. 3.2.12. (Calculus) We generalize Example 5. A line y = mx + b is an oblique asymptote of a function y = f (x) provided that limx→±∞ ( f (x) − (mx + b) ) = 0. (a) Verify that the hyperbola x 2 /a 2 − y 2 /b2 = 1 has asymptotes y = (b/a)x and y = −(b/a)x. Show that the product of the linear equations when rearranged gives a degenerate conic. Compare the graphs of x 2 /a 2 − y 2 /b2 = 1 and x 2 /a 2 − y 2 /b2 = −1 for the same values of a and b. Describe the difference between these two curves. *(b) Use calculus to find the asymptotes of the hyperbola y = x + (1/x). Verify that when you write the hyperbola’s equation in the general form of a conic, ac − b2 < 0 and that the determinant is nonzero. How does this form of the hyperbola relate to the product of the linear equations of its asymptotes? (c) Verify that all hyperbolas y = x + (k/x), k ̸= 0, have the same asymptotes as the hyperbola in part (b). Graph two hyperbolas, one with k > 0 and one with k < 0, on the same axes with the asymptotes. How does changing k change the graph? Show that every point not on the asymptotes is on just one of the hyperbolas. (d) Pick two intersecting lines and find the equation of the degenerate conic they form. Use the equation to give the equation of the family of hyperbolas for which the lines are the asymptotes. 3.2.13. Find the equation of the ellipse with foci at (−1, 0) and (1, 0) passing through the point (z, 0), where z > 1. Hints: What is the sum of the distances from (z, 0) to the foci? 2

2

Use it to determine a in ax 2 + by2 = 1 from Example 2. Now find a point on the y-axis and the ellipse to determine b. 3.2.14. Apollonius found the following locus characterization of conics, except circles: the locus of points P so that the distance from P to a focus F is k times the distance of P to a directrix m. (a) What type of conic is this when k = 1? (b) Let F be the point (k, 0) and m be the line x = −1. Use analytic geometry to solve the locus problem. (c) For what values of k is the locus in part (b) an ellipse? an hyperbola? (d) How does the locus problem fit with circles? (Apollonius didn’t think of circles as conics.) 3.2.15. The Loran navigation system enables someone on a ship to determine its position. An instrument on the ship receives precisely timed signals from three transmitters and uses the differences in arrival times to determine the ship’s position. Explain how the definition of a hyperbola is essential to the determination. Explain why the ship needs to receive signals from three transmitters, rather than just two.

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3.2.2 Pierre de Fermat Although a lawyer and political councillor by profession, Pierre de Fermat (1601–1665) is best remembered for the mathematics he pursued in his spare time. He corresponded extensively with other mathematicians, rather than seeking publication. Indeed, his work in analytic geometry, even though done before Descartes, wasn’t published until after Fermat died. He investigated the shapes of curves from their equations. This approach complemented Descartes’ work, as Descartes emphasized finding the equation of a curve defined by some geometric process. Fermat explicitly discussed first- and second-degree equations and explored many new curves with higher degree equations. Fermat also contributed extensively and profoundly to number theory. He is best known for his conjecture, known as Fermat’s last theorem: a n + bn = cn has no nontrivial integer solutions if n > 2. This statement holds the record for the greatest number of incorrect published “proofs,” but Andrew Wiles finally proved it 1994. This simple-looking statement has led to extensive and profound investigations in number theory. Fermat’s theorems in number theory concentrate on primes, divisibility, and powers. For example, Fermat’s little theorem, an important tool in abstract algebra and coding theory, states that if a is not a multiple of a prime p, then p divides a p−1 − 1. Other areas of mathematics also benefited from Fermat’s creativity. The founding of probability as a mathematical subject grew out of letters between Fermat and Blaise Pascal (1623– 1662). They corresponded at some length and eventually solved a problem on how to distribute fairly the wagers of an interrupted game of chance. Fermat’s method was an important precursor of calculus for finding maxima and minima. In addition, he understood the rules that we now describe as differentiating and integrating polynomials.

3.3 Further Topics in Analytic Geometry [Barycentric coordinates] replaced the need for genius, upon which no one can depend, with the ability to solve problems mechanically. —Carl Friedrich Gauss

The great flexibility of algebra and vectors fosters many geometric and applied variations on the analytic geometry model. We present a few here.

3.3.1 Parametric Equations Leonhard Euler introduced parametric equations in his influential 1748 textbook on analytic geometry. Separate functions for the x- and y-coordinates allow graphs of complicated curves. The functions, x(t) and y(t), are in terms of a third variable, t, which we can think of as representing time. The point (x(t), y(t)) traces out a curve as t varies. As y does not depend on x, the curve can double back on or even cross itself. With three or more parametric equations we can describe curves in three or more dimensions. Example 1. (Calculus) For x(t) = t 3 − t, y(t) = t 2 − t 4 , and −1.1 < t < 1.1, the point (x(t), y(t)) traces out the curve shown in Figure 3.19. We can think of the graph starting

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3.3 Further Topics in Analytic Geometry

at the lower left for t = −1.1, heading up to the right, looping around, and ending at the lower right for t = 1.1. The function y(t) controls the heights of points. Setting the √ derivative y ′ (t) = 2t − 4t 3 equal to 0 shows that the curve reaches its maxima when t = ± 2/2 and its local minimum when t = 0. ♦ 0.2 0.1 –0.4

–0.2

0.2

0.4

–0.1 –0.2

Figure 3.19 Graph of x(t) = t 3 − t, y(t) = t 2 − t 4 . Exercise 3.3.1.∗ (Calculus) For which values of t does the curve in Example 1 cross itself at (0, 0)? What happens to the curve when x ′ (t) = 0? Example 2. A cycloid is the curve traced by a point on the circumference of a circle as the circle rolls along a line. Find parametric equations for a cycloid for a circle of radius r . Solution. We solve with r = 1. Let a circle of radius 1 roll along the x-axis with the point starting at (0, 0) when t = 0. Figure 3.20 illustrates a general point on the cycloid. As the circle rolls, the center moves along the line y = 1. When the circle has turned an angle of t radians, the center is at (t, 1). The point rotates around the center, so its coordinates will be (t + f (t), 1 + g(t)) for some functions f and g. Further, we know the radius is 1 so f (t)2 + g(t)2 = 1, which means that we can represent f and g with some variation on sine and cosine. For some values of t we can determine the coordinates of the point: if t = 0, the point is (0, 0) = (0 + 0, 1 − 1); if t = π /2, we get (π/2 − 1, 1 + 0); and when t = π , we have (π, 2) = (π + 0, 1 + 1). In general, trigonometry gives the coordinates (t − sin t, 1 − cos t) and so the parametric functions are x(t) = t − sin t and y(t) = 1 − cos t. Note that 1 − cos t = 0 only when t is a multiple of 2π , or when the circle has rolled a whole number of turns. How is the equation altered if the radius is r ? ♦ 2 1.5 1 0.5 2

4

Figure 3.20 A cycloid.

6

8

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Analytic Geometry

Example 3. Figure 3.21 illustrates a helix, a three-dimensional curve sometimes mistakenly called a spiral. Its equations are x(t) = cos t, y(t) = sin t, and z(t) = t/4. The equations for the x- and y-coordinates by themselves trace out a circle. Thus when seen from the top, the coils of the helix cover a circle. ♦

–0.5 –1.0

0.0

–1.0 1.0 –0.5 0.0 0.5

0.5

1.0

3

2

1

0

Figure 3.21 A helix.

3.3.2 Polar Coordinates Jakob Bernoulli (1645–1705) developed polar coordinates, an alternative analytic model of Euclidean geometry (Figure 3.22). The first coordinate of a point gives its distance from the origin (positive or negative), and its second coordinate gives the angle made with the x-axis. A point in polar coordinates is an ordered pair of real numbers (r, θ ), but two ordered pairs (r, θ ) and (r ′ , θ ′ ) can represent the same point if (1) r = r ′ = 0, (2) r = r ′ and θ and θ ′ differ by a multiple of π (180◦ ). By the multiple of 2π (360◦ ), or (3) r = −r ′ and θ and θ ′ differ by an odd! law of cosines the distance between two points (r, θ) and (a, φ) is r 2 + a 2 − 2ra cos(θ − φ), provided r and a are positive (Figure 3.23). (r, θ)

(r, θ)

(a, φ) a

r r

θ

Figure 3.22 Polar coordinates.

θ

φ

Figure 3.23 Distance in polar coordinates.

As with Cartesian coordinates we use functions and equations to represent sets of points, including lines. Lines through the origin make a constant angle with the x-axis, so satisfy θ = α

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3.3 Further Topics in Analytic Geometry

(r, θ) r

(a, φ)

a φ

Figure 3.24 Lines in polar coordinates. for some angle α. Other lines require some trigonometry, illustrated in Figure 3.24, where (r, θ) is a point on the line. There is a point ( p, α) on the line closest to the origin (0, 0), where the line forms a right angle with the line from the origin. The right triangle with hypotenuse of length r and one side of length p gives us cos(θ − α) = rp . We can solve for r to get a function of r in terms of the variable θ and the constants p and α: r = p sec(θ − α). Polar coordinates exhibit an advantage with curves having easy formulas for r in terms of θ but difficult formulas in Cartesian coordinates, where we express y in terms of x. Fortunately, graphing devices can give graphs in polar coordinates as well as Cartesian coordinates. Example 4. The easiest equations for the radius set r equal to a positive constant, say c. The graph of r = c is a circle of radius c and center (0, 0). Another easy formula, r = kθ, has the spiral shown in Figure 3.25 for its graph when θ ≥ 0 and k > 0. ♦ 2 1 –3

–2

–1

1

2

3

4

–1 –2 –3

Figure 3.25 A spiral. Exercise 3.3.2.∗ What does the graph r = kθ from Example 4 look like when θ < 0? How is that graph modified when we add a positive constant to get r = kθ + c? Example 5. In 1694 Bernoulli investigated a curve he called a lemniscate, given by r 2 = cos(2θ) whose graph appears in Figure 3.26. We need only consider angles where cos(2θ ) ≥ 0. By the periodicity of the cosine, we can restrict the angles to just − π4 ≤ θ ≤ π4 . The radius will satisfy −1 ≤ r ≤ 1, which explains why the curve remains within the unit circle. Since limθ→π/4 cos(2θ ) = 0, the curve has a tangent of θ = π4 , which is a line through the origin.

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Similarly, θ = − π4 is tangent at the origin, which helps to explain the figure-eight shape of the curve. ♦ 0.3 0.2 0.1 –1.0

–0.5

–0.1 –0.2 –0.3

0.5

1.0

Figure 3.26 A lemniscate. Exercise 3.3.3.∗ How does the lemniscate of Example 5 change when we consider r 2 = cos(θ )?

3.3.3 Barycentric Coordinates Barycentric coordinates, devised by August M¨obius in 1827, have a number of applications, including representing centers of gravity, his original motivation. He later developed the more important homogeneous coordinates in projective geometry from barycentric coordinates. (See Chapter 7.) Statisticians now use barycentric coordinates in trilinear plots, as in Example 6. Example 6. Some political polls classify potential voters as “leaning Republican,” “leaning Democratic,” and “independent,” which includes those with other party affiliations as well as those without leanings. According to a 2011 Gallup poll, 40% of U.S. voters identify themselves as leaning Republican, while 43.7% identify themselves as leaning Democratic. However, Presidential elections are determined state by state. In Figure 3.27 each point represents a state or the District of Columbia, positioned according to the percentages in the three categories. The distribution of the dots conveys quickly differences and similarities of political identification among the states without the need for a long table of numbers. For example, Washington, DC, labelled DC, has 11.7% of voters identifying with the Republican party and 78.6% identifying with the Democratic party, leaving just 9.7% classified as independent. Utah (marked with UT) has the highest percentage identifying as Republican: 59.5%, with 25.6% identifying as Democrats. The axis on the left and the dashed horizontal lines help determine for each dot the percentage identifying as Republican. Similarly, the axis on the bottom and the dashed lines with a negative slope indicate the percentage identifying as Democrats. The remaining axis and lines do the same for independents. Since two of the percentages for a state determine the third, the information is essentially two-dimensional. Nevertheless, there are three barycentric coordinates for each dot. In the case of Utah, the coordinates are (0.595, 0.256, 0.149). The upper region bounded by solid lines indicates states where a plurality of voters identify as Republican. Similarly, the lower left region indicates the Democratic leaning states. However, in any election, independent voters can determine the outcome, so states close to the dividing line are considered swing states. (The two-dimensional map overlooks the sizes of the different states, which greatly affect the outcome of Presidential elections. Source: Gallup Party Affiliation: /www.gallup.com/poll/125066/state-states.aspx.) ♦

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90% 80% 70% Republican Leaning Percentage 60%

10% 20% Independent 30% Percentage

UT

40%

50%

50%

40%

60%

30%

70%

20%

80% DC

10%

90%

90% 80% 70% 60% 50% 40% 30% 20% 10% Democratic Leaning Percentage

Figure 3.27 A trilinear plot of political identification by state. Barycentric coordinates relative to two points rather than three provide an easier setting to describe the idea. Suppose that two points P and Q have weights w p and wq that add to 1. (Thus they give the proportion of the total weight each point has.) From Archimedes’ law of the lever, their center of mass is at point S, between P and Q, which divides their distance in the same proportion as the weights with S closer to the heavier point. M¨obius defined (w p , wq ) to be the barycentric coordinates of the center of gravity of P and Q. (“Baro” is Greek for weight or heavy.) If P and Q have equal weights, S is the midpoint with coordinates (0.5, 0.5). If P has all the weight, S = P = (1, 0). While a point has two coordinates, the information is one-dimensional since wq = 1 − w p . Figure 3.28 illustrates the case for two points using distances, without reference to weights. (However, without reference to weights, the switch of the order of the coordinates and distances may seem counterintuitive.) For a point S between P and Q , where P Q = 1, the coordinates of S are (a, b), where P S = b and S Q = a. P = (1,0)

S = (a,b)

Q = (0,1)

b a a+b = 1

Figure 3.28 Barycentric coordinates in one dimension. Figure 3.29 extends the idea to barycentric coordinates relative to three points P, Q, and R. Here the coordinates (a, b, c) of a point S satisfy a + b + c = 1. As before, if the corner points have weights a, b, and c, then the center of gravity has coordinates (a, b, c). We use ratios

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and the two-dimensional situation to avoid the concept of weight. The line segment through R and S divides each segment T V parallel to P Q into segments T U and U V with the same ratio of lengths, here a to b. In particular, we can use the barycentric coordinates for the point on the segment P Q to determine the ratio of a to b. Similarly, the segment including Q and S determines the ratio between the first and third coordinates, a to c. In the same manner, the segment including P and S determines the ratio between the second and third coordinates, b to c. R = (0,0,1)

T

U

(a/(a+c), 0, c/(a+c)

V

(0, b/(b+c, c/(b+c) S = (a, b, c) Q = (0,1,0)

(a/(a+b), b/(a+b), 0) P = (1,0,0)

a+b+c = 1

Figure 3.29 Barycentric coordinates in two dimensions. This same reasoning extends to any number of points in any number of dimensions. If the weights at the original points are a, b, c, . . . and add to 1, the center of gravity is (a, b, c, . . .). We can extend barycentric coordinates to points outside the convex region determined by the original points, but some of the coordinates then will be negative and the interpretation as the center of gravity fails. We use directed distances and ratios to do this extension.

3.3.4 Other Analytic Geometries We can form an analytic geometry with many algebraic structures besides the real numbers. The other models lead to different geometries because only the real plane R2 satisfies all of Hilbert’s axioms in Appendix B. The model Q2 based on Q, the rational numbers, loses the continuity of the real numbers. A rational point is an ordered pair (x, y), where x, y ∈ Q. A rational line is the set of rational points {(x, y) : ax + by + c = 0}, where a, b, c ∈ Q and a and b are not both zero. Other interpretations remain the same as in the usual model, R2 , and all of Euclid’s postulates hold. However, Euclid’s construction of an equilateral triangle, proposition I-1, does not hold in Q2 . Without loss of generality, let the given√points be √ (0, 0) and (1, 0). Verify that the third vertex of the equilateral triangle must be ( 21 , ± 23 ). As 3 is irrational, neither vertex is a rational point. Hence there is no equilateral triangle in Q2 and proposition I-1, among others, is logically independent of Euclid’s postulates. Exercise 3.3.4.∗ Explain why, if two lines with rational coefficients have a point of intersection, it is a rational point in Q2 . Section 1.2, especially Exercise 1.2.25, considers which lengths are constructible using an initial unit length, a straightedge, and a compass. They can be written using whole numbers and any combination of addition, subtraction, multiplication, division, and square roots. If we use

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this set F of constructible numbers instead of the reals or the rationals, all Euclid’s constructions and theorems hold. However, Hilbert’s final axiom fails, which is thus independent of his other axioms. See a standard calculus text, such as Edwards and Penney [6, Chapters 10 and 12] for more on parametric equations and polar coordinates. Eves [7] has more varied information.

3.3.5 Exercises for Section 3.3 3.3.5. *(a) Explain why the parametric equations x = cos t and y = sin t determine a circle. How do they relate to x = cos 2t and y = sin 2t? (b) Graph the figure-eight given by x = cos t and y = sin 2t. (c) Graph the spiral given by x = t cos t and y = t sin t, where t ≥ 0. *(d) Compare the curve given by x = f (t) and y = g(t), where p ≤ t ≤ q with the curve given by x = f (kt) and y = g(kt), where p/k ≤ t ≤ q/k and k ̸= 0. Hint: See part (a). 2

2t 2t 3.3.6. (Calculus) (a) Verify that the parametric equations x(t) = 1+t 3 and y(t) = 1+t 3 satisfy the equation for the folium of Descartes: x 3 − 2x y + y 3 = 0. *(b) The parametric equations in part (a) are undefined at t = −1. Find limt→−1− x(t), limt→−1− y(t), limt→−1+ x(t), and limt→−1+ y(t). Explain what happens to the graph of (x(t), y(t)) for t near −1. *(c) Find limt→−∞ x(t), limt→−∞ y(t), limt→∞ x(t), and limt→∞ y(t). Explain what happens to the graph of (x(t), y(t)) as t goes to ±∞. ′ ′ 3.3.7. (Calculus) The ! tangent vector to (x(t), y(t)) is (x (t), y (t)). The tangent vector’s length ∥(x ′ , y ′ )∥ = (x ′ (t))2 + (y ′ (t))2 is the speed of the point along the curve at time t. Suppose that a projectile takes the path x(t) = 12t meters/sec, y(t) = −4.9t 2 + 14.7t meters/sec.

(a) Determine the initial velocity of the projectile (at t = 0). (b) The projectile stops when its height y(t) is again 0. Find the value of t when it stops, its velocity at that time, and determine how far it travelled horizontally. (c) The projectile reaches its highest point when y ′ (t) = 0. Find the value of t at its highest point as well as its velocity and height at that time. *3.3.8. (Calculus) Use Exercise 3.3.7 and measure angles in radians. (a) Find the tangent vectors at a value of t for the two curves in exercise 3.3.5(a). Compare their lengths and explain how these curves are related. (b) Find the tangent vector for the curve in Exercise 3.3.5(b). Compare the positions of the points and the directions of the tangent vectors when t = π /2 and t = 3π /2. (c) Find the tangent vector for the curve in Exercise 3.3.5(c). Compare the direction and length of the tangent vector when t = 2π and when t = 4π . Describe the speed of the point along the spiral S as t increases. (d) Find the points and the tangent vectors for the curves in Exercise 3.3.5(d) for t = c and t = c/k, for the first curve and the second curve, respectively. Compare the tangent vectors and their lengths.

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3.3.9. (Calculus) (a) Find the tangent vector (see Exercise 3.3.7) of the cycloid of Example 2 and use it to find the velocity at t = 2π and t = π . Use your answer to explain why, in a photograph of a moving bicycle, the spokes of the wheels nearest the ground are always in focus but the ones at the top are often blurred. (b) Find parametric equations for a modified cycloid similar to that in Example 2, but with the point inside the circle at (0, k), 0 < k < 1. Find the tangent vector for the curve. Compare the velocity at t = 2π with that in part (a). (c) Repeat part (b) with the point outside the circle (k < 0). Graph the curve. *3.3.10. Describe the paths of these three-dimensional curves and, if available, use a computer to graph them. (a) x(t) = t√cos(3t), y(t) = t sin(3t),√z(t) = t. (b) x(t) = 1 − t 2 cos(3t), y(t) = 1 − t 2 sin(3t), z(t) = t, where −1 ≤ t ≤ 1. (The curve is called a loxodromic spiral.) 3.3.11. Graph the curves having the following polar equations. (a) r 2 = sin(2θ). Explain how the curve relates to the lemniscate of Bernoulli. (b) r = sin 3θ and r = sin(4θ). Explain why the number of “petals” in the two curves are so different. (c) r = 1 + cos θ (“cardioid”) and r = 1 + 2 cos(θ) (“limac¸on”) Discuss the similarities and differences of the curves. For what values of θ does the limac¸on intersect itself? 3.3.12.

(a) Devise formulas to convert Cartesian coordinates (x, y) to polar coordinates (r, θ ) and vice versa. (b) Use part (a) to explain why the spiral in Exercise 3.3.5(c) is one of the spirals in Example 4. (c) Verify that x 2 − x + y 2 = 0 is a circle with radius 0.5 and center (0.5, 0). Convert the equation to polar coordinate form and simplify. *(d) Convert the polar coordinate formula of the lemniscate in Example 5 to Cartesian coordinates. Hint: use the double angle formula for cos(2θ ). *(e) Convert the polar coordinate formula in Exercise 3.3.11(a) to Cartesian coordinates. Hint: use the double angle formula for sin(2θ). (f) We can convert the polar coordinate form of a line, r = a sec(θ − α), for a ̸= 0, to r cos(θ − α) = a. Use the formula for cos(θ − α) and part (a) to verify this gives the equation of a line in Cartesian coordinates. (g) Use the law of cosines (Exercise 1.2.23) to justify the distance formula in polar coordinates. Explain how to modify the distance formula when one or both of the points’ first coordinates are negative. (h) Explain why polar coordinates are a model of Euclidean geometry.

3.3.13. Draw a triangle △ABC. Mark the following points on the triangle and give their barycentric coordinates with respect to A, B, and C in that order: the midpoints of

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each side, the intersection of the medians, and the points A, B, and C. Explain why the intersection of the medians is considered the center of gravity of the triangle. Mark the point in the triangle having the barycentric coordinates (0.25, 0.25, 0.5). Describe the possible barycentric coordinates of points on the median from A. 3.3.14. The points X = (1, 0), Y = (0, 1), and O = (0, 0) give particularly nice barycentric coordinates. (a) Explain why a point inside △O X Y with Cartesian coordinates (a, b) has barycentric coordinates (a, b, 1 − a − b) with respect to X , Y , and O. (b) If a point has barycentric coordinates (a, b, c) with respect to X , Y , and O, explain a b , a+b+c ). why its Cartesian coordinates are ( a+b+c 3.3.15. (a) Draw a trilinear plot representing the data below giving the percent of U.S. men (M) and women (F) at least 25 years old with different levels of education for different years. Label the points in some way to indicate trends over time. (Sources: 120 Years of American Education, ed. T. Snyder, U.S. Dept. of Educ., 1993 and U.S. Census, 2000 and 2010.) (b) Discuss to how well you think the trilinear plot conveys the information in the table, compared with the actual data. Year/Sex

< 12 years

High School graduation

>12 years

1940 M 1940 F 1950 M 1950 F 1960 M 1960 F 1970 M 1970 F 1980 M 1980 F 1990 M 1990 F 2000 M 2000 F 2010 M 2010 F

77.4 73.7 67.5 63.9 60.5 57.4 45.0 44.6 30.8 31.8 22.3 22.6 19.1 19.3 13.4 12.4

12.2 16.4 18.2 23.2 21.2 27.8 30.1 37.5 32.7 40.4 35.5 41.0 28.4 29.6 31.8 30.7

10.4 9.9 14.3 12.9 18.3 14.8 24.9 17.9 36.5 27.8 42.2 36.4 52.5 51.1 54.8 56.9

3.3.16. We investigate barycentric coordinates (a, b, c, d) of points in the plane based on the four special points at the vertices of the unit square X = (1, 0), Y = (0, 1), O = (0, 0), and U = (1, 1). As with Exercise 3.3.14, a point with Cartesian coordinates (s, t) has barycentric coordinates (a, b, c, d) satisfying (s, t) = a(1, 0) + b(0, 1) + c(0, 0) + d(1, 1) and a + b + c + d = 1. While barycentric coordinates are defined with more than three points in two dimensions, they have some peculiarities.

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*(a) Draw a picture of the square and find the point with barycentric coordinates (0.25, 0.25, 0.25, 0.25). Explain what this means in terms of the center of gravity. *(b) Repeat part (a) with the barycentric coordinates (0.5, 0.5, 0, 0) and (0, 0, 0.5, 0.5). Explain why we get this outcome and find the other barycentric coordinates that have this property. *(c) Find the family of barycentric coordinates that represent the point U . *(d) What property must the barycentric coordinates (a, b, c, d) of points on the line ← → X Y have? *(e) Find the point with barycentric coordinates (1, 2, 3, −5). (f) Which barycentric coordinate(s) must be negative for a point below the x-axis? (g) Repeat part (f) for points to the left of the y-axis. 3.3.17. Consider the analytic geometry model Q2 , where Q is the set of rational numbers. Give suitable interpretations for Hilbert’s undefined terms. Which of Hilbert’s axioms (Appendix C) are false in the model? Explain why they fail. 3.3.18. Consider the analytic geometry model Z2 , where Z is the set of integers. Give suitable interpretations for Hilbert’s undefined terms. Which of Hilbert’s axioms (Appendix C) are true in the model Z2 ? *3.3.19. Assign coordinates to the points on a sphere, using their longitudes and latitudes (Figure 3.30). Thus (x, y) represents a point provided that −90◦ ≤ y ≤ 90◦ . Every point (x, y) has multiple representations, such as (x + 360◦ , y). (a) (b) (c) (d)

Describe all coordinates for the north and south poles. Describe the curves on the sphere for y = a and x = b. Describe the curves, called loxodromes, on the sphere for y = mx + b. If you interpret point as a point on the sphere and line as any of the curves of parts (b) and (c), which of Hilbert’s axioms from group I (Appendix C) are true? Explain.

(0, 30) (0, 0)

(60, 0)

Figure 3.30 Coordinates on a sphere. 3.3.20. Assign coordinates to the points on the surface of a torus (a doughnut) using longitude and latitude measured in degrees (Figure 3.31). Every point (x, y) has multiple representations, such as (x + 360◦ , y + 720◦ ). Interpret point as a point on the

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3.3 Further Topics in Analytic Geometry

surface of a torus and line as the set of points satisfying an equation of the form ax + by + c = 0. (a) Find values of a, b, and c so that the line ax + by + c = 0 is finitely long. Find values so that that line is infinitely long. (b) Find a line that intersects n times with y = 0. Find a line that intersects y = 0 infinitely many times. (c) Which of Hilbert’s axioms in group I and IV (Appendix C) hold in the model? Explain.

(0, 0 )

(0, 30) (60, 0)

Figure 3.31 Coordinates on a torus. 3.3.21. Taxicab geometry is an alternative analytic model with a distance function corresponding to the distances taxicabs travel on a rectangular grid of streets. The taxicab distance between A = (a, b) and S = (s, t) is dT ( A, S) = |a − s| + |b − t|. The geometry has quite different properties from Euclidean geometry. (See Section 2.3 Example 3.) *(a) How are two points that have the same Euclidean and taxicab distances related? If the distances are different, which is larger? If the smaller distance is 1, how large can the other distance be? Explain and illustrate. *(b) A taxicab circle is the set of points at a fixed taxicab distance from a point. Describe taxicab circles. How can two taxicab circles intersect? Illustrate. (c) A taxicab midpoint M of A and B satisfies dT ( A, M) = dT (M, B) = 12 dT ( A, B). Illustrate the different sets of midpoints two points can have. (d) The taxicab perpendicular bisector of two points consists of all points equidistant from the two points. Describe and illustrate the different types of taxicab perpendicular bisectors. (e) In the plane find four points whose taxicab distances from each other are all the same. What is the maximum number of points in the plane all the same distance from one another using ordinary Euclidean distance? (f) We can extend Euclidean and taxicab distances to three dimensions. Find the largest number of points in three dimensions all the same distance from one another using Euclidean distance and then using taxicab distance. Explain or better justify your answers. 3.3.22. Another distance formula on R2 uses the maximum of the x and y distances: For A = (a, b) and S = (s, t), d M ( A, S) = Max{|a − s|, |b − t|}. Redo Exercise 3.3.21, using d M in place of dT .

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3.4 Curves in Computer-Aided Design Computer graphics depend heavily on analytic geometry. The computer stores coordinates of key points and an algorithm about how to combine them to make curves, surfaces, and more. Some applications just connect the dots with line segments, a computationally simple process with analytic geometry. We will discuss briefly in this section the more sophisticated way that computer-aided design (CAD) produces smooth curves based on analytic geometry and elementary calculus. An extension of these ideas enables computers to draw smooth surfaces as well. (See Prautzch [12] for more information on computer aided design.) Computer programs often approximate curves with B´ezier curves and, for more complicated curves, splines. Computer scientists generally choose cubic (third degree) polynomials to build the curves because they are sufficiently flexible, unlike second degree polynomials (parabolas), ´ and more easily computed than higher degree polynomials. Pierre Etienne B´ezier (1910–1999), a French engineer working for the automobile company Renault, patented the method behind the curves named after him. He, along with others, developed and popularized the approach, which revolutionized design in the automobile industry in the 1960s and thereafter in many other areas. Let’s start with a na¨ıve approach to finding a smooth curve to motivate why computers need a more sophisticated approach. The na¨ıve approach finds a polynomial that goes through a sequence of points. Example 1. Find the parabola y = ax 2 + bx + c through the points (0, 2), (1, 3), and (3, −7). Solution. When we replace x and y by the values for each point we get three first-degree equations in a, b, and c: 2 = c, 3 = a + b + c, and −7 = 9a + 3b + c. Solving the system gives a = −2, b = 3, and c = 2 or y = −2x 2 + 3x + 2. More points lead to more linear equations and higher degree polynomials. ♦ Theorem 3.4.1. Given n + 1 points P j = (x j , y j ), for j = 0 to n, with no two x j equal, there is exactly one nth degree polynomial y = a0 + a1 x + · · · + an x n such that for each j, y j = a0 + a1 x j + · · · + an x nj . Proof. See Strang [15, 80]. ! Example 2. For the seven points shown in Figure 3.32, Figure 3.33 shows a freehand curve connecting them with no bumps. Figure 3.34 shows the graph of the polynomial

Figure 3.32 Seven points.

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3.4 Curves in Computer-Aided Design

Figure 3.33 Curve through the points of Figure 3.32. 80 60 40 20 –3

–2

–1

1

2

3

–20 –40

Figure 3.34 Sixth degree polynomial through the points of Figure 3.32. y = 35.93x 2 − 12.97x 4 + 0.9978x 6 , which goes through the points, as promised by Theorem 3.4.1. The polynomial fits the freehand curve poorly. ♦ The method of Theorem 3.4.1 has drawbacks, starting with the poor fit of Example 2. A second drawback is that the addition of another point or the shift of one point necessitates completely recomputing the polynomial. Also, with a large number of points, the computations become time-consuming and prone to round-off errors. Furthermore, functions of y in terms of x cannot curve back on themselves or represent space curves. B´ezier curves and more generally splines solve all these problems. Instead of one polynomial going through all the points, we use a combination of parametric equations. For most applications it is not important to specify intermediate points on the curves, just the endpoints and the intermediate tendencies. A B´ezier curve uses a pair of parametric cubic polynomials. We

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specify it with four control points. The first and last points are the starting and ending points. The middle points tell us, in effect, the direction and “speed” of the curve at the start and the end. Figures 3.35, 3.36, and 3.37 show three curves all starting at P0 = (0, 0) with an initial slope of 1 and ending at P3 = (0, 1) with a terminating slope of −1. The curves differ because of the choice of the interior points. For Figure 3.35, P1 = (0.4, 0.4) and P2 = (0.4, 0.6). (See Exercises 3.4.5 and 3.4.6 for the values of P1 and P2 for the other two curves.) In addition to the curve each figure has line segments connecting the control points. The first and last line segments are tangent to the curve. Figures 3.36 and 3.37 illustrate how flexible B´ezier curves can be, even when we keep the same endpoints and slopes. 1.0 1.0 0.8 0.8 0.6 0.6 0.4

0.4

0.2

0.2

0.2

0.1 0.2 0.3 0.4

0.4

0.6

0.8

1.0

Figure 3.36 A B´ezier curve with a cusp.

Figure 3.35 A B´ezier curve.

1.0 0.8 0.6 0.4 0.2

0.2

0.4

0.6

0.8

1.0

Figure 3.37 A B´ezier curve crossing itself. The computer algorithm sums four third degree polynomials to define the parametric equations defining a B´ezier curve. The same basic four polynomials, the Bi below, are used for the x and y equations, with each Bi weighted by the coordinate of the appropriate point. For the

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3.4 Curves in Computer-Aided Design

curve of Figure 3.35, we use the x and y coordinates of P0 , P1 , P2 , and P3 separately to find x(t) = 0B0 (t) + 0.4B1 (t) + 0.4B2 (t) + 0B3 (t) and

y(t) = 0B0 (t) + 0.4B1 (t) + 0.6B2 (t) + 1B3 (t), with 0 ≤ t ≤ 1.

The cubic polynomials are B0 (t) = (1 − t)3 , B1 (t) = 3t(1 − t)2 , B2 (t) = 3t 2 (1 − t), and B3 (t) = t 3 , whose graphs are in Figure 3.38. We need more than their graphs to explain why they fulfill the requirements for B´ezier curves. 1.0 0.8 0.6 0.4 0.2 0.2

0.4

0.6

0.8

1.0

Figure 3.38 The basic cubic polynomials.

Exercise 3.4.1. Use calculus to verify that the curves y = B1 (t) and y = B2 (t) have their local at t = 1/3 and t = 2/3, respectively. Verify that for each of the four curves " 1 maximums 1 0 Bi (t)dt = 4 . Also, verify that at t = 1 we have B0 (t) = 0, B1 (t) = 0 and their derivatives are also 0 there, giving a local minimum. Similarly, at t = 0 we have B2 (t) = B3 (t) = B2′ (t) = B3′ (t) = 0 and a local minimum. In general, if the coordinates of the point Pi are (bi , ci ), then x(t) = b0 B0 (t)+ b1 B1 (t) + $3 ci Bi (t). Let’s analyze just x(t), since y(t) is b2 B2 (t) + b3 B3 (t) and similarly, y(t) = i=0 similar. At t = 0, B0 (0) = 1 and B1 (0) = B2 (0) = B3 (0) = 0. Thus x(0) = b0 and the curve starts at the correct value. Similarly, x(1) = b3 , the correct ending value. As the value of t increases, the curves Bi receive different weights, with their weights heaviest nearest t = 3i , as the graph and Exercise 3.4.1 indicate. Exercise 3.4.1 also shows that the factor of 3 for the middle two polynomials makes the areas of all four curves the same. Thus, over the entire domain of t, they each contribute the same total amount, although more near where it should. The derivatives of B2 (t) and B3 (t) are zero at t = 0, guaranteeing that the derivative x ′ (t) at 0 depends only on the first two terms. Similarly, x ′ (t) at t = 1 depends only on the derivatives of the last two terms. Example 3 verifies that the slopes of the B´ezier curve at the start and the end match the slopes of the first and final line segments. Example 3. Verify that the slopes of the B´ezier curve determined by the four points P0 = (b0 , c0 ), P1 = (b1 , c1 ), P2 = (b2 , c2 ), and P3 = (b3 , c3 ) match the slopes of the segments P0 P1 and P2 P3 , assuming that b0 ̸= b1 and b2 ̸= b3 .

0 . The slope of the B´ezier curve with points (x(t), y(t)) at Solution. The slope of P0 P1 is bc11 −c −b0 y ′ (w) ′ t = w is x ′ (w) . Now B0 (t) = −3(1 − t)2 , B1′ (t) = 3 − 12t + 9t 2 , B2′ (t) = 6t − 9t 2 , and B3′ (t) =

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3t 2 . Because y(t) = c0 B0 (t) + c1 B1 (t) + c2 B2 (t) + c3 B3 (t), y ′ (0) = c0 (−3) + c1 (3) + c2 (0) + c3 (0) = 3c1 − 3c0 and similarly x ′ (0) = 3b1 − 3b0 , giving a 0 2 at t = 0. In the same manner, the slope at t = 1 is bc33 −c and the slope of P2 P3 is slope of bc11 −c −b0 −b2 c3 −c2 also b3 −b2 . ♦ Exercise 3.4.2. Verify that the B´ezier curves of Figures 3.35, 3.36, and 3.37 have the same slopes at their starting and ending points. 8

6

4

2 2.0 1.5 1

2

3

4

5

–2

Figure 3.39 A letter built with B´ezier curves.

1.0 1

2

3

4

5

Figure 3.40 A spline.

Computer fonts are built by pasting together a collection of B´ezier curves, as Figure 3.39 illustrates. At the top and bottom of the “s,” we only need to have the common endpoints match since there are corners there. To make smooth curves where two B´ezier curves meet at the middle of the “s,” we need to match the slopes at their common endpoint. That is, to have a common tangent at the common point of two curves we need three control points to be collinear: the next-to-the-last control point for the earlier curve, the common endpoint of the two curves, and the second control point for the next curve. To change the font size, the computer only needs the quick computation of multiplying the coordinates of the various control points by the appropriate scalar. Splines use a different approach to obtain a curve with multiple twists and turns. Computing a spline involves multiple quick calculations, easy for computers to do, but labor-intensive for humans. The process is more complicated to explain than is appropriate for this text. See Prautzch [12] for more information. Figure 3.40 shows a spline with control points P0 = (0, 1), P1 = (1, 2), P2 = (2, −1), P3 = (3, 2), P4 = (2, 1), P5 = (1, 0), P6 = (0, 3), and P7 = (5, 2). As with a B´ezier curve, the first and last control points are the endpoints of the curve and their

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tangents are determined by the endpoint and the adjacent control point. Figure 3.41 illustrates a related three-dimensional spline curve.

0 2 4

0.5 0.0 –0.5 –1.0 –1.5 2.0 1.5 1.0 0.5

Figure 3.41 A three-dimensional spline.

3.4.1 Exercises for Section 3.4 Many of the problems depend on computer algebra systems. Consult the help menu of your software for the appropriate commands. *3.4.3. Find the values of a, b, and c in terms of p, q, and r so that the parabola f (x) = ax 2 + bx + c goes through the points (0, p), (1, q), and (2, r ). 3.4.4. (Calculus) *(a) Find the unique cubic satisfying the conditions f (0) = 1, f (1) = 0, f ′ (0) = 1, and f ′ (1) = 0. Graph it. This illustrates that there is a unique cubic f (x) = ax 3 + bx 2 + cx + d going through two points with different x-coordinates and with specified slopes at the points.

*(b) Find and graph a B´ezier curve with starting point (0, 1), starting slope 1, ending point (1, 0), and ending slope 0. (c) Modify the second and third control points of the B´ezier curve in part (b) to approximate the polynomial in part (a). (d) Explore the family of B´ezier curves with starting point (0, 1), starting slope 1, ending point (1, 0), and ending slope 0. *3.4.5 The B´ezier curve in Figure 3.36 has a cusp, a point where the two branches of the curve meet at a sharp point and have a common tangent. This problem investigates how this can occur even though the parametric equations x(t) and y(t) are smooth. (a) Use the control points (0, 0), (1, 1), (1, 0), and (0, 1) to find the exact equations of x(t) and y(t). (b) Find x ′ (t) and y ′ (t) and evaluate them at t = 0.5. (c) Use part (b) to explain how smooth parametric curves can have a graph with sharp point. 3.4.6. We investigate the B´ezier curve in Figure 3.37, which crosses itself. 1 1 1 , 1 + 12 ), (1 + 12 , (a) Verify that the B´ezier curve with control points (0, 0), (1 + 12 1 − 12 ), and (0, 1) matches Figure 3.37. Find the equations of x(t) and y(t) for them.

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(b) Verify that the three solutions of y(t) = 12 are t = 12 , t = 13 , and t = 23 . Find the values x( 13 ) and x( 23 ) to verify that the two values of t give the same point. 1 appearing in the control points of part (c) In general we can replace the value 12 (a) with any positive number k. Pick several values for k and graph the resulting B´ezier curves to verify that they cross themselves. Describe what happens to the curve as k increases. 3.4.7.

(a) Graph the B´ezier curves A and B given by the control points A: (0, 0), (2, 0), (1.5, 1.5), and (2, 2) and B: (2, 2), (3, 1), (3, 0), and (4, 0). *(b) Where do the curves A and B meet? Replace (3, 1), the second control point of B, with the control point (3, y). Find the value of y that makes A and this new B have a common tangent where they meet. Graph this B´ezier curve with A to show that it extends A smoothly. *(c) Find a B´ezier curve C that smoothly extends the curve from part (b) and ending at the point (5, 3) with a slope of 2. *(d) Find a B´ezier curve D that completes a smooth closed curve with A, B, and C. Graph all four curves on the same axes.

3.4.8. Choose a letter or shape with at least two curves and approximate it using B´ezier curves. 3.4.9. Compare Taylor polynomials and two B´ezier curves as approximations for y = sin x, as follows. *(a) Graph from −π to π y = sin(x) and the third-degree Taylor polynomial for sin x centered at 0. Use the same scale for both axes. *(b) Repeat part (a) for higher degree Taylor polynomials. *(c) Find the slopes of the endpoints of the B´ezier curve matching y = sin(x) at x = −π and at x = 0. Repeat for the B´ezier curve matching y = sin(x) at x = 0 and at x = π. Describe the possible control points for the two curves. (d) Find and graph the two B´ezier curves with the control points (−π , 0), (−π + 1.3, −1.3), (−1.3, −1.3), (0, 0), and (0, 0), (1.3, 1.3), (π − 1.3, 1.3), and (π , 0). (e) Adjust the second and third control points of the B´ezier curves in part (c) to make their heights at t = 0.5 fit y = sin(x) match more closely there. (f) Explain how you could approximate y = sin(x) beyond the range of −π to π by using additional B´ezier curves, modified from the two in part (d). (g) Compare the advantages and disadvantages of the preceding ways to approximate y = sin x. In particular, consider increasing the range of x-values, as considered in part (f). 3.4.10. The computation time for curve fitting is an important practical consideration. In general, additions and subtractions are faster than multiplications and divisions. So

3.5 Higher Dimensional Analytic Geometry

133

computer scientists often estimate computation time by the number of multiplications and divisions involved. *a) Gaussian elimination, an efficient and widely used way to solve a system of k linear equations in k unknowns, generally requires (k 3 + 3k 2 − k)/3 multiplications. Find the number of multiplications for the method of Theorem 3.4.1, which uses a system of n + 1 linear equations in n + 1 unknowns. *b) Given the eight coordinates of the control points of a B´ezier curve, how many multiplications are required to find its parametric equations? *c) How many multiplications are required to find the parametric equations for w B´ezier curves? d) Compare the efficiency of using B´ezier curves with the method of Theorem 3.4.1 as the number of points involved increases. (Actual designs have dozens or hundreds of points.) 3.4.11. We can make three-dimensional B´ezier curves by having three coordinates for our control points. Find several three-dimensional B´ezier curves with common tangents where their endpoints meet and that together make an interesting space curve.

3.5 Higher Dimensional Analytic Geometry The algebraic language of analytic geometry has enabled mathematicians to model three and higher dimensions as easily as two dimensions. The first investigations of geometry beyond three dimensions by Arthur Cayley and others starting in 1843 seemed puzzling and even nonsensical to most people, including many mathematicians. However, the variety and importance of applications in mathematics, physics, economics, and other fields since the nineteenth century have provided convincing evidence of the significance and naturalness of higher dimensions in geometry. We briefly develop analytic geometry, especially in R3 , before discussing polytopes, the higher dimensional analog of the polyhedra we considered in Section 1.5.

3.5.1 Analytic Geometry in Rn We have clear intuitions of what points, lines, and planes are in three dimensions. The following definitions give a uniform way of extending the ideas to higher dimensions. The definition treats a line as a path starting at one point and going in a specified direction, as well as going in the opposite direction. Similarly, a plane is a surface spreading out in two directions from a point. The formula for distance comes from the generalization of the Pythagorean theorem in two and three dimensions. So a key theorem in two and three dimensions becomes a definition characterizing Euclidean higher dimensional spaces. Definition. By a point in n-dimensional Euclidean geometry Rn , we mean an ordered list of → v , which is also real numbers (v 1 , v 2 , . . . , v n ). We’ll abbreviate the point (v 1 , v 2 , . . . , v n ) as − − → → called a vector. The origin is the point 0 = (0, 0, . . . , 0). For a real number k and a point − v = − → − → − → (v 1 , v 2 , . . . , v n ), the scalar multiple k v is the point (kv 1 , kv 2 , . . . , kv n ). The sum u + v of → → v = (v 1 , v 2 , . . . , v n ) is (u 1 + v 1 , u 2 + v 2 , . . . , u n + v n ). two points − u = (u 1 , u 2 , . . . , u n ) and − → → → → By a line we mean a set of points {α − u +− v : α ∈ R}, where − u and − v are fixed points

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− → → → → → → and − u ̸= 0 . A point − w is on the line if and only if for some α, − w = α− u +− v . Two lines − → − → − → − → → − → α u + v and β s + t are parallel if and only if s is a nonzero scalar multiple of − u . The − → − → distance between two points u and v is ! −−−→ → → d(− u ,− v ) =∥ u − v ∥= (u 1 − v 1 )2 + (u 2 − v 2 )2 + · · · + (u n − v n )2 . − → → → → → → → A plane is the set of points {α − u + β− v +− w : α,β ∈ R}, where − u and − v are not 0 , − v is − → − → not a scalar multiple of u and w is any fixed point. → → → Remarks. In the formula α − u +− v for a line, − u tells us its direction, analogous to the slope m in y = mx + b. The scalar multiple α plays the role of x: as α varies, we get different points → u : α ∈ R} goes through the origin. on the line. Figure 3.42 illustrates lines in R3 . The line {α − − → − → By adding the point v we have a line through v . The interpretation of a plane is similar to the one for a line, but uses scalar multiples of two points to make it two-dimensional. As Figure 3.43 illustrates, the scalars α and β act like the coordinates of a point (x, y). The set → → → {α − u + β− v : α,β ∈ R} is a plane through the origin. Adding the point − w translates it to the − → 3 parallel plane through w . In R , a plane can also be represented as the set of points (x, y, z) satisfying a linear equation ax + by + cz + d = 0, provided a, b, and c are not all zero. In general, a linear equation in n variables, such as a1 x1 + a2 x2 + · · · + an xn + d = 0, represents an (n − 1)-dimensional hyperplane in Rn , provided that the ai are not all zero. 3u + v 2u + v z

3u

u+v

2u

v u

(0,0,0)

y

-u + v

-u

x

→ → Figure 3.42 The line a − u +− v.

3u + 2v z

u

v

(0,0,0) x

3u + 2v + w

y

u+w v+w w

→ → → Figure 3.43 The plane a − u + b− v +− w.

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3.5 Higher Dimensional Analytic Geometry

− → → Exercise 3.5.1. Explain why the line through two points − s and t consists of points of the −−→ − → form α(s − t) + t . Hint: Consider α = 0 and α = 1. Explain why the plane through points −−→ −−→ − → − → − → → r ,− s , and t consists of points of the form α(r − t) + β(s − t) + t . → → → → u +− v , where − u = (2, 3) and − v = Example 1. (a) In R2 , convert the formula for the line α − (4, 5), to the more familiar equation of a line. → → → u,− v , and − w so (b) Given the equation 2x + 3y + 4z + 1 = 0 for a plane in R3 , find points − − → − → − → that the formula α u + β v + w gives the same plane. Solution. (a) We can rewrite the formula α(2, 3) + (4, 5) as (2α + 4, 3α + 5). That is, a point on the line has the property that when x = 2α + 4, we have y = 3α + 5. Solve in the first equation to find α = 12 x − 2. Substitute for α in the second equation and simplify to find y = 32 x − 1. When α = 0, we have the point (4, 5) and when α = −2, we have the point (0, 1), both of which satisfy y = 32 x − 1.

→ → → (b) The formula α − u + β− v +− w gives a general point in the plane. When we put that point’s x-, y-, and z-coordinates in for the x, y, and z of 2x + 3y + 4z + 1, the sum needs to be 0. That is, regardless of the values of α and β, we always have 2(αu 1 + βv 1 + w1 ) + 3(αu 2 + βv 2 + w2 ) + 4(αu 3 + βv 3 + w3 ) + 1 = 0.

Let’s start with the easiest situation: α = β = 0, which gives us 2w1 + 3w2 + 4w3 + 1 = 0. → One point satisfying the equation is − w = (0, 0, − 14 ). We have left the first two coordinates → → v = (0, 1, v 3 ) to work with these coordinates separately free. We can try − u = (1, 0, u 3 ) and − and in the process figure out what u 3 and v 3 need to be. The equation now becomes 2α + 3β + 4(αu 3 + βv 3 + w3 ) + 1 = 0. We can pick u 3 so that 2α + 4αu 3 = 0. That is, u 3 = − 12 . Similarly, for 3β + 4βv 3 = 0 we need v 3 = − 34 . So the formula α(1, 0, − 12 ) + → u, β(0, 1, − 34 ) + (0, 0, − 14 ) gives us the plane 2x + 3y + 4z + 1 = 0. Other choices of − − → − → v , and w can also work. Exercise 3.5.6 generalizes this example. ♦ Example 2. Use analytic geometry and linear algebra to verify Postulate 8 of the SMSG axioms: In three-dimensional geometry, if two distinct planes intersect, they intersect in a line. − → → → → → → Solution. Let’s begin with two distinct planes α − u + β− v +− w and γ − r + δ− s + t . Then the coordinates of the six points are known, but the values of the four numbers α, β, γ , and δ can → vary. If the point − a = (a1 , a2 , a3 ) is on both planes, the following system of three equations in the four unknowns α, β, γ , and δ has a solution. αu 1 + βv 1 + w1 = γ r1 + δs1 + t1

(= a1 )

αu 3 + βv 3 + w3 = γ r3 + δs3 + t3

(= a3 )

αu 2 + βv 2 + w2 = γ r2 + δs2 + t2

(= a2 )

However, as there are more unknowns than equations, there are infinitely many solutions. The planes are distinct, so the solutions must correspond to the points on a line, not a whole plane. ♦ Conics. We can generalize the conics of Section 2.2 to conic surfaces in three dimensions. (See Figure 3.44.) Some of them, for example the ellipsoid x 2 /4 + y 2 + z 2 = 1, the hyperboloid of two sheets x 2 − y 2 − z 2 = 1, and the paraboloid z = x 2 + y 2 , can be defined using foci (and a

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plane as a directrix for the paraboloid) in the same way as conics are defined. Analytic geometry allows us more generally to define a conic surface as the set of points satisfying a second-degree equation in x, y, and z. (The terms x y, x z, and yz are considered second degree, as are x 2 , y 2 , and z 2 .)

1.0

2

1

0

–1 –2

1.0 0.5

–1.0 0.0–0.5

2.0

2

1.5

0.5

0

0.0

1.0 0.5

–2

–0.5 –1.0

–1.0–0.5 0.0 0.5

–2 1.0

–2 0

0 2

2

0.0 –1.0 –0.5

0.0

0.5

1.0

Figure 3.44 An ellipsoid, a hyperboloid and a paraboloid. Exercise 3.5.2. For the paraboloid z = x 2 + y 2 verify that the distance from a point (x, y, z) on the paraboloid to (0, 0, 0.25) is the same as the distance from (x, y, z) to the point (x, y, −0.25), which is the point in the plane z = −0.25 nearest to (x, y, z). That is, the point (0, 0, 0.25) is the focus and the plane z = −0.25 is the directrix of the paraboloid. Figure 3.45 shows the graph of x 2 + y 2 − z 2 = 1, a hyperboloid of one sheet. Hyperboloids of one sheet have the unusual property that certain straight lines lie entirely on the surface, even though the surface is curved. They also do not have foci. –2

0

2 4

2

(0, y, z) 0

(x, y, z)

–2

–2

0

2

–4

Figure 3.45 A hyperboloid of one sheet.

(x, y, 0)

Figure 3.46 Projecting a point and a solid.

Gaspard Monge (1746–1818) developed descriptive geometry, a method of representing three-dimensional geometric constructions by means of two (or more) two-dimensional projections. This method radically improved engineering design. Figures 3.46 and 3.47 illustrate

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137

Figure 3.47 Descriptive geometry representation of a point and a solid. how descriptive geometry enabled engineers to design on paper exact plans for their threedimensional constructions. A point (x, y, z) in space is represented by the two points (x, y, 0) and (0, y, z) on the separate planes. Similarly the solid has two projections in Figure 3.46. Monge rotated the two projections to be in the same plane, one above the other, as in Figure 3.47. Similarly, a line in space is represented by its two projections on the horizontal and vertical planes. Computer-aided design now supersedes this hand-drawn approach but is based on the same analytic geometry.

3.5.2 Regular Polytopes Mathematicians have studied polygons in two dimensions and polyhedra in three dimensions for millennia. Augustus M¨obius broached the idea of four-dimensional geometry before 1830. By 1850 some geometers started studying their higher dimensional analogs, called polytopes. A polyhedron is constructed by attaching polygons at their edges. By analogy, for a fourdimensional polytope we attach polyhedra at their faces. As few of us can visualize this, coordinates help us analyze polytopes. Here we consider only some of the regular polytopes, first described for four dimensions in 1852 by Ludwig Schl¨afli (1814–1895). A four-dimensional convex polytope is regular provided that all the polyhedra in it are the same regular polyhedron and the same number of polyhedra meet at every vertex. Higher dimensional polytopes can be defined similarly. (See Coxeter [3, Chapter 22].) There are five regular polyhedra in three dimensions. Schl¨afli and some other nineteenth century mathematicians discovered independently that there are six regular polytopes in four dimensions, but in dimensions greater than four, there are only three regular polytopes. These three families of regular polytopes are the analogs of the cube, the tetrahedron, and the octahedron, and fortunately we can represent the vertices of the polytopes in the families with simple Cartesian coordinates. The Cube and Hypercubes. Figure 3.48 illustrates one convenient way to place a square and a cube in Cartesian coordinates. The coordinates of the points are 1 or −1. Furthermore, for the cube, every possible combination appears among the 8 = 23 points. This suggests a way to generalize to four dimensions. Consider the 16 = 24 points in R4 whose coordinates are

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(-1,-1,1) (1,1)

(-1,1)

(-1,1,1)

(1,-1,1) (1,1,1) (-1,-1,-1)

(-1,1,-1) (-1,-1)

(1,-1)

(1,-1,-1) (1,1,-1)

Figure 3.48 A square and a cube. (-1,1,1,1)

(-1,-1,1,1)

(1,-1,1,1)

(1,1,1,1)

(-1,-1,1,-1) (-1,1,1,-1)

(1,1,1,-1)

(1,-1,1,-1) (-1,-1,-1,-1) (-1,-1,-1,1)

(1,-1,-1,-1)

(-1,1,-1,-1) (1,1,-1,-1) (-1,1,-1,1)

(1,1,-1,1)

(1,-1,-1,1)

Figure 3.49 A hypercube. (0,0,1)

(0,1)

(-1,0,0)

(0,-1,0) (1,0)

(-1,0)

(1,0,0)

(0,-1)

(0,1,0) (0,0,-1)

Figure 3.50 A square and an octahedron. all ±1, as in the two-dimensional representation of Figure 3.49. The points are the vertices of a hypercube. The distance between two points joined by an edge in Figure 3.50 is 2. Each point has four neighbors √ √ at this distance. The other possible distances between points on the hypercube are 2 2, 2 3, and 4. Each square face is the intersection of two of the cubes in the hypercube.

3.5 Higher Dimensional Analytic Geometry

139

Exercise 3.5.3.∗ Describe the coordinates √ of√the points on the hypercube at a distance of 2 from (1, 1, 1, 1). Repeat for the distances 2 2, 2 3, and 4. The construction extends readily to higher dimensions. The vertices (v 1 , v 2 , . . . , v n ) of a hypercube in Rn satisfy the condition that each v i is either +1 or −1, and the vertices use all possible combinations of them. The Octahedron and Cross Polytopes. The placement of the square and the regular octahedron in Figure 3.50 suggests a way to generalize them. In R4 , consider the eight points with 0 for three of their coordinates and 1 or −1 for the other coordinate. We designate them as the vertices of the regular four-dimensional analog, called a cross polytope. Connect two vertices with an edge if their nonzero values appear in different coordinates. Thus (0, 1, 0, 0) and (0, 0, −1, 0) are connected with an edge, but (0, 1, 0, 0) and (0, −1, 0, 0) are not. Exercise 3.5.4.∗ Find the two possible distances between vertices for the four-dimensional analog of the octahedron. Verify that each vertex is on six edges. In higher dimensions the vertices (v 1 , v 2 , . . . , v n ) of the cross polytope satisfy the condition that all but one of the v i are 0, that remaining coordinate is either +1 or −1, and the vertices use all possible combinations of these numbers. The Tetrahedron and Regular Simplexes. A regular tetrahedron has four vertices all the same distance from √ one another. The regular octahedron in Figure 3.50 gives us three vertices a a point equidisdistance of 2 from one another: (1, 0, 0), (0, 1, 0), and (0, 0, 1).√By symmetry ! 2 + x2 + x2 = (1 − x) 2 = tant from them has coordinates (x, x, x) for some x. That is, √ 2 3x − 2x + 1. One solution is x = 1, giving the point (1, 1, 1). We call the analogs in higher dimensions regular simplices. In four dimensions we need five points all equidistant from one another. (Figure 3.51 gives a two-dimensional representation.) Start with (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), and (0, 0, 0, 1) from the cross polytope. Similar to ! the tetrahedron case, the √re√ maining point has coordinates (x, x, x, x) and satisfies 2 = (1 − x)2 + 3x 2 , or x = 1±4 5 . Higher dimensional regular simplexes can be found in the same way: The n-dimensional regular simplex has n + 1 points equidistant from each other. We can use n vertices of the form (v 1 , v 2 ,√. . . , v√ n ), where all but one of the√v i is 0 and that one is +1, together with (x, x, . . . , x), where 2 = nx 2 − 2x + 1 or x = 1± n1+n .

Figure 3.51 A regular 4-dimensional simplex. The proof that there are at most six regular four-dimensional polytopes is similar to the proof in Exercise 1.5.18 that there are just five regular polyhedra. For polyhedra, we computed the angle defect. For example, a square has a 90◦ angle, so the three squares around a vertex of a cube

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give a total of 270◦ , less than 360◦ . Thus we can fold them together in three dimensions. We can’t do that with three regular hexagons since each has an angle of 120◦ , for a total of 360◦ , giving no angle defect. The analogous idea in four dimensions puts polyhedra around an edge. Again, we can fit three cubes around an edge and still have a gap since the dihedral angle between the faces of a cube is 90◦ . (The dihedral angle, as shown in Figure 3.52, is the angle between two rays in the planes that are perpendicular to the intersection of the planes.) Thus, we can theoretically fold the cubes in four dimensions to create a corner of a hypercube. Exercise 3.5.17 shows the dihedral angle of a tetrahedron is arccos(1/3) or approximately 70.53◦ . Thus we can fit three, four, or five tetrahedra around an edge and still have a gap since 5(70.53◦ ) = 352.65◦ < 360◦ . The options give three regular polytopes. Three tetrahedra around an edge form a regular simplex and four tetrahedra around an edge form a cross polytope. Exercise 3.5.18 shows the dihedral angle for the octahedron is arccos(−1/3) or approximately 109.47◦ . The dihedral angle for the dodecahedron is approximately 116.57◦ . So we can fit three of them around an edge and still have a gap. Thus the two options give two more regular polytopes. However, the dihedral angle of an icosahedron is over 138◦ , too large to fit three around an edge, let alone have a gap for folding in the fourth dimension. By Exercise 1.5.18 there are just five regular polyhedra and we have determined the options for each regular polyhedron. Thus our understanding of dihedral angles in three dimensions shows that there are at most six regular polytopes. Exercises 3.5.19 to 3.5.23 investigate them further. A

D C

B

E ← → Figure 3.52 The dihedral angle ∠ABC, where the planes intersect at D E, AB ⊥ D E, and BC ⊥ D E.

We have discussed polytopes primarily to develop intuition for higher dimensional geometry. However, polytopes serve important purposes outside of geometry. For example, businesses use the simplex method extensively to find optimal solutions to many applied problems. In the simplex method in linear programming, the potential solutions are points in a polytope in many dimensions. One can prove that the optimal solution is always a vertex, so the computer can limit its search to vertices. Hypercubes provide a useful model in coding theory, where some of the vertices correspond to possible code words. (See Section 8.3.) Higher dimensional simplices form fundamental building blocks in topology. Coxeter’s books [3] and [4] go into more depth on polytopes.

3.5.3 Exercises for Section 3.5 → → → *3.5.5. (a) Find a representation in the form α − u + β− v +− w for the plane through (0, 0, 0), (1, 2, 4), and (2, 0, 0). (b) Find an equation for the plane through (0, 0, 0), (1, 2, 4), and (2, 0, 0).

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→ → 3.5.6. (a) Find a representation in the form α − u +− v for the line in R2 with equation y = mx + b. Repeat for the equation of a vertical line, x = b. (b) Find the equation of the form ax + by + c = 0 for the line in R2 of the form → → → → v = (v 1 , v 2 ). Hint: First consider α− u +− v , where − u = (u 1 , u 2 ) ̸= (0, 0) and − the case u 1 ̸= 0. What is the slope of the line? What happens when u 1 = 0? → → → (c) Find a representation in the form α − u + β− v +− w for the plane in R3 with equation ax + by + cz + d = 0. First consider the case c ̸= 0. Then explain how to modify it when c = 0 but one of the other coefficients is nonzero. (d) Find an equation of the form ax + by + cz + d = 0 for the plane in R3 of the form α(2, 0, 1) + β(0, −3, 1) + (3, 8, 6). Hint: Use several simple values of α and β to get several equations in a, b, c, and d. If ax + by + cz + d = 0 is an answer, so is 2ax + 2by + 2cz + 2d = 0 or any other nonzero multiple. So you can choose any one of the nonzero coefficients. → → *3.5.7. (a) Find a representation of the form λ− u +− v for the line of intersection for the

planes x + 2y + 3z = 0 and 4x + 5y + 6z = 0. (b) Determine whether the lines α(1, 2, 3) + (4, 5, 6) and β(7, 8, 9) + (10, 11, 12) intersect, are parallel, or are skew. (Two lines are skew if and only if they are not parallel and do not have any point in common.) (c) Repeat part (b) for the lines α(1, 2, 3) + (0, 1, 4) and β(2, 3, 4) + (1, 0, 5). → → → → → → → → (d) Determine conditions on − u,− v ,− w , and − z so that α − u +− v and β − w +− z are parallel.

3.5.8. If two planes in R4 intersect in at least one point, must they intersect in more than that one point? If so, show why. If not, find an example of two planes in R4 intersecting in exactly one point. 3.5.9. (Descriptive geometry) In descriptive geometry we represent a point by its projections onto the x y-plane and the yz-plane; that is, a point P = (a, b, c) will project to P1 = (a, b) and P2 = (b, c), and a line k has two projections, k1 and k2 (Figure 3.53). The yz-projection directly above the x y-projection indicates that the y-coordinates match. *(a) Redraw Figure 3.53 and include the line parallel to k through the point P. Explain why your line is parallel to k. *(b) Find the 3-dimensional distance between points A and B if the projections of A are (1, 2) and (2, 3) and the projections of B are (3, −1) and (−1, 9). *(c) Suppose the projections of a line are y = 2x − 1 and y = z + 2 and the projections of another line are y = x + 2 and y = −z + 8. Find the three-dimensional coordinates of their point of intersection. Explain why it is their point of intersection. *(d) Consider Figure 3.54. Decide whether lines k and m are parallel, intersect, or neither. Explain. *(e) Describe what the projections of two parallel lines look like. Repeat for two lines that intersect. Repeat for skew lines, lines that neither intersect nor are parallel. (f) Find a convex solid other than a cube whose projections onto the x y-plane and the yz-plane are squares.

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P1 k1

P2

k2

Figure 3.53 A point and a line in descriptive geometry.

Figure 3.54 Two lines in descriptive geometry.

(g) Suppose that in addition to the projections in part (e), the projection of the convex solid on the x z-plane is also a square. Must the solid be a cube? If so, explain why; if not, give an example. 3.5.10. (a) Define an ellipsoid, a hyperboloid of two sheets, and a paraboloid, based on the definitions from Definition 2.2.1. (b) Derive the text’s equations for the conic surfaces of part (a). 2 + z 2 = 1 and verify that 3.5.11. Select a variety of points (x, y, z) on the ellipsoid x 2 /4 + y√ √ the sum of their distances from the points ( 3, 0, 0) and (− 3, 0, 0) is constant. That is, verify that they are the foci of the ellipsoid.

3.5.12. (a) Suppose that you know the distances (x, y) from (0, 0), (1, 0), and (0, 1). Explain why no other point (x, y) in the plane R2 can be at the same respective distances from them. (b) Explain why part (a) holds if you replace (0, 0), (1, 0), and (0, 1) with three points not on a line. What happens if the three points are on the same line? (c) Generalize part (b) to three dimensions. How many points are needed? (d) Redo part (c) in n dimensions. *3.5.13. Plot with a computer graphics package the following surfaces that are given parametrically. Describe the shapes in words. √ √ (a) x(u, v) = 1 − u 2 cos(v), y(u, v) = 1 − u 2 sin(v), and z(u, v) = u, for −1 ≤ u ≤ 1. √ √ (b) x(u, v) = 2 1 − u 2 cos(v), y(u, v) = 2 1 − u 2 sin(v), and z(u, v) = u, for −1 ≤ u ≤ 1. 2 (c) x(u, v) = u√cos(v), y(u, v) = u sin(v), √ and z(u, v) = u , for 0 ≤ u ≤ 2. 2 2 (d) x(u, v) = 1 + u cos v, y(u, v) = 1 + u sin v, and z(u, v) = u. 3.5.14. We verify that the hyperboloid x 2 + y 2 − z 2 = 1, graphed in Figure 3.45, has lines lying on its surface. √ Start with a line through (1, 0, 0) and a point on the hyperboloid at a height of z = 3. √ (a) Explain why points on the hyperboloid with z = 3 can be written as √ (2 cos θ , 2 sin θ , 3). Explain √ why the line through the point and (1, 0, 0) is given by α(2 cos θ − 1, 2 sin θ , 3) + (1, 0, 0).

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(b) Explain why, if the line in part √ (a) is on the surface of√the hyperboloid, it is also on (2 cos −θ , 2 sin −θ , − 3) = (2 cos θ , −2 sin θ , − 3). Show that θ satisfies cos θ = 12 . Find the two values of sin θ and determine the two lines. (c) Show that every point on the lines in part (b) lies on the hyperboloid. (d) Explain why an appropriate rotation of the two lines you found will give two lines through every point on the hyperboloid. (e) Make a model of the hyperboloid by stretching strings between two circles. When one circle is rotated over the other, the strings lean to form a hyperboloid (Figure 3.55).

Figure 3.55 A string model of a hyperboloid of one sheet. *3.5.15. (a) Explain why the equation of the unit sphere is x 2 + y 2 + z 2 = 1. (b) Explain why we can use ax + by + cz = 0 as the equation of a great circle on the unit sphere, where not all a, b, and c are zero. (See Section 1.5.) (c) Find the two points of intersection of the great circles 2x + 2y + z = 0 and 2x − 2y + z = 0. , 1 ). (d) Find the great circle through the points ( 32 , 23 , 13 ) and ( 23 , −2 3 3 (e) Explain why the vector ( p, q, r ) is perpendicular to the plane px + qy+ r z = 0. 3.5.16. (Linear algebra) The dot product enables us to determine angles. For nonzero vectors − → → → → u and − v , if − u ·− v = 0, then they are perpendicular (orthogonal). (a) Explain why the plane 2x + 3y − 4z = 0 is perpendicular to the line through (2, 3, −4) and the origin in three dimensions. (b) Is the plane 2x + 3y − 4z + 5 = 0 perpendicular to the line in part (a)? Explain. (c) Describe all lines perpendicular to the plane in part (a). (d) Explain why every plane through the line in part (a) is perpendicular to the plane in part (a). What conditions do perpendicular planes satisfy? Hint: the plane and the line go through the origin. *3.5.17. Use Figure 3.56 and the law of cosines to determine the dihedral angle of a tetrahedron. 3.5.18. Use Figure 3.57 and the law of cosines to determine the dihedral angle of a octahedron.

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P

A

D

Q

B E C

Figure 3.56 ∠AE D is a dihedral angle of the tetrahedron.

R

Figure 3.57 ∠P Q R is a dihedral angle of the octahedron.

3.5.19. *(a) An edge, a square, a cube, and a hypercube are analogous shapes in one to four dimensions. For each, count the relevant items of the following: edges per vertex, faces per vertex, cells per vertex, faces per edge, cells per edge, and cells per face, as well as the total number of vertices, edges, faces, and cells. (The polyhedra making up a polytope are called its cells.) Make a table of this information. *(b) Look for patterns, and explain any you find. (c) Find the values corresponding to part (a) for the five-dimensional and sixdimensional hypercubes. (d) Write formulas for the different categories counted in parts (a) and (c), based on only the dimension. 3.5.20. Repeat Exercise 3.5.19 for an edge, a square, an octahedron, and the cross polytope. *3.5.21. Repeat Exercise 3.5.19 for the sequence leading to the four-dimensional simplex. 3.5.22. The octahedron is the dual of the cube: when you connect the centers of the cube’s faces, you get an octahedron. Thus the octahedron has the same number of vertices as the cube has faces, and vice versa. (a) How are the edges of the cube and its dual related? (b) In what way is the four-dimensional cross polytope the dual of the four-dimensional hypercube? Use this duality to find the number of vertices, edges, faces, and octahedra of a four-dimensional cross polytope. (c) Repeat part (b) for the five-dimensional cross polytope. 3.5.23. Schl¨afli devised what are now called Schl¨afli symbols to describe regular polygons, polyhedra, and polytopes. He used {n} for a regular n-gon. He wrote {n, k} for a polyhedron with k n-gons at each vertex. He denoted a polytope with j polyhedra of type {n, k} around an edge as {n, k, j}. *(a) Match the name of each of the regular polyhedra with its Schl¨afli symbol. (b) Match the description in the text of each regular 4-dimensional polytope with its Schl¨afli symbol. (c) Describe how we can build the three regular 5-dimensional polytopes and give their Schl¨afli symbols. Hint: In 3-dimensions, we fit faces around a vertex and in 4-dimensions we fit cells around an edge. (d) Redo part (c) for higher dimensions. (For more information, see Coxeter [4].)

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3.5.4 Gaspard Monge It is not often that a mathematical subject becomes a state secret. At age 19 Gaspard Monge (1746–1818) started developing the subject we now call descriptive geometry. He became an instructor at the Royal School of Engineers in France soon after, and his idea was quickly seen to be vital for military planning and declared a state secret. Descriptive geometry provides accurate two-dimensional drawings of three-dimensional figures. In effect, it combines analytic geometry and thee-dimensional geometry into two-dimensional drawings. His geometrical brilliance and his gifts as a teacher were recognized throughout his career, and he advanced to become a Royal professor of physics as well as mathematics by 1775. The French Revolution of 1789 and subsequent rise of Napoleon vastly changed the political landscape, but not the need for Monge’s teaching and his mathematical talent. He supported the revolution and made important contributions to it and to Napoleon’s regime that followed. Monge served in various posts in the new government, including a ministerial post. His greatest ´ contribution was the founding in 1795 and the sustaining of the Ecole Polytechnique. Monge contributed greatly to our understanding of education with the organization of this renowned school, which became a military academy under Napoleon, and with his own teaching. Monge and Napoleon became friends. Napoleon had Monge direct some of the planning for the French invasion of Egypt and later appointed Monge a senator. Monge was so tied to Napoleon that upon Napoleon’s exile in 1815, Monge lost his positions. Three years later he died and the government of the new King tried to downplay his funeral. However, many of his students, former officers, and some notable mathematicians came, mourning the loss of a great teacher and influential mathematician. In addition to his development of descriptive geometry, Monge championed geometric methods in mathematics, following a long period of dominance by algebra and calculus. He recognized the need for both analytic and synthetic approaches and revived interest in projective geometry. He also made major contributions to what we now call differential geometry, making it into a separate branch of mathematics.

3.5.5 Projects for Chapter 3 1. Pick’s theorem gives the area of many lattice polygons, that is, polygons whose vertices have integer coordinates, called lattice points. (a) For each lattice polygon shown in Figure 3.58, find its area and count the number B of lattice points on the boundary of the polygon and the number I of lattice points in its interior. Find an equation relating the area with the numbers B and I . Does this equation hold for the polygons shown in Figure 3.59?

Figure 3.58 Lattice polygons.

Figure 3.59 Nonconvex lattice polygons.

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(b) Prove your equation for lattice rectangles parallel to the axes. (c) Do the lattice polygons shown in Figure 3.60 satisfy your equation? Explain how they differ from those in Figures 3.58 and 3.59. State Pick’s theorem, incorporating any needed hypotheses.

Figure 3.60 More lattice polygons.

Figure 3.61 Building a lattice polygon.

(d) Add an interior edge to divide a lattice polygon into two smaller lattice polygons (Figure 3.61). How do B and I for the smaller polygons compare to those for the original polygon? How far can you carry this process of dividing? What can you say about the smallest lattice polygons? (See Coxeter [3, 209] for a proof of Pick’s theorem.) (e) Develop, state, and prove a restriction of Pick’s theorem in three dimensions to rectangular boxes parallel to the axes (Figure 3.62). Does the restriction work for pyramids? (Figure 3.63.) Reeve [13] proves a general three-dimensional version.

Figure 3.62 Lattice rectangular boxes.

Figure 3.63 A lattice pyramid.

2. “Napoleon’s theorem.” On each side of a triangle, construct an equilateral triangle lying outside (or inside) it. The centers of the triangles form an equilateral triangle. (There is no historical evidence to connect Napoleon with the theorem, although it has carried his name for over a hundred years.)

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(a) Show√that in an equilateral triangle with side x, the distance from the center to a vertex is x/ 3. Also, rewrite cos( A + 60◦ ) by using the formula cos(A + B) = cos A cos B − sin A sin B. Without loss of generality let the vertices of the triangle be S = (0, 0), T = ( p, 0), and U = (q, r ). Call the centers of the equilateral triangles V , W , and X (Figure 3.64). Napoleon’s theorem states that V W , V X , and W X are equal. (b) Explain why showing that V W = W X is sufficient. (c) Find ST , SU , and T U and then SV , SW , T W , and T X . (d) Explain why ∠V SW is 60◦ wider than ∠U ST and similarly why ∠X T W is 60◦ wider than ∠U T S. 2 2 2 2 2 (e) Use part √ (a) and the law of cosines to verify that V W and W X equal ( p + q + r − pq − 3 pr )/3. (f) Write a proof of Napoleon’s theorem.

V

U

X

S

T W

Figure 3.64 Construction for Napoleon’s theorem. 3. (Calculus) The folium of Descartes (x 3 − 2x y + y 3 = 0, shown in Figure 3.1) is a challenge to graph by hand, even with the aid of calculus. (a) Use implicit differentiation to find dy/d x. (b) Find the values of x and y for which dy/d x = 0. (c) When is dy/d x undefined? There the curve has a vertical tangent. How are x and y related to those in part (b)? (d) Find the points on the curve where y = x. Why is the curve symmetric with respect to the line y = x? (e) Graph the curve, using the previous information and Newton’s method to plot points. (f) Graph the folium of Descartes parametrically, using the following tricky substitution. In x 3 − 2x y + y 3 = 0, replace y by t x and solve for x to find x(t). Then find y(t) = t x(t). Verify that the graph matches the results in parts (a) through (e). 4. Use a computer algebra system to investigate graphs of equations that can’t readily be written as functions. For example, find the graphs of some of the conchoids in the family (x − 1)(x 2 + y 2 ) = ax 2 investigated by Ren´e Walter in 1662 without the aid of calculus. Other curves include the astroid with equation (x 2 + y 2 − 1)3 + 27x 2 y 2 = 0 and the cruciform curve x 2 y 2 = x 2 + y 2 . Some software such as Mathematica can use contour plots to graphs the equations. 5. Solve the following locus problems. It may help to place the items conveniently. (a) The locus of points forming the third vertex of a right triangle whose hypotenuse is fixed.

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(b) The locus of points C forming a triangle with fixed points A and B so that m∠AC B = 60◦ . What happens when we alter the measure of ∠AC B? Hint: Use Exercise 1.2.17, rather than coordinates. (c) The locus of centers of circles tangent to a given line and going through a given point. (d) The locus (called a lemniscate) of points P so that the product of the distances P F1 and P F2 is a 2 , where 2a is the distance between the foci F1 and F2 . (e) The locus of points P where the distance from P to the origin (0, 0) is 2 more than the distance from P to the line y = 1. Hint: Consider the case when P is above the line separately from the case when it is below the line. (f) Investigate other locus problems. See Maurer [11]. 6. Define and investigate taxicab conics and other topics in taxicab geometry. (See Krause [10].) Different positions of the foci and different slopes of the directrix lead to differentlooking ellipses, hyperbolas, and parabolas. 7. Find data sets, like the one in Example 6 of Section 3.3, where a trilinear plot can provide understanding and make the plots. 8. Investigate computer-aided design. (See Prautzsch et al [12].) 9. Investigate descriptive geometry. (See Douglass and Hoag [5].) 10. The regular polytope {3, 4, 3} in Schl¨afli symbols has its vertices at the midpoints of the edges of the cross polytope. (a) Find the coordinates of the midpoints of the cross polytope discussed in the text. (b) Determine the number of vertices of {3, 4, 3}. (c) Determine the possible distances between vertices for the {3, 4, 3} polytope. (d) The edges of the {3, 4, 3} polytope are between vertices that are the smallest distance apart. Determine how many edges come from each vertex and how many edges altogether. (e) What polygons are the faces of the {3, 4, 3} polytope? Determine the total number of faces, the number of faces per vertex, and the number of faces per edge for it. (f) What polyhedra are the cells of the {3, 4, 3} polytope? Determine the total number of cells, the number of cells per vertex, and the number of cells per edge for it. 11. Determine the number of vertices, edges, faces, and cells for the regular polytopes {3, 3, 5} and {5, 3, 3}. Hint: they are dual polytopes. 12. Investigate Schl¨afli’s generalization of Euler’s formula to polytopes (proved by Poincar´e). See Coxeter [4]. 13. (Calculus) (a) The cross section of a sphere is a circle. For the sphere with center at (0, 0, 0) and radius r , determine the area of a vertical cross section in terms of x, the value of its x-coordinate. (b) Write and evaluate an integral for the volume of the whole sphere based on the cross sections of part (a). (c) Explain why the equation for a four-dimensional hypersphere centered at the origin and with radius r is x 2 + y 2 + z 2 + w 2 = r 2 . (d) Use part (c) to explain why the cross section of a four-dimensional hypersphere is a three-dimensional sphere. (e) Repeat part (a) for cross sections of the four-dimensional hypersphere. (f) Write and evaluate an integral for the hypervolume of the hypersphere in part (c). Hint: It may help to use a computer algebra system to evaluate the integral.

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14.

15.

16. 17.

18. 19.

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(g) Repeat parts (c), (d), (e), and (f) for a five-dimensional sphere. (h) Investigate hypervolume more generally in higher dimensions. Investigate further topics in polytopes and four-dimensional geometry, including generalizing Euler’s formula. (See Coxeter [3, Chapter 22], Coxeter [4, Chapter 9], and Rucker [14].) Julius Pl¨ucker (1801–1868) extended the idea of a coordinate system enormously when he realized that entities other than points could have coordinates. For example, circles in the plane have “circular coordinates” (a, b, r ), where (a, b) is the center and r > 0 is the radius of the circle. Thus the set of circles in the plane is, in a sense, three dimensional. (a) Find the circular coordinates for the circle x 2 + y 2 + 4x − 6y − 3 = 0. (b) Find the equation of a circle having circular coordinates (5, −4, 7). (c) Devise a set of “linear coordinates.” How many “dimensions” does the set of lines in the plane have? Find, in your linear coordinates, the coordinates of the line whose equation is y = 3x + 5. Can two distinct pairs of your linear coordinates represent the same line? Are there any lines that do not have coordinates in your system? (Consider x = 4.) Are there any subsets of coordinates in your system that do not represent lines? (See Eves [7] for more information.) Investigate algebraic geometry and elliptic curves. (See Hulek [9].) Write an essay exploring the meaning of geometry in four or more dimensions. Suggestions: Compare multidimensional with plane geometry in a world that is strictly three dimensional. (See Rucker [14] also.) Write an essay comparing analytic and synthetic geometry as ways to do and understand geometry. Write an essay comparing the certainty of algebraic derivations of geometric properties with the certainty of proving those properties in an axiomatic system. (See Grabiner [8].)

3.5.6 Suggested Readings 1. Boyer, C., History of Analytic Geometry, New York: Scripta Mathematica, 1956. 2. Broman, A., and L. Broman, Museum exhibits for the conics, Mathematics Magazine, 1994, 67(3): 206–209. 3. Coxeter, H., Introduction to Geometry, 2nd ed. New York: John Wiley & Sons, 1969. 4. Coxeter, H., Regular Polytopes, New York: Macmillan, 1963. 5. Douglass, C., and A. Hoag, Descriptive Geometry, New York: Holt, Rinehart and Winston, 1962. 6. Edwards, C., and D. Penney, Calculus with Analytic Geometry, Englewood Cliffs, NJ: Prentice Hall, 1994. 7. Eves, H., A Survey of Geometry, vols. I and II. Boston: Allyn & Bacon, 1965. 8. Grabiner, J., The centrality of mathematics in the history of Western thought, Mathematics Magazine, 1988, 61(4): 220–230. 9. Hulek, K., Elementary Algebraic Geometry, Providence, RI: American Mathematical Society, 2003. 10. Krause, E., Taxicab Geometry, Mineola, NY: Dover, 1986. 11. Maurer, S., Some unusual locus problems, The College Mathematics Journal, 1983, 14(2): 146–153.

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12. Prautzch, H., W. Boehm, and M. Paluszny, B´ezier and B-Spline Techniques, New York: Springer, 2002. 13. Reeve, J., On the volume of lattice polyhedra, Proceedings of the London Mathematical Society, 1957, 3(7): 378–395. 14. Rucker, R., Geometry, Relativity and the Fourth Dimension, Mineola, NY: Dover, 1977. 15. Strang, G., Linear Algebra and its Applications, 2nd ed. New York: Academic Press, 1980.

4

Non-Euclidean Geometries

Figure 4.0 A design in the Poincar´e model of hyperbolic geometry. The repeated insects in this design clearly follow some rule, although the geometry behind the pattern may seem mysterious. M. C. Escher made several of these designs by hand. Douglas Dunham, a geometer with an interest in computer graphics, has combined his knowledge of non-Euclidean geometries and computers to produce a large variety of these designs.

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The fundamentals of non-Euclidean geometries are basic to an understanding of the mathematics of these designs. The most suggestive and notable achievement of the last century is the discovery of non-Euclidean geometry. —David Hilbert To this interpretation of geometry I attach great importance for should I have not been acquainted with it, I never would have been able to develop the theory of relativity. —Albert Einstein

4.1 Overview and History The classical understanding of axioms (postulates) as self-evident truths was shattered in mathematics by the introduction and development of non-Euclidean geometries during the nineteenth century. Mathematics students deserve to know the important role that non-Euclidean geometries played in the history of ideas. They illustrate the need to transcend the intuitive models of elementary mathematics, allowing us to think successfully about the more abstract concepts of modern mathematics. In this chapter we focus primarily on the non-Euclidean geometry now called hyperbolic geometry. Sections 4.2 to 4.4 develop it as an axiomatic system. In the last section we briefly consider spherical and single elliptic geometries. Before 1790 no one questioned the truth of Euclidean geometry, although many tried to remove a perceived blemish in Euclid’s masterpiece. His fifth postulate about parallels was hardly self-evident. (See Section 1.3.) Mathematicians tried repeatedly to prove the postulate from the others. Mostly, though, they either explicitly or implicitly used equivalent assumptions, such as Playfair’s, now used in geometry books. (His axiom reads “Through a given point not on a given line m there passes at most one line which does not intersect m.”) Before 1800, the Italian mathematician Giovanni Girolamo Saccheri (1667–1733) came closest to realizing the improvability of the fifth postulate from the others. He started from two ways to negate the fifth postulate, looking for contradictions. From one he found a contradiction. From the one leading to hyperbolic geometry he deduced increasingly bizarre results, such as the angle sum of a triangle is less than 180◦ and the existence of straight lines that approach each other but never cross. However, he found no explicit contradiction. He just concluded that “the hypothesis . . . is absolutely false, because it is repugnant to the nature of the straight line.” Bonola [1,43]. He could plausibly receive credit for developing a non-Euclidean geometry. However, he could more easily be seen as remaining inside the world view of asserting Euclidean geometry to be true and the geometry of the physical world. That world view, accepted for centuries, found even more support in the 1700s. First, Immanuel Kant (1724–1804), the most influential philosopher of the time, held that the truths of mathematics differed fundamentally from most facts about the world, such as the earth has one moon. Mathematics, Kant taught, was a priori true—necessarily true, even before any experience we have. Geometry, as developed by Euclid, seemed a compelling example of how humans could obtain absolute knowledge about the world. In addition Kant argued that we

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needed an essentially inborn understanding of geometry and space before we could experience anything in space. (Studies of blind people and people with recovered sight indicate that Kant’s argument is incorrect; sight is essential to develop the usual conception of space. See Sacks [13, 124].) Kant also thought that geometry had to be Euclidean. In the eighteenth century, the Age of Enlightenment, mathematicians and philosophers built on the perceived absolute truth of mathematics in a second way. The astounding success of Newton’s calculus and physics convinced his successors that mathematics wasn’t just true in some metaphysical sense, but also in a tangible sense. The ideal world of mathematics, it seemed, was the real world. The physical meaning of much of the mathematics developed in the eighteenth century was so convincing that the rigorous deductive methods of Greek geometry seemed superfluous. The shock of the radically different mathematical results of the nineteenth century, starting with non-Euclidean geometries, forced mathematicians to reintroduce careful proofs. The first person to break from the world view of Euclidean geometry, its unquestionable truth, and its applicability to the physical world was Carl Friedrich Gauss (1777–1855). Despite his fame, Gauss never published anything on non-Euclidean geometries because he feared ridicule. Nicolai Lobachevsky (1793–1856) and J´anos Bolyai (1802–1860), the two young mathematicians who did publish on non-Euclidean geometry, were greeted with silence for years after their publications in 1829 and 1832, respectively. Indeed, only with the publication of Gauss’s notes after his death did the wider community of mathematicians start investigating non-Euclidean geometries.

4.1.1 The Advent of Hyperbolic Geometry Hyperbolic geometry, the non-Euclidean geometry that Gauss, Lobachevsky, and Bolyai developed, retains Euclid’s first four postulates and changes the fifth postulate to the following axiom. In Sections 4.2, 4.3, and 4.4, we investigate hyperbolic geometry in much the same way that they did. However, to make the process clearer, we make explicit certain logically necessary assumptions, overlooked until the end of the nineteenth century. (See Sections 2.1 and 2.2.) Hyperbolic geometry is sometimes called Lobachevskian geometry to honor Lobachevsky’s priority in publishing. Felix Klein called the geometry hyperbolic in his classification of geometries, which we discuss in Section 7.5. Characteristic Axiom of Hyperbolic Geometry Given a line k and a point P not on k, there are at least two lines on P that do not intersect k (See Figure 4.1). P

k

Figure 4.1 The characteristic axiom of hyperbolic geometry. Various consequences follow from this change, including the many that Saccheri found. The most startling is the theorem that the measures of the angles of a triangle do not add up to π (or 180◦ ), as they do in Euclidean geometry. The goal in our study of hyperbolic geometry,

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Theorem 4.1.1, goes even further, relating the angle sum to the area of the triangle. Figure 4.2 illustrates that the greater the area of the triangle is, the smaller its angle sum is. (We state the theorem using radians to emphasize the analogy with Theorem 1.5.3. Otherwise we will measure angles in degrees.) Consequently, there is a constant k so that each triangle’s area is less than k. Since the sides of triangles can become indefinitely long, this consequence seems paradoxical.

Figure 4.2 The angle sum of a hyperbolic triangle depends on its area.

Theorem 4.1.1. In hyperbolic geometry the difference, π − (m∠A + m∠B + m∠C), between π and the angle sum of a triangle is proportional to the area of the triangle. Exercise 4.1.1.∗ Compare Theorem 4.1.1 with Theorem 1.5.3. Restate the theorem using degrees. Theorem 4.1.1 suggests that we could decide which geometry is true by measuring real triangles. Since at least 1890, mathematicians and physicists have realized the futility of empirically deciding which geometry is correct because physical assumptions must be made to test mathematical relations. For example, we would have to assume that the path of a light ray is a straight line to measure even a moderately large triangle. The first notable response to the advent of hyperbolic geometry came in 1854 when Georg Riemann (1826–1866) gave his introductory lecture to the faculty at G¨ottingen University. Only Riemann’s mentor, the aging Gauss, apparently caught its point, entitled “On the Hypotheses which underlie Geometry.” However, the talk, published after Riemann’s death, focused geometric thought on a new field, differential geometry, and spurred an active debate on non-Euclidean geometries. Riemann had realized that the work of Gauss, Lobachevsky, and Bolyai was more than playing abstractly with a postulate. He recognized that the revised postulate implied that space had to be shaped differently than what Euclid’s fifth postulate implied. He then articulated how infinitely many different geometries could be created, each with its own shape. Differential geometry, the field pioneered by Riemann, Gauss, and others, investigates geometries by looking at how they behave in small regions and, in particular, how they curve. Gauss had shown how to measure the curvature of surfaces. Where the flat Euclidean geometry had zero curvature and a sphere had positive curvature, hyperbolic geometry has a uniform negative curvature, similar to the saddle shape of Figure 4.3. Riemann defined curvature in higher dimensions and envisioned geometries in any number of dimensions with changing curvatures

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throughout. The general theory of relativity builds on Riemann’s work on the curvature of space. Einstein used a four-dimensional geometry (three spatial dimensions interacting with time) curved by the gravitational forces. Light waves travel along geodesics, paths of shortest length following the curvature of the surrounding space. Einstein’s theory settled the nineteenth century question of the empirical truth of Euclidean and hyperbolic geometries in a surprising way: both are false! (See Chapter 9 as well as do Carmo [3], Kline [8], McCleary [9], Rucker [12], and Weeks [16] for further information.)

Figure 4.3 A saddle shape with negative curvature. The abstraction throughout mathematics and the strangeness of the new geometries led mathematicians to search again for certainty in their mathematical arguments. No longer could they rely on an intuitive model to reveal the essential idea behind an argument. They examined the axioms for geometry, culminating in the modern axioms for Euclidean geometry, including Hilbert’s (see Section 2.2). Hilbert also obtained an axiomatic system for hyperbolic geometry by replacing Playfair’s axiom with the characteristic axiom above. (See Bonola [1] and Kline [8, Chapter 36] for further historical information.)

4.1.2 Models of Hyperbolic Geometry In the latter part of the nineteenth century, various mathematicians developed models of hyperbolic geometry. We partially treat some of them here. Each has disadvantages, but they all help give a feel for the geometry. They are based in part on a Euclidean plane or space, with suitable interpretations of the undefined terms. For the Poincar´e model, we first need a definition: Two (Euclidean) circles are orthogonal if and only if their radii at the circles’ points of intersection are perpendicular, illustrated in Figure 4.4. In all the models the terms on and between have their usual meaning. In the language of Section 2.3, the formal models show the relative consistency of hyperbolic geometry based on Euclidean geometry.

Figure 4.4 Orthogonal circles.

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Non-Euclidean Geometries

Informal Models. A saddle shape, as in Figure 4.3, provides a direct feel for a small part of hyperbolic geometry, although the negative curvature of actual saddles varies considerably from point to point. A physical version of this model has the advantage of facilitating actual measurements on its surface. Crocheted surfaces, as in Figure 4.5, developed by Daina Taimina, provide a more extended realization of approximately constant negatively curved surfaces. (See Henderson and Taimina [6, 49 ff].) The pseudosphere, represented in Figure 4.6 and discussed below, has constant negative curvature except on its “equator.” In all the models, lines are interpreted as geodesics—intuitively shortest paths along the surface. Distances are measured along them and angles are measured using tangents to them where they meet. The surfaces give direct insight into hyperbolic geometry, at least in the region where the curvature is a constant negative.

Figure 4.5 An extended crocheted surface with negative curvature.

Figure 4.6 “pseudosphere” illustrating the characteristic axiom. The Pseudosphere (1868). Ferdinand Minding (1806–1885) investigated this surface of constant negative curvature, shown in Figure 4.6, as early as 1840. Eugenio Beltrami (1835–1899) was the first to realize that it and all other surfaces of constant negative curvature provide (local) models of hyperbolic geometry. Beltrami’s work suggested the relative consistency of hyperbolic geometry for the first time. The models now named for Felix Klein (1849–1925) and Henri Poincar´e (1854–1912) resolved logical concerns about this geometry.

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The Klein Model (1868). Point means a point in the interior of a Euclidean circle. Line means the portion of a Euclidean line in its interior. Figure 4.7 illustrates that the characteristic axiom holds in the model. Its biggest advantage is the matrix representation of its transformations, considered in Chapter 7. However, distances and angle measures are distorted and complicated in it.

P k

Figure 4.7 The characteristic axiom in the Klein model. The Poincar´e Model (1882). Point means a point in the interior of a Euclidean circle. Line means the portion interior to the circle of one of its diameters or of a Euclidean circle orthogonal to it. The chief advantage of this model is that angles are measured as they would be for curves in Euclidean geometry. In Figure 4.8 the curved sides of the triangle in bold show visually that the angle sum is less than 180◦ . The picture at the beginning of this chapter uses the Poincar´e model. Because the angle measures are Euclidean, the repeated objects in this design are recognizably the same, although somewhat distorted to our eyes. Their hyperbolic size is the same, which gives a sense of how distances are measured using the same complicated formula as the Klein model. In Section 5.6 we consider transformations for this model and their connections with complex numbers. There are software programs, such as NonEuclid, that facilitate constructions in this model.

Figure 4.8 A triangle in the Poincar´e model. The Half-Plane Model (1882). Point means a point in the upper half of the Euclidean plane; that is, points (x, y) with y > 0. Line means the portion above the x-axis of either a vertical line or a circle with its center on the x-axis. In Figure 4.9, △ABC has an angle sum less than 180◦ , although that fact may not be as clear as in Figure 4.8. Henri Poincar´e developed the model and

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the closely related Poincar´e model at the same time. It measures angles as Euclidean angles and, in addition, its lines have equations that are easy to find. However, distances appear even more distorted than in the previous two models.

B A

C

Figure 4.9 A triangle in the half plane model. Unless explicitly stated, the figures in the rest of this chapter are not based on any of the models. This will encourage thinking of the geometry as an entire system, not an isolated exercise applying only to an artificial model. Many of the lines drawn in the figures will be curved so that they will not appear to intersect. This corresponds to one of the findings of Saccheri, mentioned earlier, that lines can approach one another in this geometry without intersecting.

4.1.3 Exercises for Section 4.1 Throughout the problems “line” refers to a hyperbolic line in the relevant model. Euclidean objects are identified as Euclidean. 4.1.2. Use a physical saddle-shaped surface, as in Figure 4.3, to explore basic properties of hyperbolic geometry. (The curves shown in that figure are not geodesics; they are the intersections of the surface with vertical planes to convey the three-dimensional curving.) An L-shaped piece of flexible plastic with straight sides and a right angle will lie on the surface closely enough to draw approximate perpendicular hyperbolic lines (geodesics). Use the protractor of Figure 2.16 to find the approximate measure of angles. (a) Construct different sized triangles on the surface and measure their angle sums. Verify Theorem 4.1.1. (b) Construct a quadrilateral ABC D with right angles at A and B and with sides AD and BC congruent. Measure the angles at B and C. What can you say about them? Compare the lengths of AB and C D. (c) Make a large right triangle. Measure the lengths of its sides. Compare the squares of their lengths with the equation of the Pythagorean theorem (Theorem 1.1.2). (d) Draw lines on the surface to illustrate the characteristic axiom of hyperbolic geometry. *4.1.3 Use the Klein model of hyperbolic geometry to investigate how many lines through a point P do not intersect a line k that is not on P. Explain your answer. 4.1.4. Use the points inside the Euclidean unit circle x 2 + y 2 = 1 for the Poincar´e model. *(a) Show that the circle C : (x − 1.25)2 + y 2 = (0.75)2 is orthogonal to the unit circle. Graph the unit circle and the part of C in the unit circle on the same axes.

4.1 Overview and History

159

(b) We form a hyperbolic triangle with vertices P = (0, 0), Q the intersection of the circle C with y = (9/13)x (inside the unit circle), and R the intersection of the circle C with y = −(9/13)x (inside the unit circle). Assume that x > 0. Verify that Q and R have x-coordinate 0.65. Find their y-coordinates. Include △P Q R in the graph for part (a). (c) Verify visually that the angle sum of the triangle in part (b) is less than 180◦ . *(d) Find the measures of the angles of the triangle in part (b) and verify that their sum is less than 180◦ . Hints: The angle made by a Euclidean line and a Euclidean circle is the angle made by the Euclidean line and the tangent to the circle at their intersection. The slope of a Euclidean line equals tan α, where α is the angle that line makes with the x-axis. The arctangent function converts slopes to angles. *4.1.5 Although three-dimensional Euclidean space satisfies the characteristic axiom, it doesn’t satisfy all the axioms of plane geometry. Find which of Hilbert’s axioms in Appendix C fail when we interpret a plane as three-dimensional Euclidean space and a line as a Euclidean line. 4.1.6. Use the half-plane model of hyperbolic geometry. *(a) Find the equation of the line (Euclidean half circle) that is on the points (1, 1) and (4, 2). (b) ! Explain why the equations of lines in the model are either x = c or y = r 2 − (x − c)2 , for appropriate constants c and r . (c) Use a Euclidean argument about Euclidean circles and lines to explain why every two points in this model have one (hyperbolic) line that! is on both of them. *(d) Verify that the point (−4, 10) is not on the line y = 25 − (x!+ 4)2 . Find the equations of two lines on (−4, 10) that do not intersect y = 25 − (x + 4)2 . Graph the lines. (e) Use a Euclidean argument to explain why the characteristic axiom always holds in this model. Hint: Consider√the two kinds of lines ! separately. *(f) Verify that the lines y = 25 − x 2 and y = 16 − (x − 1)2 do not have a point of the model in common, although they intersect in a Euclidean point as curves in the Euclidean plane. Find the Euclidean point. We will call these lines sensed parallels. *(g) Describe the type of intersection of the lines in part (f). (h) Verify that every √ point in the model has at least one line that is a sensed parallel to the line y √ = 25 − x 2 in the sense that the only Euclidean intersection of the line with y = 25 − x 2 is the Euclidean point you found in part (f). Explain why, for a hyperbolic line k represented by a semicircle, that there are exactly two sensed parallels to k through a point not on k. How can we interpret sensed parallels so this property remains correct if a vertical line represents k? 4.1.7. (Calculus) In the half-plane model of hyperbolic geometry, consider the triangle formed ! ! 2 by the lines y = 5 − (x − 3) , y = 10 − (x − 4)2 , and x = 4. Find the vertices of the triangle and graph it. Use the method of Exercise 4.1.4(d) to find the measures of the angles of this triangle. Hint: Use calculus and trigonometry. The angle sum is approximately 161.6◦ .

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√ 4.1.8. *(a) In the half! plane model, verify that!the line y = 16 − x 2 is perpendicular to the 2 lines y = 9 − (x − 5)2 and y = 9 − (x √ + 5) at their points of intersection. *(b) Find the intersections the line y = 34 − x 2 makes with the lines y = ! 9 − (x ± 5)2 and the angles it makes with them. (See Exercise 4.1.4(d) and 4.1.7 for finding the angles.) (c) Compare the angles in part (b) with Exercise 4.1.2(b). 4.1.9. A curious property of hyperbolic geometry is that two lines that do not intersect and are not sensed parallels (described informally in Exercise 4.1.6) have a common perpendicular. Use the√half-plane model of hyperbolic geometry to illustrate this property with the lines y = 1 − x 2 and x = 2. Graph the lines first. Then use relevant Euclidean concepts, including orthogonal circles, to find the hyperbolic line perpendicular to both of them.

4.1.4 Carl Friedrich Gauss Carl Friedrich Gauss (1777–1855) dominated mathematics during the first half of the nineteenth century, making fundamental contributions to virtually every area of mathematics. The prince of his small German state sponsored his education after learning of his prodigious abilities as a child. By age five he had found an arithmetic error in his father’s accounts. In another story his grade school teacher made the class add the numbers from one to one hundred. After some thought Gauss found a formula and simply wrote down the correct answer, whereas his classmates toiled and made mistakes. Gauss soon won acclaim as a mathematician. At 18 he developed the method of least squares, which is used extensively in statistics. A year later he constructed with straightedge and compass a regular 17-sided polygon and later characterized all constructible regular polygons, the first such advances since the ancient Greeks. He earned his Ph.D. at 22 by giving the first proof of the fundamental theorem of algebra, which says that every nth degree polynomial with real or complex coefficients has n roots, counting repetitions and these roots are all complex numbers. Two years later, he published his first major treatise on number theory, rejuvenating that ancient area. That same year, 1801, Gauss astonished astronomers by determining the orbit of the first asteroid, Ceres, based on only a few observations and after it had been lost from view owing to weather conditions. To do so he had to generalize his method of least squares. Gauss subsequently became a professor of astronomy and director of the observatory in G¨ottingen. By 1800 Gauss had become convinced that Euclid’s parallel postulate could not be proven, and in installments he developed what he called non-Euclidean geometry. Although none of this work was published until after his death, he did correspond with a number of mathematicians about it. Gauss also made seminal contributions to differential geometry, including the curvature of a surface and geodesics. He showed that Gaussian curvature determined all the properties of a surface not related to how it is placed in a surrounding space. Gauss’s contributions on complex numbers persuaded mathematicians that complex numbers were essential in mathematics. He represented complex numbers geometrically, as some others before him had. Additionally, he extended number theory to complex integers, and

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4.2 Properties of Lines and Omega Triangles

contributed fundamentally to what is now called complex analysis. He advanced knowledge in astronomy, magnetism, optics, and other applied fields. Gauss’s contemporaries revered him as the “Prince of Mathematicians,” and he is widely considered the greatest mathematician since Newton. He was also one of the last who could claim to know all the mathematics in existence at his time. Gauss’s work can be seen both as crowning the great expansion of mathematics in the sixteenth and seventeenth centuries and as igniting the explosion of specialized, abstract mathematics since then.

4.2 Properties of Lines and Omega Triangles We develop hyperbolic geometry following the path of Saccheri, Gauss, Lobachevsky, and Bolyai, but add a modern foundation, due to Hilbert and others. We can thus accentuate the geometric intuition of earlier work while having the logical basis required to avoid the flaws discovered later. Our undefined terms are point, line, on, between, and congruent. Our axioms are Hilbert’s axioms for plane geometry (see Appendix C), with the exception of axiom IV-1 (Playfair’s axiom), which we replace with the characteristic axiom of hyperbolic geometry. (See Section 4.1.) We also avail ourselves of Euclid’s first 28 propositions, the ones he proved without using the parallel postulate. (See Appendix A.) They historically provided the key properties about lines and triangles needed to development of hyperbolic geometry. They are provable in Hilbert’s axiomatic system without either Axiom IV-1 or the characteristic axiom. We will also make use of some of the SMSG postulates that hold in hyperbolic geometry. We assume the SMSG postulates in Appendix B, except 16, 20, 21, and 22. (They are false in hyperbolic geometry.) Like the developers of hyperbolic geometry, we use proofs by contradiction for many theorems in this chapter, including Theorem 4.2.1. Until 1868, when Beltrami built the first model of hyperbolic geometry, mathematicians wondered whether the characteristic axiom itself was potentially a contradiction. The heavy use of proofs by contradiction increased mathematicians’ unease before Beltrami’s model. A rigorous proof of even Theorem 4.2.1 requires considerable development from Hilbert’s axioms. For example, in Figure 4.10 we would need to show that B can always be chosen so that the line m enters △ABC at P. (By “enter” we mean that the line through P also has a point inside of the triangle on it, following Pasch’s axiom.) The use of real numbers from the SMSG postulates does simplify the proof. Moise [10, Chapter 24] gives a careful development of the basics from Hilbert’s axioms, including the justification of right- and left-sensed parallels in the definition below.

B

m P l

C

X D

k

Z

Y?

A

Figure 4.10 Constructing non-intersecting lines.

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Theorem 4.2.1. Given a point P not on a line k, there are infinitely many lines on P that have no points on k. Proof. Let the two lines indicated by the characteristic axiom be l and m. Pick points A on k and B and C on l such that line m enters △ABC at P. By Pasch’s axiom, m intersects the triangle on another side, without loss of generality at D on AB. SMSG postulate 3 permits us ← → to match the points on AB with the real numbers, with 0 at B and 1 at D. Now, for every point ← → X on the segment B D, we can draw P X . (See Figure 4.10.) ← → ← → We claim that P X does not intersect k. Suppose, for a contradiction, that P X and k had the point Y in common. Then the line m would enter triangle △X Y A at D, and by Pasch’s axiom m ← → would have to intersect the triangle again on X Y or on Y A. However, m already intersects P X at P, eliminating X Y . Furthermore, Y A is part of k, which has no point in common with m, which is a contradiction. The points X between B and D match with numbers between 0 and 1. ← → Since there are infinitely many such numbers, there are infinitely many lines P X that have no point in common with k. ! For the same k and P as in the preceding proof, the lines through P split into two categories, those that intersect k and those that do not. In Figure 4.10, for Z on the segment AD, some of ← → ← → ← → the lines P Z , such as P A, intersect k, but others don’t, such as P D. The points Z match with numbers greater than 1, but less than the number matched with A. By the continuity of the real numbers, some number w must separate the numbers of points with lines intersecting k from those not intersecting k. Let W be the point matching w, so W must separate these lines. Does ←→ P W intersect k? Draw a figure illustrating the following argument by contradiction showing ←→ ←→ that P W cannot intersect k. If T is the supposed intersection of P W and k, consider a point S ← → ←→ on k with T between A and S. Now draw P S, which clearly intersects k at S. Therefore P W would not be the last line, contradicting the assumption about W . A similar situation occurs on the left side of Figure 4.10. Definition. Given a point P not on line k, the first line on P in each direction that does not intersect k is the (right- or left-) sensed parallel to k at P. Other lines on P that do not intersect k are called ultraparallel to k. Let A be on k with A P ⊥ k. Call the smaller of the angles a sensed parallel makes with A P the angle of parallelism at P. (If the angles are equal, either is the angle of parallelism.) The goal of this section has two parts. First, we start building inequalities leading to the big inequality in Theorem 4.1.1 about the angle sum of a triangle. Corollary 4.2.3, on the angle of parallelism, provides the starting point. Theorems 4.2.8 and 4.2.9 accomplish the other goal, showing that the angle of parallelism depends only on the length of the segment A P. Theorem 4.2.2. If l and m are the two sensed parallels to k at P, they have the same angle of parallelism. ← → ← → Proof. As shown in Figure 4.11, let A P ⊥ k and P B and PC be the sensed parallels to k at P. For a contradiction, let the angles of parallelism differ, say, m∠A P B < m∠A PC. Construct

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← → ← → inside ∠A PC a new angle ∠A P D ∼ = ∠A P B. Because PC is a sensed parallel, P D intersects k, say at E, giving a triangle △ A P E. Let F be on k with AF ∼ = AE. Then △A P F ∼ = △A P E ← → ← → by SAS (Euclid I-4). But ∠A P F ∼ = ∠A P D ∼ = ∠A P B, which means that P F and P B are the ← → same line by Hilbert’s axiom III-4. However, P B is a sensed parallel to k, so it cannot intersect k at F, which is a contradiction. Hence the angles of parallelism must be the same. ! P C

B D k

E

F

A

Figure 4.11 Angles of parallelism. The proof of Theorem 4.2.2 may seem on first reading overly complicated. However, in order to use the Euclidean proposition SAS, we need “ordinary” triangles. Corollary 4.2.3. All angles of parallelism are less than a right angle. Two lines with a common perpendicular are ultraparallel. Proof. See Exercise 4.2.5. ! Theorem 4.2.4. Let P be a point not on k and let m be a sensed parallel to k at P. If S is any other point on m, then m also is a sensed parallel to k at S. Proof. Without loss of generality, let m be a right-sensed parallel to k at P. Case 1. Suppose that S is on m to the left of P and U is on m with S between P and U . Because m and k do not intersect, m is either the right-sensed parallel to k at S or m is ultraparallel to k at S. For a contradiction, suppose that l, not m, is the right-sensed parallel to k ← → at S, as in Figure 4.12. Pick T on l and on the opposite side of m from k. Then T P crosses m at T

U S

P

l

m

k C

A

B

Figure 4.12 The sensed parallel to k at P is also sensed parallel at S.

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Non-Euclidean Geometries

← → P. As m is the right-sensed parallel to k at P, T P must intersect k, say at A. Let U be on m with S between P and U and C on k with SC ⊥ k. Now l enters △U PC at S. Since l is right-sensed parallel, it must exit on PC, rather than U C. In turn l enters △C P B on C P and, by Pasch’s ← → axiom, l would need to intersect either C B or P B. But l is a right-sensed parallel of k = C B, so l intersects P B. Finally l enters △P B A. Again by Pasch’s axiom, either l intersects B A, which is k, or it intersects P A. As before, k can’t intersect l. However, l already intersected P A at T , so that also is impossible. We thus have a contradiction from assuming l was the right-sensed parallel. Hence m is the right-sensed parallel to k at S. Case 2. Let S be on m to the right of P. (See Exercise 4.2.6.) !

4.2.1 Omega Triangles In both the Klein model (Figure 4.13) and the Poincar´e model (Figure 4.14) of hyperbolic geometry m and n, the sensed parallels to k at P, “meet” k on the circular boundary. Only points interior to the boundary qualify as points in these models. Following the historical development, we say that two sensed parallels meet at an imaginary point called an omega point. Although the originators of hyperbolic geometry didn’t have models in the modern sense, they benefited greatly from thinking of sensed parallels “meeting at infinity.” They used omega triangles, which have one omega point, to prove theorems about ordinary triangles. They first showed, as do we here, that omega triangles share some key properties with regular triangles. Figures 4.15 and 4.16 illustrate an omega triangle in the Klein and Poincar´e models with one vertex (labelled -) P n

P m

n

m

k

k

Figure 4.13 Sensed parallels in the Klein

Figure 4.14 Sensed parallels in the Poincar´e

model.

model.

∏

A

∏

A

B

B

Figure 4.15 An omega triangle in the Klein

Figure 4.16 An omega triangle in the Poincar´e

model.

model.

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4.2 Properties of Lines and Omega Triangles

on the boundary. The models also suggest Theorem 4.2.5, which we accept without proof. (For a proof, see Moise [10, 321–322].) Definition. All lines right-sensed parallel to a given line are said to have the same right omega point and similarly for left omega points. An omega triangle △AB" consists of two (ordinary) −→ −→ points A and B, the segment AB, and the sensed parallel rays A" and B". Theorem 4.2.5. If m is sensed parallel to k, then k is sensed parallel to m. If m is sensed parallel to l and l is sensed parallel to k with the same omega point, then m is sensed parallel to k with the same omega point. Exercise 4.2.1. Draw an omega triangle for the half-plane model. Theorem 4.2.6. (Modified Pasch’s axiom for omega triangles) If a line k contains a point interior to △AB- and contains one of the vertices, then k intersects the opposite side of △AB-. ← → ← → Proof. Let C be in △AB-. First consider the line k = AC (Figure 4.17). Because A- is the ← → −→ ← → sensed parallel to B- at A, k intersects B-, say at D. The line BC is handled similarly. We can ← → extend the theorem by treating - as a vertex. The line C- enters the ordinary triangle △AB D ← → ← → at C, and by Pasch’s axiom C- intersects AB or B D. However, if C- intersected B D, there ← → ← → ← → ← → would be two sensed parallels to A- at that point: B- and C-. So C- intersects AB. ! B D C

k

∏

A

Figure 4.17 The modified Pasch’s axiom for an omega triangle. The following theorem transforms the inequality of the angle of parallelism into an inequality about omega triangles. In the next section, we will apply this inequality to regular hyperbolic triangles. Theorem 4.2.7. (Euclid I-16 for omega triangles) The measure of an exterior angle of an omega triangle is greater than the measure of the opposite interior angle. Proof. Let △AB- be an omega triangle and extend AB (Figure 4.18). We prove ∠C A- is greater than ∠AB- by eliminating the other two possibilities through contradictions. Case 1. For a contradiction, assume that m∠C A- < m∠AB-. Construct ∠AB Z inside ← → −→ ← → ∠AB- with ∠AB Z ∼ = ∠C A-. Then B Z intersects A-, say at D because B- is the sensed

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F B

B

E A C

Z

∏

θ

A

D

C

D

Figure 4.18 Euclid I-16 for omega triangles

Figure 4.19 Euclid I-16 for omega triangles

(Case 1).

(Case 2).

← → parallel to A- at B. But then ∠C A- is an exterior angle to an ordinary triangle, △AB D, with an opposite interior angle ∠AB D ∼ = ∠C A-. This contradicts Euclid I-16. Case 2. For a contradiction, assume that ∠C A. ∼ = ∠AB.. As in Figure 4.19, let E be ← → the midpoint of AB and draw the perpendicular D E to A.. By Corollary 4.2.3, the angle of ← → parallelism is acute, so D is not A. Construct F on B. so that F B ∼ = AD and F and D are ← → on opposite sides of AB. By our extra assumption in this case ∠F B E ∼ = ∠D AE. So by SAS, △F B E ∼ = △D AE. In particular the angles at E are congruent, ensuring that D, E, and F are on the same line, by Euclid I-14. Because ∠AD E is a right angle, ∠B F E is also. But then the ← → ← → angle of parallelism for F. to D. would be a right angle, which contradicts Corollary 4.2.3. Hence the assumption that ∠C A. ∼ = ∠AB. must also be wrong. Since both alternatives are impossible, the theorem holds. ! We can extend Euclid’s concept of congruent triangles to omega triangles. However, the lengths of two of the sides of an omega triangle are infinite, and we can hardly measure the imaginary angle at the omega point. Hence there are only two angles and the included side to consider. Definition. Two omega triangles △AB" and △C D/ are congruent, written △AB" ∼ = △C D/, ∼ ∼ if and only if AB ∼ C D, ∠B A" ∠DC/, and ∠AB" ∠C D/. = = = We show that if two omega triangles have two of the three parts congruent, their third parts are congruent. Theorem 4.2.8. In omega triangles △AB" and △C D/, if AB ∼ = C D and ∠AB" ∼ = ∼ ∠C D/, then △AB" = △C D/. Exercise 4.2.2. Illustrate the proof of Theorem 4.2.8. Proof. For a contradiction, assume that ∠B A" is not congruent to ∠DC/, and without loss of generality ∠DC/ has a smaller measure. Construct ∠B A P inside ∠B A" with ∠B A P ∼ = ← → −→ ∠DC/. Then A P intersects B", say at E. From the ordinary triangle △AB E we now construct −→ a congruent one in the other omega triangle. Let F be the point on D/ with D F ∼ = B E. Then △D FC ∼ = △B E A by SAS. However, this means that ∠DC F ∼ = ∠B AE ∼ = ∠DC/, which ← → ← → implies that C/ intersects D/, a contradiction. Hence the two angles, and so the two omega triangles, are congruent. !

4.2 Properties of Lines and Omega Triangles

167

Theorem 4.2.9. If corresponding angles of omega triangles are congruent, then the omega triangles are congruent. Proof. See Exercise 4.2.11. ! If ∠AB" is a right angle in an omega triangle △AB", then ∠B A" is the angle of parallelism. Theorem 4.2.8 implies that the size of the angle depends only on the length AB. Conversely, if we know the angle of parallelism, Theorem 4.2.9 says that only one length has that angle of parallelism. A comparison with Euclidean geometry reveals the oddity of this situation. In Euclidean geometry the angles of a shape tell us nothing about its size. Similar shapes have the same angles but different sizes. In hyperbolic geometry, the angles of shapes determine their sizes. In both geometries we can measure angles absolutely by comparing them with a right angle. However, Theorem 4.2.9 suggests that in hyperbolic geometry we can, in principle, measure lengths in an absolute way by using the angle of parallelism.

4.2.2 Exercises for Section 4.2 *4.2.3. a) Suppose that m is a left-sensed parallel to k and l is a right-sensed parallel to k. Use the Klein model to determine whether m and l can (i) intersect. Repeat for (ii) being sensed parallel and for (iii) being ultraparallel. Explain. b) Draw ultraparallel lines k and n in the Klein model. Draw all lines that are sensed parallel to both k and n. 4.2.4. a) √ In the half plane model the point P = (0, 4) is not on the line k given by y = 4 − x 2 . Find the equations of the two sensed parallels to k through P. b) Repeat part (a) replacing P with Q = (2, 4). c) Repeat part (a) replacing k by the line m with equation x = 4. (d) Verify Theorem 4.2.7 in the half plane model with the △AB", √ omega triangle ← → ← → ← → 2 where AB has!equation x = 0, A" has equation y = 64 − x , and B" has the equation y = 25 − (x − 3)2 . (Exercise 4.1.4(d) describes how to find the angle a Euclidean circle makes with a line.) *4.2.5. Prove Corollary 4.2.3. Hint: What are the options for the angle of parallelism? Use a proof by contradiction. 4.2.6. Prove case 2 of Theorem 4.2.4 Hint: Pick T on l between m and k. *4.2.7. Relate Euclid I-27 and I-28 to Corollary 4.2.3. 4.2.8. State and prove Pasch’s axiom (Hilbert II-4) for omega triangles. ← → ← → ← → ← → ← → *4.2.9. Suppose that AC ⊥ C", B is between A and C, and that A", B", and C" are all right-sensed parallels. Theorem 4.2.9 implies that the angles of parallelism for △AC" and △BC" cannot be equal. Which is larger? Prove your answer. ← → ← → ← → ← → *4.2.10. Let AC and B D be sensed parallels, AB ⊥ B D, and C D ⊥ B D. Suppose, as Figure 4.20 suggests, that C D is shorter than AB. Use your answer to exercise 4.2.9 to compare 2π (or 360◦ ) with the angle sum of the quadrilateral AB DC.

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Non-Euclidean Geometries

A C

B

D

Figure 4.20 Approaching sensed parallels?

4.2.11. Prove Theorem 4.2.9. 4.2.12. If M is the midpoint of AB in △AB", prove that ∠A ∼ = ∠B if and only if ∠AM" is a right angle. 4.2.13. In omega triangles △AB" and △C D/, if ∠A ∼ = C D, = ∠B, ∠C ∼ = ∠D, and AB ∼ ∼ must △AB" = △C D/? If so, prove it. If not, explain. 4.2.14. Define a double omega triangle △A-# as a line k with omega points - and #, a −→ −→ regular point A, and the two sensed parallel rays A- and A# to k from A. The only thing we can measure in a double omega triangle is the angle at A. We define two double omega triangles to be congruent if and only if their angles are congruent. a) Suppose that △A-# and △B$. are congruent double omega triangles. (See ← → ← → Figure 4.21.) Let C be the point on -# so that AC ⊥ -# and, similarly, D be the ← → ← → point on $. so that B D ⊥ $.. Prove that the omega triangles △AC-, △AC#, △B D$, and △B D. are congruent. A

∏

Σ

C B

∆

D

θ

Figure 4.21 Congruent double omega triangles. b) Prove the following converse of part (a). Suppose that △A-# and △B$. are ← → ← → ← → double omega triangles, C is on -# so that AC ⊥ -#, and D is on $. so that ← → B D ⊥ $.. Further suppose that AC ∼ = B D. Prove the double omega triangles are congruent.

169

4.3 Saccheri Quadrilaterals and Triangles

4.2.3 Nikolai Lobachevsky and J´anos Bolyai . . . [F]rom nothing I have created another wholly new world. —J´anos Bolyai

Nikolai Lobachevsky (1793–1856) was a mathematics professor at the University of Kazan in Russia, where he first published on hyperbolic geometry in 1829. His extensive development included its trigonometry, corresponding to the trigonometry of a sphere of imaginary radius. (The hyperbolic cosine function, cosh(x), and related functions, such as sinh(x), appear in applications and in calculus texts. For example, a cable hanging from its ends follows the curve of cosh(x).) He thought that hyperbolic geometry might be relevant in astronomy because he realized that over large distances it differed greatly from Euclidean geometry. J´anos Bolyai (1802–1860) was a Hungarian army officer with a reputation for dueling. One day he reportedly fought with several officers, playing his violin between duels. Bolyai found basically the same results as Lobachevsky. In 1832 his publication The Science of Absolute Space also reported his investigations of the properties common to both Euclidean and hyperbolic geometries. The fame of Lobachevsky and Bolyai rests entirely on their work in hyperbolic geometry. Surprisingly, the wider mathematical community ignored their publications during their lifetimes, which deeply disappointed them. Certainly the original languages of Russian and Hungarian deterred readers. However, in 1840 Lobachevsky published a book on his research in German, then the foremost language of mathematics and science. Perhaps a more important reason for the neglect was that geometric research then focused on projective geometry, which didn’t consider parallel lines. Also, the trigonometric formulations may have impeded people interested in the philosophical and geometric implications. Only after the posthumous publication of Gauss’s notebooks in 1855 did an active study of non-Euclidean geometry begin. Because of the number of distinguished mathematicians who had failed to prove Euclid’s fifth postulate, Lobachevsky and Bolyai should have found an interested, critical audience for their radical answer to this fundamental question in geometry.

4.3 Saccheri Quadrilaterals and Triangles We now turn from unbounded sensed parallels and omega triangles to bounded regions, starting with the quadrilaterals at the center of Saccheri’s investigations. Much earlier the Persian mathematician Omar Khayyam (circa 1048–1131) pursued similar research with quadrilaterals and proved some of the results in this section. So the quadrilaterals could reasonably be renamed Khayyam quadrilaterals. However, modern mathematicians learned of Khayyam’s work after tradition had attached Saccheri’s name to them. Definition. A Saccheri quadrilateral has two opposite congruent sides perpendicular to one of the other sides. The common perpendicular is the base, and its opposite side is the summit. Our first result about Saccheri quadrilaterals holds in both Euclidean geometry and hyperbolic geometry. Exercise 1.2.9 asked for a proof of a more general result based on Euclidean

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propositions common to both geometries. Theorem 4.3.2, however, only holds in hyperbolic geometry. Exercise 4.3.1.∗ In Euclidean geometry what type of quadrilateral is a Saccheri quadrilateral? Theorem 4.3.1. The summit angles of a Saccheri quadrilateral are congruent. The base and summit are perpendicular to the line on their midpoints. Proof. Let AB be the base of a Saccheri quadrilateral ABC D and AC and B D the diagonals (Figure 4.22). Then △ABC ∼ = △B AD by SAS, and by SSS △ADC ∼ = △BC D. Hence the summit angles, ∠ADC and ∠BC D, are congruent. For the second statement of the theorem, let E and F be the midpoints of AB and C D, respectively (Figure 4.23). Draw D E and C E. Then △D AE ∼ = △C B E by SAS, and △D E F ∼ = △C E F by SSS. Thus ∠D F E ∼ = ∠C F E. Since the angles form a straight angle, E F ⊥ C D. Similarly, E F ⊥ AB. ! D

C

A

B

D

F

A

E

C

B

Figure 4.22 A Saccheri quadrilateral with

Figure 4.23 A Saccheri quadrilateral with

its diagonals.

midpoints.

Theorem 4.3.2. The summit angles of a Saccheri quadrilateral are acute. ← → ← → Proof. On a Saccheri quadrilateral ABC D with base AB, let C- and D- be left-sensed ← → parallels to AB (Figure 4.24). By Theorem 4.2.7, the exterior angle ∠E D- of omega triangle △DC- is larger than ∠DC-. Since AD ∼ = BC and ∠-AD ∼ = ∠-BC, Theorem 4.2.8 gives congruent omega triangles and so the angles of parallelism ∠AD- and ∠BC- are congruent. Hence ∠E D A is larger than the summit angle ∠DC B. The summit angles are congruent, so ∠E D A is bigger than ∠C D A. The angles form a straight line, so the smaller, a summit angle, must be acute. ! E

D

C

∏ A

B

Figure 4.24 A Saccheri quadrilateral with sensed parallels.

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4.3 Saccheri Quadrilaterals and Triangles

A direct consequence of Theorem 4.3.2 is that the angle sum of a Saccheri quadrilateral is always less than 2π (360◦ ). This extends the angle inequalities from Section 4.2 to at least some bounded region. Theorem 4.3.3 converts this fact to the corresponding inequality about triangles by dissecting a triangle to form a Saccheri quadrilateral (Figure 4.25). From the dissection we find that the sum of the measures of the summit angles of the Saccheri quadrilateral equals the angle sum of the triangle. In the next section we develop the concept of area enough to strengthen Theorem 4.3.3 to the full statement of Theorem 4.1.1. Theorem 4.3.3. The angle sum of a triangle is less than π . Proof. In a triangle △ABC, let D be the midpoint of AB and E be the midpoint of AC. Construct the perpendiculars AF, BG, and C H to D E. We claim that either D is between F and G or D = F = G. (See Exercise 4.3.5.) As a consequence the construction looks like one of the options shown in Figures 4.25, 4.26, or 4.27 or their mirror reflections. A

A

E

G D

H

E

F

H

D=F=G

B

C

C

B

Figure 4.25 Dissecting a triangle, case 1.

Figure 4.26 Dissecting a triangle, case 2.

A 3 4 D G

E

H

F 1 2 B

5

6 C

Figure 4.27 Dissecting a triangle, case 3. Cases 1 and 2 correspond to Figures 4.25 and 4.26, respectively. (See Exercise 4.3.6.) Case 3. Let D lie between F and the other points G, E, and H on the line (Figure 4.27). In △AF D and △BG D, we know that AD ∼ = B D, ∠AF D ∼ = ∠BG D, and ∠AD F ∼ = ∠B DG. ∼ ∼ Hence, by AAS, △AF D = △BG D. Similarly, △ AF E = △C H E. Then BG ∼ = AF ∼ = C H. The congruences, together with the right angles at G and H show that G H C B is a Saccheri quadrilateral. Thus both summit angles, ∠C BG and ∠BC H , are acute. Finally we need to show

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Non-Euclidean Geometries

that the angle sum of the triangle equals the sum of the measures of the two summit angles. Following the labeling in Figure 4.27, we have m∠ABC + m∠B AC + m∠AC B

= m∠2 + m∠1 + m∠4 + m∠5 = m∠2 + m∠3 + m∠4 + m∠5 = m∠2 + m∠6 + m∠5 = m∠G BC + m∠H C B. !

Exercise 4.3.2.∗ If we dissect a Euclidean triangle as in Theorem 4.3.3, what type of quadrilateral results? Exercise 4.3.3.∗ Explain why the argument in the preceding proof wouldn’t show that Euclidean triangles have angle sums less than π . Corollary 4.3.4. The angle sum of a quadrilateral is less than 2π . ← → Proof. In a quadrilateral ABC D draw B D. If A and C are on opposite sides of B D, then ABC D is the union of △AB D and △C B D, as in Figure 4.28. In this case, the angle sum of ABC D is the sum for the two triangles. We use Theorem 4.3.3 twice to see the angle sum of ← → ABC D is less than 2π . Otherwise, A and C are on the same side of B D. By the definition of a polygon in Exercise 1.2.13 the edges can’t intersect between their endpoints. In Exercise 4.3.8 you are asked to show that either A is in the interior of △C B D or C is in the interior of △AB D. Either way ABC D is the union of △ABC and △ADC. As before, two applications of Theorem 4.3.3 give the result. ! A

D

B

C

Figure 4.28 A quadrilateral. The next theorem shows that the powerful Euclidean notion of similar triangles plays no role in hyperbolic geometry. In hyperbolic geometry similar triangles are always congruent. As Saccheri and others before him noted, the existence of noncongruent but similar triangles is equivalent to Playfair’s axiom. Theorem 4.3.5. If two triangles have corresponding angles congruent, then the triangles are congruent. Proof. Let △ABC and △A′ B ′ C ′ have corresponding angles congruent (Figure 4.29). If one pair of corresponding sides is congruent, then by ASA the triangles would be congruent. For a contradiction, assume without loss of generality that AB is longer than A′ B ′ . Construct D −→ between A and B such that AD ∼ = A′ B ′ and construct E on the ray AC so that AE ∼ = A′ C ′ .

173

4.3 Saccheri Quadrilaterals and Triangles

Then by SAS, △AD E ∼ = △A′ B ′ C ′ . By properties of betweenness, either E is between A and C or C is between A and E, as illustrated in Figures 4.29 and 4.30, respectively. Exercise 4.3.9 asks you to derive a contradiction in each case. The contradictions show that AB cannot be longer than A′ B ′ , and so the triangles are congruent. !

A'

A A'

B

A D B

B'

E

C'

D

B'

C'

C

C

E

Figure 4.29 Triangles with congruent angles,

Figure 4.30 Triangles with congruent angles,

case 1.

case 2.

4.3.1 Exercises for Section 4.3 4.3.4. (a) In the Poincar´e model using the unit circle graph the Saccheri quadrilateral described as follows. The base is on y = 0, the two sides are on [x ± (5/3)]2 + y 2 = (4/3)2 , and the summit is on x 2 + [y − (5/3)]2 = (4/3)2 . Explain why the sides are perpendicular to the base. Why can you expect the sides to be congruent? Verify visually that the summit angles are acute. (b) In the half plane model, described as follows. √ graph the Saccheri quadrilaterals ! − x 2 , the two sides are on y = 7.52 − (x ± 12.5)2 , The base is on y = 102 √ and the summit is on y = 72 − x 2 . Explain why the sides are perpendicular to the base. Why can you expect the sides to be congruent? Verify visually that the summit angles are acute. √ (c) Repeat part (b), replacing the equation for the summit with y = 152 − x 2 . For which part of the problem are the summit angles most clearly acute? 4.3.5. (Proof of the claim in Theorem 4.3.3.) If D, F, and G are the same point or D is between F and G, the claim holds. Suppose, for a contradiction, some other situation occurs. That is, either D coincides with just one of the other points or that F and G are both on the same side of D. Find a contradiction for each. Hint: Use Euclid I-16 for the second option. 4.3.6. Prove the first two cases of Theorem 4.3.3, based on Figures 4.25 and 4.26. −→ *4.3.7 (a) Prove that the supposition in Exercise 4.2.10 is correct. Hint: Consider X on DC such that AB D X is a Saccheri quadrilateral. Use Theorem 4.3.2. (b) Explain why the supposition you showed in part (a) implies, as Saccheri proved, that sensed parallels approach one another. 4.3.8. Illustrate the second case in Corollary 4.3.4. Use the definition of a polygon to explain why either A is in the interior of △BC D or C is in the interior of △AB D.

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Non-Euclidean Geometries

4.3.9. (a) Finish the proof of Theorem 4.3.5. Hints: Use Corollary 4.3.4 for one case and Euclid I-16 for the other. Be sure to show in the second case that D E and BC intersect. (b) Explain why the first case of part (a) doesn’t give a contradiction in Euclidean geometry. 4.3.10. Assume that D is between B and C, as shown in Figure 4.31. By Theorem 4.3.3, △ABC and △AB D have angle sums less than 180◦ . Which triangle, if any, has the larger sum? Prove your answer. A C D B

Figure 4.31 Triangles. *4.3.11. Generalize Corollary 4.3.4 to convex polygons with n sides. Prove your generalization by using induction. Does the generalization hold if the polygon isn’t convex? Explain. 4.3.12. In Figure 4.23 it appears that the sides AD and BC are longer than the segment E F. Is that accurate or misleading? Prove your answer. 4.3.13. In Section 4.4 we will define the defect of a triangle to be the difference between π and the angle sum. Prove that the defects of triangles are additive in the following sense. If S lies between Q and R in △P Q R, show that the defect of △P Q R equals the sum of the defects of △P Q S and △P S R.

4.3.14. From the curved arc for the summit in Figure 4.22 we might guess that the summit is longer than the base. Is that accurate or misleading? Prove your answer.

4.3.2 Omar Khayyam A Book of Verses underneath the Bough, A Jug of Wine, a Loaf of Bread–and Thou Beside me singing in the Wilderness— Oh, Wilderness were Paradise enow [enough]. —Omar Khayyam

Omar Khayyam (1048–1131) is most widely known today in the West as a poet. In 1859 Edward Fitzgerald translated a selection of Khayyam’s poems, which Fitzgerald called The Rubiyat. This popular book introduced English readers to what they considered to be exotic oriental thought. The quote above became the most famous of these verses. However, Khayyam’s renown during his lifetime and in the Islamic world rests on his scholarship. A noted astronomer, he calculated the length of the year to five decimal place accuracy and instituted calendar reform based on it. Khayyam was an even more influential mathematician, particularly through his work in algebra. (“Algebra” comes from the Arabic “al jebr,” meaning the “reunion of broken parts,” part of their process of solving algebraic problems.) His algebra text refined earlier Arabic texts

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4.3 Saccheri Quadrilaterals and Triangles

and went beyond their presentation of what we call first and second degree equations. Before the advent of algebraic notation he geometrically solved a variety of third degree equations using intersections of conic sections. He worked on a general theory and realized that some problems could have more than one solution. He stated that they couldn’t be solved with straightedge and compass alone, a result whose rigorous proof needs abstract algebra, developed 700 years after his death. In his geometry text Khayyam attempted to prove Euclid’s parallel postulate, making use of what are today called Saccheri quadrilaterals. In it he also extended Euclid’s work on ratios. Unfortunately, his geometric work was generally forgotten well before the advent of hyperbolic geometry in nineteenth century. Only in the twentieth century did historians connect his work with the work of Saccheri and others.

4.3.3 Giovanni Girolamo Saccheri The Italian priest Giovanni Girolamo Saccheri (1667–1733) taught theology, philosophy, and mathematics. However, he is now best remembered for his book Euclid Freed from Every Flaw, published just after his death. Eugenio Beltrami rescued the book from obscurity in 1868, around the time the wider mathematical community realized the significance of non-Euclidean geometry. It seems that Gauss, Lobachevsky, and Bolyai were unaware of Saccheri’s work. Similarly, there is no evidence that Saccheri knew of Khayyam’s earlier results. The history of mathematics has many instances of people developing mathematics unaware of previous work. Saccheri recast the long-standing question about Euclid’s parallel postulate in terms of the quadrilaterals that bear his name. He realized that the summit angles in Theorem 4.3.1 could logically be right, obtuse, or acute. The hypothesis of a right angle he showed was equivalent to Euclidean geometry. So he pursued each of the other options in long proofs by contradiction. Now we see the hypothesis of an obtuse angle, illustrated in Figure 4.32, as leading to spherical geometry, discussed in Section 4.5. However, Saccheri correctly realized that this hypothesis contradicts Euclid’s Proposition I-16, something he and others believed was fully proven using only Euclid’s uncontroversial first four postulates. Thus he disposed of that option fairly quickly. D

C

A

B

Figure 4.32 A Saccheri quadrilateral in spherical geometry. The hypothesis of the acute angle, the content of Theorem 4.3.2, proved to be a more daunting starting point. We know now that Saccheri couldn’t have found a contradiction because hyperbolic geometry is logically consistent, assuming the consistency of Euclidean geometry. However, the consequences of the hypothesis run counter to strongly held intuitions about geometry. In the end Saccheri rejected this option with the statement “It is repugnant to the nature of a straight line.” He did admit, however, that he hadn’t reached a logical contradiction.

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Non-Euclidean Geometries

The inability of Saccheri to see a new geometry unfolding from his theorems suggests how much mathematics has changed. Modern mathematics readily accepts hyperbolic geometry and other systems as legitimate alternatives to Euclidean geometry.

4.4 Area, Distance, and Designs Areas in hyperbolic geometry do not have easily remembered formulas, such as A = 12 bh for a Euclidean triangle. In place of that, Theorem 4.1.1 asserts that the area of a triangle is proportional to the defect of the triangle, or the amount by which its angle sum is less than π . Without formulas for areas, we need to build our understanding of hyperbolic areas from the SMSG postulates 17, 18, and 19 in Appendix B. We restate the postulates about area, which hold in Euclidean, hyperbolic, and spherical plane geometry. 17. Each polygonal region has an area, which is a unique positive real number. 18. Congruent triangles have the same area. 19. Suppose that a region R is the union of S and T . If S and T intersect in at most a finite number of points and line segments, then the area of R equals the sum of the areas of S and T . Exercise 4.4.1.∗ Explain why the area of a point or line segment is zero. Example 1. Verify that a triangle and the associated Saccheri quadrilateral from Theorem 4.3.3 have the same area. Solution. We consider case 1 of Theorem 4.3.3, as shown in Figure 4.25. In the proof we showed that △AF D ∼ = △BG D and △AE F ∼ = △C E H . The quadrilateral B D EC is congruent to itself. By postulate 18 the pieces of △ABC have the same areas as the corresponding pieces of the Saccheri quadrilateral. Further, the intersections of the pieces are line segments and so have zero area by Exercise 4.4.1. Thus the areas of the pieces add to the same value, showing △ABC and its Saccheri quadrilateral BG H C have equal areas. See Exercise 4.4.2 for Cases 2 and 3. ♦ In Section 4.3 we showed that the angle sum of a triangle equaled the sum of the measures of the summit angles of the associated Saccheri quadrilateral. Example 1 equates their areas. The following sequence of theorems build on this link to achieve our goal of connecting the area of a triangle with its angle sum. Theorem 4.4.4 generalizes Example 1. Theorems 4.4.1 and 4.4.2 justify the definition of equivalent polygons. The notion of equivalent polygons applies in many geometries. For example, W. Bolyai and P. Gerwien showed that if two Euclidean polygons have the same area, they are equivalent. Theorem 4.4.3 enables us to show when two triangles have the same area by comparing their Saccheri quadrilaterals. Theorem 4.4.1. If a set S is the union of a finite number of sets A1 , A2 , . . . , An and the intersection of two sets Ai and A j is a finite number of points and line segments, then the area of S is the sum of the areas of A1 , A2 , . . . , An . Proof. Exercise 4.4.3 asks you to prove this using induction. !

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4.4 Area, Distance, and Designs

Definition. Two polygons A and B are equivalent if and only if there are finitely many triangles A1 , A2 , . . . , An and B1 , B2 , . . . , Bn such that A is the union of the Ai ; B is the union of the Bi ; for each i, Ai ∼ = Bi ; and the intersection of Ai and A j or of Bi and B j is at most a line segment. Theorem 4.4.2. If the polygons A and B are equivalent and the polygons B and C are equivalent, then A is equivalent to C. Proof. Divide A and B into families of congruent triangles, Ai and Bi . Divide B and C into families of congruent triangles, B ′j and C ′j , where Bi and B ′j can be different (Figure 4.33). We subdivide the triangles Bi and B ′j into smaller triangles Bi jk so that we can reassemble them into either A or C. The intersection Bi j of the convex sets Bi and B ′j is convex by Theorem 2.1.4. Because Bi j is a convex polygon, it can be subdivided into triangles Bi jk . Then the union of the triangles Bi jk must give B and, by the assumptions of equivalence, can be rearranged to form both A and C. This result shows that A and C can be divided into triangles Ai jk and Ci jk to make A and C equivalent. ! B

B B'1

B1

A

C

B'3

C'1

B2

A2

B3

B'2

A3

B'4

C'2 C'3

A1 C'4

A

B

C

Figure 4.33 Equivalent polygons. Theorem 4.4.3. Two Saccheri quadrilaterals with congruent summits and congruent summit angles have congruent sides and bases and so are congruent. Proof. Let ABC D and E F G H be two Saccheri quadrilaterals with C D ∼ = G H , ∠ADC ∼ = ∠E H G, and bases AB and E F. We need to show that their sides and bases are congruent. Without loss of generality suppose, for a contradiction, that BC > F G (Figure 4.34). Hence there is a point B ′ on BC such that B ′ C ∼ = F G. Similarly, there is an A′ on AD such that ′ ∼ A D = E H . From the midpoints P and Q of the summits C D and G H draw the line segments A′ P, P B ′ , E Q, and Q F. Parts (a) and (b) of Exercise 4.4.4 ask you to show the congruences △A′ D P ∼ = △E H Q and △A′ B ′ P ∼ = △E F Q. Part (c) asks you to prove that A′ B ′ C D is a

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Non-Euclidean Geometries

Saccheri quadrilateral. Part (d) asks you to derive a contradiction. Thus BC ∼ = F G. Part (e) ∼ ∼ F G that AB E F. By part (f), ABC D E F G H. ! asks you to show from BC ∼ = = = D

P

C

H

G

A'

B'

E

F

A

B

Figure 4.34 Saccheri quadrilaterals with congruent summits and summit angles. Definition. The defect of a triangle is the difference between π and its angle sum. Theorem 4.4.4. Triangles with the same defect have the same area. Proof. Case 1. Suppose that triangles △ABC and △A′ B ′ C ′ have the same defect and, further, they have a pair of congruent corresponding sides, BC ∼ = B ′ C ′ . We construct the associated Sac′ ′ ′ ′ cheri quadrilaterals, G H C B and G H C B , using BC and B ′ C ′ as the summits (Figure 4.35). From Example 1 each triangle is equivalent to its Saccheri quadrilateral. Further the angle sums of the summit angles of the Saccheri quadrilaterals equal the angle sum of the corresponding triangles by Theorem 4.3.3. These sums are equal, so the summit angles of the Saccheri quadrilaterals must be congruent. For now we are assuming that their summits are congruent. Thus by A

G

A'

F'

D

F

E

B

D' G'

B'

H

C

E'

H'

C'

Figure 4.35 Triangles with a common side and defect.

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4.4 Area, Distance, and Designs

Theorem 4.4.3 the Saccheri quadrilaterals are congruent and equivalent. Then Theorem 4.4.2 shows the two triangles are equivalent and so have the same area. Case 2. For the general case, we suppose only that △ABC and △A′ B ′ C ′ have the same defect. We construct a third triangle with one side congruent to one side of the first triangle and another side congruent to one side of the other triangle and with the same defect as the given triangles. Then we use case 1 twice to conclude that △ABC and △A′ B ′ C ′ are both equivalent to the third triangle. Hence we need only to construct it and prove that it satisfies the needed properties (Figure 4.36). Let B ′ C ′ be the longest of the six sides of the two triangles and let X ′ ←→ be the midpoint of B ′ C ′ . There is a point X on line G H , which includes the base of the Saccheri quadrilateral G H , such that XC ∼ = X ′ C ′ . (We can find X because EC is no longer than X ′ C ′ .) ← → Construct P on XC with X the midpoint of P and C. Then △P BC is the candidate for the third triangle. P

Y

A

Z GD

E

H

F X

B

A'

B'

C

X' C'

Figure 4.36 Triangles with the same defect. ←→ Next, we show that the intersection Z of P B with G H is the midpoint of P B so that △P BC has G H C B for its Saccheri quadrilateral. Let Y be the point where the perpendicular from P ←→ meets G H . Then △PY X ∼ = C H , which is congruent to = △C H X by AAS, implying that PY ∼ ∼ BG. Thus △PY Z = △BG Z , by AAS, and Z is indeed the midpoint of P B. Finally, as △P BC has the same Saccheri quadrilateral as △ABC, it has the same defect. We can now use case 1 twice. ! Theorem 4.4.5 (Restatement of Theorem 4.1.1). There is a real number k such that, for every triangle △ABC, its area is k(Defect of △ABC). Proof. The defect and the area are functions of the triangle. First, Exercise 4.3.13 shows the defect function to be additive: for D between B and C, the defects of △AB D and △AC D add to the defect of △ABC. Postulate 19 shows that the area function also is additive. Both functions are continuous. A result from calculus states that a continuous function f that satisfies

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Non-Euclidean Geometries

f (x + y) = f (x) + f (y) must be of the form f (x) = cx, for some constant c. (See Stein [15, 54–55].) Both the defect function and the area function for triangles have this form, so one is a constant multiple of the other: the area of △ABC is k · (Defect of △ABC). ! Theorem 4.4.6. The area of a convex polygon is proportional to its defect. Proof. See Exercise 4.4.7(b). ! The tie between the area of a polygon and its defect leads to curious possibilities. In Euclidean geometry, all equilateral triangles have 60◦ angles, regardless of the length of their sides. In hyperbolic geometry, however, the angles of an equilateral triangle, though congruent, are less than 60◦ . As the sides of the triangle lengthen continuously, the angle measure decreases continuously, by Theorem 4.1.1 (Figure 4.37). Thus for some length we obtain an equilateral triangle whose angles measure exactly 45◦ , so we can fit eight equilateral triangles around a point. We can extend the pattern to cover the entire plane, as partially illustrated in Figure 4.38 for the Poincar´e model. We can create a corresponding pattern with seven or more equilateral triangles around a point. Similarly, we can create patterns with five (or even more) hyperbolic squares, that is, a figure with four congruent sides and four congruent angles of 360◦ /5 = 72◦ .

Figure 4.37 Equilateral triangles with differing angles.

Figure 4.38 Part of a hyperbolic tessellation with triangles.

With the help of the geometer H. S. M. Coxeter, the Dutch artist M. C. Escher created several imaginative patterns, building on basic hyperbolic designs like that shown in Figure 4.38. More recently, Douglas Dunham [4] has used geometry and computer graphics to create more varied

4.4 Area, Distance, and Designs

181

Figure 4.39 A hyperbolic tessellation with angels and devils.

Figure 4.40 A hyperbolic tessellation with lizards. patterns relatively quickly (Figures 4.39 and 4.40). From one copy of the repeating motif the computer constructs congruent copies to fill the plane. Of course, the computer needs to be programmed to use a model of hyperbolic geometry instead of Euclidean geometry. Inversions, which we discuss in Section 5.6, provide one key to drawing congruent shapes for the Poincar´e

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model. We also need to compute distances between points in the model. Let A and B be hyperbolic points. The hyperbolic line on them intersects the boundary of the Poincar´e model in the two omega points of that line, say - and #. Poincar´e found the hyperbolic distance between A and B in terms of the Euclidean distances among the four points A, B, -, and #. (See Figure 4.41.) The same formula gives distances in the Klein model. (See Figure 4.42.) The formula in the definition below involves the cross-ratio, discussed in Chapter 7. (We drop the constant usually included in the formula since it doesn’t affect our considerations. See Hilbert and Cohn-Vossen [7, 256].) Σ

Σ

B

B A

∏

Figure 4.41 Poincar´e model.

A

∏

Figure 4.42 Klein model.

Definition. In the Poincar´e or Klein models, the hyperbolic distance between A and B is 0 & '0 0 A" A/ 00 d H ( A, B) = 00log ÷ , B" B/ 0

where X Y is the Euclidean distance from X to Y and " and / are the two omega points of the line AB.

Example 2. Verify that neighboring points Pi in Figure 4.43 are equally spaced along the x-axis in either the Poincar´e or Klein model. The x-coordinates of the points are P0 = 0, P1 = 13 , , P5 = 31 , P−i = −Pi , # = −1, and - = 1. P2 = 35 , P3 = 79 , P4 = 15 17 33 Solution. The Euclidean distances between the points are the differences of their x-coordinates. Then (P0 #/P1 #) ÷ (P0 -/P1 -) = (1/(4/3)) ÷ (1/(2/3)) = 1/2. Similarly, (P1 #/P2 #) ÷ (P1 -/P2 -) = ((4/3)/(8/5)) ÷ ((2/3)/(2/5)) = 1/2. All the corresponding quotients equal 1/2 or 2. In turn, the absolute values of their logarithms are all the same. A diameter is a line in both the Poincar´e and Klein models. ♦

P-4 P-2 Σ

P-5P-3 P-1

P0

P2 P1

P4

P3P5

∏

Figure 4.43 Equally spaced points.

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4.4 Area, Distance, and Designs

4.4.1 Exercises for Section 4.4 4.4.2 Verify Example 1 for cases 2 and 3 of Theorem 4.3.3. 4.4.3. Prove Theorem 4.4.1 using induction on the number of sets. 4.4.4. Complete the proof of Theorem 4.4.3 as follows. (a) (b) (c) (d) (e) (f)

Prove that △A′ D P ∼ = △E H Q. Prove that △A′ P B ′ ∼ = △E Q F. Prove that A′ B ′ C D is a Saccheri quadrilateral. Prove that AB B ′ A′ has four right angles, a contradiction. From BC ∼ = F G prove that the bases, AB and E F, must be congruent. Use the definition of congruent quadrilaterals in Exercise 1.2.24 to show that ABC D ∼ = E FGH.

*4.4.5. Prove that there is a real number K such that the area of every triangle is less than K . (The smallest such K is, in effect, the area of a “triple omega triangle,” made from three mutually sensed parallel lines with three omega points for vertices.) *4.4.6. Unlike Euclidean triangles, hyperbolic triangles with the same height and congruent bases don’t necessarily have the same area. In Figure 4.44 what does Theorem 4.1.1 tell us about the area of △AB Bi as i → ∞? What happens to the area of △ABi Bi+1 as i → ∞? Explain. A

B1 B2 B3 B4 B5

Figure 4.44 Triangles with congruent bases. 4.4.7. Recall that the angle sum of a convex Euclidean polygon with n sides is (n − 2)π or (n − 2)180◦ . (a) Prove by induction that the angle sum of a convex hyperbolic polygon with n sides is less than (n − 2)π. (b) The defect of a convex hyperbolic polygon with n sides is the difference between (n − 2)π and its angle sum. Prove that the area of the polygon is proportional to its defect. *4.4.8. Generalize Exercise 4.4.5 to show polygons with n sides have a largest area. Find the relationship between the least upper bound K n of the areas of polygons with n sides and K = K 3 , the least upper bound for the areas of triangles. 4.4.9. We can construct the inner circle of equilateral triangles shown in Figure 4.38 with the help of Figure 4.45. (a) Construct the unit circle, and the diameters on the x-axis, the y-axis, and y = ±x.

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Non-Euclidean Geometries

(b) The remaining side of each of the eight triangles is an arc of a circle orthogonal to the unit circle. Explain why the circles all have their centers on the lines forming ◦ angles of 22 12 + k · 45◦ with the x-axis. Explain why they all have their centers the same distance, x = OC, from the origin and have the same, as yet unknown, radius r . (c) Explain why, in Figure 4.45, O A and O B equal r . (d) Use the Poincar´e model to explain why, in Figure 4.45, x 2 = 1 + r 2 . (e) The law of cosines gives a second equation in x and r , namely, x 2 = r 2 + r 2 − ◦ 2r 2 cos 22 12 . Find x and r . (f) Finish constructing the eight equilateral triangles from part (a). A

r x

O

C

450 r B

Figure 4.45 Constructing equilateral triangles with 45◦ angles. 4.4.10. (a) Verify that C : (x − 53 )2 + y 2 = ( 34 )2 is orthogonal to the unit circle. (b) Find the intersections " and / of C with the unit circle. Verify that P = ( 13 , 0), Q = √

√

( 21 , 615 ), and R = ( 25 , 1539 ) are on C. Find d H (P, Q), d H (Q, R), and d H (P, R). Why would we expect the sum of the two smaller distances to equal the larger distance? Verify that it does.

4.4.11. Show the following properties for d H in the Klein model. (a) For all points A and B, d H ( A, B) = d H (B, A) ≥ 0. (b) For all points A and B, d H ( A, B) = 0 if and only if A = B. (c) For all collinear points A, B, and C, if B is between A and C, then d H ( A, B) + d H (B, C) = d H ( A, C). Hint: Assume, without loss of generality, that the order of the points is /, A, B, C, ". What does that imply about the sign of the logarithms? 4.4.12. (a) Find the general pattern for the x-coordinates of the points Pi in Example 2. (b) Use part (a) to show that d H (Pi , Pi+1 ) = d H (P0 , P1 ). 4.4.13. Define one khayyam, a made up unit of hyperbolic area, to be the area of a Saccheri quadrilateral with base and sides of length 1. (a) Decide whether a Saccheri quadrilateral with sides of length 1 and base of length 2 has an area of greater than or less than 2 khayyams. Prove your answer. (See Figure 4.46.) 1

1 1

1

1 2

Figure 4.46 Saccheri quadrilaterals.

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4.5 Spherical and Single Elliptic Geometries

(b) Redo part (a) with a Saccheri quadrilateral with sides of length 2 and base of length 1. Make an appropriate figure. (c) Construct or describe how to construct a quadrilateral with area of exactly 2 khayyams.

4.5 Spherical and Single Elliptic Geometries In one sense, mathematicians have studied the geometry of the sphere for millennia. However, before Bernhard Riemann in 1854 no one had thought of spherical geometry as a separate geometry, but only as properties of a Euclidean figure. In spherical geometry two lines (interpreted as great circles) always intersect in two points. (See Section 1.5.) Felix Klein in 1874 modified spherical geometry into a geometry matching the familiar notion of Euclidean and hyperbolic geometries, where two points determine a line. One common model identifies opposite points on the sphere as the same point and studies this “collapsed” geometry, which Klein called single elliptic geometry. Thus the characteristic axiom of single elliptic geometry is that two distinct lines intersect in exactly one point. (Klein called spherical geometry double elliptic geometry because two lines intersect in two points.) Spherical and single elliptic geometries share many theorems in common, such as the angle sum of a triangle is greater than π. Single elliptic geometry possesses some unusual features worth noting. We can represent single elliptic geometry as the half of a sphere facing us so long as we remember that a line (or curve) that leaves the part facing us immediately reappears directly opposite because opposite points are identified. A line in either geometry has many of the same properties as a circle in Euclidean geometry. First, we can’t determine which points are “between” two points because there are two ways to go along a line from one point to another. Instead, we can use two points to “separate” two other points (Figure 4.47). Also, the total length of a line is finite. A single elliptic line has another, more unusual property: it doesn’t separate the whole geometry into two parts, unlike lines in Euclidean, hyperbolic, and spherical geometries. Figure 4.48 indicates how to go along a path on a line connecting the points P and Q not on a line k so that the path does not cross k. More formally, in single elliptic geometry the separation axiom fails, even though Pasch’s axiom holds. The proof in Section 1.2 showing the equivalence of the two axioms relies on one of Hilbert’s betweenness axioms (II-2), which also fails in single elliptic geometry.

A C

B D

Figure 4.47 Points A and B separate C and D.

P Q

k

Figure 4.48 Line k doesn’t separate P and Q in single elliptic geometry.

In certain ways, Euclidean geometry is intermediate between spherical and single elliptic geometries on the one hand and hyperbolic geometry on the other hand. The angle sum of a

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Non-Euclidean Geometries

Euclidean triangle always adds to π. As we know in hyperbolic geometry, the sum falls short of π in proportion to the area of the triangle. In spherical and single elliptic geometries, the sum is always more than π and the excess is proportional to the area of a triangle. (Theorem 1.5.3 shows this condition for Euclidean spheres.) Triangles can have three obtuse angles, so the sum can approach 3π . In our development of hyperbolic geometry we assumed that Euclid’s first 28 propositions hold, for they used only Euclid’s first four postulates, but not the fifth postulate. The first fifteen, including two of the triangle congruence theorems (SAS and SSS), continue to hold in spherical and single elliptic geometries. However, many of the propositions after I-15, including AAS, do not hold, even though they do not depend on the fifth postulate. Exercise 4.5.1.∗ Show that AAS fails. That is, find spherical triangles △ABC and △D E F so that ∠ABC ∼ = D E, but the two triangles are not = ∠D E F, ∠BC A ∼ = ∠E F D, and AB ∼ congruent. Figure 4.49 illustrates Euclid’s approach to showing, as I-16 states, that in a triangle an exterior angle, such as ∠BC D, is larger than either of the other two interior angles, ∠ABC and ∠B AC. From the midpoint E of BC, Euclid extended AE to F so that E F ∼ = E A. Then by SAS △EC F ∼ = △E B A. He then concluded that ∠BC D is larger than ∠EC F, which is congruent to the interior angle ∠E B A. Figure 4.49 supports this conclusion, but the similar situation shown in Figure 4.50 for single elliptic geometry reveals that the conclusion depends on implicit ← → assumptions. In Figure 4.50, the part of AE that looks like segment AE covers more than half ← → the length of the line. Hence the corresponding part of E F overlaps the apparent segment. Euclid implicitly assumed that lines extend infinitely in each direction. Postulate 2 only says, “to produce a finite straight line continuously in a straight line.” The overlapping “segments” AE and E F in Figure 4.50 satisfy the letter and, within reason, the spirit of postulate 2. Nevertheless, I-16 is false here because ∠BC D can be smaller than ∠EC F. B B

F

A

E A

C

E F C

D

D

Figure 4.49 Euclidean diagram for I-16.

Figure 4.50 Corresponding single elliptic diagram.

Exercise 4.5.2. Draw the figure in spherical geometry corresponding to the situation depicted in Figure 4.50. We partially develop single elliptic geometry as we did hyperbolic geometry. (For a more thorough development, see Gans [5].) Theorems 4.2.1–4.2.8 do not relate to geometries where two lines always intersect. Therefore we start with Theorem 4.3.1 concerning Saccheri quadrilaterals, which we repeat here.

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4.5 Spherical and Single Elliptic Geometries

Theorem 4.5.1. The summit angles of a Saccheri quadrilateral are congruent. The base and summit are perpendicular to the line on their midpoints.

Exercise 4.5.3. Verify that the proof of Theorem 4.3.1 holds in spherical and single elliptic geometry. In hyperbolic geometry, Theorem 4.3.2 showed that the summit angles of a Saccheri quadrilateral were acute. Theorem 4.5.3 shows they are now obtuse. Theorem 4.5.2 provides a key step. Exercise 4.5.4.∗ In Euclidean geometry what can you say about the summit angles of a Saccheri quadrilateral? Theorem 4.5.2. In single elliptic geometry, all lines perpendicular to a given line intersect in one point. Proof. From Theorem 4.5.1 we know that, in a Saccheri quadrilateral ABC D, E F is perpendic← → ← → ular to both AB and C D. By the characteristic property of single elliptic geometry, AB and C D intersect in a unique point, say, P. We claim that d( A, P) = d(B, P) and d(C, P) = d(D, P). (In a model, the distance is along the shortest path. As shown in Figure 4.51, the shortest path from B to P goes “around behind” because we can “jump” from the right edge to the left edge.)

D

F

C P′

P A

E

B

Figure 4.51 Intersection of common perpendiculars. ← → For a contradiction, construct P ′ on AB so that A P ∼ = B P ′ and P and B separate A and P ′ . ′ (Intuitively P is on the other side of B, as depicted in Figure 4.51.) Then △AD P ∼ = △BC P ′ by SAS. Thus ∠BC P ′ , which is congruent to ∠AD P, is supplementary to ∠BC D. Then by ←→ ← → ← → Euclid I-14, C P ′ is the same line as C D. This result would give two points of intersection of AB ← → with C D, which is a contradiction. So P ′ = P, showing that d(A, P) = d(B, P) and similarly that d(C, P) = d(D, P). Furthermore, the distance from E to P is the same by way of A as by way of B, which is the maximum distance two points can be separated. The same holds for the distance from F to P. Since the angles at E and F are right angles, △E F P is isosceles and so d(E, P) = d(F, P) by Euclid I-6. Moreover, d(E, P) and d(F, P) do not depend on the length

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← → of E F. That finishes the proof. We have also shown that for any point G on E F the segment P G has the same length as P E. ! Theorem 4.5.3. In single elliptic geometry, the summit angles of a Saccheri quadrilateral are obtuse. ← → Proof. In a Saccheri quadrilateral ABC D construct DG perpendicular to AD and let Q be ← → ← → the intersection of DG and AB (Figure 4.52). Then d( A, P) is less than or equal to d(A, Q) because Q is the point furthest from A by Theorem 4.5.2. If the distances were equal, we’d have P = Q. In turn, we’d have A = E and so A = B, giving a trivial Saccheri quadrilateral. ← → Hence d( A, P) < d( A, Q). Thus D P enters △Q D A at D, and so ∠P D A must be smaller than ∠Q D A, which is a right angle. Hence ∠ADC, the summit angle and supplementary to ∠P D A, is obtuse. !

G

D

C

F

Q P

A

E

B

Figure 4.52 Saccheri quadrilateral in single elliptic geometry.

Figure 4.53 Nonorientable nature of single elliptic geometry.

Once we have Theorem 4.5.3, the development of single elliptic geometry follows the lines of hyperbolic geometry. (See Project 5.) Single elliptic geometry has another property worth discussing. Consider what happens to an asymmetrical object in single elliptic geometry as it moves along the path of a line (Figure 4.53). It actually has two opposite representations. As the front one slides out of view, it is replaced by the rear one, which is fundamentally different. The rear representation won’t match the front representation because it has an opposite orientation. That is, the representations are mirror images. Now matter how you twist and turn two objects with opposite orientations in Euclidean geometry, as long as you stay in the plane you cannot make them coincide. In single elliptic geometry the representations are the same, so we have the curious property that moving an object around in it can switch its orientation. Single elliptic geometry is topologically a nonorientable surface, whereas spherical, Euclidean, and hyperbolic geometries are orientable. On a nonorientable surface there is no consistent way to define a clockwise direction.

4.5 Spherical and Single Elliptic Geometries

189

4.5.1 Exercises for Section 4.5 *4.5.5. Call a triangle with two right angles a doubly right triangle. (a) Prove in single elliptic geometry that two doubly right triangles are congruent if their sides between the right angles are congruent. (b) Prove in single elliptic geometry that two doubly right triangles are congruent if their third angles are congruent. 4.5.6. Euclid I-26 has two congruence theorems: AAS and ASA. *(a) Explain why Theorem 4.5.2 shows that AAS is not a triangle congruence theorem in single elliptic and spherical geometries. (b) Prove that ASA is a triangle congruence theorem in single elliptic and spherical geometries, using SAS and a proof by contradiction. *4.5.7. Do Exercise 4.5.5 in spherical geometry. 4.5.8. Reformulate and prove Theorem 4.5.2 for spherical geometry. *4.5.9. Prove Theorem 4.5.3 in spherical geometry. 4.5.10. Assume Theorem 1.5.3, the analog to Theorem 4.1.1 in single elliptic and spherical geometries: The area of a triangle is proportional to the excess of the angle sum of a triangle over π. Also assume that the measure of each angle of a triangle is less than π . *(a) What would be the triangle if all three angles had a measure π ? What would be its area if the sphere has radius r ? (b) On a sphere consider a triangle △N S A, where N is the North pole, S is the South pole, and A is any other point. Why is the side not including A not determined? What can you say about the area of the region? Why can’t this situation occur in single elliptic geometry?

4.5.2 Georg Friedrich Bernhard Riemann Bernhard Riemann (1826–1866) showed obvious mathematical ability early in his life. Nevertheless, he started studying theology at age 20 at the request of his father, a Lutheran minister. Within a year, though, he had turned to mathematics and finished his Ph.D. at age 25 under Gauss’s direction. He became a professor at G¨ottingen University in Germany when he was 27 and remained there until his health deteriorated. Riemann suffered from tuberculosis for nearly the last four years of his life, working when he was well enough and dying before the age of 40. Although Riemann is also remembered for the Riemann sums and integrals that appear in calculus texts, his major work focused on physics and more advanced mathematics. Riemann made profound contributions to complex analysis and what later became topology and differential geometry. He blended deep physical and geometric intuition with insightful arguments. Some of his contemporaries criticized his proofs for a lack of rigor. However, his approaches, conjectures, and results have shaped all the areas he investigated. The Riemann hypothesis, a technical statement about complex functions, remains unproved. Mathematicians have used it to give provisional proofs of a number of interesting results. Such provisional proofs led Hilbert

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Non-Euclidean Geometries

to list it as one of the 23 most important problems facing mathematics in 1900. One hundred years later, the Clay Institute offered $1,000,000 for a proof or disproof of it, one of the seven Millennium Prize Problems. Barely ten years after mathematicians started exploring n-dimensional Euclidean geometry, Riemann’s introductory lecture as a professor in 1854 developed a concept of space far more general. His vision of what we now call differential geometry included Euclidean, hyperbolic, spherical, and elliptic geometries, in any number of dimensions, as special cases. He showed how to base the concept of space on theoretical or physical measurements of distances. How those measurements of a space differed from the corresponding Euclidean measurements based on the Pythagorean theorem described how that space was curved. Sixty years after Riemann’s death his work became the foundation for Einstein’s general theory of relativity, which related those measurements to the effects of gravity.

4.5.3 Projects for Chapter 4 1. Construct an empirical model of hyperbolic geometry, using two regular mirrors and one convex mirror, as illustrated in Figure 4.54. To make the curved mirror, lay a flexible reflective surface, such as mylar, on the outside of a cylindrical frame. Place a design inside the mirrors. You should see replicas of your design as though it were a part of hyperbolic geometry. As the angles at A, B, and C change, you will get different numbers of copies of your design around the points. B

A C

Figure 4.54 Mirrors to model hyperbolic geometry. 2. Use some software representing the Poincar´e model of hyperbolic geometry to investigate various properties. For example, NonEuclid is available for free on the internet. 3. Use some software for Euclidean geometry (for example, Geometer’s Sketchpad or Geogebra) to create the Poincar´e model or half plane model of hyperbolic geometry. Investigate how the angle sum of a hyperbolic triangle changes as the size of the triangle changes. 4. Use the interior of the unit circle x 2 + y 2 = 1 as the Poincar´e model of hyperbolic geometry and construct a hyperbolic square as follows. (a) Suppose one of the orthogonal circles has center (a, 0) and radius r . Find r in terms of a and give its equation. (See Figure 4.55.) (b) Find the equations of the three other circles shown as dashed arcs in Figure 4.55. (c) Find the coordinates (x, y) of the corner of the hyperbolic square shown in Figure 4.55.

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4.5 Spherical and Single Elliptic Geometries

(d) Explain why the hyperbolic quadrilateral outlined in bold in Figure 4.55 deserves the name hyperbolic square. (e) Construct on graph paper several hyperbolic squares using different values of a. What happens to the angles at the corners as a increases? (f) Find the minimum value of a so that the circles with dashed arcs intersect. What is the corresponding hyperbolic figure?

1

r (x,y)

(0,0)

(a,0)

Figure 4.55 A hyperbolic square. 5. Investigate ultraparallel lines in hyperbolic geometry. (Cederberg [2, 301] proves that two ultraparallel lines have a common perpendicular.) 6. Investigate limiting curves or horocycles, the points at a constant distance from a hyperbolic line. The sets act like circles of infinite radius. (See Smart [14, 390].) 7. In Euclidean geometry develop and prove the analogs, if any, to Theorems 4.3.3–4.4.5 using any of Euclid’s propositions in Appendix A. How do they differ from the nonEuclidean cases? 8. In spherical geometry develop and prove the analogs to Theorems 4.3.3–4.4.5 using Exercise 4.5.8, excepting Theorem 4.3.5. 9. In single elliptic geometry develop and prove the analogs to Theorems 4.3.3–4.4.5 using Theorem 4.5.3, excepting Theorem 4.3.5. 10. Investigate patterns in spherical geometry. Find all “regular patterns,” or those where the same number of the same regular spherical polygons fit around every point. How do they compare with polyhedra and regular patterns in Euclidean and hyperbolic geometries? 11. In spherical geometry we could let a side of a triangle be longer than half of the circumference of the sphere and, correspondingly, an angle could be greater than π (180◦ ), as illustrated in Figure 4.56. How does the formula for the area of a spherical triangle in Theorem 1.5.3 extend to triangles with angles greater than π ? Explain and prove your extension. Hint: A great circle divides the surface of a sphere into two equal areas. 12. Investigate the historical development of non-Euclidean geometry. (See Bonola [1], Kline [8, Chapter 36], and Richards [11].) 13. Investigate other non-Euclidean geometries. (See Yaglom [17].) 14. Investigate the concept of space. (See, for example, Rucker [12] and Weeks [16].)

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Non-Euclidean Geometries

∠BAC A B

C

Figure 4.56 A spherical “triangle” with an angle greater than 180◦ . 15. Write an essay comparing Euclidean and hyperbolic geometry. Address your essay to someone at the level of calculus. 16. Write an essay comparing how models and synthetic development have helped you understand hyperbolic geometry. Discuss how convincing the proofs in hyperbolic geometry are to you. 17. Write an essay discussing the difference between spherical geometry as a separate geometry and the geometry of the sphere as a part of Euclidean geometry. 18. Write an essay discussing whether you think that mathematics is discovered or invented. Use your understanding of non-Euclidean geometries as an example. (See for example, Kline [8, 1032–1038].)

4.5.4 Suggested Readings 1. Bonola, R., Non-Euclidean Geometry., Mineola, NY: Dover, 1955. 2. Cederberg, J., A Course in Modern Geometries, New York: Springer-Verlag, 1989. 3. do Carmo, M., Differential Geometry of Curves and Surfaces, Englewood Cliffs, NJ: Prentice-Hall, 1976. 4. Dunham, D., Hyperbolic symmetry, Comp. and Maths. with Appl., 1986, 12(1 and 2): 139-153. 5. Gans, D., An Introduction to Non-Euclidean Geometry, New York: Academic Press, 1973. 6. Henderson, D. and D. Taimina, Experiencing Geometry, 3rd ed., Upper Saddle River, NJ: Prentice Hall, 2005. 7. Hilbert, D. and S. Cohn-Vossen, Geometry and the Imagination, New York: Chelsea, 1952. 8. Kline, M., Mathematical Thought from Ancient to Modern Times, New York: Oxford University Press, 1972. 9. McCleary, J., Geometry from a Differentiable Viewpoint, Port Chester, NY: Cambridge University Press, 1994. 10. Moise, E., Elementary Geometry from an Advanced Standpoint, Reading, MA: AddisonWesley, 1974. 11. Richards, J., Mathematical Visions, San Diego: Academic Press, 1988. 12. Rucker, R., Geometry, Relativity and the Fourth Dimension, Mineola, NY: Dover, 1977. 13. Sacks, O., An Anthropologist on Mars, New York: Knopf, 1995.

4.5 Spherical and Single Elliptic Geometries

14. 15. 16. 17.

193

Smart, J.. Modern Geometries, 5th ed., Monterey, CA: Brooks/Cole, 1988. Stein, S., Calculus and Analytic Geometry, New York: McGraw-Hill, 1987. Weeks, J., The Shape of Space, New York: Dekker, 1985. Yaglom, I., A Simple Non-Euclidean Geometry and Its Physical Basis, (transl. A. Shenitzer), New York: Springer-Verlag, 1979.

5

Transformational Geometry

Figure 5.0 Computers store and reproduce intricate shapes such as the fractal pictured, relying on only a few numbers to encode all the information of the shape. To do so they depend on geometric transformations to represent the shapes, applying transformations thousands of times to create an image. Computer-aided design (CAD) depends on transformations to present different views of an object. Transformational geometry underlies the applications and many other aspects of mathematics. Geometry is the study of those properties of a set which are preserved under a group of transformations on that set. —Felix Klein A mathematician, like a painter or a poet, is a maker of patterns. If his patterns are more permanent than theirs, it is because they are made with ideas. —G. H. Hardy

5.1 Overview and History Moving geometric figures around is an ancient and natural approach to geometry. However, the Greek emphasis on synthetic geometry and constructions and, later, the development of analytic geometry overshadowed transformational thinking. The study of polynomials and their roots in the early nineteenth century led to algebraic transformations and groups, a vital area of abstract algebra. At the same time, Augustus M¨obius began studying geometric transformations.

195

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Transformational Geometry

In the last third of the nineteenth century, Felix Klein and Sophus Lie showed the central importance of groups and transformations for geometry. This approach enabled Klein and others to unify geometry at a time when new and different geometries seemed to be splitting the ancient discipline into competing theories. Transformations remained the dominant approach to geometry for fifty years. Transformations underlie the modern understanding of symmetry, which is essential in physics and chemistry as well as mathematics. (See Chapter 6.) Early in the twentieth century physicists realized the power of transformations, starting with Einstein’s theory of relativity and then with quantum mechanics. Although many geometric topics now transcend transformational geometry, it remains vital for understanding geometry and other fields. More recently, the Common Core State Standards call for high school geometry courses to place transformations in a central role for understanding congruence, similarity, and symmetry. We discuss the connection between congruence and isometries in Section 5.2, while postponing similarity and similarities until Section 5.4. We first investigate isometries—transformations preserving distance. The proofs in Sections 5.1 and 5.2 are based on a synthetic approach, although we freely use coordinates in examples. The inclusion of linear algebra starting in Section 5.3 enables us to extend the types of transformations and to work in higher dimensions. Appendix D provides a quick overview of the aspects of linear algebra needed for this chapter and Chapter 7. Finally Section 5.6 discusses inversions, a topic rich in geometric ideas and connected to complex analysis. We will use small Greek letters for transformations to distinguish them from the points and sets being transformed. Example 1. The top and bottom of the design depicted in Figure 5.1 are mirror images. Switching points with their mirror images is one type of transformation, a mirror reflection. Like all Euclidean transformations, it acts on the entire Euclidean plane R2 . Set the mirror on the x-axis and call the reflection µ. We can describe it algebraically by µ(x, y) = (x, −y). The points on the x-axis remain fixed by µ: µ(x, 0) = (x, 0). If we perform the transformation twice, a point’s image is mapped back to the original point. That is, µ(µ(x, y)) = µ(x, −y) = (x, y). We say that the shape in Figure 5.1 is stable under µ because µ maps it to itself. ♦

Figure 5.1 A Mexican design with mirror symmetry. No amount of turning and twisting can turn a left hand into a right hand, even though they mirror each other. This condition is caused by the different orientation of an object and its mirror image. An (asymmetrical) three-dimensional object and its mirror image can’t be superimposed on each other, even though the object and its image are congruent. We particularly

5.1 Overview and History

197

notice this change of orientation if we try to read the mirror image of a book. Rotations do not change orientation. The same effect happens in two dimensions. No rotation around a point in two dimensions can replicate a two-dimensional mirror reflection. A two-dimensional mirror reflection switches the orientation of points in the plane. However, the situation is more complicated when we mix dimensions. A two-dimensional mirror reflection, as described in Example 1, can be accomplished by a three-dimensional rotation of 180◦ of the space around the axis of fixed points. Definitions. A transformation τ on a set S is a function from S to itself that is one-to-one and onto. That is, (i) (τ is a function)] for every point P of S there is a unique point Q that is the image of P under τ : τ (P) = Q, and (ii) (τ is one-to-one and onto) for every point Q of S there is a unique point P for which Q is the image of P under τ . Two transformations α and β of a set S are equal if and only if for all P ∈ S, α(P) = β(P). A point P is a fixed point of the transformation τ if and only if τ (P) = P. The image of a subset T of S for the transformation τ is τ [T ] = {τ (P) : P ∈ T }. A subset T of points in S is stable under the transformation τ if and only if τ [T ] = T , even if individual points of T move to other points in T . (Stable sets are often called invariant sets.) Note that a transformation is a mapping of the entire space. However, we often focus on the transformation’s effect on a portion of the space. In particular, the fixed points and stable sets of a transformation tell us important information about the transformation. For example, in Section 5.2 we match the type of an isometry with its fixed points and stable lines. Symmetry, the topic of Chapter 6, involves the study of transformations and their stable sets more deeply. In dynamical systems, a relatively new field of mathematics with important applications, fixed points and stable sets help investigate a broader family of functions than we present here. (See Abraham and Shaw [1].) Remark. We don’t need to separate the properties of one-to-one and onto here, although the difference is often important in mathematics. (See Sibley [10, 124 ff] for a discussion of the concepts.) Exercise 5.1.1.∗ In Example 1 verify that vertical lines are stable under µ even though points on them move. Verify that the x-axis is the only horizontal stable line under µ, and no line with a nonzero slope is stable. Verify the image of the parabola y = x 2 + 1 is the parabola y = −x 2 − 1. What circles are stable under µ? Example 2. On the Euclidean plane R2 define ρ(x, y) = (y + 2, 2 − x) (Figure 5.2). Show that ρ is a transformation. (Later we will see ρ as a rotation.) Solution. Each point (x, y) has a unique image (y + 2, 2 − x), so ρ is a function. To show that it is one-to-one and onto, we start with a point (u, v) and show that there is a unique point (x, y) that ρ sends to (u, v). When we solve u = y + 2 and v = 2 − x, we find the solution: (x, y) = (2 − v, u − 2). Because there is only one solution, ρ is by definition one-to-one and onto. ♦ Exercise 5.1.2.∗ In Example 2 verify that ρ fixes the point (2, 0). Why does ρ have no other fixed point? Assume that ρ is a rotation of 270◦ around (2, 0). Why are circles with center (2, 0)

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4 2 -4

-2

2 -2 -4

Figure 5.2 The images of some points under ρ. stable under ρ? Find the images of several points on the line y = x to verify that its image is y = −x + 4. Is any line stable under ρ? Does ρ switch orientation? Example 3. On R2 define ψ(x, y) = (e x , sin y). Although ψ is a function, it is neither one-toone nor onto. No matter what value of x we choose, e x is positive, so ψ cannot be onto all of R2 . Furthermore, the sine function is periodic, so two different points can map to the same point, such as ψ(0, 0) = (1, 0) = ψ(0, π ), demonstrating that ψ isn’t one-to-one. ♦ Example 4. The biologist D’Arcy Thompson used the idea of transformations in his study of comparative anatomy. Figure 5.3 reproduces one of his illustrations, depicting how he compared features of related species. The actual transformations he used go beyond the level of our study. (See Thompson [12].) ♦ d c b a 0

1

2

3

4

5

Figure 5.3 Comparing skulls of a human, a chimpanzee, and a baboon using transformations distorting coordinates. Definition. For functions f and g from a set S to itself, the composition of g followed by f , f ◦ g is defined by f ◦ g(P) = f (g(P)). Example 5. On R2 let ρ(x, y) = (y + 2, 2 − x) and µ(x, y) = (x, −y). The composition ρ ◦ µ is given by ρ ◦ µ(x, y) = ρ(µ(x, y)) = ρ(x, −y) = (−y + 2, 2 − x) = (2 − y, 2 − x). As in Example 2, we can show ρ ◦ µ is a transformation. ♦ Exercise 5.1.3.∗ Verify in Example 5 that ρ ◦ µ leaves the line y = 2 − x stable. Describe this transformation. Is µ ◦ ρ the same transformation?

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199

Composing functions enables us to build and study a wide variety of functions in calculus, geometry, and other areas of mathematics. Theorem 5.1.1 shows that the composition of two transformations on a set is again a transformation on the set. This property is the first of several general properties about transformations that nineteenth century mathematicians realized were crucial. They form the definition of an algebraic group, a fundamental algebraic idea. In later sections we use the properties and groups to enable us to understand and prove geometric properties. Theorem 5.1.1. If α and β are transformations on S then α ◦ β is a transformation on S. Proof. Let α and β be transformations on S and P ∈ S. For α ◦ β to be a function, there must be a unique R ∈ S such that α ◦ β(P) = R. Now β is a transformation, so there is a unique Q ∈ S such that β(P) = Q. Similarly, there is a unique R ∈ S such that α(Q) = R. Hence R is the unique image of P under α ◦ β. Let V ∈ S. For α ◦ β to be one-to-one and onto, there must be a unique T ∈ S such that α ◦ β(T ) = V . Because α is a transformation, there is a unique U ∈ S such that α(U ) = V . Similarly, there is a unique T ∈ S such that β(T ) = U . This implies that T is the only element of S which α ◦ β takes to V . Thus α ◦ β is a transformation. ! Example 6. For a set S, define the identity transformation by ι(P) = P for P ∈ S. For a transformation α on S, α ◦ ι = α = ι ◦ α. The identity may seem of little importance by itself, but its presence simplifies investigations about transformations, just as the number 0 simplifies addition of numbers. Every point of S is fixed by ι, and every subset of S is stable. ♦ Example 7. On R2 , if we compose ρ(x, y) = (y + 2, 2 − x) and σ (x, y) = (2 − y, x − 2), then ρ ◦ σ (x, y) = ρ(2 − y, x − 2) = ((x − 2) + 2, 2 − (2 − y)) = (x, y). Thus ρ ◦ σ = ι, the identity transformation: ρ undoes what σ did. ♦ Definition. A transformation β is the inverse of a transformation α if and only if α ◦ β = ι and β ◦ α = ι. We write α −1 for the inverse of α. Exercise 5.1.4. In Example 7, verify that σ ◦ ρ is also the identity. In Example 5, verify that ρ ◦ µ is its own inverse. Theorem 5.1.2. Every transformation α on a set S has a unique inverse, α −1 , the transformation satisfying α −1 (Q) = P if and only if α(P) = Q. Proof. Parts (i) and (ii) of the definition of a transformation are closely related. From the relationship α −1 , as defined in the statement of Theorem 5.2.1, is a function because α is oneto-one and onto. Similarly, α −1 is one-to-one and onto because α is a function. Hence α −1 is a transformation. Further, the definition of α −1 guarantees that α ◦ α −1 = ι = α −1 ◦ α. To show uniqueness, we suppose that β also is an inverse of α and show that β = α −1 . Let Q be an element of S. Because α is a transformation, there is a unique P such that α(P) = Q. By the definition of an inverse β(Q) = P = α −1 (Q). As α −1 and β agree everywhere, by definition they are equal. !

200

Transformational Geometry

Klein found that many important geometric properties correspond with sets of transformations that form groups, which are structures of great importance in mathematics. We define transformation groups because they are the only groups we will consider. (Associativity plays a key role in group theory, but it always holds for composition of functions. (See Exercise 5.1.13.) Hence we can omit it from the definition. (See Sibley [10, Chapter 7].)) Definition. A set T of transformations on a set S is a transformation group if and only if the following properties hold. (i) (closure) The composition of two transformations in T is in T. (ii) (identity) The identity transformation is in T. (iii) (inverses) If a transformation is in T, then its inverse is in T. Theorem 5.1.3. The set of all transformations on a set is a transformation group. Proof. See Theorems 5.1.1 and 5.1.2 and Example 6. !

5.1.1 Exercises for Section 5.1 *5.1.5. On R, the real numbers, define α(x) = x 3 and β(x) = 2x − 1. (a) (b) (c) (d)

Show that α and β are transformations. Find α ◦ β and β ◦ α. Show that α ◦ β ̸= β ◦ α. Find the fixed points of α and the fixed point of β. A function f of R has a fixed point at x = w provided that the graph of y = f (x) crosses y = x at x = w. Use this to determine the fixed point of α ◦ β and the fixed point of β ◦ α. (e) Find α −1 and β −1 . Graph α and α −1 together. Repeat for β and β −1 . Describe the relationship between the graph of a transformation on R and the graph of its inverse. (f) Show that a transformation and its inverse have the same fixed points. (g) Find α −1 ◦ β −1 and β −1 ◦ α −1 . Which is the inverse of α ◦ β? Explain your choice. 5.1.6. (a) Explain why the function f (x) = x 2 is not a transformation on R. (b) Show that if we restrict the function f (x) = x 2 to the set [0, 1] = {x : 0 ≤ x ≤ 1}, then it is a transformation. (c) For the restricted function in part (b), find its fixed points and its inverse. *5.1.7. On R2 define θ by (x, y) to θ(x, y) = (4 − x, 6 − y). (a) On a graph show what θ does to various points. Describe how it moves points. (b) Show that θ is a transformation. (c) Find the inverse, the fixed point(s), and the stable line(s) of θ . 5.1.8. (a) For ρ ◦ µ in Example 5 show that the fixed points are the points on the line y = 2 − x. (b) Use part (a) and Exercise 5.1.1 to describe the stable lines of ρ ◦ µ. (c) Find the formula for ρ ◦ ρ ◦ µ and use it to determine its fixed points and stable lines. (d) Repeat part (c) for ρ ◦ ρ ◦ ρ ◦ µ.

201

5.2 Isometries

(e) Describe the relationship among the sets of fixed points of the transformations in parts (a), (c), (d), and µ. (f) Repeat part (c) for µ ◦ ρ.

*5.1.9. For each of the following functions on R2 , illustrate on a graph what it does to various points, show that it is a transformation, and find its fixed point(s) and stable line(s). (a) ω(x, y) = (2 − x, 4 − y) (b) λ(x, y) = (y − 1, x + 1) (c) γ (x, y) = (2x, 2y) (d) ψ(x, y) = ( 12 x + 1, 12 y − 1) 5.1.10. If α and β are transformations on a set S, prove that α −1 ◦ β −1 and β −1 ◦ α −1 are transformations. Which is the inverse of α ◦ β? Prove your answer.

5.1.11. For each of the following functions on R2 , illustrate on a graph what it does to various points, show that it is a transformation, find its inverse, and find its fixed point(s) if any. Describe its stable line(s). (a) α(x, y) = (x + 3, y − 2) (c) γ (x, y) = (y, x + 2)

(b) β(x, y) = (x + y, y)

5.1.12. (a) On R2 define σ by σ (x, y) = ( x2 + 2, 2y − 1). Find the fixed point of σ and call it F. (b) Let P0 be a point in R2 and define the sequence {P0 , P1 , P2 , . . .} by Pn+1 = σ (Pn ). Graph it for several choices of P0 . What happens in each case? (c) Repeat parts (a) and (b) for φ(x, y) = (2x + 2, 2y − 4). (d) Repeat parts (a) and (b) for δ(x, y) = (3 − y, x − 1). The study of dynamical systems involves finding the long-term result of repeated application of a function. The fixed point of σ is called an attracting (or stable) fixed point because σ takes all points closer to it. The fixed point of φ is called a repelling (or unstable) fixed point because φ sends other points farther from it. The transformation δ is said to be periodic of period 4 since four applications of δ give the identity. These terms describe the dynamics of many familiar transformations. (See Abraham and Shaw [1].) (e) The transformation γ (x, y) = (x + 2, −y) doesn’t fit any of the preceding dynamics. Describe its long-term dynamics. 5.1.13. Prove composition is associative. That is, for three transformations α, β, and γ on a set S, show that α ◦ (β ◦ γ ) = (α ◦ β) ◦ γ . Hint: Use the definition for the equality of two functions.

5.2 Isometries Members of the most important family of transformations, isometries, do not change the distance between points as the transformations move them. Isometries are the dynamic counterpart of the Euclidean notion of congruence. (“Iso” and “metric” have Greek roots meaning “equal” and “measure.”)

202

Transformational Geometry

Definition. A transformation σ is an isometry on a set S with a distance function d if and only if for points P and Q in S, d(P, Q) = d(σ (P), σ (Q)). If d is the usual distance on the Euclidean plane, then σ is a Euclidean plane isometry. Example 1. A rotation of the plane is one type of Euclidean isometry and has exactly one fixed point, except for rotations of a multiple of 360◦ . In Figure 5.4 A is the fixed point or center of rotation of the rotation ρ. Most rotations have no stable lines. For the special case of a rotation of 1800 the stable lines are the lines through the center of rotation. For example, in Figure 5.5 ← → the line AB through the center A also goes through ρ(B), the image of B. ♦ C C D

A

B A

B

E

D

Figure 5.5 A rotation of 180◦ .

Figure 5.4 A rotation.

Example 2. As shown in Figure 5.6, σ doubles x-coordinates and halves y-coordinates. The points (1, 2) and (−1, 2) become (2, 1) and (−2, 1), respectively. Since the distance d((1, 2), (−1, 2)) = 2 does not equal d(σ (1, 2), σ (−1, 2)) = 4, it is not an isometry, even though it is a transformation. The only fixed point is (0, 0). The axes are the stable lines. ♦ 4 (-1,2)

(1,2) (2,1)

(-2,1) -4

4 -2 -4

Figure 5.6 a non-isometry. Exercise 5.2.1.∗ Use the definitions of a circle and an isometry to explain why an isometry takes a circle to a circle. Example 3. Mirror reflections are a type of isometry. We say that a mirror reflection µ in the plane is over a line k, which contains the fixed points of µ (Figure 5.7). The stable lines are the line of reflection and lines perpendicular to it. ♦

203

5.2 Isometries

k

Q

P

Figure 5.7 A mirror reflection. Exercise 5.2.2.∗ In Example 3 explain why m is stable provided that m ⊥ k or m = k. Example 4. The isometry τ depicted in Figure 5.8, called a translation, adds 3 to the x-coordinate and 2 to the y-coordinate of a point. A translation has no fixed points. Its stable lines are parallel to the direction of the translation. ♦ P

Q

Figure 5.8 A translation. Definition. A Euclidean plane isometry τ is a translation if and only if, for distinct points P and Q, the points P, Q, τ (Q), and τ (P) form a parallelogram or, as a special case, P, Q, τ (P), −−−−→ and τ (Q) are collinear. (See Figure 5.8.) We call Pτ (P) the vector of the translation τ . A Euclidean plane isometry ρ is a rotation of r ◦ if and only if there is a point O such that ρ(O) = O and, for other P, m(∠P Oρ(P)) = r ◦ , where angle measure is directed, implying that m(∠ρ(P)O P) = −r ◦ . (See Figure 5.4.) A Euclidean plane isometry µ is a mirror reflection over the line k if and only if, for every point P on k, µ(P) = P and for every point Q not on k, k is the perpendicular bisector of the segment Qµ(Q). (See Figure 5.7.) Exercise 5.2.3.∗ Why is the identity both a translation and a rotation?

5.2.1 Classifying Isometries Theorems 5.2.3 and 5.2.5 give important geometric descriptions of isometries, leading to the classification of the possible types in Theorem 5.2.7. Their proofs depend on the algebraic ideas of transformation groups and on geometric properties. You might benefit from first exploring these ideas visually with actual mirrors or using dynamic geometrical software. (See Projects 1-4 at the end of the chapter.) Theorem 5.2.1. The isometries of a set form a transformation group. Proof. For closure, we let α and β be isometries on a set S and show that α ◦ β is an isometry. Let P and Q be points in S. Then d(P, Q) = d(β(P), β(Q)) = d(α(β(P)), α(β(Q))), showing

204

Transformational Geometry

α ◦ β to be an isometry. Next, the identity, which fixes every point, preserves distance and so is an isometry. Finally, for an isometry α we show that its inverse α −1 is also an isometry. For P and Q in S, let α −1 (P) = U and α −1 (Q) = V . We must show that d(P, Q) = d(U, V ). Because α is an isometry and the inverse of α −1 , d(U, V ) = d(α(U ), α(V )) = d(P, Q). Thus the isometries form a transformation group. ! Theorem 5.2.2. A Euclidean plane isometry that fixes three noncollinear points is the identity. Proof. Let α be an isometry, and A, B, and C be three noncollinear points fixed by α, and D be any other point. We must show that α(D) = D. By definition of an isometry, d( A, D) = d(α( A), α(D)), which equals d( A, α(D)) since A is fixed. Thus wherever α(D) is, it is on the circle with center A and radius AD. Similarly, D is on the circle with center B and radius B D and also on the circle with center C and radius C D (Figure 5.9). Because A and B are distinct, the first two circles intersect in at most two points, one of which is D. If D is the only ← → intersection, we are done. But suppose that there is another point, say E. Then AB is the ← → perpendicular bisector of D E by SSS. However, C is not on AB. Thus C cannot be the same distance from D and E. Hence α(D) cannot be E, forcing α(D) = D. !

D

A

C

B

E

Figure 5.9 Locating D by its distances from A, B, and C. Theorem 5.2.2 is a special, easily proven case and so by itself not particularly important. The properties of a group, shown in Theorem 5.2.1, enable us in the next theorem to begin to characterize all Euclidean plane isometries from this special case. The following proof exemplifies the power of using an algebraic approach in geometry. Theorem 5.2.3. A Euclidean plane isometry is determined by what it does to any three noncollinear points. Proof. Let A, B, and C be three noncollinear points, α and β be isometries such that α(A) = β(A), α(B) = β(B), and α(C) = β(C). By Theorem 5.2.1, β −1 ◦ α is an isometry and β −1 ◦ α( A) = A, β −1 ◦ α(B) = B, and β −1 ◦ α(C) = C. By Theorem 5.2.2, β −1 ◦ α is the identity: β −1 ◦ α = ι. When we compose both sides on the left with β, we get α = β. That is, there is just one isometry taking three noncollinear points to their images. !

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5.2 Isometries

Theorem 5.2.4. For two distinct points P and Q in the Euclidean plane, there is exactly one mirror reflection that takes P to Q. Proof. Given distinct points P and Q we can use Euclid I-10 and I-11 to construct a perpendicular bisector. It is unique by SMSG postulates 3 and 12. Then the definition of a mirror reflection gives a unique mirror reflection, taking one point to the other. ! We use algebraic thinking again to build from the special case of one type of isometry in Theorem 5.2.4 to all isometries in Theorem 5.2.5. Theorem 5.2.5 shows that mirror reflections are the basic building blocks of isometries. Theorem 5.2.5. Every Euclidean plane isometry can be written as the composition of at most three mirror reflections. Proof. Let α be a Euclidean plane isometry and A, B, and C be noncollinear points in the plane. By Theorem 5.2.3 we only have to find a composition of mirror reflections that together take A, B, and C to the same images as α does, say, P, Q, and R, respectively (Figure 5.10). We will consider the general case first. Suppose A ̸= P. By Theorem 5.2.4, there is a mirror reflection µ1 such that µ1 ( A) = P. Let µ1 (B) = B ′ and µ1 (C) = C ′ . Again for the general case we suppose B ′ ̸= Q and repeat this process, finding µ2 , mapping B ′ to Q. However, we need to prove that µ2 leaves P fixed. Note that d(P, Q) = d( A, B) = d(P, B ′ ). As in Theorem 5.2.2 by SSS P is on the perpendicular bisector of Q B ′ , which means that µ2 (P) = P. Hence µ2 ◦ µ1 moves A to P, B to Q, and C to some point C ′′ . Finally, we need to move C ′′ to R. As before for the general case, we assume C ′′ ̸= R, use a mirror reflection µ3 taking C ′′ to R, and verify that µ3 leaves P and Q fixed. Thus, in the general case, µ3 ◦ µ2 ◦ µ1 maps A to P, B to Q, and C to R. If A = P, B ′ = Q, or C ′′ = R, then we omit µ1 , µ2 , or µ3 , respectively. The only case not covered by the argument is the identity transformation, ι. However, ι = µ ◦ µ for any mirror reflection µ. Hence every Euclidean plane isometry can be written as the composition of three or fewer mirror reflections. ! B'

C'

B

C'' Q

P

C A

R

Figure 5.10 Three mirror theorem. We can use Theorem 5.2.5 to classify the four types of Euclidean plane isometries, which are summarized in Theorem 5.2.7. We have already discussed three kinds of isometries: mirror reflections, rotations, and translations. We will discuss the other kind of isometry, glide reflections, shortly. First, let’s relate the three kinds of isometries we know to the composition of mirror reflections. The “composition” of just one mirror reflection must be that mirror reflection.

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Transformational Geometry

For the composition of two mirror reflections there are three cases for the lines of reflection: the two lines are the same, they are distinct but parallel, and they intersect in a unique point. The first case gives the identity since µ ◦ µ = ι. The second case, illustrated in Figure 5.11 and shown in Exercise 5.2.11, gives a translation twice as long as the distance between the lines and in a direction perpendicular to them. The third case gives a rotation around the intersection of the two lines, where the angle is twice the angle between the lines, illustrated in Figure 5.12 and proven in Exercise 5.2.12. A mirror reflection switches orientation. Because the identity, translations, and rotations are composed from two mirror reflections, they do not switch orientation and we say that they are direct isometries. d C'

A'

C

k

A

G

m

F' G'

B'

D

r E

F H

B

Figure 5.12 Two intersecting mirrors.

Figure 5.11 Two parallel mirrors.

The remaining option, the composition of three mirror reflections, switches orientation three times and so is an indirect isometry, like a mirror reflection. Theorem 5.2.6 shows that three mirror reflections result either in a mirror reflection or a glide reflection, defined as follows. Definition. A Euclidean plane isometry γ is a glide reflection if and only if there is a line k such that γ is the composition of the mirror reflection over k and a translation in the direction parallel to k. Example 5. Figure 5.13 indicates that a glide reflection, denoted γ , can be written as the composition of a mirror reflection and a translation. ♦

γ γ γ

Figure 5.13 Repeated images of a quadrilateral under a glide reflection.

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5.2 Isometries

Exercise 5.2.4.∗ Explain how a mirror reflection is a special case of a glide reflection. Explain why no other glide reflections have fixed points. Explain why only the line k is stable in other glide reflections. Theorem 5.2.6. The composition of three mirror reflections is either a mirror reflection or a glide reflection. Proof. By Theorem 5.2.3, the images of three noncollinear points determine an isometry. Let β be the composition of three mirror reflections taking A to A′ , B to B ′ , and C to C ′ . Construct the midpoint M A of the segment A A′ and the midpoint M B of B B ′ . ←−−→ Case 1. Assume that M A ̸= M B . Let µk be the mirror reflection over the line k = M A M B (Figure 5.14). Let X be the intersection of k with the line through A and µk ( A). Then △AM A X ←−−−→ is similar to △A A′ µk ( A) by Theorem 1.4.4. Theorem 1.4.1 then implies that µk ( A) A′ is parallel ← − − − → to k. Similarly, △B M B Y ∼ △B B ′ µk (B) and µk (B)B ′ is parallel to k, where Y is the intersection ←−−−→ ←−−−→ of k with Bµk (B). Thus µk ( A) A′ is parallel to µk (B)B ′ . Also △A′ B ′ C ′ ∼ = △µk ( A)µk (B)µk (C) because both are congruent with △ABC. C' A'

k C A B

Q

MA

B'

MB X Y

Figure 5.14 Composition of three mirrors. Let τ be the translation taking µk ( A) to A′ . We claim that τ takes µk (B) to B ′ and µk (C) to C , so β = τ ◦ µk , a glide reflection. For the moment, let τ take µk (B) to B ′′ and µk (C) to C ′′ . Then µk (B), B ′ , and B ′′ are on the ←−−−→ ←−−−→ same line parallel to µk ( A)P. Only two points on µk (B)B ′ have a distance from A′ of d( A′ , B ′′ ): B ′′ or the point Q, for which M A is the midpoint of B Q. However, we assumed for case 1 that M A ̸= M B , so B ′ ̸= Q. Because d( A′ , B ′ ) = d( A′ , B ′′ ), B ′ = B ′′ . Similarly, there are only two places for C ′′ to be, one of which is C ′ . However, the triangle △A′ B ′ C ′′ has the same orientation as △A′ B ′ C ′ , so C ′′ = C ′ . Thus β is the composition of µk with the translation τ . If µk ( A) = A′ , then the translation is the identity and β = µk . Otherwise β is a glide reflection. Case 2. Assume that M A = M B . If the midpoint of CC ′ is this same point, then by Exercise 5.2.10 the isometry is a rotation of 180◦ around the point. But a rotation is a direct isometry and so is not β. Hence we can assume that the midpoint of A and A′ differs from the midpoint of C and C ′ . Now we can apply the reasoning of case 1. ! ′

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Theorem 5.2.7. There are four types of Euclidean plane isometries: mirror reflections, translations, rotations, and glide reflections. Proof. From Theorem 5.2.5 every isometry can be written as the composition of one, two, or three mirror reflections. Exercises 5.2.11 and 5.2.12, Theorem 5.2.6, and the preceding discussion cover all the possibilities for isometries. !

5.2.2 Congruence and Isometries Congruence theorems, such as SSS, provide one way to determine when two triangles are just alike. An isometry mapping one triangle to another triangle gives an alternate understanding of when they are just alike. Are these ideas equivalent? They are in Euclidean geometry and in hyperbolic and spherical geometries. (See Project 22 for extending the theorems of subsection 5.2.1 to these other geometries.) In our proofs of the preceding theorems we used SSS explicitly and implicitly because, by definition, isometries preserve distances. Conversely in this subsection we use isometries to prove congruence theorems, as suggested by the Common Core State Standards. We need to prove postulate 15 of the SMSG axioms (SAS) for this discussion, rather than assume it. We’ll use the reasoning of Theorem 5.2.5 for a different purpose. By definition of congruent triangles, their corresponding sides are congruent and their corresponding angles are congruent. Our definition of an isometry only considers distances and so applies only to the sides of triangles. We need to assume additionally that isometries preserve angles. Also, to avoid circular reasoning, we need to assume some earlier results that depended on the congruence theorems we are trying to justify. Here is a more explicit list of the assumptions in this subsection: (i) (ii) (iii) (iv)

Isometries preserve distance and angle measure. Theorem 5.2.3. Theorem 5.2.4. A point X is on the perpendicular bisector of Y Z if and only if X Y ∼ = X Z.

∼ P Q, ∠ABC ∼ Theorem 5.2.8 (SAS). For △ABC and △P Q R if AB = = ∠P Q R, and BC ∼ = ∼ ∼ ∼ ∼ Q R, then △ABC = △P Q R, that is, ∠B AC = ∠Q R P, AC = P R, and ∠BC A = ∠Q R P. Proof. Given △ABC and △P Q R, by Theorem 5.2.3 there is a mirror reflection µ1 taking A to P. Then △ABC ∼ = △Pµ1 (B)µ1 (C) by (i). If µ1 (B) = Q, skip the rest of this paragraph, replacing µ2 in the next paragraph by the identity. For µ1 (B) ̸= Q, again by Theorem 5.2.3 there is a mirror reflection µ2 taking µ1 (B) to Q. By the first “S” in SAS, P Q ∼ = AB. Since Pµ (B). So P is on the perpendicular bisection of Qµ1 (B) by µ1 is an isometry, by (i) P Q ∼ = 1 (iv). Hence µ2 (µ1 ( A)) = µ2 (P) = P and µ2 (µ1 (B)) = Q. (See Figure 5.10.) If µ2 (µ1 (C)) = R, then △ ABC ∼ = △P Q R by (i). Otherwise, by Theorem 5.2.3 there is a mirror reflection µ3 taking µ2 (µ1 (C)) to R. As in the preceding paragraph, from the second “S” of SAS, (BC ∼ = Q R) and the fact that µ2 ◦ µ1 is an isometry Q R ∼ = BC ∼ = Qµ2 (µ1 (C)). So µ3 leaves µ2 (µ1 (B)) = Q fixed. We need to prove that P is fixed. To do so we will need to use the “A” of SAS, which gives ∠P Q R ∼ = ∠ABC ∼ = ∠P Qµ2 (µ1 (C)). By definition, there ← → is a mirror reflection over P Q. We will show that mirror reflection is µ3 , but for now call it µ. Then µ(P) = P and µ(Q) = Q. Then ∠P Q R ∼ = ∠P Qµ(µ2 (µ1 (C))). = ∠P Qµ2 (µ1 (C)) ∼

5.2 Isometries

209

← → Further R and µ(µ2 (µ1 (C))) are on the same side of P Q. By SMSG postulate 12 the rays −−−−−−−−−−→ −→ Q R and Qµ(µ2 (µ1 (C))) are the same. Also, Q R ∼ = Qµ2 (µ1 (C)) ∼ = Qµµ2 (µ1 (C))). Postulate 3 guarantees that µ(µ2 (µ1 (C))) = R. Theorem 5.2.4 gives µ = µ3 . Thus µ3 ◦ µ2 ◦ µ1 takes △ABC to △P Q R and △ABC ∼ = △P Q R. ! Exercise 5.2.16 asks you to use similar reasoning to prove SSS using isometries. Alternatively, since we have SAS, we could use Exercise 1.2.18 to prove SSS and similarly Exercise 1.2.17 for ASA.

5.2.3 Klein’s Definition of Geometry Felix Klein in his famous Erlanger Programm, given in 1872, used groups of transformations to give a definition of geometry: Geometry is the study of those properties of a set that are preserved under a group of transformations on it. Klein realized that we can, for example, investigate the properties of Euclidean geometry by studying isometries. Thus he would say that the area of a triangle is a Euclidean property because it is preserved by isometries. That is, for a triangle △ABC and an isometry σ , △ABC and △σ ( A)σ (B)σ (C) have the same area. Under Klein’s definition, congruence and measures of lengths and angles are Euclidean properties, as is the shape of a figure. However, the orientation of figures isn’t a Euclidean property because mirror reflections and glide reflections switch orientation. Also, verticality isn’t a Euclidean property because some isometries, such as a rotation of 45◦ , tilt vertical lines. If we wanted to study orientation or verticality, we would need to use different groups of transformations, and, according to Klein, we would be studying a different geometry. Other geometries with a notion of distance have corresponding groups of isometries. The non-Euclidean geometries of Chapter 4 have interesting isometry groups. Section 5.5 discusses isometries in spherical geometry as a lead-in to higher dimensional Euclidean space. Project 22 in this chapter and Section 7.5 investigate isometries in single elliptic and hyperbolic geometries. The list of properties preserved by isometries in them looks like the list in Euclidean geometry. They include lengths, angles, area, and in general congruence of shapes. Properties like orientation and verticality don’t make the lists. Klein’s approach allowed mathematicians to see relationships among geometries. From Klein’s point of view the group for projective geometry includes the groups of transformations of Euclidean geometry and some non-Euclidean geometries. (See Section 7.5.)

5.2.4 Exercises for Section 5.2 *5.2.5. (a) Suppose that an isometry α takes (0, 0) to (−1, 2), (1, 0) to (−1, 3), and (0, 1) to (−2, 2). Find the images of (−1, −1) and (2, 2) and of (x, y). Draw a figure showing the five points and their images. Describe what type of isometry α is. (b) Repeat part (a) for the isometry β taking (1, 0) to (−1, 2), (2, 0) to (0, 2), and (0, 2) to (−2, 4). (c) Repeat part (a) for the isometry γ taking (1, 0) to (−1, 2), (2, 0) to (−1, 3), and (0, 2) to (1, 1). (d) Repeat part (a) for the isometry δ taking (1, 0) to (−1, 0), (2, 0) to (−1, −1), and (0, 2) to (1, 1).

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5.2.6. Outline the original placement of a small rectangular piece of paper on a larger piece of paper. Label the corners of both the small rectangle and the outline A, B, C, and D so that you can determine the rectangle’s movements. The centers of rotation are on the outline and do not move. (a) Rotate the small rectangle 180◦ around A and then 180◦ around C on the outline. Describe the resulting transformation. (b) Return the small piece of paper to its starting position and repeat part (a) but switch the order of the rotations. Describe how this new transformation differs from the one in part a). (c) Repeat parts (a) and (b) but use rotations of 90◦ at A and C. (d) Repeat part (c) but rotate the rectangle 90◦ around A followed by a rotation of −90◦ around C. (e) Repeat part (c) with various angles and centers of rotations. Make a conjecture about the resulting transformations. *5.2.7. On graph paper, draw the points A = (2, 0), B = (3, 0), and C = (4, 1). (a) Find the images µ( A), µ(B), and µ(C) for the mirror reflection µ over the line y = x. Draw and label the points on the graph paper. (b) Find the images τ (A), τ (B), and τ (C) for the translation τ along the line y = x that takes (0, 0) to (2, 2). Draw and label the points on the graph paper. (c) We can compose µ and τ in either order. Compare the images τ (µ( A)) and µ(τ ( A)). Repeat for B and C. How are τ ◦ µ and µ ◦ τ related? (d) By definition, τ ◦ µ is a glide reflection. Describe the isometry resulting from composing it with itself. To do so, find τ (µ(τ (µ( A)))) and repeat for B and C. Draw and label them on the graph paper. 5.2.8. (a) If µk is a mirror reflection over the line k and τ is a translation of length d in the direction of k, investigate whether µk ◦ τ = τ ◦ µk and justify your answer. (b) What is the composition of a glide reflection with itself? Justify your answer. *5.2.9. Let µx be the mirror reflection over the x-axis, µ y be the mirror reflection over the y-axis and µ3 be the mirror reflection over the line y = −x + 1. (a) (b) (c) (d) (e)

Find formulas for them. Find a formula for µ y ◦ µx . Describe the isometry. Repeat part (b) for µ3 ◦ µx and for µx ◦ µ3 . Repeat part (b) for µ3 ◦ µ y ◦ µx . Repeat part (b) for µ y ◦ µ3 ◦ µx . How are the lines of reflections for the glide reflections in parts (d) and (e) related?

5.2.10. Let ρ be a rotation of 180◦ about a point P. For a point A show that P is the midpoint of A and ρ(A). What theorem guarantees that a Euclidean plane isometry satisfying this property is this rotation? *5.2.11. Let k and m be parallel with a perpendicular distance of d between them and let µk and µm be the mirror reflections over the lines. We prove that µk ◦ µm is a translation of length 2d in the direction perpendicular to k and m.

5.2 Isometries

211

(a) Suppose, as in Figure 5.11, that A and B are points on m and C and D are the points on k so that AC ⊥ k and B D ⊥ k. Why does µk ◦ µm ( A) = µk ( A) and similarly µk ◦ µm (B) = µk (B)? Use congruent triangles to show for A and B that µk ◦ µm matches the translation of length 2d in the direction perpendicular to k and m. (b) Show △C AB ∼ = △ µk ◦ µm (C)C D. Why does = △µm (C) AB and △µm (C)C D ∼ this show that µk ◦ µm matches the translation of length 2d in the direction perpendicular to k? Use Theorem 5.2.3 to finish the proof. (Figure 5.11 denotes the images of A, B, and C under µk ◦ µm as A′ , B ′ , and C ′ , respectively.) (c) Prove that µm ◦ µk and µk ◦ µm are inverses. 5.2.12. Let k and m intersect at E and form an angle of r ◦ and let µk and µm be the mirror reflections over these lines. We prove that µk ◦ µm is a rotation of 2r ◦ around E. (a) Why is µk ◦ µm (E) = E? (b) Let F be any point on m other than E and let G be the point on k so that F G ⊥ k. Follow the reasoning of 5.2.11(a) to show that F ′ in Figure 5.12 is the image of F both for the rotation and for µk ◦ µm . (c) Let H be the point on m so that G H ⊥ m. Follow the reasoning of 5.2.11(b) to complete the proof that µk ◦ µm is a rotation, where G ′ is the image of G. (d) Prove that µm ◦ µk and µk ◦ µm are inverses. *5.2.13. Let Q be between P and R on a Euclidean line. Explain why, for an isometry α, α(Q) is between α(P) and α(R) and they are on a line. 5.2.14. Let ρ 1 and ρ 2 be rotations. Prove that their composition ρ 1 ◦ ρ 2 is a translation, a rotation, or the identity. Find the conditions that are necessary and sufficient for it to be a translation. *5.2.15. Let τ 1 and τ 2 be translations and let P and Q be two points. How are τ 2 ◦ τ 1 and τ 1 ◦ τ 2 related? Draw a figure showing P, Q, τ 1 (P), τ 1 (Q), τ 2 (τ 1 (P)), and τ 2 (τ 1 (Q)). Prove that the composition τ 1 ◦ τ 2 is a translation. Hint: Use SAS. 5.2.16. For △ABC and △P Q R if AB ∼ = P Q, AC ∼ = P R, and BC ∼ = Q R, use mirror reflec∼ tions to prove that △ABC = △P Q R. 5.2.17 (a) If α is an isometry that fixes two points, prove that α is the identity or the mirror reflection over the line through them. (b) If α and β are isometries such that α( A) = β( A) and α(B) = β(B), prove that ← → α = β or α = β ◦ µ, where µ is the mirror reflection over the line AB. 5.2.18. Prove that D, the set of all direct isometries of the Euclidean plane, is a transformation group. In addition to preserving all Euclidean properties D preserves orientation. 5.2.19. *(a) Let V be the set of Euclidean plane isometries that take vertical lines to vertical lines. Describe the isometries in V and prove that V is a transformation group. (b) Show that D ∩ V is a transformation group. Describe it. (c) Let T be the set of Euclidean plane translations. Prove that T is a transformation group.

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(d) Describe the set of Euclidean plane isometries that leave the x-axis stable. (e) Describe the set of Euclidean plane isometries that leave the unit circle, x 2 + y 2 = 1, stable. (f) For a subset S of the Euclidean plane, let S be the set of Euclidean plane isometries that leave S stable. Prove that S is a transformation group. (In Chapter 6 we call S the group of symmetries of S.) 5.2.20. Define sets A = { Ai : i ∈ I } and B = {Bi : i ∈ I } to be congruent, written A ∼ = B, if and only if for all i, j ∈ I, d( Ai , A j ) = d(Bi , B j ). (a) Why are triangles congruent under this definition also congruent under the usual definition? Hint: Consider their vertices. (b) Why are any two lines congruent under this definition? 5.2.21. Define two sets A and B to be isometric iff there is an isometry α such that α(A) = B. (a) Prove that isometric sets are congruent using the definition in Exercise 5.2.20. (b) Suppose △A1 A2 A3 ∼ = △B1 B2 B3 . Explain why the proof of Theorem 5.2.8 guarantees that these triangles are isometric. (c) Use part (b) and the proof of Theorem 5.2.2 to prove that in Euclidean geometry, for congruent Euclidean plane sets A = { Ai : i ∈ I } and B = {Bi : i ∈ I }, there is an isometry taking A to B. Hint: First suppose that each of these sets have three points that are not all on the same line.

5.2.5 Felix Klein At the age of 23 Felix Klein (1849–1925) gave his inaugural address as a professor at the University of Erlangen, the talk for which he is best remembered today. In it, he presented the Erlanger Programm, which raised transformation groups in geometry from an important concept to a unifying theme. Using transformation groups, Klein showed that the various nonEuclidean geometries, projective geometry, and Euclidean geometry were closely related, not competing subjects. When Klein was 17 he became the assistant to Julius Pl¨ucker, a physicist and geometer. Inspired by Pl¨ucker’s approach, Klein always emphasized the physical and intuitive aspects of mathematics over rigor and abstraction. After Pl¨ucker’s death in 1868, Klein went to Berlin to finish his graduate work. There he met Sophus Lie, who became a close friend. They went to Paris in 1870 for further studies. Both men were deeply influenced there by the possibility that group theory could unify mathematics. Klein’s Erlanger Programm in 1872 is a direct outgrowth of this inspiration. Klein developed numerous theoretical models in geometry, including the Klein bottle, a curious two-dimensional surface with no inside that requires four dimensions to realize it. Figure 5.15 illustrates a three-dimensional representation of a Klein bottle, which intersects itself, unlike the theoretical four-dimensional shape. He distinguished single elliptic geometry from spherical geometry and investigated its models and transformations. To connect projective, Euclidean, and non-Euclidean geometries by using transformations, Klein developed the model of hyperbolic geometry named for him. He started to develop what we now call the Poincar´e model, but he failed to see its connection to inversions that Henri Poincar´e (1854–1912) found.

5.3 Algebraic Representation of Transformations

213

Figure 5.15 A Klein bottle. Klein was impressed with Poincar´e’s work and corresponded with him. However, their common interests soon became a fierce rivalry. Both produced important mathematics, but Klein suffered a nervous breakdown from the intense strain and felt that he had lost the contest. After recovering, Klein produced some mathematics and wrote several books. However, he focused on new tasks that called on his superior administrative abilities, building up mathematics research and education at his university, throughout Germany, and even in the United States.

5.3 Algebraic Representation of Transformations How can a computer display various viewpoints, zooming in and rotating as the user desires? Computer graphics software uses vectors, matrices, and linear algebra extensively to compute geometric transformations. (Section 7.6 explains how the software provides perspective through projective geometry.) The algebraic representation helped mathematicians, physicists, and others long before the advent of computers. Matrices and linear algebra also give a deeper insight into geometric transformations. For convenience we often identify a transformation with its matrix. (See Appendix D for a brief summary of the parts of linear algebra relevant to our study.) We use vectors to represent points. If M is a matrix and P a vector, then Q = M P indicates that Q is the image of P by the transformation represented by M. This notation fits well with that of functions, y = f (x), and follows standard1linear 2 algebra notation. However, it means that points need to be column vectors—for example 23 , rather than the familiar ordered pair (2,⎡3). ⎤Unfortunately, column vectors waste space in a text, so we will write a column x1

vector ⎣ x···2 ⎦ as (x1 , x2 , . . . , xn ). xn

Example 1. Recall that

3

a d

b e

4 3 4 3 4 x ax + by = . y d x + ey

2 1 Thus the matrix 1da be 2defines a (linear) transformation taking (x, y) to (ax + by, d x + ey). represents the rotation of 90◦ around the origin (Figure 5.16). The The matrix R1 = 012 −1 0 2 0 matrix D = 0 2 doubles each point’s distance from the origin (Figure 5.17). The matrix 1 2 0.8 M = 0.6 represents the mirror reflection over the line y = 12 x (Figure 5.18). ♦ 0.8 −0.6

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Figure 5.16 A rotation.

Figure 5.17 A dilation.

Figure 5.18 a mirror reflection. Exercise 5.3.1. In Example 1 select various points, such as (−1, 1) and (1, 2), and find their images under these matrices to verify their descriptions. The preceding examples indicate that 2 × 2 matrices can represent a variety of plane transformations. However, they have a drawback for our purposes: they all fix the origin (0, 0). Thus they cannot represent nontrivial translations and many other transformations. For instance, the translation τ (x, y) = (x + 3, y + 2) takes (0, 0) to (3, 2). We have a way around this problem by using the plane z = 1 in R3 . Since it is a plane, it has the same geometric properties as R2 , which, in effect, is the plane z = 0. However, z = 1 has the key algebraic advantage that all its points (x, y, 1), including the new origin (0, 0, 1), can be moved by 3 × 3 matrices. The third coordinate of (x, y, 1) does not really do anything. For instance, the distance between two points still depends only on their first two coordinates. Interpretation. By a point of the Euclidean plane we mean any ! triple (x, y, 1), where x and y are real numbers. The distance between (x, y, 1) and (u, v, 1) is (x − u)2 + (y − v)2 .

Exercise 5.3.2. Verify that the matrix

⎡

a ⎣d 0

b e 0

⎤ 0 0⎦ 1

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5.3 Algebraic Representation of Transformations

maps (x, y, 1) to (ax + by, d x + ey, 1) and so corresponds to the same transformation given in 2 1 Example 1 as da be . Verify that the matrix ⎡

1 ⎣0 0

0 1 0

⎤ 3 2⎦ 1

represents the translation τ (x, y, 1) = (x + 3, y + 2, 1). It is no accident that the bottom row5 of both matrices in Exercise 5.3.2 is [0 0 1]. It 6 a b c ensures that a general 3 × 3 matrix M = d e f maps the plane z = 1 to itself. That is, g h i M · (x, y, 1) equals ( , , 1). This forces M to have g = h = 0 and i = 1. A theorem of linear algebra states that a linear transformation is one-to-one and onto the whole space if and only if its matrix is invertible or, equivalently, its determinant is not zero. Hence an invertible 3 × 3 matrix represents a plane transformation provided that its bottom row is [0 0 1]. We call the transformations for the plane z = 1 affine transformations to distinguish them from linear transformations, which leave the origin of the space fixed. Definition. By a (plane) affine matrix we mean an invertible 3 × 3 matrix whose bottom row is [0 0 1]. that an affine matrix leaves (0, 0, 1) fixed if and only if the last column Exercise 5.3.3.∗ 5 Verify 6 0 of the matrix is 0 . Rewrite the matrices of Example 1 as affine matrices. Verify that ae − bd 1

is the determinant of

⎡

a ⎣d 0

b e 0

⎤ c f ⎦. 1

Linear algebra provides an elegant interpretation of lines. In analytic geometry (as in Chapter 3), a line is the set of points (x, y) satisfying an equation of the form ax + by + c = 0. However, with triples (x, y, 1) for points, the 1 has a natural place in that equation: ax + by + c · 1 = 0. If we replace the left-hand side by the product of a row vector and a column vector, the equation ⎡ ⎤ x [a, b, c] ⎣ y ⎦ = 0 1

suggests that we can represent a line by three coordinates [a, b, c]. Exercise 5.3.4 indicates how to convert between the vector form and the familiar equation of a line when b ̸= 0. If b = 0 and a ̸= 0, we have a vertical line. (If b = 0 = a, the equation reduces to c = 0, which is not a line.)

Exercise 5.3.4. Verify that the familiar equation y = mx + b corresponds to [m, −1, b]. Verify that vertical lines x = c correspond to [1, 0, −c]. You may recall that y = 2x + 1 and 2y = 4x + 2 represent the same line since the same points satisfy them. In vector form, the lines [2, −1, 1] and [4, −2, 2] represent the same

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line. In general for any nonzero k, [ka, kb, kc] represents the same line as [a, b, c] because [ka, kb, kc](x, y, 1) = k[a, b, c](x, y, 1). Technically, then, a line is the equivalence class of all row vectors differing by a factor of a nonzero scalar. However, for convenience, we use a row vector [a, b, c] as the name of a line. Interpretation. A line is a row matrix [a, b, c] such that not both a and b are 0. The point (x, y, 1) is on the line [a, b, c] if and only if their product is 0: ax + by + c · 1 = 0. Two row vectors represent the same line if and only if one is the product of the other by a nonzero scalar. Exercise 5.3.5. Verify that the line [−2, −1, 1] is on the points (−1, 3, 1) and (2, −3, 1). Verify similarly that the point (−2, −1, 1) is on the lines [−1, 3, 1] and [2, −3, 1]. There is a close relationship between points and lines in this model, which we explore more deeply in Chapter 7. Our interpretation of lines ensures that affine transformations map lines to lines since a 3 × 3 matrix takes a row vector to a row vector. For the same reason, a transformation of the Euclidean plane taking lines to lines must be an affine transformation. We know that an affine matrix M moves a point P to M P, but where M takes a given line isn’t obvious, as Example 2 illustrates. Theorem 5.3.1 provides the answer. 51 0 36 Example 2. In Exercise 5.3.2 the translation T = 0 1 2 takes a point (x, y, 1) to (x + 0

0

1

3, y + 2, 1). In Figure 5.19 the line y = −x or [−1, −1, 0] goes to the line y = −x + 5. It might be natural to expect that the image of [−1, −1, 0] would be [−1, −1, 0]T , which is [−1, −1, −5]. However, that line is y = −x − 5, which is incorrect. Theorem 5.3.1 shows that we need to use the inverse of T . ♦

(3,2)

y= -x+5

(0,0) y= -x

Figure 5.19 The image of a line by a translation. Theorem 5.3.1. The affine matrix M takes [a, b, c] to the line [a, b, c]M −1 . Proof. We need to show that, for any point (x, y, 1), its image M(x, y, 1) is on the proposed image of [a, b, c], namely, [a, b, c]M −1 if and only if (x, y, 1) is on [a, b, c]. The new point M(x, y, 1) is on [a, b, c]M −1 if and only if their product is 0. Now [a, b, c]M −1 · M(x, y, 1) = [a, b, c]I (x, y, 1) = [a, b, c](x, y, 1), where I is the identity matrix. Thus the old and new

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5.3 Algebraic Representation of Transformations

products are equal. That is, the new point is on the new line if and only if the original point was on the original line, finishing the proof. !

5.3.1 Isometries A 3 × 3 matrix is determined by where it takes three noncollinear points. For ease we use the reference points O = (0, 0, 1), X = (1, 0, 1), and Y = (0, 1, 1), as in Figure 5.20. Theorem 5.3.2 uses them to describe which affine matrices are isometries. Y' Y = (0,1,1)

X'

O' direct isometry X'

O = (0,0,1) X = (1,0,1)

O'

indirect isometry Y'

Figure 5.20 A reference triangle and its images. Exercise 5.3.6. Verify that the images of the points O, X , and Y under the matrix ⎤ ⎡ a b c M = ⎣d e f ⎦ 0 0 1

are O ′ = (c, f, 1), X ′ = (a + c, d + f, 1), and Y ′ = (b + c, e + f, 1). Theorem 5.3.2. An affine matrix ⎡

a M = ⎣d 0

is an isometry if and only if for some angle θ, ⎡ ⎤ cos θ − sin θ c M = ⎣ sin θ cos θ f ⎦ or 0 0 1

b e 0

⎤ c f⎦ 1 ⎡

cos θ M = ⎣ sin θ 0

sin θ − cos θ 0

⎤ c f ⎦. 1

Proof. (⇒) If M is an isometry, then, for the three points O ′ , X ′ , √and Y ′ of Exercise 5.3.6, △O ′ X ′ Y ′ ∼ = △O X Y (Figure 5.20). The distance d(O ′ , X ′ ) is a 2 + d 2 . As M is 2 = 1. Then for some angle θ, a = cos θ and d = sin θ . Similarly, the an isometry, a + d 2 √ distance d(O ′ , Y ′ ) = b2 + e2 forces b = cos φ and e = sin φ, for some angle φ. Furthermore, m∠X ′ O ′ Y ′ = 90◦ . Thus φ = θ ± 90◦ . As Figure 5.20 illustrates, when φ = θ + 90◦ ,

218

Transformational Geometry

the isometry is direct and when φ = θ − 90◦ , the isometry is indirect. Trigonometry gives sin(θ ± 90◦ ) = ± cos θ and cos(θ ± 90◦ ) = ∓ sin θ. (⇐) See Exercise 5.3.13. ! Exercise 5.3.7. Verify that the determinant of ⎡ cos θ − sin θ ⎣ sin θ cos θ 0 0

is +1 and that the determinant of

is −1.

⎡

cos θ ⎣ sin θ 0

sin θ − cos θ 0

⎤ c f⎦ 1 ⎤ c f⎦ 1

Isometries split naturally into two classes, as given in Theorem 5.3.2. Exercise 5.3.7 shows that determinants identify them. The first class, with determinant +1, contains the direct isometries, for which θ is the angle of rotation. If θ = 0, the direct isometry is a translation. An isometry of the second class, with determinant −1, is an indirect isometry. Exercise 5.3.14 shows that its line of reflection makes an angle of θ /2 with the x-axis. Example 3. Find the center of rotation of

50 1 0

−1 0 0

26 3 . 1

Solution. The center of this 90◦ rotation is a fixed point, say, (u, v, 1), satisfying ⎡ ⎤⎡ ⎤ ⎡ ⎤ u 0 −1 2 u ⎣1 0 3⎦⎣v ⎦ = ⎣v ⎦. 1 0 0 1 1

The equation reduces to −v + 2 = u and u + 3 = v, or u = − 12 and v = 2 12 . Thus the fixed point is (− 12 , 2 12 , 1). ♦

Example 4. Find the line of reflection of ⎡ Which lines are stable under M?

0.6 M = ⎣ 0.8 0

0.8 −0.6 0

⎤ 2 −4 ⎦ . 1

Solution. We can find the fixed points of the mirror reflection M as in Example 3, obtaining the pair of equations −0.4u + 0.8v + 2 = 0 and 0.8u − 1.6v − 4 = 0. The second is a multiple of the first, so we get an infinite family of fixed points, (u, 12 u − 2 12 , 1). That is, all the fixed points are on the line of reflection [ 21 , −1, −2 12 ] or, more familiarly, y = 12 x − 2 12 . Finding stable lines needs a more general approach because for any nonzero multiple λ, λ[a, b, c] is the same line as [a, b, c]. We need to solve [a, b, c]M −1 = λ[a, b, c], but first have to find the possible values of λ. We need [a, b, c]M −1 = λ[a, b, c] = λ[a, b, c]I or,

219

5.3 Algebraic Representation of Transformations

equivalently, [a, b, c](M −1 − λI ) = 0. From a result in linear algebra, the matrix (M −1 − λI ) has a determinant of zero. For ⎡ ⎤ ⎡ ⎤ 0.6 0.8 2 0.6 − λ 0.8 2 M = ⎣ 0.8 −0.6 −4 ⎦ = M −1 , M −1 − λI = ⎣ 0.8 −0.6 − λ −4 ⎦ 0

0

1

0

0

1−λ

and the determinant is (1 − λ)[(0.6 − λ)(−0.6 − λ) − (0.8)(0.8)] = (1 − λ)(λ2 − 1). Thus the possibilities are λ = 1 (as a double root) and λ = −1. ⎡ ⎤ 0.6 0.8 2 For λ = 1, we get [a, b, c] ⎣ 0.8 −0.6 −4 ⎦ = [a, b, c]. This becomes 0 0 1 0.6a + 0.8b = a 0.8a − 0.6b = b . 2a − 4b + c = c Each equation reduces to a = 2b, with c undetermined. This outcome gives a family of parallel lines [2, 1, c] or, equivalently, y = −2x − c, all perpendicular to the line of reflection. A similar computation with the root λ = −1 gives the equation [a, b, c]M −1 = −1[a, b, c] = [−a, −b, −c]. Its solution is the line of reflection, [ 21 , −1, −2 12 ]. The double root gives a family of stable lines, whereas the single root gives just one line. The fact that λ = 1 is a double root for M ensures that there is a family of fixed points, as we found previously. Exercise 5.3.17 shows that the only values of λ for an isometry are 1 and −1. In linear algebra the λ are called eigenvalues, and the solutions [a, b, c] are called eigenvectors. ♦ Example 5. Find the matrix S representing a rotation of θ around the point (u, v, 1). 51 0 u6 Solution. We build the rotation around (u, v,51) from the translation T = 0 1 v that moves 6 cos θ

− sin θ

0

0

0

1

(0, 0, 1) to (u, v, 1) and the rotation R = sin θ cos θ 0 of θ around (0, 0, 1). In effect, 0 0 1 we first move the point (u, v, 1) to (0, 0, 1), rotate there, and move back (Figure 5.21). That is, we claim that S = T RT −1 . First, Theorem 5.2.1 guarantees that S is an isometry. Further, S(u, v, 1) = T RT −1 (u, v, 1) = T R(0, 0, 1) = T (0, 0, 1) = (u, v, 1), so the isometry fixes (u, v, 1), the center of rotation. Next, from linear algebra, the determinant of S is the product S (u,v,1)

T

T-1

R (0,0,1)

Figure 5.21 The rotation S equals T RT −1 .

220

Transformational Geometry

of the determinants of T , R, and T −1 , which are all 1. Thus S has determinant 1 and is a direct isometry. Hence it is a rotation by Theorem 5.2.7. Finally, we determine the angle of rotation of S. Now S(u, v, 1) = (u + cos θ , v + sin θ , 1), so the point is rotated through an angle of θ . Then T RT −1 is indeed the desired rotation. ♦

5.3.2 Exercises for Section 5.3 *5.3.8. (a) Find the matrix form of translations along the line y = 0. (b) Find the matrix for the mirror reflection over the line y = 0. (c) Use parts (a) and (b) to find the matrix form for glide reflections over the line y = 0. (d) Find the matrix of the mirror reflection over the line y = x. (e) Find the matrix form of glide reflections over the line y = x. 5.3.9. *(a) Find the matrix for the rotation of 30◦ with a center of rotation of (0, 0, 1). *(b) Find the matrix for the rotation of 30◦ with a center of rotation of (2, 3, 1). (c) Repeat part (b) for the rotation of 60◦ with the same center (2, 3, 1). The right column of this matrix differs from the one in part (b). (d) Find the matrix of the mirror reflection over the line [1, −1, 0] (that is, y = x). (e) Repeat part (d) for the line [1, −1, 4] (or y = x + 4). Hint: Find a point on y = x + 4 and use Example 5. (f) Use a different point on y = x + 4 and repeat part (e) to verify that you obtain the same matrix. *5.3.10. For each matrix, decide whether it is a translation, a rotation, a mirror reflection, or a glide reflection and find its fixed points and stable lines. ⎡

0.8 A = ⎣ 0.6 0

−0.6 0.8 0 ⎡

⎤ 2 0⎦ 1

0.8 D = ⎣ 0.6 0

⎡

1 0 B = ⎣0 1 0 0

⎤ 0.6 2 −0.8 1 ⎦ 0 1

⎤ −2 7 ⎦ 1

⎡

0.8 C = ⎣ 0.6 0

⎡√ √2/2 E = ⎣ 2/2 0

⎤ −1/3 1 ⎦ 1

0.6 −0.8 0

√ ⎤ −√ 2/2 2 2/2 0 ⎦ 0 1

5.3.11. Repeat Exercise 5.3.10 for the matrices A · B, B · A, B · C, and C · D. 5 0.6 0.8 2 6 5.3.12. (a) Show that G = 0.8 −0.6 1 is a glide reflection over the line y = 12 x or 0

0

1

[0.5, −1, 0]. State what the translation part of G is.

*(b) Determine the relationship of the values c and f in H =

5 0.6

0.8 0

0.8 −0.6 0

c f 1

6

so that

H is a glide reflection over the line y = 12 x. Explain your answer. (c) Find the matrix form for all glide reflections over the line y = 12 x + 2. Hint: Find a point on y = 12 x + 2 and use Example 5.

221

5.3 Algebraic Representation of Transformations

5.3.13. Let ⎡

cos θ M1 = ⎣ sin θ 0

− sin θ cos θ 0

⎤ c f ⎦, 1

⎡

cos θ M2 = ⎣ sin θ 0

sin θ − cos θ 0

⎤ c f ⎦, 1

P = (u, v, 1), and Q = (s, t, 1). Show that d(P, Q) = d(M1 P, M1 Q) = d(M2 P, M2 Q). Hint: cos2 (θ ) + sin2 (θ) = 1. 5 cos θ sin θ 0 6 5.3.14. (a) Show that sin θ − cos θ 0 switches (1, 0, 1) and (cos θ , sin θ , 1). Explain why 0

0

1

the isometry must be the mirror reflection over the line through the origin that makes an angle of θ /2 with the x-axis. (b) Let M and N be matrices of mirror reflections over lines through the origin with different variables for their angles. Use trigonometric formulas to show that the product M N is a rotation around the origin. How is N M related to M N ? (c) Let K and L be indirect isometries with the same angle θ. Use the matrix form of indirect isometries in Theorem 5.3.2 with variables in the last column to represent K and L. What can you say about their product K L? Interpret the conclusion geometrically.

5.3.15. (a) Let R and S be matrices of rotations with center (0, 0, 1) and of angles θ and φ, respectively. Use trigonometric formulas to show that RS and S R both represent a rotation of θ + φ around the origin, unless θ + φ is a multiple of 360◦ . What happens in this situation? (b) Let T be the matrix of a rotation of angle θ with variables in the last column. Similarly let U be a rotation of angle φ. Verify that T U does not generally equal U T . Use trigonometric formulas to show that T U and U T both represent rotations of θ + φ, unless θ + φ is a multiple of 360◦ . What happens in this situation? 5.3.16. The central symmetry with respect to A takes a point P to the point P ′ , where A is the midpoint of P P ′ (Figure 5.22). *(a) Find the matrix for the central symmetry with respect to (0, 0, 1). *(b) Repeat part (a) for (u, v, 1). Hint: Use graph paper to find the image of (0, 0, 1). *(c) Verify that the matrix in part (b) has a determinant of +1 and its square is the identity. Explain what the algebraic properties mean geometrically. *(d) Find the stable lines of the general central symmetry from part (b). *(e) Find the composition of two central symmetries, one with respect to (u, v, 1) and the other with respect to (s, t, 1). Identify the type of isometry for the composition and explain what is happening geometrically. What happens when you switch the order? (f) Prove that a central symmetry sends every line to a line parallel to itself. (g) Prove that an isometry sends every line to a line parallel to itself if and only if it is either a central symmetry or a translation. (h) Show that the set of translations and central symmetries form a transformation group. (Part (g) shows the isometries preserve the direction of a line.)

222

Transformational Geometry

R' Q S

S'

P P'

A

Q' R

Figure 5.22 A central symmetry. 5.3.17. (a) Prove that λ = 1 is an eigenvalue of an affine matrix. (b) Prove that an indirect isometry has λ = 1 as a double eigenvalue and λ = −1 as an eigenvalue. (c) Prove that a matrix representing a direct isometry satisfies one of the following, depending on the value of θ: (i) λ = 1 is a triple eigenvalue (θ is a multiple of 360◦ ); (ii) λ = 1 is an eigenvalue, and λ = −1 is a double eigenvalue (θ is an odd multiple of 180◦ ); or (iii) λ = 1 is an eigenvalue, and the other eigenvalues are complex (all other values of θ ). Thus the only possible real eigenvalues of an isometry are λ = 1 and λ = −1. 5.3.18. (a) Prove that an invertible matrix M never has λ = 0 as an eigenvalue. (b) If λ ̸= 0 is an eigenvalue of an affine matrix M, prove that 1/λ is an eigenvalue of M −1 . 5.3.19. Let T be the matrix for a translation and M the matrix for an isometry. We consider Example 5 more generally. The isometry is called the conjugate of M by T and is the matrix for a transformation closely related to M. (a) If M represents a rotation of m ◦ with center P, prove that T M T −1 is a rotation with the same angle. Give the center of T M T −1 in terms of P and the matrix T . (b) If M represents the mirror reflection over the line k, prove that T M T −1 represents a mirror reflection by showing that its determinant is −1 and when multiplied by itself it gives the identity. Explain why the algebraic properties force T M T −1 to be a mirror reflection. (c) What can you say about T M T −1 if M is a translation? 5a b c 6 *5.3.20. Matrices d e f for isometries leaving [0, −1, 0] (the x-axis) stable have signifi0

0

1

cant restrictions on the variables. Determine the possible values of a, b, c, d, e, and f . Explain how the values correspond to the types of isometries leaving the x-axis stable.

5.3.21. (a) Give the matrix form for a translation and show the set of these matrices to be a transformation group. (b) Give the matrix form for a rotation fixing (0, 0, 1) and show the set of translations is a transformation group. (c) Show the set of matrices for rotations fixing a point A is a transformation group. Hint: Use part (b) and Example 5.

5.4 Similarities and Affine Transformations

223

(d) Show the set of isometries fixing a given point is a transformation group. (e) Show the set of isometries leaving a line stable is a transformation group. (f) Show that the indirect isometries are not a transformation group. Which properties for a transformation group fail?

5.4 Similarities and Affine Transformations 5.4.1 Similarities Intermediate between isometries and affine transformations are similarities, the transformations corresponding to similar figures, discussed in Section 1.4. The group of similarities includes more than the group of isometries, the transformations matching congruence. So Klein’s approach to geometry considers similarity and congruence as properties belonging to different but related geometries. Although the separation seems artificial from the point of view of high school geometry, non-Euclidean geometry provides reason for it. In both hyperbolic geometry and spherical geometry, similar figures are necessarily congruent figures. So it makes sense to study similarities apart from isometries. Definition. A transformation σ of the Euclidean plane is a similarity if and only if there is a positive real number r such that for all points P and Q in the plane, d(σ (P), σ (Q)) = r · d(P, Q). The number r is the scaling ratio of σ . Example 1. In Figure 5.23, we transform the smaller triangle into the larger triangle by rotating it 90◦ , translating it and scaling it by a ratio of r = 1.5. The transformations compose to give one similarity. ♦

Figure 5.23 Similar triangles

Theorem 5.4.1. The set of similarities forms a transformation group.

Proof. See Exercise 5.3.9. ! We next investigate the matrix form for similarities in Theorem 5.4.2 and Exercise 5.4.1. Then we match the transformations with the familiar properties of similar figures in Theorem 5.4.3. Theorem 5.4.4 gives an insight about them and similar figures not readily available without the power of linear algebra. It allows us to classify geometrically the types of similarities, as Theorem 5.2.7 did for isometries.

224

Transformational Geometry

5r 0 06 Example 2. The matrix S = 0 r 0 takes (x, y, 1) to (r x, r y, 1). Thus the origin (0, 0, 1) is 0 0 1 fixed, and all points expand or contract with respect to the origin by a scaling ratio of r . That is, the distance between (0, 0, 1) and (r x, r y, 1) is r times the distance between the original points (0, 0, 1) and (x, y, 1). To show S is the matrix for a similarity, we need to show the scaling works for any two points, not just for the origin and one other. Consider the two points (a, b, 1) and ( p, q, 1). Then ! d((ra, r b, 1), (r p, rq, 1) = (ra − r p)2 + (r b − rq)2 ! = r 2 (a − p)2 + r 2 (b − q)2 ! = r (a − p)2 + (b − q)2 Thus S gives a similarity. ♦

= r d((a, b1), ( p, q, 1).

Theorem 5.4.2. An affine matrix M represents a similarity if and only if ⎡ ⎤ r cos θ ∓r sin θ a M = ⎣ r sin θ ±r cos θ b ⎦ for some r > 0. The determinant is ±r 2 . 0 0 1 Proof. See Exercise 5.4.10. ! Exercise 5.4.1. Show that the matrix of a similarity can be written as the product of the matrix of an isometry and a matrix in the form given in Example 2. Theorem 5.4.3. Similarities preserve angle measures and the proportions of distances. If a similarity has a scaling ratio of r , then the area of the image of a convex polygon is r 2 times the area of the original polygon. Proof. We use the theorems of Section 1.4. See Exercise 5.4.11. ! Theorem 5.4.4. A similarity with a scaling ratio of r ̸= 1 has a unique fixed point. Proof. Let M be the similarity matrix and (u, v, 1) be a candidate for a fixed point. From Theorem 5.4.2 we have one of the following systems of two equations in the two unknowns u and v: r cos(A)u − r sin( A)v + a = u r sin( A)u + r cos(A)v + b = v

The first system becomes

or

r cos(A)u + r sin( A)v + a = u . r sin( A)u − r cos(A)v + b = v

(r cos A − 1)u − r sin( A)v = −a , r sin( A)u + (r cos A − 1)v = −b which has a unique solution when its determinant is not zero. The determinant equals (r cos A − 1)2 + r 2 sin2 A = r 2 − 2r cos A + 1. By the quadratic formula, r 2 − 2r cos A +

225

5.4 Similarities and Affine Transformations

√ 1 = 0 if and only if r = (2 cos A ± 4 cos2 A − 4)/2. The value under the radical sign is negative except when cos2 A = 1, which forces r = ±1. As r > 0, the determinant can be 0 only when r = 1. Hence, if r ̸= 1, there is a unique fixed point. Similarly, the second system reduces to (r cos A − 1)u − r sin( A)v = −a . r sin( A)u − (r cos A + 1)v = −b Since cos2 A + sin2 A = 1, the determinant reduces to 1 − r 2 . For r > 0 and r ̸= 1, we have a nonzero determinant and so a unique fixed point. ! Theorem 5.4.4 guarantees a fixed point, but it gives no geometrical insight as to where it is. Project 7 explores this question. The theorem does, however, allow us to describe the types of similarities geometrically, rather than algebraically with matrices. Dilations, defined below and illustrated in Figure 5.24, generalize Example 2 by allowing any point P to be the center around which we scale everything. From Section 5.2 the isometries fixing a point P are either rotations around P or mirror reflections over a line going through P. The isometries give rise to the two types of similarities. The first, a composition of a dilation and a rotation, as in Figure 5.25, creates a spiraling effect. As with rotations, it preserves orientation, has a positive determinant, and is direct. The second, composed of a dilation and a mirror reflection, as in Figure 5.26, looks like a scaled glide reflection. It switches orientation, has a negative determinant, and is indirect.

P

P

Figure 5.24 A dilation.

Figure 5.25 A similarity composing a dilation and a rotation.

P

Figure 5.26 A similarity composing a dilation and a mirror reflection. Definition. A dilation δ with center P and scaling ratio r takes every point X to its image δ(X ) −→ that is the point on the ray P X so that d(P, δ(X )) = r d(P, X ).

226

Transformational Geometry

5.4.2 Affine Transformations In Section 5.3 we defined affine transformations by their matrix form. This approach provides little geometric insight, which Theorems 5.4.5 and 5.46 and Example 4 address. Affine transformations allow considerable distortion, as Example 3 illustrates. However, the properties from the theorems and example restrict the type of distortion. 52 1 06 Example 3. The matrix T = 1 1 0 transforms △ABC and the circle of the left of Fig0

0

1

ure 5.27 into △A′ B ′ C ′ and the ellipse shown on the right. Affine transformations don’t need to preserve angle measure, proportions of lengths, or overall shape, unlike similarities. ♦ C

B'

B

C'

A Y

A'

X

Y' X'

Figure 5.27 An affine transformation.

Theorem 5.4.5. The set of affine transformations is a transformation group. Affine transformations preserve lines, parallelism, betweenness, and proportions on a line. Proof. In Section 5.3 we noted that the image of a line under an affine transformation was a line. Here we prove betweenness and proportions on a line. (See Exercise 5.4.16 for proofs about the group and parallelism.) From Section 3.5, the segment P Q contains a point R (considered as a vector) if and only if R = P + r (Q − P) for some real number r between 0 and 1. For r = 0, the equation gives R = P and for r = 1, it gives R = Q. The values of r give the coordinates for the points on the line through P and Q with origin at P and P Q the unit length. An affine transformation α is a linear transformation, which preserves addition and scalar multiplication. So α(R) = α(P) + r (α(Q) − α(P)). Since the same r relates P, Q, and R and their images, α preserves betweenness and proportions on a line. ! Example 4. Verify that the affine matrix

53

0 2 0

0 0 1

6

triples every x-coordinate, doubles every y-coordinate, has a determinant of 6, and increases the area of every triangle by a factor of 6. 0 0

Solution. We leave all but the last part to you. For the last part, note that every triangle △ABC without a horizontal side can be split into two triangles, each with a horizontal side (B D in Figure 5.28). Then the smaller triangles have their bases tripled and their heights doubled. Hence the areas of the smaller triangles and so the entire triangle are multiplied by 6. As the example suggests, the absolute value of the determinant of an affine matrix is the scaling factor for areas, as is the case for similarities. ♦

227

5.4 Similarities and Affine Transformations

A D

B

C

Figure 5.28 Example 5. The affine matrix S =

51 0 0

1 1 0

0 0 1

6

is an example of a shear transformation, where

all points move in parallel, but have different displacements (Figure 5.29). The determinant of S is 1, so S preserves area. Geologic shears display the same effect on parts of the earth’s crust as geometric shears do on the plane. ♦

Original

Image by shear

Figure 5.29 The effect of a shear transformation. Theorem 5.4.5 considered properties based on straight lines. Convexity, considered in Theorem 5.4.6, applies to general, nonlinear shapes. The importance of convexity in some applications, such as optimization, increases the significance of Theorem 5.4.6. While circles can distort to ellipses, as in Example 3, they can’t be distorted to other shapes. The converse of Theorem 5.4.6 holds as well: transformations of the Euclidean plane preserving convexity must be affine. Theorem 5.4.6. An affine transformation preserves convexity. Proof. Suppose that α is an affine transformation and A is a convex set. We must show that the image α[A] is convex. That is, for points X ′ and Y ′ in α[ A] and Z ′ , a point between them, we must show that Z ′ is in α[ A]. By the definition of α[A], there are X and Y in A such that α(X ) = X ′ and α(Y ) = Y ′ . Furthermore, there is a unique Z such that α(Z ) = Z ′ . We need only show that Z is between X and Y because then the convexity of A will guarantee that Z is in A and so Z ′ is in α[ A]. Consider α −1 , the inverse of α, which is also an affine transformation. By Theorem 5.4.5, α −1 preserves betweenness. As Z ′ is between X ′ and Y ′ , Z is between X and Y , implying that α[ A] is convex. ! Given two points A and A′ we can find an isometry, in fact a translation, taking A to A′ . This says that every point looks like every other point. With similarities, Exercise 5.4.8 shows that we can take any pair of distinct points to any pair of distinct points, so that every segment looks like every other segment from the point of view of similarities. Of course, only congruent segments look alike using isometries. We can extend the comparison to affine transformations

228

Transformational Geometry

using triangles. Example 3 suggests and Exercise 5.4.19 shows that from the point of view of affine transformations, every triangle looks like every other triangle. For similarities, only similar triangles are, well, similar. For isometries, only congruent triangles are alike. As we increase the set of transformations, we increase the amount of freedom of movement and the amount of likeness. At the same time, we decrease the set of properties satisfied by the transformations.

5.4.3 Iterated Function Systems Whereas the affine image of a convex set is convex, iterated function systems (abbreviated IFSs) combine affine transformations in an unusual way to yield highly nonconvex sets called fractals. The points in fractals are the limits of infinitely many applications of the affine transformations in all possible orders. While formal proofs require analysis, we can understand the basic ideas of IFS without analysis. The figure introducing this chapter shows a fractal drawn with the aid of five matrices. A simpler example reveals the idea behind this approach to fractals, introduced by Michael Barnsley. Example 6. Matrices A, B, C, and D shrink the unit square U shown in Figure 5.30 to the four smaller squares shown, where ⎡ ⎡ ⎤ ⎤ cos 60◦ /3 − sin 60◦ /3 1/3 1/3 0 0 B = ⎣ sin 60◦ /3 cos 60◦ /3 A = ⎣ 0 1/3 0 ⎦ , 0 ⎦, 0 0 1 0 0 1 ⎡

⎤ cos 300◦ /3 − sin 300◦ /3 √1/2 C = ⎣ sin 300◦ /3 cos 300◦ /3 3/6 ⎦ , 0 0 1

and

⎡

1/3 0 ⎣ D= 0 1/3 0 0

⎤ 2/3 0 ⎦. 1

Thus a starting point P in U maps to some point P ′ in one of the small squares in Figure 5.30. In turn, P ′ is mapped to some point P ′′ in the sixteen even smaller squares in Figure 5.31. If we continue the process infinitely often, any point will end up somewhere along the convoluted Koch curve shown in Figure 5.32, which fits into all the nested squares. ♦

U

BU AU

CU

DU

Figure 5.30 The images of the unit square under A, B, C, and D.

Figure 5.31 The next iteration of the unit square.

229

5.4 Similarities and Affine Transformations

Figure 5.32 The Koch curve. For the practical goal of drawing interesting pictures, we don’t need to iterate the process of Example 6 infinitely many times. Even six iterations will shrink the square to a multitude of squares so small that they look like points. Each iteration quadruples the number of squares and shrinks their sides by a factor of one-third. Thus, after six iterations, there will be 46 = 4096 squares, each with a side of length 1/36 = 1/729. Barnsley realized that it is faster still to apply a random sequence of the defining transformations to one point and plot the resulting images. The images quickly approximate the theoretical curve as closely as the eye can see. Not every affine matrix can be part of an IFS, for the matrices must contract all distances to ensure limit points to form the fractal. Similarities with a scaling ratio r < 1 contract distances, as do some other affine matrices. Example 7. Figure 5.33 shows a distorted Koch curve made from the matrices ⎡ ⎤ ⎡ 1/3 1/6 0 cos 60◦ /3 cos 60◦ /6 − sin 60◦ /3 B ′ = ⎣ sin 60◦ /3 sin 60◦ /6 + cos 60◦ /3 A′ = ⎣ 0 1/3 0 ⎦ , 0 0 1 0 0 ⎤ ⎡ cos 300◦ /3 − cos 300◦ /6 − sin 300◦ /3 1/2 ⎥ ⎢ √ ◦ ◦ ◦ C′ = ⎢ 3/6 ⎥ ⎦, ⎣ sin 300 /3 − sin 300 /6 + cos 300 /3 0 0 1 and

⎡

1/3 D′ = ⎣ 0 0

−1/6 1/3 0

⎤ 2/3 0 ⎦. 1

Figure 5.33 A distorted Koch curve.

⎤ 1/3 0 ⎦, 1

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Figure shows 6 the images of the unit square for each of them, which are products of the 5 1 5.34 ±1/2 0 shears 0 1 0 with the matrices of Example 6. Shears shift rectangles to parallelograms, 0 0 1 as illustrated in Figure 5.29. ♦ U

B ′U

A ′U

C ′U

D ′U

Figure 5.34 The images of the unit square for the distorted Koch curve. Definition. An affine transformation κ is a contraction mapping if and only if there is a real number r with 0 < r < 1 such that for all points P and Q, d(κ(P), κ(Q)) ≤ r · d(P, Q). An iterated function system (IFS) is a finite set of contraction mappings. The set of limit points resulting from infinitely many applications of the contraction mappings in all possible orders is an IFS fractal. Theorem 5.4.7. Every contraction mapping has a unique fixed point. An IFS fractal is a closed and bounded set. Proof. See Barnsley [2] for a proof, which uses analysis. The reasoning of Exercise 5.4.12(a) applies to a contraction mapping to give a unique fixed point. As in Example 6, the fractal is bounded by the sequence of parallelograms. ! Exercise 5.4.2.∗ Explain why the set of contraction mappings does not form a transformation group. Figures 5.0, 5.32, and 5.33 were made by using two-place decimals for each matrix entry. An affine matrix is determined by the six entries in its top two rows. In other words only twelve digits per matrix times the number of matrices are needed to encode any such fractal. Five matrices, and so 60 digits, contain all the information needed to produce the intricacies of the fractal of Figure 5.0 introducing this chapter. Barnsley has patented a way to replace an entire picture of a real scene with a collection of IFSs. It allows him to compress pictorial information into a small set of numbers. A computer program can quickly recover the picture from the set of numbers. Currently, one CD-ROM encyclopedia encodes its 10,000 pictures as IFS. (See Barnsley [2] for a detailed development of IFSs.)

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5.4 Similarities and Affine Transformations

5.4.4 Exercises for Section 5.4 5.4.3. For each similarity below, find its fixed point and stable line(s), if any. ⎡ ⎤ ⎡ ⎤ 0 −0.5 3 2 0 2 ∗(b) B = ⎣ 0.5 *(a) A = ⎣ 0 2 4 ⎦ 0 1⎦ 0 0 1 −3 0 (c) C = ⎣ 0 −3 0 0 ⎡

0

0 1 ⎤ 0 2 4 (d) D = ⎣ 2 0 −2 ⎦ 0 0 1

⎤

−6 6 ⎦ 1

⎡

*5.4.4. Find the matrices of the two similarities fixing (1, −3, 1) and taking (2, −3, 1) to (1, 1, 1). Identify which is direct and which is indirect. For the direct one, what is the angle of rotation? For the indirect one, what is the line of reflection? For the indirect one, find the other stable line and explain how it relates to the line of reflection. 5 0 −2 0 6 *5.4.5. (a) Let M = 2 0 0 and X = (1, 0, 1). Find and plot on a graph the points X , 0

0

1

M X , M 2 X , and M 3 X . Describe the shape of the curve that appears to go through the points. (b) You can fill in some of the points on the curve part (a) 6 by finding a matrix S 5 1 of −1 0 that is the square root of M. Verify that S = 1 1 0 satisfies S 2 = M. Find 0 0 1 and graph on the same axis as part (a) the points S X , S 3 X , and S 5 X . Do they fill in the curve you described in part (a)? (c) For √ ⎡√ ⎡ ⎤ ⎤ r cos θ −r sin θ 0 r cos(θ/2) − r sin(θ/2) 0 √ √ M = ⎣ r sin θ r cos θ 0 ⎦ and S = ⎣ r sin(θ/2) r cos(θ/2) 0 ⎦ , 0

0

1

0

0

1

use trigonometry to prove that S 2 = M. Explain how S and M move points. (d) For M in part (c), find a “cube root” C and an “nth root” N and explain why they qualify as a cube root and an nth root of M. 50 2 16 5.4.6. (a) Let F = 2 0 1 , O = (0, 0, 1), X = (1, 0, 1), and W = (0, 2, 1). Find and 0

0

1

plot on a graph the points O, X , W, and their images under F, F 2 , and F 3 . Describe what happens to △O X W under repeated applications of F. (b) From the graph of part (a) predict what the images of △O X W will look like under F −1 , F −2 , and F −3 . (c) Find the matrix F −1 and use it to plot the images of O, X , and W under F −1 , F −2 , and F −3 . Do the images fit with your prediction in part (b)? 5 0 √3 2 1/(1+ √3 2+ √3 4) 6 √ √ √ (d) Let C = 3 2 0 1/(1+ 3 2+ 3 4) . Verify that C 3 = F. Explain why it makes sense 0

0

1

for F to have a cube root, but not a square root.

*5.4.7. (a) Find the matrices of the two similarities taking (0, 0, 1) to (−2, 4, 1) and taking (1, 0, 1) to (0, 2, 1).

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(b) Find the matrices of the two similarities taking (1, 1, 1) to (3, 4, 1) and taking (2, 1, 1) to (0, 0, 1). 5.4.8. (a) For a point P = ( p, q, 1), show that there are two similarities fixing (0, 0, 1) and taking (1, 0, 1) to P. (b) For S = (s, t, 1) and V = (v, w, 1), show that there are two similarities taking (0, 0, 1) to S and (1, 0, 1) to V . Hint: First redo part (a) with P = V − S. Then compose the results with an appropriate isometry taking (0, 0, 1) to S. (c) Use part (b) and group properties to show that for two distinct points A and B there are two similarities taking A to (0, 0, 1) and B to (1, 0, 1). (d) Use parts (b) and (c) to show for distinct points A and B and distinct points S and V that there are two similarities taking A to S and B to V . 5.4.9. Prove Theorem 5.4.1. Hint: Modify the proof of Theorem 5.2.1 to incorporate the fact that similarities can alter distances by a factor. 5.4.10. Prove Theorem 5.4.2. Hint: See Theorem 5.3.2. 5.4.11. Prove Theorem 5.4.3. Hint: Use Exercise 5.4.1 and Section 1.4. 5.4.12. Figure 5.35 suggests an alternative proof for Theorem 5.4.4. Let α be a similarity with a scaling ratio r ̸= 1 and P = P0 be a point. Then without loss of generality assume that α(P) ̸= P. Find the fixed point of α as follows. (a) (Calculus) Case 1: r < 1. Let P1 = α(P) = α(P0 ) and, in general, Pn+1 = α(Pn ). What happens to d(Pn+1 , Pn ) as n → ∞? Explain why there is a point Q = limn→∞ Pn . Explain why Q is a fixed point of α. (A proof of this uses analysis and so is beyond the level of this book.) (b) Case 2: r > 1. Show that the scaling ratio for α −1 is 1/r < 1. By part (a), α −1 has a fixed point. Prove that α has the same fixed points as α −1 . (c) Prove that if r ̸= 1, there cannot be two distinct fixed points of α. P = P0 P1

P2

P4

P3

Figure 5.35 The similarity images of a point approaching the fixed point. *5.4.13. (a) Show that an indirect similarity with scaling factor r has eigenvalues of 1, r, and −r . (b) Let σ be a direct similarity with scaling factor r and an angle of 0◦ . Show that it has an eigenvalue of 1 and a double eigenvalue of r .

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(c) Let σ be a direct similarity with scaling factor r and an angle of 180◦ . Show that it has an eigenvalue of 1 and a double eigenvalue of −r . (d) Let σ be a direct similarity with scaling factor r and an angle that is not a multiple of 180◦ . Show that it has an eigenvalue of 1 and two complex eigenvalues. Hint: See Exercise 5.3.17. 5.4.14. Let P be the set of affine transformation α such that for a line k, k∥α(k). (a) Prove that P is a transformation group. (b) Show that lines parallel to [s, t, u] have the form [r s, r t, r w], for r ̸= 0. 5r 0 p6 (c) Show that, if M = 0 r q , where r ̸= 0, then M ∈ P. What type of transfor0

0

1

mation is M when r = 1? Repeat for when r ̸= 1. (d) Show the converse of part (c). Hint: Consider the images of the lines [1, 0, 0], 5a b c 6 [0, 1, 0], and [1, 1, 0] under an affine matrix K such that K −1 = d e f and 0

find conditions on the variables in K −1 .

5.4.15. Find the fixed several points cally. ⎡ 2 *(a) A = ⎣ 1 0 ⎡ 2 (c) C = ⎣ 0 0

0

1

point(s) and stable line(s), if any, for the affine transformations. Plot and their images and try to describe the transformation geometri−1 0 0

⎤ 0 0⎦ 1 ⎤

2 0 1 0⎦ 0 1

⎡

⎤ −2 −1 2 *(b) B = ⎣ 1 0 0⎦ 0 0 1 ⎡ ⎤ −1 2 2 (d) D = ⎣ 0 1 2 ⎦ 0 0 1

5.4.16. (a) Find the product of two affine matrices and verify that its bottom row is [0, 0, 1]. Use this and properties of matrices to show closure for affine transformations. (b) Find the product of an affine matrix and a general 3 × 3 matrix. Show that the bottom row of the general 3 × 3 matrix must be [0, 0, 1] for the product to be the identity matrix. Use this and properties of matrices to prove the inverse of an affine transformation is an affine transformation. (c) Prove that affine transformations form a group, using parts (a) and (b). (d) Let k = [ p, q, r ] and m be another line. What conditions must m satisfy so that k ∥ m? Let α be an affine transformation. Prove that α(k) ∥ α(m). Hint: what does part (b) tell us about the general form of α −1 ? (e) Prove that affine transformations map rays to rays. 5.4.17. Prove that an affine matrix with determinant d changes the area of convex polygons by a factor of |d|, as follows. 51 p 06 (a) Show that a matrix A = 0 q 0 changes the area of a triangle by a fac0

0

1

tor of |q|. Hint: Show that A takes horizontal lines to horizontal lines and use Example 4. 5r 0 06 (b) Show a claim similar to part (a) for a matrix B = s 1 0 . 0

0

1

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(c) Show that an affine matrix M can be written as the product T AB (or T B A if the center entry of M is 0), where T is a translation, A is a matrix of the form in part (a), and B is a matrix of the form in part (b). (d) Use part (c) to extend parts (a) and (b) to any affine matrix. Hint: The determinant of a product is the product of the determinants. (e) Extend part (d) from triangles to convex polygons. 5 −1 2 0 6 5.4.18. Show that M = −1 1 0 maps the circle x 2 + y 2 = 1 to an ellipse as follows. 0

0

1

For (x, y, 1) on x 2 + y 2 = 1, show that (u, v, 1) = M(x, y, 1) satisfies 2u 2 − 6uv + 5v 2 = 1. Use Section 3.2 to verify that 2u 2 − 6uv + 5v 2 = 1 is the equation of an ellipse. Graph it.

5.4.19. Show that a triangle △ABC can be transformed to any other triangle △P Q R by some affine transformation as follows. (a) Give general coordinates for P, Q, and R. Find the matrix M that takes O = (0, 0, 1) to P, X = (1, 0, 1) to Q, and Y = (0, 1, 1) to R. Explain why M is a transformation provided that P, Q, and R aren’t collinear. (b) Let A, B, and C be noncollinear points. Use part (a) to show that there is an affine matrix N that takes A to O, B to X , and C to Y . (c) Prove that there is an affine matrix taking A to P, B to Q, and C to R. 5.4.20. Consider the IFS with just two matrices, ⎡ ⎤ 1/3 0 0 A = ⎣ 0 1/3 0 ⎦ and 0 0 1

⎡

1/3 D=⎣ 0 0

0 1/3 0

⎤ 2/3 0 ⎦. 1

(a) On a graph, show the unit square U and its images AU , DU , A AU , ADU , D AU , and D DU . (b) Describe and, as best as you can, draw the fractal of this IFS. (It is a very disconnected set called the Cantor set.) *(c) The matrices A and D of this problem are two of the four matrices in Example 5. Explain how the limit set of this IFS relates to the one shown in Figure 5.32.

*5.4.21. For ease of programming, IFSs are often restricted to maps on the unit square; that is, the points (x, y, 1) satisfying 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. Find restrictions on the coefficients of an affine matrix so that it will be a contraction mapping that sends the unit square into itself.

5.4.5 Sophus Lie The Norwegian Sophus Lie (1842–1899) achieved international status in mathematics for his profound work in continuous transformation groups. Lie met Felix Klein in Berlin, and they quickly became close friends. Together they went to Paris in 1870 and studied group theory, but the outbreak of the Franco–Prussian War in 1871 ended their studies. Lie decided to spend his enforced vacation hiking in the Alps. As a tall, blond stranger with poor French he was soon arrested as a spy. He spent a month in prison, working on mathematics problems. When he was freed, in part owing to a French mathematician’s efforts, he continued his hiking tour.

5.5 Transformations in Higher Dimensions; Computer-Aided Design

235

Both Lie and Klein were deeply influenced by the possibility that group theory could unify mathematical thinking. Lie first applied transformation groups in differential equations, classifying solutions. He searched more broadly for an understanding of all continuous transformation groups, a goal that hasn’t yet been achieved. He pioneered the study of the groups, now called Lie groups. He revealed their geometric structure and went on to develop algebras, also named for him, that matched them. Lie groups and Lie algebras are essential ideas in quantum mechanics, a part of physics, as well as in mathematics. Lie applied his profound understanding of transformations to solve in 1893 a geometry problem posed and partially solved by Hermann von Helmholtz. Helmholtz sought to characterize all “continuous homogeneous geometries”—intuitively, geometries in which rigid bodies could be freely moved. Lie used transformations to prove that the only such spaces have constant zero, negative, or positive curvature, corresponding to Euclidean, hyperbolic, spherical, and single elliptic geometries, respectively, in any number of dimensions.

5.5 Transformations in Higher Dimensions; Computer-Aided Design Transformations in three and more dimensions illustrate the power of linear algebra. We utilize the same algebraic approach we used in two dimensions to move the origin: we add an extra coordinate to the vectors and represent transformations by the correspondingly enlarged matrices. Our visualization may struggle with higher dimensions, but the similarity of the structure enables us to think geometrically. Isometries of spherical geometry enable us to generalize the definition of isometries to three and more dimensions. Interpretation. By a point in three-dimensional affine space, we mean a column vector (x, y, z, 1). By a three-dimensional affine matrix we mean an invertible 4 × 4 matrix whose bottom row is [ 0 0 0 1 ]. 31 0 0 p4 Example 1. The translation T = 00 10 01 qr moves points by p in the x-direction, q in 0

0

0

1

the y-direction, and r in the z-direction. The final column gives the image of the origin (0, 0, 0, 1). ♦ Exercise 5.5.1. Apply the reasoning from Section 5.3 to explain why the bottom row of an affine transformation must be [ 0 0 0 1 ].

5.5.1 Isometries of the Sphere Any affine transformation that maps the unit sphere to itself necessarily maps the origin to itself. Hence spherical isometries can be represented as 3 × 3 matrices, in effect the upper left corner of the 4 × 4 affine transformations. The isometries give insights about isometries in all dimensions and the symmetries of polyhedra. This generalizes the two-dimensional situation where the upper left 2 × 2 corner of isometries determined the type of isometry. 4 3 0 −1 0 Example 2. The transformation R = 0 0 −1 is a rotation of the sphere (and all of R3 ). 1

0

0

Points of the form (a, −a, a) = a(1, −1, 1) are fixed by ρ and so form the axis of rotation,

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which is the line through the origin and (1, −1, 1). (See Section 3.5 for a general description of lines in more than two dimensions.) Furthermore, R(a, b, c) = (−b, −c, a), R(−b, −c, a) = (c, −a, −b), and R(c, −a, −b) = (a, b, c), so the composition of R three times will take every point back to itself, showing the angle of rotation to be 120◦ . Equivalently, we can use matrix multiplication to see that R 3 = I , the identity matrix. In particular, the x-, y-, and z-axes map to one another, although the positive and negative directions intermix (Figure 5.36). The determinant of the matrix R is 1, just like the situation with two-dimensional rotations. ♦ Z

Y X

Figure 5.36 A rotation in 3 dimensions.

Exercise 5.5.2. Verify that the transformation M =

3

1 0 0

0 0 −1

0 −1 0

4

has determinant −1 and that

M 2 is the identity, indicating that M is a mirror reflection. It leaves the plane y + z = 0 fixed (Figure 5.37). z

y x

Figure 5.37 A mirror reflection in 3 dimensions. We can compose rotations and mirror reflections to form the only other isometries of the sphere, rotatory reflections, illustrated in Example 3 and explored in Project 5. Translations and glide reflections have no fixed points, so they aren’t isometries of the sphere.

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5.5 Transformations in Higher Dimensions; Computer-Aided Design

Example 3. The rotatory reflection ⎡ ⎤ ⎡ ⎤ ⎡ 0 −1 0 1 0 0 0 ⎣1 0 0 ⎦ = ⎣0 1 0 ⎦ · ⎣1 0 0 −1 0 0 −1 0

⎤ −1 0 0 0⎦ 0 1

is the composition of a rotation of 90◦ around the z-axis followed by a mirror reflection over the plane z = 0 (the equator) (Figure 5.38). The eigenvalues of the matrix are −1, i, and −i, which show that there is no fixed point. The antipodal North and South poles (0, 0, 1) and (0, 0, −1) are mapped to each other. ♦

Figure 5.38 A rotary reflection. Definition. A rotation in three-dimensional Euclidean geometry fixes the points on one line, called the axis of rotation, and rotates other points through an angle around it. A mirror reflection over a plane S in three-dimensional Euclidean geometry maps a point P to the point Q such that S is the perpendicular bisector of P Q. A rotatory reflection in three-dimensional Euclidean geometry is the composition of a rotation with a mirror reflection in a plane perpendicular to the axis of rotation. A spherical isometry maps the mutually perpendicular unit basis vectors (1, 0, 0), (0, 1, 0), and (0, 0, 1) to mutually perpendicular unit vectors. In turn, any mapping taking them to mutually perpendicular unit vectors is a spherical isometry. The geometric description leads to the algebraic description of spherical isometries in Theorem 5.5.1. Vectors ( p, q, r ) and (s, t, u) are perpendicular (orthogonal) if and only if ( p, q, r ) · (s, t, u) = ps + qt + r u = 0. ! p 2 + q 2 + r 2 . The transpose of a matrix M, written M T , The length of ( p, q, r ) is switches the rows and columns of M. (See Appendix D for a summary of linear algebra.) Theorem 5.5.2 uses linear algebra to give a more geometric description of spherical isometries. Theorem 5.5.1. The following statements are equivalent. (i) A 3 × 3 matrix M is an isometry of the unit sphere. (ii) M −1 = M T , the transpose of M. (iii) The columns of M form an orthonormal basis of R3 .

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Transformational Geometry

Proof. We show the equivalence of the first two parts and leave their connection with part 5a b c 6 (iii) to Exercise 5.5.12. (⇒) First suppose that M = d e f is a spherical isometry. Then g

h

i

M(1, 0, 0) = (a, d, g), M(0, 1, 0) = (b, M is a spherical ! e, h), and M(0, 0, 1)√= (c, f, i). Since ! 2 2 2 2 2 2 isometry, the images have unit length: a + d + g = 1 = b + e + h = c2 + f 2 + i 2 . Further, they are mutually perpendicular. Thus (a, d, g) · (b, e, h) = ab + de + gh = 0. Similarly, ac + d f + gi = 0 and bc + e f + hi = 0. Now ⎡ ⎤ ⎡ ⎤ a d g a b c MT · M = ⎣ b e h ⎦ · ⎣ d e f ⎦ c

⎡

f

i

a2 + d 2 + g2 = ⎣ ab + de + gh ac + d f + gi

g

h

i

ab + de + gh b2 + e2 + h 2 bc + e f + hi

⎤ ac + d f + gi bc + e f + hi ⎦ . c2 + f 2 + i 2

By the earlier deductions, this equality reduces to M T · M = I and so M T is the inverse of M, as claimed. ⎡ ⎤ a b c (⇐) Suppose M = ⎣ d e f ⎦ satisfies M −1 = M T . Then we have g h i ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ a d g a b c 1 0 0 I = MT · M = ⎣ b e h ⎦ · ⎣ d e f ⎦ = ⎣ 0 1 0 ⎦ . c f i g h i 0 0 1

The diagonal elements are the squares of the lengths of the image vectors (a, d, g), (b, e, h), and (c, f, i). Since the product is the identity, whose diagonal elements are all 1, the lengths of the images are 1. Furthermore, the off-diagonal entries of the identity matrix are 0, which shows that the image vectors are mutually perpendicular. Thus M is a spherical isometry. ! Theorem 5.5.2. Every isometry of the sphere has at least two antipodal points on it that are either fixed or are mapped to each other. The circle on the sphere midway between them is stable.

Proof. When we use eigenvalues to find fixed points (and stable lines), we obtain an equation in λ called the characteristic equation. The characteristic equation of a 3 × 3 matrix involves a third-degree real polynomial. All odd-degree real polynomials have a real root, so every isometry of the sphere has at least one real eigenvalue. For it, an eigenvector of length 1 is a point on the sphere. Eigenvectors are mapped to scalar multiples of themselves. The matrix is a spherical isometry, so the point will be mapped back onto the sphere. Thus the image also has length 1, forcing the real eigenvalue to be either 1 or −1. For an eigenvalue of 1, the eigenvector (point) is fixed, as is its negative (antipodal point). For −1, the antipodal points change places. In either case, the circle midway between them is stable because the transformation is an isometry. !

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239

5.5.2 Transformations in Three and More Dimensions We obtain all isometries of R3 by combining in our general 4 × 4 matrix a 3 × 3 submatrix representing an isometry of the sphere and the final column of the matrix, representing a translation. The schematic equation below represents this, where the letters s represent the values for the spherical isometry and t the values for the translation. In effect, we build an isometry by composing these two special cases. The upper left 3 × 3 submatrix describes the type of transformation, up to a translation. ⎡

s ⎢s ⎢ ⎢ ⎣s

s s s

s s s

⎤

t t t

0 0 0 1

⎡

1

⎥ ⎢0 ⎥ ⎢ ⎥=⎢ ⎦ ⎣0 0

0

0

1 0 0 1 0 0

⎤ ⎡

⎤ s s s 0 ⎢ ⎥ t⎥ ⎥ ⎢s s s 0⎥ ⎥·⎢ ⎥ t ⎦ ⎣s s s 0⎦ 1 0 0 0 1 t

In addition to the rotations, translations, mirror reflections, and glide reflections from two dimensions, there are two other types of three-dimensional isometries. Rotatory reflections, defined earlier as isometries of the sphere, are also isometries of the entire space. Figure 5.39 and Example 4 illustrate the other type of isometry, screw motions.

Figure 5.39 A screw motion.

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Transformational Geometry

Example 4. The matrix ⎡

cos θ ⎢ sin θ R=⎢ ⎣ 0 0

− sin θ cos θ 0 0

⎤ 0 0 0 0⎥ ⎥ 1 0⎦ 0 1

represents a rotation of θ around the z-axis. A point (0, 0, k, 1) on the z-axis is fixed by R. The plane z = k is stable under R because R(x, y, k, 1) has k for its third coordinate. A screw motion is the composition of a rotation and a translation in the direction of the axis of rotation—for example, ⎡ ⎤ ⎡ ⎤ cos θ − sin θ 0 0 1 0 0 0 ⎢ sin θ ⎢ ⎥ cos θ 0 0 ⎥ ⎥ , where Z = ⎢ 0 1 0 0 ⎥ . RZ = ⎢ ⎣ 0 ⎦ ⎣ 0 1 z 0 0 1 z⎦ 0 0 0 1 0 0 0 1 Verify that R Z = Z R. ♦

Exercise 5.5.3.∗ Give definitions for translations and glide reflections in three dimensions. 30 1 0 2 4 Example 5. Determine what type of isometry M = 10 00 01 −2 is and find its fixed points 0 0 0 0 1 and stable planes. Solution. The determinant of M is −1 and M 2 = I . So, as in the two-dimensional case, M is a mirror reflection, which we confirm with the fixed points. For the fixed points we solve ⎡ ⎤⎡ ⎤ ⎡ ⎤ 0 1 0 2 x y+2 ⎢ 1 0 0 −2 ⎥ ⎢ y ⎥ ⎢ x − 2 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣0 0 1 0 ⎦⎣ z ⎦ = ⎣ z ⎦, 0

0

0

1

1

1

which simplifies to x − y − 2 = 0. In two dimensions, this would be a line, but in three dimensions it is a vertical plane since the z-coordinate is free to vary. Because its fixed points form a plane, M is a mirror reflection. The general equation of a plane in R3 is ax + by + cz + d = 0. As with two-dimensional lines, we write the three-dimensional plane as the row vector [a, b, c, d]. For example, the plane of fixed points is written [1, −1, 0, −2]. We determine stable planes the way we found stable lines: the plane [a, b, c, d] is stable if and only if [a, b, c, d]M −1 = λ[a, b, c, d], where λ is an eigenvalue of M −1 . Now M −1 = M and its eigenvalues are ±1. Consider first λ = −1, which gives [a, b, c, d]M −1 = −[b, a, c, 2a − 2b + d] = [−b, −a, −c, −2a + 2b − d]. The first two coordinates give b = −a. The third coordinate forces c = 0. The last coordinate becomes d = −2a + 2b − d or 2d = −4a. As with two-dimensional lines, scalar multiples of a row vector represent the same plane. If we pick a = 1, we get [1, −1, 0, −2], the plane of fixed points. Now consider λ = 1, which gives [a, b, c, d]M −1 = [b, a, c, 2a − 2b + d]. Then a = b and c and d can be any values. The conditions give a family of planes [a, a, c, d], perpendicular to the plane of fixed points. ♦

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241

Theorem 5.5.3. Every three-dimensional Euclidean isometry can be written as the composition of at most four mirror reflections. There are three types of three-dimensional direct Euclidean isometries: translations, rotations, and screw motions and three indirect types: mirror reflections, glide reflections, and rotatory reflections. Proof. See Project 12 and Coxeter [4]. ! The definition and form for three-dimensional similarities mimics the form from twodimensions: we multiply an isometry by a dilation, which has a scaling factor r . (See Exercise 5.5.13.) As in the two-dimensional case, if r ̸= 1, there is a unique fixed point of the similarity. The argument in Exercise 5.4.12 of Section 5.4 for a fixed point applies equally well to three and higher dimensions. The determinant of a three-dimensional similarity with scaling factor r is ±r 3 . Exercise 5.5.13 considers what the size of the determinant tells us. By now the pattern may be clear: for points in n-dimensional affine space, use column vectors with n + 1 coordinates, the last of which is 1. The corresponding row vectors are, in general, called hyperplanes and are (n − 1)-dimensional.1 The affine transformations are 2 (n + 1) × (n + 1) invertible matrices whose bottom row is 0 . . . 0 1 . The upper left n × n corner tells us, up to a translation, what type of a transformation we have. For the entire matrix to represent an isometry in n dimensions, the upper left n × n corner must be a spherical isometry in n dimensions. Theorem 5.5.1 leads to the definition of isometries in higher dimensions. Definition. 1 An (n + 1) × 2 (n + 1) invertible matrix is an affine matrix if and only if its bottom row is 0 0 . . . 1 . An n × n matrix M is orthogonal if and only if M −1 = M T . An (n + 1) × (n + 1) affine matrix is a Euclidean isometry if and only if its upper left n × n submatrix is orthogonal. Exercise 5.5.4. Verify that two-dimensional isometries satisfy the definition. Exercise 5.5.5.∗ Define translations in n-dimensional space. What does the matrix of a translation in n dimensions look like?

5.5.3 Computer-Aided Design and Transformations A computer-aided design (CAD) program stores the reference points of a design as the columns in a matrix. In industrial applications, architecture and animated movies, there can be hundreds of them. Then for a rotation, zoom, or other change of the appearance of the design, the computer only needs to multiply the matrix of reference points by the matrix of the appropriate transformation. To make the appearance more realistic, the programs generally employ steps beyond the transformations of this chapter. For example, programs use matrices to show the design in perspective, discussed in Section 7.6. We will see there how the bottom row of the transformation matrix connects with perspective. In addition, the programs need to hide invisible parts of the design and color and shade surfaces, as in Figure 5.40. See Mortenson [9] for these and other technical aspects of CAD.

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Figure 5.40 Rotated view of the building design in Figure 3.0. Exercise Draw6 the quadrilateral in the plane whose four corners have the6 columns of 5 1 5.5.6. 5 0.6 −0.8 0 −2 −1 1 A = 0 2 1 −3 for their coordinates. For the rotation ρ = 0.8 0.6 0 of approxi1 1 1 1 0 0 1 5 1.6 −0.6 −1 2.8 6 ◦ mately 53 around (0.5, 1, 1), the product ρ A = 0.8 1.2 −1 −2.6 gives the matrix whose 1

1

1

1

columns are the images under ρ of the corners of the quadrilateral. Draw it on the same axes as the original one. Theorem 5.5.4. Let α be an n-dimensional affine transformation and A be an (n + 1) × k matrix whose columns A1 , A2 , . . . , Ak are k points in n-dimensional affine space. Then the columns of α A are α( A1 ), α( A2 ), . . . , α( Ak ).

Proof. See Exercise 5.5.14. ! As a result of Theorem 5.5.4, once the computer has been given the images of the reference points, it can redraw the lines, curves, and surfaces among them in the same manner as originally. The analytic geometry of Chapter 2 combined with the linear algebra of this chapter underlie the graphics of CAD. Computers also use matrices to present three-dimensional designs as two-dimensional graphics displays and printouts. The matrices, illustrated in Example 6, aren’t transformations since they are not one-to-one. (See Mortenson [9].) Example 6. The matrix

31 0 0 0

0 1 0 0

0 0 0 0

0 0 0 1

4

, called a projection mapping, sends a point onto the

x y-plane, which has the equation z = 0. Two points with the same x- and y-coordinates are mapped to the same point. Thus the matrix has no inverse and so is not a transformation. Its determinant is 0. ♦

5.5 Transformations in Higher Dimensions; Computer-Aided Design

243

5.5.4 Exercises for Section 5.5 *5.5.7. (a) Explain why ⎡

−1 ⎢ 0 ⎢ ⎣ 0 0 (b) (c)

(d) (e) (f)

0 −1 0 0

0 0 1 0

⎤ 0 0⎥ ⎥ 0⎦ 1

is a 180◦ rotation around the z-axis. Find the matrices for rotations of 180◦ around the x- and y-axes. What is their product? What is the product of one of them with the matrix in part (a)? Verify your answers in part (b) physically by rotating a cube 180◦ around the centers of two opposite faces, followed by a 180◦ rotation around the centers of two other opposite faces. Mark several points on the cube so that you can recognize their starting and ending positions. Repeat part (c) with 90◦ rotations and describe the resulting transformation. Find the matrices for the 90◦ rotations in part (d) and multiply them. Describe the product. Describe the matrix for a rotation of θ around the x-axis.

5.5.8. The central symmetry with respect to Q takes each point P to the point P ′ , where Q is the midpoint of P P ′ . In two dimensions, the isometry is a rotation of 180◦ , often called a half-turn. (a) Find the matrix for the central symmetry with respect to (0, 0, 0, 1). Find its determinant and decide what type of an isometry it is. Explain your answer. (b) Repeat part (a) for (u, v, w, 1). Hint: Find the image of (0, 0, 0, 1). (c) Describe the composition of two three-dimensional central symmetries. Verify your answer by multiplying two central symmetries with different centers. (d) Repeat parts (a), (b), and (c) for four and more dimensions. 5.5.9. *(a) Find the matrix for a screw motion made from a rotation of θ around the y-axis followed by a translation in the y-direction by a length of w. *(b) Do you get the same screw motion as in part (a) if you first translate and then rotate? Explain your answer. *(c) Let S be the matrix for the screw motion rotation of 120◦ around the y-axis followed by a translation in the y-direction by a length of 2. Describe the transformation given by S 3 . (d) Let T be the matrix for the screw motion rotation of 360◦ /n around the yaxis followed by a translation in the y-direction by a length of k. What is the smallest positive exponent p so that T p is the type of transformation S 3 was in part (b)? (e) Is a rotation of θ around the y-axis followed by a translation in the x-direction a rotation or a screw motion? Explain your answer. 5.5.10. (a) Find the matrix for a rotary reflection made from a mirror reflection over the plane x = 0 followed by a rotation of θ around the x-axis.

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(b) Do you get the same rotary reflection as in part (a) if you rotate first and then reflect? Explain your answer. (c) Let R be the rotary reflection of part (a) with an angle of θ = 90◦ . What is the smallest power of R equal to the identity matrix? Explain your answer. (d) Repeat part (c) using an angle of θ = 120◦ . (e) Repeat part (c) using an angle of θ = 360◦ /n. Hint: You need two cases, depending on n. 5.5.11. For each isometry, determine its fixed point(s) and stable plane(s), if any. Describe the type of isometry. For rotations, determine the angle of rotation. ⎡ ⎤ ⎡ ⎤ 0 0 −1 2 1 0 0 0 ⎢ 0 1 0 0⎥ ⎢ 0 0 −1 4 ⎥ ⎥ ⎥ *(a) A = ⎢ (b) B = ⎢ ⎣ −1 0 0 2 ⎦ ⎣0 1 0 2⎦ 0 0 ⎢1 *(c) C = ⎢ ⎣0 0 ⎡

0

0 0 1 0

1 0 0 0

0 1 ⎤ 0 0⎥ ⎥ 0⎦ 1

⎡

0 ⎢0 (d) D = ⎢ ⎣1 0

0 0 0 1 1 0 0 0 0 0

5.5.12. Finish the proof of Theorem 5.5.1.

0 1 ⎤ 0 0⎥ ⎥ 0⎦ 1

*5.5.13. (a) Define a three-dimensional similarity with a scaling ratio of r . (b) Explain why a three-dimensional similarity can be written as the product of an isometry and a dilation centered at the origin, represented by ⎡

⎤ r 0 0 0 ⎢0 r 0 0⎥ ⎢ ⎥ ⎣0 0 r 0⎦. 0 0 0 1 (c) What are the possible determinants of a three-dimensional similarity with a scaling ratio of r ? What does the determinant tell you about a three-dimensional object, such as a cube, and its image under a similarity? (d) Find the fixed point and stable plane of the similarity ⎡

0 ⎢ 0 ⎢ ⎣ −2 0

2 0 0 0

0 2 0 0

⎤ 0 4 ⎥ ⎥. −2 ⎦ 1

5.5.14. Prove Theorem 5.5.4. 5.5.15. Rotations in four dimensions extrapolate properties of rotations in two and three dimensions. For convenience, use 4 × 4 orthogonal matrices so that the origin (0, 0, 0, 0) is fixed. (a) Describe what is fixed by a rotation in two dimensions and by a rotation in three dimensions. What should be fixed by a rotation in four dimensions?

245

5.6 Inversions and the Complex Plane

(b) Verify the following matrices are orthogonal with determinants of +1. ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 0 −1 0 0 0 0 −1 0 1 0 0 0 ⎢1 0 0 0⎥ ⎢ ⎥ ⎢ ⎥ ⎥ , B = ⎢ 0 1 0 0 ⎥ , and C = ⎢ 0 1 0 0 ⎥ . A=⎢ ⎣0 0 1 0⎦ ⎣1 0 0 0⎦ ⎣ 0 0 0 −1 ⎦ 0 0 0 1 0 0 0 1 0 0 1 0

(c) The preceding matrices are rotations of the four-dimensional sphere. Find their fixed points (“axes”) and angles of rotation. (d) Find the products and fixed points of AB and AC. Describe how they differ. One of the products is again a rotation. Decide which product it is and find its angle of rotation. What can you say about the other product?

5.5.16. (a) Prove that the set of orthogonal n × n matrices is a transformation group. (b) Use the definition of an orthogonal matrix to prove that its determinant must be either +1 or −1. Why does the proof guarantee that the determinant of an n-dimensional isometry must also be +1 or −1? (As in two dimensions, the direct isometries have a determinant of +1, whereas the indirect isometries have a determinant of −1.) 5.5.17. The standard basis of Rn is the set {(1, 0, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, . . . , 0, 1)}. (a) Prove that an n × n matrix is orthogonal if and only if it maps the standard basis vectors to an orthonormal basis of Rn . (b) Explain why orthogonal n × n matrices will be isometries of the n-dimensional unit sphere. (c) Explain why the matrices we defined to be n-dimensional isometries fulfill the conditions of being isometries.

5.6 Inversions and the Complex Plane Affine transformations map lines to lines, but some important classes of transformations do not do so. Inversions form one important family of such “nonlinear” transformations. They appear in complex analysis and in the Poincar´e model of hyperbolic geometry. An inversion switches points on the inside of a circle with points on the outside of a circle (Figure 5.41). The center of the circle has no Euclidean point for its image, and no Euclidean point can be mapped to the center. Transformations must be one-to-one onto functions, so we add a point to the plane that can switch places with the center. Intuitively, the extra point must be at infinity, so we call it ∞. The new point is defined to be on every line. P Q'

Q

O

P'

R R'

Figure 5.41 An inversion.

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Transformational Geometry

Definition. The inversive plane is the Euclidean plane with an additional point, denoted ∞. An inversive line is a Euclidean line together with the point ∞. Let a circle C with center O and radius r be given. The inversion ν c with respect to C maps a Euclidean point P(P ̸= O) to ν c (P) −→ on the ray O P, where d(O, P) · d(O, ν c (P)) = r 2 . We define ν c (O) = ∞ and ν c (∞) = O. The center is O. Exercise 5.6.1.∗ Illustrate the inversion with respect to the unit circle x 2 + y 2 = 1. Verify that a line through the origin is mapped to itself, as is the unit circle. Explain why a circle with center (0, 0) is mapped to another circle with the same center. How are their radii related? Exercise 5.6.2.∗ Explain why an inversion is its own inverse. Theorem 5.6.1. Let k be a line not through the center of inversion O of ν c . Then the image of k is a circle through O. Conversely, the image of a circle through O is a line not through O. Proof. Let the perpendicular from O to k intersect k at A and let A′ = ν c ( A). Figure 5.42 illustrates the two cases, depending on whether k intersects the circle of inversion C. We show that the circle D with diameter O A′ is the inversive image of k. To do so, let P be a point on k ← → and P ′ the second intersection of O P with the circle D, besides O. (Explain why there must be a second point of intersection.) By Exercise 1.2.17, since O A′ is a diameter, ∠O P ′ A′ is a right angle. From Theorem 1.4.2, △O P ′ A′ ∼ △O A P. By the proportionality of the sides, there is some p such that d(O, P ′ ) = p · d(O, A) and d(O, A′ ) = p · d(O, P). Then d(O, P ′ ) · d(O, P) = p · d(O, A) ·

1 d(O, A′ ) = d(O, A) · d(O, A′ ). p

Since the product is constant and A′ = ν c ( A), then P ′ = ν c (P), as claimed. Thus D is the inversive image of k. For the other direction, from Exercise 5.6.2 P ′ = ν c (P) if and only if ν c (P ′ ) = P. Hence the construction shows the inversive image of a circle D through O must be the corresponding line k. !

k C

P P'

O

A' D

A

D

C O

A'

A P

k P'

Figure 5.42 The inversion of a line.

247

5.6 Inversions and the Complex Plane

Exercise 5.6.1 is a special case of Theorem 5.6.2. Another special case of that theorem, Theorem 5.6.3, is important in the Poincar´e model of hyperbolic geometry. Although the proof of Theorem 5.6.2 isn’t difficult, it requires lemmas from Euclidean geometry that would sidetrack us. Theorem 5.6.2. Let D be a circle that does not pass through the center of inversion of ν c . Then the inversive image of D is another circle that does not pass through the center of inversion. Proof. See Eves [6, 78]. ! Definition. Two circles are orthogonal if and only if their radii at their points of intersection are perpendicular. Theorem 5.6.3. If a circle D is orthogonal to the circle of inversion C, then D is its own image: ν c [D] = D. Proof. With D orthogonal to C, the radii of C that go to the intersections P and Q are tangents ← → to D (Figure 5.43). Because P and Q are on C, they are fixed by the inversion. Lines O P ←→ and O Q are stable. By Theorem 5.6.2, ν c [D] is a circle through P and Q. Furthermore, lines ← → ←→ O P and O Q must still be tangent to ν c [D], for each has one point of intersection with the circle. Thus the perpendiculars to the lines through P and Q intersect in the center of the circle. However, the center is also the center of D, which shows that ν c [D] = D. ! C P D

O Q

Figure 5.43 The inversion of an orthogonal circle. In the Poincar´e model of hyperbolic geometry, arcs of orthogonal circles are used as lines. (See Section 4.1.) Henri Poincar´e had the key insight that certain inversions correspond to mirror reflections in the model now named for him. In Figure 5.44, the points inside circle H are the points of the hyperbolic plane. Circle C is orthogonal to H , so its arc inside H is a hyperbolic line. By Theorem 5.6.3, the inversion with respect to the circle C maps H to itself. In particular, the inversion maps the interior of H to the interior of H . Hence it is a transformation: it switches points inside H that are inside the circle C with points outside C, but still inside H . ←−→ ← → For example, it switches the hyperbolic lines P Q and P ′ Q ′ . Using the definition of distance given in Section 4.5, Poincar´e showed that the transformation is actually a mirror reflection. Euclidean mirror reflections over diameters of circle H also are hyperbolic mirror reflections.

248

Transformational Geometry

Hyperbolic mirror reflections resemble Euclidean mirror reflections in two ways. First, they switch the orientation of figures. Second, all hyperbolic plane isometries can be written as compositions of three or fewer hyperbolic mirror reflections, analogous to Theorem 5.2.5. (See Project 22.)

H

P' Q P Q' O C

Figure 5.44 Inversions and hyperbolic mirror reflections. To understand hyperbolic isometries in the Poincar´e model we need to consider compositions of inversions, which generally are not inversions. Complex numbers provide a convenient way to find formulas for inversions and M¨obius transformations, which we discuss shortly. (See Durbin [5, 149-153] for a discussion of complex numbers.) Inversion is a two-dimensional analog of a function such as f (x) = 1/x (Figure 5.45), which inverts the real numbers with respect to a circle of radius 1 centered at 0. The real function f (x) = 1/x considers just one line through the center of inversion, and 1 and −1 are the only points on the circle. Complex numbers, of the form a + bi, provide a way to represent inversions on the plane. While we can represent the points of the plane with ordered pairs in R2 , we can’t multiply such two-dimensional numbers. Complex numbers do have multiplication, which we need to represent inversions. However, Example 1 shows that we need a more sophisticated function than just 1/x to represent inversions in the plane. Just as we needed to add a point, ∞, to the Euclidean plane to make inversions transformations, we need to extend the complex numbers. We write C# for the extended complex numbers, or the usual complex numbers and ∞, which is the limit of a + bi as a → ∞ or b → ∞ or both.

–2

–1

–1 2

0

–1 2

1

Figure 5.45 Inversion on the number line.

2

249

5.6 Inversions and the Complex Plane

Example 1. Consider the function f defined on the complex numbers by f (z) = 1/z. Figure 5.46 shows the images of various complex numbers. Verify that 1/(a + bi) = (a − bi)/(a 2 + b2 ) by multiplying by a + bi. Verify that f ( 21 i) = −2i, f (2 − 2i) = 14 + 14 i, and f (−1 − 2i) = − 15 + 25 i. Points inside the unit circle are mapped to points outside it, and vice versa. However, a point and its image are not on a line with the origin. Instead, there is a mirror reflection over the xaxis (real axis) in addition to the inversion. Fortunately, a simple complex function acts as a mirror reflection over that axis: complex conjugation. Recall that a + bi = a − bi. A complex number z and its conjugate z are mirror images with respect to the x-axis. Hence we can define the inversion with respect to the unit circle as ν(z) = 1/z. Verify that v(a + bi) = (a + bi)/(a 2 + b2 ), a positive scalar multiple of a + bi. ♦ 2i

–1 + i 1 2 –1 –1i 2 2

2

–1i 2

Figure 5.46 f (z) = 1/z is not an inversion. Theorem 5.6.4. The inversion with respect to the circle of radius r and center w is given by ν(z) = r 2 /(z − w) + w, for z ̸= w. Proof. First, we show that the formula works in the special case when w = 0 + 0i = 0. Call the function in this case ν 0 (z), which reduces to r 2 /z. The product z · r 2 /z has a length of r 2 , showing that r 2 /z is the correct distance from the origin. To complete the case, we show that 0, z, and r 2 /z are on the same ray. Let z = a + bi. Then 1/z = (a − bi)/(a 2 + b2 ), and r 2 /z = r 2 (a + bi)/(a 2 + b2 ) = r 2 z/(a 2 + b2 ), a positive real scalar multiple of z. We use the method in Example 4 of Section 5.3 to develop the general formula. The addition of w to a complex number is a translation τ of the complex plane. Then the formula of τ is τ (z) = z + w and similarly, τ −1 (z) = z − w. The formula ν(z) = r 2 /(z − w) + w gives the composition τ ◦ ν 0 ◦ τ −1 , which is an inversion because translations do not alter distance. To verify that it is the correct inversion, let z be a point on the circle of radius r and center w. Then z − w is a point on the circle of radius r at the origin. This implies that r 2 /(z − w) = z − w from the first part. The addition of w now takes this point back to z. Thus every point on the desired circle of inversion is fixed, proving the theorem. !

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Transformational Geometry

Example 2. Find the transformation of the extended complexes C# that is the composition of the inversions 1/z followed by 4/z. Solution. When we replace the z of 4/z with 1/z, we get 4/(1/z) = ((4z)) = 4z, a similarity (dilation) with a scaling factor of 4. ♦ Compositions of inversions can be similarities (as in Example 2), hyperbolic isometries, or other transformations. The transformation group that contains the inversions and their compositions is called the M¨obius transformations, after Augustus M¨obius, one of the first mathematicians to understand the importance of transformations in geometry. The M¨obius transformations leaving a circle H stable are the isometries for the Poincar´e model of hyperbolic geometry. All M¨obius transformations, as Theorem 5.6.6 shows, preserve angles, an important fact both in complex analysis and in geometry. In complex analysis, transformations preserving angles are called conformal. Definition. A M¨obius transformation is a function of the extended complex numbers C# that is one-to-one, onto, and has one of the two forms f (z) = ( pz + q)/(r z + s) or f (z) = ( pz + q)/(r z + s), where z ̸= −s/r . We define f (−s/r ) = ∞ and f (∞) = p/r , or, in the second form, f (∞) = p/r . The constants must satisfy ps − qr ̸= 0. Exercise 5.6.3.∗ Do the following to investigate the conditions in the definition above. Pick values of p, q, r , and s so that ps = qr . Why does the formula for f (z) no longer give a transformation? Use limits to explain why the definition f (−s/r ) = ∞ makes sense. Similarly, explain why f (z) = ( pz + q)/(r z + s) should go to a limit of p/r as z goes to infinity. Explain how to rewrite the inversions of Theorem 5.6.4 as M¨obius transformations. Theorem 5.6.5. The set of M¨obius transformations is a transformation group. Proof. See Exercise 5.6.11. ! Theorem 5.6.6. M¨obius transformations preserve angle measure. Proof. A general M¨obius transformation can be written as a composition of similarities and inversions. (See Exercise 5.6.12.) We already know from Theorem 5.4.3 that similarities preserve angles. Hence we only need to show that inversions preserve angles. To simplify Figure 5.47, we will start with the angle between two intersecting lines k and m. (One or both of k and m can be a circle. Then we use the tangent line(s) to form the angle. Thus this case represents the general situation.) The angle between two curves is the angle that their tangents make at their intersection. As shown in Figure 5.47, we are given the angle at P and are to show that the corresponding angle at P ′ has the same measure. In Figure 5.48 we first focus on just one line k and its image ← → ν c [k]. We compare the angles that the image and its tangent at P ′ make with O P. We show that ∠P P ′ Q ′ ∼ = ∠O P A. In Figure 5.48, the image υ c [k] is a circle with center O ′ . Then ∠O P ′ A′ is a right angle ←−→ because O A′ is a diameter. Also ∠O ′ P ′ Q ′ is a right angle because P ′ Q ′ is tangent to the circle

251

5.6 Inversions and the Complex Plane

Vc[m]

m

P

P' O Vc[k] C

k

Figure 5.47 Angles and inversions.

C

P P' Q' O

O'

A'

A

k'

k

Figure 5.48 Restricted to line k and its image. and O ′ P ′ is a radius. Thus ∠O P ′ O ′ ∼ = ∠A′ P ′ Q ′ . Also ∠P ′ O O ′ ∼ = ∠O P ′ O ′ because △O O ′ P ′ ′ ′ ′ ∼ ′ ′ is isosceles. Thus ∠A P Q = ∠P O O . Angles ∠O P A and ∠P ′ O O ′ are complementary because they are in a right triangle. Angles ∠P P ′ Q ′ and ∠A′ P ′ Q ′ together form a right angle, so they are complementary. Hence ∠P P ′ Q ′ ∼ = ∠O P A, as required. We prove the other ← → congruence similarly. Since the inversion preserves both of the angles with O P, it preserves their sum, finishing the proof. ! The fundamental concept of a transformation has great significance in mathematics, linking geometry and algebra. Transformations and their groups are important throughout mathematics

252

Transformational Geometry

and many of its applications. Other areas use the transformations we have studied, as well as others. For example, topology considers far more general transformations focusing on continuity, allowing bending, but not cutting.

5.6.1 Exercises for Section 5.6 *5.6.4. Use the circle with equation x 2 + (y − 2)2 = 4 as the circle for the inversion ν. Find the coordinates of the images of the following points. Graph the circle, the points, and their images on graph paper. A = (0, 1), B = (2, 0), C = (0, 0), D = (1, 3). 5.6.5. Let C be the circle with equation x 2 + y 2 = 4 and D be the circle with equation x 2 + y 2 = 1. *(a) Find the coordinates of ν D (1, 0), ν C (ν D (1, 0)), ν C (1, 0), and ν D (ν C (1, 0)). Are ν D ◦ ν C and ν C ◦ ν D equal? *(b) Find the coordinates of ν D (0, 3), ν C (ν D (0, 3)), ν C (0, 3), and ν D (ν C (0, 3)). (c) Graph the circles, the points and their images on graph paper. Label the points and their images. (d) Describe what the transformations ν D ◦ ν C and ν C ◦ ν D are. 5.6.6. Let C be the circle with equation x 2 + y 2 = 4 and D be the circle with equation (x − 3)2 + y 2 = 4. (a) Find the intersection P and Q of these circles. What are the images ν D (P), ν C (ν D (P)), ν D (Q), and ν C (ν D (Q))? Explain. (b) Find the coordinates of ν D (1, 0), ν C (ν D (1, 0)), ν D (5, 0), and ν C (ν D (5, 0)). (c) Graph the circles, the given points and their images on graph paper. Label the points and their images. (d) The points P, Q, (1, 0), and (5, 0) are on the circle D. Verify that their images are on a circle. Find its equation. *5.6.7. (a) Find distinct circles C and D for which ν C (4, 0) = ν D (4, 0) = (1, 0). (b) Explain why the center of a circle C or D for part (a) must be on the xaxis. Can every point on the x-axis be the center of such a circle of inversion? Explain. (c) Find a formula for the radius of the circle from part (b) in terms of the center. Hint: Look first at centers to the left of (1, 0). Consider the distances from the center to (1, 0) and to (4, 0). 5.6.8. (a) Let P be a point outside the circle of inversion C with center O. Draw the tangents from P to C and let their points of intersection with C be Q and R. Prove that the ← → ← → inversive image of P is the point of intersection of O P and Q R. Illustrate your proof. (b) Reverse the process of part (a) to give a construction of the inversive image of a point that is inside the circle of inversion. Explain your construction. Prove it correct. ← → ← → 5.6.9. In Exercise 5.6.8 (a) the line Q R is perpendicular to O P and through the inversive image of P. Such a line is called the polar of P, and P is called the pole of the line

5.6 Inversions and the Complex Plane

253

← → Q R with respect to the circle C. We can use the definition even when P is not outside of the circle. *(a) For the circle x 2 + y 2 = 4, find the polars s and t of the points S = (2.5, 0) and T = (1.6, 0). Graph the circle, the points and their polars. (b) Prove that, if a point U is on the polar of a point P, then P is on the polar of U . Draw a diagram for your proof. 5.6.10. Suppose that a circle D passes through a point P and the inversive image P ′ of P with respect to the circle C, where P ̸= P ′ . Prove that D is orthogonal to C. Hint: Assume that three noncollinear points determine a circle and that a path from inside a circle to outside the circle must intersect the circle. 5.6.11. Let P ′ , Q ′ , and R ′ be the inversive images of P, Q, and R, respectively, with respect to a circle C. Does Theorem 5.6.6 imply that △P Q R ∼ △P ′ Q ′ R ′ ? Explore this question using the circle x 2 + y 2 = 25 and the points P = (3, 0), Q = (0, 4), and R = (3, 4). Find the images of the points and m∠P R Q. Verify visually that m∠P ′ R ′ Q ′ is much greater than 90◦ . Explain what angle in the image is congruent to ∠P R Q. *5.6.12. This problem investigates isometries and similarities as M¨obius transformations using complex numbers. (a) Describe the similarity f (z) = z + (a + bi). (b) If r is a nonzero real number, what similarity is f (z) = r z? If p + qi is on the unit circle, what similarity is f (z) = ( p + qi)z? If a + bi is a nonzero complex number, what similarity is f (z) = (a + bi)z? Hint: Write a + bi as r ( p + qi). (c) What similarity is f (z) = z? (d) Prove that f (z) = ( pz + q)/s and f (z) = ( pz + q)/s are similarities for complex numbers p, q, and s, where p and s are not 0. (e) Explain how to write a M¨obius transformation as the composition of a similarity and at most one inversion. *5.6.13. M¨obius transformations are used to convert the half-plane model to the Poincar´e model and vice versa. As Poincar´e knew, the procedure implies that the hyperbolic isometries of the half-plane model are M¨obius transformations. √ (a) Find the function for the inversion ν C with respect to a circle with radius 2 and center at −2i. (b) Find the function for the inversion ν D with respect to a circle with radius 2 and center at −3i. (c) Find a circle E that is the image of the real axis under ν C . Verify that the points above the axis (a + bi, with b > 0) are mapped to the interior of the circle. Draw a picture. (d) Verify that the unit circle is the image of circle E under ν D . Verify that the points inside E are mapped to points inside the unit circle. Draw a picture. (e) The composition ν D ◦ ν C maps the half-plane model to the Poincar´e model. Find the image of (0, 1) = i from the half-plane model in the Poincar´e model.

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(f) Find a composition of inversions that converts the Poincar´e model to the half-plane model. 5.6.14. (a) Show that the composition of two M¨obius transformations is a M¨obius transformation. Hint: First leave out the complex conjugates. Then describe how complex conjugates affect the compositions. (b) Show that the inverse of a M¨obius transformation is a M¨obius transformation. Hint: solve w = ( pz + q)/(r z + s) for z. (c) Prove that the M¨obius transformations form a transformation group. 5.6.15. In complex analysis C# is often represented on the surface of a sphere. Imagine a sphere of radius 0.5 with its south pole on the origin of the complex numbers. Then match complex numbers with the points on the sphere as follows. Draw a line from the north pole to a complex number. Where the line intersects the sphere (besides the north pole) is the matching point. The north pole acts as the infinity point ∞. (In reverse, the process is called the stereographic projection of the sphere.) (a) Illustrate the process of matching complex numbers to the points on the sphere. *(b) What on the sphere corresponds to the unit circle? What spherical isometry corresponds to the inversion with respect to the unit circle?

5.6.2 Augustus M¨obius Augustus M¨obius (1790–1868) earned his living as an astronomer in Leipzig, Germany, but achieved international recognition as a geometer. His absentminded concentration on mathematics often caused him to forget his keys or other things. He was very shy, which may have led him to avoid a controversy at the time between geometers about which approach, synthetic or analytic, was superior. From a modern vantage point, the quarrel seems pointless because the approaches complement one another. His work built on both. In 1827 he invented barycentric coordinates and later developed them into homogeneous coordinates for projective geometry. (See Sections 3.3 and 7.3.) Julius Pl¨ucker (1801–1868) developed the coordinates into the first analytic model for projective geometry, then the cutting edge of synthetic geometry research. Forty-five years before Klein’s Erlanger Programm, M¨obius investigated transformations in geometry. Even without the powerful framework of group theory, he was able to initiate the study of isometries, similarities, affine transformations, and the transformations of projective geometry. He also developed the geometry of inversions and investigated the complex transformations that now bear his name. Later in life, he initiated the study of topological transformations. At age 68, he discovered the M¨obius strip, a mathematical model with the curious topological property that it has just one side and one edge, as shown in Figure 5.49.

5.6.3 Projects for Chapter 5 1. Place two mirrors at an angle facing each other with a shape in their interior. Investigate the multiple images of the shape with the aid of a protractor. (a) How do the images move as you move the original shape closer to one of the mirrors? (b) For mirror angles of 90◦ , 60◦ , 45◦ , and smaller, count the number of images (plus the original shape) that you can see. Find a formula relating the angle and the number of images.

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Figure 5.49 A M¨obius strip. (c) Use an asymmetric shape so that you can distinguish the orientation of the images. Describe the orientation of successive images. For various mirror angles, measure as best as you can with a protractor the angle between the original shape and the first image having the same orientation. How does the angle relate to the mirror angle? 2. Place two mirrors parallel and facing each other with a shape in their interior. Investigate the multiple images of the shape. (a) How do the images move as you move the shape closer to one of the mirrors? (b) Use an asymmetric shape so that you can distinguish the orientation of the images. Describe the orientation of successive images. For various distances between the mirrors, measure as best as you can with a ruler the distance between the original shape and the first image having the same orientation. Relate the distance between the mirrors and the distance between the original and the image. 3. Place three mirrors facing each other to make three sides of a square with an asymmetric shape inside the square. Investigate the multiple images of the shape. (a) Make a diagram showing the various images and their orientations. (b) Which images have the same orientation as the original shape and which have the opposite orientation? (c) Indicate on the diagram in part (a) which isometry produces each image. 4. Use dynamical geometric software, such as Geometer’s Sketchpad or Geogebra, to explore the effects of isometries. (a) Make △ABC. Create other points D and E as centers of rotation and lines j, k, l, and m for mirror reflections, where j ⊥ k, k ∥ l, but m is not perpendicular or parallel to any of the other lines. (b) Find the image of △ABC under a 180◦ rotation around D followed by a 180◦ rotation around E. Repeat switching the order of rotation. What isometries are these compositions? How are they related to each other? (c) Repeat part (b) using 90◦ rotations. (d) Find the image of △ABC under pairs of reflections using lines j, k, l, and m, in different orders. What isometries are the compositions? How are two compositions in the opposite order related to each other? (e) Find the image of △ABC under the composition of three reflections over the lines j, k, and l in different orders. Which compositions give the same isometry? What type of isometry is the composition? (f) Investigate the isometry resulting from composing a rotation and a mirror reflection.

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5. Place a mirror face up on a horizontal surface and place two mirrors vertically on it, facing each other and forming an angle of 60◦ . Place an asymmetric object in the trihedral angle formed by the mirrors. You should see eleven images of this object, in addition to the object itself. Some of the images are mirror images of the original, some are rotations of it and some are rotary reflections of it. Identify which type of isometry gives each image. Repeat with other angles, including 45◦ , between the two vertical mirrors.

Figure 5.50 6. On a transparency, draw axes and randomly insert numerous small dots (Figure 5.50). Photocopy it and align the transparency on top of the copy. (a) Rotate the transparency by a small angle relative to the paper around their common origin. Describe the pattern that the two sets of dots form. Translate the transparency relative to the paper and describe the resulting pattern. Try this procedure with other translations and small rotations. What can you say about the composition of a rotation followed by a translation? (b) Switch the order and repeat part (a). What can you say about the composition of a translation followed by a rotation? Do you get the same transformation regardless of the order? (c) Each composition in part (a) has a fixed point. Keep the initial angle of rotation the same and describe what happens to the fixed point as you use increasingly long translations in the same direction. (d) Repeat part (c), using the switched order of part (b). Compare the fixed points with those in part (c). (e) Experiment with two rotations of small angles around different points. Based on the two centers and the angles of rotation, can you predict where the new center of rotation will be?

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7. Use dynamical geometric software, such as Geometer’s Sketchpad or Geogebra, to explore the effects of similarities. (a) Make a triangle, dilate it about a center C and then rotate the image around a different center D. Try visually to find the fixed point of the composition of the dilation and rotation. (b) Switch the order of the transformations in part (a). Is the center of the new composition the same as the center of the similarity in part (a)? (c) Redo parts (a) and (b) using a mirror reflection instead of a rotation. 8. Let µ1 , µ2 , and µ3 be the mirror reflections over the angle bisectors of a triangle. Explain why µ1 ◦ µ2 ◦ µ3 ◦ µ1 ◦ µ2 ◦ µ3 is the identity. Generalize to angle bisectors of other convex polygons. Hint: First consider polygons with an odd number of sides. (See Berglund and Taylor [3].) 9. Investigate geometric applications of inversions. (See Eves [6, 84–91] and Greenberg [7, 257–262].) 10. Investigate transformations in CAD programs. (See Mortenson [9].) 11. Investigate iterated function systems (IFSs). (See Barnsley [2] and look online for free software to make designs. For example, /ecademy.agnesscott.edu/˜lriddle/ifskit/.) 12. Prove the three-dimensional analogs of Theorems 5.2.2–5.2.5. (a) A three-dimensional Euclidean isometry fixing four points not all in the same plane is the identity. Hint: What can the intersection of two spheres be? (b) A three-dimensional Euclidean isometry is determined by where it maps four points not all in the same plane. (c) Given two distinct points in three-dimensional Euclidean space, there is a unique mirror reflection switching them. (d) A three-dimensional Euclidean isometry can be written as the composition of at most four mirror reflections. 13. State the analogs of Theorems 5.2.2–5.2.7 in n-dimensional Euclidean geometry. 14. State the analogs of Theorems 5.4.1–5.4.3 for three-dimensional similarities and prove them. 15. Define similarities in n dimensions. State the analogs of Theorems 5.4.1–5.4.3 for ndimensional similarities. 16. Investigate Theorem 5.4.6 on convex sets in n dimensions. 17. Investigate stereographic projections and other mappings of a sphere to a plane. (See Hilbert and Cohn-Vossen [8, 248–263].) 18. Investigate dynamical systems. (See Abraham and Shaw [1].) 19. Investigate topological transformations. Describe properties preserved by transformations that are more general than affine ones. (See Smart [11, Chapter 8].) 20. Investigate the use of transformations in biology. (See Thompson [12].) 21. (Calculus) Vertical lines play a special role in calculus because of the vertical line rule: functions have just one y-value for any x-value. (a) Prove5that an affine 6 matrix maps vertical lines to vertical lines if and only if it is of the form

a d 0

0 e 0

c f 1

. Prove such matrices form a transformation group.

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(b) Investigate what happens to the points (x, x 2 , 1) on the parabola y = x 2 under the 52 0 16 matrix 1 1 0 . Graph the resulting curve. Does the minimum point on the original 0

0

1

curve get mapped to the minimum point on the image? Find the equation of the new function. Hint: If w = 2x + 1, write y = f (w) in terms of x and then replace x by (w − 1)/2. Explain why this hint works. (c) Show that the affine matrix in part (a) transforms the function y = g(x) to the function h(ax + c) = d x + e · g(x) + f . (d) If the function g in part (c) has a derivative at every point, what can you say about the derivative of the function h? Can you find its relative maxima and minima? Be sure that your answers match with what you found in part (b). Experiment with other polynomials for g. What can you say about the second derivatives of g and h from part (c)? 22. We generalize Theorems 5.2.2 to 5.2.5 to include spherical, single elliptic, and hyperbolic geometries as well as Euclidean geometry. (See Chapter 4 for more on these geometries.) (a) For the proof of Theorem 5.2.2, we need to show that two distinct circles in any of the geometries intersect in at most two points. Suppose that P and Q are intersections of the circles with centers A and B. Prove △A P B ∼ = △AQ B. Let M be the midpoint of P Q. Prove AB is the perpendicular bisector of P Q. Now suppose, for a contradiction, that the circles had a third point R of intersection. What can you say about P R and Q R? Finish generalizing the proof of Theorem 5.2.2. (b) Why do the arguments in part a) not hold in taxicab geometry? (See Section 2.3 for more on taxicab geometry.) (c) Explain why the proofs of Theorems 5.2.3 and 5.2.4 remain valid in the other geometries. (d) Verify the proof of Theorem 5.2.5 remains valid in them. (e) In spherical and single elliptic geometry, any pair of lines intersect. Use this fact to classify the types of isometries in these geometries. (f) In hyperbolic geometry, two lines can intersect, be sensed parallel, or ultraparallel. Use this to list the possibilities for types of isometries in hyperbolic geometry. Remark. There are actually only five types of hyperbolic isometries: (1) A composition of one mirror reflection is a mirror reflection. (2) The composition of two mirror reflections whose lines intersect is a rotation. (3) The composition of two mirror reflections whose lines are sensed parallel is a horolation. It acts like a rotation whose center is an ideal point. The repeated images of a point under such a composition lie on a horocycle. (4) The composition of two mirror reflections whose lines are ultraparallel can be called a hyperbolic translation. It is a translation for points on the common perpendicular of the lines. Points not on the common perpendicular must remain the same distance from the common perpendicular but their repeated images do not lie on a line, but rather on hypercycles. (5) The composition of three mirror reflections is either a mirror reflection or a hyperbolic glide reflection, built from a mirror reflection and a hyperbolic translation along

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23.

24.

25. 26.

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the line of reflection. (See Greenberg [7] for more on hyperbolic geometry and its isometries.) Determine the isometries and similarities of taxicab geometry as follows. (See Example 3 of Section 2.3 and Exercise 2.3.16.) (a) Verify that all Euclidean translations are taxicab isometries. (b) Suppose a taxicab isometry τ fixes (0, 0). Consider the points (±1, 0) and (0, ±1), which are all a distance 1 from (0, 0). Hence τ must take them to the taxicab unit circle. Find the distances between any two of the points (±1, 0) and (0, ±1). Verify that no other set of four points on the taxicab unit circle has those distances. (c) Use part (b) to describe all isometries fixing (0, 0). (d) Use parts (a) and (c) to give the matrix form of all taxicab isometries. (e) Use part (d) to give the matrix form of all taxicab similarities. Recall that an equivalence relation, ≡, on a set S is reflexive, symmetric, and transitive. (See Sibley [10, 150 ff].) (a) Suppose that G is a group of transformations on a set S and that, for a, b ∈ S, we define a ≡ b whenever there is α ∈ G such that α(a) = b. Prove that ≡ is an equivalence relation. (b) Suppose that ≡ is an equivalence relation on a set S and that we define G to be the set of all transformations α on S such that, for all a ∈ S, a ≡ α(a). Prove that G is a transformation group. (c) Let S be the set of all lines in the Euclidean plane and interpret ≡ as parallel. What is G in part (b)? (d) Let G be the transformation group of isometries of the Euclidean plane and S be the set of all line segments. Describe ≡. (e) Repeat part (d) with S being the set of all triples of points. (f) Describe other groups of transformations and equivalence relations. Investigate the history of transformational geometry. (See Yaglom [13].) Write an essay discussing Klein’s definition of geometry in light of the variety in groups of transformations presented in this chapter.

5.6.4 Suggested Readings 1. Abraham, R., and C. Shaw (eds.), Dynamics—The Geometry of Behavior (4 vols.), Santa Cruz, CA: Ariel Press, 1982–1988. 2. Barnsley, M., Fractals Everywhere, New York: Academic Press, 1988. 3. Berglund, J. and R. Taylor, Bisections and reflections, Mathematics Magazine, 2014, 87(4): 284–290. 4. Coxeter, H., Introduction to Geometry, 2nd ed. New York: John Wiley & Sons, 1969. 5. Durbin, J., Modern Algebra: An Introduction, 6th ed., Hoboken, NJ: Wiley, 2009. 6. Eves, H., Fundamentals of Modern Elementary Geometry, Boston: Jones and Bartlett, 1992. 7. Greenberg, M., Euclidean and Non-Euclidean Geometries, San Francisco: W. H. Freeman, 1980. 8. Hilbert, D., and S. Cohn-Vossen, Geometry and the Imagination, New York: Chelsea, 1952.

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Mortenson, M., Geometric Modeling, 3rd ed., New York: Industrial Press, 2006. Sibley, T., Foundations of Mathematics, Hoboken, NJ: Wiley, 2009. Smart, J., Modern Geometries, 3d Ed. Pacific Grove, CA: Brooks/Cole, 1988. Thompson, D., On Growth and Form, New York: Cambridge University Press, 1992. Yaglom, I., Felix Klein and Sophus Lie, Boston: Birkh¨auser, 1988.

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Symmetry

Figure 6.0 The rules of symmetry restrict how an artist can fit copies of a motif (such as this Iranian design) together to make a design. Archaeologists and anthropologists have started using symmetry in their study of designs to provide greater insight into cultures. Chemists and physicists use symmetry to organize new discoveries, to analyze empirical evidence, and to suggest fruitful lines for future inquiries. Knowing the fundamental concepts and mathematics of symmetry increases understanding in many subjects. Symmetry, as wide or as narrow as you may define its meaning, is one idea by which man through the ages has tried to comprehend and create order, beauty, and perfection. —Hermann Weyl The investigation of the symmetries of a given mathematical structure has always yielded the most powerful results. —Emil Artin

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6.1 Overview and History Repeated patterns abound in nature, and artists in virtually every culture and time have used repeated patterns in their designs. The repetition of a motif underlies symmetry, whether in a bilateral Mexican design (Figure 6.1) or the intricate atomic structure of a diamond crystal (Figure 6.2). Symmetry combines aesthetic and practical values which happily augment one another.

Figure 6.1 A Mexican Design.

Figure 6.2 The lattice structure of a diamond crystal. The bodies of most animals illustrate bilateral symmetry; that is, a mirror reflection interchanges the two sides of the animal. Hunting lions as well as hunted antelopes need the ability to turn left as readily as right and to hear from each side equally well. However, feet are useful only underneath an animal, so there is no evolutionary advantage to a symmetry between up and

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down. Similarly, running backward isn’t important for either predator or prey and we find no symmetry between the front and the back of animals. Hence the practical needs of most animals only require symmetry between right and left, but no other symmetry. Exercise 6.1.1.∗ Some animals, such as jellyfish, have more than bilateral symmetry. What makes extra symmetry advantageous for these animals? Evolution can explain symmetry in animals, but it provides no insight into the aesthetic appreciation people have for symmetry. The unity and balance of symmetric objects seems to appeal to people of all cultures. In his classic book on symmetry [23, 3] Hermann Weyl writes “Symmetric means something like well-proportioned, well-balanced, and symmetry denotes that sort of concordance of several parts by which they integrate into a whole. Beauty is bound up with symmetry.” Human beings have used symmetry in art for thousands of years. Albrecht D¨urer (1471– 1528) and others studied symmetry in art. However, a mathematical approach to symmetry required a shift from a static viewpoint to a dynamic one. Group theory and transformational geometry provided the mathematics needed to study symmetry. The symmetries of a figure are the transformations under which the figure is stable, and they always form a group. (See Section 5.1 for the definitions of stable, transformation, and group.) Thus the evolution of the concept of a group in the nineteenth century is inseparable from symmetry in algebra and geometry. Joseph Louis Lagrange (1736–1813) started investigating transformations (symmetries) of ´ the roots of polynomials in 1770. Evariste Galois (1811–1832) pursued these ideas and in the process developed many important theorems and concepts of group theory. Chemistry provided another impetus for the study of symmetry, especially the classification in 1830 of all possible types of chemical crystals. The German professor of mineralogy J. H. C. Hessel (1796–1872) developed the classification long before x-ray crystallography in 1912 confirmed his geometrical conclusions. Indeed, his classification predicted some crystal types discovered later. Auguste Bravais (1811–1863) used transformations and symmetry to deepen this understanding. A third strand leading to the mathematical study of symmetry comes from the geometric transformations pioneered by Augustus M¨obius (1790–1868). Camille Jordan (1838–1922) united the ideas in the first book on group theory in 1870. Groups organized and simplified all of these areas of research and more in important ways. Felix Klein (1849–1925) and Sophus Lie (1842–1899) extended the use of transformations and groups throughout geometry. Before 1905 Hendrik Lorentz and Henri Poincar´e determined a group of symmetries matching some recent experimental and theoretical findings in physics. In 1905 Einstein’s theory of special relativity provided a physical explanation unifying the ideas. In the following decades symmetry groups became essential in the development of quantum mechanics, another major area of physics. The classification of symmetry groups has supplied scientists and others with a clear understanding of the patterns that can be found in their areas. Other insights about symmetry by twentieth century mathematicians, including H. S. M. Coxeter, have affected many disciplines. Symmetric patterns, especially in physics, are often formal rather than visual. Even so, the same geometric intuition underlies symmetry. Questions from other disciplines have stretched the notion of symmetry and raised new mathematical questions. The beauty of the mathematics of symmetry and the beauty of symmetric objects have inspired the study of symmetry.

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Definition. A transformation σ is a symmetry of a subset T of a geometric space if and only if σ [T ] = T . (Individual points in T can move to other points in T , but T is stable.) A motif of a design is a basic unit from which the entire design can be obtained as its images under the symmetries of the design. The possible symmetries of a figure are usually some of the isometries of the larger space. Exercise 6.1.10 and Section 6.6 consider some other possibilities. In any case, as Theorem 6.1.1 shows, the symmetries of a set form a transformation group, called the symmetry group of that set. Theorem 6.1.1. The symmetries of a subset T form a transformation group. Proof. Recall from Section 5.1 that a group of transformations needs to have closure, identity, and inverses. If α is a symmetry of T , then α[T ] = T . For closure, let α and β be symmetries of T . Then α ◦ β[T ] = α(β[T ]) = α[T ] = T . Thus α ◦ β is also a symmetry of T . The identity ι clearly takes any subset T to itself. Finally, let α be a symmetry of T and α −1 its inverse, as guaranteed by Theorem 5.1.2. Then α −1 [T ] = α −1 (α[T ]) = ι[T ] = T , showing α −1 also is a symmetry of T . ! Example 1. The symmetry group of the design shown in Figure 6.1 contains only the identity and a mirror reflection. Either half of the design can be considered as the motif. The symmetry group of the diamond crystal shown in Figure 6.2 has infinitely many symmetries, if we assume that the design continues forever in all directions. For example, some translations will slide the entire crystal over to coincide with itself. Rotations of 120◦ around the lines representing the bonds also are symmetries, as well as some mirror reflections. The motif in this case is one carbon atom and the bond attaching it to one its neighbors. ♦ Exercise 6.1.2.∗ Describe the symmetries of the Iranian design of Figure 6.0 at the beginning of this chapter. The variety of artistic motifs is limited only by artists’ imaginations. However, they can fit symmetrically in relatively few ways, which we will classify. In this chapter we discuss symmetry in two and three dimensions and present applications of symmetry in the sciences and other fields. In Section 6.6 we investigate fractals, a new area of mathematics and science that stretches the idea of symmetry in an intriguing direction. (Stewart [19] and Yaglom [24] provide further historical information; Weyl [23] is a classic study of the ideas of symmetry.)

6.1.1 Exercises for Section 6.1 *6.1.3. (a) Classify the letters of the alphabet in terms of their symmetries. Consider uppercase (capital) and lowercase letters in any font you choose. (Possible symmetries include vertical mirror reflections, horizontal mirror reflections, and 180◦ rotations.) (b) Find the longest word you can that has vertical bilateral symmetry (a vertical mirror reflection). Repeat for horizontal bilateral symmetry. Repeat for 180◦ rotational symmetry.

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(c) Make up a sentence that is a palindrome; that is, it has the same order of letters backward as forward. How does the symmetry of a palindrome differ from vertical bilateral symmetry? 6.1.4. If a figure has both horizontal and vertical mirror reflections as symmetries, must it have any other symmetry? Illustrate and explain. *6.1.5. Describe the symmetries of each design shown in Figure 6.3.

Figure 6.3 6.1.6. Find the next few shapes in the sequence shown in Figure 6.4.

Figure 6.4 *6.1.7. In many cultures some symmetry patterns include two interchangeable colors. For each design in Figure 6.5 identify the symmetries that keep the colors the same in it. Identify the symmetries that switch colors.

Figure 6.5 6.1.8. For each design in Exercise 6.1.7 determine whether the symmetries keeping the colors the same form a group. Repeat for the symmetries switching colors. Explain. 6.1.9. *(a) The designs shown in Figure 6.6 are intended to continue forever in a line. They represent infinite designs that have translations as symmetries. Describe the types of symmetries of each.

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(b) The symmetry groups for the designs shown in Figure 6.6 differ. Create other designs having translational symmetry that have different symmetry groups. How many different types of designs can you create?

Figure 6.6 6.1.10. (a) The designs shown in Figure 6.7 have symmetries that are affine transformations but not isometries. Describe their symmetries. (b) Create designs that have symmetries besides isometries. Identify the symmetries.

Figure 6.7 6.1.11. *(a) Make two squares of the same size. Attach one corner of one of the squares to the center of the second square in such a way that the first square rotates in the plane freely around the center of the other (Figure 6.8). Find the maximum and minimum percentages of the area of second square covered by the first square. Explain your answer. Hint: Symmetry is the key.

Figure 6.8

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*(b) Repeat part (a) for two regular hexagons. (c) Repeat part (a) for two equilateral triangles. Explain why the answer differs from the previous parts. 6.1.12. Let T and U be subsets of a space S. (a) Prove that the set of transformations of S that are symmetries for both T and U form a group of transformations. (b) Does the set of transformations of S that are symmetries for T or for U (or both) form a group of transformations? If so, prove it; if not explain why not and give an example. (c) Can two distinct subsets T and U have the same set of symmetries? If so, give an example. If not, explain why not.

6.2 Finite Plane Symmetry Groups Johannes Kepler (1571–1630) noted that different snowflakes possess the same symmetries, even though they are not exactly alike. Figure 6.9 illustrates the six rotations and six mirror reflections that form the group of symmetries of a snowflake. The swastika (Figure 6.10), a religious symbol in ancient India long before the Nazis appropriated it, has four symmetries, all rotations of multiples of 90◦ . In this section we classify the finite groups of plane symmetries. Leonardo da Vinci (1452–1519) realized that, in modern terms, all designs in the plane with finitely many symmetries have either both rotations and mirror reflections like those in Figure 6.9 or just rotations like those in Figure 6.10. The two types of symmetry groups in the classification presented in Theorem 6.2.2 are called dihedral and cyclic. Dihedral means “two faces” and refers to the fact that the symmetries can be found by using two mirrors at an angle. (Project 1 of Chapter 5 explores this idea.) The argument in Theorem 6.2.2 uses algebraic reasoning to turn da Vinci’s geometric intuition into a proof.

Figure 6.9 A snow flake and its symmetries.

Figure 6.10 A swastika.

Definition. The cyclic group Cn contains n rotations, all with the same center. The angles of rotation are the multiples of 360◦ /n, where n is a positive integer. The dihedral group Dn contains the n rotations of Cn and n mirror reflections over lines passing through the center of the rotations. The angles between the lines of the mirror reflections are multiples of 180◦ /n. (See Figure 6.9.)

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Exercise 6.2.1.∗ What are the symmetry groups for Figures 6.9 and 6.10? Exercise 6.2.2.∗ Find the symmetry group of a regular n-sided polygon. Theorem 6.2.1. The isometries of a finite plane symmetry group fix a point and so are rotations around it or mirror reflections over lines through it. Proof. We adapt a proof from Gallian [6, 404]. Let G be a finite symmetry group of plane isometries and assume that the plane has coordinates. For a point A, let S = {γ ( A) : γ ∈ G}; that is, S is the set of images of A under the isometries of G. Because G is finite, so is S, whose elements can be listed as (x1 , y1 ), (x2 , y2 ), . . . , (xn , yn ). We claim that the center of gravity of n n , y1 +y2 +···+y ), is fixed by γ in G. Each (xi , yi ) is the image of the n points, (x, y) = ( x1 +x2 +···+x n n A by at least one γ i in G: (xi , yi ) = γ i ( A). Now, for γ in G, γ ◦ γ i is another element of G, so γ will move the points of S around among themselves. Thus their center of gravity, (x, y), is fixed. From Section 4.2, we know that the only plane isometries that fix a point are rotations around it and mirror reflections over lines through it. ! Theorem 6.2.2. A finite symmetry group containing only Euclidean plane isometries is either a cyclic group or a dihedral group. Proof. We need to show that the possible rotations and mirror reflections from Theorem 6.2.1 always fit exactly as cyclic and dihedral groups require. First, consider the rotations. If there is only one, it is the identity, a rotation of 0◦ . Otherwise, let the smallest positive angle of rotation be A◦ . From Chapter 5 the composition of rotations of B ◦ and C ◦ is a rotation of (B + C)◦ . Thus by closure there are rotations by all multiples of A◦ . The number of rotations is finite, so A divides some multiple of 360. We’ll show that A divides 360. Let k A be the largest multiple of A less than 360. Then 360 ≤ (k + 1)A < 360 + A. If (k + 1)A◦ is greater than 360◦ , it is the same angle as ((k + 1)A − 360)◦ . However, this last angle would be positive and smaller than A◦ , which is a contradiction. Hence, A divides 360, say A = 360/n. Thus we have at least the n rotations whose angles are multiples of A◦ . We claim that there are no others. Suppose that there were a positive rotation of B ◦ , not a multiple of A◦ . Let j A be the largest multiple of A less than B. Then there would be a rotation of (B − j A)◦ , which would be less than A◦ , the smallest angle, which is a contradiction. Hence the rotations form Cn . Next consider the mirror reflections of the symmetry group. If there are none, we have Cn . If there is at least one, its compositions with the n rotations give n different mirror reflections. Exercise 5.2.12 showed that the composition of two mirror reflections over lines meeting at an angle of C ◦ is a rotation of 2C ◦ . As the angles of rotation must be multiples of A◦ , the angles between the lines must be multiples of 12 A◦ = (180/n)◦ . Since there are n such angles, there are n lines and the symmetry group is Dn . ! The next two theorems apply more generally than only to Euclidean plane geometry. Theorem 6.2.3 gives the total number of symmetries of a design as a product, without finding them individually. It is an application of Lagrange’s theorem in group theory, but we prove it directly. (For those who have studied abstract algebra, the classes in the proof are the cosets of G P , which is a subgroup of G. The subgroup is the stabilizer of the point, and the points to

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which the point can be moved are its orbit. Theorem 6.2.3 is often called the orbit-stabilizer theorem.) Example 1. Count the symmetries of a pentagonal bipyramid. Solution. The pentagonal bipyramid shown in the left of Figure 6.11 has seven vertices, but they can’t all be mapped to one another. The middle of Figure 6.11 pictures the polyhedron from the top vertex. The symmetries fixing the top vertex form the dihedral group D5 , so ten symmetries fix it. The only other vertex symmetric to the top vertex is the bottom. Hence, when we apply Theorem 6.2.3 using the top vertex, we find a total of 10 × 2 = 20 symmetries. We can arrive at this same number of symmetries by using one of the vertices on the equator. From the side view in Figure 6.11, there are four symmetries fixing one of these vertices, which form the dihedral group D2 . With five such vertices, Theorem 6.2.3 again yields 4 × 5 = 20 symmetries. ♦

top view

side view

Figure 6.11 Three views of a pentagonal bipyramid. Theorem 6.2.3. The number of symmetries of a figure, if finite, equals the product nk, where n is the number of symmetries of the entire figure that leave a point fixed and k is the number of points, including that point, to which it can be moved by symmetries. Proof. Let P be a point of the figure, G the symmetry group, G P be the set of symmetries fixing P, and let G p have n elements. We collect the symmetries of G into disjoint classes, show the classes to be the same size, n, and count the number of classes, k. Define two symmetries α and β to be in the same class if and only if they map P to the same point Q: α(P) = β(P) = Q. The number of classes, k, is the number of points to which P can be moved, including P itself. To complete the proof we need to show all the classes to be the same size, n, as G P , the class that maps P to P. Suppose that α(P) = Q and consider [α], the class of α. For every γ ∈ G P , α ◦ γ is another element of [α] because α(γ (P)) = α(P) = Q. Hence any other class has at least as many elements as G P . Conversely, for β in [α], α −1 ◦ β is in G P because α −1 (β(P)) = α −1 (Q) = P. (By Exercise 6.2.8, the symmetries α ◦ γ and α are distinct, as are α −1 ◦ β and α −1 .) Hence the classes are all the same size, n. Thus the number of symmetries in G is nk. ! Theorem 6.2.4. In a finite symmetry group, either all the isometries are direct or exactly half of them are direct. Proof. Let D be the set of the direct isometries and I the set of the indirect isometries, if any, in the symmetry group. If D is the entire symmetry group, we are done. If γ ∈ I, then γ D = {γ ◦ δ : δ ∈ D} is a subset of I because γ switches orientation but δ does not. Furthermore,

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distinct δ give distinct products γ ◦ δ by Exercise 6.2.8. Hence I has at least as many elements as D. A similar argument with γ I = {γ ◦ β : β ∈ I} shows that D has at least as many elements as I. Hence they have the same number of elements. ! Example 2. Describe the symmetries of a pentagonal bipyramid. Solution. By Theorem 6.2.4, we know that half of the twenty symmetries are rotations, including the identity. Four rotations around the axis through the top and bottom vertices have angles of rotation that are multiples of 72◦ . (See Figure 6.11(a).) The five remaining rotations, each of 180◦ , are around axes that go through one of the remaining five vertices on the pentagonal equator. There are five vertical mirror reflections and one horizontal mirror reflection. Theorem 6.2.4 guarantees four more indirect isometries, which are rotary reflections. (See Section 5.5.) They can be written as compositions of the horizontal mirror reflection with the rotations around the vertical axis. ♦

6.2.1 Exercises for Section 6.2 6.2.3. *(a) Classify the symmetry group of the designs shown in Figure 6.12. *(b) Classify the different types of triangles by their symmetry groups. *(c) Classify the different types of quadrilaterals (parallelogram, kite, rhombus, and so on) by their symmetry groups.

Figure 6.12

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6.2 Finite Plane Symmetry Groups

(d) Draw hexagons, each with a different symmetry group from the others. Classify each symmetry group. How are the groups related? (e) Find examples of designs with different finite symmetry groups. 6.2.4. (a) Explain why the symmetries of the rectangle shown on the left in Figure 6.13 and the symmetries of the triangle shown on the right in Figure 6.13 are symmetries of the surrounding hexagons. (b) Draw an analogous design that combines two other dihedral groups. (c) Draw an analogous design that combines two cyclic groups. (d) Draw an analogous design that combines a cyclic group and a dihedral group. (e) For each of the preceding designs give the two symmetry groups. What pattern did you find between each pair of symmetry groups? Make a conjecture about the symmetry groups and try to prove it.

Figure 6.13 6.2.5. This problem introduces the idea of color symmetry. (Because this is a black and white book, we use stripes when we need a third color.) *(a) For the two-color design in Figure 6.14, describe the color preserving symmetries—the symmetries that take each region to another region of the same color. Do they form a transformation group? If so, classify it. Explain. *(b) Describe the color switching symmetries of this design. Do they form a transformation group? If so, classify it. Explain. *(c) Call the union of the symmetries from part (a) and part (b) the color symmetries of the design. Do they form a transformation group? If so, classify it. Explain. (d) The symmetries in two of the three previous parts form groups. How are they related? (e) Make other two-color designs having different symmetries. Repeat parts (a), (b), (c), and (d) for each of these designs. (f) Repeat parts (a), (b), (c), and (d) for the three-color design in Figure 6.15.

Figure 6.14

Figure 6.15

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Symmetry

(g) Make a design having at least three colors. Repeat parts (a), (b), (c), and (d) for this design. Be sure that your color-switching symmetries take all regions of one color to the regions of a second color so that the underlying relationships of the colors are preserved. (h) Make a conjecture based on the results obtained in parts (a)–(g). 6.2.6. Count the symmetries of each polyhedron named. *(a) A triangular prism and a square prism. *(b) A prism with regular n-gons for the top and bottom and n rectangles for the sides. *(c) The five regular polyhedra: cube, tetrahedron, octahedron, dodecahedron, and icosahedron. (See Figure 1.66.) (d) The regular tetrahedron has half as many symmetries as the cube. Are all the symmetries of the tetrahedron also symmetries of the cube? Explain. (e) The regular tetrahedron has one fifth as many symmetries as the dodecahedron. Are all the symmetries of the tetrahedron also symmetries of the dodecahedron? Explain. 6.2.7. If symmetry is changed to rotation throughout Theorem 6.2.3, is the theorem still correct? If so, prove the revised theorem. If not, explain why it is false. 6.2.8. Show for symmetries α, β, and γ , if β ̸= γ , then α ◦ β ̸= α ◦ γ and β ◦ α ̸= γ ◦ α. Hint: Use α −1 . 6.2.9. Alter and prove Theorem 6.2.1 for isometries in three dimensions. What types of symmetries are possible? 6.2.10. Theorem 6.2.3 can be modified to address faces rather than points. For each of the polyhedra in Exercise 6.2.6, pick a face, count the symmetries taking it to itself and the number of faces to which that face can go. Verify that the product of the numbers is the same as the number of symmetries that you found in Exercise 6.2.6. 6.2.11. Modify Theorem 6.2.3 and its proof to count the symmetries of polyhedra, using faces. (See Exercise 6.2.10.) 6.2.12. Relate the group of symmetries of a circle to the dihedral groups Dn .

6.3 Symmetry in the Plane Although real designs cannot contain infinitely many copies of a motif, many designs from various cultures convey that impression. To analyze them we assume that the motif repeats infinitely often, with translations either in just one direction (Figure 6.16) or in more than one direction (Figure 6.17). The classification of the finitely many symmetry groups for frieze and

Figure 6.16 A frieze pattern.

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wallpaper patterns provides anthropologists and archaeologists a way to analyze designs of other cultures that transcends cultural boundaries.

Figure 6.17 A wallpaper pattern. Definition. Repetitions of a bounded motif are discrete if and only if there is a minimum positive distance between copies of it. A frieze pattern is a discrete plane pattern that has translations in just one direction as symmetries. A wallpaper pattern is a discrete plane pattern that has translations in more than one direction as symmetries.

6.3.1 Frieze Patterns To classify frieze patterns we first find which isometries can be symmetries of frieze patterns and then determine how they combine to form groups of symmetries. To simplify we assume without loss of generality that the translations are horizontal. By Exercise 6.3.12, frieze patterns always have a horizontal line that is stable under every symmetry of the frieze. This line, which we call the midline, provides one way to limit the possible isometries for frieze patterns. The frieze pattern in Figure 6.16 has all the symmetries described in Theorem 6.3.1, which are thus all possible. Theorem 6.3.1. The only symmetries of a frieze pattern with horizontal translations are horizontal translations, vertical mirror reflections, rotations of 180◦ with centers on the midline of the frieze pattern, and glide reflections and mirror reflections over the midline of the frieze pattern. Exercise 6.3.1. Draw figures to illustrate the proof of Theorem 6.3.1. Proof. We show that all other isometries map the midline to a different line and so cannot be symmetries of the frieze pattern. Translations in a direction other than horizontal lift or lower the midline, so they are eliminated. Rotations other than 180◦ (or 0◦ , the identity) tilt the midline. Rotations of 180◦ whose centers are not on the midline shift it to a different horizontal line. Mirror and glide reflections over lines that are not vertical or horizontal also

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tilt the midline. Mirror and glide reflections over horizontal lines other than the midline, as well as glide reflections over vertical lines, shift the midline to a different horizontal line. By Theorem 5.2.7, we have eliminated all other isometries. ! The analysis of how the possible symmetries fit to form groups requires a deeper understanding of groups than needed in Section 6.2 because now the groups are infinite. (We often say group instead of symmetry group or transformation group.) We need to find a small number of symmetries for each frieze pattern from which we can generate all the others. Definition. A subgroup H of a group G is a subset of G that is a group in its own right, using the same operation as G. The elements g1 , g2 , . . . , gn of G generate a subgroup H if and only if the elements are in H and every element h of H can be written in terms of them and their inverses, in some order and with any finite number of repetitions. We write ⟨g1 , g2 , . . . , gn ⟩ to indicate the subgroup generated by them. Example 1. Show that the dihedral group Dn is generated by two neighboring mirror reflections, µ1 and µ2 . Solution. The composition of µ1 and µ2 gives the smallest rotation: µ1 ◦ µ2 = ρ. Repetitions of it, for example, µ1 ◦ µ2 ◦ µ1 ◦ µ2 = ρ 2 , give other rotations. (See Figure 6.9.) We generate the mirror reflections as compositions of the form µ1 ◦ ρ i = µ1 ◦ (µ1 ◦ µ2 )i , for some power i. ♦ Exercise 6.3.2.∗ Explain why one translation generates the group of symmetries T of the frieze pattern shown in Figure 6.18.

Figure 6.18 Example 2. Explain why a translation, a vertical mirror reflection, and the horizontal mirror reflection generate the group of symmetries of the frieze pattern shown in Figure 6.16. Solution. We use properties from Chapter 5 to analyze compositions. The smallest translation generates all the others. The compositions of a vertical mirror reflection with the translations give all the vertical mirror reflections. Similarly, the horizontal mirror reflection and the translations generate the glide reflections. The composition of the vertical and horizontal mirror reflections is a rotation of 180◦ . The other rotations are obtained by composing it with the translations. ♦ Theorem 6.3.1 reveals that the symmetry group for the frieze shown in Figure 6.16 is in some sense the largest such group. The group in Exercise 6.3.2 must in the same sense be the smallest, because every frieze pattern must have translations and this frieze pattern has only translations. To be rigorous, we need to be more careful. Let T ′ be the translations that shift

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6.3 Symmetry in the Plane

p1

p1m1

p11g

p2mm

p211

p11m

p2mg

Figure 6.19 The seven types of frieze patterns from pottery designs of the San Ildefonso pueblo. a motif an even number of positions in Figure 6.18. Then T ′ is a group of symmetries even smaller than group T in Exercise 6.3.2. However, T ′ doesn’t differ in any substantial way from T . In algebraic terms, the groups are isomorphic. (See Section 2.3.) The geometric difference is that the distance between repetitions using T ′ is twice the distance between repetitions using T , which is irrelevant for finding different types of frieze patterns. To classify types of frieze patterns, we assume without loss of generality that any two friezes have the same smallest translation to the right. Theorem 6.3.2 shows that there are just seven types of frieze patterns. Example 3. Figure 6.19 shows the seven types of frieze patterns, each having a group of symmetries different from the others. The patterns are traditional patterns from the pottery of the San Ildefonso pueblo in New Mexico. (See Crowe and Washburn [3].) Beside each pattern is the name of the group of symmetries for it. With one exception the names of the groups

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Symmetry

have the form prvh and tell us what symmetries they have. If r = 2, there are rotations of 180◦ . (Note that 2 · 180◦ = 360◦ , a complete rotation.) If v = m, there is a vertical mirror reflection. If h = m there is a horizontal mirror reflection (as well as glide reflections). If h = g, there are glide reflections, but not a horizontal mirror reflection. A 1 in any of these positions indicates that the group doesn’t have this type of symmetry. The exception is the frieze group p1, which has only translational symmetry. (In effect, p1 could be written out fully as p111.) ♦ Theorem 6.3.2. There are exactly seven groups of symmetries for frieze patterns, up to geometric isomorphism. Proof. Example 3 shows that there are at least seven frieze groups. To show there are no others, we consider the possible sets of generators for frieze groups chosen from the isometries described in Theorem 6.3.1. We use τ for the smallest translation to the right, ρ for a rotation, η for the horizontal mirror reflection, ν for a vertical mirror reflection, and γ for a glide reflection. By Exercise 6.3.12 and the following observations, we do not need to consider every possible such set. All possible rotations are generated by any one rotation and τ . All possible vertical mirror reflections are generated by any one vertical mirror reflection and τ . The horizontal mirror reflection (or a glide reflection) and τ generate all the glide reflections. The composition of a vertical mirror reflection and a rotation can give two different types of symmetry. If the center of rotation is on the line of reflection, the composition is the horizontal mirror reflection. Otherwise it is a glide reflection. Thus we need to consider generators of the form ⟨τ , ?⟩, where ? is replaced by some of the following general symmetries: ρ, ρ ′ , ν, η, or γ , where we assume the center of ρ to be on the line of reflection of ν but the center of ρ ′ not to be. (If ν is not one of the generators, it does not matter ⟨τ , ρ ⟩ = p211, ⟨τ , ν ⟩ = p1m1, ⟨τ , η⟩ = p11m, and whether we use ρ or ρ ′ .) Then ⟨τ ⟩ = p1, : 9 : 9 ⟨τ , γ ⟩ = p11g. We obtain p2mg from τ , ρ ′ , ν , ⟨τ , ρ, γ ⟩, ⟨τ , ν, γ ⟩, or τ , ρ ′ , ν , γ . We obtain p11m from ⟨τ , η, γ ⟩ or ⟨τ , η⟩. In Exercise 6.3.14 you are asked to show that all other sets of generators yield p2mm. ! Remark. Some of the seven frieze groups are algebraically isomorphic, though they differ geometrically. For example, p1g1 is generated by a glide reflection, and it is algebraically isomorphic to p1. However, p1g1 has both glide reflections (indirect isometries) and translations (direct isometries), whereas p1 has just translations, so the groups are geometrically distinct.

6.3.2 Wallpaper Patterns The classification of wallpaper patterns is more complicated than that of the frieze patterns. Figure 6.20 gives designs with different angles of rotation, not only the 180◦ option for frieze patterns. Fortunately, Theorem 6.3.3 restricts the angles of rotation to those illustrated in Figure 6.20. The result, first proved in 1822 for three dimensions, is called the crystallographic restriction because it is crucial in the classification of three-dimensional crystals. Theorem 6.3.3. The Crystallographic Restriction. The minimal positive angles of rotations that can be symmetries of a wallpaper pattern are 60◦ , 90◦ , 120◦ , 180◦ , and 360◦ . All other angles of rotation for a given wallpaper pattern are multiples of the minimum angle.

6.3 Symmetry in the Plane

277

Figure 6.20 Wallpaper patterns from (a) Mexico, (b) Spain, (c) Mongolia, (d) Morocco, and (e) Iraq with different angles of rotation. Proof. Let A be a center of rotation for a wallpaper pattern and let B be a point closest to A for which a symmetry takes A to B. Then B is, by symmetry, also a center of rotation for the wallpaper pattern with the same angles as at A. No two other images of A can be any closer together than are A and B. Without loss of generality assume that A is to the left of B. Let ρ be the smallest positive rotation with center at A. Then the angle of φ, the least negative rotation with center at B, is the negative of ρ. Now consider A′ = φ(A) and B ′ = ρ(B), as in Figure 6.21. By symmetry, A′ and B ′ are centers of rotation like A. If ρ rotates less than 60◦ , as in the left part of Figure 6.21, then d(B, B ′ ) will be less than d(A, B), which is impossible. If

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Symmetry

the angle for ρ is between 60◦ and 90◦ , as in the right of Figure 6.21, then d( A′ , B ′ ) < d( A, B), which is impossible. Similar reasoning (see Exercise 6.3.13) eliminates other minimum angles except 90◦ , 120◦ , 180◦ , or 360◦ . Hence only the specified angles are compatible with wallpaper patterns. ! A'

A'

A

B'

B'

B

A

B

Figure 6.21

Theorem 6.3.4. There are exactly seventeen groups of symmetries for wallpaper patterns, up to isomorphism. Proof. See Crowe [2]. ! The Russian chemist and mathematician Vyatseglav Fedorov in 1891 first stated and proved Theorem 6.3.4, but his proof wasn’t widely noted. Several other mathematicians, including Felix Klein, independently found and proved the classification, which is based on group theory. Our understanding of dihedral groups in Section 6.2 leads us to suspect geometrically that mirror and so glide reflections can fit particular angles of rotation in only finitely many ways. (Exercise 6.3.16 provides a more algebraic way to determine that there are only finitely many types of wallpaper patterns.) While the number of types of wallpaper patterns is finite, the number seventeen remains somewhat mysterious. The flowchart presented in Figure 6.22 compresses the mathematics of the proof into a methodical way of classifying wallpaper patterns. The names of the wallpaper groups in Figure 6.22 need some explanation. The numbers 2, 3, 4, and 6 refer to the maximum number of rotations around a center of rotation. The letters m and g refer to mirror and glide reflections. The letter c stands for a rhombic lattice, instead of a rectangular lattice. The difference is explained in Example 4. Then Example 5 illustrates two groups that are often as difficult to distinguish as are their names, p31m and p3m1. Example 4. Classify the patterns shown in Figure 6.23. Solution. Both patterns have 180◦ rotations. Following that branch in Figure 6.22, we note that both have mirror reflections and perpendicular mirror reflections. In the Chinese design, the motifs stack like boxes, so from the flowchart it has the group pmm. However, the Japanese motifs stack like bricks in a wall, so the group is cmm. The Japanese pattern, unlike the Chinese one, has vertical glide reflections along lines halfway between adjacent lines of reflection. ♦

6.3 Symmetry in the Plane

279

Figure 6.22 A flowchart for the wallpaper groups. Example 5. Classify the patterns shown in Figure 6.24. Solution. Both designs have rotations of 120◦ but not 60◦ . Each has some mirror reflections that pass through centers of rotation. The design on the left of Figure 6.24 has mirror reflections through every center of rotation, such as the centers of hexagons and the centers of the Y shapes.

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Symmetry

Figure 6.23 Designs from (a) China and (b) Japan.

Figure 6.24 Two Japanese designs. Hence its group is p3m1. However, the centers of the black triangles of the design on the right of Figure 6.24 don’t have mirror reflections, so that pattern has group p31m. ♦ The anthropologist Dorothy Washburn teamed with the geometer Don Crowe to pioneer the use of symmetry groups in cross-cultural studies. Before their work, researchers had tried to analyze the varying motifs of different cultures. Although the motifs in Figure 6.23 clearly differ, it is easy to imagine a range of variations of each. The great variety of motifs made their analysis culturally subjective. However, the symmetry groups are independent of culture. For many cultures, the frieze and wallpaper designs used by their artists remain the same over long periods of time. The stability provides people studying cultures another marker in the study of societies and their interactions. A new design can indicate the influence of trade with another region or the conquering of one group by another. Cultures with weaving tend to make designs in other media, such as ceramics, which have the symmetry patterns of weaving. For an ancient culture for which no traces of weaving may remain, patterns on ceramics, which can endure

6.3 Symmetry in the Plane

281

millennia, can provide indirect evidence about weaving. (The Spanish and Mongolian patterns in Figure 6.20 artistically vary the basic weaving pattern.) Many cultures create patterns with two-color symmetry or multiple-color symmetry. Exercise 6.2.5 introduced the mathematics of color symmetry. Color symmetry involves two groups: the color-preserving group and the color group, a larger group that also includes the colorswitching symmetries. We can analyze each just as we do any other symmetry group. (If needed, stripes represents a third color.) Example 6. Classify the colored patterns shown in Figure 6.25. Solution. First consider the color-preserving symmetries of the two-color Peruvian design in Figure 6.25. All the black staircases are upright, implying no color-preserving rotations. The rows of black staircases alternate facing left and right, indicating no mirror reflections, but indicating glide reflections that preserve colors. From the flowchart (Figure 6.22), the colorpreserving group is pg. All the white staircases are upside down, so 180◦ rotations switch colors. Horizontal mirror reflections between rows switch colors, but no vertical mirrors work. Thus the color group is pmg. We write the pair of groups with the color group on top: pmg/pg.

Figure 6.25 A two-color Peruvian design and a three-color design. Consider the three-color frieze pattern in Figure 6.25. The only symmetries preserving all three colors are translations and glide reflections, so the color-preserving group is p1g1. The marked vertical mirror reflection preserves the color of black triangles but switches the white and striped colors. Just to the right of the marked mirror is a center of rotation switching the white and black, but preserving the striped. The color-switching symmetries include vertical mirror reflections and rotations, but not horizontal mirror reflections. So the color group is pmg2, and the classification is pmg2/p1g1. Some translations and glide reflections, half the length of the color preserving ones, permute all three colors. ♦

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Symmetry

Definition. A symmetry of a design is color-preserving if and only if every repetition A of the motif in the design is mapped to a copy that is the same color as A. A symmetry κ of a design is color-switching if and only if, whenever the repetitions A and B of the motif are the same color, then κ( A) and κ(B) are the same color, but a different color from A and B. A color symmetry of a design is either color-preserving or color-switching symmetry. Theorem 6.3.5. The color-preserving symmetries of a design form a subgroup of the color symmetries of the design. Proof. See Exercise 6.3.16. ! The widespread use of frieze and wallpaper patterns in many cultures testifies to symmetry’s appeal across time, language, and race. The geometric understanding needed to join motifs to make the patterns is also global. Patterns with rotations of 180◦ and 90◦ are common worldwide. Fewer cultures developed patterns with angles of 120◦ and 60◦ . I know of only one site before 1890, the Alhambra of Moorish Spain, that has all seventeen wallpapers represented. However, there is no evidence that any individual or society explicitly considered finding all types of these patterns, let alone proving there were no more, until the late nineteenth century. Training in formal mathematics, especially group theory, led naturally to the idea of classification. Mathematics provides new ways to see the world, enriching understanding.

6.3.3 Exercises for Section 6.3 *6.3.3. Classify the frieze patterns in Figure 6.26.

Figure 6.26 Designs from (a) Morocco, (b) France, (c) Indonesia, (d) Borneo, (e) Peru, and (f) China. 6.3.4. Classify the two-color frieze patterns shown in Figure 6.27. The designs represent all seventeen types of two-color frieze patterns. The first fourteen designs (parts (a)–(n)) are from the pottery of the San Ildefonso pueblo. Crowe and Washburn [3] designed the last three (parts (o)–(q)) to complete the set. *6.3.5. Classify the wallpaper patterns shown in Figure 6.28. 6.3.6. Classify the two-color wallpaper patterns of Figure 6.29. 6.3.7. (a) Make a flowchart to classify frieze patterns. (b) Describe which symmetry groups for frieze patterns are subgroups of the others.

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6.3 Symmetry in the Plane

a)

b)

c)

d)

e)

f)

g)

h)

i)

j)

k)

l)

m)

n)

o)

p)

q)

Figure 6.27 The 17 types of two-color friezes. 6.3.8. Find as many types of wallpaper patterns as you can with a motif of a rectangle twice as long as it is wide. The rectangles need not all line up the same way, although they shouldn’t overlap or have gaps. Which symmetry groups can’t be realized with the motif? Explain. 6.3.9. *(a) Find all regular wallpaper patterns. To be regular, the motif must be a regular polygon, there can be no gaps between or overlaps of the polygons, and two polygons with more than a point in common must share an entire edge. (b) Find all eight semiregular wallpaper patterns. A semiregular pattern differs from a regular pattern in that the motif must be two or more regular polygons and all vertices must have the same pattern of polygons around them. (Johannes Kepler (1571–1630) was the first to find the patterns.) Hints: The angles at each vertex must sum to 360◦ . There are at least three polygons at each vertex.

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Symmetry

Figure 6.28 Designs from (a) Congo, (b) Borneo, (c) Spain, (d) Mexico, (e) Rome, (f) Congo, (g) Celtic, (h) Afghanistan, (i) Borneo, (j) Borneo, (k) Spain, (l) Italy.

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6.3 Symmetry in the Plane

Figure 6.28 (cont.)

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Figure 6.29 Two-color designs from (a) Morocco, (b) Brazil, (c) Egypt, (d) Iran, (e) U. S. A., (f) Congo, (g) Spain, (h) Hawaii, (i) U.S.A.

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6.3.10. Find ways to color the regular and semiregular wallpaper patterns of Exercise 6.3.9 so that each has a color group differing from the color-preserving group. Hint: For a semiregular pattern to have color switching, some polygons with the same numbers of sides have to be colored differently. *6.3.11. Draw the design for a plain weave, where the horizontal threads alternately go over and under the vertical threads. Classify the symmetry group of the design. Verify that all the symmetries of the Spanish design in Figure 6.20 are symmetries of a plain weave. Repeat for the Mongolian pattern of Figure 6.20. 6.3.12. Prove that in every frieze pattern with horizontal translations there must be at least one stable horizontal line. Hints: Suppose that no stable line existed. Case 1. A horizontal line goes to a non-horizontal line. Why is there a translation along the other line? Case 2. A symmetry σ takes a horizontal line k to a different horizontal line m. Let l be line halfway between k and m. What types of symmetry can σ be? Which leave l stable? For those moving l show that there is a translation in some other direction. 6.3.13. Complete the proof of Theorem 6.3.3. 6.3.14. List the possible sets of generators of a frieze pattern. Show that each not listed in the proof of Theorem 6.3.2 generates pmm2. 6.3.15. This problem investigates formulas in R2 for the symmetries of frieze patterns. We use τ (x, y) = (x + 1, y) as the formula for the translation of one step to the right. *(a) Find the formula for a general translation τ n of n steps. (If n > 0, the translation goes to the right; if n < 0, it goes to the left.) *(b) Find the formula for the horizontal mirror reflection over the x-axis. (c) Find the formula for a general glide reflection over the x-axis with a glide of n horizontally. *(d) Find the formula for a vertical mirror reflection over the line x = d. (e) Find the formula for a 180◦ rotation around the center (e, 0). (f) Find the formula for the composition of the vertical mirror over x = d with the rotation around (e, 0). What type of symmetry is it? (g) Find the formula for the composition of two vertical mirrors, one over x = d and one over x = g. What type of symmetry is it? (h) Find the formula for the composition of two rotations with centers (e, 0) and (h, 0). What type of symmetry is it? 6.3.16. (a) For Theorem 6.3.5, prove that both sets of symmetries form groups. (b) (Group theory.) Prove that the color-preserving group is a normal subgroup of the color group. (c) (Group theory.) Prove that the translations form a normal subgroup T of a wallpaper group G. Remark. The quotient group G/T describes the local behavior of the group. It is isomorphic to one of the cyclic or dihedral groups C1 , C2 , C3 , C4 , C6 , D1 , D2 , D3 , D4 , or D6 . We can deduce that there are only finitely many types of wallpaper patterns.

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6.3.17. (Group theory.) (a) Prove that the frieze group p1m1 is generated by two adjacent vertical mirror reflections. Hint: Use the formulas in Exercise 6.3.15. (b) Repeat part (a) for the group p211 using two adjacent rotations. (c) Repeat part (a) for the group p2mg using a vertical mirror and an adjacent rotation. (d) Explain why parts (a), (b), and (c) show that the groups are algebraically isomorphic. Explain why they are not geometrically isomorphic. (e) Show that the group p11m is not algebraically isomorphic to the groups in parts (a), (b), and (c). Hint: Use the formulas in exercise 6.3.15 to compare the compositions η ◦ τ , τ ◦ η, ν ◦ τ , τ ◦ ν, ρ ◦ τ , and τ ◦ ρ, where τ is a translation, η is the horizontal mirror reflection, ν is a vertical mirror reflection, and ρ is a rotation. 6.3.18. (Group theory.) Investigate the subgroup relations of the seventeen wallpaper groups.

6.3.4 M. C. Escher The Dutch graphics artist Maurits Cornelis Escher (1898–1972) combined high artistic ability with a fascinating geometrical imagination. He is well known for over one hundred lively designs related to wallpaper patterns. Instead of abstract motifs, he filled his designs with cleverly interlocking animals, birds, fishes, and more. His life-long passion for what he called “regular division of the plane” started with his visit to the Alhambra at age 24. He made sketches of many of the intricate repeating abstract patterns there. It took him years of experimentation to work out how to incorporate the symmetry of wallpaper papers with realistic shapes. His art developed empirically and intuitively without formal mathematical training. In addition to imaginative Euclidean wallpaper patterns, Escher created other mathematically related works. Without knowing anything about hyperbolic geometry, he tried to develop designs along the lines of Figures 4.39 and 4.40. After several dead ends, he consulted with the famous geometer H. S. M. Coxeter, who showed him the Poincar´e model of hyperbolic geometry. Escher than created his circle limit etchings. Other Escher etchings featured imaginative renderings of M¨obius strips, spherical geometry, and other geometrical themes. He also was intrigued by representing impossible structures that nevertheless look realistic in small areas. The designs often depend on a careful manipulation of perspective, showing an intuitive insight into the mathematics behind that artistic technique. Escher always insisted he didn’t understand mathematics, by which he meant formal mathematics. Nevertheless, mathematicians realize how much a deep intuitive understanding of mathematics infuses many of his designs. His art represents an important bridge between art and geometry, two essential ways of seeing the world.

6.4 Symmetries in Higher Dimensions 6.4.1 Finite Three-Dimensional Symmetry Groups Mathematicians and others since the time of ancient Greece have enjoyed polyhedra, especially the five regular ones with their elegant beauty. However, not until the nineteenth century was polyhedral symmetry analyzed. Given the variety of polyhedra, the types of symmetry they can possess is surprisingly limited. Auguste Bravais first published the classification of the finite three-dimensional symmetry groups (Theorem 6.4.2 below) in 1848, although Johann Hessel had found it in 1830.

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Figure 6.30 A rectangular prism. Some of the three-dimensional symmetry groups extend the families Cn and Dn discussed in Section 6.2. A rectangular prism with regular polygons for bases (Figure 6.30) has twice as many symmetries as its polygonal base. Its symmetry group is called Dnh in Sch¨onflies’ notation, where n is the number of vertices of one of the bases. Theorem 6.2.4 guarantees that half the symmetries in Dnh are rotations. They form the group Dn , algebraically isomorphic to the two-dimensional dihedral group of the same name. In place of the n mirror reflections in the plane, we have n rotations of 180◦ around horizontal axes. (See Project 9 for the various possible subgroups of Dnh .) In addition to the infinite family of Dnh and its subgroups, there are surprisingly few other finite three-dimensional symmetry groups. They come from the symmetries of the regular polyhedra. Although there are five regular polyhedra, they determine only three groups of symmetries. (See Figure 1.66.) The tetrahedron has 24 symmetries, forming the tetrahedral group, T. They include rotations of 120◦ and 180◦ around different axes. The cube and octahedron have another group of 48 symmetries, the octahedral group, W. The angles of rotation in it are multiples of 90◦ and 120◦ . The symmetries of the icosahedron and dodecahedron form the third group of symmetries, the icosahedral group, P, which has 120 symmetries. The angles of rotation in it include multiples of 72◦ , 120◦ , and 180◦ . Thee groups T, W, and P have various subgroups, most notably their subgroups of rotations T, W, and P. Many interesting polyhedra, including eleven of the thirteen Archimedean solids have for their symmetry groups T, W, or P. The two remaining Archimedean solids have only rotations for symmetries, and their groups of symmetries are W and P. (See Wenninger [22] for pictures.) Example 1. Find the group of symmetries of the great stellated dodecahedron (Figure 6.31).

Figure 6.31 The great stellated dodecahedron.

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Solution. This polyhedron has axes of rotation other than those of a prism. The axis facing out allows rotations of 72◦ , so the group can’t be T or W or their subgroups. Because the polyhedron has mirror reflections, by Theorem 6.4.2 below, it must have P for its group of symmetries. Indeed, the polyhedron has twenty triangular pyramids built on the triangular faces of an icosahedron. ♦ Theorem 6.4.1. The three-dimensional isometries in a finite symmetry group must fix a point and so form a subgroup of the symmetries of a sphere. Proof. See Exercise 6.4.9. ! Theorem 6.4.2. A finite group of three-dimensional isometries must be one of the following or one of their subgroups: Dnh , T, W, and P. Proof. See Weyl [23, 149ff]. !

6.4.2 The Crystallographic Groups The beauty of crystals, especially gems, has fascinated people for centuries. However, prior to 1830 no chemical explanation existed for the regularities and other properties of crystals. The geometric classification by Hessel in 1830 of the 32 types of crystals spurred the study of geometric arrangements of atoms in crystals. (Chemists sometimes say that there are 33 types; two are mirror reflections.) The crystals are the three-dimensional analogs of frieze patterns and wallpaper patterns. A mathematical crystal is a discrete pattern having translations in at least three directions, not all in the same plane. The subgroup of translations takes a point to a threedimensional lattice of points (Figure 6.32). To classify crystals, Hessel used the crystallographic restriction (Theorem 6.3.3) that also gives the rotations possible for two-dimensional wallpaper patterns. Theorem 6.3.3 applies to crystals as follows. Consider a three-dimensional rotation about an axis k and a point Q in the lattice not on k. The rotation moves Q to another point in the plane perpendicular to k and through Q. The points of the lattice lying in the plane form a wallpaper pattern. Thus the rotation must be a rotation of the wallpaper pattern. Hence Theorem 6.3.3 applies to mathematical (and chemical) crystals. However, crystals can combine

Figure 6.32 A lattice of points.

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the possible rotations in more ways than wallpaper patterns can. The enriched possibilities for crystals over wallpaper patterns corresponds to the enriched rotations of polyhedra compared with polygons. We consider symmetry in crystals more in Section 6.5. (See Senechal [17] for more depth.)

6.4.3 General Finite Symmetry Groups Finite groups appear in many contexts, many of which don’t correspond in any obvious way to isometries in Euclidean geometry in any number of dimensions. Nevertheless, a finite group may be thought of as a symmetry group of some figure. We may think of the group as a transformation group moving points around. Theorem 6.4.4 allows us to think of the points as embedded in Euclidean space of an appropriate dimension and the transformations of the group as isometries. Definition. The group of transformations on a set of n distinct points is the symmetric group Sn . If n is a positive integer, n! (read “n factorial”) is the product of the numbers from 1 to n. Theorem 6.4.3. For a positive integer n, the group Sn has n! elements. Proof. See Exercise 6.4.10. ! Theorem 6.4.4. The group of isometries of the (n − 1)-dimensional regular simplex is Sn . Proof. See Exercise 6.4.12. (A regular simplex is defined in Section 3.5.) ! Example 2. The symmetry group of an equilateral triangle is D3 by Theorem 6.2.2 and is S3 by Theorem 6.4.4. The groups are algebraically isomorphic, that is, structurally identical. Similarly, the regular tetrahedron illustrates that the groups T and S4 are isomorphic. ♦

6.4.4 Exercises for Section 6.4 *6.4.1. Explain why the symmetries of a rectangular box form a subgroup of the symmetries of a square prism, which form a subgroup of the symmetries of a cube. Draw figures. Use Theorem 6.2.3 to find the number of symmetries of each. Describe the symmetries of a rectangular box, which form the group D2h . 6.4.2. An antiprism has two regular polygons for its bases, but, unlike a prism, the vertices of the top polygon alternate with the vertices on the bottom so that the sides are isosceles triangles (Figure 1.77). Count the number of symmetries of an antiprism with regular n-gons for its bases. Compare the symmetries of an antiprism with the symmetries of the corresponding prism. (See Project 9 to classify the symmetry group of an antiprism with regular n-gons for bases.) *6.4.3. Figure 6.33 illustrates one of the family of prisms whose 2n-gon bases are truncated regular n-gons. (Truncation removes congruent isosceles triangles from the corners,

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Figure 6.33 A prism with truncated bases. illustrated on the left of Figure 6.33.) Count the number of symmetries and classify the symmetry group of the prisms. Explain why the symmetry group is a subgroup of D2nh . 6.4.4. (a) Build a sort of antiprism whose bases, as in Exercise 6.4.3, are truncated regular n-gon, rotated so that the short edges of one base fit symmetrically with the long edges of the other base. The faces connecting the bases are trapezoids. Count the number of symmetries. Explain why this symmetry group is a subgroup of D2nh . (See Project 9 to classify the symmetry group of polyhedra.) (b) Describe or build other modifications of prisms with non-regular polygons as bases whose groups of symmetries are other subgroups of Dnh . *6.4.5. Let ζ be the three-dimensional central symmetry with respect to the origin. (See Exercise 5.5.8.) Write ζ as a 3 × 3 matrix and show that it commutes with every 3 × 3 matrix σ . That is, σ ◦ ζ = ζ ◦ σ . Spherical isometries can be written as 3 × 3 matrices, including the symmetries in the groups of Theorem 6.4.2. For which n does the group Dnh contain ζ ? Which of the groups T, W, and P contain ζ ? 6.4.6. *(a) Describe the rotations and mirror reflections of a cube. Explain why the regular octahedron has the same symmetries. How many rotary reflections does a cube have? (b) Describe the rotations and mirror reflections of a regular tetrahedron. How many rotary reflections does a tetrahedron have? (c) Repeat part (a) for a regular icosahedron and dodecahedron. 6.4.7. Show that the tetrahedral group is a subgroup of the octahedral group. Hint: Fit a tetrahedron in a cube. 6.4.8. Build or use a model of a regular dodecahedron. Locate eight vertices that are also the vertices of a cube. Count the number of symmetries of the dodecahedron that take the vertices of the cube to themselves. Describe these symmetries, which form a subgroup of both P and W. 6.4.9. Prove Theorem 6.4.1. Hint: See the proof of Theorem 6.2.1. 6.4.10. Use induction to prove Theorem 6.4.3. Hint: For the set {1, 2, . . . , n}, why are there n possible places where 1 can go? Use the reasoning followed in Theorem 6.2.3. *6.4.11. Classify the symmetry groups of the thirteen Archimedean solids (the semiregular polyhedra that aren’t prisms or antiprisms). (See Wenninger [22] for pictures of them.) 6.4.12. Prove Theorem 6.4.4.

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6.4.13. (a) Use Theorem 6.2.3 and your knowledge of the hypercube and the symmetries of a cube to count the symmetries of a four-dimensional hypercube. (See Section 3.5.) Explain why the four-dimensional cross polytope has the same symmetries as the hypercube. (b) Extend part (a) to higher dimensions. 6.4.14. (a) (b) (c) (d)

For which n is the group Dnh a subgroup of T? Repeat part (a) but replace T with W. Repeat part (a) but replace T with P. Repeat part (a) but replace T with Dkh , where n ≤ k.

6.4.5 H. S. M. Coxeter The Canadian mathematician H. S. M. (Donald) Coxeter (1907–2003) is widely considered the most distinguished geometer of the twentieth century. He made significant contributions to many areas of geometry. Among his many books, his texts on non-Euclidean geometry, projective geometry, polytopes, group theory, and especially his Introduction to Geometry have been standard references for decades. His research and teaching were centered for more than five decades at the University of Toronto, interspersed with many sojourns at universities around the world. At the age of fifteen Coxeter showed his mathematical talent and fascination with geometry: he won a prize for an essay on polytopes, figures in four or more dimensions generalizing polyhedra. After his Ph.D. at age 24 he used symmetry groups and other approaches to revitalize the study of polytopes. Symmetry and so group theory were central for much more of his mathematical research than polytopes. For example, he generalized symmetrical patterns generated by mirror reflections, leading to what are now called in his honor Coxeter groups. (Theorem 5.2.5 shows that mirror reflections generate the group of Euclidean isometries, which is thus a Coxeter group.) He was generous with sharing his knowledge to mathematicians and non-mathematicians alike. He was instrumental in developing a series of short films on geometry at a time few thought of film as an educational medium. He was friends with Buckminster Fuller and M. C. Escher, helping both connect their ideas with mathematics. Escher had struggled unsuccessfully for some time to make what he later titled his “circle limit” designs. Coxeter provided the key through the Poincar´e model of hyperbolic geometry and its tessellations.

F

B F

F

Figure 6.34 Boron trifluoride.

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6.5 Symmetry in Science 6.5.1 Chemical Structure Chemists benefit greatly from a geometric understanding of the arrangement of the atoms (and ions) in chemical compounds. (For simplicity we refer to the parts of compounds as atoms, ignoring the distinction between atoms and ions. Similarly, we avoid a deeper discussion of electrons, orbitals, and quantum mechanics, even though symmetry plays a vital role at that level.) Chemists represent the atoms in a compound by vertices and the bonds between them by edges. For example, boron trifluoride (BF3 ) has three fluorine atoms bonded to a boron atom (Figure 6.34). Atoms bonded to an atom tend to arrange themselves as far from one another as possible, while staying the appropriate distance from the atom. The arrangement shown in Figure 6.34 is as symmetric as possible: the six (two-dimensional) symmetries form the group D3 . Thus symmetry frequently allows analysis of chemical structure. Carbon usually bonds with four atoms and hydrogen with one. The simplest carbon compound, the gas methane (CH4 ), has one carbon atom and four hydrogen atoms. You might imagine that the hydrogens would arrange themselves as some chemistry textbooks display the compound for simplicity as in Figure 6.35, which has the symmetry group D4 . Although the arrangement has considerable symmetry, chemistry involves three dimensions. Figure 6.36 represents the actual placement of the hydrogens as the vertices of a regular tetrahedron with the carbon at the center, which was deduced in 1874. Hence the hydrogen atoms are farther apart than depicted in Figure 6.35. The angle of two bonds, as shown in Figure 6.36, is approximately 109.5◦ , rather than 90◦ , as in Figure 6.35. Furthermore, the tetrahedral group T has twenty-four symmetries, or more than the eight two-dimensional symmetries of Figure 6.35. (As a threedimensional shape, Figure 6.35 has sixteen symmetries, which is still fewer than the tetrahedral shape of Figure 6.36.) An increase in symmetry corresponds to a lower, more stable energy state. H

H

H

C

H

H

C H

H

Figure 6.35 An inaccurate representation of methane.

H

Figure 6.36 The 3-dimensional structure of methane. (The dashed line indicates a bond receding into the page and the triangle indicates a bond projecting in front of the page.)

One form of pure carbon, a diamond crystal, extends the symmetry of Figure 6.36. In a perfect diamond, each carbon atom is bonded to four other carbon atoms that form a regular tetrahedron (Figure 6.2). To analyze a crystal, we assume that it continues in all three dimensions forever, which is a reasonable simplification. For example, a one-carat diamond has approximately 1022 carbon atoms, which means millions of repetitions of the pattern shown in Figure 6.2

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along any axis. The symmetries of the crystal include translations in three dimensions and the local isometries of the tetrahedral group. The variety of the directions of the bonds and their uniformity makes diamond the hardest naturally occurring substance. The angles of the bonds also determine the angles at which gem diamonds can be cut. For example, you will never see a diamond cut as a cube. Carbon forms another crystal, graphite, whose very different physical and chemical properties reflect the different geometry of the crystal. The atoms of graphite form layers one atom thick with only weak bonds between layers (Figure 6.37). The weakness of the bonds corresponds to the 90◦ angle they make, which represents higher stress than the optimal tetrahedral angle of Figure 6.2. In Figure 6.37 each layer of carbon atoms forms a pattern of hexagons. The layers slide easily over one another, making graphite an excellent lubricant. The “lead” in pencils also contains graphite.

Figure 6.37 The structure of graphite. Exercise 6.5.1.∗ Classify the wallpaper pattern of a layer of graphite. Salt (NaCl) exhibits a third crystalline structure (Figure 6.38). The cubic arrangement at the atomic level ensures that salt grains always have rectangular faces. Potassium chloride (KCl)

Figure 6.38 The structure of salt. (Dots represent sodium and empty circles represent chlorine.)

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has the same crystalline form as salt and is extremely close to it chemically. Indeed, KCl is a salt substitute for those restricting their intake of sodium (Na).The presence of two different kinds of atoms in salt and potassium chloride affects the group of symmetries. For example, a translation of one bond’s length along the x-, y-, or z-axis switches the sodium and chlorine atoms. This switch corresponds to the color symmetries discussed in Section 6.3. The crystal sphalerite, made of zinc and sulfur, is a two-color version of the diamond crystal, with each zinc atom bonded to four sulfur atoms and conversely. (See Senechal [17] for more on crystals.) Exercise 6.5.2.∗ Describe a rotation of the salt crystal that switches sodium and chlorine atoms.

6.5.2 Quasicrystals In Section 6.3 we proved Theorem 6.3.3, the crystallographic restriction, describing the angles of rotation compatible with translations in two or three dimensions. This mathematical result matched chemists’ experimental data on x-ray diffraction patterns perfectly until 1984. That year a team of chemists found a compound that gave sharp diffraction patterns with the angle of 72◦ , which was inconsistent with Theorem 6.3.3. Sharp patterns had previously only been seen with crystals, so the new compounds were called quasicrystals. Mathematicians, chemists, and others have explored the rapidly growing field of quasicrystals. Chemists found that, as the crystallographic restriction assures us, the arrangement of atoms in quasicrystals is not periodic. In 2011 one of the chemists, Dan Shechtman, received the Nobel prize in Chemistry for his work on quasicrystals. Twenty years before quasicrystals were discovered, mathematicians devised aperiodic patterns covering the plane without admitting translations. By 1974 Sir Roger Penrose was able to build such patterns using only two shapes, such as the one depicted in Figure 6.39. (See Project 15.) The Penrose tiling fails to have 72◦ rotational symmetry, consistent with the crystallographic restriction. However, a rotation of 72◦ at certain points will make a small region of the tiling land on itself, which we might call a partial rotational symmetry. By the early 1980s, Penrose and others had generalized the tilings to three dimensions. The three-dimensional tilings can also have partial rotational symmetry, which corresponds mathematically to the diffraction

Figure 6.39 A Penrose tiling.

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patterns of quasicrystals. Thus Penrose tilings provide one approach to the mathematical analysis of quasicrystals. Another mathematical insight of the 1970s brought symmetry into the study of quasicrystals in another way. Mathematicians showed that a six-dimensional “hypercrystal” could be sliced into a three-dimensional cross section that would look like a quasicrystal. The symmetries (including translations) of the six-dimensional model and the angle of the cut determine the properties of the three-dimensional cross section. Hypercrystals in more than three dimensions can have other angles of rotation besides those of Theorem 6.3.3, including 72◦ . We still don’t know the chemical relevance of six mathematical dimensions, but researchers Bak and Goldman “emphasize that thinking of [a quasicrystal] as a periodic structure in six dimensions is not merely an amusing mathematical abstraction.” (Jaric [10, 146].) (See Peterson [13, 200–212] and Senechal [17] for more information on quasicrystals.)

6.5.3 Symmetry and Relativity Albert Einstein’s theory of relativity transformed physics. In geometric language, the special theory corresponds to a four-dimensional group of symmetries that preserve physical properties. Although physicists had already used time as a fourth dimension, Einstein (1879–1955) realized that space and time interacted. The simpler concept of Galilean relativity is helpful to illustrate the role of symmetry and to clarify Einstein’s contribution. Galileo Galilei (1564–1642) explained why we don’t feel the motion of the earth as it travels through space. The motion of everything on the earth includes its speed, so our measurements (and senses) detect only the differences in the speeds of objects and observers, not an object’s absolute speed. More formally, the laws of physics remain the same for observers moving at a constant velocity relative to one another. In more modern terms, Galileo’s principle of relativity says that constant velocities in any direction are symmetries for the laws of physics. Sir Isaac Newton (1642–1727) and physicists following him made another principle explicit: the measurement of time and distance is absolute. That is, clocks record the same amount of time passing and rulers measure the same lengths, regardless of the speed and direction they are traveling relative to one another. The absolute measurement of space and time leads to the additivity of velocities. For example, suppose that an observer on the ground measures a train moving at a rate of 20 meters/sec and an observer in the train observes someone else in the train walking in the direction of the train at 1.5 meters/sec. For the observer on the ground, the walker would be moving at 21.5 meters/sec. At modest speeds, the additivity of velocities matches our experience. The Michelson–Morley experiments conducted at the end of the nineteenth century revealed a contradiction from assuming the two principles described above. Their experiments sought to measure the influence of the motion of the earth on the velocity of light. Relative to the sun, the earth is moving approximately 30 km/sec, incredibly quick but still a tiny part of light’s velocity of approximately 300,000 km/sec. Michelson and Morley devised an experiment accurate enough to detect the comparatively small difference between 300,000 and 300,030 km/sec. Regardless of the direction the light was sent, they found its velocity always to be the same, contradicting the additivity of velocities. Later experiments have confirmed that the speed of light in a vacuum is constant. In 1905 Hendrik Lorentz and Henri Poincar´e determined the Lorentz transformations, the theoretical group of symmetries consistent with the experiments and Galileo’s principle of relativity. At almost the same time, Albert Einstein developed the physical theory combining

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the constancy of the speed of light and Galileo’s principle of relativity. Necessarily he dropped the absolute measurement of space and time and so the additivity of velocities. Surprisingly, the formula for combining relativistic velocities corresponds to composing hyperbolic translations. (See Project 21 of Chapter 5 and especially Section 7.5.) For ease, we write velocities as fractions of the speed of light, so 1 is the speed of light. For the simple case of velocities along a line, we replace the addition of velocities, x + y, with the following equation: x⊕y=

x+y . 1 + xy

(6.1)

Example 1. Suppose that observer A is moving at a velocity of 0.4 with respect to observer B and that an object is moving in the same direction at a velocity of 0.5 with respect to A. Then the object is moving at a velocity of 0.4 ⊕ 0.5 = 0.75 with respect to observer B, noticeably less than 0.4 + 0.5 = 0.9. The velocities of a spacecraft and the earth are much smaller, approximately 0.00003 and 0.0001. For them, 0.00003 ⊕ 0.0001 = 0.0001299999996, which for all practical purposes is 0.00013, or 0.00003 + 0.0001. Thus NASA doesn’t need to use relativity theory to plan space missions. ♦ Exercise 6.5.3. Verify that two observers observe the same speed for light (y = 1 in (6.1) regardless of their relative velocities, x. Example 2. Suppose that two people A and B find the coordinates of two points C and D by using different axes, as illustrated in Figure 6.40. The Pythagorean theorem in Euclidean geometry guarantees that they will obtain $xA2 + $yA2 = $xB2 + $yB2 for the square of the distance between the two points. ♦ yB yA

∆y

B

∆yA

C

∆x

B

D

xB ∆xA

xA

Figure 6.40 Finding distance in different coordinate systems. In the theory of relativity the measurements of elapsed time and distance by different observers are related much as distances are in Example 2. Suppose that two observers record two events taking place using suitable units. Observer A finds the difference in time between the events to be $tA and the differences in the x-, y-, and z-directions to be $xA , $yA , and $z A ,

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respectively. Similarly, $tB , $xB , $yB , and $z B are observer B’s measurements. The theory of relativity guarantees that $xA2 + $yA2 + $z A2 − $tA2 = $xB2 + $yB2 + $z B2 − $tB2

(6.2)

Hermann Minkowski developed a four-dimensional geometry by using (6.2) as a distance formula. The Lorentz transformations are the transformations of Minkowski space and so preserve (6.2). That is, the Lorentz transformations are the symmetries of special relativity. (Section 7.6 discusses them. See Taylor and Wheeler [20] for more on the theory of relativity.) Example 3. Suppose that person A flashes a light twice, 1 second apart according to A’s measurement. If person B is traveling in the x-direction at some velocity with respect to A and observes them, $xB and $tB will differ from $xA = 0 and $tA = 1, but $xB2 − $tB2 = $xA2 − $tA2 = 0 − 1 = −1. The faster B is traveling with respect to A, the larger $xB and $tB will be. If B flashes the light twice, the situation is reversed. Thus, although A and B think that their clocks are running at different rates, it doesn’t make sense to say whose clock “actually” is slower. Similarly, we can’t say that one has a shorter unit of x-distances than the other. ♦ Physicists use symmetry extensively in quantum mechanics, but this important application goes beyond the level of this text. See Project 19. Symmetry has become a vital tool for science, allowing people to investigate both visual and abstract patterns using the precision and insight of mathematics.

6.5.4 Exercises for Section 6.5 Two arrangements of the atoms for a molecule are chemically equivalent provided that a direct isometry exists that converts one arrangement to the other. If no direct isometry exists, the possible arrangements are isomers. *6.5.4. (a) Describe the two-dimensional symmetries of the ethene molecule (C2 H4 ), which has a double bond between the carbon atoms. Classify the group of symmetries of C2 H4 (Figure 6.41). Note that all six atoms lie in a plane. (b) Dichloroethene (C2 H2 Cl2 ) replaces two of the hydrogen atoms of ethene with chlorine atoms. Draw its three isomers and describe their symmetries. Verify that the three groups of symmetries are subgroups of the symmetries of ethene. H

H C

H

C H

Figure 6.41 Ethene. 6.5.5. (a) Butane, with chemical formula C4 H10 , has two isomers. Draw two-dimensional versions of them and classify their groups of symmetries. The one with more symmetries is called isobutane and is used as a more environmentally friendly replacement for the refrigerant Freon. (b) Pentane, with chemical formula C5 H12 , has three isomers. Draw two-dimensional versions of them and classify their groups of symmetries.

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6.5.6. Draw two isomers for a molecule with a central carbon atom and four different atoms (W, X, Y, and Z) attached to it. How are they different geometrically? Chemically, they polarize light differently and are called chiral or optical isomers. 6.5.7. The six carbon atoms of benzene (C6 H6 ) form a ring (Figure 6.42). They exhibit resonance, the nonlocalized sharing of electrons, illustrated as a circle. Thus only three atoms are directly attached to each carbon. Assume that the twelve atoms of benzene lie in a plane. (Actually, the molecule is three-dimensional.) *(a) Describe the two-dimensional symmetries of the model shown in Figure 6.42. Classify the group of symmetries. *(b) Repeat Exercise 6.5.4(b) for dichlorobenzene (C6 H4 Cl2 ), which replaces two of the hydrogen atoms with chlorine atoms. *(c) Repeat Exercise 6.5.4(b) for trichlorobenzene (C6 H3 Cl3 ), which replaces three of the hydrogen atoms with chlorine atoms. Consider color symmetries that switch the hydrogen and chlorine atoms, as well as those that don’t. (d) Consider dichlorodifluorobenzene (C6 H2 Cl2 F2 ). Draw its isomers and describe their symmetries, including those that switch the hydrogen, chlorine, and fluorine atoms and those that don’t. H

H C

H

C C

C C H

H

C H

Figure 6.42 Benzene. 6.5.8 Find the symmetry group of buckminsterfullerene, C60 , the third form of pure carbon. (Figure 6.43 illustrates the bonds as edges and the carbons as vertices.)

Figure 6.43 Buckminsterfullerene.

6.5 Symmetry in Science

301

*6.5.9. (a) Each layer of carbonates (CO3 ) in a crystal of calcite (CaCO3 ) looks like the structure depicted in Figure 6.44. Classify its wallpaper pattern. (The calcium atoms lie in different layers not shown. Marble is one form of calcite.)

Figure 6.44 A layer of carbonates in a calcite crystal. (b) Figure 6.45 arranges the layer of carbonates in a different way that does not form in nature. Classify its wallpaper pattern. Give an explanation based on chemistry why this very symmetrical arrangement doesn’t occur.

Figure 6.45 A non-natural arrangement of carbonates. 6.5.10. (a) Classify the wallpaper pattern of one layer of a graphite crystal (Figure 6.37). (b) Classify the two-colored wallpaper pattern of one (horizontal) layer of a salt crystal (Figure 5.38). 6.5.11. Figure 6.46 illustrates the relationship between atoms in two layers of wurtzite, a crystal made of zinc, represented by dots, and sulfur, represented by circles. The zinc atoms are in one plane and the sulfur atoms in a parallel plane somewhat above the other plane.

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Figure 6.46 Wurzite. Classify the wallpaper pattern that preserve both layers of atoms. (Color switching symmetry, which exists, involves three-dimensional symmetries, and so doesn’t correspond to a wallpaper pattern.) 6.5.12. Use the operation given in (6.1). Explain your answers. *(a) Two streams of protons are approaching each other head-on in a cyclotron, with a velocity of 0.9 of the speed of light relative to an observer. How fast is one stream moving relative to the other stream? *(b) Suppose that you observe a particle moving at a velocity of 0.7 relative to you, followed by a second particle moving in the same direction with a velocity of 0.8 relative to you. How fast is the second particle moving relative to the first? (c) Suppose that a first spaceship is moving at a velocity of 0.5 relative to you and launches a second spaceship in the same direction moving at a velocity of 0.5 relative to it. How fast is the second ship moving relative to you? (d) Suppose that we have a sequence of spaceships like the two in part (c), where each can launch the next ship in the same direction at a velocity of 0.5 relative to it. How many spaceships must there be for the last one to be moving at a velocity greater than 0.9 relative to you? Repeat for 0.99 and 0.999. Show your work. 6.5.13. Show that the numbers strictly between −1 and 1 form a group with the operation ⊕ given in (6.1) as follows. (a) What velocity is the identity for ⊕? Prove your choice correct. (b) Find the inverse velocity of x. Prove your choice correct. (c) Prove that ⊕ is associative. That is, a ⊕ (b ⊕ c) = (a ⊕ b) ⊕ c. (This operation isn’t necessarily a geometric transformation, so you need to verify associativity, a property of groups that automatically holds for composition of functions.) (d) (Calculus.) Prove closure. Hint: Let a be a constant between −1 and 1. Verify that the derivative of f (x) = (x + a)/(1 + xa) with respect to x is always positive. Why does this show that f (−1) ≤ f (a) ≤ f (1)? Now find f (1) and f (−1).

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6.5.14 (a) On a graph with axes $x and $t, draw and describe the curve of points ($x, $t) such that $x 2 − $t 2 = −1. The curve represents the measurements that person B could obtain in Example 3. Events with a negative difference are called timelike because every observer can determine which event occurred first. (b) Repeat part (a) for $x 2 − $t 2 = 1. This difference could occur if person A flashed two lights at the same time and at a distance of 1 apart in A’s measurement. Events with a positive difference are called spacelike. Describe how person B could observe either light flash first. (c) Repeat part (a) for $x 2 − $t 2 = 0. Describe how person A could send two light flashes to give this difference. Describe how person B could observe the two events as happening at the same time. Events with a difference of zero are called lightlike. Symmetries for relativity must take the curves where $x 2 − $t 2 is constant to themselves. The value of b in parts (e) and (f) depends on the velocity of B with respect to A. (d) Define cosh a = (ea + e−a )/2 and sinh a = (ea − e−a )/2. Verify cosh2 a − sinh2 a = 1. (e) Explain why you can write the coordinates for the difference between two timelike events as (r sinh a, r cosh a) for some a and r in R. For a fixed b verify that f (r sinh a, r cosh a) = (r sinh(a + b), r cosh(a + b)) preserves $x 2 − $t 2 for timelike events. Describe what f does to points on the curve of part (a). (f) Modify and repeat part (e) for spacelike events. 6.5.15. The ! terms in Example 2 gives the usual distance formula d((x1 , y1 ), (x2 , y2 )) = $x 2 + $y 2 , which satisfies the following properties of a metric: d( A, B) = d(B, A). d( A, B) = 0

d(A, B) ≥ 0.

if and only if A = B.

d( A, B) + d(B, C) ≥ d( A, C). We use (6.2) to define the formula r ((x1 , y1 , z 1 , t1 ), (x2 , y2 , z 2 , t2 )) = which can be imaginary.

!

$x 2 + $y 2 + $z 2 − $t 2 ,

(a) For events A = (x1 , y1 , z 1 , t1 ) and B = (x2 , y2 , z 2 , t2 ), prove that r satisfies the first property of a metric. (b) Give examples of events illustrating the failure of the other three properties of a metric using r . (c) Use the terminology of exercise 6.5.14 to reinterpret the second and third properties of a metric using r . (d) Investigate when the fourth property, the triangle inequality, holds with r , assuming the values are all positive. What can you say when the values are all negative?

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6.5.5 Marjorie Senechal Marjorie Senechal (1939–) is now a retired professor at Smith College, a distinguished college for women, and was a leading researcher in mathematical crystallography. Although she was fascinated with patterns from a young age, not until much later did she realize that “mathematics is the science of patterns.” She believes visual thinking to be vital for all mathematics and particularly crystallography. After completing her Ph.D. in number theory in 1965, Senechal shifted her focus on patterns to mathematical crystallography and, more generally, to discrete geometry. She taught and conducted research at Smith College for over forty years, interspersed with extended research stays in seven countries. Senechal has published six books (with several in preparation) and dozens of articles, and has contributed chapters in many books. In addition, she has organized numerous conferences, given many featured addresses, served several professional organizations as a committee or board member, and has made many other less visible contributions to modern mathematical research. Senechal eagerly responded to the discovery of quasicrystals in 1984 and the ensuing heightened activity in mathematics, chemistry, and physics. In addition to traditional research in this area, she worked with the advanced computing facilities at the Geometry Center of the University of Minnesota. Powerful computers are beginning to provide visual and analytic insight into the mathematical structure of quasicrystals, which lack the well-understood repetitions of crystals.

6.6 Fractals Historically, geometry has focused on relatively simple, ideal shapes: circles, triangles, polyhedra, and the like. However, even a cursory glance at nature reveals a vast array of shapes unrelated to them. Benoit Mandelbrot, the originator of fractals, found geometric structure underlying complicated natural shapes. In 1975 he coined the word fractal to describe the convoluted curves and surfaces that can be used to model natural shapes that had previously seemed beyond mathematical study. Mathematicians initiated the abstract study of curves related to fractals much earlier— before 1900. In 1904 Helge von Koch defined the Koch curve (Figure 5.32) as the limit of an infinite process, illustrated in Figure 6.47. We start with the line segment at the top and replace its middle third with two segments forming an inverted “v.” This is the motif of the Koch curve. Now we replace each segment of the motif with a smaller copy to get the next design. We continue, replacing each segment with a still smaller copy of the motif. The repetitive process is called iteration, an important technique in computing. We call the motif the first iteration and the succeeding ones the second and third iterations. After infinitely many iterations we arrive at the Koch curve, which has the property of self-similarity: The entire curve is similar to a part of it. None of the iterations have the property of self similarity; rather each iteration is similar to a part of the next iteration. Example 1. Show the Koch curve is infinitely long. Solution. Suppose that the original segment in Figure 6.47 has a length of 1 unit. The first iteration (the motif) has four segments of length 1/3, for a length of 4/3 ≈ 1.3 units. The

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6.6 Fractals

Figure 6.47 Iterations leading to the Koch curve. second iteration has sixteen segments and a length of 16/9 = (4/3)2 ≈ 1.8 units. In general, the nth iteration has length (4/3)n units, increasing as n increases. The length of the Koch curve is limn→∞ (4/3)n = ∞. The Koch curve is enclosed in a finite area despite its infinite length. ♦ Exercise 6.6.1.∗ Explore why the method of IFS in Section 5.4 produces the same curve as the method Koch used. Mandelbrot noticed that many real phenomena, such as coastlines, mountains, and lungs, have a roughly self-similar shape: Their smaller features have the same overall bumpiness as the larger features. Of course, no part of the coastline of France will exactly replicate the entire coastline, and no real shape can exhibit even an approximately self-similar shape at the subatomic level. Thus exact self-similarity is too limited to model nature. Mandelbrot uses the term statistical self-similarity to describe approximate similarity over a range of scales. He avoided an exact mathematical definition of it because such a definition would apply only to mathematical objects, defeating his purpose. Computers can readily draw statistically selfsimilar shapes by modifying the iterations at smaller scales with randomly generated fluctuations. The resulting graphics often look strikingly realistic and support the usefulness of statistical selfsimilarity (Figure 6.48). However, Mandelbrot built the theory of fractals on deeper mathematics so that fractals can lead to new insights, not just interesting graphics. Developed in 1919, the Hausdorff dimension provides a measure of how convoluted mathematical shapes are, and Mandelbrot modified it to measure real objects. Felix Hausdorff (1868– 1942) noted a relationship between dimensions and the growth in the number of units as the measuring scale decreases. Figure 6.49 illustrates the intuition behind Hausdorff ’s approach. If we divide each side into three equal pieces, a line segment has 3 = 31 smaller segments, a square has 9 = 32 smaller squares, and a cube has 27 = 33 smaller cubes. The dimension appears as the exponent. The smaller pieces are similar to the originals by a scaling ratio of r = 13 . (Hausdorff ’s technical definition uses analysis. See Falconer [5].) Exercise 6.6.2. Create drawings analogous to Figure 6.49, with each side divided into four smaller units, so r = 14 . Verify that the line segment has 41 smaller segments, the square has 42 small squares, and the cube has 43 smaller cubes. We can generalize Figure 6.49 and Exercise 6.6.2: the number of units is proportional to n = (1/r )d , where d is the dimension and r is the scaling ratio. If we take logarithms and solve

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Figure 6.48 A fractal mountain.

Figure 6.49 Divisions of a segment, a square, and a cube. for d, we obtain the equation for the dimension: d = log n/ log(1/r )

(6.3)

Exercise 6.6.3. Verify that (6.3) holds for the line segment, square, and cube of Figure 6.49 for any choice of r . Verify that we get the same value for d if we use natural logarithms instead of logarithms to the base 10 in (6.3). Hausdorff dimension applies to more interesting shapes than line segments, squares, and cubes, including fractals in Example 2. For shapes that are not self similar, we need to use limits, as in Example 3. Example 2. Find the Hausdorff dimension of the Koch curve. Solution. In Example 1 a scaling factor of r = 13 gives n = 4 times as many units. Then (6.3) gives d = log 4/ log 3 ≈ 1.262. Verify that a scaling ratio r = 19 = ( 13 )2 gives the same value of d. The Koch curve is too convoluted to be measured by the one-dimensional unit of length— indeed, it is infinitely long. However, with an area of 0, the curve doesn’t act as a two-dimensional object. The value log 4/ log 3 is its Hausdorff dimension, indicating that the convoluted Koch curve is something between a line and a surface. ♦

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Example 3. We verify that the Hausdorff dimension of a circle is 1. This makes sense since the circumference of a circle is a length, which is a one-dimensional measure. In Figure 6.50, we can approximate the circumference of a circle with increasingly smaller units. The table below gives data for regular polygons, starting with the number of sides in the first column. The next column gives the length of each side. The next column, the perimeter, approximates 2π , the circumference of a circle of radius 1. The final column gives the ratio of the length of an edge of the polygon to the length of the edge of the preceding polygon. (Project 3 of Chapter 1 gives a way to find the values.) Sides 3 6 12 24 48

Length √ 3 ≈ 1.732

1 ! √ 2 − 3 ≈ 0.518 ; ! √ 2 − 2 + 3 ≈ 0.261 < ; ! √ 2 − 2 + 2 + 3 ≈ 0.131

Perimeter

Ratio

5.196

—

6

0.577

6.212

0.518

6.265

0.504

6.279

0.501

When we double the number of sides in an approximating polygon, the table indicates that length and so the scaling factor are not exactly half as long, but the ratio quickly approaches one half. This is exactly what we expect. The ratio log n/ log(1/r ) approaches 1 in the limit, which gives the Hausdorff dimension of the circle. ♦

R

Figure 6.50 Approximating a circle with edges. For simple shapes, such as a circle or even the Koch curve, the limit as r approaches 0 in (6.3) determines the Hausdorff dimension. More complicated sets of points require Hausdorff ’s exact definition. Using analysis, Hausdorff proved that every nonempty subset of Rk has a unique Hausdorff dimension, which is at most k. Since we can’t take mathematical limits with physical objects, the Hausdorff dimension unfortunately doesn’t apply to real shapes. However, a modification of (6.3) gives a way to estimate what is called the fractal dimension of an object. In 1961, Lewis Richardson published what may seem a strange article that contained a study of the estimated length of coastlines according to maps of different unit lengths. The estimated lengths differed widely, but Richardson found a function of the lengths in terms of the unit lengths. Mandelbrot recognized that Richardson’s equation fits the intuition behind the Hausdorff dimension. (Richardson apparently was unaware

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of Hausdorff ’s work.) We modify Richardson’s equation to relate it to Hausdorff dimension d by replacing his original uninterpreted exponent with 1 − d. Richardson’s equation is L( f ) = c( f 1−d ), where f represents the unit length, L( f ) is the estimated length (or area) of the object for that value of f , and c is a constant depending on the amount of the object measured. Example 4. In the motif of the Koch curve (see Figure 6.47) let each segment have a length of f 1 = 13 . Then the four segments give L( f 1 ) = 4/3 as an estimated length of the Koch curve. . In the third In the second iteration when f 2 = 13 · 13 , the sixteen segments give L( f 2 ) = 16 9 , and so on. We solve for d in Richardson’s equation by iteration when f 3 = 19 · 13 , L( f 3 ) = 64 27 1 1 1−d = c( · ) = using two values of f . With f 1 we get 4/3 = c( 13 )1−d , and with f 2 we get 16 9 3 3 1 1−d 1 1−d 4 1 1−d 1 d = 3 · (3) or c( 3 ) ( 3 ) . Substituting the first equation into the second yields 3 = ( 3 ) 4 = 3d . Then d = log 4/ log 3, as before. ♦ The exponent 1 − d in Richardson’s equation isn’t as mysterious as it may seem. For any unit f , the estimated length L( f ) of a self-similar curve should be the number of units times their length. Moreover, from (6.3) the number of units is proportional to n = ( r1 )d = ( 1f )d . That is, L( f ) = c · n · f = c · (1/ f )d · f = c · f 1−d . We want to solve Richardson’s equation for d, based on empirical values for f and L( f ). With two unknowns, c and d, we need two values of f and L( f ). From L( f 1 ) = c( f 1 )1−d and L( f 2 ) = c( f 2 )1−d , we find L( f 1 )/L( f 2 ) = ( f 1 / f 2 )1−d and so d =1−

log[L( f 1 )/L( f 2 )] . log( f 1 / f 2 )

(6.4)

Example 5. Figure 6.51 shows the coastline of England and Wales from Bristol to Liverpool. If we estimate its length using f 1 = 57-mile segments, we need five segments, so L( f 1 ) ≈ 285 Liverpool

Liverpool Holyhead

f3

f2

f1

Fishguard Swansea Bristol

Figure 6.51 Part of the coastline of England and Wales.

Bristol

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6.6 Fractals

miles. For f 2 = 28.5 miles, we need 12 12 segments, so L( f 2 ) ≈ 356 miles; f 3 = 14.25 miles gives 32 segments and L( f 3 ) ≈ 456 miles. As we shorten the unit length, we can follow the contour better, taking into account more and more of the multitude of peninsulas and bays. When we use f 1 and f 2 , (6.4) gives d = 1 − log(285/356)/ log(57/28.5) ≈ 1.32. ♦ Exercise 6.6.4. Verify that d ≈ 1.36 when you use f 2 and f 3 and d ≈ 1.34 when you use f 1 and f 3 . The values of d that Richardson found for coastlines are remarkably stable over a range of scales f . (They can’t be stable over all scales unless the object is perfectly self-similar. The range of scales depends on the object studied.) Mandelbrot calls the empirical value the fractal dimension to distinguish it from Hausdorff ’s. Mandelbrot and others have estimated fractal dimensions for a variety of natural curves and surfaces. He avoids defining a fractal, but the following provisional definitions are helpful. Example 2 illustrates how self-similar shapes, such as the Koch curve, can have unexpected dimensions. Self-similarity gives an exact number of copies, n, for an appropriate scaling ratio, r . More convoluted shapes generally have more small copies at a given scale and so have higher dimensions. (Richardson’s method for estimating fractal dimension of curves is difficult for computers. Project 21 gives an alternative method.) Provisional Definition. A fractal curve has fractal dimension greater than 1. A fractal surface has fractal dimension greater than 2. Fractals provide a good model for the lungs. The trachea splits into the bronchial tubes, which split into shorter and narrower tubes. In addition, the embryonic development of the lung is an iterative process (Figure 6.52). The iterative process (grow and then split when some chemical concentration is reached) would be easier to encode in DNA than the complicated convolutions of a lung. The convoluted surface of the lung greatly increases its area while

Figure 6.52 The development of the lungs.

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Symmetry

keeping its overall volume small. The large surface is biologically essential because the amount of carbon dioxide and oxygen that the lungs can exchange is roughly proportional to their surface area. A person’s lung would fit easily in a sphere of radius 4 inches or 0.1 meter, which has a surface area of less than 0.13 square meters. Using a light microscope, biologists found approximately 80 square meters of surface area in a lung (roughly the floor space of a large classroom). The higher magnification of an electron microscope yielded an approximation of 140 square meters. The increase in area at higher magnification corresponds to the increasing length of the coastline in Example 5. Scientists have estimated the fractal dimension of a lung to be 2.17. Blood vessels, kidneys, the liver, and other organs have good fractal models. Although fractals provide insightful models, scientists are hoping for more than explanations of already known facts. Why is the fractal dimension of a lung 2.17? After all, a higher dimension would give even more surface area. Perhaps higher dimensions impede the free passage of air or blood in the lung. Questions such as this provide ample challenges for research in the applications of fractals. (See Mandelbrot [12].)

6.6.1 Exercises for Section 6.6 6.6.5. For each motif of the fractal curve, sketch several iterations. (a) A stylized tree, where a branch splits into two others half as long. (b) A stylized tree, where a branch splits into three others half as long. (c) A Cantor set, where you divide a line segment into three equal pieces, remove the middle piece, and iterate with the remaining pieces. (d) A modified Koch curve, with a square on the middle third of a line segment rather than a triangle. (e) A modified Koch curve, where you divide a line segment into fifths and construct squares on the alternate sides of the second and fourth fifths. (f) A Sierpinski gasket, where you divide a triangle into four smaller triangles by connecting the midpoints of the sides, remove the middle triangle, and iterate with the remaining triangles. (g) A Sierpinski square, where you start with a square divided into nine smaller squares and remove the center square. Iterate with each of the remaining squares. *6.6.6. Find the Hausdorff dimension of the fractals in Exercise 6.6.5. 6.6.7. Investigate what happens to the figures and the fractal dimension when you increase the number of smaller copies in the fractals in Exercise 6.6.5(b) and 1(e). *6.6.8. For each motif for a fractal surface, sketch the first and second iterations and find the fractal dimension. (a) A Koch pyramid, where you divide a triangle into four smaller triangles, as in Exercise 6.6.5(f), but construct a regular triangular pyramid on the middle triangle and iterate with the smaller triangles and the faces of the pyramid. (b) A Koch cube, where you divide a unit square into nine smaller squares, construct a cube on the middle square, and iterate as in part (a). (c) A Menger sponge, where you divide a unit cube into 27 smaller cubes, remove the center cube, and iterate with each of the 26 remaining smaller cubes.

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6.6 Fractals

6.6.9. For the Cantor set of Exercise 6.6.5(c), start with the interval [0, 1] of length 1. The first iteration removes a segment of length 1/3. The second iteration removes two segments of length 1/9 each and so on. *(a) Give the total length of the segments removed in the nth iteration. *(b) Use a geometric series to find the total length of the segments removed. Determine the length left in the Cantor set. (c) A fat Cantor set starts with the interval [0, 1], but removes the middle 1/k instead of the middle third, where k > 3. Draw the first and second iterations of a fat Cantor set with k = 4. *(d) Repeat parts (a) and (b) to determine the length left in a fat Cantor set. 6.6.10. (a) Use a geometric series to find the total area of the infinitely many squares in Exercise 6.6.5(d). (b) Repeat part (a) for the area of the squares removed in Exercise 6.6.5(g). What is the area of the fractal that is left after removing all of these squares? (c) Repeat part (a) for the total volume of cubes in Exercise 6.6.8(b). (d) Repeat part (a) for Exercise 6.6.8(c) to find the volume of the cubes removed. What is the volume remaining in the Menger sponge? 6.6.11. (Calculus.) Use the information of Example 3 to explain why Richardson’s method and (6.4) give a fractal dimension of 1 for a circle. 6.6.12. *(a) You can estimate the fractal dimension of the coastline of Norway using the map shown in Figure 6.53 by using different length line segments. Start at Oslo and

Figure 6.53 Part of the coastline of Norway.

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Symmetry

lay out first 2 inch, than 1 inch, and then 12 inch line segments along the coast to Bergen. Then use (6.4) to calculate the fractal dimension. On this map, 12 inch ≈ 13.5 miles. (b) Estimate the fractal dimension of other coastlines using other maps. (There is a conjecture that the fractal dimension of a coastline is related to the geological age of the rocks making up the coastline.)

6.6.2 Benoit Mandelbrot Intuition is not something that is given. I’ve trained my intuition to accept as obvious things which were initially rejected as absurd and I find everyone can do the same. [Fractals] provide a handle to representing nature, and intuition can be changed and refined and modified to include them. —Benoit Mandelbrot

Benoit Mandelbrot (1924–2010) was the leading proponent of fractal geometry, which he pioneered. With the rise of the Nazis, his family fled their native Poland and the chaos of World War II further disrupted his life and his education. However, his well-developed geometric intuition helped him to complete a doctorate in mathematics. His interests ranged over a variety of unusual aspects of mathematics, physics, and engineering, including noise in electrical transmissions, which others had thought was simply random. Mandelbrot found that the frequency of noise measured in second-long intervals resembled the frequency at longer intervals. He found other phenomena with a uniformity under change of scale, or what he called statistically self-similar, or fractal. He collected examples of fractal behavior much the way naturalists collect specimens. His interests ranged over many fields and he collaborated with many people in and outside of academia. Mandelbrot’s interest in the application of fractals was coupled with an intense interest in mathematical ideas, although he was much less interested in mathematical proof. He drew on the results of others, which, coupled with his remarkable visual intuition and stunning computer graphics, allowed him to build new mathematical ideas and conjectures. He helped pioneer the use of computers to draw fractals and to approach mathematics as an experimental field.

6.6.3 Projects for Chapter 6 1. Find and classify examples of cyclic and dihedral symmetry patterns. For example, company logos, hubcaps, and other objects often exhibit such symmetry. 2. Design your own wallpaper patterns by hand or use software. 3. Determine which types of frieze patterns and wallpaper patterns can be formed with mirrors. Arrange the mirrors in various ways facing one another. Place symmetric and asymmetric designs in the region between the mirrors. 4. Investigate symmetry in the art of M. C. Escher. (See Schattschneider [16].) 5. Investigate symmetry in weaving. (See Pizzuto [14].) 6. Find examples of actual wallpaper and classify them. Determine which symmetry types of wallpaper patterns predominate. Investigate the reasons, including aesthetic ones, that underlie the choice of symmetry in wallpaper.

6.6 Fractals

313

7. Investigate symmetry in the art of various cultures. (Wade [21] provides a large variety of examples. For Islamic art, see El-Said and Parman [4].) 8. Investigate tilings of the plane, which generalize wallpaper patterns. (See Gr¨unbaum and Shephard [8].) 9. In many cultures circular friezes appear as ornaments on cups and other cylindrical objects. Circular friezes replace the translations with rotations about a vertical axis. The seven frieze groups become seven subgroups of Dnh , where 360/n is the smallest angle of rotation around a vertical axis. (We use Sch¨onflies’ notation.) (a) Draw a frieze with symmetry group p2mm on a sheet of paper. Roll the paper into a cylinder to create a circular frieze. Describe how the symmetries of the frieze become symmetries of the circular frieze. Explain why the circular frieze has Dnh for its group of symmetries. What determines the value of n that a circular frieze has? Why do the symmetries of other circular friezes form subgroups of some Dnh ? (b) For the six other types of friezes, make a circular frieze pattern. Describe their rotations, mirror reflections, and rotatory reflections. (c) One of the remaining circular friezes has a horizontal mirror reflection. Its group is called Cnh . Find it. (d) Two of the remaining circular friezes have rotatory reflections of angles half as large as the rotations about the vertical axes. Find the one whose rotations are all around a vertical axis; it has group S2n . Describe the additional rotations of the other circular frieze with rotary reflections, whose group is Dnd . (e) Which of the circular friezes in (b) has the symmetry group Cn ? (f) One of the two remaining circular frieze types has only direct symmetries. Its group is Dn . Identify it. (g) The remaining type has the symmetry group Cnv . Identify it. (h) Make a table to match the frieze and circular frieze groups. (i) Classify the group of an antiprism. (See Exercise 6.4.2.) (j) Place a mirror face up on a table. Place two mirrors perpendicular to it, facing each other at various angles. Which of the circular frieze patterns can you make by placing symmetric and asymmetric designs in the region between the mirrors? 10. Investigate the symmetries of continuous (non-discrete) frieze and wallpaper patterns. Try to classify types of such patterns. 11. Investigate the symmetry of knots and braids. 12. Investigate symmetry in fugues and twelve-tone music. (See Senechal and Fleck [18].) 13. (a) Use Exercise 6.3.7(b) to list the twenty-two conceivable two-color frieze pattern types. (The color group must include all the symmetries of the color preserving group.) (b) As exercise 6.3.4 illustrates, there are exactly seventeen actual two-color frieze pattern types. Determine the five types from part (a) that don’t actually exist. Explain why they can’t exist. Hint: You may need more than two colors. 14. Borrow a set of hand bells to investigate symmetry and change ringing. (See Senechal and Fleck [18, 47].) 15. Investigate color symmetry. (See Loeb [11].) 16. Investigate symmetry in crystals. (See Senechal [17].) 17. Build a Penrose tiling by using the two types of tiles shown in Figure 6.54. Penrose proved that matching sides with dots on them ensures that the tiling will be nonperiodic.

314

Symmetry

Investigate Penrose tiles. (See Gardner [7], Gr¨unbaum and Shephard [8], and Peterson [13].)

36°

72°

Figure 6.54 Tiles for a Penrose tiling. 18. 19. 20. 21.

22. 23. 24. 25. 26.

27. 28.

Investigate quasicrystals. (See Jaric [10], Peterson [13], and Senechal [17].) Design fractals with or without software. Investigate the Mandelbrot set and Julia sets. (See Mandelbrot [12].) Estimate the fractal dimension of real shapes. Determine the scale at which the fractal nature of your examples breaks down. The fractal dimension of a surface can be estimated by adding 1 to the fractal dimension of a cross section. (See Mandelbrot [12].) The next paragraph gives an alternative method to approximate the fractal dimension of a curve. Computers can modify the approach with digital images of fractals. Make several transparencies with square grids of different scales (for example, 1/4 inch, 1/8 inch, and 1/16 inch squares). Place each transparency over the curve and count the number of squares that have parts of the curve in them. Let n s be the number of squares when the length of the square’s side is s. Explain why the fractal dimension is approximated by log(n s /n t )/ log(s/t). Investigate the role of symmetry in quantum mechanics. (Bunch [1] and Rosen [15] provide elementary expositions and bibliographies.) Investigate other ideas in symmetry. (See Bunch [1], Hargittai [9], Rosen [15], and Senechal and Fleck [18].) Write an essay considering the relationship of symmetry and culture. Is a classification of designs by their symmetries culturally objective? Write an essay considering the relationship of symmetry and beauty. Write an essay discussing the application of abstract mathematics to the world around you. For example, why do proofs about infinite, perfect mathematical crystals tell you anything about real, finite crystals? Write an essay discussing the additional insights given by a study of symmetry over the insights from transformational geometry. Write an essay discussing the notion of mathematics as an experimental science. Consider the questions: Are proofs essential to mathematics? In what ways is experimental evidence appropriate and/or convincing in mathematics?

6.6.4 Suggested Readings 1. Bunch, B., Reality’s Mirrors, New York: John Wiley & Sons, 1989. 2. Crowe, D., Symmetry, Rigid Motions, and Patterns, Arlington, MA: COMAP, 1986. 3. Crowe, D., and D. Washburn, Groups and geometry in the ceramic art of San Ildefonso, Algebras, Groups and Geometries, 1985, 2(3): 263–277. 4. El-Said, I., and A. Parman, Geometric Concepts in Islamic Art, Palo Alto, CA: Dale Seymour, 1976.

6.6 Fractals

5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24.

315

Falconer, K., The Geometry of Fractal Sets, New York: Cambridge University Press, 1986. Gallian, J., Contemporary Abstract Algebra, Lexington, MA: D. C. Heath, 1994. Gardner, M., Penrose Tilings to Trapdoor Ciphers, New York: W. H. Freeman, 1989. Gr¨unbaum, B., and G. Shephard, Tilings and Patterns, New York: W. H. Freeman, 1989. Hargittai, I. (ed.), Symmetry, Elmsford, NY: Pergamon Press, 1986. Jaric, M. (ed.), Introduction to the Mathematics of Quasicrystals, Boston: Academic Press, 1988. Loeb, A., Color and Symmetry, New York: John Wiley & Sons, 1971. Mandelbrot, B., The Fractal Geometry of Nature, New York: W. H. Freeman, 1982. Peterson, I., The Mathematical Tourist, New York: W. H. Freeman, 1988. Pizzuto, J., 101 Weaves in 101 Fabrics, Pelham, NY: Textile Press, 1961. Rosen, J., Symmetry Discovered, New York: Cambridge University Press, 1975. Schattschneider, D., Visions of Symmetry, New York: W. H. Freeman, 1990. Senechal, M., Crystalline Symmetries, Philadelphia: Bristol, 1990. Senechal, M., and G. Fleck (eds.), Patterns of Symmetry, Amherst: University of Massachusetts Press, 1977. Stewart, I., Why Beauty Is Truth, New York: Basic Books, 2007. Taylor, E., and J. Wheeler, Spacetime Physics, San Francisco: W. H. Freeman, 1966. Wade, D., Geometric Patterns and Borders, New York: Van Nostrand Reinhold, 1982. Wenninger, M., Polyhedron Models, New York: Cambridge University Press, 1971. Weyl, H., Symmetry, Princeton, NJ: Princeton University Press, 1952. Yaglom, I., Felix Klein and Sophus Lie, Boston: Birkh¨auser, 1988.

7

Projective Geometry

Figure 7.0 The artists of the Renaissance uncovered the rules of perspective, as illustrated by this detail from The Flagellation of Christ by Piero della Francesca. Projective geometry provides the mathematical foundation for perspective in art and computer graphics. We explore projective geometry and some of its connections to art and computer graphics.

317

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Projective Geometry

Metrical geometry is thus a part of [projective] geometry, and [projective] geometry is all geometry. — Arthur Cayley

7.1 Overview and History Projective geometry grew out of the study of perspective in Renaissance art. Piero della Francesca (circa 1415–1492), Leonardo da Vinci (1452–1519), Albrecht D¨urer (1471–1528), and other artists worked out the geometric rules of perspective. They weren’t particularly complicated, but they transformed the appearance of paintings. One key insight clearly contradicted Euclidean geometry: Lines parallel in the world have images meeting at what are called ideal points or vanishing points on the horizon, an ideal line infinitely far away. The painters were not challenging the truth of Euclidean geometry. (Euclid had written about such effects of perception in his work, The Optics.) The attempt to paint the way people saw the world emerged from a philosophy that took people as “the measure of all things,” rather than viewing the world from a divine point of view. Girard Desargues (1593–1662) was the first person to prove geometric properties of perspective beyond what was needed for painting. Unfortunately, his notes were hard to read and had unusual conventions, such as using flowers and trees as the names of geometric objects. Only a few contemporary mathematicians, particularly Blaise Pascal (1623–1662), built on Desargues’ efforts and saw the unifying power of what we now call projective methods. Johannes Kepler (1571–1630), who used conics effectively in his astronomical work, realized that circles stretch continuously to ellipses and then to parabolas and hyperbolas. The shadow of a lampshade cast on a wall illustrates the process, suggesting the notion of projective transformations. The progress made by the pioneers was overshadowed and mostly forgotten because of the marvelous advances of analytic geometry and then calculus, which quickly dominated mathematical thought. Led by Gaspard Monge (1746–1818), the rediscovery and investigation of projective ideas started after 1800. One of his students, Jean Victor Poncelet (1788–1867) developed the subject extensively and accelerated others’ research with his book Treatise of the Projective Properties of Figures, published in 1822. Poncelet, following Monge’s lead, emphasized a synthetic approach to geometry, investigating properties not depending on distance. Poncelet realized the importance of duality, which says that points and lines function formally in the same way. In 1822 analytic methods seemed useless in projective geometry since ideal points, where Euclidean parallel lines meet on the horizon, had no obvious coordinates. The success of the projective approach launched a rivalry between geometers who advocated an analytic approach and those, including Poncelet, who pursued projective geometry synthetically. Augustus M¨obius (1790–1868) and Julius Pl¨ucker (1801–1868) developed coordinates for projective geometry by 1830, and gradually geometers realized that the synthetic and analytic approaches complemented each other. M¨obius initiated the study of transformations in projective geometry and other geometries. Karl Van Staudt (1798–1867) showed how to derive projective coordinates independently of Euclidean geometry. Projective geometry rose to central prominence once mathematicians realized how it unified many of the areas of geometry and even other areas. Arthur Cayley (1821–1895) and Felix Klein (1849–1925) showed how to develop Euclidean, hyperbolic, and single elliptic geometries within projective geometry. Klein realized the vital role projective transformations played in unifying geometry. In the early twentieth century mathematicians showed that the geometry of special

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relativity is a part of higher dimensional projective geometry. Other applications of projective geometry include computer graphics, statistical design theory, and photogrammetry (which infers geometric properties of objects based on photographs).

7.1.1 Projective Intuitions Perspective painting depends on the idea of projecting a scene onto the plane of the easel with respect to the eye of the painter. The detail of D¨urer’s woodcut in Figure 7.1 depicts an instrument

Figure 7.1 D¨urer’s woodcut on drawing in perspective. D¨urer developed to help artists learn to draw in perspective. The artist views the scene through a lined vertical glass plate from the point of perspective indicated by the pointed marker and transfers it to the lined paper on the table. Figure 7.2 illustrates such a projection, with the parallel

Figure 7.2 A simple perspective drawing.

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sides of the road appearing to meet at a ideal point at infinity on the horizon. Axiomatically we characterize projective geometry by the axiom that two distinct lines in the plane have at least one point on both lines. The definition of a perspectivity formalizes the idea of an artist transferring a scene onto a perspective drawing. In the definition, we can think of T as the artist’s eye, the Ui as the points in the scene, and the Vi as their images in the drawing. Definition. A perspectivity with respect to a point T is a mapping of the points Ui on one line to the points Vi on another line so that Vi is the image of Ui if and only if Ui , Vi , and T are on the same line (Figure 7.3).

T V2

U1 V1

U2

V2

V1

U3

V3 T

U1 U2

U3

V3

Figure 7.3 Perspectivities with respect to a point. In Euclidean geometry, the distance between two points is a fundamental measurement. As the spacing of the telephone poles in Figure 7.2 and the drawing on the left of Figure 7.3 suggest, perspective distorts distances and proportions. Moreover, the ordering of points on a line can change, as indicated in the drawing on the right of Figure 7.3. That is, projective transformations can move points around so that one no longer needs to lie between the other two. (Artists never need to choose a perspective that switches the order.) Exercise 7.1.5 asks you to show that a pair of perspectivities can map three points on one line to any three points on another line. That is, properties involving only two or three points on a line, such as distance, betweenness or the ratio of lengths, are not projective properties. The non-right angle the telephone poles make with the side of the road in Figure 7.2 suggests that angle measure is not a projective property. At first it might seem hard to describe any geometric properties preserved in perspective. Geometers have found properties involving four or more points, built on straight-line constructions, and preserved by perspective and in projective geometry. The most basic of the properties, a harmonic set, is built from a complete quadrangle: four points, no three on the same line, such as T1 , T2 , T3 , and T4 in Figure 7.4, and the six lines they determine form a complete quadrangle. The points P, Q, R, and S in that figure form a harmonic set. Definition. Four distinct points P, Q, R, and S on a line k form a harmonic set, denoted H (P Q, RS), if and only if there is a complete quadrangle with six lines different from k such that P is on two of the lines, Q is on two other lines, R is on one other line, and S is on the remaining line.

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7.1 Overview and History

T1 T4

k4

k1 T2

k2 k3 m

P

R

T3

k5

k6

Q

S

Figure 7.4 A complete quadrangle. Example 1. From three points P, Q, and R on a line, the following construction determines the unique fourth point S to form a harmonic set, illustrated in Figure 7.4. Although not obvious, every such construction will give the same fourth point. (See Exercises 7.1.7 and 7.1.8.) Uniqueness is an axiom in Section 7.2. Let P, Q, and R be three distinct points on a line m. Draw lines k1 and k2 through P and another line k3 through R. (The lines ki are distinct and differ from m.) Let T1 and T2 be the intersections of k3 with k1 and k2 , respectively. Let k4 and k5 connect Q with T1 and T2 , respectively. Then k4 intersects k2 , say at T3 , and k5 intersects k1 , say at T4 . Finally, draw the line connecting T3 and T4 and find its intersection S with the original line m. The complete quadrangle with points Ti and lines k j satisfies the conditions of the definition, showing H (P Q, RS). ♦ Exercise 7.1.1. A complete quadrilateral is a set of four lines, no three on the same point, and the six points they determine. Draw a complete quadrilateral. Example 2. Triangles △ABC and △D E F are said to be perspective from the point P whenever ← → ← → ← → P is on the lines AD, B E, and C F. Figure 7.5 illustrates this, as well as the following closely related concept. Triangles △ABC and △D E F are said to be perspective from the line k ← → ← → ← → whenever the corresponding sides of the triangles intersect on k. That is, AB and D E, AC P

A C

U

B

T F E

S

k

D

Figure 7.5 Two triangles perspective from a point and from a line.

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Projective Geometry

← → ← → ← → and D F, and BC and E F have their three points of intersections, S, T , and U , respectively, on k. Desargues used three-dimensional Euclidean properties to show perspectivity from a point implies perspectivity from a line. The theorem is now named for him and stated below. Desargues’ theorem is sometimes taken as an axiom, and then properties of harmonic sets are proven from it. Artists use Desargues’ theorem at least implicitly when they determine the placement of shadows. Figure 7.6 illustrates this, where △ABC is the object in space, △D E F is its shadow on the ground, and P is the light source. The plane of the original object and the ground meet in the line k, the line of perspective. Three-dimensional insight lay behind Desargues’ proof of his theorem, as well as more modern proofs. ♦ P

A C

U

B

T F

D

E

S

Figure 7.6 Shadows follow perspective rules.

Theorem 7.1.1. (Desargues’ theorem). If two triangles are perspective from a point, then they are perspective from a line. Proof. See Project 4. ! See Kline [7, Chapters 14 and 35] for more on the history of projective geometry. See Dillon [4] and Frantz and Crannell [6] for more on the connections between perspective and projective geometry.

7.1.2 Exercises for Section 7.1 *7.1.2. Draw a picture modifying Figure 7.2 so that the telephone poles are spaced geometrically. For example, let the first pole be 4 inches from the ideal point on the horizon to which the poles are heading, and let the succeeding poles be 3 = 4 · 34 , 2.25 = 4 · ( 43 )2 , . . . inches from the ideal point. Do the distances between poles look

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7.1 Overview and History

constant or do they appear to be getting smaller or larger as they go toward the ideal point? Try other distance relationships between the poles. What relationship describes the placement of equally spaced items drawn in perspective, as in Figure 7.2? 7.1.3. (a) Investigate how to continue the checkerboard pattern in Figure 7.7 so that the squares appear the same size in perspective. (b) Figure 7.7 is said to be drawn in two-point perspective since there are clearly two ideal points. However, every point on the horizon line serves an ideal point. Illustrate this by indicating on a copy of the figure the location of the ideal points where the diagonals of the squares intersect on the horizon. (Figure 7.2 is an example of one-point perspective.) (c) In a drawing in two-point perspective, the intended viewpoint for the perspective drawing is on a hemisphere with diameter the segment between the two ideal points X and Y . Hold the book so that the horizon line is even with your eye and bring Figure 7.7 close enough so that your eye is approximately on a semicircle with diameter X Y . Now look at the checkerboard. Do its squares appear square to your eye? Y

X

Figure 7.7 A checkerboard pattern in perspective.

7.1.4. (a) Figure 7.8 gives a view of a rectangular box in three-point perspective. Use the method of Exercise 7.1.3(c) to determine the viewpoint for perspective. Describe where the point of perspective seems to be relative to the top and bottom corners of the box nearest to the viewer. (b) Make a three-point perspective drawing of the box where the viewpoint for perspective is even with the bottom of the box. Describe how this affects the location of the ideal points and the apparent shape of the box. (c) Make a more involved drawing with several buildings in three-point perspective. Explain why most perspective paintings employ two-point perspective, as in Figure 7.7, rather than three-point perspective. 7.1.5. *(a) For two distinct points P and Q on a line k and two distinct points P ′ and Q ′ on another line k ′ , show how to find a perspectivity from some point T that takes P to P ′ and Q to Q ′ .

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Projective Geometry

Y

X

Z

Figure 7.8 Three-point perspective. (b) For three distinct points P, Q, and R on a line k and three distinct point S, T , and U on another line m, show how to use one or two perspectivities to take P to S, Q to T , and R to U . Hint: Use part (a) to take P, Q, and R to P ′ , Q ′ , and R ′ on a third line k ′ , where R ′ = U . 7.1.6. Use a lamp with a light bulb casting sharp shadows on a wall. (a) Cut out a triangle. Can you make the shadow of any of its angles be acute? right? obtuse? What happens to the shadow angles if you hold one side of the triangle parallel to the wall and rotate the triangle about it? (b) Describe the shapes that the shadow of a cut out circle can have. (c) Experiment with other shapes. Can the shadow of a convex planar set ever be nonconvex? Can the shadow of a nonconvex planar set ever be convex (without becoming a line segment)? 7.1.7. (a) Use a dynamic geometry construction program to model the construction of Example 1. Keep P and Q fixed and move R along line k. Note how S moves in response to R. (b) Keep P, Q, and R fixed but move the lines through them. Does S move? (c) Have the program measure the distances d(P, Q), d(P, R), and d(P, S). Set d(P, Q) at some convenient distance, such as 1 or 10, and investigate the relationship of d(P, R) and d(P, S) as R moves. (d) From the same points P, Q, and R use different lines to reproduce the construction of Example 1. Do you get the same point S? (It will help if the two constructions are on different sides of the line k.)

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7.1 Overview and History

*7.1.8. Use graph paper to construct a harmonic set H (P Q, RS), where P, Q, and R are on the x-axis at coordinates 0, 1, and 3. What is the coordinate of S? Choose different initial lines and construct S from the same three points. Do your two constructions yield (nearly) the same point? (It will help if the two constructions are on different sides of the line k.) 7.1.9. (a) Prove that, if H (P Q, RS), then H (P Q, S R), H (Q P, RS), and H (Q P, S R). (b) Suppose that H (P Q, RS). Use constructions to explore which, if any, of H (P R, Q S), H (P S, Q R), and H (RS, P Q) are correct. That is, can they be harmonic sets? *7.1.10. A harmonic set of lines H ( jk, lm) relates to a complete quadrilateral in the same way that a harmonic sets of points relates to a complete quadrangle. The roles of points and lines are interchanged. (See Exercise 7.1.1.) (a) Define a harmonic set of lines. (b) Convert Example 1 to a construction of a harmonic set of lines. Carry out the construction twice, once where l is a bisector of j and k and once where the lines seem unrelated. In the first case, how is m related to any of the other lines? 7.1.11. Let P, Q, R, and S be points on the x-axis with coordinates 0, 1, a, and h(a), respectively, with a ̸= 2 and a ̸= 12 . Assume that H (P Q, RS) and look for a formula for h(a). *(a) Let l1 , l2 , and l3 be the lines x = 0, y = x, and y = −x + a. Verify that P is on l1 and l2 and R is on l3 . Find the points T1 and T2 , as denoted in Figure 7.4. (b) Verify that l4 and l5 are y = −ax + a and y = ax/(a − 2) − a/(a − 2). a a a , 1+a ) and T4 = (0, 2−a ). (c) Verify that T3 = ( 1+a *(d) Find the equation of l6 . Verify that S = (h(a), 0), where h(a) = a/(2a − 1). 7.1.12. (a) In Exercise 7.1.11, alter one of the lines (l1 , l2 , or l3 ) to show that, when a = 2, h(2) = 23 . (b) In Exercise 7.1.11, for a = 12 find the equation of l6 . What happens to S? How does this fit with the equation in Exercise 7.1.11(d)? 7.1.13. In Figure 7.9 the circle has the equation (x − 12 )2 + y 2 = 0, with P = (0, 0), Q = ← → (1, 0), R = (a, 0), and 0 < a < 1, a ̸= 12 . Find the coordinates of S, where S A and ← → S B are tangents to the circle and A and B have a for their x-coordinate. Exercise 7.1.11 shows that H (P Q, RS). A

P

R Q

S

B

Figure 7.9 A circle and a harmonic set.

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Projective Geometry

7.1.14. *(a) In Figure 7.5 verify that △AB P and △T U F are perspective from the point C and find the line from which they are perspective. (b) Find two other triangles, a point, and a line such that the triangles are perspective from both the point and the line. ← → ← → 7.1.15. In Exercise 7.1.13 the line AB is called the polar of S and S is called the pole of AB, both with respect to the circle. Poles and polars can be defined similarly with respect to any conic provided the pole is outside of the conic. (We don’t consider here the situation where the pole is in the interior of the conic. See Exercise 5.6.9.) Given a conic and a point S outside it, construct the two tangents to the conic from S. The line determined by their points of tangency is the polar of S and S is its pole. *(a) For the point S = (0, w) on the y-axis with w < 0, find the equation of the polar of S with respect to the parabola y = x 2 . Graph the parabola, S, its tangents, and its polar. *(b) Find the intersection R of the y-axis with the polar and the intersection Q of the y-axis with the parabola. How is Q related to R and S? Where would the fourth point P in the harmonic set H (P Q, RS) need to be? Relate this to Exercise 7.1.12(b). (c) Repeat√parts (a) for the point S = (0, w) with 0 < w < 1 and the hyperbola y = ± x 2 + 1. Find the intersection R of the y-axis with the polar and the intersections P and Q of the y-axis with the hyperbola. Verify that P, Q, R, and S form a harmonic set. What happens when S = (0, 0)? (d) Repeat part (c) for the point S = (0, w), where w > 1 and the ellipse y = √ ± 1 − 4x 2 . 7.1.16. In Figure 7.10 the plane of a perspective drawing is in the x z-plane (with equation y = 0) and the artist’s eye is at the point (0, −d, 0) to the left of the drawing. For a point P = (x, y, z) in space, determine the coordinates of its image Q on the painting, where y = 0. z Q Eye at (0,-d,0).

P = (x,y,z)

y x

Plane of drawing, y = 0.

Figure 7.10 Coordinates in perspective drawing. 7.1.17. The name “harmonic set” comes from relationships the ancient Greeks found between music and mathematics. The segment O P in Figure 7.11 represents a taut string on a guitar or other stringed instrument.

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7.2 Axiomatic Projective Geometry

(a) If we choose different points, such as C, E, and G in Figure 7.11, to place our finger and pluck the string between O and that point, we get notes at different pitches. The notes of the very harmonic sounding major chord C E G have relative lengths OC = 5/4, O E = 1, and OG = 5/6. Use Exercise 7.1.11(d) to verify that H (O E, C G). O

G

E

C

P

Figure 7.11 Harmonic intervals. (b) Musical harmonics relate to the number of waves a plucked string can have between its starting point and its ending point. The principal tone corresponds to one wave between O and P (labelled W 1 in the top part of Figure 7.12). A string vibrating with half of that wavelength sounds an octave higher (W 2). For one-third the length (W 3), the sound is an octave and a fifth higher, for one fourth the length (W 4), the sound is two octaves higher, and so on. The Greeks called the sequence 1, 12 , 13 , 14 , . . . harmonic because of this relationship. Verify that the harmonic sequence gives harmonic sets. That is, for the labelling of points in the bottom part of Figure 7.12 verify that H (0 12 , 13 1), H (0 13 , 14 12 ), and so on. W1 W4

W3

P

O W2

0

1/5 1/4 1/3

1/2

1

Figure 7.12 Harmonic sets.

7.2 Axiomatic Projective Geometry In this section we develop axiomatically some elementary properties of projective planes. (For a more thorough axiomatic development, see Coxeter [2] or Cederberg [1].) The first five axioms describe the minimum relations of points, lines, and harmonic sets. Recall the definition of a harmonic set in Section 7.1 in terms of points and lines. The separation axioms, which appear later in this section, provide a structure of the points on a line taking the place of order. As we will see, the structure matches the arrangement of points on a circle. The undefined terms are point, line, on, and separate. Axioms 7.2.1. (i) Two distinct points have exactly one line on them. (ii) There are at least four points with no three on the same line. (iii) Every two distinct lines have at least one point on both lines.

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Projective Geometry

(iv) Given three distinct points P, Q, and R on a line k, there is a unique point S on k, distinct from P, Q, and R, such that H (P Q, RS). (v) If H (P Q, RS), then H (RS, P Q). Theorem 7.2.1. (i) Two distinct lines have exactly one point on both lines. (ii) Every line has at least four distinct points on it. (iii) If H (P Q, RS), then H (P Q, S R). Proof. See Exercise 7.2.12 for parts (i) and (iii). For part (ii), let k be a line. By axiom (ii) there are four points A, A1 , A2 , and A3 and at least one of them, say A, is not on k. Consider the lines ki on A and Ai , for i = 1, 2, 3. By axiom (ii) the lines are distinct. By part (i) they have exactly one point in common with k, which gives three distinct points on k. Axiom (iv) guarantees the fourth point. ! Definition. The point on the lines k and l is denoted k · l. Points on the same line are collinear, and lines on the same point are concurrent. We follow Karl Van Staudt’s method of using harmonic sets to construct and coordinatize infinitely many points on a line and the plane without any dependence on distance. The subscripts reflect Exercise 7.1.12(b), which showed that two points, their Euclidean midpoint, and the ideal point formed a harmonic set. Definition. Given distinct collinear points X 0 , X 1 , and X , define X 2 to be the point such that H (X X 1 , X 0 X 2 ). Given X n and X n+1 , define X n+2 to be the point such that H (X X n+1 , X n X n+2 ). Given X a and X b , define X (a+b)/2 to be the point such that H (X X (a+b)/2 , X a X b ) (Figure 7.13).

X0 X1/2 X1 X2 X3 X4

X

Figure 7.13 Using harmonic sets to give coordinates. Exercise 7.2.1.∗ Define points X −n , for n a positive integer. Example 1. Figure 7.13 illustrates the placement of various points X a . If X is the ideal point on the horizon of a perspective painting, the construction shows how to place points X 0 , X 1 , X 2 , etc. to look equally spaced. By Exercise 7.2.10, the sequence of Euclidean distances d(X n , X ) forms a harmonic sequence, such as 1, 12 , 13 , 14 , . . . . ♦ The first five axioms don’t guarantee that the points X a are distinct. (Indeed, in Chapter 7 we construct a finite projective plane with just four points on each line and a total of thirteen

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7.2 Axiomatic Projective Geometry

points satisfying all five axioms.) Theorem 7.2.3 depends on the following separation axioms to ensure that the X a are distinct. We write P Q//RS to denote that P and Q separate R and S. For example, in Figure 7.13 we will have X 0 X 1 // X 1/2 X 2 and in general X 0 X 1 // X 1/2 X n for n > 1. From a Euclidean intuition, the points P and Q need to alternate with R and S to separate them. Separation in projective geometry takes the role of betweenness in Hilbert’s axiomatization of Euclidean geometry. Axioms 7.2.2. Separation Axioms (vi) If H (P Q, RS), then P Q//RS. (vii) If P Q//RS, then P, Q, R, and S are distinct collinear points, P Q//S R, and RS//P Q. (viii) If P, Q, R, and S are distinct collinear points, then at least one of the following holds: P Q//RS, P R//Q S, or P S//Q R. (ix) If P Q//RS and P R//QT , then P Q//ST . Theorem 7.2.2. (i) If P Q//RS, then Q P//RS, Q P//S R, RS//Q P, S R//P Q, and S R//Q P. (ii) If A, B, C, and D are distinct collinear points, then exactly one of the following holds: AB//C D, AC//B D, or AD//BC. Proof. See Exercise 7.2.14. ! Exercise 7.2.2. Interpret line as a Euclidean circle and point as a point on it. Interpret P Q//RS as the point S is not contained in the arc " P R Q, illustrated in Figure 7.14. Verify that axioms (vii), (viii), and (ix) as well as Theorem 7.2.2 hold with these interpretations. Q S E R

P

Figure 7.14 Separation on a circle: P Q//RS.

Theorem 7.2.3. If X p and X q are determined from the definition and Exercise 7.2.1 and p ̸= q, then X p and X q are distinct. Partial Proof. Let a, b, and c be nonnegative integers with a < b < c. Then X a , X b , and X c are determined from the definition. We use induction on the size of c to prove that X X b // X a X c . For the initial case, let k = 1 and c ≤ k + 1. That forces a = 0, b = 1, and c = 2. But H (X X 1 , X 0 X 2 ) by definition and the points of a harmonic set are, by definition, distinct. Further, axiom (vi) gives X X 1 // X 0 X 2 . For the induction step, assume that X X b // X a X c , where 0 ≤ a < b < c ≤ k + 1 holds for the integers k ≥ 1 and let 0 ≤ a < b < c ≤ k + 2. If

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c < k + 2, we are done by the induction hypothesis. Suppose that c = k + 2. Then, by definition, H (X X k+1 , X k X k+2 ). Axiom (vi) gives X X k+1 // X k X k+2 . If a = k, then b = k + 1 and we are done. If 0 ≤ a < k, we have X X k // X k+1 X a by the induction hypothesis. Axiom (ix), with P = X , Q = X k+1 , R = X k , S = X k+2 , and T = X a , gives X X k+1 // X k+2 X a , which Theorem 7.2.2 converts to X X k+1 // X a X k+2 . If b = k + 1, we are done. Exercise 7.2.17 considers the remaining case, a < b < k + 1. Axiom (vii) then forces the points to be distinct. To include the negative integers, we assume that −k ≤ a < b < c and use Exercise 7.2.1 and a similar induction argument on k. The situation with fractions is more complicated, but similar in spirit. In general, we show that H (X X i+ 1 , X i X i+1 ) for any integer i. Then we fill in 2 by induction with the fractions having bigger denominators. ! The final axiom, the continuity axiom, extends the strategy of Theorem 7.2.3 to ensure that a projective line includes all points X r , where r is a real number, together with the additional point X . Figure 7.15 illustrates how to match the points Pi on a circle, except the point P, with the points X i on a Euclidean line. We match P with the ideal point X of a projective line. In effect the final axiom ensures that the points on a projective line are arranged like the points on a circle. Visualizing lines in perspective drawings as circles is difficult because movement can be along the line in two directions, but only one seems to go toward the horizon. Imagine, though, standing near some railroad tracks and looking in both directions: the tracks appear to intersect at the horizon both ways. P

Pc

Pb Pa

Xc

P0 = X0

Xa

Xb

X

Figure 7.15 Matching points on a circle and a line. Theorem 7.2.1 forces each line to intersect the horizon (ideal line) in just one point, so the directions must somehow meet at the same ideal point on the horizon, completing the effect of a circle. The (topological) arrangement of points of the entire projective plane is the same as the points in single elliptic geometry. (See Section 4.5.) Axiom 7.2.3. Continuity Axiom. (x) Given three points X , X 0 , and X 1 on a projective line, there is a one-to-one correspondence between the real numbers r and the points X r on the line except X such that b is between a and c if and only if X a X c // X b X . Section 7.1 described the projective plane in terms of the familiar Euclidean plane together with an added horizon, a line of ideal points. Based on the axioms presented we can now confirm that description. From axiom (ii) we can start with four points O, X , Y , and U , no three of which ← → ← → are collinear. Think of O X and OY in Figure 7.16 as the x- and y-axes and O as the origin for the points of the Euclidean plane. Axiom (x) and the discussion preceding Theorem 7.2.2 ← → ← → ← → enable us to fill out the lines O X and OY , provided we can find three points on each. For O X

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7.2 Axiomatic Projective Geometry

← → ← → ← → ← → ← → we use O = X 0 , O X · Y U = X 1 , and X . Similarly, for OY we use O = Yo , OY · XU = Y1 , ← → ← → and Y . Thus we have all the points X a and Yb on the axes O X and OY . Next we include the points Pab corresponding to the Euclidean plane, using two numbers, just as coordinates in ←→ ←→ analytic geometry. In Figure 7.16, Pab is the intersection of X a Y and X Yb . For a point P not ← → ← → ← → ← → ← → on the line X Y , PY intersects O X at some point X a and P X intersects OY at some point Yb , ← → which gives P = Pab . With this notation, U = P11 . Thus the points not on the line X Y look like ← → ← → the Euclidean plane and X Y is the line of ideal points. However, the points on X Y don’t have natural coordinates in this procedure. We used up all possible pairs (a, b) to label the points ← → Pab not on X Y . In Section 7.3 we present homogeneous coordinates, an elegant solution to the problem, involving the use of three coordinates for each point. Y Yb Pa,b Y1

O

U

X1

Xa

X

Figure 7.16 Coordinates in the projective plane.

7.2.1 Duality Poncelet noticed that, without parallel lines, lines have the same properties as points in projective plane geometry. That is, the words point and line in any axiom or theorem of projective geometry (and any definitions) can be exchanged to get another theorem, called the dual. For example, the dual of “Every line has infinitely many points on it” is “Every point has infinitely many lines on it.” Similarly, “collinear” and “concurrent” are dual concepts, as are “complete quadrangles” and “complete quadrilaterals.” People think about points differently from lines, so the structural similarity was and is still hard to see. However, duality is more than an aesthetically pleasing property; it doubles the number of theorems available, often giving us theorems we might not have imagined. Exercise 7.2.3.∗ State the duals of axioms (i), (ii), and (iii). Theorem 7.2.4. The duals of axioms (i), (ii), and (iii) hold. Proof. See Exercise 7.2.15. ! Next we develop the duals of harmonic sets of points and separation of points. The labeling of Figure 7.17 provides the key to connecting a harmonic set of points H (P Q, RS) and a harmonic set of lines H ( pq, r s). A harmonic set of points H (P Q, RS) and a point O not on their line determine a harmonic set of the lines through O and the points. Conversely, a harmonic set of lines H ( pq, r s) and a line k not on their common point determine a harmonic

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set of points. The following definition uses the separation of points to define the separation of lines. (We assume that the definitions of harmonic sets and the separation of lines are welldefined; that is, different choices of the line k give the same result whether or not pq//r s or H ( pq, r s). See Tuller [14] for more information.) B A C O p P

r k

s

q R

Q

S

Figure 7.17 Harmonic sets of points and lines. Definition. Let p, q, r , and s be concurrent lines on O, k be any line not on O, and P = p · k, Q = q · k, R = r · k, and S = s · k. Define pq//r s if and only if P Q//RS and H ( pq, r s) if and only if H (P Q, RS). Theorem 7.2.5. The duals of axioms (iv)–(x) hold. Proof. We prove the dual of axiom (iv). (See Exercise 7.2.15 for the others, which are similar.) The dual of axiom (iv) states, “Given three distinct lines p, q, and r on a point O, there is a unique line s on O, distinct from p, q, and r , such that H ( pq, r s).” To prove it, start with the three lines p, q, and r on O. Let k be a line not on O. By axiom (iii), k intersects the lines p, q, and r in the points P, Q, and R, respectively. By Theorem 7.2.1, P, Q, and R are distinct. Axiom (iv) holds for k, so there is a unique point S on k distinct from P, Q, and R such that ← → H (P Q, RS). By Theorem 7.2.1 s = O S is distinct from p, q, and r . Then H ( pq, r s) follows from the definition. Finally, note that, as S is unique, s also is unique. ! Once we have the duals of the axioms, the duals of the theorems follow immediately. We could mechanically write the proof of a dual by switching the words point and line and so on throughout the original proof. Exercise 7.2.4.∗ Write the duals of Theorems 7.2.1, 7.2.2, and 7.2.3.

7.2.2 Perspectivities and Projectivities The notion of a perspectivity originates in perspective drawing. A perspectivity maps the points Ui of one line to the points Vi of another line, using a point T such that for each i, T , Ui , and Vi are collinear. Theorem 7.2.6 shows that harmonic sets and the relation of separation are preserved in perspectivities and so in perspective drawings.

7.2 Axiomatic Projective Geometry

333

Theorem 7.2.6. A perspectivity preserves harmonic sets of points and the relation of separation. That is, if a perspectivity from O maps the collinear points P, Q, R, and S to the collinear points P ′ , Q ′ , R ′ , and S ′ and H (P Q, RS), then H (P ′ Q ′ , R ′ S ′ ). Similarly, if P Q//RS, then P ′ Q ′ //R ′ S ′ . Proof. See Exercise 7.2.18. ! Definition. A composition of perspectivities is a projectivity. Exercise 7.2.5.∗ Explain why Theorem 7.2.6 applies to projectivities as well as perspectivities. By Exercise 7.1.5, there is a projectivity that maps any three collinear points to any three collinear points. Theorem 7.2.7 reveals that their images determine the images of all other points on the line. Theorem 7.2.7. Fundamental Theorem of Projective Geometry. A projectivity of a line is completely determined by three points on the original line and their images. Proof. Without loss of generality call the points on the original line X , X 0 , and X 1 and suppose that the projectivity takes them to Y , Y0 , and Y1 , respectively. Theorem 7.2.6 ensures that a point X p described in Theorem 7.2.3 goes to Y p . Exercise 7.2.8 and the continuity axiom extend the process, matching X r with Yr . ! Exercise 7.2.6.∗ Define and illustrate the concept of a line perspectivity taking concurrent lines to concurrent lines. Write the duals of Theorems 7.2.6 and 7.2.7.

7.2.3 Exercises for Section 7.2 *7.2.7. The points X r in Figure 7.13 correspond to Euclidean points on a line with X as the ideal point. (a) What Euclidean concept about the points X a , X b , and X c corresponds to the separation X X a // X b X c ? (b) What Euclidean concept about the points X a , X b , and X c corresponds to the harmonic set H (X X a , X b X c )? (c) Suppose the Euclidean lines j, k, l, and m go through the point P. Describe in Euclidean terms what jk//lm tells us about them. 7.2.8. Let jm be the Euclidean line with equation y = mx and j be the y-axis. All the lines go through the origin. (a) Show that the labeling of lines corresponds to the second coordinates of the points of their intersection with x = 1. *(b) Find the line jm so that H ( j j0 , j1 jm ). Hint: Use part (a) and Exercise 7.2.7(b). (c) Find the line jm so that H ( j j0 , jh jm ), where h is a nonzero slope. (d) Find the line jm so that H ( j0 j1 , ja jm ). Hint: See Exercise 7.1.13.

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*(e) In Euclidean geometry the lines j1 and j−1 are the perpendicular bisectors of the angles formed by jm and j1/m . Is H ( j−1 j1 , jm j1/m )? Hint: Find a linear function taking −1 to 0 and leaving 1 fixed. Generalize. 7.2.9. Art books give various methods of constructing equally spaced poles in a perspective painting. (See Powell [10].) Use Exercise 7.2.7(b) to explain how the following construction blends harmonic sets and Euclidean ideas. Start with the horizon line, the first vertical pole (B1 T1 ), and the base B2 of the next pole, as in Figure 7.18. Extend line b connecting the two bases to the horizon line to find the ideal point H . Draw the line t connecting H with T1 , the top of the first pole. All the poles will have their bases on b and tops on t. Draw the second pole parallel to the first pole. Draw the vertical line v through the ideal point parallel to the poles. Draw the line k1 from T1 , the top of the first pole through B2 , the base of the second pole to point P on v. The line k2 connecting T2 , the top of the second pole, with P intersects b at B3 , the base of the third pole. Continue as in Figure 7.18 to determine the other poles. T1

T2

t

H b

B2

B1

B3 k2

v

k1

P

Figure 7.18 Equally spaced poles in perspective. 7.2.10. Redo Exercise 7.2.9 with the following construction. Also, explain why the line m in Figure 7.19 should go through the middle of the poles, such as M2 and M3 . Start with the horizon line, the first vertical pole (B1 T1 ), and the base B2 of the next pole, as in Figure 7.19. Extend the line b connecting the two bases to the horizon line to find the ideal point H . Draw the line t connecting H with T1 , the top of the first pole. All the poles will have their bases on b and tops on t. Draw the second pole parallel to the T1

T2 X

M2 B2

T3 M3

t m

H b

B3

B1

Figure 7.19 Equally spaced poles in perspective.

7.2 Axiomatic Projective Geometry

335

first pole. Let X be the intersection of T1 B2 and T2 B1 . Draw X H and label M2 its ←−→ intersection with T2 B2 . The line T1 M2 intersects b at B3 , the base of the third pole. ←−→ Similarly, B1 M2 intersects t at T3 . We continue in the same manner for other poles. 7.2.11. Use the construction from Example 1 of Section 7.1 to construct the harmonic set H (P Q, RS). Illustrate axiom (v) by repeating the construction, but starting with two lines through R and one through P. Then draw the two lines through S determined by the construction. Verify that the sixth line goes through Q. 7.2.12. Let r represent the Euclidean point (r, 0). a ). Use similarity to show, for k > 0, *(a) In Exercise 7.1.11, you saw that H (0 1, a 2a−1 ak H (0 k, ak 2a−1 ). 1 1 , n1 n+2 ). *(b) Show that H (0 n+1 (c) Use part (b) to show how to construct poles that will appear equally spaced in a perspective painting.

7.2.13. Describe numbers q such that X q is determined from the definition before Exercise 7.2.1. Give an example of a rational number q so that X q is not determined by the definition. Explain why all positive real numbers r can be written as limits of them. 7.2.14. (a) Prove the rest of Theorem 7.2.1. (b) What other orderings of P, Q, R, and S form harmonic sets, assuming H (P Q, RS)? Prove your answer using the definition of a harmonic set in Section 7.1. 7.2.15. Explore axiom (ix) by letting P, Q, R, S, and T be points on a Euclidean number line with P at 0 and Q at 1. *(a) If R is at r , where 0 < r < 1, and S is at s, where can s be so that P Q//RS? Where can t, the value for T , be so that P R//QT ? Does it follow that P Q//ST ? Explain. Illustrate the placement of the points on a number line. (b) Repeat part (a), where r < 0. (c) Repeat part (a), where 1 < r . 7.2.16. Prove Theorem 7.2.2. Hint: For part (ii) use axiom (ix). 7.2.17.

(a) Prove Theorem 7.2.4. *(b) State the duals of axioms (v)–(x). (c) Prove the duals of axioms (v)–(x).

7.2.18.

(a) In a complete quadrangle with vertices T1 , T2 , T3 , and T4 , the diagonal points are ←→ ←→ ←→ ←→ the three points of intersection of the opposite sides: T1 T2 · T3 T4 , T1 T3 · T2 T4 , and ←→ ←→ T1 T4 · T2 T3 . Draw a picture to illustrate this and prove that the diagonal points aren’t collinear. *(b) Define the dual concept of part (a) and illustrate the dual theorem.

7.2.19. In Theorem 7.2.3, prove the case a < b < k + 1. Hint: Use axiom (vi) with P = X . ← → ←→ ← →← → 7.2.20. Prove Theorem 7.2.6 Hint: Use duality twice, starting with H ( O P O Q, O R O S).

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Projective Geometry

7.2.21. (a) Illustrate Pappus’s theorem. Let ABC D E F be a (self-crossing, nonconvex) hexagon so that A, C, and E are on a line and B, D, and F are on a different line. Then the three intersections of the extensions of the opposite sides are ← → ← → collinear. That is, let X be the intersection of AB and D E. Similarly, Y is the ← → ← → ← → ← → intersection of BC and E F and Z is the intersection of C D and F A. Then X , Y , and Z are on the same line. See Coxeter [2, 36-39] for a projective proof of the theorem based on projectivities. Pappus (circa 290–350) gave a complicated Euclidean proof of the theorem, provided none of the opposite sides were parallel. (b) Illustrate Pascal’s theorem. Let ABC D E F be a hexagon so that all six vertices are on a circle (or any conic). Then the three intersections of the extensions of ← → ← → the opposite sides are collinear. That is, let X be the intersection of AB and D E. ← → ← → ← → Similarly, Y is the intersection of BC and E F and Z is the intersection of C D and ← → F A. Then X , Y , and Z are on the same line. See Coxeter [2,85] for a proof. (c) State and illustrate the dual of Pappus’s theorem. (d) State and illustrate the dual of Pascal’s theorem, which is called Briachon’s theorem. Assume that the dual of a point on a conic is a tangent to a conic. Charles Julien Brianchon (1783–1864) proved this theorem while he was a student. See Coxeter [2, 83] for a proof. Two lines can be considered a type of degenerate conic. (See Section 3.2.) Thus Pappus’s theorem is a special case of Pascal’s theorem.

7.2.4 Jean Victor Poncelet Jean Victor Poncelet (1788–1867) grew up in France during revolutionary times. He studied at ´ the Ecole Polytechique under the influence of the legendary Monge. He then became an engineer and officer in Napoleon’s army in the ill-fated campaign against Russia. He spent a year during 1813 and 1814 in a Russian prison. In prison he had the opportunity to reflect and write. He reconstructed his geometric education from memory and went on to discover many new results. Poncelet worked for various French governments after the fall of Napoleon and occasionally taught. Projective geometry became a separate subject and moved to prominence with the publication in 1822 of Poncelet’s treatise on the subject, a revised and expanded version of his prison writings. Poncelet was an outspoken advocate of the synthetic approach, following the dictum of Lazare Carnot: “. . . to free geometry from the hieroglyphics of analysis.” He realized the power of the general analytic approach compared to the isolated proofs of classical geometry. However, he felt that analytic geometry gave answers without giving insight. He developed projective geometry to provide general methods within the synthetic tradition. In the process he rediscovered many of the properties found previously by Desargues and others but then forgotten. (Desargues’ work was found noticeably after Poncelet published his book.) He looked for properties of figures preserved by perspectivities. He was the first to realize the importance of duality. The principle of continuity, which he used implicitly as an axiom, provided a powerful method for generalizing many results. Thus a shape can pass continuously from circles to ellipses to parabolas to hyperbolas, much as the shadow of a lamp shade on a wall does as the lamp is tilted. The projective properties of tangents and other objects transfer along with the changing curves. For example, an asymptote of a hyperbola becomes a special type of tangent.

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7.3 Analytic Projective Geometry

He even extended the principle of continuity to explore imaginary intersections of lines and conics that don’t intersect in real points. Thus Poncelet initiated the study of complex projective geometry, although without using coordinates.

7.3 Analytic Projective Geometry We develop homogeneous coordinates to represent analytically all the points and lines in projective geometry and to emphasize their duality. Homogeneous coordinates enable us to consider projective transformations in Section 7.4. In Section 7.2 we saw that two coordinates aren’t sufficient to describe all points in the projective plane. In Chapter 4 we used three coordinates (x, y, 1) for points, enabling transformations to move all points. In the process we used row vectors [a, b, c] to represent lines, where (x, y, 1) is on [a, b, c] provided that ax + by + c · 1 = 0. Also, nonzero multiples [λa, λb, λc] of [a, b, c] represent the same line. The duality of points and lines in projective geometry suggests using triples (x, y, z) for the enlarged set of projective points. As in Chapter 4 (x, y, z) represents a column vector. Examples 1, 2, and 3 provide three ways to view them. Example 3 is the most important but most abstract representation. We leave it as an exercise to verify that the following interpretation satisfies the first three axioms of Section 7.2. Interpretation. By a point in the projective plane, we mean a nonzero column vector (x, y, z), where (x, y, z) and (λx, λy, λz) represent the same point whenever λ ̸= 0. By a line in the projective plane, we mean a nonzero row vector [a, b, c], where [a, b, c] and [λa, λb, λc] represent the same line for λ ̸= 0. A point (x, y, z) is on the line [a, b, c] if and only if ax + by + cz = 0. Exercise 7.3.1.∗ Find the point on the lines [2, −1, 3] and [4, −2, 5]. Find the line on the points (1, 3, 0) and (2, 3, 0). Example 1. We can think of an ordinary Euclidean point (x, y) as the projective point (x, y, 1) or in general (λx, λy, λ). For this example we ignore the scalar for simplicity. Figure 7.20 illustrates several projective points and lines, with the grid suggesting the Euclidean plane within the projective plane. Ideal points have 0 for their third coordinate. For example, the x-axis, y = 0, or [0, 1, 0], has the ideal point (1, 0, 0) and the y-axis, x = 0, or [1, 0, 0], has (0, 1, 0) on it. The Euclidean points on the line y = x or [1, −1, 0] are of the form (x, x, 1). (0,1,0) x = 0, [1,0,0]

(1,1,0) y = x, [1,-1,0] (2,1,1) (1,0,0)

(0,0,1)

y = 0, [0,1,0]

Figure 7.20 The projective plane.

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Projective Geometry

The ideal point on this line is (1, 1, 0), which is on [1, −1, 0] and also on all the Euclidean lines parallel to it, y = x + c, which are of the form [1, −1, c]. However, there is no reason in projective geometry to single out any line or point as ideal or different from any other. The beauty of homogeneous coordinates is that they make all points and lines equivalent. ♦ Example 2. We can relate each point of the projective plane to two antipodal points on a sphere. For each projective point (x, y, z), there are scalars λ and −λ such that (λx, λy, λz) and (−λx, −λy, −λz) have length 1 and so are on the sphere. Projective lines correspond to great circles. The great circle corresponding to [a, b, c] is in the plane perpendicular to the point (a, b, c). Figure 7.21 illustrates this initially nonintuitive fact and Example 3 provides an explanation. The identification of antipodal points on the sphere gives single elliptic geometry. (See Section 4.5.) The points and lines of single elliptic geometry match exactly the points and lines of projective geometry. Single elliptic geometry has distance and angle measure in addition to the projective notions. ♦ (a,b,c)

(-a,-b,-c)

[a,b,c]

Figure 7.21 Example 3. Three-dimensional linear algebra provides a powerful language for projective geometry. A projective point is a one-dimensional subspace (a Euclidean line through the origin). The choice of the scalar λ determines which vector along the line we are considering. A projective line is a two-dimensional subspace (a Euclidean plane through the origin). Recall that [a, b, c] contains points (x, y, z), with ax + by + cz = 0 or [a, b, c] · (x, y, z) = 0. The dot product of two vectors is 0 whenever the vectors are perpendicular. So [a, b, c] is the Euclidean plane through the origin containing all the points (x, y, z) perpendicular to the vector (a, b, c). Linear transformations take subspaces to subspaces, so we can use them to represent projective transformations. ♦ Exercise 7.3.2.∗ Use Example 3 to verify that the interpretation satisfies Axioms (i), (ii), and (iii) of a projective plane from Section 7.2. Example 4. We claim that5distinct points ( p, q, r ), (s, t, u), and (v, w, x) are collinear if and 6 p s v only if the determinant of q t w is 0. r

u

x

Solution. The points are on the line [a, b, c] if and only if the products [a, b, c]( p, q, r ), 5p s v6 − → [a, b, c](s, t, u), and [a, b, c](v, w, x) are 0. In matrix notation, [a b c] q t w = 0 r

u

x

represents a system of three homogeneous equations. From linear algebra, it has a nonzero solution [a, b, c] (that is, the points are collinear) if and only if the matrix is singular and its determinant is 0. ♦

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7.3 Analytic Projective Geometry

To relate the axioms of Section 7.2 with our model we next turn to representing harmonic sets and separation analytically. Since they focus on points of one line, we need homogeneous coordinates for the points on a line. We represent the points on a line as (u, v), where (u, v) = (λu, λ v), for λ ̸= 0. Example 5. The points on [1, −1, 2] are of the form (x, y, z), where x − y + 2z = 0 or y = x + 2z. The last equation enables us to eliminate the y-value, so the two coordinates (x, z) are sufficient to describe which point on this line we are considering. From the Euclidean point of view, the point (x, 1) = (λx, λ) corresponds to the real number x and (1, 0) = (λ, 0) corresponds to the ideal point. By solving the equation y = x + 2z for x or z, we could eliminate either of the variables. The methods of determining two homogeneous coordinates (and others) are compatible; they correspond in linear algebra to changes of coordinates. ♦

7.3.1 Cross Ratios The ancient Greeks initially explored the concept of the cross ratio of four collinear points for Euclidean geometry. Nineteenth century geometers realized the cross ratio provided the analytic key for both harmonic sets and separation, although it took some time to strip the cross ratio of its Euclidean meaning. To simplify the general formula of the cross ratio, we write | A| for the determinant of the matrix A. We use the notationally easy form of the cross ratio given in Example 6 whenever possible. Theorem 7.3.1 shows that our analytic interpretation matches axiom (x), the most powerful separation axiom. Exercises 7.3.16 through 7.3.19 consider the remaining axioms from Section 7.2. Definition. The cross ratio of four distinct collinear points P and W = (w, x) is 0 0 0 0 p u0 0s 0 0 0 0q v 0 0t R(P, S, U, W ) = 0 0÷0 0 p w0 0s 0 0 0 0q x 0 0t

= ( p, q), S = (s, t), U = (u, v), 0 u 00 v0 0. w 00 x0

If q = t = v = x = 1, this formula reduces to the special case of Example 6. p−u ÷ Example 6. For P = ( p, 1), S = (s, 1), U = (u, 1), and W = (w, 1), R(P, S, U, W ) = p−w s−u . When we can replace homogeneous coordinates by cartesian coordinates, we’ll write s−w −3 −2 ÷ 2/5 = −1. In general, from Exercise 7.1.11, R( p, s, u, w). For example, R(0, 1, 3, 35 ) = −3/5 x 1 x )= H (0 1, x 2x−1 ), provided x ̸= 0, 2 , or 1. Using the cross ratio we have R(0, 1, x, 2x−1 0−x 1−x ÷ 1−(x/(2x−1)) , which simplifies to −1 since none of the denominators is 0. ♦ 0−(x/(2x−1))

Interpretation. For collinear points P, S, U , and W , H (P S, U W ) if and only if R(P, S, U, W ) = −1 and P S//U W if and only if R(P, S, U, W ) < 0. Cross ratios, harmonic sets, and separation for lines through a point are given dually. Exercise 7.3.3.∗ Suppose that four real numbers satisfy a < c < b < d. Use Example 6 to verify that R(a, b, c, d) < 0 and so X a X b // X c X d .

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Example 7. Let P = (1, 1), S = (2, 1), U = (3, 1), and W = (w, 1). Investigate what happens to R( p, s, u, w) as w varies. 1−3 2−3 ÷ 2−w = 4−2w . Figure 7.22 gives the graph of f (w) = 4−2w . Solution. R( p, s, u, w) = 1−w 1−w 1−w From it we see that 1 and 2 separate 3 and w if and only if 1 < w < 2. Also, as w → ± ∞, the cross ratio tends to 2. While the cross ratio is undefined when there aren’t four distinct points, the formula gives R(1, 2, 3, 3) = 1 and R(1, 2, 3, 2) = 0. The formula is undefined for R(1, 2, 3, 1) since R(1, 2, 3, w) goes to ±∞ as w goes to 1. ♦

5 4 3 2 1 –2

–1

–1

2

Figure 7.22 f (w) =

3

4

5

4−2w . 1−w

Theorem 7.3.1. Axiom (x) holds in the analytic projective plane. Proof. First, when λ ̸= 0, match (aλ,λ ) with X a and (λ, 0) with X . This accounts exactly for all the points in the axiom. Let (s, t) be a point in our model. If t = 0, then s ̸= 0 and so this matches with X . Otherwise, t ̸= 0 and (s, t) is the same point in our model as ( st , 1), which matches with X s/t . Hence we have the one-to-one correspondence of the axiom. Now let a < b < c be real numbers. Then the homogeneous coordinates of the four points of the axiom are X a = (a, 1), X b = (b, 1), X c = (c, 1), and X = (1, 0). Then 0 0 0 0 0a b0 0c b0 0 0 0 0 01 10 01 10 a−b a−b c−b ÷ = . R(X a , X c , X b , X ) = 0 0÷0 0 = 0a 10 0c 10 −1 −1 c−b 0 0 0 0 01 00 01 00 Since a < b < c, the fraction is negative and so X a X c // X b X , as required. !

7.3.2 Conics Projective geometry provides a unified way to study conics. Figure 3.7 illustrates the conics as cross sections of a cone. We can view the apex of the cone as the point of perspective and so the cross sections are perspective views of one another. That is, projective geometry doesn’t distinguish between circles, ellipses, parabolas, and hyperbolas. Section 3.2 gave the general equation of a conic in cartesian coordinates: ax 2 + 2bx y + 2 cy + 2d x + 2ey + f = 0. In homogeneous coordinates it becomes ax 2 + 2bx y + cy 2 + 2d x z + 2eyz + f z 2 = 0. Now every term has second degree in the variables x, y, and z, so we say that it is a homogeneous second-degree equation. (Homogeneous first-degree equations,

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ax + by + cz = 0, represent lines.) Because a conic has a second-degree equation, it can intersect a line in zero, one, or two points. Linear algebra makes the homogeneous equation even more useful, as Exercise 7.3.4 illustrates. The matrix form explains why we choose to have the factors of 2 in the general equation. Not every such matrix gives a conic, as Exercise 7.3.5 illustrates. Quadratic forms in linear algebra are closely tied to conics and their generalizations. (See Fraleigh and Beauregard [5, Chapter 7].) Exercise 7.3.4. Show that a point P = (x, y, z) is on the conic with equation ax 2 + 2bx y + cy 2 + 2d x z + 2eyz + f z 2 = 0 if and only if P T C P = 0, where P T is the transpose of P and ⎡ ⎤ a b d C = ⎣b c e ⎦. d e f Exercise 7.3.5. Verify that ⎡

2 −3/2 C = ⎣ −3/2 1 1 −1/2

⎤ 1 −1/2 ⎦ 0

has a determinant of 0 and represents the product of the two lines 2x − y = 0 and x − y + z = 0. As in Section 3.2, when the determinant is 0 we call the conic degenerate. The interpretation below excludes degenerate conics. Interpretation. By a conic we mean a symmetric invertible 3 × 3 matrix. Two matrices represent the same conic if and only if one is the multiple of the other by a real number λ ̸= 0. A point P is on a conic C if and only if P T C P = 0. Although projective geometry doesn’t distinguish different types of conics, the approach of Example 1 enables us to discuss them. If, as in Figure 7.23, we designate one line k as an ideal line, hyperbolas intersect it in two points, parabolas intersect it in one point, and circles and ellipses do not intersect it. (To distinguish circles from ellipses requires measures of angles or distances, which are not projective concepts.) We define a tangent to a conic to be a line with only one point on it. Exercise 7.3.6 explores this definition of tangent, which doesn’t work with Euclidean parabolas and hyperbolas. The ideal line of Figure 7.23 is tangent to the (Euclidean) parabola. From a projective point of view each Euclidean asymptote to a hyperbola intersects it at an ideal point. Thus the asymptotes become tangents in projective geometry since we don’t distinguish an ideal line.

ideal line hyperbola parabola circle or ellipse

Figure 7.23 Conics in relation to the ideal line.

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Exercise 7.3.6.∗ Verify x 2 − yz = 0 and x y − z 2 = 0 are the homogeneous equations for the conics y = x 2 and y = x1 . Explain why in Euclidean geometry a vertical line x = b, with b ̸= 0, intersects the curves in exactly one point. Find the two intersections in projective geometry of the line [1, 0, −b] for b ̸= 0 with the conics. Find the intersection of the line [1, 0, 0] and x y − z 2 = 0. What are the coordinates of the other Euclidean asymptote of x y − z 2 = 0? What is its intersection with x y − z 2 = 0? Two distinct Euclidean circles intersect in at most two points. However, as Exercise 7.3.7 illustrates, two Euclidean conics and so projective conics can intersect in more than two points. Five points, no three collinear, are required to completely determine a conic. Exercise 7.3.7.∗ Graph y = x 2 and x 2 + (y − 2)2 = 2 and find their intersections.

7.3.3 Exercises for Section 7.3 7.3.8. *(a) Find the point of intersection of the lines [m, −1, b] and [m, −1, c], where b ̸= c. Interpret this in Euclidean geometry. *(b) Find the line through the points ( p, q, 1) and (1, m, 0). Interpret this situation in Euclidean geometry. (c) Repeat part (a) for [m, −1, b] and [0, 0, 1]. (d) Verify that the line on the points ( p, q, r ) and (s, t, u) is given by 30 0q 0 0r

0 0 0p t 00 , − 00 0 u r

0 0 s 00 00 p , u0 0q

04 s 00 , t0

0 v y0 0 where 0 w z is the determinant of a 2 × 2 matrix. Use the formula to verify your answer in part (b). (This formula for a line is the transpose of the cross product of ( p, q, r ) and (s, t, u).) (e) Give a formula for the intersection of the lines [a, b, c] and [d, e, f ]. Use your formula to verify your answer in part (a). Justify your formula. 7.3.9. We think of 2x + 3y + z = 0 as the equation of the line with homogeneous coordinates [2, 3, 1]. By duality, 2a + 3b + c = 0 would be the “equation of a point.” Explain what this equation tells us about the point (2, 3, 1). *7.3.10. Let A = (1, 1), B = (2, 3), C = (2, 1), D = (3, 1), and E = (5, 3). (a) Place the points on a number line. (b) Which pairs of them separate other pairs? (If P Q//RS, then the pair {P, Q} separates the pair {R, S} and vice versa.) Use the cross ratio to verify your answer. (c) Determine all harmonic sets choosing four of the points. Show your work. (d) Give a matching of A, B, C, D, and E with the points of axiom (ix) that satisfies its hypotheses. Verify that the matching also satisfies the conclusion. Use the cross ratio to verify the hypotheses and conclusion. 7.3.11. Follow Van Staudt’s coordinatization of Section 7.2, using O = (0, 0, 1), X = (1, 0, 0), Y = (0, 1, 0), and U = (1, 1, 1). Find homogeneous coordinates for X a , Yb , and Pab . ←−→ Find homogeneous coordinates for the intersection of [0, 0, 1] and O Pab .

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343

7.3.12 Let P = (1, 1), Q = (2, 1), and R = (4, 1). *(a) Find conditions on x so that X = (x, 1) and P Q//R X . Find x so that H (P Q, R X ). (b) Repeat part (a) for y, where Y = (y, 1), PY //Q R, and H (PY, Q R). (c) Repeat part (a) for z, where Z = (z, 1), Q Z //P R, and H (Q Z , P R). (d) On a number line shade in the possible range of values for x, for y, and for z where we get the separation relation, given your answers in parts (a), (b), and (c). Also indicate the points where we get the harmonic sets. (e) Which of the separation statements P Q//R X , PY //Q R, and Q Z //P R remains true if we replace X , Y, or Z , respectively, by (1, 0)? Justify and explain your answer. 7.3.13. Let A = (1, 0), B = (0, 1), C = (1, 1), and D = (r, 1). (a) Verify that R( A, B, C, D) = r . (b) There are 24 orderings of the four points A, B, C, and D, but there are only six different cross ratios, such as R(B, A, C, D) compared with R( A, B, C, D) = r . Use the coordinates above to find the five other cross ratios. List those orderings with a cross ratio of r . How are they related? (c) Explore whether, for four distinct collinear points, the relationship of the six different cross ratios of part (b) holds. (d) Explore what values the cross ratio R( A, B, C, D) can have if we extend the definition for three distinct points, with one of them repeated. 7.3.14. *(a) Use Example 1 to plot the points (0, 0, 1), (6, 0, 1), (3, 3, 1), and (8, 4, 1). Verify that no three of them are collinear. Find the diagonal points of the complete quadrangle they form both graphically and analytically to verify that they are not collinear. (See Exercise 7.2.18.) *(b) Pick two of the diagonal points of part (a), say A and B, and find the coordinates of the line k on them. Find the intersections C and D of k with the other sides of the complete quadrangle. Explain why the points A, B, C, and D form a harmonic set. Use the cross ratio to verify that they do. (c) Do the dual of parts (a) and (b) with the lines [0, 1, 0], [1, 1, 1], [1, −1, 0], and [2, −1, −2]. 7.3.15. *(a) Use Example 1 and the following points to illustrate Desargues’ Theorem from Section 7.1. Let P = (0, 0, 1) be the point of perspective, △ABC have vertices A = (1, 0, 1), B = (0, 1, 1), and C = (1, 1, 1), and △D E F have vertices D = (4, 0, 1), E = (0, 6, 1), and F = (2, 2, 1). Plot the points and find the line of perspective. Find analytically the coordinates of the line of perspective and the three points S, T, and U on it. (b) Use the ten points in part (a) to illustrate Desargues’ Theorem but with E as the new point of perspective. That is, find the two triangles that are in perspective from E and find the line through the remaining three points that is the line of perspective for them. (c) Repeat part (b) using F as the point of perspective.

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7.3.16. (a) Prove that axiom (vi) holds with the analytic interpretations of this section. (b) Use the definition of the cross ratio to prove for four distinct collinear points that R(P, S, U, W ) = R(U, W, P, S). (c) Prove that axiom (v) holds with the analytic interpretations of this section. (d) Use the definition of the cross ratio to determine for four distinct collinear points the relationship between R(P, S, U, W ) and R(P, S, W, U ). (e) Use part (d) to prove that if P S//U W , then P S//W U and if H (P S, U W ), then H (P S, W U ). (f) Prove that axiom (vii) holds with the analytic interpretations of this section. 7.3.17. The following parts prove that axiom (iv) holds with the analytic interpretations of this x section. From Example 6 if x differs form 0, 12 , and 1, then R(0, 1, x, 2x−1 ) = −1 and x so H (0 1, x 2x−1 ). (a) For x different from 0, 12 , and 1, show that R(0, 1, x, y) = −1 if and only if x . This shows axiom (iv) for the special case P = 0, Q = 1, and R = x. y = 2x−1 (b) For any number z and distinct numbers p, s, u, and w prove that R( p + z, s + z, u + z, w + z) = R( p, s, u, w). Explain why, if P, Q, and R are not (1, 0), we may now assume that P = 0. (c) For a number z and distinct numbers p, s, u, and w prove that R( pz, sz, uz, wz) = R( p, s, u, w). Explain why, if P, Q, and R are not (1, 0), we may now assume that P = 0 and Q = 1. (d) If R is the Euclidean midpoint of P Q, prove that S = (1, 0) makes H (P Q, RS) and the choice of S is unique. Why do parts (a) through (d) prove axiom (iv) when none of P, Q, and R is the point (1, 0)? (e) Show how to use part (d) to prove the existence and uniqueness of a harmonic set when one of the points P, Q, or R is (1, 0). 7.3.18. The following parts prove that axiom (viii) holds with the analytic interpretations of this section. Let P, Q, R, and S be distinct points on the line k. In this model k has infinitely many points, so we can choose a fifth point on k to fulfill the role of X in axiom (x). So we may assume that P = ( p, 1), Q = (q, 1), R = (r, 1), and S = (s, 1). (a) Then the coordinates p, q, r, and s have some ordering, say, a < b < c < d, where a, b, c, and d are p, q, r, and s in some order. Prove that ac//bd using the cross ratio. (b) Why are there then 24 possible ways to match the coordinates p, q, r, and s with a, b, c, and d? Use axiom (vii) and Theorem 7.2.2 to show that each matching, together with ac//bd, yields one of the three options P Q//RS, P R//Q S, or P S//Q R. 7.3.19. The following parts prove that axiom (ix) holds with the analytic interpretations of this section. Let P, Q, R, S, and T be distinct points on the line k. In this model k has infinitely many points, so we can choose a sixth point on k to fulfill the role of X in axiom (x). So we may assume that P = ( p, 1), Q = (q, 1), R = (r, 1), S = (s, 1), and T = (t, 1). From Exercise 7.3.17 parts (a) and (b), explain why we may assume that P = (0, 1) and Q = (1, 1).

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345

(a) Suppose A = (a, 1), B = (b, 1), C = (c, 1), and D = (d, 1) are collinear points. If a < b < c, prove that AB//C D if and only if a < d < b. Find and prove similar conditions on d so that AC//B D and so that BC//AD. For axiom (ix), assume that P Q//RS and P R//QT . (b) Case 1. Suppose that 1 < r . Use part (a) to give conditions on s and t. Use part (a) again to prove that P Q//ST . (c) Repeat the reasoning of part (b) for Case 2, where r < 0. (d) Repeat the reasoning of part (b) for Case 3, where 0 < r < 1. 7.3.20. *(a) In ordinary analytic geometry y = x/(x + 1) is a hyperbola with asymptotes y = 1 and x = −1. Convert the equations to homogeneous equations. Find the ideal point of each line and verify that it is the intersection of the line and the conic. (b) Find the asymptotes of x 2 − y 2 = 1 and repeat part (a). (c) Convert the equation of the parabola y = x 2 − x to homogeneous form and verify that the ideal line [0, 0, 1] has one point of intersection with it. 7.3.21. Explain why a point on x 2 + y 2 − z 2 = 0 (the unit circle) is of the form (cos α, sin α, 1). (Why can you always choose z = 1?) Use analytic geometry to find the homogeneous coordinates of the line tangent to the circle at this point. Hint: The coordinates of the tangents and their points of tangency have a remarkable property. (The set of tangents forms a line conic, the dual of an ordinary conic, which can be called a point conic.) 7.3.22.

(a) Show that ax 2 + cy 2 + f z 2 = 0 is the equation for conics through the four points (1, 1, 1), (1, −1, 1), (−1, 1, 1), and (−1, −1, 1), where a + c + f = 0. *(b) Use Example 1 to graph the three degenerate conics through the four points and find their equations. (Each is a pair of lines.) *(c) Graph the Euclidean conic if a = 1, c = 1, and f = −2. Hint: replace z by 1. (d) Repeat part (c) for a = 4, c = 1, and f = −5. *(e) Repeat part (c) for a = 2, c = −1, and f = −1. (f) Repeat part (c) for a = −1, c = 2, and f = 1. (g) Explain why every point ( p, q, r ), except the four given points, is on just one of the conics in the family of part (a), including the degenerate ones of part (b). (h) What types of Euclidean conics are in the family of part (a)?

7.3.23. (Calculus) *(a) Write the equation y = x 2 in homogeneous coordinates. *(b) Show that the lines [a, b, c] tangent to the equation in part (a) satisfy the equation of the line conic a 2 − 4bc = 0 as follows. First, pick x0 , a real number, and find the coordinates of the point (x0 , y0 , 1) on the conic. Next use calculus to find the equation of the tangent line. Now convert the line to the form [a, b, c] and verify that it satisfies the equation a 2 − 4bc = 0. (c) Graph y = x 2 and selected tangents.

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7.3.24. (Calculus) Repeat Exercise 7.3.23 for the hyperbola y = x1 and the corresponding line conic 4ab − c2 . Explain the similarities in the problems from a projective viewpoint. 7.3.25. Suppose that P is a point on a nondegenerate conic C. We claim that the line l = P T C is tangent to C at P. (a) Verify the claim by using the matrix form for conics in Exercises 7.3.23 and 7.3.24. (b) Prove in general that P is on the line l of the claim. (c) Suppose that P and Q are two points on the conic C and that l and k are their corresponding lines in the claim. Show that l = k if and only if P = Q. Explain why the equivalence shows that the lines in the claim intersect the conic in only one point. Write a proof of the claim. (d) Show that all tangents l satisfy the equation of the line conic lC −1l T = 0.

7.3.4 Julius Pl¨ucker While synthetic geometry enjoyed a resurgence in the first half of the 1800s, Julius Pl¨ucker (1801–1868) strongly advocated the use of analytic methods. He is best remembered for his development of projective geometry independent of Euclidean geometry. He developed, along with M¨obius, what we now call homogeneous coordinates in a book he published at age 30. Shortly afterwards he became a lower level professor at the University of Berlin, then the most prestigious university in what is now Germany. However, at about the same time Jacob Steiner (1796–1863) was appointed to the chair of mathematics there and Steiner was a leading advocate for the synthetic approach. Their personalities clashed as well as their approaches to geometry and they soon became scientific rivals. Pl¨ucker, as the junior faculty member, decided he needed to leave, ending up at the University of Bonn. In addition to the familiar idea of coordinate systems of points, Pl¨ucker devised coordinate systems for the lines in three-dimensional space. He then proved properties based on them. As an analogous example, we can describe a Euclidean circle with center (a, b) and radius r by the coordinates (a, b, r ). In an ironic twist, in 1835 Pl¨ucker was the first to develop and investigate what are now called Steiner triple systems. (They are examples of balanced incomplete block designs, discussed in Section 8.3.) In addition to his important geometry research, Pl¨ucker did research in physics and became a professor of physics as well as mathematics. Among other areas, he investigated the spectral lines different chemicals emit when heated. Near the end of Pl¨ucker’s life, Felix Klein became his assistant when only age 17. Klein’s strong grounding in geometrical and physical intuition owes much to Pl¨ucker’s influence.

7.4 Projective Transformations The usefulness of homogeneous coordinates becomes apparent in terms of transformations. A transformation is a one-to-one function from a space onto itself and that points are column vectors and lines are row vectors. Thus projectivities (transformations of a line to itself) and collineations (transformations of all the points of the projective plane) correspond to invertible

7.4 Projective Transformations

347

matrices. If we use two homogeneous coordinates for points on a line, a projectivity maps points (u, v) to points (s, t) and can be represented as an invertible 2 × 2 matrix. Interpretation. A projectivity is represented by an invertible 2 × 2 matrix. Two such matrices that differ by a nonzero constant represent the same projectivity. In linear algebra a 2 × 2 matrix is determined by where it send two points, most commonly (1, 0) and (0, 1). Exercise 7.1.5 shows that a projectivity can map three collinear points to any three collinear points. Theorem 7.4.1 shows the surprising fact that 2 × 2 matrices can mimic the extra flexibility projectivities require. Theorem 7.4.1. For distinct collinear points (u 1 , v 1 ), (u 2 , v 2 ), and (u 3 , v 3 ) and distinct collinear points (s1 , t1 ), (s2 , t2 ), and (s3 , t3 ), there is a unique projectivity mapping (u i , v i ) to (si , ti ). Proof. First we show that a projectivity can map the collinear points (1, 1 0), (0, 2 1), and (1, 1) to any distinct collinear points (s1 , t1 ), (s2 , t2 ), and (s3 , t3 ). The matrix st 1 st 2 maps (1, 0) to 1 2 (s1 , t1 ) and (0, 1) to (s2 , t2 ). If we multiply the first column of the matrix by a scalar λ, (1, 0) the second column by µ maps (0, 1) still goes to (s1 , t1 ) = (λs1 , λt1 ) and similarly 2 1 λs µsmultiplying 1 2 to (s2 , t2 ) = (µs2 , µt2 ). For the matrix λt µt the image of (1, 1) is (λs1 + µs2 , λt1 + µt2 ), 1 2 which we set equal 0 s tos(s03 , t3 ) and solve for λ and µ. Because (s2 , t2 ) is not a multiple of (s1 , t1 ) the determinant 0 t 1 t 2 0 ̸= 0. Thus there is a solution. In fact, the solutions form a family that 1 2 are scalar multiples of one another. Thus a unique projectivity M takes (1, 0), (0, 1), and (1, 1) to distinct points (s1 , t1 ), (s2 , t2 ), and (s3 , t3 ). Similarly, a unique projectivity K takes (1, 0), (0, 1), and (1, 1) to (u 1 , v 1 ), (u 2 , v 2 ), and (u 3 , v 3 ). Then the projectivity represented by M K −1 uniquely satisfies the theorem’s conditions. ! Theorem 7.4.2. Cross ratios, harmonic sets, and separation are preserved under projectivities. 2 1 Proof. Let X i = (u i , v i ) for i = 1, 2, 3, 4 be collinear points and M = ab dc be a projectivity. We first show the cross ratios R(X 1 , X 2 , X 3 , X 4 ) and R(M X 1 , M X 2 , M X 3 , M X 4 ˙) to be equal. A 0cross ratio 0 involves four determinants of the coordinates of the points.0 For example, 0 0 u 1 u 3 0 and the determinant of the images M(u 1 , v 1 ) and M(u 3 , v 3 ): 0 Mu 1 Mu 3 0 = consider Mv 1 Mv 3 0 u u v10 v3 |M|0 v 1 v 3 0. The extra factor of |M| appears in each of the determinants in the cross ratio of 1 3 the images. The two factors of |M| in the numerator cancel the two in the denominator. Thus R(X 1 , X 2 , X 3 , X 4 ) = R(M X 1 , M X 2 , M X 3 , M X 4 ). Harmonic sets and separation are defined in terms of the cross ratio, so they also are preserved under projectivities. ! Interpretation. A collineation of the projective plane is represented by an invertible 3 × 3 matrix. Two matrices represent the same collineation if and only if one is a nonzero scalar multiple of the other. A point P is fixed by a collineation α if and only if α(P) = P. A line k is stable under a collineation α if and only if every point on k is mapped by α to a point on k.

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All the affine transformations from Chapter 5, including isometries and similarities, are collineations because affine transformations have invertible matrices of the form ⎡ ⎤ a b c ⎣d e f ⎦. 0 0 1

Theorem 7.4.3 describes how to find the images of lines and conics under collineations. Although points, lines, conics, and collineations have multiple representations, for simplicity we use one representation in the examples. The proof of Theorem 7.4.2 holds for collineations. Thus collineations preserve cross ratios, harmonic sets, and separation, even when the collineation maps one line to another one. Theorem 7.4.3. The image of the line [a, b, c] under the collineation M is [a, b, c]M −1 . The image of the conic C under M is M −1T C M −1 . Proof. See Exercise 7.4.5. ! Theorem 7.4.4. The set of projectivities of a line to itself forms a group of transformations. The set of collineations forms a group of transformations. Proof. See Exercise 7.4.6. ! Example 1. Consider the Euclidean translation ⎡ 1 0 ⎣ T = 0 1 0 0

⎤ 3 −2 ⎦ . 1

In Chapter 5 we saw that a translation has no fixed affine points (x, y, 1). However, we now have more points (x, y, z). Because λ(x, y, z) represents the same point as (x, y, z), we need to consider the general eigenvector problem T (x, y, z) = λ(x, y, z). Eigenvectors for a nonzero eigenvalue represent fixed points. We obtain three equations: x + 3z = λx, y − 2z = λy, and z = λz. The last equation forces λ = 1 or z = 0. Then the first two equations force both λ = 1 and z = 0. Thus every ideal point (x, y, 0) is fixed by a translation. Ideal points are where Euclidean parallel lines meet and that translations take a line to a line parallel to itself. The affine stable lines of a translation, as we saw, are parallel—in this case [− 23 , −1, c]. As a collineation, there is one more stable line: [0, 0, 1], the ideal line z = 0. All the stable lines can a, −a, c] and go through the point (3, −2, 0). ♦ be written in the form [ −2 3 Example 2. The ideal line [0, 0, 1]1is stable for 2 all similarities and affine transformations, for the bottom row of their inverses is 0 0 1 . Consider ⎡ ⎤ 0 2 3 A = ⎣2 0 0⎦, 0 0 1

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7.4 Projective Transformations

which reflects over the line y = x − z ([1, −1, −1]) and expands by a factor of 2 around the fixed point (−1, −2, 1). There are two other fixed projective points, which we find by solving the eigenvector problem A(x, y, z) = λ(x, y, z). There are three values of λ such that the determinant of A − λI is zero: 1, 2, and −2. The fixed points are (−1, −2, 1), (1, 1, 0), and (1, −1, 0), respectively. To find the stable lines we need to solve the eigenvector problem [a, b, c] A−1 = λ[a, b, c]. Verify that ⎡ ⎤ 0 0.5 0 A−1 = ⎣ 0.5 0 −1.5 ⎦ 0

0

1

and that the eigenvalues are 1, 0.5, and −0.5, the multiplicative inverses of the eigenvalues of A. The stable lines are [0, 0, 1], [−1, −1, −3], and [1, −1, −1]. Verify that they are stable and intersect in the fixed points. ♦

Because collineations are 3 × 3 matrices, we can send the points X = (1, 0, 0), Y = (0, 1, 0), and Z = (0, 0, 1) to any three noncollinear points. However, as in Theorem 7.4.1, we can do even more. We can send X , Y , Z , and U = (1, 1, 1) to any four points, provided no three of them are collinear. Exercise 7.4.1. Verify that the collineation ⎡ 6 ⎣9 3

⎤ −1 0 1 0⎦ −1 3

takes X , Y , Z , and U to (2, 3, 1), (1, −1, 1), (0, 0, 1), and (1, 2, 1), respectively. Example 3. Figure 7.24 portrays the transformation of a checkerboard pattern into a perspective 5 9 1 20 2 view using the matrix 7 7 4 . The first column gives the image X ′ of X = (1, 0, 0), the 1

1

4

ideal point on the x-axis. Similarly, the second column gives the image Y ′ of Y = (0, 1, 0) and the third column gives the image Z ′ = (5, 1, 1), scaled by a factor of 4, of the Euclidean origin, Z = (0, 0, 1). The image U ′ of U = (1, 1, 1) is (9 + 1 + 20, 7 + 7 + 4, 1 + 1 + 4) = (30, 18, 6), which is the same point as (5, 3, 1). You can check that the matrix transforms

7

Y (0,4,1)

(3,4,1) T

U X Z = (0,0,1)

Y' = (1,7,1) D' = (5,7,1)

6 5 D 4 3 2 1

(3,0,1) (0,0,1) 1

X'= (9,7,1)

T' U' Z' = (5,1,1) 2 3 4

5 6 7

8 9

Figure 7.24 A perspective view of a checkerboard.

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T = (2, 2, 1) to T ′ = (5, 4, 1) and (0, 4, 1) to (3, 4, 1), among other points. The ideal point D = (1, 1, 0) on the line through Z , U, and T becomes D ′ = (5, 7, 1). The reader can verify that R(Z , T, U, D) = R(Z ′ , T ′ , U ′ , D ′ ) = −1 and so H (Z T, U D) and H (Z ′ T ′ , U ′ D ′ ), illustrating that collineations preserve cross ratios and harmonic sets. The viewpoint for the perspective appears to be directly perpendicular to D ′ at a distance Y ′ D ′ in the second figure. With regard to the original checkerboard, the viewer’s eye appears to be above the plane of the pattern approximately over the point (−2, −2, 1) in the third quadrant. In fact, the matrix takes (−2, −2, 1) to the ideal point (0, 1, 0) for its image. ♦ While plane collineations match perspective drawings of two-dimensional drawings, as in Example 3, they are not enough for the mathematical basis of perspective in art. Artists convert a three-dimensional scene into a two-dimensional representation. We discuss the mathematics for the conversion on Section 7.6. Theorem 7.4.5. Let P1 , P2 , P3 , and P4 be four points, no three of which are collinear, and Q 1 , Q 2 , Q 3 , and Q 4 be any four points. Then a unique collineation takes P1 , P2 , P3 , and P4 to Q 1 , Q 2 , Q 3 , and Q 4 , respectively, if and only if no three of Q 1 , Q 2 , Q 3 , and Q 4 are collinear. Proof. See Exercise 7.4.9. ! In Chapter 5 we were able to distinguish different types of isometries by their fixed points and stable lines. For example, rotations of any angle other than 180◦ and 0◦ have one fixed point and no stable lines. Translations have no fixed points and a family of parallel stable lines. Projective geometry has an additional ideal line and the ideal points on it, which radically alter the existence of fixed points and stable lines. As collineations, all isometries and, more generally, all affine transformations leave stable the ideal line. Translations fix every ideal point. Theorem 7.4.6 relies on linear algebra to ensure the existence of fixed points and stable lines. Theorem 7.4.6. Every collineation of the projective plane has at least one fixed point and at least one stable line. Proof. To solve the eigenvector problem A(x, y, z) = λ(x, y, z) we find the values of λ for which the determinant of A − λI is zero. As A − λI is a 3 × 3 matrix, with λ appearing in the three diagonal entries, the determinant (characteristic polynomial) is a third-degree real polynomial in λ. Every third-degree polynomial crosses the x-axis and so has at least one real root. Thus every collineation has an eigenvalue. Since the matrix is invertible, the eigenvalue is nonzero, has a nonzero eigenvector, and so a fixed point. The same argument applies to the inverse matrix, giving a stable line. ! Example 4. The matrix

50 1 0

−1 0 0

0 6 0 −1

is a rotary reflection when thought of as a spherical

isometry or three-dimensional Euclidean isometry. It rotates the sphere 90◦ around the z-axis and reflects it over the equator (x y-plane). As a spherical isometry, it fixes no Euclidean point. However, it maps (0, 0, 1) and (0, 0, −1) to each other, which are the same projective point.

7.4 Projective Transformations

351

The fixed point is for the only real eigenvalue, −1. The only stable line is [0, 0, 1], the sphere’s equator or the Euclidean x y-plane, z = 0. ♦ As mentioned in Section 7.3, a conic is determined by five points, no three collinear. Theorem 7.4.5 asserts that collineations are determined by four points and their images. Surprisingly, despite the disparity, we can map every conic to every other conic, although we can’t always specify where points on one conic map onto the other conic. Exercise 7.4.10 illustrates this flexibility.

7.4.1 Exercises for Section 7.4 *7.4.2. (a) Describe all 2 × 2 matrices that send (1, 0) to itself. Repeat for the point (0, 1) and the point (1, 1). Use the results 1 to 2explain why the only 2 × 2 matrices leaving the points fixed are of the form λ0 λ0 . (b) Repeat part (a) with 3 × 3 matrices and the points (1, 0, 0), (0, 1, 0), (0, 0, 1), and (1, 1, 1). x ). 7.4.3. Let x = (x, 1) and ∞ = (1, 0). Recall from Section 7.1 that H (0 1, x 2x−1 1a c 2 *(a) Find the projectivity b d taking 0 to 0, 1 to 1, and 2 to 32 . Verify that the harmonic set H (0 1,12 23 ) goes 2 to a harmonic set. a c (b) Find the projectivity b d taking 0 to 0, 1 to 1, and 2 to 12 . Why does the harmonic set? set H (0 1, 2 23 ) go to 1a harmonic 2 (c) Find the projectivity ab dc taking 0 to 0, 1 to 1, and 2 to x ̸= 12 . Verify that the x to H (0 1, x 2x−1 ). Hint: Pick a = x. harmonic set H (0 1, 12 23 ) goes 2 a c (d) Find the projectivity b d taking 0 to 0, 1 to ∞, and 2 to k. Find the image of 23 . Explain why the image forms a harmonic set with 0, ∞, and k.

7.4.4. We consider views of the checkerboard pattern of Figure 7.24 from different viewpoints. *(a) Find the collineation taking the corners of the unit square (0, 0, 1), (1, 0, 1), (0, 1, 1), and (1, 1, 1) to (0, 1, 1), (1, 2, 1), (−1/3, 2, 1), and (0.5, 2.5, 1), respectively. Make a drawing of the image of the checkerboard pattern of Figure 7.24, similar to the perspective drawing there. Include the images of X and Y and the ideal line. Describe approximately where the viewpoint of this drawing is, with respect to the original drawing and the perspective drawing. (b) Find the matrix and drawing as in part (a), where the images of the corners are now (2, −1, 1), (5, 0, 1), (0, 1, 1), and (1, 2, 1), respectively. Describe approximately where the viewpoint of this drawing might be with respect to the original drawing. Is it a plausible perspective drawing? (c) Find the matrix and drawing as in part (a), where the images of the corners are now (1, −2, 1), (4, 4, 1), (−3, 6, 1), and (1.5, 3, 1), respectively. Explain why the resulting drawing doesn’t represent a perspective drawing from any reasonable viewpoint. 7.4.5. Prove Theorem 7.4.3. Hint: See Theorem 5.3.1. 7.4.6. Prove Theorem 7.4.4.

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7.4.7. *(a) Give the matrix form of the family of collineations fixing P = (0, 0, 1) and fixing each point on the line [0, 0, 1]. Describe these collineations as a type of transformation from Chapter 5. Hint: Find conditions on a matrix fixing P and the points (1, 0, 0), (0, 1, 0), and (1, 1, 0) on [0, 0, 1]. Then explain why this fixes every point on [0, 0, 1]. *(b) Pick a matrix from part (a) that is not the identity and find the image of △ABC, where A = (1, 2, 1), B = (3, 0, 1), and C = (2, −2, 1). The new triangle and △ABC are perspective from P. Verify analytically that [0, 0, 1] is the line of perspective for △ABC and its image. Make a graph. (c) Repeat parts (a) and (b) for P = (1, 0, 0) and the line [1, 1, 1]. Hints: Use points on [1, 1, 1] with a coordinate 0 and another coordinate 1. Reduce the form of the 5a b c 6 matrices d e f so that i = 1 and the other entries are constants or in terms g

h

i

of a.

7.4.8. Recall Desargues’ theorem from Section 7.1: if △ABC and △A′ B ′ C ′ are perspective from a point P, then they are perspective from a line. Let A = (1, 0, 1), B = (1, 1, 1), C = (0, 1, 1), A′ = (2, 0, 1), B ′ = (4, 4, 1), C ′ = (0, 3, 1), and P = (0, 0, 1). ← → ←→ (a) Graph the points. (See Example 1, Section 7.3.) Find the intersections AB · A′ B ′ , ← → ←→ ← → ←→ AC · A′ C ′ , and BC · B ′ C ′ and verify that they are collinear. 5a d g6 (b) Find the collineation α = b e h that fixes P and takes A to A′ , B to B ′ , and c

f

i

′

C to C . Hint: Pick a = 12. Verify that the line you found in part (a) is stable under α. Show further that every point on this line is fixed by α.

7.4.9. Prove Theorem 7.4.5. Hint: See Theorem 7.4.1. 5 1/2 0 0 6 7.4.10. *(a) Let M = 0 −1 −1 . Find M −1 and the image of the unit circle x 2 + y 2 − 0

(b) *(c) (d)

(e)

1/2

−1/2

z 2 = 0 under M. Explain why [cos θ , sin θ , −1] are lines tangent to unit circle. Convert the image in part (a) to nonhomogeneous coordinates (using z = 1). What type of Euclidean conic is the image? (Calculus) Use derivatives to find the tangent to the conic in part (c) when x = k to show that [4k, −1, −2k 2 ] is the tangent. Show that the ideal line [0, 0, 1] is also a tangent. of θ in part (b), verify that M maps the tangents of For the values 0, π2 , π , and 3π 2 part (b) to tangents of the form in part (d).

7.4.11. If a line k is tangent to a conic C and α is any collineation, prove that the image of k under α is tangent to the image of C under α. Hint: See Exercise 7.3.25. 7.4.12. We investigate the effect of the family of collineations Mw = 5 1 0 −1 6 C = 0 1 0 (the circle x 2 + y 2 − 2x z = 0). −1

0

0

51

0 w

0 1 0

06 0 1

on the conic

7.4 Projective Transformations

353

*(a) Find Mw−1 and Mw−1T and the image Mw−1T C Mw−1 of C under Mw . *(b) For the following values of w, convert Mw−1T C Mw−1 to the equation of a conic in nonhomogeneous coordinates (z = 1): w = 12 , − 14 , − 12 , and −1. (c) Graph the original circle and the conics of part (b). *(d) Which range of values of w in part (a) takes the circle to Euclidean ellipses? to hyperbolas? to parabolas? (e) Verify that the lines [1, 0, −2] and [0, 1, −1] are tangent to the original circle by graphing them with the circle. *(f) Find the images of the lines of part (e) for the values of w in part (b) and graph them with the corresponding conics to verify that they are tangents. 7.4.13. A point is exterior to a Euclidean circle if and only if there are two distinct tangents to the circle through it. (See Example 4 of Section 1.2.) In projective geometry define a point P to be exterior to a conic C if and only if there are exactly two tangents to C through P. *(a) Find the two tangents to the unit circle C : x 2 + y 2 − z 2 = 0 through the point P = (2, −1, 1). (b) Convert the conic C ′ with equation x 2 − yz = 0 to nonhomogeneous coordinates, and graph it. Describe the points exterior to it, including any ideal points. (c) (Calculus) Find the two tangents to the C ′ through the point P ′ = (1, 0, 1) and graph them on the graph from part (b). (d) Find the matrix M ′ taking (1, 0, 1) to (1, 1, 1), (−1, 0, 1) to (−1, 1, 1), (0, −1, 1) to (0, 0, 1), and (0, 1, 1) to (0, 1, 0). Verify that M ′ takes C to C ′ , takes P to P ′ , and takes the tangents of part (a) to the tangents of part (c). (e) Repeat part (b) using the conic C ′′ with equation x 2 − y 2 + z 2 = 0. (f) Repeat part (c) using the conic C ′′ and P ′′ = (−2, −1, 1). (g) Find the matrix M ′′ taking (1, 0, 1) to (1, 1, 0), taking (−1, 0, 1) to (−1, 1, 0), and leaving (0, 1, 1) and (0, −1, 1) fixed. Verify that M ′′ takes C to C ′′ , takes P to P ′′ = (−2, −1, 1), and takes the tangents of part (a) to the tangents to C ′′ at P ′′ . Hint: M ′′ has a simple form. (h) Define a projective point to be in the interior of a conic if and only if it is not on the conic nor is it exterior to the conic. Equivalently, a point is interior if and only if every line through it intersects the conic in two points. On the graphs in parts (b) and (e) shade in the interiors of the conics. On a new graph, draw C ′′ and the conic D with equation x 2 + y 2 − 3z 2 = 0. Shade in the region(s) interior to both C ′′ and D. Discuss how well the definition of the interior of a conic fits with your intuition. 7.4.14. *(a) Find the two tangents to the conic C (Euclidean parabola) with equation x 2 − 2yz + z 2 = 0 from the point P = (0, 0, 1). Find the points of intersection A and B of them with the conic and the equation of the line k through A and B. *(b) Find the two points of intersection, S and T , of the conic of part (a) and the line [1, 0, 0]. Also find the intersection Q of [1, 0, 0] with the line k. Find the cross ratio R(P, Q, S, T ) and explain what it tells us.

354

Projective Geometry

(c) Let M =

5m 0 0

0 m m−1

06 0 , where m 1

̸= 0 Find the image of the conic C under M, and

the images of the points A, B, P, Q, S, and T . (Several of these points are fixed.) Verify that the cross ratio of the images of P, Q, S, and T remains the same. (d) Graph C and its tangents and the associated points. (e) For m = 2, graph the images C, its tangents and the associated points. Hint: The image is a Euclidean ellipse. (f) Repeat part (e) for m = 12 . Hint: The image is a Euclidean hyperbola.

7.5 Subgeometries Mathematicians originally thought of projective geometry as an extension of Euclidean geometry. Over time they figured out how to establish projective geometry independently. Then in 1859 Arthur Cayley turned the tables, constructing Euclidean distance and angle measure within projective geometry. That is, he showed Euclidean geometry to be a subgeometry of projective geometry. He also related distance in spherical geometry to projective geometry but was unaware of any other geometries at that time. Felix Klein in 1871 built on Cayley’s construction and other insights to show that hyperbolic and single elliptic geometries (discussed in Chapter 4) were subgeometries of projective geometry. Klein’s construction led him to pick the names for them. (Under his classification Euclidean geometry was a “parabolic” geometry.) The unification of many geometries under projective geometry emphasized the usefulness of projective geometry and the central importance of modern, abstract mathematics. A geometry is a subgeometry of another one in two regards. First, from Klein’s Erlanger Programm (see Section 5.2), if all transformations of one geometry are transformations of a second geometry, the first is a subgeometry of the second. For example, all Euclidean isometries are similarities, which are affine transformations and these are all collineations. Second, the undefined terms, such as points and lines, of the subgeometry must be defined from the terms of the encompassing geometry. Fortunately, the conditions reinforce each other. Figure 7.25 shows relationships of geometries in terms of their transformation groups. Collineations

Affine transformations

Similarities

Hyperbolic isometries

Euclidean isometries

Single elliptic isometries

Figure 7.25 Some subgeometries of projective geometry.

7.5 Subgeometries

355

Projective properties of points on a line must involve at least four points because three points can be mapped to any three points by Theorem 7.4.1. Cross ratios, harmonic sets, and separation, involving four points, are preserved under all projectivities by Theorem 7.4.2, as are all projective properties. However, distance is a relation involving just two points. Cayley realized that he could define the distance between two points A and B relative to two other points " and / on a line using the cross ratio. To obtain " and / he used the intersection of the line with a fixed conic, which he called the absolute conic. Cayley similarly defined the angle between two lines by using two other concurrent lines tangent to the absolute conic. We develop this idea fully only for distance in Klein’s model of hyperbolic geometry. The other situations are more complicated. The transformations for various geometries are characterized as projective collineations that take the absolute conic to itself. (See Cederberg [1, Section 4.12] for more information.)

7.5.1 Hyperbolic Geometry as a Subgeometry Chapter 4 used the points in the interior of the unit circle for the Klein model of hyperbolic geometry. For simplicity, we use the circle here, although from a projective point of view any (real, nondegenerate) conic would work. Thus the unit circle is the absolute conic in Cayley’s terms. Any projective line through an interior point intersects the circle in two points. We defines a point to be in the interior of a conic if and only if every line on that point intersects the conic in two distinct real points. Cayley based the formula for hyperbolic distance between A and B on their cross ratio with the intersections " and / of the line they determine with the conic. Collineations preserve the cross ratio, so automatically Klein’s distance formula is preserved under whichever collineations are hyperbolic. (The Poincar´e model from Chapter 4 also uses the unit circle. Poincar´e defined hyperbolic distance in his model in the same way we do here, although our definition of the cross ratio needs to be generalized to points on a circle. Poincar´e recognized that the cross ratio is preserved under inversions and collineations. See Sections 4.4 and 5.6.) To simplify the distance formula we use Euclidean distance and ignore the constant of the general formula. Interpretation for Hyperbolic Plane Geometry. By absolute conic we mean x 2 + y 2 − z 2 = 0. By point we mean the points (x, y, z), with x 2 + y 2 < z 2 , interior to the absolute conic. By line we mean the nonempty set of points interior to the absolute conic that are on a projective line [a, b, c]. The intersections of a line with the absolute conic are the omega points of the 0 0 line. A" B" 0 ÷ B/ ) , where The distance between A and B is d H ( A, B) = |log(R( A, B, " , /)| = 0log( A/ X Y is the Euclidean distance between X and Y , and " and / are the two omega points of line AB. By hyperbolic isometry we mean a collineation that leaves the absolute conic stable. Example 1. Verify that the adjacent points Pi and Pi+1 shown in Figure 7.26 have the same distance between them. The x-coordinates of the points are P0 = 0, P1 = 13 , P2 = 35 , P3 = 79 , , P5 = 31 , P−i = −Pi , " = −1, and / = 1. P4 = 15 17 33 Solution. The Euclidean distances between the points are the differences of their xcoordinates. Then (P0 "/P0 /) ÷ (P1 "/P1 /) = (1/1) ÷ ( 34 / 23 ) = 12 . Similarly, (P1 "/P1 /) ÷ (P2 "/P2 /) = ( 34 / 23 ) ÷ ( 58 / 25 ) = 12 . All the corresponding products equal 12 or 2. The absolute values of their logarithms are all the same. Hence the distances all are the same. ♦

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Projective Geometry

P-3 P-2 P-1

P1 P2 P3

P0

Figure 7.26 Equidistant points in the Klein model of the hyperbolic plane. Example 2. Euclidean rotations and mirror reflections fixing the origin map the unit circle to itself and so are hyperbolic isometries. A matrix X x analogous to a translation shifting (0, 0, 1) along the x-axis to (x, 0, 1) has for its last column (x, 0, 1), with −1 < x < 1. The translation must leave the omega points (1, 0, 1) and (−1, 0, 1) fixed. That is, ⎡

a ⎣ Xx = d g

b e h

⎤ x 0⎦ 1

must satisfy X x (1, 0, 1) = (a + x, d, g + 1) = λ(1, 0, 1) and X x (−1, 0, 1) = (−a + x, −d, −g + 1) = λ(−1, 0, 1). One way to satisfy the equations is a = 1, d = 0, and g = x. Figure 7.27 suggests that X x should shift (0, 1, 1) and (0, −1, 1) to the points on√the unit circle directly above and below (x, 0, 1). This condition forces b = 0 = h and e = 1 − x 2 . Exercise 7.5.1 gives the final form of the matrix X x . ♦ (0,1,1)

(-1,0,1)

(0,0,1)

(x,0,1)

(1,0,1)

(0,-1,1)

Figure 7.27 A hyperbolic “translation.” Exercise 7.5.1. Let X x =

51 0 x

0 √ 1−x 2 0

x6 0 . 1

Verify that C =

51 0 0

0 1 0

0 6 0 , −1

the unit circle, is sta-

ble under a scalar multiple of X x , and verify that the inverse of X x is a scalar multiple of 5 1 0 −x 6 √ 0 . Explain why X x is a hyperbolic isometry and why, in effect, X −x is the 1−x 2 0 −x

0

inverse of X x .

1

357

7.5 Subgeometries

Theorem 7.5.1. The hyperbolic isometries form a group of transformations. The set {λX x : −1 < x < 1, λ ̸= 0} forms a subgroup. Proof. See Exercise 7.5.10. ! Surprisingly the matrices X x correspond to the transformations of velocities in the special theory of relativity discussed in Section 6.5. In Section 7.6 we explore the connection, including representing the geometry of relativity as a subgeometry of four-dimensional projective space. The geometry, called Minkowski geometry, combines features of hyperbolic geometry and Euclidean geometry. We next derive the general matrix form of hyperbolic isometries, which is quite similar to spherical isometries. (See Section 5.5.) A hyperbolic isometry M must take C to itself. By Theorem 7.4.3, M −1T C M −1 = λC, for λ ̸= 0. The special form of C simplifies the equation further. To emphasize the relationship of the isometries with spherical isometries, define the h-inner product of two vectors (r, s, t) ·h (u, v, w) to be r u + sv − tw. Only the minus sign distinguishes it from the usual definition. Define the h-length of a vector (r, s, t) to be (r, s, t) ·h (r, s, t) and two vectors to be h-orthogonal if and only if their h-inner product is 0. (The definitions do not fulfill all the usual properties of inner products and lengths. For example, nonzero vectors can be h-orthogonal to themselves and so have zero h-lengths.) Example 3. Find conditions on a 3 × 3 invertible matrix M so that M is a hyperbolic isometry. Solution. For M to be a hyperbolic isometry, we must have M −1T C M −1 = λC for some λ ̸= 0. Multiply by M T on the left and M on the right to get C = M T λC M. We can factor out λ and move it to the other side to get λ1 C = M T C M. If we write P for the first column, Q for the second column, and R for the third column of M, then ⎡ ⎡ ⎤ ⎤ P ·h P P ·h Q P ·h R 1 0 0 M T C M = M T ⎣ 0 1 0 ⎦ M = ⎣ Q ·h P Q ·h Q Q ·h R ⎦ 0

0 −1

R ·h P

R ·h Q

R ·h R

which is supposed to equal

⎡ 1/λ 1 C =⎣ 0 λ 0

0 1/λ 0

⎤ 0 0 ⎦. −1/λ

Then the columns of M must be h-orthogonal to each other to give 0 off the main diagonal. Furthermore, the first two columns have the same h-length, which is the negative of the h-length of the third column. We summarize the example as Theorem 7.5.2. ♦ Theorem 7.5.2. A projective collineation is a hyperbolic isometry if and only if it can be written as a 3 × 3 matrix whose columns are h-orthogonal to each other with the first two columns having the same h-length and the third column having the negative h-length of the first two. Proof. See Example 3. !

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Exercise 7.5.2. Verify that the X x satisfy the conditions of Example 3. Compare the conditions in Example 3 with the conditions for a spherical isometry in Section 5.5.

7.5.2 Single Elliptic Geometry as a Subgeometry Single elliptic geometry, discussed in Section 4.5, needs all projective points and lines. Thus the absolute conic must contain no projective points. For the absolute conic we pick the imaginary conic x 2 + y 2 + z 2 = 0, whose only real solution is (0, 0, 0), which isn’t a projective point. An 51 0 06 isometry M is a collineation that takes this degenerate conic C = 0 1 0 to itself. Because 0

−1T

0

1

−1

C M = λC for λ ̸= 0 reduces to M −1T M −1 = λI . C = I , the identity, the equation M Orthogonal matrices, the isometries for spherical geometry (see Section 5.5), satisfy the similar equation M T = M −1 , or M T M = I . Theorem 7.5.3. A collineation M is an isometry of single elliptic geometry if and only if M T = λM −1 for some nonzero real number λ. The isometries form a group of transformations. Proof. See Exercise 7.5.10. !

7.5.3 Affine and Euclidean Geometries as Subgeometries The projective plane can be thought of as the affine or Euclidean plane and one extra ideal line, [0, 0, 1]. Cayley realized that [0, 0, 1], thought of as the degenerate conic z 2 = 0, functioned as the absolute conic. As affine geometry has no notions of distance or angle measure, we can directly give the interpretation of the affine plane as a subgeometry of projective geometry. Interpretation for Affine Plane Geometry. By absolute conic we mean z 2 = 0. By point we mean a point not on the absolute conic: (x, y, 1) = (λx, λy, λ). By line we mean a line other than the absolute conic: [a, b, c] = [λa, λb, λc], where not both a and b are zero. An affine transformation is a collineation that leaves the absolute conic stable. Exercise 7.5.3. Verify that the affine transformations of Chapter 5 are affine transformations in the preceding interpretation. Explain why the interpretation includes matrices whose bottom right entry is a nonzero number. To derive Euclidean distance, Cayley needed to pick two specific complex points on the absolute conic z 2 = 0, the circular points at infinity, or I = (1, i, 0) and J = (1, −i, 0). Their name comes from the fact, noted by Poncelet, that they are on every Euclidean circle. Example 4 reveals that all Euclidean isometries map I and J to themselves, showing the set {I, J } to be stable under them. The cross-ratio definition of distance (and angle measure) Klein employed in hyperbolic and single elliptic geometries works for Euclidean angle measure but not for Euclidean distance. Cayley utilized a more complicated method, which we omit. Example 4. Show that Euclidean isometries map I and J to themselves.

359

7.5 Subgeometries

Solution. From Section 5.3, Euclidean isometries are of the form ⎡ ⎤ ⎡ cos θ − sin θ c cos θ sin θ ⎣ sin θ ⎣ sin θ − cos θ or cos θ f⎦ 0 0 1 0 0 The first form, for direct isometries, takes I = (1, i, 0) to

⎤ c f ⎦. 1

(cos θ − i sin θ, sin θ + i cos θ, 0) = (cos θ − i sin θ, i(cos θ − i sin θ), 0) = (1, i, 0). Similarly, J maps to itself. The other matrices, for indirect isometries, switch I and J . ♦ Klein gave the names of hyperbolic and single elliptic geometry in part from an analogy with Figure 7.23. A hyperbola intersects the ideal line in two points. Each line of the Klein model of hyperbolic geometry intersects the absolute conic in two points. An ellipse doesn’t intersect the ideal line at all. Similarly, no line of the projective model of single elliptic geometry has a real intersection with its absolute conic. (Klein added “single” to the name of single elliptic geometry to distinguish it from spherical geometry, in which two lines intersect twice.) A parabola has one point of intersection with the ideal line. Each line in the affine model has one point on the ideal line, leading Klein to call affine geometry and its subgeometries “parabolic.” However, the name didn’t catch on, unlike the other two. In 1800 there was just one geometry—Euclid’s. Then there were several mutually contradictory geometries. Klein’s and Cayley’s insights brought a longed-for unity back to geometry: projective geometry subsumed Euclidean, hyperbolic, and single elliptic geometries as subgeometries.

7.5.4 Exercises for Section 7.5 7.5.4. (a) (b) (c) (d)

Verify that the coordinate for each Pi in Example 1 is (2i − 1)/(2i + 1). Verify that X 1/3 , as defined in Exercise 7.5.1, takes Pi to Pi+1 . Verify that X 1/3 , as defined in Exercise 7.5.1, takes −Pi+1 to −Pi . Verify that X 1/3 · X 1/3 is a scalar multiple of X 3/5 .

7.5.5. For the points A = (a, 0, 1) and B = (b, 0, 1), verify the identity d H ( A, B) = d H (X x ( A), X x (B)), where X x is the hyperbolic translation of Exercise 7.5.1. Hint: Simplify the cross-ratio for X x ( A), X x (B), " = (1, 0, 1), and / = (−1, 0, 1) by factoring. 7.5.6. *(a) In Figure 7.28 let b = 0.5 and d = 0.6. Find the coordinates of S and T . Then compare the hyperbolic distance between (0, 0, 1) and (b, 0, 1) with the hyperbolic distance between (0, d, 1) and (b, d, 1). (b) Repeat part (a) with b = 0.5 and d = 0.8. (c) For a fixed b make a conjecture about the hyperbolic distance between (0, d, 1) and (b, d, 1) as d increases. Find the coordinates of S and T in terms of a general value d. Give a formula for the hyperbolic distance between (0, d, 1) and (b, d, 1) for general b and d. Test your conjecture for specific values of b and d. 7.5.7. Prove that the composition of the hyperbolic translations X a and X b is a scalar multiple of X a⊕b , where a ⊕ b = (a + b)/(1 + ab) is the addition of velocities in the special theory of relativity. (See Section 6.5.)

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Projective Geometry

T (0,d,1) (b,d,1)

S (-1,0,1)

(0,0,1) (b,0,1)

(1,0,1)

Figure 7.28 7.5.8. (a) Let

M0.6

⎡

0 0.8 0

−1 ⎣ = 0 −0.6

⎤ 0.6 0 ⎦ 1

and

⎡

⎤ −1 0 0 M0 = ⎣ 0 1 0 ⎦ . 0 0 1

Find M0.6 · M0.6 and relate M0.6 to its inverse as collineations. Verify that both leave the unit circle stable, so both are hyperbolic isometries. They are mirror reflections, which we investigate. (b) Verify that M0 fixes every point (0, y, 1) on the y-axis and switches (1, 0, 1) with (−1, 0, 1). Similarly verify that M0.6 fixes every point ( 31 , y, 1) on the line x = 13 and switches (1, 0, 1) with (−1, 0, 1). (c) Find M0.6 · M0 and M0 · M0.6 . Describe the products as a hyperbolic isometries and how they relate to each other. 51 0 06 *(d) Verify that H0 = 0 −1 0 is a hyperbolic mirror reflection, following parts (a) 0

0

1

and (b). What line is its line of reflection? What type of hyperbolic isometry do you think H0 · M0 and H0 · M0.6 are? Justify your answer.

7.5.9.* Find the matrix for a hyperbolic translation Yb along the y-axis. Verify that, in general, X a · Yb ̸= Yb · X a . How are X a · Yb and Yb · X a related as matrices? Let (c, d, 1) be a point inside the unit circle. Find hyperbolic translations X a and Yb such that X a · Yb (0, 0, 1) = (c, d, 1). 7.5.10. (a) Prove Theorem 7.5.1. (See Exercise 6.5.13.) (b) How do the single elliptic isometries differ from the spherical isometries of Section 4.5? (c) Prove Theorem 7.5.3. Hint: See Theorem 5.5.1. 7.5.11. (a) Show that a conic ax 2 + 2bx y + cy 2 + 2d x z + 2eyz + f z 2 = 0 has the points (1, i, 0) and (1, −i, 0) on it if and only if it is a Euclidean circle: That is, a = c and b = 0. (b) Show that similarities take the circular points at infinity to themselves. 7.5.12. Find the properties of an inner product and of length in a linear algebra text and explain which hold and which fail for the h-inner product and the h-length. (See, for example, Fraleigh and Beauregard [5].)

7.6 Projective Space

361

7.5.13. (a) Show that a single elliptic isometry that is simultaneously an affine transformation must be a similarity. Are there any other restrictions on them? Hint: What are the images of the points O = (0, 0, 1), X 1 = (1, 0, 1), and Y1 = (0, 1, 1) under a similarity? (b) Describe collineations that are simultaneously single elliptic isometries, hyperbolic isometries, and Euclidean isometries. Justify your answer and show that they form a group.

7.5.5 Arthur Cayley The most fundamental mathematical contributions of Arthur Cayley (1821–1895) came during the fifteen years he practiced law following his mathematical education at Cambridge University. Cayley was first in his class and started teaching there, but soon left because he didn’t want to be ordained as a priest, then required of all Cambridge professors. Later, as a renowned mathematician, he taught at Cambridge (without ordination) from 1863 until his death, except for a year in the United States. Cayley published over two hundred papers while practicing law and over nine hundred in his lifetime, largely on algebra and geometry. At the age of 20, Cayley started publishing on algebraic invariants, a subject now displaced by topics in algebra and geometry that he and others developed from invariants. While in his early 20s he published some of the first works on n-dimensional geometry. In 1845, at age 24, he found an 8-dimensional algebra generalizing the complex numbers and Hamilton’s recently published 4-dimensional quaternions. In 1849, he proposed the abstract definition of a group and later published the theorem on groups named after him. Algebraists had studied determinants for some time, but Cayley’s work on invariants led him to be the first, in 1855, to study matrices and their properties. He used matrices to represent systems of equations and transformations. He defined matrix multiplication so that it would correspond to the composition of transformations. His crowning achievement in what we now call linear algebra came in 1858 with publication of the Cayley–Hamilton theorem on the characteristic equation of a matrix. Invariants also led Cayley in the following year to his key derivation of Euclidean geometry within projective geometry relative to an absolute conic. He had long used projective geometry in his study of algebraic invariants, which included conics as second-degree invariants. His method of finding metrical properties inside projective geometry led to the unification of several geometries. Cayley learned of hyperbolic geometry only after publishing this important paper.

7.6 Projective Space Homogeneous coordinates and collineations can be readily extended to higher dimensional projective spaces, which provide important models for a variety of purposes. For example, computer programmers use three-dimensional projective space to make perspective views in computer-aided design (CAD). In the language of perspective, projective space adds an ideal plane to the Euclidean points (x, y, z). Homogeneous coordinates for projective space require an extra dimension. We’ll use (x, y, z, t) as a general three-dimensional projective point. From a Euclidean point of view, the ideal points have homogeneous coordinates (x, y, z, 0) and form a plane. As before, any nonzero multiple (λx, λy, λz, λt) represents the same point as (x, y, z, t). In linear algebra terms, the one-dimensional vector space is the point. While perspective in

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art has no need of higher dimensions, other applications need more dimensions. The Lorentz transformations in the special theory of relativity are the transformations of a subgeometry of four-dimensional projective space related to hyperbolic geometry. Interpretation. By a point in n-dimensional projective space Pn we mean a one-dimensional subspace of the n + 1 dimensional vector space Rn+1 . By line we mean a two-dimensional subspace, by plane we mean a three-dimensional subspace, and so on. A point (line, and so on) is on a line (plane, and so on) if and only if the first is a subset of the second. By a collineation in Pn we mean an invertible (n + 1) × (n + 1) matrix. A nonzero scalar multiple of a matrix represents the same collineation. Example 1. Two points in P3 (or any Pn ) have a unique line on them. In terms of linear algebra, two distinct one-dimensional subspaces are spanned by a unique two-dimensional subspace. However, in P3 two lines can fail to intersect, which corresponds to skew lines in Euclidean space. For example, the lines (two-dimensional subspaces) {(x, y, 0, 0) : x, y ∈ R} and {(0, 0, z, t) : z, t ∈ R} intersect only at the origin (0, 0, 0, 0), the zero-dimensional subspace, which isn’t a projective point. However, as we show, two distinct planes (three-dimensional subspaces) in P3 intersect in a line (a two-dimensional subspace): each plane has three basis vectors, giving six possible vectors. The whole space needs only four basis vectors, so there must be an overlap of at least two. Because the planes are distinct, the overlap is exactly two—a line. We used a row vector [a, b, c, d] in Section 5.5 to represent a plane in three-dimensional Euclidean space. This interpretation holds in P3 as well. A point (x, y, z, t) is on the plane [a, b, c, d] if and only if ax + by + cz + dt = 0. The roles of column vectors (points) and row vectors (now planes) are dual. So in P3 points and planes are duals. Lines are said to be “selfdual.” Thus the dual statement in P3 of “two points determine a line” is “two planes determine a line.” Similarly the dual of “three points not on the same line determine a plane” is “three planes not on the same line determine a point.” In general, the dual of an i-dimensional subspace in Pn is an (n + 1 − i)-dimensional subspace. ♦

7.6.1 Perspective and Computer-Aided Design A general 4 × 4 collineation in P3 can be broken into component parts, some of which appear in affine transformations. From Section 5.5, the rightmost column of an affine matrix describes where the origin goes and corresponds to a translation. The upper left 3 × 3 submatrix determines the type of affine transformation: rotation, reflection, shear, dilation, and so on. The bottom row 1 2 of an affine matrix is 0 0 0 1 . The bottom row of a collineation provides additional flexibility lacking in affine transformations. In a CAD system the change of the lower right entry from a 1 to 1/r magnifies or shrinks the entire picture by a factor of r . The alteration is a faster and easier way to make a change of scale for a similarity than changing the twelve upper entries by a factor of r . The first three entries of the bottom row determine perspective views in each dimension. The matrix ⎤ ⎡ a a a tx ⎢ a a a ty ⎥ ⎥ ⎢ ⎣ a a a tz ⎦ px p y pz 1/r

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7.6 Projective Space

summarizes this discussion, where a stands for affine, t for translation, p for perspective, and 1/r for the scaling ratio. (See Penna and Patterson [9] for more information on projective geometry and computer graphics.) Example 2. We illustrate the effects of the perspective entries in a collineation by projecting a cube using several related matrices. (For simplicity we ignore the translation and scaling entries.) By Theorem 5.5.4, we need only consider what the collineations do to the 4 × 8 matrix ⎡

0 ⎢0 V =⎢ ⎣0 1

0 0 1 1

0 1 0 1

0 1 1 1

1 0 0 1

1 0 1 1

1 1 0 1

⎤ 1 1⎥ ⎥ 1⎦ 1

containing the eight vertices of the cube. The matrix ⎡

0 ⎢0 P=⎢ ⎣0 0

0 1 0 0

0 0 1 0

⎤ 0 0⎥ ⎥ 0⎦ 1

represents a projection parallel to the x-axis; it isn’t a collineation because it collapses threedimensional objects to two dimensions. For example, P maps both points (0, 0, 0, 1) and (1, 0, 0, 1) to (0, 0, 0, 1). On a computer screen in nonhomogeneous coordinates the eight vertices in P V , together with their edges, would appear as a square in the yz-plane (Figure 7.29). z

y

Figure 7.29 A flat projection of a cube. To see the three-dimensional form of a cube we need to have a different viewing angle. The matrix R rotates the cube −30◦ around the z-axis and then −30◦ around the (original) y-axis. ⎡

cos −30 ⎢ 0 R=⎢ ⎣ sin −30 0 ⎡ 3/4 ⎢ −1/2 √ =⎢ ⎣ − 3/4 0

0 − sin −30 1 0 0 cos −30 0 0 √ √3/4 1/2 3/2 √ 0 −1/4 3/2 0 0

⎤⎡ 0 cos −30 ⎥ ⎢ 0 ⎥ ⎢ sin −30 0⎦⎣ 0 1 0 ⎤ 0 0⎥ ⎥. 0⎦ 1

− sin −30 cos −30 0 0

0 0 1 0

⎤ 0 0⎥ ⎥ 0⎦ 1

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Projective Geometry

z

y x

Figure 7.30 A non-perspective view of a cube. Then P RV gives the familiar, non-perspective view of the cube (Figure 7.30). Without perspective, all the faces in Figure 7.30 are parallelograms. Let ⎡

and

1 ⎢ 0 Px = ⎢ ⎣ 0 −1/5

⎤ 0 0 0 1 0 0⎥ ⎥, 0 1 0⎦ 0 0 1

Px yz

⎡

1 ⎢ 0 =⎢ ⎣ 0 −1/5

Px y

⎡

1 ⎢ 0 =⎢ ⎣ 0 −1/5

0 1 0 −1/5

0 1 0 −1/5

⎤ 0 0 0 0⎥ ⎥, 1 0⎦ 0 1

⎤ 0 0 0 0⎥ ⎥. 1 0⎦ −1/5 1

Then P R Px V gives the one-point perspective view in Figure 7.31, P R Px y V gives the twopoint perspective view in Figure 7.32, and P R Px yz V gives the three-point perspective view in Figure 7.33. Vx z

x

y

Figure 7.31 A one-point perspective view of a cube. In Figure 7.31, we have one-point perspective: the sides of the cube parallel to the x-axis meet at an ideal point denoted Vx . The other sides remain parallel as in Euclidean geometry. We can compute the coordinates of the vanishing point by finding √ the projection of (1, 0, 0, 0), the ideal point along the x-axis: P R Px (1, 0, 0, 0) = (0, − 12 , − 43 , − 15 ) = (0, 2.50, 2.17, 1), or,

365

7.6 Projective Space

Vx z Vy

y x

Figure 7.32 A two-point perspective view of a cube. in the nonhomogeneous coordinates of the figure, (2.50, 2.17). Figure 7.32 includes the ideal points for both the x- and y-axes, Vx and Vy . Actually, Px y automatically gives vanishing points ←−→ for all directions in the x y-plane, which appear on the ideal line Vx Vy shown in Figure 7.32. For example, the diagonals connecting (0, 0, 1, 1) to (1, 1, 1, 1) and (0, 0, 0, 1) to (1, 1, 0, 1) meet at the ideal point located at (−0.92, 1.71), the projection of the ideal point (1, 1, 0, 0). In three-point perspective (Figure 7.33) every direction has a vanishing point. Thus every point of the projection plane can be an ideal point, as well as the image of many regular points. ♦ z

Vx

Vy y x

Vz

Figure 7.33 A three-point perspective view of a cube. Projective geometry, like single elliptic geometry, is not oriented. (See Section 4.5.) There is no consistent way to define clockwise rotations around a point or direction on a line. One consequence of nonorientability is that collineations can alter the appearance of a figure beyond what an artist or computer operator needs. For example, projectivities and so collineations can map three collinear points to any three collinear points. Therefore they can alter the betweenness relations of points on a line. However, no perspective view of a real object will ever turn it inside out. Stolfi [12] has devised an oriented projective geometry for CAD by not allowing negative scalars. We place a tilde (˜) over a point to indicate an oriented point. An oriented point is still → → v and − w represent the same point if and only if there is a a vector in R4 , but two such vectors − − → − → positive real number k such that w = k · v . Each projective point P is split into two opposite = and − P. = This corresponds exactly to the relationship between spherical and oriented points P

366

Projective Geometry

single elliptic geometry: two antipodal points in spherical geometry represent the same point in single elliptic geometry. All invertible matrices are still collineations in oriented projective geometry, although their effect changes and two matrices represent the same collineation if and only if they differ by a positive scalar. In oriented projective geometry, we can define betweenness and convex sets. As Theorem 7.6.1 shows, collineations interact with these concepts in valuable ways. = is between P = and Q = if and only if there are positive real numbers a and b such Definition. R =Q = is the set { R =: R = is between P = and Q, = R = = P, = or == aP = + b Q. = The line segment P that R = = = = = = R = Q}. A set S of oriented points is convex if and only if, for P and Q in S, P Q is a subset of S. # # # Example 3. The oriented points between (0, 0, 1) and (1, 0, 1) are (x, 0, 1), where 0 < x < 1. # # # Proof. Let (u, v, w) be between (0, 0, 1) and (1, 0, 1). Then there are positive reals a and b # # # , 0, 1). Thus x = b/(a + b) such that (u, v, w) = a (0, 0, 1) + b(1, 0, 1) = (b, # 0, a + b) = ( b# a+b

is between 0 and 1 because a and b are positive. Without loss of generality we can pick a and # # b so that a + b = 1. Then (a, b) are the barycentric coordinates of (x, 0, 1) in terms of (0, 0, 1) # # # # and (1, 0, 1). Conversely, we can write (x, 0, 1) as (1 − x)(0, 0, 1) + x (1, 0, 1). (See Section 3.3 for a discussion of barycentric coordinates.) !

If we used the same definitions with regular projective points P and Q and nonzero scalars, ← → the entire line P Q would be between P and Q and so would be a line segment. Theorem 7.6.1 would still be provable if we change the word positive to nonzero. However the result would not be particularly helpful since the only convex sets would be single points, (projective) lines, and the whole plane. Theorem 7.6.1. Collineations in oriented projective geometry preserve betweenness and convexity. = and Q = be two points. Then R == aP = + b Q, = for a and Proof. Let γ be a collineation and P = and Q. = Because γ is a linear transformation, γ R = = γ (a P = + b Q) = = b positive, is between P = + bγ Q, = which shows that γ R = is between γ P = and γ Q. = Convexity is defined in terms of aγ P betweenness, so collineations preserve convexity by the argument used in Theorem 5.4.6. !

Oriented projective geometry has some peculiarities. For example, two oriented projective lines intersect in two oriented points, which forces an oriented line in effect to have two copies of the real line on it (Figure 7.34). Also, as Exercises 7.6.12 and 7.6.13 illustrate, the image of a convex set is not necessarily what we think it will be.

–X –1

–X 0

–X 2

–X ∞

X0 X1

Figure 7.34 An oriented projective line.

X∞

7.6 Projective Space

367

7.6.2 Subgeometries of Projective Space Euclidean, hyperbolic, and single elliptic geometries of n dimensions are subgeometries of the projective geometry of the same number of dimensions. The transformation groups are entirely analogous to the corresponding groups in two dimensions. We briefly consider three-dimensional hyperbolic space to lead into Minkowski geometry, used in the special theory of relativity. In three dimensions, Cayley’s absolute conic becomes an absolute quadric surface. (See Exercise 7.6.5 for more on quadric surfaces.) The points of three-dimensional hyperbolic space are the points in the interior of the unit sphere x 2 + y 2 + z 2 − t 2 = 0, which we take as the absolute quadric surface. Hyperbolic isometries leave the surface stable. Two 4 × 4 matrices represent the same three-dimensional hyperbolic isometry if and only if they differ by a nonzero factor. Exercise 7.6.1.∗ Extend the definitions from Section 7.5 of h-inner product, h-orthogonal, and h-length to vectors in R4 . Theorem 7.6.2. A 4 × 4 nonsingular matrix represents a hyperbolic isometry if and only if any two of its columns are h-orthogonal, the first three have the same h-length, and the last column has the opposite h-length. Proof. See Exercise 7.6.9. ! In Section 6.5 we discussed the Lorentz transformations, which preserve $xA2 + $yA2 + − $tA2 , a value closely related to the equation of the absolute quadric surface for hyperbolic space. The equation $xA2 + $yA2 + $z A2 − $tA2 = $xB2 + $yB2 + $z B2 − $tB2 expressed the invariance of the distance and time measurements between two events from two frames of reference moving at a constant velocity relative to each other. Transformations preserving all values k = $xA2 + $yA2 + $z A2 − $tA2 clearly leave x 2 + y 2 + z 2 − t 2 = 0 stable and so are hyperbolic isometries. Actually, the four-dimensional geometry for relativity, called Minkowski geometry, needs all points (x, y, z, t). As with affine geometry, an extra coordinate is needed to permit movement of the origin. Hence the points are written (x, y, z, t, 1). Thus Minkowski geometry is a subgeometry of P4 , rather than P3 . The transformations of Minkowski geometry are called Lorentz transformations. $z A2

Definition. A Lorentz transformation is a collineation of P4 represented by a 5 × 5 matrix whose upper left 4 × 4 submatrix is a three-dimensional hyperbolic isometry and whose bottom row is [ 0 0 0 0 1 ]. Corollary 7.6.3. The2 Lorentz transformations are collineations of P4 , where the bottom row 1 is 0 0 0 0 1 and the upper left 4 × 4 submatrix is a hyperbolic isometry, with the first three columns having h-length = ±1 and the fourth column having h-length = ∓1. Proof. See Exercise 7.6.10. ! In Section 6.5 we showed that the value k = $xA2 + $yA2 + $z A2 − $tA2 could be zero if the two events were lightlike. For example, the differences in space and time coordinates of two

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Projective Geometry

points on the path of a light beam would have k = 0. In effect, the h-length of the difference of two lightlike events is zero. Similarly, two events can have k positive or negative depending on whether their relationship is spacelike or timelike, respectively. (For further information on the special theory of relativity see Taylor and Wheeler [13].) Projective geometry encompasses even more geometric ideas than Cayley and Klein envisioned. Though projective geometry falls far short of “all geometry,” as Cayley exclaimed in the quote opening this chapter, it has proven its worth in classical geometry, CAD systems, relativity theory, and other applications.

7.6.3 Exercises for Section 7.6 7.6.2. *(a) Find the plane [a, b, c, d] in P3 through the points (1, 0, 0, 0), (0, 1, 0, 0), and (0, 0, 1, 0). *(b) Repeat part (a) for the points (2, 2, 1, 1), (2, 1, 2, 1), and (1, 2, 2, 1). (c) Find the family of planes (in terms of p), where a given plane is through the points ( p, 1, 1, 1), (1, p, 1, 1), and (1, 1, p, 1). Explain why in Euclidean space with points (x, y, z, 1) the planes would be parallel. In P3 they intersect in a line. Find two distinct points lying on all the planes in this family. (d) Determine the image of a plane [a, b, c, d] in P3 under a collineation M and prove your answer is correct. Hint: See Theorem 7.4.3. 7.6.3. *(a) Find a condition similar to the one in Example 4 of Section 7.3, describing when four points of P3 are in the same plane. (b) Use part (a) to show that the points (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), and (0, 0, 0, 1) are not in the same plane nor is (1, 1, 1, 1) with any three of the preceding points. (c) Prove the condition in part (a). 7.6.4. *(a) Find a projective space collineation taking (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1), and (1, 1, 1, 1) to (1, 2, 0, 0), (0, 1, 2, 0), (0, 0, 1, 2), (0, 0, 0, 1), and (2, 3, 0, 5), respectively. Hint: See the proof of Theorem 7.4.1. (b) For five points A1 , A2 , A3 , A4 , and A5 in P3 , where no four are coplanar, prove there is a projective space collineation taking (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1), and (1, 1, 1, 1) to A1 , A2 , A3 , A4 , and A5 , respectively. (c) For five points A1 , A2 , A3 , A4 , and A5 in P3 , where no four are coplanar and five points B1 , B2 , B3 , B4 , and B5 in P3 , where no four are coplanar, prove there is a projective space collineation taking each Ai to Bi . 7.6.5. *(a) Use the definition of a conic in Section 7.3 to define its three-dimensional analogue, called a quadric (or quadratic) surface in P3 . Also define when a point is on a surface. *(b) Rewrite the equation of the Euclidean unit sphere x 2 + y 2 + z 2 = 1 in homogeneous coordinates and as a matrix. (c) Rewrite the equation of the Euclidean paraboloid z = x 2 + y 2 + 1 in homogeneous coordinates and as a matrix. Find the ideal point (with t = 0) on the paraboloid. Use a computer graphing program to graph the surface. (d) Repeat part (c) for the hyperboloid of one sheet z 2 = x 2 + y 2 − 1. Find its ideal points in the form ( , , 1, 0).

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7.6 Projective Space

7.6.6. Let S be a quadric surface S and M a matrix representing a collineation in P3 . Find the matrix form of the image of S under M and prove your answer. 7.6.7. Decide for which values of n the sets in parts (a)–(c) intersect in Pn . Justify your answers. *(a) a line and a plane *(b) two planes (c) a k-dimensional subspace and a j-dimensional subspace (d) What happens to the intersection in part (c) as n decreases? 7.6.8. *(a) Draw a three-point perspective image of the cube in Example 2, using −0.25 for the first three entries in the bottom row. Find the vanishing points Vx , Vy , and Vz . What effect does changing the entries from −0.2 to −0.25 have? (b) Repeat part (a), using +0.2 for the first three entries in the bottom row. Find the vanishing points Vx , Vy , and Vz . What effect does changing the entries from −0.2 to +0.2 have? 7.6.9. Prove Theorem 7.6.2. Hint: See Example 3 of Section 7.5. 7.6.10. Prove Corollary 7.6.3. 7.6.11. Prove that the intersection of a collection of convex sets in oriented projective geometry is again a convex set. *7.6.12. (a) Find a collineation M fixing (0, 0, 1), (1, 0, 1), and (0, 1, 1) and taking (1, 1, 1) to ( 13 , 13 , 1). Hint: Use (1, 1, 3) for the image. On a graph draw the Euclidean quadrilateral whose vertices are the three fixed points and ( 13 , 13 , 1), which is not convex in Euclidean geometry. However, the original points form a square, a convex set in Euclidean geometry. (b) In Euclidean geometry the segment from (0, 0, 1) to (1, 0, 1) includes the points (x, 0, 1) for 0 < x < 1. Use M to find the image of the points (x, 0, 1) for 0 < x < 1. Indicate where they are on the graph in part (a). Do the same for the points (x, 1 − x, 1), for 0 < x < 1, which in Euclidean geometry is the diagonal between (0, 1, 1) and (1, 0, 1). What does M do to the Euclidean interior of the square? In oriented projective geometry the image is convex. Explain how it can be convex in oriented projective geometry. 7.6.13. (a) Graph the Euclidean circle x 2 + y 2 = 4 and hyperbola x 2 − y 2 = 1 and shade in the interiors of the circle x 2 + y 2 < 4 and of the hyperbola x 2 − y 2 > 1. The intersection of their interiors has two separate regions, which therefore can’t form a convex Euclidean set. # (b) Explain why the oriented points (x, y, z), with x 2 + y 2 < 4z 2 and z > 0, are interior to one oriented representation of the circle in part (a) and why they form a convex set. 5 0 0 16 (c) Verify that the collineation 1/2 0 0 takes the circle of part (a) to the hyper0

1/2

0

bola of part (a). Verify that points interior to the circle are taken to points interior to the hyperbola. (d) Explain how to resolve the following seeming contradiction between Theorem 7.6.1, Exercise 7.6.11, and the preceding parts of this problem. The oriented

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Projective Geometry

interior of the circle is convex, the circle is mapped to the hyperbola, and Theorem 7.6.1 implies that the hyperbola’s oriented interior is convex. Exercise 7.6.11 says that the intersection of two such convex sets must be convex, yet the intersection in part (a) is clearly not convex. 7.6.14. (a) Verify that ⎡

1

⎢ ⎢0 ⎢ ⎢ Xw = ⎢ 0 ⎢ ⎢w ⎣ 0

√

0 1 − w2 0

0 0

0

w

0

0

√ 1 − w2

0

0

1

0

0

0

⎤

⎥ 0⎥ ⎥ ⎥ 0⎥ ⎥ 0⎥ ⎦ 1

is a Lorentz transformation. It represents how to shift frames of reference from one object to another object moving at a velocity w relative to the first object in the positive x-direction. We measure velocities in terms of the speed of light. So w = 1 is the speed of light. (b) Let object B be moving at a velocity of w with respect to A and let C be moving at a velocity of v with respect to B, both in the x-direction. Find the velocity of C with respect to A by multiplying X v · X w and adjusting the upper left corner to fit the form in part (a). Verify that your answer fits with formula (6.1).

7.6.4 Projects for Chapter 7 1. Investigate how artists make perspective drawings. (See Frantz and Crannell [6] and Powell [10].) 2. Relate homogeneous coordinates and barycentric coordinates. (See Section 2.3.) 3. Determine the point of perspective in D¨urer’s woodcut St. Jerome in His Study and use it to determine whether D¨urer depicted the table correctly. (See Crannell et al [3].) 4. Use parts (a) and (b) below and Figure 7.6 to explain why Theorem 7.1.1 (Desargues’ theorem) holds in three-dimensional projective space. Desargues’ theorem: if two triangles are perspective from a point, then they are perspective from a line. (a) First let triangles △ABC and △D E F be in two nonparallel planes and perspective from P. Explain why the plane through P, A, and B must include D, E, and S. Repeat similarly for the plane through P, A, and C as well as for the plane through P, B, and C. Why must S, T, and U all be on the intersection of the plane through A, B, and C with the plane through D, E, and F? Finish the case by explaining why the triangles ← → must be perspective from the line ST . (b) Explain how to use part (a) twice to prove Desargues’ theorem if the triangles are in the same plane. Hint: Use a third triangle and two other points of perspective. (c) State the converse of Desargues’ theorem and explain how you could prove it. 5. Jacob Steiner defined conics using projectivities as follows. Let ki be the family of lines on point P, let m i be the family of lines on point Q, and assume that ki is related to m i by a projectivity of lines that isn’t a perspectivity. Then the points Ri , which are the intersections of ki and m i , form a conic.

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7.6 Projective Space

6. 7.

8.

9.

(a) Explore this method with graph paper. Let P = (5, 12), Q = (10, −7), lines ki intersect the y-axis at (0, i), and lines m i intersect the x-axis at (i, 0). Find points Ri and sketch the conic. (b) Repeat part (a) with P = (−10, −6) and Q = (6, 10). (c) Identify the types of Euclidean conics you obtained in parts (a) and (b). Explain why points P and Q are on the conic. Experiment with other placements of P and Q and other ways to relate the families of lines. (d) Reproduce the construction using a dynamic software program and vary the placement of P and Q. Must a three-dimensional collineation have a fixed point and a stable plane? If so, prove it; if not, give a counterexample. Generalize to higher dimensions. Use the following outline to explain why every hyperbolic plane isometry has a fixed point or a fixed omega point. With regard to the absolute conic, a point is either on the conic, exterior to it, or interior to it. Use Theorem 7.4.6 with each case. When the point is exterior to the conic, use Exercise 7.1.15 and analyze what can happen. In Section 2.2 we defined an ellipse by using two foci. When we transform an ellipse to another ellipse by an affine transformation, do the old foci map to the new foci? Investigate with the ellipses x 2 /a 2 + y 2 /b2 = 1 and affine transformations that map the x- and y-axes to themselves. 5 −1 0 w6 To generalize Exercise 7.5.8, let Mw = 0 √1−w2 0 , for −1 < w < 1. −w

10.

11.

12. 13. 14. 15. 16.

0

1

(a) Verify that Mw is a hyperbolic isometry, its square is the identity collineation, it switches (1, 0, 1) and (−1, 0, 1) and it switches (0, 0, 1),√and (w, 0, 1). (b) Prove that√(v, y, 1) is fixed, where v = (1 − 1 − w2 )/w and so the (vertical) line x = (1 − 1 − w2 )/w is the line of reflection. Use the distance formula to prove that (v, 0, 1) is the midpoint of (0, 0, 1) and (w, 0, 1). (c) Prove that M0 · Mw = X w . Explain how this and parts (a) and (b) fit with the result of Exercise 5.2.11. (d) Describe how to find the matrix for a hyperbolic mirror reflection over a hyperbolic line by using Mw and a rotation fixing (0, 0, 1). A correlation is a transformation that maps points to lines and lines to points. Investigate correlations and how they relate to duality, poles, and polars. (See Cederberg [1] or Coxeter [2].) Investigate the types of eigenvalues and eigenvectors of a two-dimensional collineation. Find the sets of fixed points and stable lines. How does the set of fixed points relate to the set of stable lines? Investigate how computer-aided design uses projective geometry. (See Penna and Patterson [9].) Investigate oriented projective geometry. (See Stolfi [12].) Investigate the special theory of relativity. (See Taylor and Wheeler [13].) Investigate photogrammetry, the process of deducing distances and angles form photographic evidence. (See Linder [8].) Convert various ordinary functions, such as y = x 2 /(x 2 − 1), to homogeneous coordinates, in this case x 2 y − x 2 z − yz 2 = 0. (Each term needs to be of the same degree, here the third degree.) Find their ideal points and relate them to the graphs and asymptotes of the

372

Projective Geometry

original equations. Algebraic geometry uses homogeneous coordinates to explore polynomial equations. (See Smith [11].) 17. Write an essay discussing Klein’s definition of geometry (see Section 4.2) in light of the various subgeometries of projective geometry. 18. Write an essay comparing the advantages and disadvantages of synthetic and analytic approaches to projective geometry. Do you agree with Poncelet that analytic geometry gives answers without insight? Explain.

7.6.5 Suggested Readings 1. Cederberg, J., A Course in Modern Geometries, New York: Springer-Verlag, 1989. 2. Coxeter, H., Projective Geometry, Toronto: University of Toronto Press, 1974. 3. Crannell, A., M. Frantz, and F. Futamura, D¨urer: disguise, distance, disagreements, and diagonals!, Math Horizons, Nov. 2014, 8–11, 25. 4. Dillon, M., Projective geometry for all, College Math. J., 45 # 3 (May 2014), 169–178. 5. Fraleigh, J., and R. Beauregard, Linear Algebra, Reading, MA: Addision-Wesley, 1987. 6. Frantz, M., and A. Crannell, Viewpoints, Princeton, NJ: Princeton University Press, 2012. 7. Kline, M., Mathematical Thought from Ancient to Modern Times, New York: Oxford University Press, 1972. 8. Linder, W., Digital Photogrammetry, New York: Springer, 2009. 9. Penna, M., and R. Patterson, Projective Geometry and its Applications to Computer Graphics, Englewoods Cliffs, NJ: Prentice Hall, 1986. 10. Powell, W., Perspective, Tustin, CA: Walter Foster, 1989. 11. Smith, K., An Invitation to Algebraic Geometry, New York: Springer, 2000. 12. Stolfi, J., Oriented Projective Geometry, Boston: Academic Press, 1991. 13. Taylor, E., and J. Wheeler, Spacetime Physics, New York: W. H. Freeman, 1966. 14. Tuller, A., A Modern Introduction to Geometries, New York: Van Nostrand Reinhold, 1967.

8

Finite Geometries

Figure 8.0 Experimenters testing varieties of seed, fertilizers, or pesticides need to guard against other factors accidentally favoring one variety over another. For example, the drainage or quality of the soil will vary from one part of a field to another. Thus differences in the success of different varieties might be due to extraneous factors rather than the varieties themselves. Planting the same variety of seed in a row, as in Figure 8.0, risks confounding variables such as drainage with sloping land. This type of planting isn’t actually used in the research phase of testing seeds. Instead finite geometries and other designs provide a way to minimize such potential bias in these experiments. Other experiments, such as taste testing, also use design theory. Mathematicians have also used finite geometries to develop error correcting codes. The moving power of mathematical invention is not reasoning but imagination. —Augustus DeMorgan

8.1 Overview and History Finite geometries as a subject arose from the investigations of geometric axioms at the end of the nineteenth century. The advent of hyperbolic and other geometries prompted a renewed

373

374

Finite Geometries

interest in axioms. In particular, geometers sought models satisfying certain axioms but not others. Often the models had finitely many points, whence the term finite geometries. The first explicitly finite geometry was a three-dimensional example with fifteen points developed in 1892 by Gino Fano. Geometers soon realized that finite geometries and their axiom systems were interesting for their own sake. Looking back, they found earlier mathematical designs, results, and problems anticipating finite geometry as a subject. As so often happens in mathematics, significant applications arose from the blend of interesting problems and mathematical structure. The applications have enriched mathematics by suggesting new areas to explore. In the 1920s, statistical design theory drew on finite geometries and expanded the types of geometric structures and theorems. After World War II coding theory required other structures and posed different problems related to finite geometries. More recently computers have greatly influenced the study of finite geometries, by considering more complicated examples and meeting the need for sophisticated designs. Brute computing force may solve particular problems but, as Example 1 illustrates, more general problems require mathematical insight. Example 1. Euler (approximately pronounced “oiler”) posed the “36-officer problem” in 1782. He imagined six military regiments, each with the same six ranks and supposed one officer of each rank from each regiment. Euler sought a square array of them so that no two officers of the same rank or from the same regiment were in the same row or column. He was convinced that this problem had no solution but was unable to give a proof. (The impossibility was proved in 1900 by an exhaustive search.) He was able to construct a solution to the corresponding problem for 52 = 25 officers, among others. (Table 8.1 gives one solution, where the ranks are given by letters and the regiments by numbers.) He also conjectured that similar problems with n 2 officers, where n = 4k + 2, would be impossible. Euler’s intuition was uncharacteristically flawed. In 1960 after extensive research, mathematicians showed that they could solve the n 2 officers problem for every value of n except 2 and 6. ♦ Table 8.1 A1

B2

C3

D4

E5

D3

E4

A5

B1

C2

B5

C1

D2

E3

A4

E2

A3

B4

C5

D1

C4

D5

E1

A2

B3

Euler’s fame as a mathematician provided one reason to tackle the general problem in Example 1, but a modern application provided another impetus. In the 1920s Sir Ronald A. Fisher developed statistical design theory. He and others wanted to test, for example, how the interactions of varieties of fertilizers and pesticides affected the growth of a plant. They needed to test each variety of fertilizer with each variety of pesticide. For n varieties of each, there are n 2 combinations. Previously, others had arranged different varieties of fertilizer in the rows and different varieties of pesticide in the columns. However, fields in nature, unlike in mathematics,

375

8.1 Overview and History

aren’t uniform. Hence some rows or columns may provide better drainage, soil nutrients, or other growing conditions than others. To control for such confounding variables, Fisher realized that he needed a design where each type of fertilizer and each type of pesticide appeared once in each row and column. Thus Euler’s 36-officer problem became the prototype for Fisher’s design problems. Fortunately, geometers had already solved many of these types of problems by using affine planes, which we develop in Sections 8.2 and 8.4. Fisher used the solutions, starting a vital interaction between finite geometries and design theory. Example 2. In 1850, the Rev. Thomas Kirkman posed and solved the “15-schoolgirl problem.” He supposed that the girls took walks every day in an artificially regimented style of five rows of three each. He asked for daily arrangements of them so that two girls walked in the same row just once a week. The solution in Table 8.2 makes use of Fano’s finite three-dimensional projective geometry, discussed in Section 8.4. The girls are points of the geometry and the groups of three girls are the lines. Table 8.2 Sunday

1, 2, 3

5, 11, 14

6, 9, 15

7, 10, 13

4, 8, 12

Monday

1, 4, 5

3, 9, 10

7, 11, 12

2, 13, 15

6, 8, 14

Tuesday

1, 6, 7

4, 9, 13

5, 10, 15

3, 8, 11

2, 12, 14

Wednesday

1, 8, 9

2, 5, 7

3, 13 14

4, 11, 15

6, 10, 12

Thursday

1,10, 11

3, 12, 15

5, 8, 13

7, 9, 14

2, 4, 6

Friday

1, 12, 13

7, 8, 15

2, 9, 11

3, 5, 6

4, 10, 14

Saturday

1, 14, 15

6, 11, 13

3, 4, 7

5, 9, 12

2, 8, 10

Which of the numerical conditions in Kirkman’s problem imply the others? The total number of girls (fifteen) and the size of each row (three) clearly determine the number of rows (five). We use the condition that each pair of girls occurs just once to determine the number of days possible. For any girl the fourteen other girls must be put with this girl in pairs so that there can only be seven nonrepeating days. But are there enough different triples to fill out all seven days? Five rows and seven days require 35 rows. Each of the fifteen girls can be in seven rows, seemingly making 105 rows. However, that product counts each row three times—once for each girl in a row. Dividing 105 by 3 yields 35, the exact number needed. This reasoning is a example of combinatorics, a branch of mathematics that provides insight into numerical relationships. ♦ Current finite geometry research benefits from the cross-fertilization of geometry, algebra, and combinatorics in mathematics and areas outside of mathematics. Transformational geometry and group theory give powerful insights about finite geometries just as they have since the nineteenth century for traditional geometries. Combinatorics provides essential insights into finite and other areas of geometry. For example, the consequences of Euler’s formula (see Chapter 1) were identified through combinatorial reasoning. Error-correcting codes from coding

376

Finite Geometries

theory, which we discuss in Section 8.3, are used in electronic data transmission and benefit from the interaction of these areas of mathematics.

8.1.1 Exercises for Section 8.1 8.1.1. *(a) Find a solution to the 9-officer problem. That is, arrange nine officers, where there are three of each of three ranks and three of each of three regiments so that just one officer of each rank and each regiment is in each row and column. (b) Repeat part (a) for the 16-officer problem, with four ranks and four regiments. (c) Explain why there can be no solution to the 4-officer problem. 8.1.2. *(a) Describe how the As, Bs, and other letters in Table 8.1 relate to each other. Similarly describe the placement of the numbers. (b) Find a different solution to the 25-officer problem and describe any patterns you find in the placement of the letters and numbers. (c) Does Table 8.1 avoid the risk of confounding variables that Fisher needed for his statistical work? Repeat for your solution in part (b). Discuss your answers. *8.1.3. Modify Example 2 so there are nine girls. Find daily arrangements of nine girls in rows of three so that each pair of girls appears in the same row just once. How many days can they go walking before they repeat partners? 8.1.4. Consider generalizing Example 2 to a v-schoolgirl problem, where there are still three girls in each row and every two girls are in the same row just once. (a) Show that v must be an odd multiple of 3, say v = 6n + 3. (b) For v = 6n + 3, find the number of days that the girls can go for walks without any girl walking with another girl twice. (Combinatorics won’t guarantee the existence of such arrangements of the 6n + 3 girls, but the existence was proven in 1965 for all n.) 8.1.5. Modify Exercise 8.1.4 so that there are four girls in each row but that two girls are still in the same row just once. (a) Show that v, the total number of girls, must satisfy v = 12k + 4. *(b) Repeat Exercise 8.1.4(b) for v = 12k + 4. (c) Find a solution to the 16-schoolgirl problem with the number of days you found in part (b). 8.1.6. Generalize parts (a) and (b) of Exercise 8.1.5, where there are n girls in each row. 8.1.7. (a) Each of eight small cubes has all faces painted with same one of four colors. The cubes are stacked to make a bigger cube. Find an arrangement of the colors of the cubes so that each face of the larger cube has all four colors. The left part of Figure 8.1 illustrates the stack of cubes with one cube colored. (b) Modify part (a), where we have 27 small cubes each painted with one of nine colors and stacked to form a larger cube. The right part of Figure 8.1 illustrates the stacked cubes, with one colored cube. Find an arrangement of the colors so that each of the

377

8.1 Overview and History

faces and each of the interior parallel layers has all nine colors. (It’s convenient to use numbers instead of colors.) (c) Modify part (b), replacing 27 = 33 with 64 = 43 and replacing 9 = 32 with 16 = 42 . (d) Generalize part (b) by replacing 27 = 33 with n 3 and replacing 9 = 32 with n 2 .

Figure 8.1 stacked cubes. 8.1.8. A geometric figure G satisfies the following conditions. (i) (ii) (iii) (iv)

G is made of points, some pairs of which are adjacent. Every point of G is adjacent to two other points. No three points of G are mutually adjacent. In G, for any non-adjacent points A and B exactly one point is adjacent to A and B.

*(a) Find a figure satisfying all four conditions. How many points are there? (b) Repeat part (a) with three in condition (ii). (This figure is called the Petersen graph.) (c) Answer the following questions and then find a formula for the total number of points in a figure satisfying all four conditions, where we use a general value of k in (ii) instead of two or three. Hint: It may help to answer these questions for k = 2 and k = 3 before doing the general case. If there are n points and each point is adjacent to k points, how many adjacent pairs of points are there? How many pairs of points are there, adjacent or not? How many non-adjacent pairs are there? How many non-adjacent pairs can any one point connect to? (d) For various values of k, find other figures that satisfy conditions (i), (ii), and (iv) but not (iii).

8.1.2 Leonhard Euler The Swiss mathematician Leonhard Euler (1707–1783) was the most prolific mathematician of all time—it took the St. Petersburg Academy 47 years to finish publishing the more than two hundred papers left when he died. Euler finished his university degree in theology at age 15. Meanwhile he had started studying mathematics with Johann Bernoulli and published his first paper at age 18. He followed Bernoulli to the St. Petersburg Academy in Russia and stayed there until he was 34. Euler then went to Berlin for 25 years at the request of Frederick the Great. Euler returned to Russia in 1766 blind in one eye and soon became completely blind. However, his production of mathematics didn’t diminish. He dictated his papers and used his phenomenal memory to accomplish even the most complicated computations in his head. (Euler created the 36-officer problem, with its strong visual appeal, when he was blind.) Euler knew and made major contributions to all of mathematics and many of its applications in physics.

378

Finite Geometries

Euler, like mathematicians of his time, focused on particular problems leading to deeper, general insights. Euler’s solution of the K¨onigsberg Bridge problem in 1735 marks the beginning of graph theory, an area of mathematics closely tied to combinatorics. Euler’s formula and its generalizations, discussed in Section 1.5, are key theorems in graph theory and topology. Euler derived his formula in 1751. Euler’s many textbooks established the importance of functions in analytic geometry and calculus. His 1748 textbook introduced parametric equations (see Section 3.3). The problem of representing the curved surface of the earth on flat maps spurred many of Euler’s investigations. More generally, Euler made important contributions to differential geometry. He published on space curves, curvature of surfaces, and geodesics—shortest paths on surfaces.

8.1.3 Rev. Thomas Kirkman Nineteenth Century England produced a number of noted amateur mathematicians, including the Rev. Thomas Kirkman (1806–1895). Although his teachers recognized his strong intellectual potential, Kirkman’s father forced him to drop out of school at age 14. Nine years later Kirkman countered his father’s wishes, going to Ireland for his undergraduate education. He returned to England and became a pastor, serving a parish well into his eighties. Given the light demands of the job in those days, he could devote considerable time to his hobby and passion, mathematics. Kirkman published widely on topics broadly related to geometry. He wrote major papers on polyhedra and on transformation groups acting on sets. Kirkman had a fascination for individual problems and a flair for finding problems leading to more general insights. For example, he investigated what are now called Hamiltonian paths in graph theory before Hamilton did his work. He is best remembered for the schoolgirl problem discussed in Example 2. It was part of his investigations of structures that were later called Steiner triple systems, discussed in Section 8.3. He was upset that Steiner got credit for them, even though Kirkman had published on them six years before Steiner. At age 78 Kirkman tackled a new topic in mathematics—knot theory. A misguided idea by the famous physicist Lord Kelvin that atoms differed because they were different types of knots in space sparked interest in classifying knots. In addition to other questions about knots, Kirkman worked with Peter Guthrie Tait (1831–1901) to classify knots with up to ten crossings. While the physical motivation soon disappeared, knots have continued to fascinate mathematicians for their own sake. The theory has more recently found applications in biology and chemistry, for example to analyze knotted DNA and other molecules.

8.2 Affine and Projective Planes 8.2.1 Affine Planes Focusing on a few of the undefined terms and axioms of Euclidean or other geometries helps us understand them better. The restriction also leads naturally to geometries having finitely many points. With affine planes we explore relations of points and lines, especially parallelism, but

379

8.2 Affine and Projective Planes

not other Euclidean concepts, such as betweenness, distance, and congruence. (Of course, mathematicians have explored them in other geometric systems.) The geometries possess interesting regularities not relevant in Euclidean geometry. Axiomatic System: Affine Plane. Undefined Terms: point, line, on. Axioms: (i) Every two distinct points have exactly one line on them both. (ii) There are at least four points with no three on the same line. (iii) For every line and point not on that line, there is a unique line on that point that has no point in common with the given line. Exercise 8.2.1. Verify that the Euclidean plane is an affine plane. Definition. Two lines, k and l, are parallel, written k∥l, if and only if either k = l or no point is on both k and l. Definition. An affine plane with n points on each line is of order n. Example 1. Figure 8.2 gives two models of an affine plane of order 2, both with dots representing the four points and line segments representing the six lines. In the lefthand one, only the vertices of the square are points and the diagonals don’t intersect in a point of the model. That is, the diagonals are parallel, as are the pair of horizontal segments and the pair of vertical segments. Figure 8.3 depicts a model of order 3 with nine points and twelve lines. Some of the lines in Figure 8.3 have curved parts. In each figure, lines parallel to one another are drawn with the same kind of line for clarity. We can use Figure 8.3 to obtain solutions to the nine-schoolgirl problem in Exercise 8.1.3. For example, for the first day the rows could be the horizontal lines, for the second day the rows could be the vertical lines, and so on. ♦

Figure 8.2 An affine plane of order 2.

Figure 8.3 An affine plane of order 3.

After proving some initial, elementary properties of affine planes in Theorems 8.2.1 and 8.2.2, in Theorem 8.2.3 we show that all finite affine planes have an order, as stated in the definition. In Theorem 8.2.4 we use a combinatorial argument to reveal relationships between

380

Finite Geometries

the numbers of points and lines. However, counting arguments can’t guarantee the existence of affine planes. Theorem 8.2.1. In an affine plane (i) (ii) (iii) (iv) (v)

two distinct lines have at most one point in common, for every point there is a line not on that point, for every line there is a point not on that line, every line has at least two points on it, and every point is on at least three lines.

Proof. For parts (iii) and (iv), let k be a line. Let A, A1 , A2 , and A3 be the four points guaranteed by axiom (ii). Again by axiom (ii), at least one of the points, say A, is not on k, showing part (iii). From axiom (i) there are lines ki that are on A and Ai . Axiom (ii) ensures that they are distinct. By axiom (iii) at most one of them is parallel to k, and hence the other two intersect k. They already have A in common, so they must intersect k in different points by part (i). Thus k has at least two points on it. See Exercise 8.2.7 for the remaining parts. ! Theorem 8.2.2. Parallelism is an equivalence relation for lines in an affine plane. That is, for lines k, l, and m, three properties hold: reflexive, k∥k; symmetric, if k∥l, then l∥k; and transitive, if k∥l and l∥m, then k∥m. Proof. See Exercise 8.2.8. ! Theorem 8.2.3. If a line of an affine plane has n points on it, then each line has n points on it and each point has n + 1 lines on it. Exercise 8.2.2. Draw diagrams to illustrate the proof of Theorem 8.2.3. Proof. Let k be a line with n points on it, say, P1 , . . . , Pn , where n ≥ 2. First, let l be a line not parallel to k. By axiom (ii), there is a point Q on neither l nor k. By axiom (i), Q has exactly n lines on it that intersect k. In addition, by axiom (iii), Q has a parallel to k, for a total of n + 1 lines. Theorem 8.2.1 guarantees all are distinct lines. Axiom (iii) forces all but one of the lines to be on l, so l must have exactly n points on it. Now suppose that a line m is parallel to k. Let A be on k, B be on m, and j be the line on A and B. Then j is not parallel to either k or m. Now, from the first part of the proof, we know that j must have the same number of points as k and as m. Hence m has the same number of points as k. The preceding argument for the point Q shows that any point not on line k has n + 1 lines on it. For a point on line k, the same reasoning holds with a line not on that point. ! Exercise 8.2.3.∗ What happens in the proof of Theorem 8.2.3 for a line with infinitely many points on it?

381

8.2 Affine and Projective Planes

Theorem 8.2.4. In an affine plane if some line has n points on it, then there are n 2 points and n 2 + n lines, and each line has n lines parallel to it, including itself. Proof. Let P be a point. Then there are n + 1 lines on P and each has n − 1 points on it other than P by Theorem 8.2.3. Thus there are (n + 1)(n − 1) points besides P, giving a total of (n + 1)(n − 1) + 1 = n 2 points. See Exercise 8.2.9 for the rest of this proof. ! Example 2. There is no affine plane of order six. Solution. We show that an affine plane with six points on a line would provide a solution to Euler’s 36-officer problem, which we assume to be impossible. So, for a contradiction, suppose that there were an affine plane of order 6. We choose two nonparallel lines to determine the rows and columns of the officers and fix a point P (See Figure 8.4). Pick a line k through P other than a row or column. Put the officers from the first regiment on k and from each other regiment on one of the parallels of k. Then no two officers from the same regiment would be in the same row or column by part (i) of Theorem 8.2.1. Next, pick a line m through P other than k or a row or a column. (By Theorem 8.2.3 there are seven lines through each point, so we can find k and m.) Put the officers of each rank on one of the parallels of m. Again, we would have one officer from each rank in each row and column. This outcome contradicts the fact that there is no solution to Euler’s problem. Hence there can be no affine plane of order 6. For a direct proof, see Anderson [1, 91]. ♦

m

P k

Figure 8.4 Lines in a hypothetical affine plane of order 6. The idea behind Example 2 enabled Fisher to turn affine planes into a solution of the statistical design problems discussed in Section 8.1. But for which orders are there affine planes? By 1896, geometers had shown algebraically that affine planes (and projective planes) of order n exist if n = p k , where p is a prime number. In Section 8.4 we develop analytic geometries with this approach. No other orders of n are known to give planes, and many geometers believe that no other finite orders are possible. A theoretical result eliminates infinitely many other values of n. (See Dembowski [5, 144] for the statement of the Bruck-Ryser Theorem.) An extensive computer search in 1988 revealed that there is no affine (or projective) plane of order 10, the smallest order not previously decided. The possibilities for a computer to check increase exponentially with the possible order of a plane. So even the next few unknown orders, 12, 15, and 18, would require unreasonably long computer searches. Indeed, mathematicians need some pattern to find examples of finite geometries, except in small cases where a computer search is feasible. Proving which orders of n give planes is one of the key open problems in finite geometry. (See Mullen [8].)

382

Finite Geometries

8.2.2 Projective Planes Axiom (iii) for affine planes characterizes parallelism. Projective geometry focuses on a contrasting idea from perspective drawings in art, whereby parallel lines appear to meet at a point on the horizon. The revised axiom (iii) below characterizes projective planes, and is the only change from the axioms of affine planes. (The axioms here are the first three axioms of the real projective plane in Section 7.2.) Axiomatic system: Projective Plane. Undefined Terms: point, line, on. Axioms 8.2.2 (i) Every two distinct points have exactly one line on them both. (ii) There are at least four points with no three on the same line. (iii) Every two lines have at least one point on them both. Exercise 8.2.4.∗ Explain why spherical geometry isn’t a model of the axioms but single elliptic geometry from Section 4.5 and the projective plane from Section 7.3 are. Example 3. Figure 8.5 gives a model of a projective plane with three points on each line. It has seven dots, representing points, and seven segments or arcs, representing lines. Gino Fano developed the model in 1892. ♦

Figure 8.5 A projective plane with 7 points.

Theorem 8.2.5. In a projective plane (i) (ii) (iii) (iv)

every two distinct lines have exactly one point on them both, there are at least four lines with no three on the same point, every line is on at least three points, and every point is on at least three lines.

Proof. See Exercise 8.2.13. ! Projective planes have an important and aesthetically pleasing property called duality. Every theorem about points and lines remains a theorem when we switch the words point and line. The original theorem and the switched theorem are called duals of each other. The first two parts of Theorem 8.2.5 contain the duals of axioms (i) and (ii). (The dual of axiom (iii) is weaker than axiom (i), so the dual holds.) Because the duals of the axioms are all theorems, whenever we prove a theorem in projective geometry we immediately obtain its dual by switching the words point and line in the proof.

8.2 Affine and Projective Planes

383

Theorem 8.2.6. If one line of a projective plane has n + 1 points on it, then all lines have n + 1 points on them and all points have n + 1 lines on them. Proof. See Exercise 8.2.14. ! Theorem 8.2.7. If one line of a projective plane has n + 1 points on it, there are n 2 + n + 1 points and n 2 + n + 1 lines. Proof. See Exercise 8.2.15. ! An important and natural connection exists between an affine plane with n points on a line and a projective plane with n + 1 points on a line. The following definition builds on it, explained in the construction following the definition. Definition. A projective plane of order n has n + 1 points on a line. Theorems 8.2.4 and 8.2.7 show that a projective plane of order n has one more line and n + 1 more points than an affine plane of the same order. The following construction converts an affine plane into a projective plane of the same order by adding one line and its points. The result corresponds to the addition of a horizon line in perspective drawing. For an affine plane of order n, we collect the lines parallel to each other in a class. For each of the n + 1 classes of parallel lines, we add a new point that we define to be on each of them. We also define the n + 1 new points all to be on the same new line. Figure 8.6 illustrates the process, starting from the affine plane of order 3 on the left. Each of the horizontal segments is extended to intersect the new, rightmost point in the model on the right. We leave it as an exercise to verify that the affine points and lines, together with the new points and line, satisfy the axioms of a projective plane. The other direction works also. Take a projective plane and delete a line and the points on it. The remaining points and lines form an affine plane, where lines become parallel if they formerly intersected on the deleted line. (For more on affine and projective planes see Karteszi [7].)

Figure 8.6 Constructing a projective plane from an affine plane.

384

Finite Geometries

Exercise 8.2.5. Convert the affine plane in Figure 8.2 to the projective plane in Figure 8.5 using the construction given above.

8.2.3 Exercises for Section 8.2 8.2.6.

(a) Construct an affine plane of order 5 using a 5 × 5 array of dots. Figure 8.7 suggests one way to start constructing a set of lines parallel to the diagonal direction. *(b) Explain how to use an affine plane of order n to give a solution to a n 2 -schoolgirl problem with n girls in a row, as proposed in Exercise 8.1.6.

Figure 8.7 Parallel lines. 8.2.7. Prove the rest of Theorem 8.2.1. 8.2.8.

(a) Prove Theorem 8.2.2. *(b) If a line in an affine plane intersects one of two parallel lines, prove that it intersects the other.

8.2.9. Prove the rest of Theorem 8.2.4. 8.2.10. *(a) In the models illustrated in Figures 8.2 and 8.3 find the maximum number of points so that no three are on the same line. (b) In an affine plane of order n, prove there can be at most n + 2 points such that no three are on the same line. (c) Repeat part (b) for a projective plane of order n. 8.2.11. (a) For the thirteen-point projective plane in Figure 8.6 pick three collinear points to be P, Q, and R. Carry out the construction of S, the fourth point of the harmonic set described in Example 1 of Section 7.1. Do you get the fourth point on the line through P, Q, and R? (b) Find the three diagonal points of the complete quadrangle you constructed in part (a). Are they collinear? (See Exercise 7.2.18.) 8.2.12. (a) Use the affine plane of order 5 from Exercise 8.2.6 to construct a projective plane of order 5. (b) Repeat Exercise 8.2.11 for the projective plane of part (a). 8.2.13. Prove Theorem 8.2.5. 8.2.14. Prove Theorem 8.2.6. Hint: Modify the proof of Theorem 8.2.3. Also use duality. 8.2.15. Prove Theorem 8.2.7. Hint: Modify the proof of Theorem 8.2.4. Also use duality.

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8.3 Design Theory

8.2.16. *(a) For which axiom of affine planes is its dual provable? Prove it. (b) For the other axioms, show that their duals must be false. *(c) For which parts of theorems of affine planes are the duals provable? Prove them. 8.2.17. A weak projective plane satisfies projective axioms (i) and (iii) and the following replacements for axiom (ii). (ii′ ) There exist a point and a line not on that point. (ii′′ ) Every line has at least two points on it. *(a) Find a model of a weak projective plane with two points on each line. (b) Prove that duality holds in the revised axiomatic system. (c) Find a model of a weak projective plane with two lines having different numbers of points on them. (d) Give another model for part (c) with a different total number of points than your model in part (c). (e) Suppose we replace axiom (ii) for affine planes with the axioms (ii′ ) and (ii′′ ). Show that every such weak affine plane is an affine plane.

8.3 Design Theory Design theory encompasses many notions besides affine and projective geometries. Design theory developed from mathematicians’ natural inclination to generalize and explore as well as to meet the needs of applications, especially designs for statistical experiments. We concentrate on balanced incomplete block designs. Example 1. Imagine that a food company wants a taste test to compare eleven products, some potential new ones and some current products. Research has shown that individual tasters can’t reliably compare so many different tastes. So each taster will compare just five of the products. Yet every pair of products needs to be compared. Also, the organizer of the trials must guard against biases caused by which varieties are tasted together. Table 8.3, with eleven tasters (A to K) and eleven products (1 to 11), satisfies all the conditions. Each pair of products is compared by two tasters, who compare them with different combinations of products. Table 8.3 A: 1, 3, 4, 5, 9

B: 2, 4, 5, 6, 10

C: 3, 5, 6, 7, 11

D: 1, 4, 6, 7, 8

E: 2, 5, 7, 8, 9

F: 3, 6, 8, 9, 10

G: 4, 7, 9, 10, 11

H: 1, 5, 8, 10, 11

I: 1, 2, 6, 9, 11

J: 1, 2, 3, 7, 10

K: 2, 3, 4, 8, 11

Determine how many testers taste a given product. Look for a pattern to the arrangement of the products for the tasters. We’ll see in Example 5 a way to generate this and other designs. ♦ Fisher and others developed designs in which each block (for example, an individual tester) has the same number k of varieties and all pairs of varieties are compared the same number of times, λ. In Example 1, k = 5 and λ = 2. Such designs are called balanced incomplete block

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designs (BIBD). Balanced refers to the uniformity of the arrangement, and incomplete refers to the fact that no block includes every variety. Fisher proved a number of results about BIBDs. (See Anderson [1, Chapter 6].) Definition. A balanced incomplete block design (BIBD) is an arrangement of v varieties into blocks (subsets), each of size k, so that each pair of varieties appears in λ blocks. We write (v, k, λ) to describe the numerical type of a BIBD. We use b for the total number of blocks. Example 2. From Theorem 8.2.4, an affine plane of order n is a (n 2 , n, 1) BIBD, with the points as varieties and the lines as the blocks. The condition λ = 1 corresponds with axiom (i): “Every two distinct points have exactly one line on them both.” There are b = n 2 + n blocks altogether and each point (variety) is on the same number, r = n + 1, of lines (blocks). ♦ Example 3. From Theorem 8.2.7, a projective plane of order n is a (n 2 + n + 1, n + 1,1) BIBD, with the points as varieties and the lines as the blocks. There are b = n 2 + n + 1 lines and each point is on the same number, r = n + 1, of lines. ♦ Example 4. Given a (v, k, 1) BIBD we can make a (v, k, λ) BIBD by repeating each block of the original BIBD λ times. However, as in Example 1, experimenters often need to avoid repetition to gain additional information. Table 8.4 gives blocks for a (7, 3, 2) BIBD with no repetitions of blocks. The two lines are different (7, 3, 1) BIBDs. ♦ Table 8.4 124

235

346

457

156

267

137

126

237

134

245

356

467

157

Checking even the short table of Example 4 to verify that it is a (7, 3, 2) BIBD is tedious. It would be much worse with a larger design and harder still to find such a design by trial and error. Indeed, not every set of values (v, k, λ)—even those values which satisfy the combinatorial relations of Theorem 8.3.1—has a BIBD. Exercise 8.3.1.∗ Explain why there can be no (6, 3, 1) BIBD. Mathematicians and statisticians have found efficient ways to construct BIBDs and other designs. In Section 8.4 we use finite analytic geometries to construct many affine and projective spaces in any number of dimensions. Although other designs don’t have such representations, symmetry often aids their construction. Example 5. Figures 8.8 and 8.9, together with symmetry, provide the key to the BIBDs of Examples 4 and 1, respectively. The regular seven-sided polygon of Figure 8.8 has three distances between vertices, which we take as our varieties. The triangle in bold, △124, has each of them. When we rotate △124 to the seven possible positions, such as △235, shown with thin lines, we get seven blocks that form a (7, 3, 1) BIBD. The design is a projective plane of order 2. Further, △157, shown with dashed lines, is the mirror image of △124, so the same reasoning applies to it and rotations of it. The fourteen blocks of Example 4 are the rotations and mirror reflections

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of △124. The construction and the three different lengths of sides ensure that an edge (pair of vertices) appears in exactly two triangles. 1 11

2

1 7

3

10

2

9

4

3

6

8 5

5 7

4

6

Figure 8.8 Building a (7, 3, 2) BIBD using

Figure 8.9 Building an (11, 5, 2) BIBD using

rotation and reflection.

rotation.

The regular 11-gon of Figure 8.9 has five different distances between the vertices, which we take as our varieties. As you can verify, the pentagon 13459 and its diagonals together have each of these distances twice. We rotate it to get the eleven blocks of the (11, 5, 2) BIBD of Example 1. (We don’t need to know the actual distances between points. It is easier to use the differences, going either clockwise or counterclockwise. For example, the difference between 9 and 4 is 5, as is the difference between 9 and 3.) ♦ Before attempting to construct a BIBD with certain values, it pays to determine whether the values are compatible with the conditions of a BIBD. Theorem 8.3.1 gives necessary relations among the values for a BIBD. However, there is no guarantee, as Exercise 8.3.2 indicates, that values satisfying the relations always have a corresponding BIBD.

Theorem 8.3.1. In a (v, k, λ) BIBD, each variety is in the same number r of blocks, r (k − 1) = λ(v − 1) and v · r = b · k. Exercise 8.3.2. Verify that the values in Example 2 for an affine plane of order n satisfy Theorem 8.3.1. However, there is no affine plane of order 6. Proof. Let W be a variety of a BIBD and r W be the number of blocks containing W . Each block has k − 1 other varieties. So r W (k − 1) counts all appearances of the other varieties in blocks containing W . The value λ counts the number of times another variety appears with W . Thus λ(v − 1) also counts appearances of other varieties in blocks containing W . Hence r W (k − 1) = λ(v − 1) for every W . As v, k, and λ are constant, all varieties must be in the same number r W = r of blocks, and the first equation of the theorem holds. Similarly, vr = bk counts the number of times that a variety appears in a block in two ways: each of the v varieties appears r times, and each of the b blocks contains k varieties. !

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As mentioned in Example 2, a BIBD with λ = 1 satisfies the geometric axiom “Every two distinct points have exactly one line on them both.” In the 1800s several mathematicians investigated such systems as geometries long before the definition of a BIBD. A BIBD with three points per line (k = 3 and λ = 1) is called a Steiner triple system. In 1852, Jacob Steiner (1796–1863), for whom the BIBDs are named, investigated such systems. As too often happens in mathematics, the name doesn’t mean Steiner was first. Rev. Thomas Kirkman published on them in 1846 and Julius Pl¨ucker, a rival of Steiner, had published on them even earlier, by 1835. (See Anderson [1], Berman and Fryer [3], and Karteszi [7] for more on BIBDs and Steiner systems.) Theorem 8.3.2. There is a Steiner triple system (v, 3, 1) if and only if v = 6n + 1 or v = 6n + 3. Proof. Exercise 8.3.5 shows v must equal either 6n + 1 or 6n + 3. (See Anderson [1,112] for the existence of such systems.) ! Example 6. Is there a finite “hyperbolic” plane with three points on a line? Solution. The characteristic axiom of hyperbolic geometry says, “Given any line m and any point P not on m, there are at least two lines on P which do not intersect m.” Such a BIBD with three points per line would be a Steiner triple system in which each point (variety) has at least two more lines than there are points on a line. That is, r ≥ k + 2 = 5. A projective plane of order 2 gives a Steiner triple system with v = 7 and an affine plane of order 3 gives a Steiner triple system with v = 9. By Theorem 8.3.2, the smallest possible hyperbolic plane with three points on a line would have v = 13 points, giving a (13, 3, 1) BIBD. Theorem 8.3.1 in turn forces r = k + 3 = 6 lines on every point and a total of b = 26 lines. However, combinatorics give no clue about how to construct such a design. The vertices of a regular 13-gon (Figure 8.10) have six different distances or, more simply, differences among them. Figure 8.10 illustrates one way to split them between two types of triangles, such as △138 and △569. As in Example 5, we can use rotations to obtain all 26 lines for this geometry. I found the triangles by starting with the six differences. One way to split them up is {1, 3, 4} and {2, 5, 6}. The first triple corresponds 1

13

2 3

12

4

11

5

10 9

6 8

7

Figure 8.10 Building a (13, 3, 1) BIBD through rotation.

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to triangles like △569. For the second triple, note that 2 + 5 + 6 = 13, so the differences cycle around the polygon. ♦ But is the BIBD a plane? After all, 3-dimensional Euclidean space satisfies the characteristic axiom of hyperbolic geometry, given above, but isn’t a plane. A 3-dimensional space ought to have a proper subset qualifying as a 2-dimensional plane. (The 1-dimensional subsets are the lines, each with three points on it.) How could we tell whether the thirteen point design had a planar subset? Such a subset would need to be a smaller Steiner triple system and so have seven or nine points by Theorem 8.3.2. It would be tedious to check every combination of seven or nine points in the design to see whether they formed a projective or affine plane. Suppose, for a contradiction, that some proper subset S of seven points were a projective plane. Let P be a point not in S. There would need to be seven lines, each connecting P with one of the points of S. Each of the lines would have a third point, giving at least fifteen points. But the design has only thirteen points, a contradiction. So there can be no projective plane. A similar argument eliminates any affine plane. So the thirteen point Steiner triple system deserves to be called a hyperbolic plane.

8.3.1 Error-correcting Codes Advances in computers and their availability have vastly increased the amount of data transmitted as strings of 1s and 0s. Although computers are far more precise than human beings, errors in transmissions occasionally occur because of static or for other reasons. With the use of words in real language, we can often correct errors because the context of the message or the word helps us recognize where the error must be and what the correction should be. However, a string such as 0010110111 provides no clue by itself about possible errors. Instead we need to build into the transmission a code to aid in finding and correcting errors; BIBDs have important connections with such codes. One way to obtain error correction is to send each 0 or 1 three times, so we would have two code words: 000 or 111. The preceding message then would be sent as 000000111000111111000111111111. If just one of the three repetitions is accidentally altered, the receiving computer can correct the error by majority rule. If we assume that two or three errors in a triple are extremely unlikely compared to none or one, the code allows us to correct almost all errors. We can visualize this simple code by using the vertices of a cube, as in Figure 8.11. The code words are at opposite (0,0,1)

(0,1,1)

(1,0,1) (1,1,1) (0,0,0) (0,1,0) (1,0,0)

(1,1,0)

Figure 8.11 Representing potential code words.

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vertices, (0, 0, 0) and (1, 1, 1). An error in transmission would give us a different vertex of the cube. Three of them are a distance of 1 from (0, 0, 0) and further from (1, 1, 1). The other three are a distance of 1 from (1, 1, 1) and further from (0, 0, 0). We can correct the error by going to the closest code word. However, the price of error correcting with this code is using of three times as much data as the message requires. We can convert a BIBD into a matrix of 0s and 1s to construct a list of code words. The columns of the matrix, called an incidence matrix, are the varieties of the BIBD and the rows are the blocks. We put a 1 in an entry if the variety is in the block. Otherwise we put in a 0. For example, the top row of Example 3 has the incidence matrix in Table 8.5. Table 8.5 1

1

0

1

0

0

0

0

1

1

0

1

0

0

0

0

1

1

0

1

0

0

0

0

1

1

0

1

1

0

0

0

1

1

0

0

1

0

0

0

1

1

1

0

1

0

0

0

1

If we used the seven rows as code words, we would be able to detect a single error among the seven digits received. For example, 0010010 differs from 0011010 only in the fourth digit, but it differs from all the others in three or more digits. In the language of Definition 8.3.2, the Hamming distance between any two of the seven code words is 4, whereas 0010010 and 0011010 have a Hamming distance of 1. Definition. A code is a set of n-tuples of 0s and 1s, each of which is a code word. The Hamming distance between two n-tuples is the number of places in which they differ.

Theorem 8.3.3. A code can detect as many as k errors if the Hamming distance between two code words is at least k + 1. A code can correct as many as k errors if the Hamming distance between any two code words is at least 2k + 1. Proof. Suppose that all the code words are n-tuples. The number of errors in a received n-tuple is the number of places it differs from the code word sent, or the Hamming distance between them. Suppose that the Hamming distance between any two code words is at least k + 1. If between 1 and k errors occur, then the received n-tuple is not a code word and so will be detected as an error. Similarly, suppose that the Hamming distance between code words is at least 2k + 1. Then a received n-tuple with between 1 and k errors will have a smaller Hamming distance from the original code word than from any other code word, enabling us to correct it. !

8.3 Design Theory

391

Example 7. We can extend Table 8.5 to have more code words with a Hamming distance of at least 3 between any two of them. We can obtain seven more code words by switching the 0s and 1s of the code words in Table 8.5. For example, 1101000 becomes 0010111. In addition to the fourteen code words, we can also add in 0000000 and 1111111. The sixteen code words have a distance of at least 3 from the others. Could we do any better? Not with 7-tuples: each code word has seven other 7-tuple neighbors at a distance of 1 away. That makes 16 · 7 = 112 neighbors plus the sixteen code words, for a total of 128. With two choices in each of the seven coordinates of a 7-tuple, there are 27 = 128 possibilities, so every one is either a code word or a neighbor. ♦ Coding theory combines abstract algebra with design theory to develop more efficient codes than the incidence matrices of BIBDs. However, many of the codes are related to BIBDs. (See Anderson [1, Chapters 6 and 7] and Gallian [6, Chapter 31] for more information.)

8.3.2 Exercises for Section 8.3 8.3.3. *(a) Find a BIBD satisfying the values v = 5, k = 2, and λ = 1. Find b, the number of blocks, and r , the number of blocks on each variety. (b) Repeat part (a) with v = 6, k = 2, and λ = 1. (c) Explain why there is a BIBD for any value of v > 2 with k = 2 and λ = 1. Give formulas for b and r in terms of v. 8.3.4. *(a) For a Steiner triple system (v, 3, 1) give formulas for b and r in terms of v. (b) For a (v, k, λ) BIBD find a formula for the number of blocks b in terms of v, k, and λ. (c) Suppose λ = 1 and r = jk, a multiple of k. Find the values of v and b in terms of k and j so that (v, k, 1) satisfies the conditions of Theorem 8.3.1. (d) Repeat part (c) with λ = 2 and a suitably modified value of r as a multiple of k. 8.3.5. Fisher showed that in a BIBD that if k < v, then v ≤ b. (See Anderson [1, 85].) *(a) If you assume that v ≤ b, what else does Theorem 8.3.1 tell you? *(b) If b = v and λ = 1, find v, b, and r in terms of k. Now set n + 1 = k and find v in terms of n. Relate this result to projective planes. (BIBDs with b = v are called symmetric.) (c) Assume that k ≥ 3 and prove axioms (i) and (ii) of a projective plane for a symmetric BIBD (v, k, 1). Hint: For axiom (ii), show that there are three varieties (points) A, B, and C not all on the same block and count the number of varieties on the blocks (lines) they determine. Compare this number with v and recall that k ≥ 3. (d) Prove axiom (iii) of a projective plane for a symmetric (v, k, 1) BIBD, with k ≥ 3. Hint: Count the number of blocks meeting a given block in one of its varieties and then the total number of blocks meeting this block. (e) Find the value of v in terms of k for a symmetric (v, k, 2) BIBD. 8.3.6. *(a) In Theorem 8.3.2 prove that v = 6n + 1 or v = 6n + 3. Hint: Use factoring and Theorem 8.3.1 twice, first to show that v must be odd and then to eliminate v = 6n + 5.

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(b) Show that v = 6n + 1 and v = 6n + 3 are compatible with all conditions of Theorem 8.3.1. *8.3.7. Modify Example 5 to find a design using a regular 13-gon for a projective plane of order 3. Hint: You need to choose four vertices. How many different distances are there in a 13-gon? 8.3.8. (a) Modify Example 5 to find a Steiner triple system for v = 19. Hint. How many distances does a regular 19-gon have? How many sets of three vertices do you need? (b) Modify Example 5 to find a Steiner triple system for v = 21. Hint: Consider triangles △01x, △02y, △03z, and another triple. How many different distances are there? (c) Find a different Steiner triple system with v = 21 from the one in part (b). Hint: Consider triangles △013, △04x, △05y, and another triple. 8.3.9. In a round-robin tournament each player (or team) plays every other player (or team). The organizer wants to schedule as many matches at once as possible. If the number of players is even, say 2n, n matches at a time conceivably can be scheduled and thus the tournament arranged so that no one has to wait while others play. (a) Find a schedule for a round-robin tournament with four players and three rounds. (b) Find the number of rounds in a round-robin tournament with 2n players if there are n matches at a time. (c) Show for all even numbers 2n that there is a tournament schedule with no waiting as suggested in what follows. Start with a regular polygon with 2n − 1 vertices and the center of the polygon to obtain 2n points. Use an edge of the polygon and all of its parallel diagonals to make n − 1 matches. What is the nth match for that round? How many rounds can you make this way? How many do you need? Now describe the entire schedule for a tournament with 2n teams. 8.3.10 (a) A Steiner quadruple system is a (v, k, λ) BIBD, with k = 4 and λ = 1. Show that v = 12n + 1 or 12n + 4. Hint: See Exercise 8.3.6 and suppose v = 12n + j for some j from 0 to 11. Eliminate all values of j except 1 and 4. (b) Define a Steiner k-tuple system and find the possible values of v in terms of k. (c) Show that there is a Steiner double system for each v ≥ 2. 8.3.11. A projective plane of order 3 and an affine plane of order 4 are Steiner quadruple systems. What is the smallest Steiner quadruple system that could be 3-dimensional in the sense of Example 6? That is, it has a subset that is a projective plane. Justify your answer. What sizes of Steiner quadruple systems, if they exist, give hyperbolic planes? 8.3.12. (a) Repeat Exercise 8.3.11 for Steiner quintuple systems. (b) Generalize part (a) to Steiner k-tuple systems. (See Exercise 8.3.10.) 8.3.13. *(a) Write the eleven code words in the code based on the (11, 5, 2) BIBD of Table 8.3. How many errors can this code detect? How many can it correct? (b) Repeat part (a) for the code based on the affine plane of order 3, a (9, 3, 1) BIBD.

8.3 Design Theory

393

8.3.14. You can add rows to the incidence matrix from Exercise 8.3.13 (b) in a way that doesn’t give a BIBD, as Example 7 did for Table 8.5. For example, you can add rows that have four or more 1s. By Theorem 8.3.3, if the Hamming distance between two (new and old) code words is at least three, you will still be able to correct a single error while having more code words. (a) Explain why, if you want to be able to correct single errors, there is no point in adding code words with one, two, or three 1s. Hint: Consider axiom (i) of an affine plane. (b) As in Example 7, make new code words by switching 0s and 1s. What is the minimum Hamming distance of the code words from the other new or old code words? Justify your answer. Does your value hold if you add the all 0 and all 1 code words? Why or why not? (c) What is the total number of 9-tuples of 0s and 1s? Suppose that you have a set S with k code words, each a 9-tuple of 0s and 1s, and that each code word in S is a Hamming distance of at least 3 from every other code word. Let N be the set of neighbors of code words in S. That is, each 9-tuple in N is a distance of 1 from some code word in S. Find the size of N in terms of k. Given that the size of S plus the size of N must be smaller than the total number of 9-tuples, find the the maximum size of S. (d) Generalize part (c) for n-tuples of 0s and 1s.

8.3.3 Sir Ronald A. Fisher Sir Ronald A. Fisher (1890–1962) was one of the key people in establishing statistics as a mathematical discipline. Throughout his career he blended his interests in biology and statistics, building on his strong mathematical ability and insight. In 1911, as an undergraduate at Cambridge University he gave a talk marking him as one of the first to see how to combine Darwin’s theory of natural selection with Mendel’s recently rediscovered ideas in genetics. He strongly advocated eugenics, the movement to improve people’s genetic inheritance. (His involvement was innocent of and prior to the terrible use of eugenics ideas for political and racist ends in various countries, most notoriously Nazi Germany. Nevertheless, eugenics in general has morally unacceptable consequences that Fisher and others in those early years didn’t adequately consider.) From an early age his eyesight was poor, but he compensated using his geometrical intuition. For example, in high school Fisher studied spherical trigonometry entirely orally. His extraordinary visualization skills enabled him to perform all the three-dimensional thinking and computing without either a text or paper and pencil. His geometric intuition also characterized his statistical thinking. His first result, as well as many others, depended on considering a statistical sample of n points as a vector in n-dimensional Euclidean space. Many other statisticians, lacking Fisher’s geometric abilities, found his reasoning difficult to follow and felt he depended too much on intuition. He made fundamental contributions in many areas of statistics, including tests of hypotheses and the analysis of variance. He combined his statistical, mathematical, and biological interests throughout his career. In 1919, he became the statistician at Rothamsted Experimental Station, where daily practical

394

Finite Geometries

problems led him to a wide variety of theoretical discoveries in statistics. It was here that he realized the need for randomized sampling and the careful design of experiments to study the interaction of variables. His biological experiments led to development of statistical design theory.

8.4 Finite Analytic Geometry We can imitate the analytic geometry and transformations presented in Chapters 5 and 7 for finite geometries whenever we have algebraic structures corresponding to the arithmetic of the real numbers. Finite fields provide the analog that we use, although a more general approach is possible. (See Blumenthal [4].) Fields are number systems having many of the familiar properties of the four usual operations of addition, subtraction, multiplication, and division. Although there are other fields, we concentrate on the fields of integers modulo a prime number. (See Gallian [6] for more information on fields.) Example 1. Let Z3 be the set {0, 1, 2} together with addition and multiplication modulo 3. That is, after doing the usual arithmetic, we subtract multiples of 3 until we get back to a number in Z3 . For example, 2 + 2 = 4 becomes 1 (mod 3) because 4 − 3 = 1. We write 2 + 2 ≡ 1 (mod 3). Think of the three numbers placed around a circle (Figure 8.12). Then addition corresponds to clockwise rotation: for 2 + 2, we start at the 2 on the circle and go two steps clockwise around the circle to 1. Multiplication is, as in ordinary arithmetic, repeated addition. Tables 8.6 and 8.7 give additions and multiplications. Z3 is a field. ♦ 0

2

1

Figure 8.12 A representation of Z3 . Table 8.6

Table 8.7

+

0

1

2

0

0

1

2

1

1

2

2

2

0

×

0

1

2

0

0

0

0

0

1

0

1

2

1

2

0

2

1

Example 2. Let Z5 be the set {0, 1, 2, 3, 4} with addition and multiplication modulo 5, given in Tables 8.8 and 8.9. For example, 3 × 4 ≡ 2 (mod 5) because 3 × 4 = 12 ≡ 12 − 5 − 5 = 2 (mod 5). Z5 is a field. As in Example 1, we can place the numbers of Z5 clockwise around a circle, as in Figure 8.13, and then addition corresponds to clockwise rotation. The numbers 1 and 4 are additive inverses or, more informally, negatives of each other. Adding 4 is the same as subtracting 1. Similarly, 2 and 3 are additive inverses of each other. ♦

395

8.4 Finite Analytic Geometry

Table 8.9

Table 8.8 +

0

1

2

3

4

0

0

1

2

3

4

1

1

2

3

4

2

2

3

4

3

3

4

4

4

0

×

0

1

2

3

4

0

0

0

0

0

0

0

1

0

1

2

3

4

0

1

2

0

2

4

1

3

0

1

2

3

0

3

1

4

2

1

2

3

4

0

4

3

2

1

0 4

1

3

2

Figure 8.13 A representation of Z5 . Definition. By Zn we mean the set {0, 1, . . . , n − 1} with addition and multiplication modulo n. That is, after doing the usual arithmetic, we subtract multiples of n until we get back to a number in Zn . Example 3. Although Z4 satisfies many familiar algebraic properties, it misses one of the defining properties to qualify as a field. The number 2 doesn’t have a multiplicative inverse. That is, none of the numbers in the set {0, 1, 2, 3} when multiplied by 2 and reduced modulo 4 give us 1, the multiplicative identity. Thus we can’t divide by 2. We leave the addition and multiplication tables for Z4 as an exercise. ♦ Theorem 8.4.1. Z p is a field if and only if p is a prime. There is, up to isomorphism, exactly one field with pk elements, where p is a prime number. Proof. See Gallian [6, 213 and 328]. !

8.4.1 Finite Analytic Planes We can use any field to form an analytic model of an affine or projective geometry, just as we did with the real numbers in Chapters 5 and 7. The number of elements in a field is its order and equals the order of the corresponding affine or projective plane. Using algebra, we can develop matrices with entries from a field to describe the affine transformations and collineations of the corresponding geometries. As in Chapter 5, we need three coordinates for points in a plane so that the transformations can move the origin. For a field F, F3 is the three-dimensional vector

396

Finite Geometries

space over F. Readers familiar with abstract algebra can show that the affine planes defined (and projective planes defined later) satisfy the axioms of Section 8.2. The proofs that they satisfy the axioms follow the proofs for the usual analytic models using the real numbers. (See Karteszi [7].) Definition. Given a field F, define AF2 , the affine plane over F, as follows. The points of AF2 are column vectors (x, y, 1) of F3 , and the lines are row vectors [a, b, c] from F3 , where a and b are not both 0. Two row vectors represent the same line if and only if one is a scalar multiple of the other by a nonzero element of F. A point (x, y, 1) is on a line [a, b, c] if and only if ax + by + c1 = 0 in F. Example 4. The 25 vectors of AZ25 form an affine plane of order 5. Figure 8.14 illustrates the five lines of the form [m, 4, 0] through the origin, more familiarly known as y = mx, as well as the vertical line x = 0 or [1, 0, 0]. The thin line segments indicate horizontal line [0, 4, 0], the diagonal line [1, 4, 0], and the vertical line [1, 0, 0]. The bold line indicates the points on [2, 4, 0], the dashed line outlines [3, 4, 0], and the double line indicates [4, 4, 0]. Every point (x, y, 1) has six lines on it, each parallel to one of them. Lines that are neither vertical nor horizontal cycle in both directions, as the numbers in Z5 cycle. The surface of an inner tube, more formally known as a torus, as in Figure 8.15, helps visualization of the cycling. [1,0,0] [1,4,0] [2,4,0] [3,4,0] [4,4,0] [0,4,0]

Figure 8.14 Lines in an affine plane of order 5.

Figure 8.15 A representation of an affine plane on a torus.

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8.4 Finite Analytic Geometry

To find the line through points (2, 2, 1) and (1, 4, 1), we solve the system of two equations The solution is [3, 4, 1] or any of its multiples, modulo 5, such as [2 · 3, 2 · 4, 2 · 1] ≡ [1, 3, 2] (mod 5). (The Euclidean line through (2, 2) and (1, 4) is y = −2x + 6, which becomes [−2, −1, 6] and is equivalent to [3, 4, 1] (mod 5).) ♦ a · 2 + b · 2 + c · 1 ≡ 0 (mod 5) a · 1 + b · 4 + c · 1 ≡ 0 (mod 5) .

Exercise 8.4.1.∗ Give algebraic conditions for two lines [a, b, c] and [a ′ , b′ , c′ ] to be parallel in an affine plane. Pick two different parallel lines in AZ25 and show that they have no point in common. As in Chapter 5, we represent affine transformations by 3 × 3 invertible matrices whose bottom row is [0 0 1]. For a field of order n, there are at most n 6 of them because only six entries aren’t fixed, but not all are invertible. If the determinant of a matrix is not 0 (mod n), then the matrix is invertible. We use a combinatorial argument to find the number of invertible matrices with a bottom row of [0 0 1]. Again, as in Chapter 5, the images of points O = (0, 0, 1), X = (1, 0, 1), and Y = (0, 1, 1) determine an affine transformation. The condition that the matrix is invertible implies that the points must be mapped to three distinct points not all on the same line. For the field of order n, the affine plane has n 2 points and (0, 0, 1) can be mapped to any of them. Once we know where (0, 0, 1) goes, (1, 0, 1) has n 2 − 1 places to go. For (0, 1, 1) there remain n 2 − n places to go because it can’t be mapped to any of the n points on the line through the other two points. Thus there are n 2 (n 2 − 1)(n 2 − n) = n 6 − n 5 − n 4 + n 3 affine plane transformations over the field of order n. Definition. For a field F, define PF2 , the projective plane over F, as follows. The points of PF2 are the nonzero column vectors (x, y, z) of F3 , where two vectors represent the same point if and only if one is a scalar multiple of the other by a nonzero element of F. The lines of PF2 are nonzero row vectors [a, b, c] from F3 , where two row vectors represent the same line if and only if one is a scalar multiple of the other by a nonzero element of F. A point (x, y, z) is on a line [a, b, c] if and only if ax + by + cz = 0. Example 5. Figure 8.16 adds coordinates to Figure 8.6 to illustrate the projective plane PZ23 . The vector space Z33 has 33 = 27 vectors, but there are two nonzero scalars, 1 and 2. So the (0,1,0) (1,2,0)

(1,1,0)

(0,2,1) (1,2,1)

(0,1,1)

(2,2,1)

(1,1,1) (2,1,1)

(0,0,1) (1,0,1) (2,0,1)

Figure 8.16 Coordinates in PZ23 .

(1,0,0)

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Finite Geometries

26 nonzero vectors pair up to give thirteen points. The arching line on the top has coordinates [0, 0, 1] corresponding to the ideal line z = 0 of the real projective plane from Section 7.3. The third coordinate of the points on it is 0, matching the equation z = 0. The other nine points have a nonzero third coordinate. As there are four points on each line, we can define both H (AB, C D) and AB//C D from Chapter 7 if and only if A, B, C, and D are distinct collinear points. With these definitions, the model satisfies axioms (i)–(viii) of Chapter 7. ♦ Exercise 8.4.2.∗ Verify that axioms (ix) and (x) of Chapter 7 fail in PZ23 . The transformations of the projective plane, called collineations, can be represented by invertible 3 × 3 matrices. As before, two matrices represent the same collineation if one is a nonzero scalar multiple of the other. As in Chapter 7, a collineation is determined by where it sends four points, no three of which are collinear. Exercise 8.4.10 shows that (n 2 + n + 1)(n 2 + n)(n 2 )(n − 1)2 = n 8 − n 6 − n 5 + n 3 collineations exist for the projective plane over a field of order n. Theorem 8.4.2. The set of affine transformations for AF2 and the collineations for PF2 form a group. Proof. We replace the specific field R with the general field F in the proofs of Theorem 5.4.5 and Theorem 7.4.4. !

8.4.2 Ovals in Finite Projective Planes The preceding paragraphs indicate how well finite planes mimic many familiar geometric concepts. Mathematicians also have explored other traditional geometric concepts in a finite setting. Geometers use the word oval for the finite analog of a conic. We start from the Euclidean property that no three points on a conic are collinear. Figure 8.17 shows a set of six points in AZ25 with no three on the same line. Some exploring will reveal that no other point can be added to this set. In AZ2p and in PZ2p , with p > 2, we can never find more than p + 1 points with no three points collinear. (When p = 2, all four points in AZ22 form a set with no three points collinear. We omit this special case.) Conveniently, the sets in PZ2p correspond to second-degree equations. For example, the six points in Figure 8.17 satisfy the equation 2x 2 + y 2 + 2x z + yz + 4z 2 = 0. A second-degree equation in an affine plane might give fewer than the expected number of

(1,3,1) (0,2,1) (1,1,1)

(3,3,1) (4,2,1) (3,1,1)

Figure 8.17 An oval in AZ25 .

8.4 Finite Analytic Geometry

399

points because an affine plane can be seen as the corresponding projective plane without a line and its points. Therefore ovals, the analog of conics, are usually defined for projective planes. (See Beck et al. [2] and Karteszi [7, 110–119] for more information on ovals in projective planes and finite analytic geometry.) Definition. In a projective plane of order n, with n odd, an oval is a set of n + 1 points, no three of which are collinear. A line is tangent to an oval if and only if the line and the oval have just one point in common. Example 6. Find the tangent at each point of the oval 2x 2 + y 2 + 2x z + yz + 4z 2 = 0 in PZ25 . Solution. From Figure 8.17 we can guess that the tangents to this oval through the points (0, 2, 1) and (4, 2, 1) are vertical. Indeed, the lines [1, 0, 0] and [1, 0, 1] are tangents at these points. For the point (1, 1, 1) we can eliminate the lines through it and any of the other points on the oval. From Figure 8.17 the lines have slopes 0, 1, −1 ≡ 4 (mod 5), and 2, as well as the vertical line. So the slope must be 3 and the tangent through (1, 1, 1) is [3, 4, 3]. We can find the other tangents similarly or use symmetry. The tangent through (1, 3, 1) is [2, 4, 1], through (3, 3, 1) is [3, 4, 2], and through (3, 1, 1) is [2, 4, 0]. ♦ Exercise 8.4.3.∗ Find the five affine points of AZ25 on the conic x 2 + 4yz = 0, more familiarly written as y = x 2 . Find the sixth point of the conic in PZ25 . Theorem 8.4.3. Every point on an oval in a projective plane of odd order has exactly one tangent to the oval. Proof. Let P0 be a point on an oval and P1 , . . . , Pn be the other points on it. The definition of an oval guarantees that the lines P0 Pi , for i > 0, are distinct, accounting for n of the n + 1 lines through P0 . The remaining line cannot intersect the oval except at P0 , so it must be a tangent. !

8.4.3 Finite Analytic Spaces As in Sections 5.5 and 7.6 we can build higher dimensional finite spaces using higher dimensional vector spaces over finite fields. It is easier to define projective spaces first and then derive affine spaces from them. Definition. The d-dimensional projective space over the field F, written PFd , has for points one-dimensional subspaces of Fd+1 , lines two-dimensional subspaces, planes three-dimensional subspaces, and so on. We can represent a point as (x1, x2 , . . . , xd , xd+1 ), where not all of the xi are 0 and two such (d + 1)-tuples represent the same point if one is a nonzero multiple of the other. A collineation is an invertible (d + 1) × (d + 1) matrix over F, where two matrices represent the same collineation when one is a nonzero multiple of the other. Example 7. Figure 8.18 illustrates how to represent the fifteen points of PZ32 on a tetrahedron. The points are the four vertices, the six midpoints of edges, the four centers of faces, and the center of gravity of the entire tetrahedron. The vertices have a 1 in just one coordinate. The midpoint between two vertices has 1s in the corresponding coordinates. Algebraically, the midpoint is the sum of the two vertices it is between. Because the arithmetic is (mod 2), one of

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the vertices is also the sum of the midpoint and the other vertex. The points are on a line. Other line segments of the tetrahedron with three points on them are lines in the projective space, as are some other combinations of three points, for a total of 35 lines. The center of a face has 1s in the coordinates corresponding to the vertices on it. The four faces are four of the fifteen planes of the space. The center of gravity has coordinates (1, 1, 1, 1). The plane x1 = 0 faces the reader in Figure 8.18 and has all its seven points labeled and seven lines drawn in. The plane x4 = 0 is set at an angle to the reader with all its points labeled and six of its lines drawn in. This projective space is a (15, 3, 1) BIBD. ♦ (0,1,0,0) (1,1,0,0) (0,1,0,1)

(0,1,1,0)

(1,1,1,0) (1,0,0,0)

(0,1,1,1) (0,0,0,1)

(0,0,1,1)

(1,0,1,0) (0,0,1,0)

Figure 8.18 PZ32 as a tetrahedron. Example 8. The solution in Example 2 of Section 8.1 to Kirkman’s schoolgirl problem comes from the projective space of Example 7. We write the numbers from 1 to 15 in base 2 and convert them into coordinates, partially represented in Table 8.10. Table 8.10 number base 2

1 12

2

3

4

8

15

102

112

1002

10002

11112

point

(0, 0, 0, 1)

(0, 0, 1, 0)

(0, 0, 1, 1)

(0, 1, 0, 0)

(1, 0, 0, 0)

(1, 1, 1, 1)

Then the 35 rows with three girls each correspond to lines in the geometry. The five lines for each day are skew in this space—no two intersect. The collineation ⎡ ⎤ 0 1 0 0 ⎢1 0 1 0⎥ ⎢ ⎥ ⎣1 0 0 0⎦ 0 0 0 1

takes the lines making up Sunday to the lines of Monday. (The order of the days is a bit mixed up so that the first column of numbers can have a simple pattern.) ♦ Definition. The d-dimensional affine space over the field F, written AFd , has points (x1, x2 , . . . , xd , 1). Its lines are the nonempty intersections of lines of PFd with AFd . Theorem 8.4.4. For d ≥ 2, an affine space AZdn is a (n d , n, 1) BIBD.

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8.4 Finite Analytic Geometry

Proof. See Project 14. ! Example 9. Use an affine space to model the game of Set. Solution. In the game of Set each card has objects on it characterized by four properties: the number of objects, their color, their shape, and their shading. For each property there are three choices, so a full deck with one card of each possibility has 34 = 81 cards. The four-dimensional affine space AZ43 over Z3 has 81 vectors and provides the natural geometric setting to analyze the game. We represent a card by a point (x, y, z, t) in the space, where x corresponds to the number of objects on the card, y to their color, z to their shape, and t to their shading. A set is a collection of three cards so that for each property the three cards agree on it or they all disagree on it. For example, one possible set has three cards all with solidly colored ovals, but one card has one red oval, another has two green ovals, and the third has three purple ovals. We use the elements from Z3 to designate the options for the properties. For example, use x = 1, 2, and 0 when there are one, two or three objects. Similarly, for purple y = 1, red y = 2, green y = 0, for oval z = 0, and for solid t = 2. Then the set matching the previous description would be {(1, 2, 0, 2), (2, 0, 0, 2), (0, 1, 0, 2)}. What isn’t allowed is a collection of cards where for one of the properties, say color, two of the three cards have the same color and the third card has a different color. The sets in the game of Set correspond to lines in the space. How many sets are possible in this game? By Theorem 8.4.3 the space is a (81, 3, 1) BIBD. From Theorem 8.3.1 the number of sets (lines) is b = vr/k = 81(40)/3 = 1080. ♦

8.4.4 Exercises for Section 8.4 *8.4.4. Use the affine plane AZ23 in this problem. (a) (b) (c) (d)

Find the line on (1, 2, 1) and (2, 0, 1). Find the intersection of the line in part (a) with [2, 2, 2]. Find the coordinates of the line parallel to [2, 2, 2] through the point (2, 2, 1). Redo parts (a), (b), and (c) in ordinary analytic geometry, using the real numbers. Do the coordinates of the line you get convert (mod 3) to the answer in part (a)?

8.4.5. Repeat Exercise 8.4.4, using the affine plane AZ25 .

50

8.4.6. *(a) Draw a picture to illustrate the effect of the affine transformation

1 0 0

1 0

0 0 1

6

on

the affine plane AZ23 . Is it similar to the effect of the same matrix on the real affine plane? Explain. 5 0 −1 2 6 50 2 26 *(b) Repeat part (a) for 1 0 0 . Compare with the real matrix 1 0 0 , making 0

0

1

use of the congruence 6 (mod 3). 5 2 ≡ −1

(c) Repeat part (a) for (d) Repeat part (a) for

2 0 5 01 0 0

0 2 0 0 1 0

0 0 , 16 1 1 . 1

and use the real matrix

0

5 −1 0 0

0 −1 0

0

1

6 0

0 1

.

8.4.7. Repeat Exercise 8.4.6, using the affine plane AZ25 . For parts (b) and (c), replace each 2 in the matrices with 4 since 4 ≡ −1 (mod 5). 8.4.8. *(a) Use calculus and real numbers to find the tangents to y = x 2 for x = 0, 1, 2, 3, 4, 5, and 6 in usual analytic geometry.

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Finite Geometries

*(b) Convert the equations from part (a) into coordinates of lines in AZ25 . Are the lines tangent to the oval in Exercise 8.4.3? Note that 5 ≡ 0 (mod 5) and 6 ≡ 1 (mod 5). (c) Repeat part (b) for AZ27 . Make a graph in AZ27 of the points on the oval and their tangents. 8.4.9. Suppose that you try to form an affine plane using Z4 , which isn’t a field. *(a) *(b) (c) (d) (e)

Find a line [a, b, c] that doesn’t have four points on it. Find two distinct lines with more than one point of intersection. Find two lines with different slopes but no points of intersection. Which of the axioms of an affine plane fail when you try to use Z 4 ? Repeat parts (a)–(d) for Z6 , replacing four with six in part (a).

8.4.10. Use the projective plane PZ25 in this problem. *(a) Find the point on lines [2, 3, 1] and [3, 1, 4]. *(b) Find the line through (1, 3, 1) and the point from part (a). (c) The four points A = (0, 0, 1), B = (1, 0, 1), C = (2, 2, 1), and D = (3, 1, 1) are the vertices of a complete quadrangle. Find the six lines they determine and find the three diagonal points. Are the diagonal points collinear? (See Section 7.1 for complete quadrangles and Exercise 7.2.18 for diagonal points.) *(d) For the collinear points P = (0, 0, 1), Q = (1, 0, 1), and R = (1, 0, 0), find a point S on the line through them so that H (P Q, RS), as defined in Section 7.1. Is S unique? 8.4.11. Repeat Exercise 8.4.10 using the projective plane PZ27 . 52 1 16 8.4.12. Let M = 1 2 1 be a collineation in PZ25 . 1

1

2

*(a) Find the image under M of the lines in part (a) of Exercise 8.4.10. Find the image under M of the point in part (a). Verify that the image of the point is on the image of these lines. Hint: What is M · M? Also, see Theorem 7.4.3. (b) Find the image under M of the point and line of part (b) of Exercise 8.4.10. 51 5 56 8.4.13. Let T = 5 5 1 be a collineation in PZ27 . 5

1

5

(a) Find the image under T of the lines in part (a) of Exercise 8.4.10. Find the image under T of the point in Exercise 8.4.11 part (a). Verify that the image of the point is on the image of these lines. Hint: What is M · M? Also, see Theorem 7.4.3 and recall that a nonzero scalar multiple of M · M is the same collineation as M · M. (b) Find the image under T of the point of part (b) of Exercise 8.4.10 and the line of part (b) of Exercise 8.4.11.

8.4.14. Define a projectivity in PF2 . (See Section 7.4.) Count the number of projectivities if the field has order n. Hint: See Theorem 7.4.1. 8.4.15. Show that (n 2 + n + 1)(n 2 + n)(n 2 )(n − 1)2 = n 8 − n 6 − n 5 + n 3 collineations exist in a projective plane over a field of order n.

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8.4 Finite Analytic Geometry

50 1 06 8.4.16. *(a) Find the fixed points and stable lines of M = 1 0 0 in PZ25 . Compare your 0 0 1 answers with the answer you’d get thinking of the matrix as an isometry from Chapter 5. 54 0 06 (b) Repeat part (a) for the matrix R = 0 4 0 . Recall: −1 ≡ 4 (mod 5). 0 0 1 5 6 2 0 2 8.4.17. (a) Find the fixed points and stable lines of S = 0 2 0 in PZ27 . Compare your 0

0

1

0

0

1

answers with the answer you’d get thinking of the matrix as a similarity from Chapter 5. 50 1 26 (b) Find the fixed points and stable lines of G = 1 0 0 in PZ27 . Compare your

answers with the answer you’d get thinking of the matrix as an isometry from Chapter 5.

8.4.18. *(a) Find the four points on the oval x 2 + y 2 − z 2 = 0 in PZ23 . (For simplicity, assume throughout that z = 0 or z = 1.) Graph these points. (b) Repeat for x 2 + y 2 − 2z 2 = 0 in PZ23 . (c) In non-homogeneous real coordinates we think of x 2 + y 2 − w = 0 as a circle of √ radius w. Do the ovals in parts (a) and (b) seem like concentric circles? What would be their center? Describe the points not on either of the ovals in parts (a) and (b). How do the degenerate ovals x 2 + y 2 = 0 and z 2 = 0 relate to the points not on the ovals of parts (a) and (b)? 8.4.19. (a) For each value of w from 1 to 6, find the eight points on the oval x 2 + y 2 − wz 2 = 0 in PZ27 . (b) Repeat Problem 13 part (c) for the ovals in part (a). *8.4.20. Find the points (x, y, 1) in the affine plane AF25 on the ovals x 2 + y 2 = 1, x 2 + 4y = 0, and x 2 + 3y 2 = 1. Explain the difference in numbers of points. 8.4.21. A plane in PF3 can be represented as a row vector [a, b, c, d], where the point (x, y, z, t) is on the plane if and only if ax + by + cz + dt = 0. Justify your work for each part. *(a) Find the plane in PZ35 through the points (1, 0, 0, 1), (0, 1, 0, 1), and (0, 0, 1, 1). (b) Repeat part (a) for the points (1, 1, 0, 1), (1, 2, 1, 1), and (3, 3, 3, 1). (c) The planes of parts (a) and (b) intersect in a line. Find at least two points on it. 8.4.22. *(a) (b) (c) (d)

Find the number of lines in AZ35 and PZ35 . Find the number of lines in AF3 and PF3 , where F is a field with n elements. Find the number of lines in the d-dimensional spaces AZd5 and PZd5 . Find the number of lines in the d-dimensional spaces AFd and PFd , where F is a field with n elements.

8.4.23. *(a) In the game of Set find the third card making a set with (0, 1, 1, 2) and (1, 0, 1, 2). (b) Repeat part (a) with the cards (2, 0, 1, 1) and (1, 1, 2, 0). (c) What property does a + e + i satisfy if (a, b, c, d), (e, f, g, h), and (i, j, k, l) form a set? Repeat for the other coordinates. (d) Use part (c) to prove that for every two cards there is exactly one other card forming a set with them.

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Finite Geometries

(e) The cards (1, 1, 2, 0), (1, 0, 2, 1), and (2, 2, 0, 1) do not form a set. Find the affine plane of nine cards they determine, where a card (a, b, c, d) is in this plane if and only if it forms a set with two cards already in the plane. Relate this idea of a plane to the geometrical notion of a plane. (f) Find largest collection of cards you can that do not form a set.

8.4.5 Projects for Chapter 8 1. (a) In a BIBD explain why, if λ > 0, then k ≥ 2. Explain why BIBDs with k = 2 are easy to construct but not particularly interesting. (b) Use Theorem 8.3.1 to find all triples (v, k, λ) with v ≤ 25 for which there could be a (v, k, λ) BIBD with v > k > 2 and λ = 1. Hint: Explain why k ≤ 5. (c) For each triple (v, k, λ) in part (b), try to construct a BIBD. Explore if different (nonisomorphic) BIBDs can have the same values for v, k, and λ. (d) Try to construct (v, k, 2) BIBDs for v ≤ 19 that aren’t built from (v, k, 1) BIBDs by simple repetition. 2. Explore the affine and projective planes AF24 and PF24 , where F4 is the field of order 4 with the addition and multiplication given in tables 8.11 and 8.12. Table 8.11 + 0 0 0 1 1 a a b b

1 1 0 b a

a a b 0 1

Table 8.12 b b a 1 0

× 0 1 0 0 0 1 0 1 a 0 a b 0 b

a 0 a b 1

b 0 b 1 a

3. (a) Define planes in the affine space AF3 , where F is a field of order n. How many points are in them? How many lines are in them? (b) For a given line in AF3 , count the number of planes on it. (c) Count the number of planes in AF3 . (d) Count the number of planes through a given point in AF3 . 4. Let F be a field with n elements. (a) Count the number of affine transformations in AF3 . (b) Count the number of collineations in PF3 . (c) Find a formula for the number of affine transformations in AFd , the d-dimensional space over a field with n elements. (d) Find a formula for the number of collineations in PFd . 5. Program a computer to search for projective planes of orders 4, 5, and higher, using the symmetries of a regular polygon as in Section 8.3. 6. Define a point to be exterior to an oval if two tangents to the oval are on it. Define a point to be interior to an oval if no tangents to the oval are on it. Choose ovals in PZ2p and find their interior and exterior points. Look for formulas counting the number of exterior and interior points for an oval in PZ2p . Prove them.

8.4 Finite Analytic Geometry

405

7. Explore using the interior points of an oval as the points for a Klein model of a finite hyperbolic plane. (See Project 6.) 8. Define and explore ovoids, surfaces in a finite projective space PZ3p analogous to quadrics. (See Problem 4 of Section 7.6.) 9. Explore Desargues’ theorem (see Section 7.1) in PZ2p for various p > 2. (Explain why Desargues’ theorem is not interesting in PZ22 .) (Smart [9] has an axiomatization of Desargues’ configuration, which is shown in Figure 7.5.) 10. Develop and explore a finite analog to spherical geometry. (The analog might not be a BIBD.) 11. The axiomatic system for a weak hyperbolic plane has the undefined terms point, line, and on and the following axioms. (i) Every two distinct points have exactly one line on them both. (ii) There are at least four points with no three on the same line. (iii) Given a line m and a point P not on m, there are at least two lines on P with no point in common with m. (a) Show that a set of n ≥ 5 points is a weak hyperbolic plane if you take the lines to be all subsets of two elements. (b) Find a model of a weak hyperbolic plane with some lines having only one point on them. (c) Find a model of a weak hyperbolic plane with six points, every line with at least two points on it, and two lines with different numbers of points on them. Define a hyperbolic plane to be a weak hyperbolic plane satisfying this stronger version of axiom (iii): (iii′ ) Given a line with n points on it and a point not on that line, there are exactly n lines through that point which do not have any point in common with the given line. (d) Prove: If one line of a hyperbolic plane has n points on it, then all do. Find the number of points and lines in terms of n and prove your answers. (e) Are your results in part (c) consistent with Theorem 8.3.1? Explain. (f) Use a computer to find a model of a hyperbolic plane similar to that in Example 5 of Section 8.3, but with k = 5. 12. We use the following notion of distance to explore distance and isometries in finite affine planes. (a) For a prime p with p = 4n + 3, define the distance in AZ2p between (s, t, 1) and (u, v, 1) to be (s − u)2 + (t − v)2 (mod p). For p = 3 and p = 7, verify that two distinct points have a nonzero distance between them. How many points are at each distance from a given point? Do the points at a distance from a point form an oval (“circle”)? If so, what is its equation? (b) Determine which affine transformations of AZ2p are isometries; that is, they preserve the distances of part (a). Relate the isometries to the isometries of Section 5.3. Hint: In Section 5.5, an affine matrix was defined to be an isometry if and only if its upper

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Finite Geometries

(c)

(d)

(e) (f)

left submatrix was orthogonal. Count the number of isometries in AZ2p . Prove that the isometries form a group. For a prime p not of the form 4n + 3, the definition of distance in part (a) has the following curious property. There are distinct points with a distance of 0 between them. Verify this property for the primes 2, 5, 13, and 17. You can modify the distance formula for p = 4n + 1 by multiplying the term (t − v)2 by a nonzero scalar of Z p . Experiment with different scalars. Do the points at a specific distance from a given point form an oval? If so, what is its equation? How does the equation relate to the scalar? Investigate isometries for the planes with these distances. Count the number of isometries. Use different definitions of distance to explore parts (a) and (b). Do they change the number of isometries? Is there a common formula for the number of isometries for AZ2p ? Define perpendicular in AZ2p for odd primes p. Do you need different definitions for different primes? Explore similarities in AZ2p , for p = 4n + 3 and p = 4n + 1. Count the number of similarities for AZ2p and show that they form a group.

13. Let potential code words be vectors of 0s and 1s of length n. (a) Find their total number. (b) How many vectors are at a Hamming distance of 1 from a given vector? (c) For a code to be able to correct one error, each pair of code words must be a Hamming distance of at least 3 apart. Use parts (a) and (b) to determine the maximum number of code words possible if each pair must be a distance of at least 3 apart. For various values of n, look for codes (sets of code words) that can correct one error and have as many code words as possible. (d) Redo part (c) with codes that can correct two (or more) errors. Hint: The binomial theorem may be helpful. 14. Prove that projective and affine spaces over finite fields are BIBDs. Hints: Do projective spaces first. Let PFd be the d-dimensional projective space over the field with n elements. Then F has n − 1 nonzero scalars. Show that a projective line has n + 1 distinct points on it. Use linear algebra to show that two distinct projective points have a unique line on them. Now show that PFd is a BIBD with λ = 1. The affine space AFd has the points not on the hyperplane xd+1 = 0. Prove: every projective line intersects the hyperplane xd+1 = 0 in a one- or two-dimensional subspace. The affine lines are those intersecting xd+1 = 0 in one point. Use the earlier part to show that AFd is a BIBD with λ = 1. 15. Explore design theory. (See Anderson [1], Berman and Fryer [3], and Karteszi [7].) 16. Explore coding theory. (See Anderson [1] and Gallian [6].) 17. Explore finite geometries. (See Blumenthal [4] and Dembowski [5].) R 18. Explore geometric aspects of the game Set⃝ . (Search the web for sources using “game of set” and “affine.”)

8.4 Finite Analytic Geometry

407

19. Write an essay discussing the analogies between finite affine (or projective) planes and the Euclidean (or real projective) plane. What insights can finite geometries provide?

8.4.6 Suggested Readings 1. Anderson, I., A First Course in Combinatorial Mathematics, 2nd ed. New York: Oxford University Press, 1989. 2. Beck, A., M. Bleicher, and D. Crowe, Excursions into Mathematics, New York: A K Peters, 2000. 3. Berman, G., and K. Fryer, Introduction to Combinatorics, New York: Academic Press, 1972. 4. Blumenthal, L., A Modern View of Geometry. Mineola, NY: Dover, 1961. 5. Dembowski, P., Finite Geometries, New York: Springer-Verlag, 1968. 6. Gallian, J., Contemporary Abstract Algebra, 7th ed. Belmont, CA: Brooks/Cole, 2010. 7. Karteszi, F., Introduction to Finite Geometries, Amsterdam: North Holland, 1976. 8. Mullen, G., A candidate for the “next” Fermat problem. The Mathematical Intelligencer, 1995, 17(3): 18–22. 9. Smart, J., Modern Geometries, Pacific Grove, CA: Brooks/Cole, 1988.

9

Differential Geometry

Earth Moon

Star

Sun (not to scale)

Figure 9.0 In 1919 Sir Arthur Eddington (1882–1944) spectacularly confirmed a key prediction of Einstein’s general theory of relativity during a solar eclipse. Astronomers had determined that the sun should have blocked a starlight’s path if it followed a Euclidean straight line from the star to earth. Instead because the sun’s gravity distorted the shape of space, Eddington saw the star when the moon blocked the sun’s light during the eclipse. The figure greatly distorts sizes to illustrate the setting of this famous experiment. In the language of differential geometry, the light from the star followed a geodesic—a locally shortest path. The gravitational field of the sun induced enough curvature to space to make an observable difference between the geodesic and a Euclidean line. While relativity theory is beyond the level of this chapter, we will investigate curvature and geodesics.

9.1 Overview and History A plane flying from Tokyo to Los Angeles flies far north of either city, depicted in Figure 9.1. However, the two cities lie at virtually the same latitude. Why don’t planes fly straight east from Tokyo to Los Angeles? Actual airline routes minimize the distance on the curved surface of the earth. A straight eastward route follows a circle of latitude, which doesn’t minimize distances (except the equator). Differential geometry provides the language and theory to understand the geometry of curved surfaces. We will investigate how to find the curvature of a surface and geodesics, which are the locally straight paths on them. Intuitively, Euclidean straight lines do not bend at all, while circles bend at constant rates depending on their radii. (See Figure 9.2.) The ancient Greeks realized that other curves, such as the parabola in Figure 9.3, bend more at some points (for example, A) than others (such as B or C). However, real advances in defining and measuring curvature, the amount of bending, needed the language of calculus. By 1671 Sir Isaac Newton (1642–1727), one of the originators of calculus, had already figured out the correct formulation for the curvature of a plane curve. (See Theorem 9.2.1.) However, Newton delayed publishing that work and much else for decades. By that time Gottlieb Wilhelm Leibniz (1646–1716), the other founder of calculus, and his followers

409

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Differential Geometry

Russia

Alaska

Canada USA

Japan

China

Pacific Ocean Hawaiian

Philippines Palau

Fed. States of Micronesia

Papua New Guinea Solomon

Australia

Islands Vanuatu Fiji

Islands

Equator

New Caledonia New Zealand

Figure 9.1 A polar air route. had published on many of the same topics, including curvature. Jakob Bernoulli (1654–1705), a follower of Leibniz, determined geodesics on surfaces of revolution in 1698. 2.0 C

1.5 1.0 0.5 –1.5 –1.0 –0.5

Figure 9.2

A

B 0.5 1.0 1.5

Figure 9.3 Curvature on a parabola.

In the 1730s Alexis-Claude Clairaut (1713–1765) and Leonhard Euler (1707–1783) considered space curves, such as the helix of Figure 9.4. They analyzed how much such curves twisted out of a plane, now called the torsion of a curve. (See Project 1.) More central to our focus, they and others investigated general surfaces and geodesics. The inherent distortions of a flat map of portions of a curved earth motivated many mathematicians from Euler to Carl Friedrich Gauss (1777–1855). In fact, as part of his employment, Gauss spent some years surveying the state of Hanover, which spurred his profound investigations in differential geometry. His major paper in 1827 shaped the direction of this subject. Gauss studied a space on its own terms, not dependent on how it is embedded in Euclidean space. For example, measurements of a portion of the surface of the earth could determine its curvature there. One of his profound theorems (Theorem 9.4.3) generalizes Theorem 1.5.3 and Theorem 4.1.1. They link the area of a triangle in spherical and hyperbolic geometry to their angle sums. In the generalization Gauss used geodesics for the sides of the triangles. Then, as

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9.1 Overview and History

–0.5 –1.0

0.0

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3

2

1

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0 0.5 1.0

Figure 9.4 A helix. he proved, the difference between the angle sum of the triangle and π was determined by the area of the triangle and the curvature within it. The sphere has constant positive curvature and the hyperbolic plane has constant negative curvature, accounting for the difference in Theorems 1.5.3 and 4.1.1. In 1854 Georg Riemann (1826–1866) generalized Gauss’s results to any number of dimensions and so transformed differential geometry again. Many mathematicians developed Riemann’s insights into rigorous mathematics with computational methods. Albert Einstein (1879–1955) built on Riemann’s profound insights to revolutionize physics with the general theory of relativity. Einstein needed a four-dimensional space built from three spatial dimensions and time, all with varying curvature corresponding to the strength of gravitational fields at different points.

9.1.1 Exercises for Section 9.1 *9.1.1. Use a computer graphing capability to find the circle best approximating the bend of y = x 2 at x = 0 as follows. (a) Explain why the circle should have a common tangent with y = x 2 at x = 0. On what line does the center of the circle lie? Use your answers to simplify the general equation of a circle: (x − a)2 + (y − b)2 = r 2 . (b) Solve the equation in part (a) for y and graph the bottom half along with y = x 2 for various radii. Which center and radius appear to give the best fit? 9.1.2. *(a) Repeat Exercise 9.1.1 for the function y = cos(x) at x = 0. (b) Repeat Exercise 9.1.1 for the function y = x 4 at x = 0.

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Differential Geometry

2

(c) Repeat Exercise 9.1.1 for the ellipse x4 + y 2 = 1 at x = 0 and again at x = 2. Hint: Graph the ellipse first to see what is happening at x = 2. √ (d) Repeat Exercise 9.1.1 for the hyperbola y = ± x 2 + 1 at x = 0. (e) Repeat Exercise 9.1.1 for y = cosh(x) = (e x + e−x )/2 at x = 0. 9.1.3. On a sphere of radius R consider the circle of points a distance of r from the north pole, as measured along the surface of the sphere. (See Figure 9.5.) Find the circumference of the circle in terms of R and r . Find a formula to compare the circumference to the circumference of a Euclidean circle of radius r .

r

R

Figure 9.5 A circle on a sphere. 9.1.4. Analogous to Exercise 9.1.3 explain how you think the circumference of the inner tube circle of radius r with center at point A in Figure 9.6 compares to the circumference of a Euclidean circle with the same radius. Repeat when the center is at point B and point C in Figure 9.6. Experiment on an actual inner tube. To draw an inner tube circle, use a narrow strip of plastic with a small hole near each end. Put a pencil through one hole to act as the center of the circle. Put a marker in the other hole to draw the circle. Make sure the strip of plastic is stretched out on the tube’s surface as you rotate it around the center. (You may need another person to help.) The radius of your circles is the distance between the holes on the strip of plastic. Chose the radius as large as possible but so that the entire circle around A is on the part of the tube bulging out, and the entire circle around B is on the saddle part of the tube. Now roll a wheel around the circumference of the circle and determine how many times around it rotates. Compare this with rolling the wheel

C B

A

Figure 9.6 A circle on a torus.

9.2 Curves and Curvature

413

around a Euclidean circle (or cylinder) of the same radius. Do your predictions for the circumferences around A, B, and C match with your experimental data? What happens to the ratios of the inner tube and Euclidean circumferences as the radius gets smaller? 9.1.5. Repeat Exercise 9.1.4 for other surfaces. Experiment with a cylinder. *9.1.6. This problem provides partial confirmation of the claim in Section 1.5 that great circles are the shortest paths on the surface of a sphere. Let c be a number with 0 ≤ c < π2 . Show all your work. (a) Verify that A = (cos(c), 0, sin(c)) and B = (− cos(c), 0, sin(c)) are on the unit sphere. Describe their locations. Hint: We use latitude and longitude to describe locations on earth. (b) Verify that the curve given by x(t) = cos(t), y(t) = 0, z(t) = sin(t) is a great circle of the unit sphere containing A and B. What is its radius? (c) Verify that the parametric equations x(t) = cos(c) cos(t), y(t) = cos(c) sin(t), z(t) = sin(c) give a horizontal circle through A and B lying on the unit sphere. What is its radius? (d) Find the distance d(c) from A to B following the circle in part (c) in terms of c. (e) Find the distance D(c) from A to B following the great circle in part (b) in terms of c. d(c) from parts (d) and (e) to verify that the great circle path is shorter (f) Graph y = D(c) than the other path unless c = 0 or c = π2 . What happens to the circle in part (c) in these cases? 9.1.7. (a) Use a narrow plastic strip with straight sides to draw approximate geodesics on an inner tube. Put the edge of the strip flush to the surface and draw a curve on the inner tube following the strip. Slide the strip along the curve to extend the curve. For which directions does the curve come back on itself? Describe what happens for other directions. (b) Repeat part (a) for a cylinder. (c) Repeat part (a) for other surfaces. 9.1.8. (a) Estimate the radius of curvature at various points on a non-spherical fruit or vegetable. It might help to cut off a region and scoop out the inside, leaving the skin. (b) Repeat part (a) for parts of an inner tube. Discuss any difficulties you encounter. (c) Repeat part (a) with other curved but non-spherical surfaces.

9.2 Curves and Curvature The circles in Figure 9.7 share the x-axis as a common tangent with the parabola at the point (0, 0). However the circle marked C visually fits the bending of the parabola better than the other circles. We say it has the same curvature as the parabola at that point. By inspection other points of the parabola have different curvatures. Mathematicians have given different geometric approaches to curvature yielding equivalent definitions for this key concept of differential geometry. We’ll start with a more geometric idea before introducing more general and easily calculated ones based more directly on calculus.

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Differential Geometry

1.4

B

1.2 1.0

C

0.8 0.6

D

0.4 0.2 –1.0

–0.5

0.5

1.0

Figure 9.7 Comparing circles and the curvature of a parabola. Example 1. Consider the circle through the three points (0, 0), (a, a 2 ), and (−a, a 2 ) on the parabola. From Exercise 1.2.15, the circle’s center is on the perpendicular bisectors of the 2 segments of Figure 9.8. Their equations are x = 0 and y = ± ax + a2 + 12 , which intersect at 2 (0, a2 + 12 ). In the limit as a → 0, the center is (0, 12 ) and the radius is 12 . Leibniz called this best approximating circle, circle C in Figure 9.7, the osculating circle. (“Osculate”comes from the Latin for “kiss.”) Since we think of small circles as bending more than large circles, we say a circle with radius r has curvature 1/r . So at x = 0, the parabola y = x 2 has a curvature of 2. In Figure 9.7 circles with radii bigger than 12 , such as B, intersect the parabola in three points. Circles, such as D, with smaller radii only intersect at (0, 0), but visually bend too much to match the parabola there. ♦ (0, a2/2 + 1/2)

y = -x/2a + a2/2 + 1/2

(a,a2)

(-a,a2) (a/2,a2/2) (0,0)

Figure 9.8 While perpendicular bisectors involve only elementary geometry, the computations become more involved with points on more complicated curves and at less symmetrical places. Instead it is easier to bring calculus in earlier and use normals to the curve. The curve can be given in various forms. We’ll use the familiar function form y = f (x) when possible, but parametric equations, discussed in Section 3.3, provide a more general setting since y = f (x) can be

415

9.2 Curves and Curvature

parametrized as (x, f (x)). To use calculus, we need the curves to have a certain amount of smoothness, given in terms of their derivatives. When we say “curve”we mean a smooth plane curve, as defined below, or a smooth space curve, as defined in Exercise 9.2.13 (a). Exercise 9.2.11 explores the last condition in the definition of a curve. As Exercises 9.2.6 and 9.2.12 explore, different parameterizations can determine the same set of points in the plane, but they give the same geometry of the curves. → Definition. A smooth plane curve is a function − c from a real interval into the plane, written −→ ′ ′ ′′ ′′ c(t) = (x(t), y(t)) so that x , y , x , and y exist, are continuous, and x ′ (t) and y ′ (t) are not −→ simultaneously both 0. Given a point P on the image of a curve c(t) in the Euclidean plane, the −→ −→ normal line to c(t) at P is the line through P perpendicular to the tangent to the image of c(t) at P. Figure 9.9 depicts two normals to a curve y = f (x) at nearby points A = (a, f (a)) and B = (a + h, f (a + h)). In general the normals will intersect in a point Pa,h depending on a and h. We fix a and let h → 0. If there is a limit P = limh→0 Pa,h , we consider it to be the center of the osculating circle through A; that is, the one with the same curvature as y = f (x) at A. Then we define the distance P A to be the radius of curvature of y = f (x) at A and P1A to be the curvature of y = f (x) at A. If the point P doesn’t exist, we say the curvature is 0. Straight lines have a curvature of 0 at every point: the perpendiculars (normals) to a straight line are parallel to each other and so never intersect. 1.5

1.0 B 0.5

A

0.5

1.0

1.5

2.0

Pa,h

Figure 9.9 Approximating the radius of curvature. Newton derived the first equation in Theorem 9.2.1 for the radius of curvature shortly after he worked out the basics of the calculus. Exercise 9.2.8 (c) extends the equation to allow f ′ (x) = 0. Although the radius of curvature is visually intuitive, the concept of curvature connects to deeper ideas. Therefore later in this section we give two additional formulas for curvature, (9.3) and (9.4), useful in different settings.

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Differential Geometry

Theorem 9.2.1 (Newton, 1671). If f ′ (x) ̸= 0 and f ′′ (x) ̸= 0, the radius of curvature of y = f (x) is 0 0 0 (1 + f ′ (x)2 )3/2 0 0 0 r (x) = 0 (9.1) 0 f ′′ (x)

and the curvature is

0 0 κ(x) = 00

0 0 f ′′ (x) 0. ′ 2 3/2 (1 + f (x) ) 0

(9.2)

Example 2. For the parabola f (x) = x 2 , Newton’s formula gives r (x) = (1 + 4x 2 )3/2 /2 for the radius of curvature. Even though f ′ (0) = 0, by Exercise 9.2.8 (c) the formulas are valid there and r (0) = 12 and κ(0) = 2. In Figure 9.7, circle C has radius 12 . As x increases, the radius r (x) quickly increases and the curvature κ(x) rapidly shrinks towards 0, indicating the parabola doesn’t bend much away from the origin. The second derivative, measuring concavity or acceleration, is a constant 2, which doesn’t convey this information about curvature. ♦ 2

12x Example 3. The curvature of y = x 4 from Newton’s formula is κ(x) = (1+16x 6 )3/2 . At x = 0, the curvature is κ(0) = 0, which may seem counterintuitive. In Figure 9.10 the circle x 2 + (y − 1)2 = 1, labelled C, intersects the curve in five points, but isn’t flat enough to fit y = x 4 at x = 0. For a > 0, the circle x 2 + (y − a)2 = a 2 will lie above the curve between x = 0 and the (first) intersection on either side. Contrast this with Figure 9.7, where the circle B lies below the parabola between x = 0 and the other intersections. ♦

2.0

C

1.5

1.0

0.5

–1.0

–0.5

0.5

1.0

Figure 9.10 Curvature of y = x 4 . Derivation. We find the formula for the radius of curvature; the formula for curvature follows immediately by its definition. The tangent to y = f (x) at x = a in point-slope form is y − f (a) = f ′ (a)(x − a) and since f ′ (a) ̸= 0, the normal is y − f (a) = f−1 ′ (a) (x − a). We solve this equation for y to get y=

−x + a + f (a) · f ′ (a) . f ′ (a)

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9.2 Curves and Curvature

Similarly, the normal when x = a + h is y=

−x + (a + h) + f (a + h) · f ′ (a + h) . f ′ (a + h)

Set these expressions equal, cross multiply and put the terms with x on the same side to get x[ f ′ (a + h) − f ′ (a)] = a[ f ′ (a + h) − f ′ (a)] − h f ′ (a) + f ′ (a) · f ′ (a + h)[ f (a) − f (a + h)]. Solve for x: x =a−

h f ′ (a) + f ′ (a) · f ′ (a + h)[ f (a + h) − f (a)] . [ f ′ (a + h) − f ′ (a)]

This expression is the x-coordinate of Pah , which may look intimidating. However, the terms in brackets are the numerators of difference quotients, which will soon become derivatives. To get the difference quotients, we divide by h. Also, we want the radius of curvature, so we focus on the difference between a and x. Then we take the limit as h → 0. We use the assumption f ′′ (a) ̸= 0 when we take the limit. a−x = = So

h f ′ (a) + f ′ (a) · f ′ (a + h)[ f (a + h) − f (a)] [ f ′ (a + h) − f ′ (a)]

(h/ h) f ′ (a) + f ′ (a) · f ′ (a + h)[( f (a + h) − f (a))/ h] . [( f ′ (a + h) − f ′ (a))/ h] lim (a − x) =

h→0

f ′ (a) + ( f ′ (a))3 . f ′′ (a)

Next we need to include the difference of the y-coordinates. Let $x be the limit above. The $x slope of the normal at A is f−1 ′ (a) . So the y-distance is the absolute value | f ′ (a) |. The Pythagorean theorem gives the radius of curvature as > 0 ! 0 & '2 > 2 ′ 2 $x ( f (a) + 1) 00 $x 1 + f ′ (a)2 00 $x 2 = =0 r = $x + 0 0 0 f ′ (a) f ′ (a)2 f ′ (a) 0 0 0 [ f ′ (a) + ( f ′ (a))3 ]!1 + f ′ (a)2 0 00 (1 + f ′ (x)2 )3/2 00 0 0 0 0. =0 ! 0=0 0 ′′ ′ 0 0 f (a) · f (a) f ′′ (x)

The expression !(1 + f ′ (x)2 )3/2 in Newton’s curvature formula may seem strange, but a closely related one, 1 + f ′ (x)2 , occurs in the arc length formula of calculus texts. Given our abbreviated investigation, we simply state the parameterized form for curvature for a general −→ parameterization after a ! short explanation. For a curve c(t) = (x(t), y(t)), the arc length formula from calculus involves x ′ (t)2 + y ′ (t)2 , relating to the denominator of (9.3). Also, note that ? ′ @′ ′′ ′ ′ (t)x ′′ (t) the quotient rule gives xy ′ (t) = y (t)x (t)−y , whose numerator is the numerator of (9.3). (t) x ′ (t)2 (See Oprea [7, 33].) −→ Parametric version of Theorem 9.2.1. The curve c(t) = (x(t), y(t)) has curvature 0 ′′ 0 0 y (t)x ′ (t) − y ′ (t)x ′′ (t) 0 0 0. κ(t) = 0 (9.3) (x ′ (t)2 + y ′ (t)2 )3/2 0

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Differential Geometry

−→ Example 4. We call the curve c(t) = (t cos(t), t sin(t)) for 0 ≤ t the spiral of Archimedes, depicted in Figure 9.11, because Archimedes proved several theorems about it. We use the formula 2+t 2 cos2 (t) + sin2 (t) = 1 to simplify the formula for curvature, obtaining κ(t) = (1+t 2 )3/2 . At t = 0, 1 we have κ(0) = 2, so the radius of curvature at the origin is 2 . Figure 9.12 illustrates how well the circle with radius 12 fits the spiral at the origin. As t increases, the curvature goes to 0. ♦

1 –3

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Figure 9.11 The spiral of Archimedes.

1 –3

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Figure 9.12 Approximating circle to the spiral. Rather than use the tangent and normal lines of calculus, it is generally more useful −→ −→ to use vector notation and the unit tangent vector. The length of a vector c(t) is ∥c(t)∥ = ; ! −−→ −→ −→ c(t) · c(t) = x(t)2 + y(t)2 and we write c′ (t) for (x ′ (t), y ′ (t)). (It is easier to compute the unit tangent than the unit normal and, since they are perpendicular, they convey the same

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9.2 Curves and Curvature

information in two dimensions. The tangents to surfaces are planes, so in Section 9.3 for the curvature of surfaces the unit normal will be easier to use.) −→ Definition. For a curve c(t), the unit tangent vector is T (t) =

−− → c′ (t) . −− → ∥c′ (t)∥

−−→ Exercise 9.2.1.∗ Use the definition of a curve to explain why ∥c′ (t)∥ is never 0 and so T (t) is always defined. −→ Theorem 9.2.2. For a curve c(t), T ′ (t) is orthogonal to T (t). Proof. By definition, ∥T (t)∥ = 1, and so T (t) · T (t) = 1. The product rule gives (T (t) · T (t))′ = T ′ (t) · T (t) + T (t) · T ′ (t), which is 0, the derivative of a constant. Thus T ′ (t) · T (t) = 0, forcing T (t) and T ′ (t) to be orthogonal. ! ? @ sin(t) sin(t)+t cos(t) √ √ Example 5. Figure 9.13 includes some unit tangents T (t) = cos(t)−t , with the 2 2 1+t 1+t −− → spiral of Figure 9.11. The vector c′ (t) = (x ′ (t), y ′ (t)) = (cos(t) − t sin(t), sin(t) + t cos(t)) is −−→ the velocity vector of the spiral. At t = 0, c′ (0) = (1, 0), which has unit length. However, ; √ −′−→ −−→ 2 c ( π2 ) = (− π2 , 1) has length π4 + 1 ≈ 1.862 and c′ (π) = (−1, −π ) has length 1 + π 2 ≈ 3.297, indicating an increasing velocity of a point moving along it. Since t measures angles in radians, as t increases equivalent angles sweep out more distance along the curve, forcing the velocity to increase. ♦

1 –3

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2

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Figure 9.13 Tangents to the spiral. Example 5 illustrates how the unit tangent vector eliminates the distraction of how we parameterize the curve. Differential geometry texts generally define curvature in terms of how quickly a curve bends in comparison to the arc length. We can measure how quickly the curve " t −−→ bends by the change of T (t). The arc length from a starting value t = a is s(t) = a ∥c′ (u)∥du.

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Differential Geometry

Then κ(t) = ∥ dT ∥ = ∥ dT · ds dt s(t), we obtain

dt ∥ ds

=

∥T ′ (t)∥ . |s ′ (t)|

By the fundamental theorem of calculus applied to

κ(t) =

∥T ′ (t)∥ . ∥c′ (t)∥

(9.4)

The elegant simplicity of (9.4) disguises the often complicated form T ′ (t) can take. You can check this using the formula for T (t) in Example 5 in comparison with Example 4. −→ This section applies to space curves c(t) = (x(t), y(t), z(t)) as well, where (9.4) can be an easier way to find curvature than a generalization of (9.3). In addition to having curvature, space curves can twist out of the plane determined by the tangent and normal vectors to the curve at a point. Mathematicians call the measure of the twisting torsion. (See Project 1.) −→ Example 6. For a helix, as in Figure 9.14, c(t) = (R cos(t), R sin(t), mt). The arrows −− → ′ m) mark T (t) at various places √= (−R sin(t), R cos(t), √ on this curve. √We have c (t) ′ 2 2 2 2 2 2 and so T (t) √ = (−R sin(t)/ R +√m , R cos(t)/ R + m , m/ R + m ). Thus T (t) = the curve (−R cos(t)/ R 2 + m 2 , −R sin(t)/ R 2 + m 2 , 0), which points horizontally √ 2 from R /(R 2 +m 2 ) ∥T ′ (t)∥ R towards the z-axis. The curvature function is constant: κ(t) = ∥c′ (t)∥ = √ R 2 +m 2 = R 2 +m 2. 2 2 All the helices lie on the surface of a vertical cylinder of radius R with equation x + y = R 2 . We will see in Section 9.4 that they are geodesics for the cylinder. ♦

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0.5

0 1.0

Figure 9.14 Tangents of a helix. Curvature provides valuable information about curves not directly apparent from the familiar first and second derivatives of calculus. For more on curves, see Opera [7, Chapter 1].

9.2.1 Exercise for Section 9.2 9.2.2.

(a) Find the formula of the radius of curvature r (x) for y = sin(x); graph y = sin(x). *(b) Explain why a circle tangent to y = sin(x) at x = π2 has its center on the vertical line x = π2 . Find the equation of the osculating circle at x = π2 . That is, the circle

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421

that is tangent to y = sin(x) at x = π2 and has the same radius of curvature there. Include it on the graph in part (a). have its center? Find (c) On what line must a circle tangent to y = sin(x) at x = 3π 2 3π the equation of the osculating circle to y = sin(x) at x = 2 . Include it on the graph in part (a). (d) If we omit the absolute value for r (x) in Theorem 9.2.1, what does the sign of r (x) tell us about the circles in parts (b) and (c)? √ 9.2.3. (a) Graph y = x, a half parabola isometric to y = x 2 , for x ≥ 0. √ (b) Find r (x) for y = x and simplify the formula. √ (c) Verify that the radii of curvature for y = x and y = x 2 are equal at corresponding points. √ (d) Graph y = x 2 + 2x, which is half of a hyperbola, together with the half parabola from part (a). Describe how you think the curvatures of the curves relate at x = 0 and as x becomes √ large. (e) Find r (x) for y = x 2 + 2x. Compare the curvature of the parabola and hyperbola for x = 0 and for x = 2. Compare the values with your description in part (d). 9.2.4. *(a) Find r (x) for y = e x . Graph y = e x . *(b) Find the normal line to y = e x at x = 0 and find r (0). Then find the center and equation of the osculating circle to y = e x at x = 0. Graph it on the graph from part (a). *(c) Find the value of x giving the minimum radius of curvature for y = e x . Verify that it is, indeed, smaller than the radius at x = 0. x −x . (d) Find r (x) for y = cosh(x) = e +e 2 (e) Graph y = cosh(x) and its osculating circle at x = 0. 9.2.5. (a) Graph y = x 3 for −2 ≤ x ≤ 2 with the same scale for both axes. Estimate visually where the largest curvature seems to be. (b) Use (9.2) to find κ(x) for y = x 3 . Explain why the formula will have an absolute maximum at some positive value of x. (c) Use calculus to find the maximum curvature of y = x 3 . −→ 9.2.6. (a) Use the parametric form c(t) = (R cos(t), R sin(t)) to verify that a circle of radius R has curvature 1/R at every value of t. −−→ (b) Explain why the parametric form cm (t) = (R cos(mt), R sin(mt)) also gives a circle −−→ of radius R for any m > 0. Verify that cm (t) also has a curvature of 1/R. −−→ (c) Find cm′ (t) in part (b) and use it to explain what effect the factor m has. (d) Find the unit tangent vector T (t) for the equation in part (b) and compare it with the unit tangent for the curve in part (a). −→ (e) Repeat parts (b), (c), and (d) for the curve g(t) = (R cos(t 3 ), R sin(t 3 )), where t > 0. −→ 9.2.7. *(a) Graph the curve c(t) = (t 2 − 4, t 3 − 4t) for −3 ≤ t ≤ 3. Estimate from the graph where the curvature looks the biggest. −→ *(b) Find the curvature of c(t) at t = 0, t = 1, t = 2, and t = 3. −−→ (c) Graph the cycloid w(t) with x(t) = t − sin(t) and y(t) = 1 − cos(t) for 0 ≤ t ≤ 2π . Estimate from the graph where the curvature looks the biggest.

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−−→ (d) For w(t) find the curvature κ(t) for t not a multiple of 2π . Find κ( π2 ) and κ(π ). What is limt→0 κ(t)? Does the value fit the shape of the cycloid at 0? 9.2.8. *(a) Use (9.2) to explain why the minimum curvature of y = f (x) always occurs at an inflection point. What can you say about the radius of curvature at an inflection point? (b) Extend your explanation in part (a) to (9.3) using the discussion before that equation. (c) Prove that (9.1) remains valid in the case where f ′ (a) = 0, but f ′′ (a) ̸= 0. Hints: The normal at a has an easy form which you can use to find the y-value of the intersections of the normals. Simplify and then subtract f ′ (a), which is 0, from the denominator. Then take a limit. −−→ −−→ −→ *9.2.9. (a) Let c(t) = (t 2 + 1, t − t 2 ). Find c′ (t) and c′′ (t) as well as T (t) and T ′ (t) in reduced − − → −− → form. Verify that c′ (t) and c′′ (t) are not orthogonal, but T (t) and T ′ (t) are. −→ (b) Find κ(t) for the curve in part (a). Find κ(0), κ(0.25), and κ(1). Graph c(t)and use calculus to find the value of t where the curvature is a maximum. −→ 9.2.10. (a) Verify that c(t) = (2 cos(t), sin(t)) for 0 ≤ t ≤ 2π is the ellipse of Exercise 9.1.2 (c). Graph it. *(b) Find the curvature κ(t). Determine the values of t where the curvature reaches its maximum and minimum. Do the values match your intuition from the graph? On the graph from part (a) draw the osculating circles for one minimum and one maximum point. −→ *(c) Graph the curve c(t) = (cos(t), sin(2t)) for 0 ≤ t ≤ 2π . Describe from the graph where the curvature looks the smallest and the largest. At what values of t does the curve cross itself? (d) For part (c) find the curvature at t = 0, t = π4 , and t = π2 . (e) For what values of t, x, and y does it look like the curve in part (c) has inflection point(s)? → 9.2.11. We explore why we require a curve − c (t) = (x(t), y(t)) to have x ′ (t) and y ′ (t) not both 0 for the same value of t. −→ (a) Graph the curve c(t) = (t 2 , t 3 ). Describe what happens to the graph at t = 0. −− → (b) The vector c′ (t) = (x ′ (t), y ′ (t)) gives the velocity of a point moving along the curve. −−→ −−→ For the curve in part (a) find the magnitude ∥c′ (t)∥ in general and ∥c′ (0)∥. What −−→ −−→ does ∥c′ (0)∥ tell us? What does ∥c′ (0)∥ imply for T (0) and T ′ (0)? (c) Verify that the curve given by

x(t) =

A

−t 3 0

if − 1 ≤ t ≤ 0 if 0 < t ≤ 1

B

and

y(t) =

A

0 t3

if − 1 ≤ t ≤ 0 if 0 < t ≤ 1

B

has continuous first and second derivatives. Then repeat parts (a) and (b) for this curve.

9.2 Curves and Curvature

423

−→ 9.2.12. (a) We can write the function y = f (x) as the curve c(t) = (t, f (t)). Verify that Equation 9.3 reduces to Equation 9.2 in this case. (b) Find the formula for the unit tangent using the equations of part (a). −−→ (c) Repeat parts (a) and (b) for cm (t) = (mt, f (mt)). 9.2.13. *(a) Modify the definition of a smooth plane curve to give a definition of a smooth space curve. −→ *(b) For a fixed θ let c(t) = (cos(θ ) cos(t), sin(θ) cos(t), sin(t)). Verify for all t that −→ −→ c(t) is on the unit sphere, which satisfies x 2 + y 2 + z 2 = 1. What curve does c(t) represent for an given value of θ ? Explain your answers. −−→ *(c) Find c′ (t), T (t), and κ(t) for the curve of part (b). Show your work. −→ −−→ (d) Let w(t) = (cos2 (t), sin(t) cos(t), sin(t)), for − π2 ≤ t ≤ π2 . Verify for all t that c(t) is on the unit sphere. Graph this curve using a computer. −−→ (e) Find w ′ (t) and T (t) for the curve in part (d). Show your work.

9.2.2 Sir Isaac Newton Nature and Nature’s laws lay hid in night: God said, “Let Newton be!” and all was light. —Alexander Pope If I have seen further than others, it is by standing on the shoulders of others. —Sir Isaac Newton

In both mathematics and physics Sir Isaac Newton (1643–1727) stands as a giant in his own right, somewhere between Pope’s intended epitaph and Newton’s own words. At 18 Newton entered Cambridge University intending to study law. In his third year he started reading science and mathematics, including works of Galileo and Kepler. He realized he needed to know more geometry and set about reading a translation of Euclid’s Elements. Soon after he studied Descartes’ Geometry and other recent mathematics. Newton received his bachelor’s degree in 1665, just before the university had to close because of an outbreak of the plague in England. Newton went home and thought for two years, working out the essentials of what are now called calculus and Newtonian mechanics in physics. Newton delayed publishing these profound results and much of his later work for years. He returned to Cambridge University, earning a master’s degree. Newton became the Lucasian professor at age 27. An entire list of Newton’s achievements in mathematics and physics would take far too much space. For this textbook, it is noteworthy that he used Euclidean geometric proofs rather than calculus in his foundational work in physics, the Principia Mathematica. He also was the first to use calculus to develop ideas now considered part of differential geometry. In later years, Newton was involved in a vitriolic rivalry with Leibniz disputing their claims of priority for calculus. Newton developed his work first, but Leibniz published sooner. Both deserve great credit, but the unfortunate consequence of their dispute was a severing of English mathematicians from continental mathematicians.

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Differential Geometry

9.3 Surfaces and Curvature The bulging of a sphere makes the angle sum of spherical triangles greater than 180◦ . Indeed, the excess of the angle sum over 180◦ depends on the area of the triangle, as Theorem 1.5.3 makes precise. The angle sum of a hyperbolic triangle also depends on the triangle’s area. But, as proven in Theorem 4.1.1, the angle sum is less than 180◦ . What kind of anti-bulging would a surface need to have to match hyperbolic geometry? Carl Friedrich Gauss, one of the founders of hyperbolic geometry, answered that question with differential geometry, although it is unclear whether he saw the connection. From Gauss’s definition of curvature, a sphere has constant positive curvature and a Euclidean plane has 0 curvature. Eugenio Beltrami (1835– 1900) realized that a surface with constant negative Gaussian curvature gave a model of (part of) hyperbolic geometry. In this section we investigate surfaces and their curvatures. To simplify the theory and calculations we will use parametric equations to embed the surfaces in R3 , Euclidean 3-dimensional space. (Gauss showed how to study the geometry of a surface independently of embedding it in R3 .) We focus on surfaces of revolution, further easing our work. We often use the formula cos2 (θ ) + sin2 (θ ) = 1 to simplify formulas for surfaces of revolution.

9.3.1 Surfaces Example 1. We can represent ! the surface of a sphere of radius R in more than one way. The pair of functions z = ± R 2 − x 2 − y 2 determine the upper and lower hemispheres. The computer drawing in Figure 9.15 uses the curves determined from them by holding one of x or y fixed and varying the other. The familiar and more useful longitude and latitude coordinates, depicted in Figure 9.16, fit with the parametric equations x(u, v) = R cos(u) cos(v), y(u, v) = R sin(u) cos(v), and z(u, v) = R sin(v), where −π < u ≤ π and − π2 ≤ v ≤ π2 . Here u gives −→ the longitude and v gives the latitude. The equations come from rotating the circle c(v) = (R cos(v), R sin(v)) around the z-axis. If we restrict v to − π2 < v < π2 , they map the points (u, v) in a rectangle to the points (x(u, v), y(u, v), z(u, v)) on the sphere, minus the north and south poles. The latitudes and longitudes in Figure 9.16 come from holding v or u fixed and varying the other. This parameterization doesn’t work well at the north and south poles, which are the images of infinitely many points (u, π2 ) and (u, − π2 ), respectively. Both representations of the sphere present difficulties at the edges of the domain—where z = 0 for the first one and where u = ±π or v = ± π2 for the second one. A more rigorous development of differential geometry covers a surface with overlapping patches, each with its own parameterization. (See Do Carmo [1, 55–57].) ♦

Figure 9.15 A sphere represented using cross sections.

Figure 9.16 Longitude and latitude lines on a sphere.

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9.3 Surfaces and Curvature

Exercise 9.3.1. Replace x and y in the first form for the sphere with their parametric formulas and reduce the expression to z(u, v) to verify that the representations of a sphere match. Exercise 9.3.2. Most texts parameterize the sphere as x(u, v) = R cos(u) sin(v), y(u, v) = R sin(u) sin(v), and z(u, v) = R cos(v), where 0 < u ≤ 2π and 0 ≤ v ≤ π . Convert it to the second one in Example 1. Example 2. We generate a torus, or more informally an inner tube, by rotating the circle (y − R)2 + z 2 = r 2 in the yz-plane around the z-axis. (See Figures 9.17 and 9.18.) The rotation replaces the y-value in the equation with corresponding points (x, y). The of (x, y) from ; distance ! ! 2 2 2 2 the origin, x + y , takes the place of y, giving the formula z = ± r − ( x + y 2 − R)2 . Figure 9.19 depicts the torus as a surface of revolution using cross sections from the parametric form ((r cos(v) + R) cos(u), (r cos(v) + R) sin(u), r sin(v)), for 0 ≤ u < 2π and . − π2 ≤ v < 3π 2 z r y x

R

Figure 9.17

Figure 9.18

Figure 9.19 Points on the bulging outside surface of the torus, where − π2 < v < π2 , have positive , have negative curvature. For them curvature. Points on the inside surface, where π2 < v < 3π 2 the surface has a saddle-like shape, curving inward for the horizontal cross section and outward for the vertical cross section. Consider the two circles of points when v = ± π2 , between the bulging points and the saddle points. Although the torus may not appear flat at them, its curvature can’t be positive or negative, and so must be 0. The situation is analogous to Example 3 of Section 9.2, where y = x 4 had a curvature of 0 at x = 0. ♦ Exercise 9.3.3. Replace x and y in the first form for the torus with the parametric formulas and reduce the expression to z(u, v) to verify that the two representations of a torus match.

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Differential Geometry

In general, a surface is a function s from a region of R2 to R3 that we represent using parametric equations: s(u, v) = (x(u, t), y(u, v), z(u, v)). There are many conditions on them to make the patches smooth enough and able to fit together. We’ll avoid the technical details and say s is a smooth surface provided it satisfies the conditions found in Oprea [7, 71–73].

9.3.2 Curvature The curvature of a surface at a point is more subtle than the curvature of curves. We use curves on the surface and through the point, in particular the cross sections, as discussed in Examples 1 and 2, to lead us to a definition of surface curvature. When we fix v, the values of s(u, v) vary ∂ y ∂z , ∂u ). Similarly, when with u and so the cross section curve has tangent vectors su = ( ∂∂ux , ∂u ∂ x ∂ y ∂z we fix u, the tangent vectors are sv = ( ∂v , ∂v , ∂v ). From multivariable calculus other tangent vectors to the surface at this point are linear combinations of them and so form a plane. However, it is easier to keep track of the unit normal, a unit vector orthogonal to the tangent plane. The cross product su × sv gives us a vector orthogonal to both tangent vectors and so it is orthogonal to the entire tangent plane. The definition of N (u, v) divides the cross product by its length to get a unit vector. We’ll use the change of the unit normal to determine how the surface bends. This is analogous to (9.4) which gave the curvature of a curve in terms of the change of the unit tangent vector. Definition. For a smooth surface s(u, v), the unit normal is N (u, v) =

su ×sv ∥su ×sv ∥ .

Example 3. For the sphere of Example 1, s(u, v) = (R cos(u) cos(v), R sin(u) cos(v), R sin(v)). Then su = (−R sin(u) cos(v), R cos(u) cos(v), 0)

and sv = (−R cos(u) sin(v), −R sin(u) sin(v), R cos(v)). We find su × sv = R 2 (cos(u) cos2 (v), sin(u) cos2 (v), cos(v) sin(v)). From the fact that cos(v) > 0 for − π2 < v < π2 we find ! ∥su × sv ∥ = R 2 cos2 (u) cos4 (v) + sin2 (u) cos4 (v) + cos2 (v) sin2 (v) ! ! = R 2 cos4 (v) + cos2 (v) sin2 (v) = R 2 cos2 (v) = R 2 cos(v).

Thus N (u, v) = (cos(u) cos(v), sin(u) cos(v), sin(v)). The vectors point directly out from the sphere at each point s(u, v). ♦ Exercise 9.3.4.∗ Find Nu =

∂N ∂u

and Nv =

∂N ∂v

and verify that Nu =

1 s R u

and Nv =

1 s . R v

Example 4. Exercise 9.3.16 shows that the normal to the torus in Example 2 is N (u, v) = (cos(u) cos(v), sin(u) cos(v), sin(v)). To determine the curvature of the surface at a point (a, b) we need to know more than the normal. We need to know how it is changing as (u, v) approaches (a, b) along the cross sections. Exercise 9.3.16 gives formulas for Nu and Nv and shows that cos(v) s and Nv = r1 sv . The ratios indicate how quickly the normal is changing in the Nu = R+r cos(v) u u and v directions. We can find the curvature of the curves of the cross sections from them. ♦

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9.3 Surfaces and Curvature

In a more complicated but analogous manner to Theorem 9.2.2 the partial derivatives Nu and Nv are linear combinations of su and sv . That is, there is a 2 × 2 matrix M with entries 1 2 1 2 1 2 0 possibly dependent on u and v so that M ssuv = NNuv . In Example 3, M = 1/R and in 0 1/R 1 cos(v) 0 2 Example 4, M = R+r 0cos(v) 1 . For a surface s(u, v), M is called the shape operator. While the r terms of M depend on the parameterization of the surface, the Gaussian curvature defined below is independent of it. (See Oprea [7, 83–105] for a more extensive exposition of shape operators and their connection with curvature.) Definition. For a point ( p, q) on the surface s(u, v), the Gaussian curvature is κ( p, q) = det M, where M is the shape operator of s(u, v) at ( p, q). Exercise 9.3.5. Use Exercise 9.3.4 to verify that at a point on a sphere of radius R the curvature is 1/R 2 . cos(v) Example 5. The curvature of the torus of Examples 2 and 4 is κ(u, v) = det M = r (R+r . For cos(v)) π π − 2 < v < 2 the torus bulges out and so should have a positive curvature, matching cos(v) > 0 1 , the product of the curvatures of and κ(u, v) > 0 for these values. For v = 0, κ(u, 0) = r (R+r ) , the torus has a the circles that are the cross sections for fixed u and v. Similarly, for π2 < v < 3π 2 saddle shape and negative curvature, matching the negative value of cos(v) and so κ(u, v). Also −1 , which is the negative of the product of the curvatures of the cross sections κ(u, π ) = r(R−r ) for fixed u and v. Finally, cos(± π2 ) = 0, giving a curvature κ(u, ± π2 ) = 0, as discussed in Example 2. ♦

Figure 9.20 A tangent plane at a saddle point. We can interpret the curvature of a surface in terms of “signed curvatures” for certain cross sections, generalizing Example 5. As noted in that example, the curvature of the torus at (u, π ) is negative. As Figure 9.20 indicates, at a saddle point, the tangent plane cuts through the surface. Thus the osculating circles for the cross sections fixing u and v have centers on opposite sides of the tangent plane at (u, π ). We pick one side of the tangent plane to carry a positive sign for curvature of curves and the other side to carry a negative sign. For points in the inside, saddle shaped part, the product of the signed curvatures will be negative. For points where the torus bulges outward, the signed curvatures have the same sign and so the product is positive. Euler noticed this relationship using the cross sections to a smooth surface through the normal at a point. Gauss proved that his definition of curvature at a point of a surface was the product of the maximum and minimum (signed) curvatures for curves formed from the cross sections.

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Differential Geometry

9.3.3 Surfaces of Revolution We derive the general equation of a surface of revolution and verify its curvature formula. Given a parametric curve ( f (v), g(v)), as in Figure 9.21, we rotate it around the axis of the second coordinate (z) to get a surface of revolution, depicted in Figure 9.22. The radius f (v) is split into its x and y components: x = f (v) cos(u) and y = f (v) sin(u). (Since f (v) is a radius, we assume that f (v) ≥ 0.) Thus the surface is s(u, v) = ( f (v) cos(u), f (v) sin(u), g(v)). 1.5 1.0 Out[2]=

0.5 0.5

1.0

1.5

2.0

2.5

3.0

–0.5

Figure 9.21 A parametric curve.

Figure 9.22 The corresponding surface of revolution. Exercise 9.3.6. Verify the following calculations. su = (− f (v) sin(u), f (v) cos(u), 0) sv = ( f ′ (v) cos(u), f ′ (v) sin(u), g ′ (v)) su × sv = ( f (v)g ′ (v) cos(u), f (v)g ′ (v) sin(u), − f (v) f ′ (v)) ! ∥su × sv ∥ = f (v) f ′ (v)2 + g ′ (v)2 . / g ′ (v) cos(u) g ′ (v) sin(u) − f ′ (v) N (u, v) = ! ,! ,! f ′ (v)2 + g ′ (v)2 f ′ (v)2 + g ′ (v)2 f ′ (v)2 + g ′ (v)2 . / . / −g ′ (v) sin(u) g ′ (v) cos(u) g ′ (v) ! ,! ,0 Nu = Nu = ! su . f ′ (v)2 + g ′ (v)2 f ′ (v)2 + g ′ (v)2 f (v) f ′ (v)2 + g ′ (v)2

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9.3 Surfaces and Curvature

In addition to the computations of Exercise 9.3.6, we also need Nv to determine the curvature of a surface of revolution. We can simplify the complicated expression for Nv using ′′ f ′ (v)−g ′ (v) f ′′ (v) . In (9.3) K is the signed curvature of the parametric the substitution K = g ((v)f ′ (v) 2 +g ′ (v)2 )3/2 curve ( f (v), g(v)), the curve we are revolving to make the surface. & '′ '′ & g ′ (v) − f ′ (v) ′ Exercise 9.3.7. Verify √ ′ 2 ′ 2 = f (v)K and √ ′ 2 ′ 2 = g ′ (v)K . f (v) +g (v)

f (v) +g (v)

From Exercise 9.3.7, Nv = K ( f ′ (v) cos(u), f ′ (v) sin(u), g ′ (v)) = K sv . The 2 1 preceding work shows that the shape operator for a surface of revolution is M = 0J K0 , where ′ √ g′ (v)2 ′ 2 and K is given above. Now the proof of Theorem 9.3.1 only requires J= f (v)

f (v) +g (v)

multiplying J and K .

Theorem 9.3.1. Suppose the surface of revolution s(u, v) = ( f (v) cos(u), f (v) sin(u), g(v)) ′ ′′ (v) f ′ (v)−g ′ (v) f ′′ (v)) . is a smooth surface. Then the curvature of s(u, v) is κ(u, v) = g (v)(g f (v)( f ′ (v)2 +g ′ (v)2 )2 Exercise 9.3.8. Verify that su and sv are orthogonal for surfaces of revolution.

9.3.4 Exercises for Section 9.3 Use computer software to make the requested graphs. 9.3.9. *(a) Explain why s(u, v) = (R cos(u), R sin(u), v) is a cylinder. Explain what curve is rotated around the z-axis to generate it. (b) Find its curvature. Explain why the formula is reasonable. (c) Verify that the helices of Example 6 of Section 9.2 are curves on the surface of the cylinder. 9.3.10. *(a) From! Example 1 we can represent the sphere as s(x, y) = (x, y, ± R 2 − x 2 − y 2 ). Find sx , s y , and N (x, y). Are sx and s y orthogonal? Are N x and N y linear combinations of sx and s y ? This part illustrates some advantages of surfaces of revolution. (b) Graph the surface s(u, v) = (u, v, u 2 − v 2 ) for −2 ≤ u ≤ 2 and −2 ≤ v ≤ 2. Describe what you think the curvature of this surface is at various points. (c) At (0, 0) the curvature of the surface of part (b) is the negative of the product of the curvatures of the cross sections when u = 0 and when v = 0. Find the curvatures of the cross sections and so the curvature of the surface at (0, 0). What does the value tell you about the curvature of the surface at other points? 9.3.11. *(a) Graph the surface (u, v, cos(u) cos(v)) for −π ≤ u ≤ π and −π ≤ v ≤ π . Describe what you think the curvature of this surface is at various points. (b) At points ( pπ, qπ ) the curvature of the surface of part (a) is the product of the curvatures of the cross sections when u = pπ and when v = qπ , for integers p and q. Find the curvatures of the cross sections and so the curvature of the surface at ( pπ , qπ ). (c) Verify that the curvatures of the cross sections u = π2 and v = π2 are 0. Explain from the graph of part (a) why the curvature at ( π2 , π2 ) is not 0.

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(d) The biggest (positive and negative signed) curvatures at ( π2 , π2 ) occur in cross sections bisecting the angles between u = π2 and v = π2 . Explain why the cross sections have equations z = cos2 (u) and z = − cos2 (u). Find their curvatures and so the curvature of the surface in part (c) at ( π2 , π2 ). 9.3.12. The surface of a paraboloid results from rotating a parabola about its axis. The parametric form is s(u, v) = (v cos(u), v sin(u), bv 2 ). (a) Explain how to derive the parametric form of the paraboloid from the equation of a parabola. (b) Graph the paraboloid with b = 1. How does increasing the value of b affect its shape? Decreasing b? . For b = 1 graph the (c) Verify that the curvature of a paraboloid is κ(u, v) = 4b24b v 4 +1 osculating sphere at v = 0 with the graph in part (b). (d) Another way to represent a paraboloid is s(x, y) = (x, y, bx 2 + by 2 ). Find sx , s y , and N (x, y). Are sx and s y orthogonal? Are N x and N y linear combinations of sx and s y ? This part illustrates some advantages of surfaces of revolution. √ 9.3.13. (a) Graph s(u, v) = (v cos(u), v sin(v), v 4 − 2v 2 ) for 0 ≤ u ≤ 2π and 0 ≤ v ≤ 2. (b) Describe the values of v for which you think the curvature of the surface in part (a) is positive. Repeat for negative curvature. (c) Graph the cone √ with formula c(u, v) = (v cos(u), v sin(u), mv) for 0 ≤ u ≤ 2π and 0 ≤ v ≤ 2. Explain why the surface is not smooth when v = 0. (d) Assume v > 0 and find the curvature of the surface in part (c). Explain why this curvature is reasonable. What happens at v = 0? −→ 9.3.14. (a) Write the equations for the surface of revolution for the curve c(t) = (t, cos(t)). (b) Graph the surface in part (a) with 0 ≤ u ≤ 2π and 0 ≤ v ≤ 3π . For which values of v does the curvature appear positive? negative? sin(v) cos(v) (c) Assume that the curvature of the surface in part (a) is κ(u, v) = v(1+sin 2 (v))2 for v > 0. Use a limit to determine κ(u, 0). For which values of v is κ(u, v) > 0? Repeat for κ(u, v) < 0. Compare the values with your answers in part (b). (d) At t = 0, cos(t) has a derivative of 0, so the surface of revolution is smooth when v = 0. Investigate as in parts (a) and (b) the surface g(u, v) made by rotating (t, sin(t)). What happens when v = 0? Graph the surface g(u, v) for 0 ≤ u ≤ 2π and 0 ≤ v ≤ 2π . −→ 9.3.15. Figure 9.23 depicts an ellipse, which has parametric form c(t) = (cos(t), a sin(t)). When we rotate the ellipse around the vertical axis, we get a spheroid. When a > 1, we call it a prolate spheroid, like a rounded football. When a < 1, as the ellipse in Figure 9.23, the spheroid is oblate. The earth is, as Newton predicted, slightly oblate, with the diameter at the equator less than 1% bigger than the distance between the poles. *(a) Find the parametric form for the surface s(u, v) of a spheroid. Graph the spheroid with a = 2. Repeat with a = 12 . *(b) Find su , sv , and su × sv .

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9.3 Surfaces and Curvature

! 2 (c) Verify that ∥su × sv ∥ = cos(v) a 2 cos2 (v) + sin ? @ (v) a cos(u) cos(v) a sin(u) cos(v) sin(v) √2 2 ,√ 2 2 ,√ 2 2 . 2 2 2 a cos (v)+sin (v)

a cos (v)+sin (v)

and

N (u, v) =

a cos (v)+sin (v)

*(d) Find Nu and write Nu as a multiple of su . (e) Use a computer algebra system to find and simplify Nv . Verify that Nv = (sin2 (v)+a 2a cos2 (v))3/2 sv and so the curvature of a spheroid is 2

κ(u, v) = (sin2 (v)+aa 2 cos2 (v))2 .

0.4 0.2 –1.0

–0.5

0.5

1.0

–0.2 –0.4

Figure 9.23 An ellipse. 9.3.16. Let s(u, v) = ((r cos(v) + R) cos(u), (r cos(v) + R) sin(u), r sin(v)), a parametric representation of the torus. *(a) Find su and sv . (b) Verify that su × sv = (r cos(u) cos(v)(R + r cos(v))

r sin(u) cos(v)(R + r cos(v)), r sin(v)(R + r cos(v)). (c) Verify that ∥su × sv ∥ = r (R + r cos(v)). (d) Find N (u, v), Nu , and Nv . cos(v) s and Nv = r1 sv . (e) Verify that Nu = R+r cos(v) u

9.3.17. Define the hyperbolic trigonometry functions tanh(x) =

e x −e−x e x +e−x

and sech(x) =

2 ee +e−x

.

(a) Verify that tanh2 (x) + sech2 (x) = 1, tanh(x)′ = sech2 (x), sech(x)′ = − tanh(x) sech(x), and sech(x)′′ = − sech3 (x) + tanh2 (x) sech(x). (b) Verify that the pseudosphere s(u, v) = (cos(u) sech(v), sin(u) sech(v), v − tanh(v)) is a surface of constant negative curvature for v ̸= 0. −→ (c) Graph the plane curve c(t) = (sech(t), t − tanh(t)). Explain why we need to require v ̸= 0 in part (b) both visually and from the definition of a curve in Section 9.2. (Sir Isaac Newton, among others in the early days of calculus, studied the −→ curve c(t) = (sech(t), t − tanh(t)), called the tractrix. Revolved it generates the pseudosphere. Beltrami realized in 1868 that the pseudosphere gave a model of hyperbolic geometry.) 9.3.18. Figure 9.24 depicts one branch of the hyperbola x 2 − v 2 = 1, for x > 0. When we rotate it around the vertical axis, as in Example 2, we get a hyperboloid of one sheet. *(a) Find the parametric form for it and graph it. (b) Use a computer algebra system to verify that the curvature of the surface in −1 part (b) is κ(u, v) = (1+2v 2 )2 .

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Differential Geometry

1.5 1.0 0.5

0.5

1.0

1.5

2.0

–0.5 –1.0 –1.5

Figure 9.24 A branch of a hyperbola.

*(c) Repeat part (a) for the hyperboloid of two sheets obtained from v 2 − x 2 = 1, for x > 0 and v ≥ 1. From the graph what can you say about its curvature?

9.4 Geodesics and the Geometry of Surfaces 9.4.1 Geodesics Imagine driving a car in hilly terrain. You can easily distinguish when the car goes straight versus turning to the right or left, independently of the curvature of the hills. Intuitively, a straight path following the terrain is a geodesic. It is a locally shortest path on the surface. Geodesics are the natural generalization to surfaces of Euclidean lines. Great circles are geodesics on a sphere, discussed in Section 1.5. And on a sphere, the shortest path between any two points follows a great circle. On a surface of constant negative curvature the geodesics become the lines of hyperbolic geometry, discussed in Chapter 4. On a torus the geodesics are more complicated. Figure 9.25 depicts some geodesics on a torus that are indeed shortest paths between two points on them. As the figure illustrates, the curvature can be positive or negative and still → the geodesic can be the shortest path. The solid curve − g between points A and B in Figure − → → 9.26 gives a geodesic, even though the curve c seems to provide a shorter path. Curve − g − → would, however, feel straight to someone traveling along it, unlike c . For nearby points on − → g , it is locally a shortest path. In this section we investigate geodesics on surfaces and their relationship to curvature. In addition we explore more deeply the geometry of a surface, leading to some profound insights of Gauss. The ideas extend to higher dimensions, which we consider briefly.

433

9.4 Geodesics and the Geometry of Surfaces

c

B

A g

Figure 9.25 Geodesics on a torus that are shortest paths.

Figure 9.26 A geodesic that is not a shortest path globally.

Imagine attaching a unit tangent vector T (t) to a point (or the car of the previous paragraph) moving along a curve on a surface. In Figure 9.27, the point is “going straight up” the surface. Here the acceleration vector T ′ (t) is perpendicular to the tangent. In this case the curve is a geodesic, following the surface. In contrast, the plane curve in Figure 9.28 is not a geodesic and here T ′ (t) is in the plane, which is the tangent plane.

T(t)

c(t)

T'(t)

T(t)

T'(t)

c(t)

Figure 9.27 A geodesic.

Figure 9.28 A non-geodesic.

−→ −−→ Suppose c(t) is a curve in the surface s(u, v). At a point c(t0 ) = s(u 0 , v 0 ) the unit tangent vector T (t0 ) to the curve and the unit normal N (u 0 , v 0 ) = N (t0 ) are perpendicular to each other. The tangent plane at this point is determined by T (t0 ) and the cross product T (t0 ) × N (t0 ) = B(t0 ), which is called the binormal to the curve. The vectors T (t0 ), N (t0 ), and B(t0 ), form a basis of R3 . So every vector from this point on the curve can be written uniquely as a linear combination of them, in particular, the acceleration vector T ′ (t0 ). By Theorem 9.2.2, T ′ is orthogonal to T , so T ′ is a linear combination of the normal and binormal vectors. The binormal component indicates the amount of turning of the curve in the tangent plane and so the deviation from the geodesic. In Figure 9.27, T ′ (t0 ) is a scalar multiple of the binormal, whereas in Figure 9.28 it is a scalar multiple of the normal. The difference motivates our definition of a geodesic. For simplicity, whenever possible we make the curve move at unit velocity so that −−−→ −−→ c′ (t0 ) = T (t0 ) and c′′ (t0 ) = T ′ (t0 ). −→ Definition. A curve c(t) on a surface s(u, v) is a geodesic if and only if for every value t, there is a scalar α(t) so that T ′ (t) = α(t)N (t).

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Differential Geometry

Example 1. The lines of longitude on a sphere run through the north and south pole and are, as proved here, geodesics. (See Figure 9.29.) A line of longitude has the form & ' & ' & '' & t t t −→ , R sin(u 0 ) cos , R sin for a fixed u 0 c(t) = R cos(u 0 ) cos R R R and the sphere is s(u, v) = (R cos(u) cos(v), R sin(u) cos(v), R sin(v)). By Exercise 9.2.13 and Example 3 of Section 9.3 we have & & ' & ' & '' −− → t t t ′ , − sin(u 0 ) sin , cos , c (t) = T (t) = − cos(u 0 ) sin R R R & & ' & ' & '' −− → 1 t t 1 1 t , − sin(u 0 ) cos , − sin , c′′ (t) = T ′ (t) = − cos(u 0 ) cos R R R R R R and & ' & & ' ' & & '' t t t t N u0, = cos(u 0 ) cos , sin (u 0 ) cos , sin . R R R R Then for every value of t we have T ′ (t) = − R1 N (u 0 , Rt ). Hence these curves are geodesics.

C(t)

T(t) B(t) N(t)

Figure 9.29 Longitude lines are geodesics. In contrast we show that the circles of latitude, depicted in Figure 9.30, are not geodesics, except the equator (when v 0 = 0). Thus we consider values v 0 strictly between − π2 and π2 , other than 0. (The north and south poles, where v 0 = ± π2 , are points, not curves.) Here −→ c(t) = (R cos(t) cos(v 0 ), R sin(t) cos(v 0 ), R sin(v 0 )) for a fixed v 0 . Then −− → c′ (t) = (−R sin(t) cos(v 0 ), R cos(t) cos(v 0 ), 0),

435

9.4 Geodesics and the Geometry of Surfaces

−−→ which is generally not a unit vector. For v 0 other than ± π2 , we can divide c′ (t) by R cos(v 0 ) to find T (t) = (− sin(t), cos(t), 0)

and

′

T (t) = (− cos(t), − sin(t), 0). However, N (t) = N (t, v 0 ) = (cos(t) cos(v 0 ), sin(t) cos(v 0 ), sin(v 0 )) and T ′ (t) is not a scalar multiple of just N (t). That is, circles of latitude are generally not geodesics. Exercise 9.4.5 asks you to show that the equator is a geodesic. ♦

C(t)

B(t) T(t)

N(t)

Figure 9.30 Latitude lines generally are not geodesics. Exercise 9.4.8 asks you to verify that great circles besides longitudes are geodesics.

9.4.2 Geodesics on Surfaces of Revolution Example 1 generalizes to a surface of revolution s(u, v) = ( f (v) cos(u), f (v) sin(u), g(v)). Curves with u 0 fixed are geodesics, as Theorem 9.4.1 shows, but curves with v 0 fixed generally are not. Surfaces of revolution have geodesics in other directions, but they can be quite difficult to determine. We add the assumption f ′ (v)2 + g ′ (v)2 = 1 in Theorem 9.4.1 only to simplify −−→ the proof. It ensures that c′ (t) will have unit length. However, the result remains correct with different parameterizations of the curve. (See Do Carmo [1, 255–258 and 356].) Exercise 9.4.1. Verify that f ′ (v)2 + g ′ (v)2 = 1 holds in Example 1. Theorem 9.4.1. Let s(u, v) = ( f (v) cos(u), f (v) sin(u), g(v)) be a smooth surface of rev−→ olution with f ′ (v)2 + g ′ (v)2 = 1 for all v. Then for all fixed u 0 , the curves c(t) = ( f (t) cos(u 0 ), f (t) sin(u 0 ), g(t)) are geodesics.

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Differential Geometry

Proof. From the assumptions, −− → c′ (t) = T (t) = ( f ′ (t) cos(u 0 ), f ′ (t) sin(u 0 ), g ′ (t))

and T ′ (t) = ( f ′′ (t) cos(u 0 ), f ′′ (t) sin(u 0 ), g ′′ (t)). By Theorem 9.2.2, T (t) · T ′ (t) = 0. That is,

f ′ (t) f ′′ (t) cos2 (u 0 ) + f ′ (t) f ′′ (t) sin2 (u 0 ) + g ′ (t)g ′′ (t) = 0 ′

′′

′

or

′′

f (t) f (t) + g (t)g (t) = 0.

For now assume that f ′ (t) ̸= 0 and rewrite the last equality as f ′′ (t) = tion 9.3.3

−g ′ (t)g ′′ (t) . f ′ (t)

From subsec-

N (t) = (g ′ (t) cos(u 0 ), g ′ (t) sin(u 0 ), − f ′ (t)). −→ For c(t) to be a geodesic, T ′ (t) must be a scalar multiple of N (t) and so the third coordinates must match. That is, we need g ′′ (t) = α(t)(− f ′ (t)) or

α(t) =

−g ′′ (t) . f ′ (t)

Now −g ′′ (t) ′ (g (t) cos(u 0 ), g ′ (t) sin(u 0 ), − f ′ (t)) f ′ (t) ' & ′ −g ′ (t)g ′′ (t) −g (t)g ′′ (t) ′′ cos(u 0 ), sin(u 0 ), g (t) = f ′ (t) f ′ (t)

α(t)N (t) =

= ( f ′′ (t) cos(u 0 ), f ′′ (t) sin(u 0 ), g ′′ (t)) = T ′ (t). −→ Thus c(t) is a geodesic. Exercise 9.4.9 considers the case f ′ (t) = 0. !

9.4.3 Arc Length on Surfaces Figure 9.31 depicts a straight line and a right triangle on a flat piece of paper. When we roll the paper into a cylinder, as in Figure 9.32, the line becomes a helix, which by Exercise 9.4.10 is a geodesic of the cylinder. Distances on the surface of the cylinder work just like Euclidean distance, as long as we don’t go too far around it. That is, in Figure 9.32 we can use the Pythagorean theorem and the lengths a and b to find c, provided the triangle is made from geodesic segments on the cylinder. In this sense, the cylinder remains flat, in spite of the bending of the paper. And, as found in Exercise 9.3.9, the cylinder has 0 curvature everywhere. The length of a curve on a surface with nonzero curvature requires a more sophisticated approach.

437

9.4 Geodesics and the Geometry of Surfaces

m

c A

b

B a C

Figure 9.32 Corresponding shapes on a

Figure 9.31 Shapes in the plane.

cylinder.

Gauss generalized the Pythagorean theorem to curved surfaces. Since general surfaces vary from point to point, his approach had to be local. He used the language of differentials, which we still use. For a surface s(u, v), Gauss expressed the distance along a geodesic in terms of the distance along the cross section curves. In Figure 9.33 we want to determine the differential (infinitesimal distance) ds along the geodesic from (u 0 , v 0 ) to (u 1 , v 1 ) in terms of the differentials du and dv. Gauss found the following amazing formula, now called the first fundamental form. (See Do Carmo [1, 92].)

ds (u0,v0) du

(u1,v1) dv (u1,v0)

Figure 9.33 A differential on a surface. ds 2 = Edu 2 + 2Fdudv + Gdv 2 ,

(9.5)

where E = su · su , F = su · sv , and G = sv · sv . Exercise 9.4.2.∗ For the cylinder s(u, v) = (R cos( Ru ), R sin( Ru ), v) find E, F, and G from equation 9.5. Explain why the values validate the text’s assertion about distances on the surface of the cylinder.

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Differential Geometry

(9.5) generalizes the distance formulas of exercises 3.1.9 and 3.1.10 when the scales of the axes differ and are at an angle to one another. In linear algebra terms, for each point of the surface (9.5) determines an inner product for the tangent plane. From Exercise 9.3.8, for surfaces of revolution su and sv are orthogonal, and so F is always 0. Exercise 9.4.3.∗ For a surface of revolution as in Theorem 9.4.1 find E(u, v) and G(u, v) to verify that ds 2 = f (v)2 du 2 + dv 2 . Theorem 9.4.2 uses (9.5) to find the length along a curve on a surface. Calculus texts derive the arc length of a plane curve through approximation with the lengths $si of line segments, as in Figure 9.34. A larger number of segments can improve the approximation. The plane curve "b! −→ c(t) = (x(t), y(t)) has arc length s = a x ′ (t)2 + y ′ (t)2 dt for a ≤ t ≤ b. On a surface we simi−→ larly approximate the length of a curve c(t) using " b geodesic " b √segments, as in Figure 9.35, giving s ≈ $ $si . In the limit we get an integral: s = a ds = a Edu 2 + 2Fdudv + Gdv 2 . From calculus du = u ′ (t)dt and dv = v ′ (t)dt. The reasoning of this paragraph gives us Theorem 9.4.2. 1.0 ∆s2 0.8 ∆x2 0.6 ∆s1

∆y3

∆s3

0.4

∆y1

0.2 ∆x1 –0.5

–1.0

∆y2

∆x3 0.5

1.0

Figure 9.34 Arc length of a plane curve. s3 s2 c(t)

s1

Figure 9.35 Arc length on a surface. −→ Theorem 9.4.2. On a smooth surface s(u, v), the length of a curve c(t) from t = a to t = b is % b! E(t)u ′ (t)2 + 2F(t)u ′ (t)v ′ (t) + G(t)v ′ (t)2 dt. (9.6) s= a

9.4 Geodesics and the Geometry of Surfaces

439

Example 2. For the sphere of radius R, s(u, v) = (R cos(u) cos( Rv ), R sin(u) cos( Rv ), R sin( Rv ). From Exercise 9.4.3, E(u, v) = R 2 cos2 ( Rv ), F = 0, and G = 1. A circle of latitude with v = −→ v 0 has c(t) = (R cos(t) cos( vR0 ), R sin(t) cos( vR0 ), R sin( vR0 )) for 0 ≤ t ≤ 2π . Here u(t) = t and v(t) = v 0 . We can substitute these values into (9.6) to obtain % 2π 0 ? @0 % 2π < 0 ? v @0 ?v @ v0 0 0 0 0 0 0 2 2 R cos R 0cos · 1 + 0 + 0dt = s= 0 dt = 2π R 0cos 0, R R R 0 0 the circumference of a circle with radius R cos( vR0 ). ♦

9.4.4 The Gauss-Bonnet Theorem Gauss coupled his insights on curvature and geodesics to relate areas of triangles and curvature. The proof of Theorem 9.4.3 goes beyond the level of this text. However, its statement unites Theorem 1.5.3 and Theorem 4.1.1 and so shows the unity of spherical and hyperbolic geometries. The notation dσ in the theorem is a shorthand for the differential of area taking into account the values of E, F, and G in (9.5). Theorem 9.4.3 (Gauss part of Gauss-Bonnet Theorem, 1827). Suppose a triangular region T of a smooth surface is bounded by geodesics and the measures of the angles made by the geodesics are α, β, and γ . Assume that T can be shrunk continuously to a point on the "" κ(u, v)dσ , where κ(u, v) is the curvature at (u, v). surface. Then α + β + γ − π = T

Proof. See Do Carmo [1, 264–270]. !

"" Example 3. A sphere of radius R has constant curvature R12 . Then T κ(u, v)dσ = R12 area(T ). Thus we can rearrange the equation of Theorem 9.4.3 to give Theorem 1.5.3 for spherical triangles: R 2 (α + β + γ − π) = area(T ). In Chapter 1 we called α + β + γ − π the spherical excess. Analogously, Theorem 4.1.1 tells us about the area of triangles in hyperbolic geometry. The hyperbolic plane has constant negative curvature, say − k1 . (See Exercise 9.3.17.) The minus sign changes Gauss’s formula to −k(α + β + γ − π ) = k(π − (α + β + γ )) = area(T ). We called π − (α + β + γ ) the defect of a triangle. In between spherical and hyperbolic geometries is Euclidean geometry, where the curvature is 0. In the Euclidean case, Theorem 9.4.3 reduces to Theorem 1.1.1, and the angle sum of a Euclidean triangle is π or 180◦ . ♦ Pierre Ossian Bonnet (1819–1892) generalized Gauss’s result in 1848 to regions bounded by curves, whether or not they are geodesics. (See Do Carmo [1, 264–270].) He also considered regions that can’t be shrunk to a point. For example, the shaded band on the torus in Figure 9.36 can’t be shrunk continuously on the surface to a point. The general Gauss-Bonnet theorem needs to take into account the topological type of the surface, using the Euler characteristic. (See Project 9. Also, Project 12 of Chapter 1 considers the generalized Euler’s formula, which relates to the Euler characteristic.)

9.4.5 Higher Dimensions Bernhard Riemann (1826–1866) boldly extended Gauss’s vision from two-dimensional surfaces to what are called n-dimensional manifolds. All our work has embedded curves and surfaces

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Differential Geometry

Figure 9.36 A band on a torus. in 2- and 3-dimensional Euclidean space. While the embedding simplified our presentation, it would have made Riemann’s advances impossible. For example, we heavily used the cross product, which is defined only in R3 . Riemann needed the results Gauss proved about surfaces independent of whether they were embedded in a space. Riemann generalized the concepts of curvature, geodesics, and more. For example, a 3-dimensional manifold m(u, v, w) has lengths generalizing the 3-dimensional Pythagorean theorem. Riemann’s version of (9.5) for the 3-manifold m(u, v, w) is ds 2 = Edu 2 + 2Fdudv + Gdv 2 + 2H dudw + 2I dvdw + J dw2 , where E = mu · mu , F = mu · mv , G = mv · mv , H = mu · mw , I = mv · mw ,

and

J = mw · mw

(9.7)

Exercise 9.4.4. Verify that the three-dimensional Pythagorean theorem (Exercise 1.5.11) corresponds to a 3-dimensional manifold with E = G = J = 1 and F = H = I = 0. Einstein’s general theory of relativity is related to 4-dimensional manifolds since physics uses three spatial dimensions and one time dimension. The values of the parameters in the formula analogous to (9.5) and (9.6) depend on the gravitational field. (However, the value corresponding to ds 2 can come out negative, so the space isn’t Riemannian. See the discussions in Section 5.5 and 6.6.) Alternatively, we could take the curvature of the manifold as determining what gravity is at each point. In any case, light travels on geodesics of it and so bends from a Euclidean straight line. In 1919 Sir Arthur Eddington provided the most well known confirmation of this bending of light, mentioned in the introduction to Chapter 9. The general theory of relativity has everyday effects as well. The calculations for global positioning systems need to compensate for relativistic effects. The effects include the difference of the earth’s gravitational field around the satellites compared with the field on the earth’s surface where we use devices to find position. The gravitational differences modify how fast the highly accurate atomic clocks on the satellites run compared to earth clocks. Hence GPS systems must compensate for relativistic effects to avoid significant errors. (See Love [5].) Differential geometry has provided profound understanding in mathematics and physics. At the heart of its insights are the geometric concepts of curvature and geodesics.

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9.4 Geodesics and the Geometry of Surfaces

9.4.6 Exercises for Section 9.4 *9.4.5. In Example 1 verify that the equator is a geodesic. 9.4.6. In parts (a), (b), and (c) find the length of the paths on s(u, v) = ((R + cos(v)) cos(u), (R + r cos(v)) sin(u), r sin(v)), the torus in Figure 9.37. Hint: The paths combine arcs of circles. (a) (b) (c) (d)

From A to D to G (a half circle). From A to C to F to I to G (three half circles). From A to B to E to H to G. Consider a path made of a geodesic from A to F together with a geodesic from F to G. Do you think it will be shorter than any of the previous paths? Explain. (e) Repeat part (d) for a path made of geodesics from A to E and from E to G.

B

H C

F

I

A

G

E D

Figure 9.37 Paths of a torus. 9.4.7. (a) Use Figure 9.38 to make a physical cone from a part of a circle of radius R, the distance from (0, 0) to A. Draw a line such as k on the paper when it is flat before rolling it into a cone. Describe the shape of the curve k becomes. ( (0, 0) will become the cone’s apex.) (b) Explain why straight lines, such as k, on the flat sheet of paper will become geodesics on the cone once the paper is rolled up. (c) The surface for the cone can be written as s(u, v) = (v cos(u), v sin(u), mv) for −α ≤ u ≤ α and 0 ≤ v. That is, a general point B on the cone has coordinates (v cos(u), v sin(u), mv). The corresponding point B on the flat paper has polar coordinates (r, θ ). Explain why u = πα θ. Find m and v in terms of r and α. Hint: The arc of the circular piece of paper becomes the circumference of an entire circle for the cone. (d) Suppose, as in Figure 9.39, that α is less than π2 . Verify physically that in this situation a geodesic on a cone can intersect itself. Imagine extending k. For what values of α can the geodesic intersect itself twice? n times? Use part (c) to find the coordinates (r, θ ) and (r, −θ) in terms of d and α, where the line k in Figure 9.39 will intersect itself the first time when rolled into a cone. If the geodesic intersects itself a second time, give the r and θ coordinates for the second intersection. Hint: use more of the circle, as indicated by the dashed lines and overlap it.

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Differential Geometry

B

(r,θ)

α

α

r

(0,0)

θ

A

(0,0)

k (d,0)

k (r,-θ)

Figure 9.38 Pattern for a cone.

Figure 9.39 A self-intersecting geodesic.

9.4.8. This problem verifies that great circles on the unit sphere through (0, 1, 0) and (0, −1, 0) are geodesics. (a) Explain why a great circle is the intersection of the sphere with a Euclidean plane of the form ax + by + cz = 0, where a, b, and c are not all 0. −→ (b) Verify that for −1 ≤ p ≤ 1 and − π2 ≤ t ≤ π2 the curve c(t) = ( p cos(t), ! sin(t), 1 − p 2 cos(t)) is on the unit sphere and goes through the points (0, 1, 0) and (0, −1, 0). Determine in terms of p which plane from part (a) con−→ tains c(t). (c) Find T (t) and T ′ (t) for the curve in part (b). (d) Use Example 3 of Section 9.3 to verify that the curve of part (b) is a geodesic. (e) Explain how to modify parts (a) through (d) to apply to all great circles on the unit sphere. (f) Redo parts (a) to (d) for a sphere of radius R. 9.4.9. Do the case f ′ (t) = 0 to finish the proof of Theorem 9.4.1. −→ 9.4.10. (a) Show that for m and b the helix c(t) = (R cos(t), R sin(t), mt + b) is a geodesic on the cylinder s(u, v) = (R cos(u), R sin(u), v). Hint: See Example 6 of Section 9.2. −−−→ −−→ (b) Verify that the helices c1 (t) = (R cos(t), R sin(t), t) and c2 (w) = (R cos(w), R sin(w), 2w) intersect infinitely often at points with t = 4π j, where j is an integer. Determine the values of w for the intersections. Is either curve the shortest path between the points where t = 0 and where t = 4π ? Explain your answer. (c) For m and w nonzero and non-equal find the values of t and w for the in−−→ finitely many intersections of the helices cm (t) = (R cos(t), R sin(t), mt) and −−−→ ck (w) = (R cos(w), R sin(w), kw + b). Explain why any two helices with different values m and k must intersect infinitely often. Explain what happens when m = k and b ̸= 0.

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9.4.11. For the surface of revolution s(u, v) = ( f (v) cos(u), f (v) sin(u), g(v)), we deter−→ mine when the curve c(t) = ( f (v 0 ) cos(t), f (v 0 ) sin(t), g(v 0 )) is a geodesic for v 0 fixed. −−→ (a) Find c′ (t), T (t) and T ′ (t). (b) Find N (t). −→ (c) What condition must hold at v 0 for c(t) to be a geodesic? What is/are the value(s) of α(t)? (d) For the torus S(u, v) = ((r cos(v) + R) cos(u), (r cos(v) + R) sin(u), r sin(v)) what values of v 0 satisfy the condition of part (c)? What are the geodesics? Do the curves fit your intuition of a geodesic? Explain. (See Example 2 of Section 9.3.) 9.4.12. Use (9.6) to find the length from the south pole to the north pole along a circle of longitude on the unit sphere. Show your work. −−→ 9.4.13. (a) The curve w(t) = (cos2 (t), sin(t) cos(t), sin(t)), for − π2 ≤ t ≤ π2 , is on the unit sphere. (See Exercise 9.2.13 (c).) Use a computer algebra system to approximate −−→ the length of w(t). −−−→ (b) Verify that the curve wm (t) = (cos(mt) cos(t), sin(mt) cos(t), sin(t)), for − π2 ≤ t ≤ π , is on the unit sphere. Use a computer algebra system to approximate the length 2 −−→ −−−→ of wm (t) for m = 0, 0.5, 2, and 4. What curve is w0 (t)? Graph the curves for m = 0.5, 2, and 4. What is the effect of increasing m? 9.4.14. We can represent the paraboloid as s(u, v) = (v cos(u), v sin(u), v 2 ) or as −→ p(x, y) = (x, y, x 2 + y 2 ). From Theorem 9.4.1 we know that the curves c(t) = (t cos(b), t sin(b), t 2 ) with b constant are geodesics on s(u, v). (a) Find E(u, v), F(u, v), and G(u, v) for s(u, v) and E(x, y), F(x, y), and G(x, y) for p(x, y). −→ (b) Let b be a constant with − π2 < b < π2 . Show that the curve c(t) = (t cos(b), ! t sin(b), t 2 ), for 0 ≤ t ≤ k 1 + tan2 (b) on the surface s(u, v) is the same curve as −−→ f (z) = (z, mz, z 2 (1 + m 2 )), for 0 ≤ z ≤ k and m = tan(b) on the surface p(x, y). Hint: Use z = t cos(b) to convert between the curves. Describe them. (c) Use a computer algebra system and Theorem 9.4.2 to find the length of the curves in part (b), using k = 1 and b = 1, b = π4 , and b = π3 . Hint: Use your description of the curves. For the same value of b, the two curves have the same length. Which way of representing the paraboloid and curve is easier to use to find the lengths? Explain. 9.4.15. For surfaces ds 2 has three terms in (9.5) and for 3-dimensional manifolds ds 2 has six terms in (9.6). (a) Give the equation for ds 2 for a 4-dimensional manifold. (b) Determine the number of terms for ds 2 in an n-dimensional manifold and describe their form. Explain your answer. 9.4.16. A hypersphere in four dimensions satisfies the equation x 2 + y 2 + z 2 + t 2 = 1.

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(a) Verify that we can write the unit hypersphere as the manifold of revolution h(u, v, w) = (cos(u) cos(v) cos(w), sin(u) cos(v) cos(w), sin(v) cos(w), sin(w)). (b) Find the values of E, F, G, H, I, and J in (9.6) for h(u, v, w) in part (a). (c) How would parts (a) and (b) be modified for a hypersphere of radius R? (d) What equation should the unit hypersphere in five dimensions satisfy? (e) Give the parametric form for the unit hypersphere in five dimensions as a manifold of revolution similar to h(u, v, w) in part (a). (f) Conjecture what you think the values should be for the terms of ds 2 for your answer in part (e). (g) Generalize parts (d), (e), and (f) to higher dimensions.

9.4.7 Albert Einstein The most incomprehensible thing about the world is that it is comprehensible. —Albert Einstein Time and space and gravitation have no separate existence from matter. —Albert Einstein

His deep physical intuition helped make Albert Einstein (1879–1955) the greatest physicist since Newton. His undergraduate training was to be a mathematics and physics teacher, but he was unable to find a position when he graduated at age 22. He worked for several years in the Swiss patent office while he worked on his doctorate in physics, completing it in 1905. While still at the patent office he published groundbreaking papers on the special theory of relativity, quantum mechanics, and Brownian motion. The first two profoundly altered the study of physics, but in 1921 he received the Nobel prize in physics for the third. By this time he had published his general theory of relativity and was internationally known for his work in relativity. Einstein became a professor in 1908 and after several posts in Switzerland, returned to Germany, where he was a professor until 1933. The rise of the Nazis caused him to renounce his German citizenship and emigrate to the United States. He was professor at Princeton and at the Institute for Advanced Study until his death. During World War II, his letter to President Franklin D. Roosevelt was instrumental in convincing Roosevelt to establish the research program resulting in the development of the atomic bomb. After the establishment of the state of Israel, he was offered the position of its first president, but declined the honor. Einstein continued to work once he became world famous. However, neither he nor anyone since then has been able to achieve his goal of a unified field theory, which would combine quantum mechanics, relativity theory, and indeed all of physics. He doubted the standard interpretation of quantum mechanics with its inherent indeterminacy. In his efforts to counter it, he suggested a number of experiments that spurred others to important discoveries. However, all these discoveries supported the standard interpretation of quantum mechanics. While Einstein may not strictly be a mathematician, his theoretical physics depended strongly on geometry and mathematics in general. And mathematics and many mathematicians have been inspired by his profound insights.

9.4 Geodesics and the Geometry of Surfaces

9.4.8 Projects Chapter 9

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−→ 1. Torsion measures the amount a curve c(t) curves out of the plane determined by T (t) and N (t), its unit tangent and normal. By definition, the torsion equals −B ′ (t) · N (t), where B(t) is the unit binormal. −→ (a) Verify that a plane curve c(t) = (x(t), y(t), 0) has torsion 0. (You may assume that x ′ (t)2 + y ′ (t)2 = 1 = x ′′ (t)2 + y ′′ (t)2 .) −→ (b) Find the torsion of the helix c(t) = (R cos( √ R 2t+m 2 ), R sin( √ R 2t+m 2 ), √ R 2m+m 2 t). Hint: It is constant. (c) Find the torsion for lines of longitude and lines of latitude on the unit sphere. (d) Investigate torsion for other curves. (See Oprea [7, 22ff].) 2. Investigate ruled surfaces—curved surfaces with straight lines lying in the surface through any point. Make models of ruled surfaces using string. (See Exercise 3.5.14 and Henderson [2, 69-72].) 3. (a) Imagine a bug crawling counterclockwise completely around a convex polygon in the plane. Use the concept of the external angles of a polygon to explain why the bug turns a total of 360◦ = 2π . (b) Does the turning of part (a) change if the polygon is not convex? Explain. (c) What is the total turning if the bug is on a sphere and goes around a spherical triangle with angles α, β, and γ ? Relate it to the area of the triangle and the square of the radius of the sphere. (d) Investigate parallel transport and holonomy in general. (See Henderson [2, 103ff] and Henderson and Taimina [3, Chapters 7 and 8].) 4. Investigate the geometry of soap bubbles, minimal surfaces, and mean curvature. (The mean curvature of a surface at a point is one half of the trace of the shape operator at that point. A surface is minimal if and only if the mean curvature is 0 at every point. Soap bubbles are physical models of surfaces with constant mean curvature. See Oprea [7, Chapter 4].) 5. Investigate map projections. (See McCleary [6, 116–130].) 6. The surface s(u, v) = (cos(u), sin(u), cos(v), sin(v)) in 4-dimensional space is often called a flat torus because it has zero curvature everywhere, yet has circular cross sections whether we fix u or v. (a) Explain why the definition of curvature in Section 9.3 only applies to surfaces embedded in R3 . (b) Use (9.5) to find ds 2 for the flat torus and explain why the adjective “flat” is appropriate. (c) Use part (b) to explain why it is reasonable to expect curves with u(t) = mt + b and v(t) = kt + d to be geodesics on the flat torus. Relate your explanation to Exercise 3.3.20. (See Oprea [7, 245–246] and Do Carmo [1, 433–435].) 7. Investigate involutes and evolutes. (See McCleary [6, 72–78].) 8. Investigate the connection between differential geometry and models of hyperbolic geometry. (See McCleary [6, 217–239].) 9. Investigate the Gauss-Bonnet theorem and the Euler characteristic of a surface. (See Weeks [9, Chapter 12] and Oprea [7, 311–319].) 10. Investigate the intuition of 3-dimensional manifolds. (See Weeks [9].) 11. Investigate the geometric intuition of four dimensions and relativity. (See Rucker [8].)

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12. Investigate the history of differential geometry. (See Kline [4, Chapters 23 and 37] and McCleary [6].) For example, study Riemann’s address On the Hypotheses which Lie at the Foundations of Geometry, found in McCleary [6, 269–278]. 13. Write an essay relating properties of hyperbolic geometry, spherical geometry, and differential geometry.

9.4.9 Suggested Readings 1. Do Carmo, M., Differential geometry of Curves and Surfaces, Englewood Cliffs, NJ: Prentice Hall, 1976. 2. Henderson, D., Differential Geometry, Upper Saddle River, NJ: Prentice Hall, 1998. 3. Henderson, D. and D. Taimina, Experiencing Geometry, 3rd ed., Upper Saddle River, NJ: Prentice Hall, 2001. 4. Kline, M., Mathematical Thought from Ancient to Modern Times, New York: Oxford University Press, 1972. 5. Love, A., GPS, atomic clocks and relativity, IEEE Potentials. vol. 13, issue 2, (1994), 11–15. 6. McCleary, J., Geometry from a Differentiable Viewpoint, New York: Cambridge University Press, 1994. 7. Oprea, J., Differential Geometry and Its Applications, 2nd ed., Upper Saddle River, NJ: Pearson Prentice Hall, 2004. 8. Rucker, R., Geometry, Relativity and the Fourth Dimension, New York: Dover, 1977. 9. Weeks, J., The Shape of Space, 2nd ed., New York: Dekker, 2002.

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Discrete Geometry

Figure 10.0 The “Ghost Map” In 1854 a terrible cholera epidemic broke out in London. At the time people believed that bad air caused epidemics. However, Dr. John Snow suspected fouled water as the cause. As Snow found, the deaths clustered around a water pump, which he convinced authorities to disable. Later Snow later published a book with a map of the houses where cholera deaths occurred with bars indicating each death and the sites of the water pumps where people got their water. The map (Figure 10.0) showed the concentration of deaths around the Broad Street pump in the center. Snow’s use of statistics and the so-called “ghost map” marked an important advance in public health. A Voronoi diagram generalizes Snow’s idea: given a set of sites (here the water pumps), a Voronoi diagram determines the regions of points closest to each site.

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10.1 Overview and Explorations Throughout the history of geometry mathematicians have explored a variety of questions and topics that are now considered part of discrete geometry. For example, what polygons can we use to cover the plane without gaps or overlaps? Can we always divide the interior of a polygon into triangular regions using only its vertices? If so, in how many ways? While polygons are traditional geometric objects, an answer to the questions won’t be found in Euclid’s text or high school geometry texts. For another example, how many different distances can a set of n points determine? A number of mathematicians in the twentieth century focused on the preceding questions and similar ones that address combinatorial aspects of points, lines, polygons, and other configurations. Gradually the questions coalesced into a separate area of geometry, called discrete geometry. Discrete geometry tends to organize around clusters of easily understood problems, rather than a unified theory. A slight alternation of a discrete geometry question can change it from an easily solvable one to one beyond the knowledge of even experts. Part of the attraction of the field lies in how the solutions of two seemingly similar problems can range so widely in difficulty. The focus on problems and particularly the large number of unsolved problems reflect the newness of this field. In response to discrete geometry’s focus on problems, this chapter centers on problems, rather than following the organization of earlier chapters. The first section gives short explorations of a number of problems. Their range may give a better sense of discrete geometry as a field than an attempt at a definition. Later sections will develop them more fully, including theorems and unsolved problems. While we prove many of the theorems, some are merely stated because their proofs go beyond the level of this introduction. All questions concern Euclidean geometry unless explicitly stated otherwise. The increasing application of computers in many fields made some discrete geometry topics suddenly useful. For example, epidemiologists use Voronoi diagrams, described in this chapter’s introduction, to study outbreaks of disease. However, there are often many possible sources, instead of the handful of pumps in Dr. Snow’s map. With a large number of potential sources, scientists turn to computers to determine the Voronoi diagram and so look for clustering. Computational geometry arose in response to the need for efficient algorithms to implement a variety of geometric ideas using computers. This chapter will focus on the discrete geometry, rather than the algorithmic aspects, even though they are closely tied together.

10.1.1 Distances between Points We call triangles with one, two, or three different lengths of sides equilateral, isosceles, or scalene, respectively. However, we lack words for the corresponding situation with quadrilaterals, let alone general polygons. Let’s rephrase the idea by considering Cn D the nonzero distances determined by sets of points. With n points, there are n(n − 1)/2 = 2 pairs of points and so up to that many C Ddistances possible. We write nk for the combinations of n things k at aCtime, D then!number of ways of choosing . See, for example, Sibley a subset of size k from a set of size n. As is well known, nk = k!(n−k)! [12, 239–243] for more on combinations.

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Exercise 10.1.1.∗ Find an example of four points in the Euclidean plane that determine exactly two nonzero distances. Repeat for exactly three, four, five, and six nonzero distances. Explain why no set of four distinct points in the Euclidean plane can determine just one nonzero distance. CD It is easy to find a set of n points determining the maximum number n2 of distances, but considerably harder to find sets determining the minimum number of distances, even in the plane. In 1946 Paul Erd os ˝ posed the general question “What is the minimum number of distances n points determine in d-dimensional Euclidean space?” We denote this minimum number of distances by Dd (n). For most values of n and d, the best results that mathematicians, including Erd os ˝ in his 1946 paper, have been able to provide are upper and lower bounds on Dd (n). The techniques used to bound Dd (n) and other discrete geometry values have proven their worth repeatedly in a variety of computer applications. Many applications can be recast as a need to solve some problem in terms of n, the number of some sort of input. The amount of computer time and so cost to solve the problem is then a function of n. Some of the functions can be determined exactly, but often upper and lower bounds give the best known results. As computer science quickly grew after World War II it benefited greatly from the techniques mathematicians developed for finding such bounds.

10.1.2 Triangulations Figure 10.1 depicts a triangulation of a polygon, a division of it into triangles all of whose vertices are vertices of it.

Figure 10.1 Triangulating a polygon. Exercise 10.1.2.∗ Find another triangulation of the polygon in Figure 10.1.1. Does it have the same number of triangles as in Figure 10.1.1? Exercise 10.1.3.∗ Describe a way to triangulate any convex polygon. Triangulations enable programmers to have computers analyze shapes in a range of applications. Algorithms for visibility, aspects of robotics, mesh generation of surfaces, and more start from triangulations. To keep our discussion elementary and focused on geometry rather than algorithms, we restrict our exploration to planar polygons.

10.1.3 The Art Gallery Problem Imagine that Figure 10.2 represents the floor plan of an imaginary art gallery or museum. We need to place guards so that they can observe every interior point. Two natural questions arise:

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What is the minimum number of guards needed for a given floor plan? Where should they be placed?

Figure 10.2 Polygon representing an art gallery. Exercise 10.1.4.∗ Find a placement of three guards that solves the art gallery problem for the floor plan of Figure 10.2. Explain why we need at least three guards. As mathematicians we state the problem more generally: find the smallest number of guards in terms of n that suffice to observe every interior point in any polygon with n vertices. Of course, to solve this problem, we will need to know where to put the guards. V´aclav Chv´atal solved the problem in 1973, soon after he heard of it. Since mathematicians found this problem interesting and others found the solution useful, people quickly posed a number of variations. For example, real buildings usually have walls at right angles. Does that restriction lower the required number of guards? How many guards are needed to observe the outside of a fortress?

10.1.4 Tilings An infinite checkerboard covers the plane with squares without gaps or overlapping regions. Figure 10.3 depicts different ways to tile the plane without gaps or overlaps using copies of one or more regions. A plane tiling is a set of tiles whose union is the entire plane and whose interiors are pairwise disjoint. (To keep our discussion elementary, we avoid technical definitions of all the terms. See Gr¨unbaum and Shephard [5, 16–19] for a more detailed exposition.)

Figure 10.3 Two plane tilings. An overly naive goal would be to describe all tiles so that isometric copies of one tile give a tiling of the plane. We call such tilings monohedral, meaning “one faced.” Section 10.3 will

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consider tilings in more depth, including an indication of why we can’t hope to determine all regions that give monohedral tilings. Exercise 10.1.5.∗ Determine the three regular polygons that give monohedral tilings of the plane. Explain why no other regular polygons can give a monohedral tiling. Exercise 10.1.6.∗ Devise a monohedral tiling of the plane with all tiles isometric to a generic triangle, like the one in Figure 10.4. (Discrete geometers use “generic” to indicate an object with no special conditions. For example, a generic triangle would have three different lengths for its sides and no special angles, so it would not be equilateral, right, or isosceles.)

Figure 10.4 Tile the plane with this triangle.

10.1.5 Voronoi Diagrams Example 1. An imaginary town has five fire stations. Figure 10.5 depicts the regions closest to each fire station. To minimize response time to a fire, each fire station should respond to fires in its region. How do we determine the boundaries of the regions?

C

B

A

D E

Figure 10.5 Regions closest to specified points.

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Solution. We first solve an easier problem: find the points equidistant from two points A and B. In Figure 10.6, the points equidistant from A and B form the perpendicular bisector of the points A and B. In Figure 10.6, the half plane determined by the bisector and containing A includes all points closer to A than B.

B

A

Figure 10.6 A Voronoi diagram with two points. To solve the problem of the five fire stations, we draw the ten perpendicular bisectors determined by the pairs of fire stations. As Figure 10.7 indicates, the lines determine the desired

BC BD

C AC

AB B

CD

A AD CE AE D

BE E

DE

Figure 10.7 A Voronoi diagram for five points.

10.1 Overview and Explorations

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regions around each fire station. Actually, only parts of eight of the ten lines, shown as heavy segments, are needed to determine the regions. ♦ Exercise 10.1.7.∗ Explain why the two other lines in Figure 10.7 are not needed to determine the five regions. Definition. Let A ∈ S, where S is a set of vertices in the plane. The Voronoi region of A is the set of points P so that for all vertices B ∈ S, d(A, P) ≤ d(B, P). The Voronoi diagram of S, V or (S), is the set of boundaries of the Voronoi regions for the vertices in S. We call the vertices of S sites. Exercise 10.1.8.∗ Explain why, if S is finite, some of the Voronoi regions must be unbounded.

10.1.6 Exercises for Section 10.1 10.1.9. ∗ (a) Suppose a convex quadrilateral in the plane has exactly two nonzero distances among its sides and diagonals. Must it be a square? If yes, justify your answer; if no, give a counterexample. (b) Can there be a non-convex quadrilateral that determines exactly two nonzero distances? If yes, give an example. If not, explain why not. 10.1.10. For the following types of quadrilaterals, give the possible number of nonzero distances it can have. Give an example for each possibility. ∗

(a) (b) ∗ (c) ∗ (d) (e) (f) ∗

Rectangle, but not a square. Rhombus, but not a square. Parallelogram, but neither a rectangle nor a rhombus. Trapezoid (but not a parallelogram). Kite but not a rhombus. None of the above.

10.1.11. ∗ (a) For four points on a line find the range of possible nonzero distances they determine. For each number of different distances give an example. Hint: All the distances can be whole numbers. (b) Repeat part (a) for five points on a line. (c) Make a conjecture about D1 (n), the minimum number of distances n points on a line can determine. Explain or, even better, justify your answer. (d) Make a conjecture about the possible range of distances n points on a line can determine. Explain or, even better, justify your answer. 10.1.12. Find the minimum number of distances four points can determine in the plane using the taxicab metric. Give an example. (Example 3 of Section 2.3 describes the taxicab metric.) ∗

10.1.13. Give your best estimate for D2 (5), D2 (6), D2 (7), and D2 (8). Make a conjecture about D2 (n). Explain. 10.1.14. Give your best estimate for D3 (4), D3 (5), D3 (6), D3 (8), and D3 (12). Explain.

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10.1.15. ∗ (a) How many diagonals does a convex 4-gon have? ∗ (b) Repeat part (a) for a convex 5-gon. (c) Repeat part (a) for a convex n-gon. Justify your answer. 10.1.16. ∗ (a) How many ways can you triangulate a convex 4-gon so that the vertices of each triangle are vertices of the original polygon? ∗ (b) Repeat part (a) for a convex 5-gon. (c) Repeat part (a) for a convex 6-gon. (d) Conjecture the number of ways of triangulating a convex n-gon. (e) Repeat part (a) for a non-convex quadrilateral. (f) What are the possible number of triangulations for a non-convex pentagon? Give an example for each. (g) Give an example of a non-convex hexagon with just one possible triangulation. (h) Give several examples of non-convex hexagons with different numbers of triangulations differing from one and from the number in part (c). 10.1.17. ∗ (a) Explain why only one guard is needed to solve the art gallery problem for a quadrilateral. ∗ (b) Find a polygon with the minimum number of sides needing two guards. (c) Repeat part (b) for three guards. (d) Make a conjecture relating the number of sides of a polygon and the number of guards needed. (e) A polygon is orthogonal if and only if all adjacent edges meet at right angles. What is the minimum number of sides for an orthogonal polygon needing two guards? Give an example. (f) Repeat part (e) for three guards. (g) Repeat part (d) for orthogonal polygons. (h) Why must an orthogonal polygon have an even number of vertices? 10.1.18. The fortress problem asks for the minimum number of guards who need to observe the exterior of a polygon (the fortress). We represent the guards as points on the polygon. ∗

(a) Solve the fortress problem when the polygon is a convex quadrilateral. Explain. ∗ (b) Repeat part (a) for a convex pentagon. (c) Repeat part (a) for a convex hexagon. (d) Repeat part (a) for a convex 7-gon. (e) Make a conjecture about the minimum number of guards when the fortress is a convex n-gon. Explain how to achieve the minimum. (f) Do the numbers in parts (a), (b), or (c) change if the polygon is not convex? 10.1.19. (a) Use graph paper to design a monohedral tiling with an asymmetric, non-convex quadrilateral, as in Figure 10.8. (b) Use graph paper to design a monohedral tiling with a pentagon having two adjacent right angles, as in Figure 10.9.

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Figure 10.8 A quadrilateral tile.

Figure 10.9 A pentagonal tile.

(c) Design a pentagon with two non-adjacent right angles that can’t be a monohedral tile. Explain why it fails to tile. (d) Design a monohedral tile whose boundary contains at least one half circle. ∗

10.1.20. The designs in Figure 10.10 are called tetrominoes and show the five different ways four squares can fit together edge to edge. Which of them give a monohedral tiling? Justify your answer.

Figure 10.10 The five tetrominoes. ∗

10.1.21. Design a tiling using a regular octagon and a square as the tiles. 10.1.22. Design a tiling using two triangular tiles whose six sides are all different lengths and are arranged in such a way that both kinds of tiles are around any point where more than two tiles have a vertex. 10.1.23. (a) Determine the Voronoi diagram for the sites given by (2, 0), (−2, 0), (0, 2), and (0, 4). Give a graph of the diagram. Show your work. (b) Repeat part (a) for the sites (2, 0), (−2, 0), (0, 2), and (0, 6). (c) Repeat part (a) for the sites (2, 0), (−2, 0), (0, 2), and (0, c), where c > 2. (d) Describe how the Voronoi diagram of part (c) changes as c increases. 10.1.24. (a) Determine the Voronoi diagram for the sites given by (2, 1), (−1, 3), (−2, −2), and (4, −2). Give a graph of the diagram. Show your work. (b) Add the site (0, 0) to those of part (a) and compute and graph the new Voronoi diagram. How does the diagram relate to the one in part (a)? (c) Add the site (2, −1) to those in part (b). Compute and graph the new Voronoi diagram. How does the diagram relate to the one in part (b)? 10.1.25. Which one of the drawings in Figure 10.11 could be part of a Voronoi diagram? For it indicate approximately where the vertex in each region shown would be. (Ignore

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the part outside of the regions shown.) For the other two drawings, explain why they can’t be part of a Voronoi diagram.

Figure 10.11 Potential Voronoi diagrams.

C B

A

D

Figure 10.12 Sites in an imaginary town. 10.1.26. Figure 10.12 depicts the very regular street grid of an imaginary town with sites located at points A = (4, 2), B = (0, 0), C = (−2, −6), and D = (−6, 6). (a) Measure distance along streets (with the taxicab metric) to determine the points equidistant from A and B. Draw them on a reproduction of Figure 10.12. (See Example 3 of Section 2.3.) (b) Repeat part (a) for the points equidistant from A and C. Describe how this set of equidistant points differs from the set in part (a). (c) Repeat part (b) for A and D. (d) Repeat part (a) for the other pairs of points. Draw the Voronoi diagram when we measure distances along city streets. (e) Draw the Voronoi diagram using the same four points, but using Euclidean distance. How does this diagram differ from part (d)?

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(f) If you were deciding the regions closest to fire stations in an actual city, would you choose to use Euclidean distance or distance along streets? Explain your choice.

10.1.7 Paul Erd˝os The prolific Hungarian mathematician Paul Erd os ˝ (1913–1996) collaborated with over 500 mathematicians on research papers, more than anyone else who ever lived. He earned his doctorate at age 21 and soon after had to leave Hungary because he was Jewish. This was the start of a lifetime of traveling, interspersed with several shorter term teaching positions at universities. For most of his life he had no home or job, staying with mathematicians eager to host him and so get to collaborate with him. He often said “another roof, another proof.” Stories abound about his many eccentricities, as Hoffman [6] engagingly recounts. Erd os ˝ focused on specific problems, rather than general theories. Nevertheless, his over 1500 papers often developed general principles and generated broader research areas. His interests centered on number theory and discrete mathematics, including discrete geometry. However, he solved problems over a wide range of mathematical areas. His broad mathematical knowledge and widespread collaborations produced unusual and fruitful connections across many areas of mathematics. He posed a multitude of challenging and intriguing problems and often offered rewards for solutions. However, the prestige of solving an Erdos ˝ problem meant that the solvers often framed the reward checks rather than cashing them. Erd os ˝ also prized elegant proofs. He often talked of a mythical Book containing the most elegant proof of every mathematical theorem. His highest praise of someone’s work was that it was a page from the Book. For example, he thought Euclid’s proof of the infinitude of prime numbers qualified for the Book.

10.2 Points and Polygons We explore more deeply three of the problems introduced in the previous section.

10.2.1 Distances between Points Denote the minimum number of nonzero distances determined by n distinct points in ddimensional Euclidean space by Dd (n). Our first theorem notes the effect of adding points or dimensions. Theorem 10.2.1. For positive integers n and d, Dd+1 (n) ≤ Dd (n) ≤ Dd (n + 1). Proof. See Exercise 10.2.3. ! As discussed in Section 10.1 the values of Dd (n) are generally difficult to determine. But some special cases are easy to find. Theorem 10.2.2 considers the case d = 1, when the points are all on one line. Another completely solved situation occurs when the minimum number of distances is the absolute minimum of one, discussed in Theorem 10.2.3.

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Theorem 10.2.2. On a Euclidean line, the numberC of nonzero distances n distinct points can nD determine can be any integer between n − 1 and 2 . Thus D1 (n) = n − 1. Proof. Consider each point ai as a real number, assuming a1 < a2 < · · · < an . Then d(ai , a j ) = |a j − ai |. Further |a2 − a1 | < |a3 − a1 | < · · · < |an − a1 |, giving at least n − 1 different nonzero distances. If ai = i, there are exactly n − 1 different distances. Exercise 10.2.4 asks you to give choices of the ai giving exactly k nonzero distances for various values of k. ! Theorem 10.2.3. For positive integers n and d, D(n, d) = 1 if and only if d ≥ n − 1. Proof. We determine the maximum number of points in d dimensions equidistant from one another. Because of the homogeneity of space, we can place the points conveniently using any distance. For ease, start with d points P1 to Pd given by P1 = (1, √ 0, . . . , 0), P2 = (0, 1, 0, . . . , 0), . . . , Pd = (0, . . . , 0, 1), any pair of which are√a distance of 2 apart. We claim that if Q = (q1 , q2 , . . . , qd ) is also at a distance of 2 from all the Pi , then all the qi are equal. First consider d(Q, P1 )2 = d(Q, P2 )2 . That is, (q1 − 1)2 + q22 + q32 + · · · + qd2 = q12 + (q2 − 1)2 + q32 + · · · + qd2 . Canceling like terms we find −2q1 = −2q2 or q1 = q2 . The same reasoning holds for any pair of coordinates, establishing the claim. Because all coordinates are the same, write Q = (q, q, . . . , q). Then d(Q, P1 )2√= 2 becomes (q − 1)2 + (d − 1)q 2 = 2 or dq 2 − 2q − 1 = 0. The two solutions are q = 1+ d1+d and √

q ′ = 1− d1+d . Either choice gives us one more point, either Q or Q ′ , for a total of d + 1 points equidistant from each other in d dimensions. Finally we show we can’t have more than d√+ 1 such √ points. The only possibility is to use both Q and Q ′ . But d(Q, Q ′ )2 = d( 1+ d1+d − 1− d1+d )2 = √ √ ) > 4. Regardless of the number of dimensions, d(Q, Q ′ ) > 2 > 2. So d( 2 d1+d )2 = 4( 1+d d there are at most d + 1 points in d dimensions equidistant from one another. Hence if n > d + 1, there must be at least two distances or, equivalently, Dd (n) = 1 if and only if d ≥ n − 1. ! We state without proof one more theorem about minimum distances, investigated in Exercise 10.2.5. We call a set of points convex provided that the smallest convex polygon containing them has all the points on its edges, not in its interior. (Several of the points can be on the same edge. For example in Figure 10.13 the smallest polygon containing the convex set of six points is the quadrilateral AD E F, since A, B, C, and D are all on AD.)

F

A

B

E

C

D

Figure 10.13 A convex set of points.

459

10.2 Points and Polygons

Theorem 10.2.4. The minimum set of n points E n F number of distances determined by a convex n in the Euclidean plane is 2 , the greatest integer less than or equal to 2 .

Proof. See Altman [1]. !

In 1946 Erd os ˝ provided bounds on Dd (n) for any number of dimensions when n is large. For a sufficiently large number n of points in the plane he proved that there is some constant c such that D2 (n) ≤ √ cn . (The value of n must be quite large for the denominator to shrink this log(n) ! ! ! number much. For example, log(10) = 1, log(10,000) = 2, and log(1,000,000,000) = 3.) For three dimensions, Erd os ˝ showed for sufficiently large n that D3 (n) ≤ cn 2/3 for some constant c. Erd os ˝ also provided lower bounds for Dd (n), although they have been improved.

10.2.2 Triangulations Figure 10.14 gives an easy way to triangulate a convex n-gon: pick a vertex and draw a diagonal from it to each vertex except the two vertices adjacent to it. Since there are n − 3 vertices not adjacent to it, we have n − 3 diagonals determining n − 2 triangles. The triangulation of a non-convex n-gon in Figure 10.15 also has n − 3 diagonals and n − 2 triangles, suggesting the values always hold. But first we need to show that every polygon has a triangulation before asking about the number of triangles. Lemma 10.2.5 starts the process by finding a first diagonal. As Theorems 10.2.6 and 10.2.7 show, polygons always have triangulations and the triangulations always have n − 2 triangles and n − 3 diagonals.

Figure 10.14 Triangulating a

Figure 10.15 A triangulation of a nonconvex

convex polygon.

polygon.

Lemma 10.2.5. A polygon with at least four vertices has a diagonal. Proof. Let A be the leftmost vertex of a polygon with at least four vertices and let B and C be those adjacent to A. (See Figure 10.16.) If two vertices are leftmost, pick the lower one of them. Draw the segment BC. If BC is entirely inside the polygon, we have the desired diagonal and we are done. Otherwise BC intersects some edge(s) of the polygon, implying there is a vertex D of the polygon inside △ABC. Without loss of generality let D be the lower leftmost such point. We claim that AD must be a diagonal. For a contradiction, suppose that some edge E F intersects AD, say at G. Now E and F are vertices of the polygon, but G need not be. Since D is in the interior of △ABC, it is to the right of A. Since G is between A and D, it is to the left of

460

Discrete Geometry

B

E? D G? A

F? C

Figure 10.16 Finding a diagonal in a polygon. D. Hence one of E and F is to the left of D, say E. Now D is the leftmost point in △ABC, so E must be outside. But then E F would intersect either AB or AC. However, the edges of polygons don’t cross, giving a contradiction. So AD is a diagonal. ! Lemma 10.2.5 splits the polygon into two polygons with fewer vertices. Informally we can now triangulate the polygon: split the smaller polygons by diagonals, continuing until every region is a triangle. Like many ideas that solve a problem by breaking it into simpler cases, the proof of Theorem 10.2.6 uses strong induction. (See Appendix F or Sibley [12, 93ff] for more on induction proofs.) Theorem 10.2.6. Every polygon has a triangulation. Proof. We use induction on n, the number of vertices. For the initial step when n = 3, we have a triangle, which is already triangulated. For the induction step, assume that every polygon with at most k vertices has a triangulation. Let P1 P2 . . . Pk+1 be a polygon with k + 1 vertices. By Lemma 10.2.5, it has a diagonal, say Pi P j with i < j. Then P1 . . . Pi P j . . . Pk+1 and Pi Pi+1 . . . P j are polygons with fewer than k + 1 vertices determined by Pi P j and the original edges. Hence each of them can, by the induction hypothesis, be triangulated. Their triangulations, together with Pi P j give a triangulation of the larger polygon, finishing the induction step. By the principle of mathematical induction, the result holds for all values of n. ! Theorem 10.2.7. A triangulation of an n-gon has n − 3 diagonals and n − 2 triangles. Proof. Again we use induction on n, the number of vertices. For the initial step when n = 3, a triangle has 0 = 3 − 3 diagonals and is “divided” into 1 = 3 − 2 triangles. For the induction step assume that every triangulation of a polygon with at most k vertices has the stated number of diagonals and triangles. Consider a (k + 1)-gon and as in Theorem 10.2.6 split it into two smaller polygons P1 . . . Pi P j . . . Pk+1 and Pi Pi+1 . . . P j that we triangulate. All the vertices of the original polygon appear in the two polygons with Pi and P j repeated. Thus if one of the smaller polygons has b vertices, the other has k − b + 3, where b is at least 3. Use the induction hypothesis to count the diagonals of the smaller polygons: b − 3 and k − b. These together with the diagonal of Lemma 10.2.5 gives a total of k − 2 = (k + 1) − 3. Similarly, there are

461

10.2 Points and Polygons

b − 2 and k − b + 1 triangles for the smaller polygons, giving a total of k − 1 = (k + 1) − 2 for the original polygon. These values prove the induction step. The principle of mathematical induction extends the property to all values of n. ! The number of triangulations of two polygons with the same number of vertices can vary enormously, depending at least partially on how many diagonals it has. Theorem 10.2.7 guarantees at least n − 3 diagonals and a triangulation in any n-gon. Convex polygons will give the maximum number of diagonals (Theorem 10.2.8) and triangulations (Theorem 10.2.9). Theorem 10.2.8. A convex n-gon has

n(n−3) 2

diagonals.

Proof. See Exercise 10.2.9. ! Example 1. Let T (n) be the number of triangulations of a convex n-gon. Quick inspection gives T (3) = 1, T (4) = 2, and T (5) = 5. A little more work yields T (6) = 14, but drawing all the possibilities quickly becomes tedious. For T (7) (and higher values) we argue recursively, using previously determined values of T (n). Edge 12 in Figure 10.17 must be part of exactly one triangle in any triangulation. From the five other vertices we have five cases. For △123, the remaining polygon is the hexagon 134567 and there are T (6) = 14 ways to triangulate it. Next, △124 determines two other polygons, △234 and the pentagon 14567, and we can triangulate them independently of one another, giving T (3)T (5) = 5 more ways. For the third case, △125 determines quadrilaterals 2345 and 1567, giving T (4)T (4) = 4 options. By symmetry, the remaining cases, △126 and △127, give T (5)T (3) = 5 and T (6) = 14 possibilities, respectively. Thus T (7) = 14 + 5 + 4 + 5 + 14 = 42. ♦ 2

1

3 7

4

6 5

Figure 10.17 A heptagon. Exercise 10.2.12 asks you to compute T (8) and T (9) recursively, as in Example 1. In general, we can determine the number of triangulations of an n-gon from the number of triangulations of smaller polygons. Computers often apply recursive processes like this. However, a formula is far more efficient when it can be found, and in this case Theorem 10.2.9 gives the formula. The sequence of numbers of triangulations of convex n-gons turns up in hundreds of contexts and is called the Catalan numbers. While we omit the somewhat involved proof, it formalizes the recursive ideas of Example 1 using mathematical induction.

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Discrete Geometry

Theorem 10.2.9. A convex n-gon has

C

D

1 2n−4 n−1 n−2

triangulations.

Proof. See Devadoss and O’Rourke [3, 9–10]. !

10.2.3 The Art Gallery Problem One guard placed anywhere in a convex polygon can observe the entire interior. For other polygons, the number of guards needed depends greatly on where they stand and the shape of the polygon. Surprisingly, we can handle the worst cases by placing guards at appropriate vertices of the polygon. Exercise 10.2.1.∗ In Figure 10.18 compare the number of guards needed when they are at vertices and when they can be anywhere.

Figure 10.18 A polygon for the art gallery problem. We use the elegant proof of the art gallery theorem (Theorem 10.2.12) published by Fisk in 1978. We give the needed definition and lemmas next, although their significance will be unclear until the proof of the theorem. Triangles △P1 P2 P3 and △P4 P5 P6 in Figure 10.19 illustrate the somewhat whimsical name “ear.” P2 P1

P5 P3 P4

P9

P6

P7 P8

Figure 10.19 An ear of a polygon. Definition. Angle ∠Pi Pi+1 Pi+2 is an ear of the polygon P1 . . . Pn if and only if Pi Pi+2 is a diagonal of the polygon.

463

10.2 Points and Polygons

Lemma 10.2.10. Every polygon with at least four vertices has at least two ears. Proof. Given an n-gon, triangulate it. If any of the triangles contains two consecutive edges and so one diagonal of the polygon, we have an ear. (For example, triangles △P1 P2 P3 and △P4 P5 P6 in Figure 10.19.) Any triangle of the triangulation has two or fewer edges of the polygon. Suppose for a moment that there were no ears. Then every triangle of the triangulation would have zero or one edge of the polygon. By Theorem 10.2.7 there are n − 2 triangles, which would account for at most n − 2 edges. Since there are n edges, at least two of the triangles have two edges and so are ears. ! Lemma 10.2.11. Fix a triangulation of a polygon. Then the vertices of the polygon can be colored with three colors so that every triangle in the triangulation has a vertex of each color. Proof. We use induction on n, the number of vertices of the polygon. For n = 3, we have a triangle and we satisfy the theorem by coloring its vertices with three colors. For the induction step, assume that any triangulation of any k-gon can be colored as specified. Consider a triangulation of a k + 1-gon. By Lemma 10.2.10, it has an ear, which we ignore for a moment. (In Figure 10.20 the ear is shaded. The letters A, B, and C indicate the colors of the vertices.) The rest is a k-gon with a triangulation and so can be colored with three colors. Two of the vertices of the ear are also adjacent vertices of the k-gon and so have different colors. Use the third color for the other vertex of the ear, coloring the entire k + 1-gon and finishing the induction step. By the principle of mathematical induction, the theorem holds for all n-gons. ! A C B C

A

B A

B C

A

C

Figure 10.20 A polygon with an ear. EnF Theorem 10.2.12. (Art Gallery Theorem) An n-gon needs at most E n F3 guards to be able to observe all interior points. For each n there is an n-gon needing 3 guards.

Proof. Triangulate the n-gon. Now use Lemma 10.2.11 to color the vertices with three colors so that each triangle has one vertex of each color. If we pick as our guards the vertices of the same color, every triangle has a guard observing it and so the entire polygon is guarded. Choose

464

Discrete Geometry

the color used the fewest number of times, which will be at most one third of the vertices.E(In F Figure 10.20 pick the three vertices marked with the letter B.) That is, we need at most n3 guards. See Exercise 10.2.13 for polygons requiring the maximum number of guards. ! The solution to the art gallery problem spurred variations. If we replace the art gallery with a fortress, the guards need to be able to observe all the points on the outside. As with many discrete geometry problems about polygons, convex polygons are easier to analyze. For each polygon there is an associated convex polygon, called its convex hull, illustrated in Figure 10.21 with the added dashed segments. H G A

M F

K

E I

L J

D C

B

Figure 10.21 Figure 10.21 The convex hull of a polygon. Definition. The convex hull of a set S of points is the smallest convex set containing S. G H Theorem 10.2.13. (Fortress Theorem). An n-gon needs at mostG Hn2 guards to be able to observe all exterior points. For each n there is an n-gon needing n2 guards.

G H Sketch of Proof. Exercise 10.2.16 shows that a convex n-gon needs n2 guards, proving the second statement of the theorem. Figure 10.21 suggests an approach to the general situation. We add dashed segments to the polygon to make its convex hull, ABC D F G H . By Exercise 10.2.16 we know how to guard the exterior of the convex hull and how many guards it takes: Gk H if the convex hull is a k-gon. The rest of the exterior forms one or more indentations— 2 for example, AH I J K L M in Figure 10.21. But now we need to guard their interiors. From Theorem 10.2.12, we generally need only about a third of the vertices for such an indented region, rather than the half we get overall in this theorem. We need the difference between the fractions to make up for a double counting problem: two of the vertices in each indentation are already counted as part of the convex hull. For example, in Figure 10.21 the convex hull has seven vertices and the indentations have three and seven vertices, adding to seventeen, even though the polygon has just thirteen vertices. It turns out that the number of guards needed for the exterior of the convex hull plus the ones needed for the indentations never exceeds the permitted number. See Exercise 10.2.20 for the rest of the analysis and the idea of the induction proof. !

10.2 Points and Polygons

465

10.2.4 Exercises for Section 10.2 10.2.2. Find the number of different distances in a regular n-gon and prove your answer is correct. Explain how it relates to Theorem 10.2.4. ∗

10.2.3. Prove Theorem 10.2.1. Hint: Start with Dd (n) as given. 10.2.4. To find the possible number of different distances determined by n points on a line we will put the points on integer coordinates of a number line. ∗

(a) How many distances are there if the points are at 1, 2, . . . , n − 1, and n + 1? (b) Generalize part (a) by using more gaps of size 2. What possible numbers of distances can you obtain this way? (c) Use different sized gaps to increase the number of different distances. (d) If the points are at 2i from i = 0 to i = n − 1 how many different distances are there? E F 10.2.5. One approach to proving in Theorem 10.2.4, that there must be at least n2 distances, is to try to find a point with that many distances to other points. ∗

(a) Consider a convex set with vertices A and B at least as far apart as any other pair of vertices. Let C be another vertex with d(A, C) < d( A, B). Prove that there can be at most one other vertex D in the set so that d( A, C) = d(A, D). E F (b) Explain why part (a) and the following conjecture would guarantee that n2 is a lower bound on the number of distances for a convex set of n points. Conjecture: there are points A and B so that d(A, B) is the maximum of the distances and there is at most one other point E so that d( A, B) = d( A, E). (A conjecture of Erd os ˝ is equivalent to this conjecture, but it remains unproven.) (c) Give an example of a convex polygon with vertices A, B, E, and F so that the maximum distance is d( A, B) = d(A, E) = d( A, F). Verify in your example that at least one of the points B, E, and F satisfies the claim in part (b). 10.2.6. Find the number of nonzero distances of the sets of points in three-dimensional rectangular grids. The coordinates are all integers. ∗ ∗

(a) {(x, y, z) : x, y, z ∈ {0, 1}} (8 points at the vertices of a 1 × 1 × 1 cube) (b) {(x, y, z) : x, y, z ∈ {0, 1, 2}} (27 points in a 2 × 2 × 2 cube) (c) {(x, y, z) : x, y, z ∈ {0, 1, 2, 3}} (64 points in a 3 × 3 × 3 cube)

10.2.7. ∗ (a) Count the number of points and the number of nonzero distances in the hexagonal grids in Figure 10.22. Complete the table. r (number of rings) 1 2 3 4 n(r ) (number of points) 7 d(r ) (nonzero distances) (b) Find a formula for n(r ), the number of points in a hexagonal grid with r rings. (c) Find an upper bound d(r ) on the number of nonzero distances in a hexagonal grid ) as r → ∞. (Erd os ˝ used an advanced with r rings. Find the limit of the ratio d(r n(r ) number theory result to find a very good upper bound d(r ) so that this limit goes to 0.)

466

Discrete Geometry

r=1

r=2

r=3

Figure 10.22 A hexagonal grid. 10.2.8. (a) Describe an example of an n-gon with n − 3 diagonals and just one triangulation. (b) Describe how to modify part (a) to give an n-gon with n − 2 diagonals. How many triangulations does it have? (c) For n > 5 repeat part (b) to get n − 1 diagonals. 10.2.9. (a) Prove Theorem 10.2.8 using induction. ∗ (b) Use the following questions to prove Theorem 10.2.8 a different way from part (a). If we connect every pair of vertices of a convex n-gon, how many segments are there? How many of them are edges of the polygon? Why are the remaining segments diagonals? (c) Discuss why the proofs in parts (a) and (b) fail for general polygons. 10.2.10. By Lemma 10.2.5 every non-convex n-gon with n > 3 has at least one diagonal. ∗

(a) Draw various non-convex pentagons and determine the number of diagonals each has. (b) Use the proof of Lemma 10.2.5 to find a diagonal in a non-convex pentagon, splitting it into two smaller polygons. How many sides do they have? Use the proof again to find a minimum number of diagonals in a pentagon. (c) Repeat parts (a) and (b) for non-convex hexagons. Compare this minimum number with Theorem 10.2.7.

10.2.11. By Theorems 10.2.7 and 10.2.8 the biggest possible range of the number of diagonals . in an n-gon is from n − 3 to n(n−3) 2 ∗

(a) For n = 5, which of the values in the range can happen? Draw examples. (b) Repeat part (a) for n = 6. Hint: Consider what happens when vertex A in Figure 10.23 moves down the dashed diagonal. You may need other examples as well. (c) Repeat part (a) for n = 7. A

Figure 10.23

467

10.2 Points and Polygons

10.2.12. (a) Use the recursive reasoning in Example 1 to find T (8) and T (9), the number of triangulations in convex 8-gons and 9-gons. ∗ (b) For each possible number of diagonals in a pentagon in Exercise 10.2.11 (a) count the number of triangulations in a pentagon with that many diagonals. (c) For hexagons with three, four, seven, or eight diagonals repeat part (b), using Exercise 10.2.11(b). Do hexagons with the same number of diagonals always have the same number of triangulations? (d) For 7-gons with four, five, six or thirteen diagonals repeat part (b) using Exercise 10.2.11 (c). Do 7-gons with the same number of diagonals always have the same number of triangulations? 10.2.13. ∗ (a) Use the added points G and H to explain why the hexagon ABC D E F in Figure 10.24 requires two guards to solve the art gallery F E problem. ∗ (b) Modify Figure 10.24 to give a 7-gon requiring 73 guards. (c) Repeat part (b) for an 8-gon. (d) Repeat part (b) for a 9-gon. (e) Explain how to design a 3k-gon requiring k guards. (f) Use parts (b) and (c) to explain how to modify the 3k-gon of part (e) handle (3k + 1)-gons and (3k + 2)-gons. B

A

E

C

D

H

G

F

Figure 10.24 A polygon requiring two guards. 10.2.14. (a) For the polygon in Figure 10.25 give three triangulations that when three-colored give different minimum numbers of guards. (b) Suggest a way to triangulate a polygon so as to try to minimize the number of guards needed. 10.2.15. (a) Modify Figure 10.24 to give an orthogonal 8-gon art gallery needing two guards. (See Exercise 10.1.17.) (b) Modify Exercise 10.2.13, part (d) to give an orthogonal 12-gon needing three guards. (c) Explain how to find an orthogonal 4k-gon needing k guards.

Figure 10.25 A polygon to three-color.

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Discrete Geometry

G H 10.2.16. Prove that a convex n-gon requires n2 , the least integer j with j ≥ (n/2), guards to solve the fortress problem. Hint: The measure of each angle is less than 180◦ . 10.2.17. (a) Draw polygons with 4, 5, and 6 vertices, each having exactly two ears. ∗ (b) Determine the range for the possible number of ears in a pentagon. Draw an example for each. (c) Repeat part (b) for hexagons. (d) Repeat part (b) for 7-gons. (e) Make a conjecture about the range of the number of ears in an n-gon. Justify your conjecture. 10.2.18. ∗ (a) Draw various orthogonal 6-gons and solve the fortress problem for them. (See Exercise 10.1.17.) (b) Repeat part (a) for 8-gons. (c) Repeat part (a) for 10-gons. (d) Make a conjecture about the minimum number of guards for a fortress in the shape of an orthogonal 2n-gon. Explain your answer. 10.2.19. The prison problem requires guards on the boundary of a polygon that can observe every point in the interior and the exterior of the polygon without looking through any edge the guard is not on. ∗

(a) Find the minimum number of guards needed to solve the prison problem for a convex n-gon. Explain your answer. (b) For each polygon in Figure 10.26 find the minimum number of guards needed. (c) Do you think the minimum number of guards can always be achieved by placing guards at vertices? Explain.

Figure 10.26 Polygons for the prison problem. 10.2.20. This problem fills out the details for an induction proof of Theorem 10.2.13, where we induct on the number of indentations. Suppose that the n-gon has a k-gon for its

469

10.3 Tilings

convex hull. First suppose that there is just one indentation with j + 2 vertices and k + j = n. F G H G H E guards. The theorem allows k+2 j (a) Explain why we need at most k2 + j+2 3 G k H E j+2 F G k+ j H guards. Determine which values of k and j make 2 + G k H 2 G .j HHints: G k+ j 3H > First show thatE if atFleast one G j Hof k and j are even, then 2 = 2 + 2 , and with . Next consider when both k and j are odd and look then compare j+2 3 2 at the size of j. F E guards. (b) Explain why in the only case from part (a) we never actually need the j+2 3 Hint: Use the fact that for every indentation, one of the two vertices that are part of the indentation and part of the convex hull is already a guard. (c) Explain how to use the preceding ideas for the induction step.

10.3 Tilings Tilings have fascinated artists and mathematicians across cultures and millennia. We used symmetry to discuss tilings in Section 6.3. Here we explore tilings from the viewpoint of the tiles, starting with ones that can tile by themselves. Theorem 10.3.1. A centrally symmetric hexagon gives a monohedral tiling of the plane. Proof. Let C be the center of symmetry of hexagon P Q R P ′ Q ′ R ′ . That is, a 180◦ rotation around C switches the pair P and P ′ and the other pairs Q and Q ′ and R and R ′ . (See Figure 10.27.) Exercise 10.3.6 shows the following three facts: (1) The angle sum of three consecutive angles, such as ∠P, ∠Q, and ∠R, is 360◦ . (2) The opposite sides are parallel and congruent so P R P ′ R ′ is a parallelogram, as are P Q P ′ Q ′ and Q R Q ′ R ′ . (3) Hence three copies of the hexagon will fit around any vertex without gaps or overlaps.

Q

P R R' Q'

C P'

Figure 10.27 A centrally symmetric hexagon. Then the translation taking P ′ to R ′ will take the original tile to an adjacent one, as indicated in Figure 10.28. Also, the translation taking Q to R ′ takes the original tile to an adjacent one. The original tile and the two images surround the point R ′ without gap or overlap because the

470

Discrete Geometry

Q

P R R'

C P'

Q'

Figure 10.28 A tiling with a centrally symmetric hexagon.

three angles add to 360◦ and the sides are parallel and congruent. Similar translations in various directions will continue the tiling, covering the plane. !

Corollary 10.3.2. Any quadrilateral gives a monohedral tiling of the plane. Any pentagon with two adjacent supplementary angles gives a monohedral tiling of the plane. Proof. See Exercise 10.3.7. ! Corollary 10.3.2 determines completely what quadrilaterals give monohedral tilings. However, you may find it surprising that adding one side turns the situation into an open question: no one has determined all the pentagons giving monohedral tilings. Even restricting the monohedral tilings to convex pentagons isn’t enough to solve the problem entirely. See Schattschneider [9]. See Exercise 10.1.17 for the definition of orthogonal polygons, used in the next corollary. Corollary 10.3.3. Any centrally symmetric orthogonal octagon gives a monohedral tiling of the plane. Proof. See Exercise 10.3.8 for an outline of the proof. ! M. C. Escher excelled in finding tiles resembling animals or other familiar objects fitting together to give tilings. Since then many others have followed his lead, including developing software for the purpose. Start with a simple tile, such as a parallelogram, possessing at least translational symmetry. Then modify corresponding sides, as in Figure 10.29, while preserving the symmetry. Thus the modified tile fills the plane in the same way as the original tile, as

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10.3 Tilings

Figure 10.30 illustrates. In Figure 10.29 the only type of symmetry preserved is translational: what is added on one side is subtracted from the opposite side. Figure 10.31 preserves numerous rotations, including 180◦ rotations around the center of each edge of the original triangle.

Figure 10.29 A tile.

Figure 10.30 A tiling with only translational symmetries.

Figure 10.31 Figure 10.31 A tile and tiling with rotational symmetries. Given the unlimited ways to modify the starting tile, there is no way to classify all possible tiles giving monohedral tilings. While some restrictions on the type of tiles have led to classifications, many more remain open questions. (See Gr¨unbaum and Shephard [5] for classifications, and unsolved problems.) The modifications Escher and others use to create tiles make clear that a polygon could have any number of sides and still give a monohedral tiling. However, restricting our attention to convex polygons fitting together edge to edge, as Theorem 10.3.4 shows, narrows the possibilities enormously.

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Theorem 10.3.4. If a convex polygon gives an edge-to-edge monohedral tiling of the plane, than the polygon has at most six vertices. Proof. Let P be a convex n-gon giving a monohedral tiling and consider a finite part of the tiling with k connected copies of P. We count the number of edges (E) and vertices (V ) in it. We count as faces (F) all the copies of P together with the outside region, so F = k + 1. (In Figure 10.32, V = 38, E = 63, and F = 27.) Since P is an n-gon, each copy of the polygon has n edges and n vertices on it. The outside face has some number, say j of each. However, E doesn’t equal nk + j since each edge has two faces. The double counting gives 2E = nk + j or E = n2 k + 2j . At least three faces meet at each vertex, except possibly at some of the j vertices on the outer region, which have at least two faces. (In Figure 10.32, nine of the outside vertices have just two faces, far fewer than j = 22.) So by similar reasoning 3(V − j) + 2 j ≤ nk + j or V ≤ n3 k + 23j . By Euler’s formula (Theorem 1.5.1) 2 = V − E + F. Substitution gives 2≤ We rearrange to get

? 2j n j n@ j n k+ − k− +k+1=k 1− + + 1. 3 3 2 2 6 6 1−

which holds for any part of the tiling.

? n@ j ≤k 1− , 6 6

(10.1)

Figure 10.32 A partial tiling. Suppose for a contradiction that n > 6, so that the right side of (10.1) is negative. Consider increasingly large, roughly circular parts of the tiling; that is, ones where the number of copies is approximately the same in each direction. As the size increases, k, the number of copies of

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the polygon, is proportional to the area. Meanwhile j, the number of vertices on the outside, is proportional to the circumference. The area grows by a multiple of the square of the radius, while the circumference grows by a multiple of the radius. For large enough k, therefore, the right side of (10.1) will be more negative than the left side, giving a contradiction. Hence n ≤ 6, as claimed. ! As the preceding discussion indicates, monohedral tilings already give a complicated family. The variety of tilings becomes bewildering when we allow more than one basic tile. We restrict ourselves to an example and some problems. See Gr¨unbaum and Shephard [5, Chapters 2 and 4] for more discussion. Example 1. Figure 10.33 depicts a tiling with two tiles, both squares of different sizes. Besides arranging squares in a more interesting way than a checkerboard pattern, the tiling suggests a proof of the Pythagorean theorem. Let the side of the small square be a and the side of the larger square be b. Call the side of the square with dashed sides c. The square with side c is divided into five pieces. Two of them fit together to give a small square with area a 2 . The other three assemble into a larger square with area b2 . Putting these facts into one equation we get the formula for the Pythagorean theorem: a 2 + b2 = c2 . ♦

Figure 10.33 A tiling with two sized square tiles. Definition. A coloring of a set of regions assigns a color to each region so that regions sharing an edge or more are different colors. Exercise 10.3.1.∗ Give a coloring of the tiling of Figure 10.33 using three colors. Explain why every coloring needs at least three colors.

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In 1976 Appel and Haken proved that any partial or complete tiling of the plane can be colored with at most four colors. Their proof and all subsequent ones use sophisticated mathematics to divide the coloring problem into a large number of irreducible cases. Then they use computers to solve each case. Many tilings or partial tilings don’t require four colors, although determining which ones require three instead of four is in general a difficult problem. Mathematicians have for centuries pondered tilings in space as well as the plane. Over 2000 years ago Aristotle incorrectly stated that copies of a regular tetrahedron would fit together to fill up space. (See Exercise 10.3.14.) Of the five regular polyhedra, only the cube can give a monohedral tiling of three-dimensional space. Mathematicians have found numerous other monohedral tilings of space. Tilings of space with translational symmetry relate to crystals in chemistry, which has led to a long interaction between chemists and mathematicians. Since the discovery of quasi-crystals in 1984 interest has rapidly increased in the study of tilings of space without translations. (See Section 6.5, Senechal [11], and Gr¨unbaum and Shephard [5, Chapter 10] for more on quasicrystals and aperiodic tilings.) Exercise 10.3.2.∗ Let P be a polygon giving a monohedral tiling of the plane. Explain how to use P to build a prism giving a monohedral tiling of space. The visual appeal of tilings quickly leads to many open problems and interesting mathematics.

10.3.1 Exercises for Section 10.3 10.3.3. (a) Use a dynamic software program to make a monohedral tiling from the tiles in Figure 10.34. (The dashed segments give the original polygon used to built each tile.) Describe the symmetries each tile possesses. (b) Classify the symmetry group of the tiling. (See Section 6.3.) (c) Design your own monohedral tilings. (d) Design tilings using two tiles.

Figure 10.34 Sample tiles. ∗

10.3.4. Determine whether a truncated equilateral triangle, as in Figure 10.35, can give a monohedral tiling. If yes, draw a partial tiling; if not, explain why not.

Figure 10.35 A truncated equilateral triangle.

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475

10.3.5. (a) Draw a partial monohedral tiling using as the tile a hexagon that is neither centrally symmetric nor orthogonal. (b) Repeat part (a) using an 8-gon. 10.3.6. ∗ (a) Prove in a centrally symmetric hexagon that the angle sum of three consecutive angles is 360◦ . (b) Prove in a centrally symmetric hexagon that opposite sides form sides of a parallelogram. (c) Prove in a centrally symmetric hexagon that three copies will fit around any vertex without gaps or overlaps. 10.3.7. (a) Prove the first sentence of Corollary 10.3.2. Hint: Rotate the quadrilateral around the midpoint of a side. (b) Use the idea of part (a) to prove the rest of Corollary 10.3.2. 10.3.8. ∗ (a) For an orthogonal octagon ABC D E F G H , prove that it has six interior angles of 90◦ and two interior angles of 270◦ . Hint: What is the angle sum? ∗ (b) In Corollary 10.3.3 why may we assume without loss of generality that the interior angles of 270◦ are at A and E? (c) Why will the translation τ taking B to E give an octagon so that it and the original octagon surround the point E without gaps or overlaps? Illustrate your answer. (d) Let β be the translation taking C to τ (H ). Why will the original octagon and its images surround the point τ (H ) without gaps or overlaps? Illustrate your answer. (e) Explain why we can continue tiling the plane with copies of the octagon, as in Theorem 10.3.1. 10.3.9. (a) Draw part of a tiling whose tiles are an equilateral triangle with side a and an equilateral triangle with side 2a. Will your tiling work with any two sized triangles? If so, explain why; if not, explain why not. ∗ (b) Draw part of a tiling whose tiles are an equilateral triangle and a regular hexagon with different length sides. Can any two lengths work? If so, explain why; if not, explain why not. (c) Draw part of a tiling whose tiles are three squares with side lengths a, 2a, and 3a. (d) Draw part of a tiling whose tiles are three equilateral triangles with side lengths a, 2a, and 3a. 10.3.10. Figure 3.1.6 in Gr¨unbaum and Shephard [5, 119] depicts a tiling of the plane with convex 7-gons. Explain how it gets around the argument concerning area and circumference in the proof of Theorem 10.3.4. The tiling is not monohedral. Explain why monohedral tilings don’t permit a similar situation. 10.3.11. ∗ (a) What is the fewest number of colors for a coloring of the tiling from a non-convex quadrilateral as in Exercise 10.3.7? (b) Repeat part (a) using a tiling with a centrally symmetric hexagon. ∗ (c) Repeat part (a) with the tiling from Exercise 10.3.9 (b) when the side of the triangle is longer than the hexagon’s side. (d) Repeat part (c) when the side of the triangle is shorter than the hexagon’s side.

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10.3.12. If isometric copies of a shape can tile a bigger copy of that shape, the shape is called a rep-tile. (“Rep” stands for “repeating.” ) Figure 10.36 depicts four copies of a polygon forming an enlarged version of the original polygon. ∗

(a) Make a drawing illustrating why any triangle is a rep-tile. How many copies of the original triangle make up the bigger copy? (b) Repeat part (a) for a parallelogram. (c) Explain why a regular hexagon can’t be a rep-tile, even though it gives a monohedral tiling. (d) Design a trapezoid that is a rep-tile. (e) Make drawings illustrating why three of the tetrominoes are rep-tiles. Explain why the other two can’t be rep-tiles. (See Exercise 10.1.20.) (f) Explain why a rep-tile gives a monohedral tiling of the plane.

Figure 10.36 A rep-tile. 10.3.13. (a) Construct several same sized models of a square pyramid similar to one with vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0), (1, 1, 0), and (0, 0, 1). Explain how the pyramid gives a monohedral tiling of space. (b) Use part (a) to describe a tetrahedron giving a monohedral tiling of space. (c) Figure 10.37 depicts a polyhedron made from four cubes glued together, one of which is hidden by the others. Explain why it gives a monohedral tiling of space. (d) Design other polyhedra giving a monohedral tiling of space.

Figure 10.37 A polyhedron that tiles space. 10.3.14. (a) Aristotle thought five regular tetrahedra would exactly fit around a common edge. Disprove his claim by finding the dihedral angle, ∠AE D in Figure 3.56. Hint: Use the law of cosines for △C D E0. See Exercise 1.2.23.

10.4 Voronoi Diagrams

477

(b) Determine the dihedral angle of two faces of a regular octahedron. (c) Use parts (a) and (b) to verify that the dihedral angles of the regular tetrahedron and octahedron are supplementary. Explain why a regular tetrahedron and a regular octahedron with sides the same length fit together face to face so that three pairs of their faces form parallelograms. (d) Build two regular tetrahedra and a regular octahedron all with sides the same length. What shape do you get when you attach the tetrahedra to opposite faces of the octahedron? Explain the tiling the polyhedra determine.

10.3.2 Branko Gr¨unbaum Over a nearly fifty year period Branko Gr¨unbaum (1929– ) has published more than two hundred research articles in geometry and discrete geometry. Through his research and his teaching he shaped the development of discrete geometry. From his native Croatia he went to the newly created Israel, earning his Ph.D. in 1957, and teaching there. He came to the United States in the 1960s and spent the rest of his career at the University of Washington. Gr¨unbaum is widely known for his efforts to revive interest in tilings, including his extensive classification results. The text Tilings and Patterns [5], written with G. C. Shephard, has become a standard source for mathematicians and non-mathematicians alike investigating this area. He originated the geometrical analysis of weaving patterns, bringing modern mathematical understanding to an ancient art. Gr¨unbaum’s geometric interests range far beyond tilings. He wrote the standard work on convex polytopes, the higher dimensional analog of polygons and polyhedra in addition to his own research in this area. He has also published numerous non-research articles to bring geometrical ideas to a wider audience of students and others.

10.4 Voronoi Diagrams Even before Dr. Snow made his map of the cholera epidemic, mathematicians started exploring the idea. Descartes informally considered the regions closest to various points in the 1600s. Peter Gustave Lejeune Dirichlet (1805–1859) made use of what we now call Voronoi diagrams. Later Georgy Voronoi (1868–1908), although not the first to study the diagrams, defined and studied them in general, including in n dimensions. Since then many scientists have developed applications of Voronoi diagrams in physics, climatology, ecology, chemistry, and computer science. In Example 1 of Section 10.1 we determined the Voronoi regions around the sites using the perpendicular bisectors of each pair. We can say more about the regions and their boundaries. For example, the regions are always convex, as Theorem 10.4.1 shows. Theorem 10.4.1. A Voronoi region is a convex set. Proof. Let A be a site in a Voronoi diagram. Its Voronoi region is the intersection of the half planes containing A and bounded by the perpendicular bisectors of A and the other sites. Because of the separation axiom (Appendix B or C), any half plane is a convex set. By Exercise 10.4.11, the intersection of a collection of convex sets is convex. !

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When can an entire line be the boundary of a Voronoi diagram? Of course, if there are only two sites, their perpendicular bisector is the entire Voronoi diagram and the regions are half planes. Consider three sites not all on one line. From Exercise 1.2.15 their perpendicular bisectors meet in a common point. In this case the Voronoi diagram consists of three rays, as in Figure 10.38.

Figure 10.38 A Voronoi diagram with three sites. Exercise 10.4.1.∗ What happens when a Voronoi diagram has three sites, all collinear? Four collinear sites? Theorem 10.4.2. A Voronoi diagram with at least three sites has an infinite line if and only if all the sites are collinear. Proof. See Exercise 10.4.13. ! Start with a point x and consider circles of increasing radii centered at x. There will be a smallest circle with at least at one site on it. If there is just one site, x is in its Voronoi region. If there are exactly two sites on the circle, x is on the boundary of their regions. Whether the boundary is finite or infinite, we call it a Voronoi edge. If three or more sites are on the circle, x is where Voronoi edges meet and we call it a Voronoi vertex. Exercise 10.1.24 considers how a Voronoi diagram changes as we add sites. Let’s consider this more generally. Given a set S = {v 1 , v 2 , . . . , v n } of sites and its Voronoi diagram Vor(S), consider an additional point v n+1 . We restrict our discussion to non-collinear sites. The new point is in just one of the Voronoi regions, say the one for v 1 , which we’ll call R1 , as in Figure 10.39, where n = 6. The perpendicular bisector k1 of v 1 and v n+1 divides R1 into two regions. One part is the region still closest to v 1 , which we rename R1′ . The rest is part of the Voronoi ′ for the added site v n+1 . There is at least one intersection of k1 with the boundary of region Rn+1 R1 , which is also the boundary of another Voronoi region, say R2 for the site v 2 . Let k2 be the perpendicular bisector of v 2 and v n+1 . As before, k2 splits R2 into two regions, the new R2′ , still ′ . We continue in this fashion to find the entire region closest to v 2 , and another piece of Rn+1 ′ Rn+1 . Computers can implement the procedure to create Voronoi diagrams, although it is not the most efficient one. See Okabe et al [8, 257] for the plane sweep method, a more efficient algorithm. Voronoi diagrams starting from a finite number of sites always have some infinite regions. Exercise 10.4.12 asks you how to determine which sites will have infinite regions. With only finitely many sites (faces), a Voronoi diagram can only have finitely many Voronoi edges and CD finitely many Voronoi vertices where the edges meet. With n sites, there are n2 perpendicular

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10.4 Voronoi Diagrams

v4

v5

k4

v1 v6

k1 v7 v2

k2

k3 v3

Figure 10.39 Adding a site. bisectors, which can intersect in only so many ways. However, Euler’s formula provides much smaller upper bounds. Theorem 10.4.3. A Voronoi diagram with n sites, where n ≥ 3, has at most 3n − 6 Voronoi edges and at most 2n − 5 Voronoi vertices. Proof. Given a Voronoi diagram, as in Figure 10.40, make a diagram by bending the infinite Voronoi edges of the diagram to meet at an extra added point, denoted P in Figure 10.41. The new diagram has F = n faces and the same number E of edges as the Voronoi diagram. Its number V of vertices is the number of Voronoi vertices in the Voronoi diagram plus 1 for P. From Euler’s formula (Theorem 1.5.1), V − E + F = 2. Each vertex (where three or more faces meet) has at least three edges. However 3V would double count the edges, so 3V ≤ 2E or V ≤ 23 E. Rearrange Euler’s formula as E = V + n − 2 and substitute to obtain E ≤ 23 E + n − 2 or E ≤ 3n − 6. A similar adjustment bounds E by 32 V ≤ E so as to eliminate E. We find 32 V ≤ V + n − 2 or V ≤ 2n − 4. Now V includes one additional vertex, so there are at most 2n − 5 Voronoi vertices. !

P

Figure 10.40 A Voronoi diagram.

Figure 10.41 A modified Voronoi diagram.

If we start with infinitely many sites, we could obtain a tiling of the plane with no infinite regions. Let’s arrange the sites to form a lattice using translations in two directions. Label the

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points with two integer subscripts, Pi, j . Without loss of generality, assume that the shortest distance is d(P0,0 , P1,0 ) = a. Similarly, the next shortest distance is d(P0,0 , P0,1 ) = b and m(∠P1,0 P0,0 P0,1 ) = θ , as in Figure 10.42. To determine the Voronoi region of P0,0 we need only consider the lattice points closest to P0,0 , which are the eight others marked in Figure 10.42. We form the perpendicular bisectors of P0,0 with each of them. At most six determine the region of P0,0 since the others are too far away. The neighbors around every point in the lattice look just like those around P0,0 . So we obtain a monohedral tiling, depicted in Figure 10.43.

P-1,1 P0,1 P1,1 b

θ

P-1.0 P0,0 a P 1,0 P-1,-1 P0,-1 P1,-1

Figure 10.42 A lattice of sites.

Figure 10.43 A Voronoi diagram with a lattice of sites. Exercise 10.4.2.∗ Determine which six of the eight perpendicular bisectors form part of the Voronoi region of P0,0 when, as in Figure 10.42, π2 < θ < π . What happens when 0 < θ < π2 ? Exercise 10.4.3.∗ If the sites of a Voronoi diagram form a lattice, explain why each tile must be centrally symmetric. Exercise 10.4.4.∗ What happens if θ =

π , 2

as in Figure 10.44? Sketch the Voronoi diagram.

Exercise 10.4.5.∗ What do the Voronoi regions look like if a = b and θ =

π ? 3

From the preceding discussion and exercises we see that for sets of lattice points the Voronoi diagram is a monohedral tiling. Further, the tiles are either centrally symmetric hexagons or rectangles.

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10.4 Voronoi Diagrams

P0,1 θ P0,0

P1,0

Figure 10.44 A lattice with right angles. See Okabe et al [8] for more on Voronoi diagrams, including applications and higher dimensional Voronoi diagrams. Discrete geometry developed rapidly in the twentieth century as mathematical interests broadened and computer science applications exploded. While elementary geometry suffices to understand most of the problems, they often lead to deep mathematics and unsolved problems. To implement discrete geometry insights requires sophisticated computer algorithms and an analysis of their efficiency.

10.4.1 Exercises for Section 10.4 10.4.6. Construct on dynamic software or graph paper the Voronoi diagram for the sites (0, 0), (2, 0), (0, 2), (−2, 0), (0, −2), (4, 4), (4, −4), (−4, 4), and (−4, −4). Hint: Use symmetry. ∗

10.4.7. Given a triangle △ABC find four sites so that the triangle will be the Voronoi region of one of these points. Illustrate your answer. 10.4.8. ∗ (a) For n = 3 give a Voronoi diagram with n sites and the maximum number Voronoi edges and Voronoi vertices, as determined in Theorem 10.4.3. ∗ (b) Repeat part (a) for n = 4. (c) Repeat part (a) for n = 5. (d) Repeat part (a) for n = 6. (e) What is the minimum number of Voronoi edges and vertices for n sites? Explain. 10.4.9. (a) Give an example of a Voronoi diagram with five sites so that one Voronoi region is a trapezoid with two adjacent obtuse angles. Hint: start with the region. (b) Repeat part (a) with a quadrilateral with three obtuse angles.

10.4.10. ∗ (a) For a general k describe the sites for a Voronoi diagram with the fewest number of sites so that the Voronoi region of a site is a regular k-gon. ∗ (b) For k = 3, 4, and 6 draw examples of Voronoi diagrams with two regular k-gons as adjacent Voronoi regions. What is the minimum number of sites needed for each? (c) Repeat part (b) for k = 5. What is the fewest number of sites needed for the Voronoi diagram?

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(d) Explain why it is impossible to have a Voronoi diagram with two adjacent regions that are regular k-gons when k > 6. Provide a figure. ) Si , is a 10.4.11. Let {Si : i ∈ I } be a collection of convex sets. Prove that their intersection, convex set.

i∈I

10.4.12. Assume that the sites of a Voronoi diagram are not collinear. Give a way to determine when a Voronoi edge is infinitely long. Justify your method. 10.4.13. Prove Theorem 10.4.2. Hint: See Exercise 1.2.15. 10.4.14. ∗ (a) Is it possible for an infinite Voronoi diagram to consist of equilateral triangles? If not, why not? Is so, do the sites form a lattice? (b) Repeat part (a) for a general acute triangle. Illustrate your answer. 10.4.15. ∗ (a) Describe the set of points in taxicab geometry that are the same distance from (0, 0) and (a, b), where 0 < a < b. Hint: The set consists of two rays and a segment. (Example 3 of Section 2.3 discusses taxicab geometry.) (b) Repeat part (a) when 0 < b < a. (c) Repeat part (a) when 0 < a = b. (The answer differs significantly from parts (a) and (b).) (d) Use taxicab geometry to draw on graph paper the Voronoi diagram for the sites (0, 0), (4, −2), and (2, 4). (e) Repeat part (d) for the sites (0, 0), (4, 4), and (8, −4). Describe any peculiarities in the diagram. 10.4.16. (a) In taxicab geometry find the Voronoi diagram for the sites A = (4, 0), B = (2, 2), C = (0, 4), D = (−2, 2), E = (−4, 0), F = (−2, −2), G(0, −4), and H = (2, −2). Some areas have two or more sites that are the closest to them. Label the regions with the names of the closest sites. What sites are closest to (0, 0)? Hint: Use symmetry. (b) Repeat part (a) for the sites J = (4, 0), K = (3, 1), L = (2, 2), M = (1, 3), and N = (0, 4). 10.4.17. Consider Voronoi diagrams on the surface of a sphere. ∗ ∗

(a) Describe the set of points equidistant from two points on a sphere. (b) Describe the Voronoi diagram for three sites spaced equally around the equator of a sphere. (c) Repeat part (b) for four sites. (d) Repeat part (b) for the four sites of part (c) together with the north and south poles. What polyhedron corresponds to the Voronoi diagram? (e) For each regular polyhedron describe a set of sites whose Voronoi diagram corresponds to it.

10.4.18. (a) Given sites in three-dimensional Euclidean space, describe what their Voronoi diagram would be and how we could determine it. Hint: A Voronoi region will have volume, whether finite or infinite. What will bound it?

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483

(b) What is the smallest number of sites in three dimensions for there to be a finite Voronoi region in their Voronoi diagram? Justify your answer. How many Voronoi vertices, edges, and faces are there in the diagram? (Distinguish between finite and infinite edges and faces.) (c) Repeat part (b) for two finite Voronoi regions. (d) Repeat part (b) for three finite Voronoi regions. (e) Make a conjecture for the minimum number of sites in a three-dimensional Voronoi diagram with k finite Voronoi regions.

10.4.2 Projects for Chapter 10 1. (a) Given n = 3 non-collinear points in the plane, find the minimum and maximum number of directions (slopes) they determine. (b) Repeat part (a) for larger values of n. (c) Make a conjecture about the minimum number of directions determined by n points in the plane, no three of which are collinear. (d) Repeat parts (b) and (c) for points in three dimensions, where no four points are coplanar. See Goodman and O’Rourke [4, 6]. (e) Explore the number of angles determined by n points in the plane, no three of which are collinear. C D 1 2n−4 2. From Section 10.2 we know that an n-gon can have from 1 to n−1 triangulations. n−2 Explore which numbers in this range can occur. 3. Investigate the art gallery problem, including polygons with holes and polyhedra. See Devadoss and O’Rourke [3]. 4. Investigate the fortress problem and the prison problem. 5. (a) Construct the twelve pentominoes, figures made of five squares fitting together edgeto-edge. (b) Determine which give monohedral tilings. (c) Determine which are rep-tiles. (See Exercise 10.3.12.) 6. In three dimensions investigate dividing a polyhedron into tetrahedra, the most basic polyhedron. As with a triangulation, all four vertices of each tetrahedron must be vertices of the original polyhedron. (a) (b) (c) (d) (e) (f)

Find a way to divide a cube into tetrahedra. How many tetrahedra do you have? Find a way to divide a cube into a different number of tetrahedra than in part (a). Repeat parts (a) and (b) for a triangular bipyramid. (See Figure 10.45.) Repeat part (a) for a triangular prism. Repeat part (a) for a regular octahedron. An n-gonal pyramid has an n-gon for a base and an apex not in the plane of the n-gon. The apex is connected to each vertex of the polygon. Explain how to divide a pyramid into tetrahedra. (g) Find an example of a polyhedron that can’t be divided into tetrahedra. (See Devadoss and O’Rourke [3].)

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Figure 10.45 A bipyramid. 7. (a) There are 21 possible arrangements of regular polygons that fit around a vertex. Illustrate all arrangements and explain why no others can occur. Hint: What must the sum of the angles be? (b) A semi-regular tiling has two or more regular polygons as tiles that fit together edge-toedge with the same arrangement of tiles at each vertex. Similarly, a regular tiling uses just one regular polygon. Only eleven of the arrangements in part (a) can be continued to give regular or semi-regular tilings. Explain why the other ten fail. Hint: Consider a polygon with an odd number of edges and the polygons surrounding it. 8. Explore the range of pentagonal monohedral tiles. See Schattschneider [9]. 9. Explore Conway’s criterion for a monohedral tiling. See Schattschneider [10]. 10. (a) Find monohedral tilings for each of the seventeen types of wallpaper patterns. Color the tilings with the fewest number of colors, following the coloring restriction of Section 10.3. Which wallpaper patterns permit tilings with only two colors? Which need three colors? Do any need more than three colors? (b) Repeat part (a) but require for two regions of the same color that there is a symmetry of the pattern taking one of them to the other. (c) Repeat part (b) but require additionally for two regions with different colors that there is a symmetry of the pattern taking one of them to the other. Section 6.3 discusses wallpaper patterns. 11. Design tilings whose tiles are regular polygons and regular star figures. (See Figure 10.46 and Project 6 of Chapter 1.)

Figure 10.46 Regular star figures.

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12. Investigate space filling polyhedra. Build enough copies of a space filling polyhedron to see how it can tile space. (See Holden [7, 154ff] and Senechal [11].) 13. Investigate the Schmitt-Conway biprism, which gives a monohedral tiling of space without having translational symmetry. 14. (a) Determine conditions on a parallelogram so that it can be a Voronoi region. (b) Repeat part (a) for a general quadrilateral. 15. Suppose we have a set of sites on the surface of a sphere. (a) Explain why each Voronoi region will be finite. (b) For a general value of n describe a way to place n sites on the surface of the sphere so that the Voronoi regions have the same area. (c) Investigate how to place n sites on the surface of a sphere so as to minimize the greatest distance from a point on the sphere to the nearest site. 16. Investigate applications of Voronoi diagrams. (See Okabe et al [8].) 17. Investigate Voronoi diagrams in three dimensions. (See Okabe et al [8].) 18. Investigate the Voronoi diagram in three dimensions where the set of sites is an infinite lattice in three dimensions. Build an example of the polyhedron that is a typical region. Consider special cases. 19. Investigate the geometry of robotic motion. (See de Berg et al [2, Chapter 13].) 20. Investigate algorithms for discrete geometry. (See de Berg et al [2], Devadoss and O’Rourke [3], and Okabe et al [8].) 21. Write an essay describing what distinguishes discrete geometry from traditional Euclidean geometry.

10.4.3 References 1. Altman, E., On a problem of P. Erd os, ˝ The American Mathematical Monthly, vol. 70 # 2 (Feb. 1963), 148–157. 2. de Berg, M., M. van Kreveld, M. Overmars, and O. Schwarzkopf, Computational Geometry, New York: Springer, 2000. 3. Devadoss, S. and J. O’Rourke, Discrete and Computational Geometry, Princeton, NJ: Princeton University Press, 2011. 4. Goodman, J. and J. O’Rourke, Handbook of Discrete and Computational Geometry, 2nd ed., NY: CRC Press, 2004. 5. Gr¨unbaum, B. and G. Shephard, Tilings and Patterns, New York: Freeman and Co., 1987. 6. Hoffman, P., The Man who Loved only Numbers, New York: Hyperion, 1998. 7. Holden, A., Shapes, Space, and Symmetry, NY: Dover, 1991. 8. Okabe, A., B. Boots, K. Sugihara, and S. Chiu, Spatial Tessellations, New York: John Wiley and Sons, 2000. 9. Schattschneider, D., Tiling the plane with congruent pentagons, Mathematics Magazine, vol. 51 # 1 (Jan. 1978), 29–44.

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10. Schattschneider, D., Will it tile? Try the Conway criterion!, Mathematics Magazine, vol. 53 # 4, (Sept. 1980), 224–233. 11. Senechal, M., Which tetrahedra fill space? Mathematics Magazine, vol. 54 # 5 (Nov. 1981), 227–243. 12. Sibley, T., Foundations of Mathematics, Hoboken, NJ: Wiley, 2009.

11

Epilogue “It is by logic we prove, it is by intuition that we invent.” “Logic, therefore, remains barren unless fertilized by intuition.” —Henri Poincar´e

For millennia geometric thinking has blossomed from the fruitful union of reasoning and intuition. For over two thousand years, geometry meant Euclidean geometry. Since 1800 new geometrical insights have given us non-Euclidean geometries, multi-dimensional geometries, projective geometry, finite geometries, differential geometry, and discrete geometry. Each required deep insight and careful reasoning. They have provided a geometrical viewpoint enriching the rest of mathematics. In addition to the advent of different geometries, we now have multiple approaches. We still honor Euclid’s logical edifice through the study of axiomatic systems. Just as vital are the algebraic formulation pioneered by Descartes and Fermat and the transformational approach of M¨obius and others. We supplement the straightedge and compass constructions of Euclid with computer software. Each approach supports the creative invention and careful arguments characteristic of geometry. While we have surveyed geometries, as the subtitle of this text suggests, the reader may still question what is the essence of the text’s title, “thinking geometrically.” What unifies the approaches and geometries? Geometric thinking lies between the purely formal reasoning of logic and algebra and the concrete insight of physical space. Consider the geometrical statement “In a Euclidean space of more than four dimensions there are exactly three regular polytopes.” Clearly this claim isn’t concrete, although we may need insight from two and three dimensions to understand it. And even if we use algebra to prove it, we need geometrical understanding to formulate the argument. It is harder, though, to detail the essence of geometrical intuition. No particular geometric content seems common to all of geometry. However, each geometric investigation retains some sense of studying space or of studying shapes and configurations in some space. For example, projective geometry is fully geometrical without distance, angle measure, or area. In principle, the undefined terms and axioms of an axiomatic system can, as Hilbert argued, be anything. Yet mathematicians don’t study arbitrary axiomatic systems, and geometric systems have geometric models. Our understanding of geometry has broadened enormously in the past 200 years and

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continues expanding without losing touch with its historical and physical roots. As Benoit Mandelbrot said, “Intuition can be changed and refined and modified. . . . ”

11.1 Topology We can illustrate geometrical thinking further with a brief description of topology. Its founder, Henri Poincar´e (1854–1912), considered more fundamental, geometric-like properties than those defined in terms of lines, distances, and angles. Topology allows not only the isometries of Euclidean geometry, but continuous deformations. For example, the sphere and the stellated octahedron of Figure 11.1 can be continuously deformed into each other. Similarly, the torus and the polyhedron of Figure 11.2 are topologically equivalent. However, the shapes of Figure 11.1 can’t be continuously deformed into those of Figure 11.2. The surfaces in Figure 11.1 share a property that the torus and polyhedron of Figure 11.2 don’t. In elementary terms, neither of the first two shapes have a hole going through it. Poincar´e defined a more exact mathematical property, called simply connected: a closed curve on the surfaces of the shapes in Figure 11.1 can be continuously shrunk to a point without leaving the surface. However, the highlighted curves on the shapes in Figure 11.2 can’t be shrunk to points without breaking the surfaces, implying that the shapes are not simply connected. Simply connected and other types of connectedness are geometric properties of shapes and spaces, even though no one from Euclid to Riemann considered them. In Section 4.5 we discussed another topological property. The surface of a sphere is orientable: if we slide the letter “b” around on the surface, we can never have it turn into the letter “d.” In terms of transformations, indirect isometries, such as mirror reflections, are completely separate from direct isometries. However single elliptic geometry is not orientable: we can slide an object around and have it switch orientation. (Figure 4.53 illustrates this.) Orientability is a geometric property, even if no one had thought of it before M¨obius devised the object named after him. Figure 5.49 depicts a one-sided M¨obius strip, which isn’t orientable.

Figure 11.1

Figure 11.2

11.2 Henri Poincar´e

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Poincar´e didn’t stop with intuitions or examples, however. He developed the mathematical approach to proving the properties and classifying spaces, now a part of algebraic topology. Topology represents a broadening of geometrical thinking, retaining the characteristic blend of intuition and reasoning. Down the centuries geometers have celebrated the fusion of intuition and proof into the beauty of geometry.

11.2 Henri Poincar´e The mathematician does not study pure mathematics because it is useful; he studies it because he delights in it and he delights in it because it is beautiful. —Henri Poincar´e

In many areas of pure mathematics as well as physics Henri Poincar´e (1854–1912) contributed fundamental insights. Poincar´e earned his Ph.D. at the age of 25 and spent the rest of his career as a professor of mathematics and physics in France. He boldly used his deep mathematical intuition to find much new and important mathematics. But he also realized the potential of intuition to mislead and so the need for careful proofs. He advocated for the fruitful interaction of theoretical and applied mathematics, and his own research bore ample testimony to this goal. At 28 Poincar´e developed his disk and half plane models of hyperbolic geometry, discussed in Section 4.1. As he showed, their isometries were the M¨obius transformations of Section 5.6. Soon after he realized that the same transformations where the key to the study of what are now called automorphic functions in complex analysis. In 1889 Poincar´e won a mathematical competition with a paper on the mutual gravitational interactions of three masses. The original version purported to solve the three body problem, but contained an error. Poincar´e’s revised paper provided the first mathematical description of what is now called chaos. While Newton had completely solved the two body problem, Poincar´e’s work revealed that the three body problem had no complete closed form solution. In hindsight his profound paper became the starting point of dynamical systems, an important branch of modern mathematics. Another major area of mathematics, topology, grew from investigations Poincar´e published starting in 1894. He characterized different kinds of two-dimensional surfaces using connectedness and orientability, described in the Epilogue. The Poincar´e conjecture extends the ideas to higher dimensions. Culminating nearly a hundred years of mathematical research, Grigori Perelman proved the final part of the conjecture in 2003. Poincar´e contributed to several areas of physics, including quantum mechanics and relativity. Most notably, he and Hendrik Lorentz determined the transformations of the special theory of relativity independently of Einstein.

A

Definitions, Postulates, Common Notions, and Propositions from Book I of Euclid’s Elements A.1 Definitions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

11. 12. 13. 14. 15. 16. 17. 18. 19.

A point is that which has no part. A line is breadthless length. The extremities of a line are points. A straight line is a line which lies evenly with the points on itself. A surface is that which has length and breadth only. The extremities of a surface are lines. A plane surface is a surface which lies evenly with the straight lines on itself. A plane angle is the inclination to one another of two lines in a plane which meet one another and do not lie in a straight line. And when the lines containing the angle are straight, the angle is called rectilineal. When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called perpendicular to that on which it stands. An obtuse angle is an angle greater than a right angle. An acute angle is an angle less than a right angle. A boundary is that which is the extremity of anything. A figure is that which is contained by any boundary or boundaries. A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure are equal to one another. And the point is called the center of the circle. A diameter of the circle is any straight line drawn through the center and terminated in both directions by the circumference of the circle, and such a straight line also bisects the circle. A semicircle is the figure contained by the diameter and the circumference cut off by it. And the center of the semicircle is the same as that of the circle. Rectilineal figures are those which are contained by straight lines, trilateral figures being those contained by three, quadrilateral those contained by four, and multilateral those contained by more than four straight lines.

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20. Of trilateral figures, an equilateral triangle is that which has three sides equal, an isosceles triangle that which has two of its sides alone equal, and a scalene triangle that which has its three sides unequal. 21. Further, of trilateral figures, a right-angled triangle is that which has a right angle, an obtuse-angled triangle that which has an obtuse angle, and an acute-angled triangle that which has its three angles acute. 22. Of quadrilateral figures, a square is that which is both equilateral and right-angled; a rectangle that which is right-angled but not equilateral; a rhombus that which is equilateral but not right-angled; and a parallelogram that which has its opposite sides and angles equal to one another but is neither equilateral nor right-angled. And let quadrilaterals other than these be called trapezia. 23. Parallel straight lines are straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either direction.

A.2 The Postulates (Axioms) Let the following be postulated: 1. 2. 3. 4. 5.

To draw a straight line from any point to any point. To produce a finite straight line continuously in a straight line. To describe a circle with any center and distance. That all right angles are equal to one another. That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.

A.3 Common Notions 1. 2. 3. 4. 5.

Things which are equal to the same thing are also equal to one another. If equals be added to equals, the wholes are equal. If equals be subtracted from equals, the remainders are equal. Things which coincide with one another are equal to one another. The whole is greater than the part.

A.4 The Propositions of Book I (Paraphrased in modern English)1 I-1. I-2. I-3. I-4.

On a given line segment, to construct an equilateral triangle. To construct a line segment. To construct a shorter line segment on a larger one. [SAS] If triangles △ABC and △D E F have AB ∼ = D E, BC ∼ = E F, and ∠ABC ∼ = ∠D E F, they will also have the remaining corresponding side and angles congruent, namely, AC ∼ = D F, ∠BC A ∼ = ∠E F D, and ∠C AB ∼ = ∠F D E.

1 (Reproduced and paraphrased from Heath, T., The Thirteen Books of Euclid’s Elements, 3 vol., Cambridge, England: Cambridge University Press, 1908. Copyright expired)

A.4 The Propositions of Book I

493

∼ BC has the two opposite I-5. [Isosceles] A triangle △ABC with two congruent sides AB = angles congruent, namely ∠B AC ∼ ∠BC A. = I-6. [Isosceles] A triangle with two congruent angles has two congruent sides: In △ABC, if ∠B AC ∼ = BC. = ∠BC A, then AB ∼ I-7. Given the lengths of the three sides, only one triangle can be formed in a given position; that is, given △ABC and points D and E so that AB ∼ = D E, there is a unique point F on one side of D E so that AC ∼ = D F and BC ∼ = E F. I-8. [SSS] If two triangles have their corresponding sides congruent, they have their corresponding angles congruent. That is, in △ABC and △D E F if AB ∼ = D E, BC ∼ = E F, ∼ ∼ ∼ and AC = D F, then ∠ABC = ∠D E F, ∠BC A = ∠E F D, and ∠C AB ∼ = ∠F D E. I-9. To bisect a given angle. I-10. To bisect a given line segment. I-11. To construct the perpendicular to a line from a point on a line. I-12. To construct the perpendicular to a line from a point not on a line. I-13. The measures of supplementary angles add to 180◦ . That is, when two lines intersect, the measures of any two of the adjacent angles they determine add to 1800 . I-14. If the measures of two adjacent angles add to 180◦ , then the two rays not shared by the angles form a line. I-15. [Vertical angles] If two straight lines cut one another, the non-adjacent angles (vertical angles) are congruent. I-16. An exterior angle of a triangle is greater than either remote interior angle. I-17. The measures of any two angles of a triangle add to less than 180◦ . I-18. In any triangle the larger side is opposite the larger angle. I-19. In any triangle the greater angle is opposite the greater side. I-20. [Triangle inequality] The lengths of any two sides of a triangle add to more than the length of the third side. I-21. If D is an interior point of △ABC, then AD is shorter than AB and AC and m(∠ ADC) > m(∠ ABC). I-22. A triangle can be constructed with sides congruent to three given segments if and only if the lengths of any two of these segments add to more than the length of the third. I-23. To construct a given angle on a given ray. I-24. If in triangles △ABC and △D E F, AB ∼ = D E, BC ∼ = E F, and m(∠ ABC) > m(∠D E F), then AC > D F. I-25. If in triangles △ABC and △D E F, AB ∼ = D E, BC ∼ = E F, and AC > D F, then m(∠ ABC) > m(∠D E F). I-26. [AAS] In △ABC and △D E F if ∠ABC ∼ = D F, = ∠D E F, ∠BC A ∼ = ∠E F D, and AC ∼ then AB ∼ = D E, BC ∼ = E F, and ∠C AB ∼ = ∠F D E. [ASA] In △ABC and △D E F if ∠ABC ∼ = E F, and ∠BC A ∼ = ∠E F D, = ∠D E F, BC ∼ ∼ ∼ ∼ then AB = D E, AC = D F, and ∠C AB = ∠F D E. I-27. For two lines cut by a transversal if alternate interior angles are congruent, the lines are parallel. (See Figure A.1.) I-28. For two lines cut by a transversal if corresponding angles are congruent, the lines are parallel. If the measures of the two interior angles on the same side of the transversal add to 180◦ , the lines are parallel. (See Figure A.1.)

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Definitions, Postulates, Common Notions, and Propositions from Book I of Euclid’s Elements

4 1 2

5 3

Figure A.1 ∠1 and ∠2 are alternate interior angles, ∠1 and ∠3 are corresponding angles, ∠3 and ∠4 are alternate exterior angles, and ∠1 and ∠5 are interior angles on the same side.

(The fifth (parallel) postulate is needed to prove propositions starting with I-29.) I-29. Two parallel lines cut by a transversal have alternate interior (and exterior) angles congruent and corresponding angles congruent, and the measures of the two interior angles on the same side of the transversal add to 180◦ . (See Figure A.1.) I-30. If k is parallel to l and l is parallel to m, then k is parallel to m. I-31. To construct a parallel to a given line through a given point. I-32. The angle sum of a triangle is 180◦ . The sum of the measures of two angles equals the opposite exterior angle. I-33. If AB is congruent and parallel to C D in quadrilateral ABC D, then AD is congruent and parallel to BC. I-34. In a parallelogram opposite sides are congruent, opposite angles are congruent, and the diagonals divide the parallelogram into congruent triangles. I-35. Parallelograms with the same base and the same height have the same area. I-36. Parallelograms with congruent bases and heights have the same area. I-37. Triangles with the same base and the same height have the same area. I-38. Triangles with congruent bases and heights have the same area. I-39. Triangles with equal areas and a common side have the same height. I-40. Triangles with equal areas and a pair of congruent sides have the same height. I-41. If a parallelogram has the same base as a triangle and the same height, the area of the parallelogram is double the area of the triangle. (In effect, the area formula A = 12 b · h for triangles.) I-42. To construct a parallelogram with the same area as a triangle and with a given angle. I-43. In any parallelogram the complements of the parallelogram about the diameter have equal area. (In Figure A.2, ABC D is a parallelogram, AC is a diameter, and E is a point on AC. The parallelograms B H E F and E G D I are complements and so have equal area.) A

F B

I

D

E G H

C

Figure A.2 I-44. To construct a parallelogram with the same area as a triangle, with a given side and a given angle.

A.4 The Propositions of Book I

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I-45. To construct a parallelogram with the same area as a polygon, with a given side and a given angle. I-46. To construct a square with a given side. I-47. [The Pythagorean theorem.] In right-angled triangles the square on the hypotenuse has an area equal to the sum of the areas of the squares on the other two sides. I-48. [The converse of the Pythagorean theorem.] If in a triangle the square on one side has an area equal to the sum of the areas of the squares on the other two sides, the angle contained by the other two sides of the triangle is right.

B

SMSG Axioms for Euclidean Geometry1 Undefined Terms. point, line, plane Postulate 1. (Line Uniqueness) Given two different points there is exactly one line that contains them. Postulate 2. (Distance Postulate) To every pair of different points there corresponds a unique positive number, called the distance between the two points. Postulate 3. (Ruler Postulate) The points of a line can be placed in a correspondence with the real numbers in such a way that (i) to every point of the line there corresponds exactly one real number, (ii) to every real number there corresponds exactly one point of the line, and (iii) the distance between two distinct points is the absolute value of the difference of the corresponding numbers. Postulate 4. (Ruler Placement Postulate) Given points P and Q of a line, a coordinate system can be chosen in such a way that the coordinate of P is zero and the coordinate of Q is positive. Postulate 5. (Existence of Points) (a) Every plane contains at least three non-collinear points. (b) Space contains at least four non-coplanar points. Postulate 6. (Points on a Line Lie in a Plane) If two points lie in a plane, then the line containing them lies in the same plane. Postulate 7. (Plane Uniqueness) Three points lie in at least one plane, and three non-collinear points lie in exactly one plane. Postulate 8. (Plane Intersection) If two planes intersect, then their intersection is a line.

1 (Reproduced with permission from School Mathematics Study Group, Geometry, New Haven: Yale University Press, 1961.)

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Postulate 9. (Plane Separation Postulate) Given a line and a plane containing it, the points of the plane that do not lie on the line form two sets such that (i) the sets are convex and (ii) if P is in one set and Q is in the other, then the segment P Q intersects the line. Postulate 10. (Space Separation Postulate) The points of space that do not lie in a given plane form two sets such that (i) each of the sets is convex and (ii) if P is in one set and Q is in the other, then the segment P Q intersects the plane. Postulate 11. (Angle Measurement Postulate) To every angle ∠B AC there corresponds a real number between 0 and 180. −→ Postulate 12. (Angle Construction Postulate) Let AB be a ray on the edge of the half plane H . For −→ every r between 0 and 180, there is exactly one ray A P with P in H such that m(∠P AB) = r ◦ .

Postulate 13. (Angle Addition Postulate) If D is a point in the interior of ∠B AC, then m(∠B AC) = m(∠B AD) + m(∠D AC).

Postulate 14. (Supplement Postulate) If two angles form a linear pair, then they are supplementary. Postulate 15. (SAS Postulate) Given a correspondence between two triangles (or between a triangle and itself), if two sides and the included angle of the first triangle are congruent to the corresponding parts of the second triangle, then the correspondence is a congruence. Postulate 16. (Parallel Postulate) Through a given external point there is at most one line parallel to a given line. Postulate 17. (Area of a Polygonal Region) To every polygonal region there corresponds a unique positive real number, called the area. Postulate 18. (Area of Congruent Triangles) If two triangles are congruent, then the triangular regions have the same area. Postulate 19. (Summation of Areas of Regions) Suppose that the region R is the union of two regions R1 and R2 . If R1 and R2 intersect at most in a finite number of segments and points, then the area of R is the sum of the areas of R1 and R2 . Postulate 20. (Area of a Rectangle) The area of a rectangle is the product of the length of its base and the length of its altitude. Postulate 21. (Volume of a Rectangular Parallelepiped) The volume of a rectangular parallelepiped is the product of the length of its altitude and the area of its base. Postulate 22. (Cavalieri’s Principle) Given two solids and a plane, if for every plane that intersects the solids and is parallel to a given plane the two intersections have equal areas, then the two solids have the same volume.

C

Hilbert’s Axioms for Euclidean Plane Geometry1 Axioms are paraphrased for ease. Parts of axioms applying only to three-dimensional geometry are placed in parentheses.

Undefined Terms. Point, line, plane, on, between, congruence Group I: Axioms of Connection I-1. Through any two distinct points A, B, there is always a line m. I-2. Through any two distinct points A, B, there is not more than one line m. I-3. On every line there exist at least two distinct points. There exist at least three points which are not on the same line.

Group II: Axioms of Order II-1. If point B is between points A and C, then A, B, C are distinct points on the same line, and B is between C and A. ← → II-2. For any two distinct points A and C, there exist points B and D on the line AC with B between A and C and C between A and D. II-3. If A, B, C are three distinct points on the same line, then one and only one of the points is between the other two. Definition. By the segment AB is meant the set of all points that are between A and B. Points A and B are called the endpoints of the segment. The segment AB is the same as segment B A. II-4. (Pasch’s axiom) Let A, B, C be three points not on the same line and let m be a line (in the plane of A, B and C) that does not pass through any of them. Then if m passes through a point of the segment AB, it will also pass through a point of segment AC or a point of segment BC. Note: This postulate may be replaced by the separation axiom.

1 (Reproduced and paraphrased from Hilbert, D., The Foundations of Geometry, 2nd ed. (transl. E. Townsend). Peru, Ill.: Open Court, 1921. Copyright expired.)

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II-4′ . (Separation axiom). A line m separates the points (of the plane) that are not on m into two sets such that if two points X and Y are in the same set, the segment X Y does not intersect m, and if X and Y are in different sets, the segment X Y does intersect m. In the first case X and Y are said to be on the same side of m; in the second case, X and Y are said to be on opposite sides of m. −→ Definition. By the ray AB is meant the set of points consisting of those that are between A and −→ B, the point B itself, and all points C such that B is between A and C. The ray AB is said to emanate from point A. A point A, on a line m, divides m into two rays such that two points are on the same ray if and only if A is not between them. Definition. If A, B, and C are three points not on the same line, then the system of three segments AB, BC, C A, and their endpoints is called the triangle △ABC. The three segments are called the sides of the triangle, and the three points are called the vertices.

Group III: Axioms of Congruence III-1. If A and B are distinct points on line m, and if A′ is a point on line m ′ (not necessarily distinct from m), then there is one and only one point B ′ on each ray of m ′ emanating from A′ such that the segment A′ B ′ is congruent to the segment AB, written AB ∼ = A′ B ′ . Also for any points A and B, AB ∼ = AB. III-2. If two segments are each congruent to a third, then they are congruent to each other. (From this it can be shown that congruence of segments is an equivalence relation; i.e., AB ∼ = AB; if AB ∼ = A′ B ′ , then A′ B ′ ∼ = C D and C D ∼ = E F, then = AB; and if AB ∼ AB ∼ = E F.) III-3. If point C is between A and B, and C ′ is between A′ and B ′ , and if AC ∼ = A′ C ′ and ′ ′ ′ ′ ∼ ∼ C B = C B , then AB = A B . Definition. By an angle is meant a point (called the vertex of the angle) and two rays (called the sides of the angle) emanating from the point. If the vertex of the angle is point A and if B and C are any two points other than A on the two sides of the angle, we speak of the angle ∠B AC or ∠C AB or simply of angle ∠A. −−→ III-4. If ∠B AC is an angle whose sides do not lie on the same line and if (in a plane) A′ B ′ is −−→ a ray emanating from A′ , then there is one and only one ray A′ C ′ on a side of line A′ B ′ , such that ∠B ′ A′ C ′ ∼ = ∠B AC. An angle (in a plane) can be laid off on a side of a ray in one and only one way. Every angle is congruent to itself. Definition. If △ABC is a triangle, then the three angles ∠B AC, ∠C B A, and ∠AC B are called the angles of the triangle. Angle ∠B AC is said to be included by the sides AB and AC of the triangle. III-5. (SAS) If two sides and the included angle of the one triangle are congruent, respectively, to two sides and the included angle of another triangle, then each of the remaining angles of the first triangle is congruent to the corresponding angle of the second triangle.

Hilbert’s Axioms for Euclidean Plane Geometry

501

Group IV: Axiom of Parallels IV-1. (Playfair’s axiom) Through a point A not on a line m there passes at most one line (in the same plane) which does not intersect m.

Group V: Axioms of Continuity V-1. (Archimedean axiom) Let A1 be any point in AB and suppose A2 , A3 , A4 , . . . are points ← → on AB satisfying A A1 ∼ = Ai Ai+1 and A is between A and A2 and Ai+1 is between Ai and Ai+2 . Then there is some n so that B is between A and An . V-2. (Axiom of linear completeness) The system of points on a line with its order and congruence relations cannot be extended in such a way that the relations existing among its elements as well as the basic properties of linear order and congruence resulting from axiom groups I, II, and III and axiom V-1 remain valid. (These two axioms may be replaced by Dedekind’s axiom of continuity: for every partition of the points on a line into two nonempty sets such that no point of either lies between two points of the other, there is a point of one set which lies between every other point of that set and every point of the other set.)

D

Linear Algebra Summary We will consider n-dimensional vectors over R, the real numbers, and matrices for them since that suffices for our needs. See a standard linear algebra text, such as Fraleigh and Beauregard, Linear Algebra, 3rd ed., Reading, MA: Addison Wesley, 1995, for more detail.

D.1 Vectors − → → A v in Rn is an ordered n-tuple. We can write a vector − v either as a column vector J I vvector 1

v2

.. , which will be denoted as (v 1 , v 2 , . . . , v n ) to save space, or as a row vector [v 1 , v 2 , . . . , v n ], . vn depending on the context. A scalar is an element of R. We add vectors componentwise: − → → v +− w = (v 1 , v 2 , . . . , v n ) + (w1 , w2 , . . . , wn ) = (v 1 + w1 , v 2 + w2 , . . . , v n + wn ). → We multiply a vector − v by a scalar s: → s− v = s(v 1 , v 2 , . . . , v n ) = (sv 1 , sv 2 , . . . , sv n ).

→ → → → The dot product − v ·− w of two vectors − v = (v 1 , v 2 , . . . , v n ) and − w = (w1 , w2 , . . . , wn ) is given by − → → v ·− w = v 1 w1 + v 2 w2 + · · · + v n wn . → → Two vectors − v and − w are orthogonal (a generalization of perpendicular) if and only if − → − → v · w = 0. → The length of a vector − v = (v 1 , v 2 , . . . , v n ) is ; K− K !→ − K→ v ·→ v = v 12 + v 22 + · · · + v n2 , v K= − which generalizes the Pythagorean theorem to n-dimensional space. In R3 we can define the cross product of two vectors:

(v 1 , v 2 , v 3 ) × (w1 , w2 , w3 ) = (v 2 w3 − w2 v 3 , −v 1 w3 + w1 v 3 , v 1 w2 − w1 v 2 ). → → → → The cross product − v ×− w is orthogonal to both − v and − w.

503

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Linear Algebra Summary

The ability to add and multiply by scalars is what makes something linear. For example, → → → → → → → → → the dot product is linear in this sense: (− u +− v )·− w =− u ·− w +− v ·− w and s(− v ·− w) = − → − → − → − → (s v ) · w = v · (s w ). Every vector (v 1 , v 2 , . . . , v n ) can be built as a linear combination of the standard basis vectors (1, 0, 0, . . . , 0), (0, 1, 0, . . . , 0), (0, 0, 1, 0, . . . , 0), . . . , (0, 0, . . . , 0, 1). That is, (v 1 , v 2 , . . . , v n ) = v 1 (1, 0, 0, . . . , 0) + v 2 (0, 1, 0, . . . , 0) + · · · + v n (0, 0, . . . , 0, 1). Similarly we can build all vectors as linear combinations from other sets of vectors. The nicest such sets, called orthonormal bases, have the property that each vector has length 1 and every → − → − → − → v is two of them are orthogonal. Suppose that { b1 , b2 , . . . , bn };is an orthonormal basis and − K K − → − → − → − → − → 2 2 2 K K a vector. If v = s1 b1 + s2 b2 + · · · + sn bn , then v = s + s + · · · + s , showing that 1

2

n

orthonormal bases work the same way as the standard basis does. The set of all linear combinations of a collection of vectors forms a subspace of the entire → space. For example, for one nonzero vector − v , the one-dimensional subspace we get contains − → all the scalar multiples: {s v : s ∈ R}, which is a line through the origin. For two nonzero → → vectors − v and − w , if neither is a scalar multiple of the other, the two-dimensional subspace − → − → {s v + t w : s, t ∈ R} is a plane through the origin.

D.2 Matrices While matrices are used in more ways, for us they primarily represent functions (linear transformations) mapping vectors to vectors. (A linear transformation need not be a transformation in the geometric meaning since it doesn’t need to be either one-to-one or onto.) An m × n matrix is a rectangular array of real numbers ⎤ ⎡ a11 a12 a13 . . . a1n ⎢ a21 a22 a23 . . . a2n ⎥ ⎥ ⎢ ⎢ . .. ⎥ .. ⎣ .. . . ⎦ am1

am2

am3

. . . amn

with m rows and n columns. If M is an m × n matrix with entries m i j , its transpose, written M T , is the n × m matrix whose rows and columns are switched with entries m ji . For example, ⎤ ⎡ 3 4T 1 4 1 2 3 = ⎣2 5⎦ . 4 5 6 3 6

by

A matrix M is symmetric if and only if M = M T . An m × n matrix maps an n-dimensional vector (v 1 , v 2 , . . . , v n ) to an m-dimensional vector ⎡

a11 ⎢ a21 ⎢ ⎢ . ⎣ ..

am1

⎤⎡ ⎤ ⎡ a1n v1 a11 v 1 + a12 v 2 + a13 v 3 + · · · + a1n v n ⎥ ⎥ ⎢ ⎢ a2n ⎥ ⎢v 2 ⎥ ⎢ a21 v 1 + a22 v 2 + a23 v 3 + · · · + a2n v n .. ⎥ ⎢ .. ⎥ = ⎢ .. . ⎦⎣ . ⎦ ⎣ .

a12 a22

a13 a23

... ... .. .

am2

am3

. . . amn

vn

am1 v 1 + am2 v 2 + am3 v 3 + · · · + amn v n

⎤

⎥ ⎥ ⎥. ⎦

Multiplication of a matrix with a vector is a special case of matrix multiplication, defined below. We will almost always be interested in n × n matrices, called square matrices since they take

505

D.3 Determinants

an n-dimensional space to itself. The n × n identity matrix I , has a 1 for each diagonal entry and a 0 for each off-diagonal entry. For example, the 3 × 3 identity matrix is ⎡ ⎤ 1 0 0 I = ⎣0 1 0⎦ . 0 0 1 Multiplication of matrices is defined so that the product G H gives the composition of G and H as functions. The product of an m × n matrix G and an n × k matrix H is an m × k matrix whose i jth entry is the dot product of the ith row of G with the jth row of H . For example, ⎡

g11 ⎢ G H = ⎣g21 ⎡

g31

g12 g22

g13 g23

g32

g33

⎡ ⎤ h 11 g14 ⎢ ⎥ ⎢h 21 g24 ⎦ ⎢ ⎣h 31 g34 h 41

⎤ h 12 h 22 ⎥ ⎥ ⎥ h 32 ⎦ h 42

g11 h 11 + g12 h 21 + g13 h 31 + g14 h 41 ⎢ = ⎣g21 h 11 + g22 h 21 + g23 h 31 + g24 h 41 g31 h 11 + g32 h 21 + g33 h 31 + g34 h 41

⎤ g11 h 12 + g12 h 22 + g13 h 32 + g14 h 42 ⎥ g21 h 12 + g22 h 22 + g23 h 32 + g24 h 42 ⎦ . g31 h 12 + g32 h 22 + g33 h 32 + g34 h 42

An n × n matrix M is invertible if and only if there is a matrix, called its inverse and written M −1 , so that M M −1 = I = M −1 M.

D.3 Determinants The determinant of an n × n square matrix A is a number det(A) with several uses. The key fact is the matrix is invertible if and only if its determinant is nonzero. The absolute value of the determinant of a 2 × 2 matrix tells us how the area of regions scale when transformed by the matrix. Similarly, with a 3 × 3 matrix, the determinant tells how volumes scale. We define the determinant recursively, building up from the 2 × 2 case. First define the minor Ai j for the i jth entry ai j of an n × n matrix A is the (n − 1) × (n − 1) matrix whose entries are from the rows and columns of A not containing ai j . For example, for ⎡ ⎤ I J a11 a12 a13 a11 a12 ⎢ ⎥ A = ⎣a21 a22 a23 ⎦ , . A23 = a31 a32 a31 a32 a33 2D C1 Then det ac db = ad − bc. For an n × n matrix with n > 2 define det( A) recursively by $ det(A) = nj=1 (−1)i+ j ai j det(Ai j ), where we can use any row i. For example, using the second row of A, det(A) = −a21 det(A21 ) + a22 det(A22 ) − a23 det(A23 ). For example, ⎛⎡ ⎤⎞ 1 2 3 det ⎝⎣4 5 6⎦⎠ = 1(5 · 9 − 6 · 8) − 2(4 · 9 − 6 · 7) + 3(4 · 8 − 5 · 7) = 0. 7 8 9

506

Linear Algebra Summary

The term (−1)i+ j forces the signs to alternate, as this chart suggests: ⎡ + ⎢− ⎢ ⎢ ⎢+ ⎢ ⎢ ⎢− ⎣ .. .

− +

+ −

− +

−

+

+

−

− .. .

...

⎤

⎥ ⎥ .. ⎥ . ⎥ ⎥. ⎥ ⎥ ⎦

+

...

Determinants can be taken using a column instead of a row.

D.4 Properties of Matrices An n × n matrix A is invertible if and only if det(A) ̸= 0. For n × n matrices A and B, AI = A = I A, (A T )T = A, ( AB)T = B T A T , and det(AB) = det(A) det(B). If in addition A and B are invertible, ( A T )−1 = (A−1 )T , (AB)−1 = B −1 A−1 , and det(A−1 ) = 1/ det(A). In general the products AB and B A are not equal.

D.5 Eigenvalues and Eigenvectors We are particularly interested in points and lines (and higher dimensional objects) that a transformation takes to themselves. In linear algebra terms, they are sent to scalar multiples of → themselves. For a square matrix M and a scalar λ, we say a nonzero vector − v is an eigenvector − → − → of the eigenvalue λ if and only if M v = λ v . (The terms characteristic value and character→ → istic vector are sometimes used.) If the vector is a row vector, we write − v M = λ− v and the mathematics is analogous. From linear algebra, we manipulate the system of equations represented by the equa→ → → → → → → tion M − v = λ− v to get M − v = λI − v or M − v − λI − v = 0 or (M − λI )− v = 0, which has a − → nonzero solution v if and only if det(M − λI ) = 0. The determinant is a polynomial with λ as the variable. The roots give the eigenvalues. For each eigenvalue λ, we can solve the system → → → M− v = λ− v to find the corresponding eigenvector(s) − v. Example. Let M =

3

2 0 0

1 3 0

4 5 1

4

. Then M − λI =

3

2−λ 0 0

1 3−λ 0

4

4 5 1−λ

. Using the bottom row,

det(M − λI ) = 0(1 · 5 − 4(3 − λ)) − 0((2 − λ)5 − 4 · 0) + (1 − λ)((2 − λ)(3 − λ) − 1 · 0) = (1 − λ)(2 − λ)(3 − λ).

The eigenvalues are λ = 1, λ = 2, and λ = 3. For λ = 3, we solve ⎡

2 ⎣0 0

1 3 0

⎤⎡ ⎤ ⎡ ⎤ 4 x x 5⎦ ⎣ y ⎦ = 3 ⎣ y ⎦ 1

z

z

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D.5 Eigenvalues and Eigenvectors

to get the system of equations 2x + y + 4z = 3x

3y + 5z = 3y

z = 3z.

The solutions are the eigenvectors (y, y, 0). Similarly, for λ = 1 the eigenvectors are (−1.5z, −2.5z, z) and for λ = 2 the eigenvectors are (x, 0, 0). ♦

E

Multivariable Calculus Summary We consider vector functions and functions of two variables. For more detail see any standard three semester calculus text, such as Rogowski, J., Calculus, 2nd ed., NY: W. H. Freeman, 2011.

E.1 Vector Functions −→ → A vector function − c : R → Rn determines a set of points {c(t) : a ≤ t ≤ b} in Rn . For example, −→ c(t) = (cos(t), sin(t), t) for 0 ≤ t ≤ 2π gives one complete rotation of a helix. Each coordinate −→ ci (t) of a function c(t) = (c1 (t), c2 (t), . . . , cn (t)) is a real function. −→ If each coordinate ci (t) of c(t) = (c1 (t), c2 (t), . . . , cn (t)) has a derivative ci′ (t), then we −− → −−→ define c′ (t) = (c1′ (t), c2′ (t), . . . , cn′ (t)). The vector c′ (t) is the derivative or velocity vector, giving −−→ −→ the direction and magnitude of change for c(t). If it exists, c′′ (t) gives the acceleration vector −−→ of c′ (t). −−→ −−→ −→ A curve c(t) is a vector function for which c′ (t) and c′′ (t) exist for all values of t in the −− → domain and for which ∥c′ (t)∥ is positive. → → → → w = (w1 , w2 , . . . , wn ) The dot product − v ·− w of two vectors − v = (v 1 , v 2 , . . . , v n ) and − − → − → is v · w = v 1 w1 + v 2 w2 + . . . + v n wn . ; −→ −→ −→ −→ The norm of a vector function c(t) is the real number ∥c(t)∥ = c(t) · c(t). −− → −→ c′ (t) The unit tangent T (t) of a curve c(t) is − − → . ′ ∥c (t)∥ → → Two vectors − v and − w are orthogonal (a generalization of perpendicular) if and only if − → → v ·− w = 0. → → c a plane curve. For a plane curve, if T ′ (t) exists, it is orthogonal If − c : R → R2 , we call − → to T (t) and so is a normal to the curve. For a space curve − c : R → Rn , with n > 2, T ′ (t) need not be in the direction of the normal. (See below.) In R3 we can define the cross product of two vectors: (v 1 , v 2 , v 3 ) × (w1 , w2 , w3 ) = → → v ×− w is orthogonal to (v 2 w3 − w2 v 3 , −v 1 w3 + w1 v 3 , v 1 w2 − w1 v 2 ). The cross product − − → − → both v and w .

E.2 Surfaces A function of two variables s : R2 → Rn determines a set of points {s(u, v) : a ≤ u ≤ b, c ≤ v ≤ d} in Rn . For example, s(u, v) = (cos(u) cos(v), sin(u) cos(v), sin(v)), for 0 ≤ u < 2π

509

510

Multivariable Calculus Summary

and − π2 ≤ v ≤ π2 gives the unit sphere. In general, each coordinate si (u, v) of s(u, v) = (s1 (u, v), s2 (u, v), . . . , sn (u, v)) is a function from R2 into R. We do not enumerate all the conditions a function of two variables needs to satisfy to be a surface. As a start, the first and second partial derivatives of the function must exist and be continuous. (See Oprea, J., Differential Geometry and Its Applications, 2nd ed., Prentice Hall, 2004 for more details.) The cross sections of a function of two variables s at a point s(u 0 , v 0 ) are the vec−−−→ −−−→ → tor functions − sv0 and − s→ u 0 given by sv 0 (u) = (s1 (u, v 0 ), s2 (u, v 0 ), . . . , sn (u, v 0 )) and su 0 (u) = (s1 (u 0 , v), s2 (u 0 , v), . . . , sn (u 0 , v)). The cross sections fix one of the variables and vary the → s→ other one. For the unit sphere, − sv0 determines the circle of latitude and − u 0 determines the circle of longitude at the point s(u 0 , v 0 ) on the surface.

E.3 Partial Derivatives − → → For a function s if a cross section − sv has a derivative, we call su = sv′ the partial derivative of s − → with respect to u. Similarly, sv = su′ is the partial derivative of s with respect to v. For the unit sphere note that the third coordinate of −−−→ sv0 (u) = (cos(u) cos(v 0 ), sin(u) cos(v 0 ), sin(v 0 )) doesn’t vary and so su = (−sin(u) cos(v), cos(u) cos(v), 0). For the unit sphere, sv = (−cos(u) sin(v), −sin(u) sin(v), cos(v)). For s to be a surface we require su and sv to exist and be linearly independent. The conditions ensure that the partial derivatives determine a plane, defined to be the tangent plane to s at a point of the surface. v . It is orthogonal to both If s : R2 → R3 , we define the unit normal to s as N (u, v) = ∥ssuu ×s ×sv ∥ partial derivatives and so orthogonal to the tangent plane.

F

Elements of Proofs Whether in an informal argument or an axiomatic system, a proof shows the logical necessity of a statement’s conclusion. In an axiomatic system, the proof can use only the statement’s hypotheses, the axioms of the system, previously proved statements and logical deduction. Informal arguments don’t require such explicit care but still require the reasoning to start from accepted statements and deduce the conclusion logically. We consider three types of proofs used in this text: direct proofs, proofs by contradiction, and induction proofs. For more information about proofs see Sibley, T., Foundations of Mathematics, New York: John Wiley and Sons, 2009.

F.1 Direct Proofs Let’s start with an example whose proof some historians credit to Thales (who lived about 600 B.C.E.), followed by a discussion of the proof. ← → ← → Theorem. (Euclid I-15, Vertical Angles). If lines AB and C D intersect at E with E between A and B and between C and D, then m∠C E B = m∠AE D. ← → ← → Proof. Let AB and C D intersect at E with E between A and B and between C and D. Since A, E, and B are in order, m∠AE B = 180◦ . Similarly, m∠C E D = 180◦ . Further, m∠AE B = m∠AEC + m∠C E B and ∠C E D = m∠C E A + m∠AE D since the straight angles are made of these angles. Then m∠AEC + m∠C E B = 180◦ = m∠C E A + m∠AE D. Subtract the common value m∠AEC = m∠C E A to get m∠C E B = m∠AE D. ! Discussion. The proof starts with the hypothesis of the theorem, the “if ” part. The hypothesis gives us the conditions we are allowed to assume at the start, which appear at the start of the proof. The body of the proof uses the definition of a straight angle to relate two angles. Then arithmetic and logical reasoning lead us to the conclusion, the “then” part, relating the two desired angles. In a direct proof, like this, the reasoning goes directly from the hypothesis to the conclusion. In an axiomatic system, we’d need to justify each step from a definition, an axiom, or a previously proved theorem. For example, the SMSG axioms in Appendix B gives postulates directly relevant to adding measures of angles (Postulates 13 and 14) and in that system a straight angle is defined to have a measure of 180◦ . Not every theorem has an explicit hypothesis. However the theorem will give a context, usually about all things of a certain type, which acts as the initial assumption. For example,

511

512

Elements of Proofs

Theorem 1.1.1 states: In Euclidean geometry the angle sum of a triangle is 180◦ . Here we are given for our hypothesis a general Euclidean triangle. Other examples of direct proofs in the text: Theorem 1.1.1, 1.3.2, 1.4.2.

F.2 Proofs by Contradiction Often it is unclear how to proceed from the hypotheses of a statement to its conclusion. In such a situation it may help to try a different approach. A proof by contradiction assumes the conclusion is false, as well as assuming the hypothesis. Then we look for a contradiction—two statements that can’t both be true. We know from logic that it is impossible to deduce a contradiction from true statements. Hence we conclude something incorrect happened. If our reasoning is correct, the only possibility is the extra assumption of the falsity of the conclusion. If it is wrong, the conclusion must be true, proving the statement. Theorem. In ABC D if AB is not parallel to C D, then the diagonals do not bisect each other. Proof. Suppose AB is not parallel to C D and, for a contradiction, that AC and B D bisect each other at, say E. Then AE ∼ = C E and B E ∼ = D E. Also, by the vertical angles theorem above, ∼ ∼ ∠AE B = ∠C E D. By SAS △AE B = △C E D, and so ∠E AB ∼ = ∠EC D. But these angles are opposite interior angles, so by Euclid I-28 AB ∥ C D, contradicting the hypothesis. Hence the diagonals do not bisect each other. ! Other examples of proofs by contradiction. Theorems 1.1.2, 2.1.1, 2.1.2

F.3 Induction Proofs Some properties apply to N, the set of natural numbers (1, 2, 3, . . .) and, in this setting, an induction proof may be an appropriate approach. (In this text exercises needing induction proofs ask you to give an induction proof.) The idea justifying this type of proof depends on a key property of the natural numbers: First, N has a starting point (1), next for each n in N, its successor, n + 1, is also in N, and finally every natural number can be reached from 1 by adding 1 multiple times. This is the principle of mathematical induction, stated more formally below. We follow the principle, which we take as an axiom, with an alternative proof for Exercise 1.1.11b). Principle of mathematical induction. For any subset S of N, if 1 ∈ S and if for all n ∈ S, we also have n + 1 ∈ S, then S = N. Theorem. The sum of the numbers from 1 to n is n(n + 1)/2. Proof. For the initial case, n = 1, we have 1(2)/2 = 1, which is the sum of 1. For the induction step, suppose for n = k that the sum of the numbers from 1 to k is k(k + 1)/2. Now consider the sum from 1 to k + 1. This new sum is k + 1 more than the previous sum: k(k + 1)/2 + (k + 1).

F.4 Other Remarks on Proofs

513

Using distributivity this sum equals (k + 1)(k/2 + 1) = (k + 1)( k+2 ) = (k + 1)(k + 2)/2. This 2 last expression is exactly the formula when n = k + 1. That is, if the formula for n = k holds, then it also holds when n = k + 1. Hence by the principle of mathematical induction, the formula n(n + 1)/2 is the sum of the integers from 1 to n for all n ∈ N. ! Discussion. Students sometimes think that the induction step of an induction proof just assumes what it is supposed to prove. The difference is we are to prove a property holds for all n, whereas the induction step assumes only that we have the property at some step k. In effect, the initial step gets us started and the induction step shows us how to keep going one step at a time. The principle of mathematical induction guarantees that these two parts of the proof are enough to guarantee we can go the whole way. An important variation of an induction proof starts at a number other than 1. If the initial case starts at, say 3, then the induction proof applies to all natural numbers from 3 on, as in the following example. Theorem. A convex n-gon (with n ≥ 3) has n(n − 3)/2 diagonals. Proof. For the initial case, n = 3, we have a triangle and it has no diagonals, as the formula indicates: 3(3 − 3)/2 = 0. Suppose for the induction case when n = k ≥ 3 that a convex k-gon has k(k − 3)/2 diagonals. Consider a convex k + 1-gon A1 A2 . . . Ak Ak+1 . Since this polygon is convex the diagonal A1 Ak is inside the polygon and so splits it into the k-gon A1 A2 . . . Ak and the triangle △A1 Ak Ak+1 . By the induction hypothesis A1 A2 . . . Ak has k(k − 3)/2 diagonals. We also have the diagonal A1 Ak and diagonals from Ak+1 . The diagonals from Ak+1 go to the vertices A2 . . . Ak−1 , giving k − 2 more diagonals. That is, there is a total of k(k − 3)/2 + 1 + k − 2 = k 2 /2 − 3k/2 + k − 1. Also, for n = k + 1, the formula gives (k + 1)(k − 2)/2 = k 2 /2 − k/2 − 1, which equals the number we found, verifying that the formula extends to k + 1. Hence by the principle of mathematical induction, the formula n(n − 3)/2 is the number of diagonals in a convex n-gon for all n ∈ N. ! Discussion. Note in the proof above we needed to derive the value (k + 1)(k − 2)/2 from the geometry, not just manipulate the assumption of k(k − 3)/3 for n = k. Examples of induction proofs: Theorems 7.2.3, 10.2.6 and 10.2.7.

F.4 Other Remarks on Proofs Proofs versus Examples. Most theorems make claims about any things satisfying certain properties and so can not be proven using examples. For example, drawing a right triangle with sides of lengths 3, 4, and 5 and noting that 32 + 42 = 52 does not prove the Pythagorean theorem. A theorem needs a general argument applying to all instances. However, one counterexample suffices to disprove a general statement. A counterexample is an example satisfying a statement’s hypothesis but not its conclusion. For example, we can show that the statement “equilateral quadrilaterals are squares” is false by exhibiting one rhombus that is not a square.

514

Elements of Proofs

If and only if For a pair of statements A and B, sometimes both “If A, then B” and “If B, then A” are theorems. We combine these two statements, saying “A if and only if B.” To prove this, we prove both “If A, then B” and “If B, then A.” Cases Sometimes we need to consider different possibilities to prove a general theorem. We explicitly state each case at the start of its proof to help the reader follow the individual arguments and to ensure that all the cases are addressed.

Answers to Selected Exercises Section 1.1 1.1.1. Throughout replace “even” by “a multiple of 3” and √ “2” by “3.” In2 the parentheses consider the cases p = 3k + 1 and p = 3k + 2. For 4 note that if p is a multiple of 4, p need only be a multiple of 2, not 4. 1.1.2. There is no simple error, but Zeno’s reasoning does point out the difficulty of reasoning about infinite processes based on finite steps. 1.1.3. Difference is ( 89 20)2 (20) − π (10)2 (20) ≈ 37.8 (units)3 , error is ( 89 2r )2 ≈ πr 2 reduces to π ≈ 3.1604938. √ √ 1.1.5. 1.4146296 ≈ 2, 42.42638 ≈ 30 2.

37.8 π (10)2 (20)

≈ 0.6%,

1.1.7. The man walks 300 units. Let x be how far the wizard flew up and z the diagonal of the wizard’s flight. Then x + z = 300 and z 2 = 2002 + (x + 100)2 . So x = 50. 1.1.9. (a) Let l and w be the length and width. Then l + w = 6.5, lw = 7.5 and l 2 − 6.5l − 7.5 = 0. So l = 5, w = 1.5. 1.1.11. (a) (b) (c) (d)

n(n + 1), where n is the number of rows. $n Divide oblong into two triangles, one upside down. Thus i=1 i = n(n + 1)/2. Pentagonal numbers: 1, 5, 12, 22, etc. Hexagonal numbers: 1, 6, 15, 28, etc. Pentagonal numbers: (3n 2 − n)/2. Hexagonal numbers: 2n 2 − n

1.1.13. (a) Add arithmetic mean to itself to get a + b. Multiply the √ geometric mean by itself to get a · b. In geometric terms, the square with side ab has the same area as a rectangle √ ! with sides a and b. (b) b = a(a + b) or b2 − ab − a 2 = 0. Then b = a( 1+2 5 ) ≈ 1.618a.

1.1.15. (a) (b) (c) (d)

72◦ , 54◦ , 108◦ , 36◦ , 108◦ , 72◦ , 72◦ , 36◦ . Use angles. △AB H ∼ △D AB ∼ △AF G. If AB = 1, BC = 1 and GC = 1, and AC = AG + GC = 1 + x. √ = x1 and x 2 + x − 1 = 0. So x = −1+2 5 or 1 + x = Since√△AB D ∼ △BGC, 1+x 1 (1 + 5)/2.

1.1.17. Angles show similar triangles △ADC and △AC B. So ac = ay , bc = bx . So cy = a 2 and cx = b2 . Also x + y = c, giving c2 = c(x + y) = a 2 + b2 .

515

516

Answers to Selected Exercises

Section 1.2 1.2.1. From the construction use SSS (I-8) to show △B AF ∼ = △B E F and so ∠AB F ∼ = ∠E B F. 1.2.2. By construction and SSS (I-8) △AB D ∼ = △G E H ∼ = △G E I and so ∠AB D ∼ = ∼ ∠G E H = ∠G E I . 1.2.3. Use the Pythagorean theorem. 1.2.4. (a) (b) (c) (d)

1, 2, 3, 9, 10, 11, 12, 22, 23, 31, 42, 44, 45, 46. 4, 5, 6, 8, 13, 15, 26, 29, 32–38, 43, 47, 48. SAA, AAA, AAS, SSA SAA is equivalent to AAS and ASA using Theorem 1.1.1. AAA is not a congruence theorem—but shows similar triangles. Neither ASS nor SSA are congruence theorems. Consider △P Q R and △P QT with R between P and T and P Q = 52, QT = 25, P T = 31.5 and P R = 16.5. Let V be between R and T with P V = 24 and use the converse of the Pythagorean theorem.

1.2.6. Suppose in △ABC that AB ∼ = AC. Now ∠B AC ∼ = ∠C AB. Then by SAS (I-4), △ABC ∼ = △AC B. Hence ∠ABC ∼ = ∠AC B.

1.2.8. Suppose that AC ⊥ B E, AE ⊥ C F and AD bisects ∠E AC.

(a) In △AD F and △AD B, ∠AF D ∼ = ∠AB D and ∠F AD ∼ = ∠B AD. By Theorem 1.1.1 (I-32) we also have ∠AD F ∼ = AD, by ASA (I-26), we = ∠AB F. Since AD ∼ have △AD F ∼ = △AD B. (b) From part (a) we have AF ∼ = AB. By ASA (I-26), △AFC ∼ = △AB E.

1.2.11. (a) For a + b on a line construct adjacent segments of lengths a and b. For a − b, make the segments overlapping with a common endpoint. (b) By similar triangles, PP QR = PPTS . (c) If P Q = 1, P R = a, and P S = b, then P T = a · b. If P Q = 1, P T = a, and P R = b, then P S = a/b. 1.2.13. (a) Pick any A on a circle with center O and construct B and C on the circle so that AB = AC = AO. Construct D and E on the circle so that B D = C E = AB. Then △AD E is equilateral. (b) Continue from (a) to construct F on the circle so that D F = B D. Then AB D F EC is a regular hexagon. (c) Pick any point A on a circle with center O and construct the diameter AC through O. Construct diameter B D perpendicular to AC at O. Then ABC D is a square. 1.2.15. (a) Let △ABC be any triangle and G be the intersection of the perpendicular bisector G D of AB and the perpendicular bisector G E of BC. Then △ADG ∼ = △B DG and △B E G ∼ = BG and BG ∼ = C G. = △C E G by SAS (I-4). Then AG ∼ (b) (Continuing from part (a).) Let F be the midpoint of AC. Then △AF G ∼ = △C F G ∼ by SSS (I-8). Thus ∠AF G = ∠C F G. Together they make a straight angle, so they are each right angles, F G ⊥ AC and F G is the perpendicular bisector of AC.

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1.2.18. Let m∠ABC = p and m∠BC A = q. Use isosceles triangles, Theorem 1.1.1 and algebra to show that 3 p = q and p = (180/7)◦ ≈ 25.7◦ . 1.2.20. m∠QT S = 180◦ − x. m∠R QT = |(180◦ − x) − 90◦ | = |90◦ − x|. 1.2.23. (a) c2 = a 2 + b2 − 2a · x. Note that x = b cos(C). (b) Let y be the length of AD and use the Pythagorean theorem. (c) No because the perpendicular from A does not intersect between B and C. So B D = x + a and c2 = a 2 + b2 + 2ax.

Section 1.3 1.3.1. (a) 55◦ . Reasoning: By I-29, m(∠F G D) = 35◦ . From Theorem 1.1.1, m(∠E F G) = 180◦ − m(∠F DG) − m(∠F G D) = 180◦ − 90◦ − 35◦ = 55◦ . (b) (i) k ∥ m by I-30. (ii) k ∥ m by I-27 (iii) k ⊥ m Use Playfair’s axiom and I-29. (c) A rectangle is a parallelogram with at least one right angle. A square is a parallelogram with at least one right angle and at least two congruent adjacent sides. A rhombus is a parallelogram with at least two congruent adjacent sides. 1.3.3. (a) By I-29 ∠R P Q ∼ = ∠P RS. Since P R ∼ = P R and P Q ∼ = RS, by SAS, △P Q R ∼ = △RS Q. ! (b) From part (a), ∠S ∼ = ∠Q and ∠R P Q ∼ = ∠P RS. From Theorem 1.1.1, m(∠S) + ◦ m(∠S R Q) = 180 . Then from I-28, P S ∥ Q R and P Q RS is a parallelogram. ! 1.3.5. (a) Proof. Suppose that ABC D is a parallelogram with diagonal AC. By I-29 ∠C AB ∼ = AC, by ASA (I-26) = ∠AC D and ∠AC B ∼ = ∠C AD. Since AC ∼ △ABC ∼ △C D A. The two triangles have the same area and together they have the = area of ABC D. So each have half of the area of ABC D. ! 1.3.6. (a) Proof. Assume the given relationships. Use I-23 to construct an angle ∠P E H congruent to ∠AB E with P on the same side of B E as D. By I-28 P E ∥ AB. By ← → Playfair’s axiom, there is only one parallel to AB through E. So P must be on D E and so ∠AB E ∼ = ∠D E H . ! 1.3.7. Proof. Let ABC D be a parallelogram with (∠ABC) = 90◦ . Since AB ∥ C D and by I-29 m∠ABC + m∠BC D = 180◦ , m∠BC D = 90◦ . The other angles are similar. !

∼ BC and C D ∼ 1.3.9. (b) Proof. Let ABC D be a kite with AB = = AD. Draw B D. By SSS ∼ △B AD ∼ △BC D and so ∠B AD ∠BC D. ! = = ← → ← → (c) Let O be intersection of B D and AC. Show △C B O ∼ = △AB O. No.

1.3.11. (a) Use the four vertices of a tetrahedron. (b) Use the points (0, 0), (1, 0), (2, 0) and (3, 0). (c) Use the “bow tie” with vertices (0, 0), (0, 1), (2, 0) and (2, 1), in that order. 1.3.13. (a) Proof. Suppose that Q is the midpoint of P R and S and T are on the same side of P R with P S ∼ = QT and Q S ∼ = RT . By SSS △P Q S ∼ = △Q RT . Then ∠S P Q ∼ = ∠T Q R and these are corresponding angles, showing P S ∥ QT .

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Answers to Selected Exercises

(b) (continuing the proof) By △P Q S ∼ = △T S Q, ST ∼ = P Q. Further, ∠P Q S ∼ = ∠T S Q. So by I-27 ST ∥ P Q.

Section 1.4 1.4.1. For k, ai and bi nonzero, kai = bi if and only if a1 · b2 = a1 · ka2 = ka1 · a2 = a2 · b1 if and only if a1 /a2 = ka1 /ka2 = b1 /b2 if and only if a1 /b1 = 1/k = a2 /b2 . 1.4.2. Let △ABG be a right triangle with AB = 13, AG = 12 and BG = 5. Let C be on AG −→ with AC = 8.25 and F be on AG with AF = 15.75. Pick A = D and B = E. 1.4.3. For △ABC let AB be the longest side. Construct rectangle AB E F with base AB and height AF equal to the height C D of △ABC. Then AB E F contains △ABC. Construct D on AB so that C D ⊥ AB. Then △ADC ∼ = △C F A and △B DC ∼ = △C E B. Hence the area of △ABC is half of the area of AB E F. 1.4.4. (a) 2 : 1 because △AF B ∼ △C F E and AB = 2 · C E. (b) 4 : 1 by Theorem 1.4.5. (c) 3 : 1. Let G be the midpoint of AB. Then △AD E has half the area of AD E G and so 14 of area of ABC D. Similarly for △BC E. Hence AD E B has 34 of the area of AD E B. (d) 3 : 4 as in part (c). 1.4.6. ratio a : b. Note that the large circle has radius a + b and the regions are made from semicircles. 1.4.8. First construct △D P Q congruent to △ABC based on Exercise 1.2.2. Draw the parallel −−→ to P Q through E. Let F be the intersection of this parallel with D Q. Then △D E F ∼ △ABC. 1.4.10. (d) is similar since it halves both the x and y-values and in general, y = sin(x) is similar to y = k1 sin(kx). 1.4.12. Use Theorem 1.4.4 to show △P SU ∼ △P Q R ∼ △S QT ∼ △U T R. Use Theorem 1.4.3 to show △T U S ∼ △P Q R. 1.4.13. (a) Proof. By Exercise 1.2.17, the measure of an inscribed angle of a given arc is half the measure of the corresponding central angle. Thus inscribed angles of the same arc of a circle are congruent. So ∠S P R ∼ = ∠S Q R and ∠P S Q ∼ = ∠P R Q. By Theorem 1.4.2 △P ST ∼ △Q RT . So P T /ST = QT /RT . Cross multiply to get P T · RT = QT · ST . ! 1.4.14. Construct E F ∥ AB with F on BC. Because B F E D is a parallelogram, Exercise 1.3.8 implies B F = D E. As with the argument in the proof, B F, FC, and BC are in the same proportions with AE, EC, and AC. 1.4.19. Assume two similar polygons P and P ′ with a ratio of k can be divided into matched similar triangles T1 ∼ T1′ , T2 ∼ T2′ , . . . , Tn ∼ Tn′ . These triangles also have the ratio k. Then $n $n $n Area(Ti ) = i=1 k 2 Area(Ti′ ) = k 2 i=1 Area(Ti ) = k 2 Area(P ′ ). the Area(P) = i=1

Answers to Selected Exercises

519

1.4.20. (a) Note that f (kz/k) = f (z), so f (x/k) stretches out f (x) by a factor of k in the x-direction. (b) Use integration by substitution where u = xk . (c) A similar volume has k 3 times the original volume. 1.4.23. (a) This definition may seem stronger because it considers all diagonals as well as sides and so all angles formed by diagonals and sides as well as angles bounded by adjacent sides. (b) Let A1 A2 A3 A4 and A′1 A′2 A′3 A′4 be two √ sides of length s and ks, respec√ squares with tively. Their diagonals have length s 2 and ks 2, respectively. All corresponding angles are 90◦ or 45◦ . (c) A square and a rhombus. (d) A rectangle and a square. (e) All sides and angles of a regular polygon and so their corresponding diagonals are congruent to each other.

Section 1.5 1.5.2. Replace B by πr 2 . 1.5.3. Use any interior point of the polyhedron as the vertex of pyramids whose bases are the faces of the polyhedron. 1.5.4. Tetrahedron: V = 4, E = 6, F = 4; cube: V = 8, E = 12, F = 6; octahedron: V = 6, E = 12, F = 8; dodecahedron: V = 20, E = 30, F = 12; icosahedron: V = 12, E = 30, F = 20. 1.5.6. 360 − 5(68.86) = 15.7 and 360 − 2(60) − 4(55.57) = 17.72. 1.5.7. Use proportions. 1.5.8. Possible angles are 60, 90, and 120. 1.5.9. (α + β + γ − 180)(π /180)r 2 . 1.5.10. Similar triangles have the same angle sum and so the same area. 1.5.11. (a) The volume is lwh. Each pyramid has volume 13 lwh. (b) The diagonal of the base is perpendicular to the edge of the height (c) Not as easily. Let △ABC and △A′ B ′ C ′ be the two triangles and use pyramids ABC A′ , A′ B ′ C ′ C and A′ B B ′ C. 1.5.13. (a) V = 2n, E = 3n, F = n + 2. √ √ 2/2, 3/2. 1.5.15. (a) 1/2, √ (b) 2 3/(3π ) ≈ 37%. (c) π /6 ≈ 52% (d) A tetrahedron, volume 1/3. √ √ 1.5.17. (c) If an edge has length l, the volume is 2l 3 2 − 8 6√1 2 l 3 = 43 l 3 2. ; √ (d) l, l/ 2, l 23 .

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Answers to Selected Exercises

1.5.19. (a) Two pyramids glued together. V = n + 2, E = 3n, F = 2n. (c) Tetrahedron is self-dual, the cube and octahedron are duals of each other, the dodecahedron and the icosahedron are duals of each other. (d) Let the subscript P indicate the original polyhedron and D its dual. Then V P = FD , E P = E D , F P = V D . √ 1.5.23. (b) Use Exercise 1.5.11.(b) for the points (1, b, 0) and (0, 1, b) to find b = (1 + 5)/2. 1.5.25. Find parallel diagonals of adjacent pentagons and their opposites. 1.5.27. (a) Use symmetry: Each angle is 90◦ and each side is the same length. (b) 195◦ = 13π /12 radians. (c) The area of △AC F is 1/48 of the sphere’s area, πr 2 /12. 1.5.29. (a) πr/2. (b) πr/2. (c) The ratio of the lengths of ( AN and ( A P equals the ratio of the measures of angles ∠AB N and ∠AB P.

1.5.33. (a) 42 = 12 + 30: the original 12 vertices plus one for each original edge. 720/42 ≈ 17.14◦ . Use Exercise 1.5.6 to find (12 · 15.7 + 30 · 17.72)/42 = 17.14. (b) 92 = 12 + 20 + 2 · 30: the original 12 vertices plus one for each original face and two for each original edge.

Section 2.1 ← → 2.1.1. P Q is the set of points R so that R is on the line P Q and additionally R is P, R is Q or R is between P and Q. 2.1.2. Undefined terms: 1, 2, 4, 5, 7. Assuming the terms used are defined, reasonable definitions are 10, 11, 12, 15, 16, 17, 18, 19, 20, 21, 22, 23. (We don’t define some of these terms this way.) 2.1.3. (a) For a contradiction, suppose B( A)C. Then by axiom (i), C( A)B, which we showed was false. Hence not B(A)C. (b) Suppose for a contradiction that A = B. Then from A(B)C we’d have B( A)C and from each of these and axiom (i) C(B) A and C( A)B. These contradict axiom (ii), showing A ̸= B. 2.1.5. (a) Let S and T be convex sets and P, Q and R points with P and R in S ∩ T . So P, R ∈ S. Because S is convex, Q ∈ S. Similarly, Q ∈ T and so q ∈ S ∩ T . 2.1.7. (a) Use axiom (iii) with s = t. (b) Consider cases: If a → b and a → c, does b → c? 2.1.9. (a) Suppose for a contradiction that x is both an apple and an orange. By axiom (iv) x likes some y. Now use axioms (ii), (iii) and (v). (b) 4 (c) 8

Answers to Selected Exercises

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Section 2.2 (2.2.1.) 2.2.2. (a) See Exercise 2.1.1 (b) See the definition in Section 2.1. −→ ← → (c) Ray P Q contains all points R on line P Q so that in postulates 3 and 4 R corresponds with a non-negative number, provided P corresponds with 0 and Q corresponds with 1. (e) Use the idea of postulate 9. 2.2.3. (a) Use Euclid’s definition 10 for a right angle and show its measure must be 90◦ . 2.2.5. (a) Use Exercise 2.2.3 part (a) and postulate 12. 2.2.6. By postulate 5 there are at least 2 points in the plane, say, O and X . By Exercise 2.2.5 ← → there is exactly one line k in this plane perpendicular to O X through O. By postulate 3 there is a point Y on k so that the distance from O to Y equals the distance from O to X . By postulate 4 we can choose this common distance to be 1. For any point A ← → of this plane, construct the parallel m to k and call its intersection with O X the point A x . Define A y similarly on k. Give A the coordinates (a, b), where a is the number corresponding to A x and b is the number corresponding to A y using postulate 4. 2.2.9. (a) The interior of △ABC is the intersection of the interiors of the angles ∠ABC, ∠BC A and ∠C AB. (b) Yes. Use Exercise 2.1.5. 2.2.10. (a) 3 points, 3 lines. (b) Two points lying on two distinct lines contradict I-2. 2.2.12. (a) I-1, I-2, I-3 (first part), II-1, II-2, II-3, III-1, III-2, III-3, V-1, V-2. (b) On every line there exist two points, say P0 and P1 by I-3. By II-2 there exists a point P2 with P1 between P0 and P2 . By II-1 these are all distinct. Similarly there are P−1 and P0.5 different from P0 and P1 with P0 between P−1 and P1 and with P0.5 between P0 and P1 . By II-3 P2 , P−1 and P0.5 must also differ from each other. (c) In III-1 let A = P0 , B = P1 and A′ = P1 . Then B ′ = P2 for the ray from P1 not including P0 . Repeat with A′ = P2 . 2.2.14. (e) Note that we have ∠AB D ∼ = ∠A′ B ′ C ′ . Use SAS.

Section 2.3 2.3.1. Yes. −→ −→ −→ 2.3.2. (a) AB = {B, C, E} = AC. AG = {D F G}. Yes. (c) I-1, I-2, first sentence of I-3, II-1, II-2, II-3. 2.3.3. (b) III-1, III-2, III-3, III-4. (c) Intersections: a point, two points, a line segment, two line segments, an entire circle.

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Answers to Selected Exercises

2.3.5. (a) 5b, 8, 10, 21, 22. (b) If we suitably alter the definition of angle to avoid rays, only 1 and 3 (ii) fail. 2.3.6. (a) Euclidean line. (c) Use part (b) and: For (i) 3 points A, B, C with just the relations A(B)C, B(C) A and C(B) A. For (ii) 3 points with all possible relations X (Y )Z . For (iii) the empty set with no relations. For (iv) two points with no relations. 2.3.8. (a) a → b, a → c, b → c, b → d, c → d, c → e, d → e, d → a, e → a, e → b. (b) First model: ai → ai+1 and ai → ai+2 , where addition is (mod 6). Second model: ai → ai+1 and ai → ai−2 , where addition and subtraction are (mod 6). In the first model, a1 and a2 both beat a3 . However, in the second model no two teams beat the same team. (e) With 6 teams, ai → ai+1 , where addition is (mod 6) and a1 → a4 , a2 → a4 , a3 → a6 , a4 → a6 , a5 → a2 , a6 → a2 . 2.3.10. (a) An infinite chessboard. (b) A cube. 2.3.12. (a) On a set with at least 4 points, interpret a circle as any subset with exactly 3 points. (d) Without loss of generality the points are A, B, C, and D. For axioms (i) and (iii) to hold there must be one circle with exactly three points on it. Without loss of generality let {A, B, C} be a circle. Now D must be on a circle with any 2 of the others, but we can’t have { A, B, C, D} as a circle because of axiom (ii). Hence each of the sets {A, B, D}, { A, C, D}, and {B, C, D} must be circles. 2.3.14. (c) Use contradiction and axiom (iii). (e) Let Q be any point. By part (d) there is a line j not on Q. Use axiom (iv). (g) For at least two parallels to any line h: By parts (d), h has at least one point P not on it. Of the three points on h at least one, say, Q is on a line with P. Let R be the third point on the line with P and Q. Then h has parallels through P and R. For exactly two parallels, suppose there were another and find a contradiction. 2.3.18. (a) II-2, II-4 (and II-4′ ), III-1, IV-1, V-1, V-2. (b) III-1, III-4, V-1, V-2. (c) V-2.

Section 3.1 3.1.1. slope: −a/b, intercept: −c/b 3.1.2. Perpendicular lines have the form bx − ay plus any constant. 3.1.3. (a) A quarter circle. C D2 (c) x 2 + y 2 = 12 .

3.1.5. (b) If !the vertices are C = (0, 0), A = (d, 0), and B = (e, f ), then cosC = e/ e2 + f 2 .

Answers to Selected Exercises

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3.1.7. (a) {(x, y) : (x − a)2 + (y − b)2 = r 2 }. (c) {(x, y) : (x − a)2 + (y − b)2 < r 2 }. ! 3.1.9. Distance formula: k 2 (x1 − x2 )2 + j 2 (y1 − y2 )2 , Circle: ((x − a)/k)2 + ((y − b)/j)2 = r 2 . 3.1.11. (c) u = ±b. Calculus gives a root at 0 as well. 3.1.13. (a) y = x 2 , y = x − x 2 , y = x 4 , y = e x , y = ln x (b) Convex functions: y = x 2 , y = x 4 , y = e x . Convex functions are “concave up” as well as enclose convex regions. Concave functions: y = x − x 2 , y = lnx. (c) Consider the second derivative. 3.1.14. (a) The parallelogram law holds for the addition of vectors. (b) They are reflections in the real axis. (c) The distance a point is from the origin.

Section 3.2 3.2.4. (a) The line that is the perpendicular bisector of the 2 points. (b) A great circle on the sphere. (c) The plane that is the perpendicular bisector of the 2 points. Parts (a) and (b) are the intersections of this plane with their respective domains. (d) A point, two points, and a line perpendicular to the plane determined by the three points. 3.2.6. (c) A hyperbola. 3.2.8. (a) (b) (c) (d) (e)

Parabola. Ellipse. Point (2, 3), degenerate ellipse. Hyperbola with asymptotes y = x and y = 2x. Ellipse.

3.2.10. For k = 1, the locus is x = 0. For k ̸= 1, it is x 2 + y 2 − (2k/(1 − k))x = 0. 3.2.11. (c) 2x y − 1 = 0, ac − b2 = −1 < 0, determinant is 1. √ √ (d) Consider, for example, the points (1, 0.5) and (1/ 2, 1/ 2). (e) y = 0 and x = 0. Use limits. 3.2.12. (b) y − x = 0 and x = 0, x 2 − x y + 1 = 0, ac − b2 = −1/4, determinant is −1/4.

Section 3.3 3.3.1. t = 0, ±1, vertical tangent. 3.3.2. Mirror reflection over the y-axis. It is rotated by an angle of c. 3.3.3. The y-axis becomes the tangent (θ = ±π /2) so the curve looks more like two adjacent circles.

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Answers to Selected Exercises

3.3.4. The algebra needed to find the intersection involves only the operations +, − , ×, and ÷, so rational values remain rational. 3.3.5. (a) They give the same graph, but a point on the second curve goes twice as quickly. (d) They give the same graph, but a point traverses the second curve k times as quickly. 3.3.6. (b) As t → −1− , the point goes to (∞, −∞). As t → −1+ , the point goes to (−∞, ∞). (c) As t → ± ∞, the values go to 0 and the point approaches the origin. 3.3.8. (a) (− sin t, cos t) and (−2 sin t, 2 cos t). The second is twice as long as the first because a point on it is moving twice as quickly. (b) (− sin t, 2 cos 2t). Both t = π /2 and t = 3π /2 give the same point (0, 0), where the curve crosses itself. (c) (cos √ t − t sin t, sin t + t√cos t). Directions: (1, 2π ) and (1, 4π ), lengths: 1 + 4π 2 ≈ 6.36 and 1 + 16π 2 ≈ 12.61. (d) Same point ( f (c), g(c)), length of second is k times as large as for the first. 3.3.10. (a) It is a sort of spiral on the surface of a cone. (b) It is a sort of spiral on the surface of a sphere. 3.3.12. (d) x 4 + 2x 2 y 2 + y 2 − x 2 + y 2 = 0. (e) x 4 + 2x 2 y 2 + y 2 − 2x y = 0. 3.3.16. (a) (b) (c) (d) (e)

Center of square Center of square, (a, a, 0.5 − a, 0.5 − a) (a, a, −a, 1 − a). c = d. (−4, −3).

3.3.19. (a) (x, 90◦ ), (x, −90◦ ). (b) Circles of latitude, great circles of longitude. (c) Spirals from the south pole to the north pole. A given spiral intersects each circle of latitude at the same angle. (d) I-1, I-3, I-4. 3.3.21. (a) They have √ the same x- or y-coordinate. Taxicab distance is greater by at most a factor of 2. (b) Squares at a 45◦ angle to the axes.

Section 3.4 3.4.3. a = p/2 − q + r/2, b = −3 p/2 + 2q − r/2, c = p 3.4.4. (a) 3x 3 − 5x 2 + x + 1. (b) For example, x(t) = 3t − 6t 2 + 4t 3 and y(t) = 1 + 3t − 9t 2 + 5t 3 . 3.4.5. (a) x(t) = 3t − 3t 2 , y(t) = 3t − 6t 2 + 4t 3 . (b) x ′ (t) = 3 − 6t, y ′ (t) = 3 − 12t + 12t 2 . x ′ (0.5) = 0 = y ′ (0.5). (c) The point has velocity 0 at t = 0.5, so it can switch directions smoothly then.

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3.4.7. (b) (2, 2), y = 3 (c) For example, use P0 = (4, 0), P1 = (5, 0), P2 = (4, 1), and P3 = (5, 3). (d) For example, use P0 = (5, 3), P1 = (6, 5), P2 = (−1, −2), and P3 (0, 0). 3.4.9. (a) T3 (x) = x − x 3 /3! (b) T5 (x) = x − x 3 /3! + x 5 /5!, etc. (c) sin′ (−π ) = −1, sin′ (0) = 1, sin′ (π ) = −1. Left half: P0 = (−π , 0), P1 = (−π + a, −a), P2 = (−b, −b), and P3 = (0, 0), where a > 0 and b > 0. Right half: P0 = (0, 0), P1 = (a, a), P2 = (π − b, b), and P3 = (π, 0), where a > 0 and b > 0. 3.4.10. (a) (n 3 + 6n 2 + 6n + 1)/3 (b) 36 (c) 36w

Section 3.5 − → → 3.5.1. At α = 1, the point is − s and at α = 0 the point is t . Since it is a linear combination − → → of 1 variable, we get a line. At α = 0 = β, we are at t , at α = 0, β = 1 we are at − s − → and at α = 1, β = 0, we are at r . A linear combination of two variables gives a plane. 3.5.3. They differ in just one coordinate from (1, 1, 1, 1). They differ in 2, 3 or all 4 coordinates. √ 3.5.4. 2 for edges and 2 for diagonals 3.5.5. (a) α(1, 2, 4) + β(2, 0, 0). (b) 2y − z = 0. 3.5.7. (a) (b) (c) (d)

α(1, −2, 1). They intersect at (3, 3, 3). skew. → − → w is a scalar multiple of − u.

3.5.9. (a) (b) (c) (d) (e)

Draw m i parallel to ki with m i going through Pi . 7. (3, 5, 3). Neither. both projections are parallel. The projections intersect at points with the same common coordinate. The projections intersect at points with different common coordinates.

3.5.13. (a) (b) (c) (d)

unit sphere ellipsoid paraboloid hyperboloid of one sheet. √ √ √ √ 3.5.15. (c) (1/ 5, 0, −2/ 5) and (−1/ 5, 0, 2/ 5). (d) x − 2z = 0. 3.5.17. C = arccos(1/3) ≈ 1.23096 radians or 70.5288◦ .

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3.5.19. (a) Let e/v represent edges per vertex, and so on. Dimension

1

2

3

4

e/v f/v c/v f/e c/e c/f v e f c

1 0 0 0 0 0 2 1 0 0

2 1 0 1 0 0 4 4 1 0

3 3 1 2 1 1 8 12 6 1

4 6 4 3 3 2 16 32 24 8

(b) For dimension n, e/v = n, f /v = n(n − 1)/2, f /e = n − 1, c/e = (n − 1)(n − 2)/2, c/ f = n − 2, v = 2n , e = n2n−1 . 3.5.21.

(a)

Dimension e/v f/v c/v f/e c/e c/f v e f c

1 1 0 0 0 0 0 2 1 0 0

2 2 1 0 1 0 0 3 3 1 0

3 3 3 1 2 1 1 4 6 4 1

4 4 6 4 4 4 2 5 10 10 5

(b) For dimension n, if n is big enough, e/v = n, f /v = n(n − 1)/2, f /e = n − 1, c/e = (n − 1)(n − 2)/2, c/ f = n − 2, v = n + 1, e = (n + 1)n/2. 3.5.23. (a) tetrahedron {3, 3}, cube {4, 3}, octahedron {3, 4}, dodecahedron {5, 3}, icosahedron {3, 5}.

Section 4.1 4.1.1. Area is proportional to how far the angle sum falls short of π in hyperbolic geometry and how far it exceeds π in spherical geometry. Replace π by 180◦ . 4.1.3. Infinitely many. See Theorem 4.2.1. 4.1.4. (a) Intersections at (0.8, ±0.6). (d) angles of 69.4◦ , 18.4◦ , and 18.4◦ . 4.1.5. II-4, III-4, and IV-1.

Answers to Selected Exercises

4.1.6. (a) (d) (f) (g)

527

! 2. y = 5 − (x − 3)! For example, y = 100 + (c + 4)2 − (x − c)2 , for −11.5 ≤ c ≤ 3.5. (5, 0). They are tangent.

4.1.8. (a) intersections: (±3.2, 2.4). (b) (±5, 3), ≈ 1.03 radians ≈ 59◦ .

Section 4.2 4.2.3. (a) All can occur. (b) There are 4 common sensed parallels. 4.2.5. Proof. By definition, the angle of parallelism is the smaller of the two angles a sensed parallel makes with the perpendicular and so is less than or equal to 90◦ . However, if the angle of parallelism were 90◦ , the two sensed parallels would be the same line, contradicting the characteristic axiom. By I-28 two lines with a common perpendicular can’t intersect and so are ultraparallel or sensed parallel. Since the angle of parallelism is acute, lines with a common perpendicular are ultraparallel. ! 4.2.7. Euclid’s parallel lines are ultraparallel in hyperbolic geometry. 4.2.9. ∠C B" is bigger than ∠C A". 4.2.10. It is less than 360◦ .

Section 4.3 4.3.1. and 4.3.2. rectangle. 4.3.3. The angle sum of the “summit angles” of a rectangle is 180◦ . 4.3.4. Use symmetry. −→ 4.3.7. (a) Assume that X is on DC with D X ∼ = B A. Then AB D X is a Saccheri quadrilateral. ← → ← → Theorem 4.3.1 and Corollary 4.2.3 show AX is ultraparallel to B D. Because AC ← → is a sensed parallel it must be under AX . (b) Part (a) implies that C is between D and X . So the perpendiculars from one sensed ← → ← → parallel AC to another B D get shorter as they “approach” the omega point. 4.3.11. The angle sum of an n-gon is less than 180(n − 2)◦ . Proof. For n = 3 use Theorem 3.3.3. For the induction step suppose for n = k the angle sum is less than 180(k − 2)◦ and we have a convex polygon with n = k + 1 sides. Divide this polygon into a triangle and a k-gon and apply the induction hypothesis and Theorem 3.3.3. This proof holds for nonconvex polygons, although one must show one can always divide the k + 1-gon into a triangle and an k-gon.

528

Answers to Selected Exercises

Section 4.4 4.4.1. Consider three congruent triangles △ABC, △B AD and △E F G, where △E F G is disjoint from the other two and these other two overlap only on the side AB. By postulate 19 R = △ABC ∪ △E F G has the same area as the quadrilateral AC B D. However, R has one more side E F then AC B D has since we don’t include AB twice in the quadrilateral. So the side E F must have area 0. Also, E F is made of the union of the point E and the rest of the segment. Since the union has area 0, so does the point E. 4.4.5. The maximum defect a triangle can have is 180◦ , so K = k × 180◦ in Theorem 4.1.1. 4.4.6. The area of △AB1 Bi is finite even as i → ∞, so the area of △ABi Bi+1 approaches 0 as i→ ∞. 4.4.8. K n = (n − 2)K .

Section 4.5 4.5.1. See Exercise 4.5.6. (a). 4.5.4. They are congruent right angles. 4.5.5. (a) From the proof of Theorem 4.5.1 common perpendiculars have the same length. Apply SSS. (b) Use Theorem 4.5.1 and SAS. 4.5.6. (a) Consider two doubly right triangles with different included sides. 4.5.7. In spherical geometry, the corresponding theorem says that all lines perpendicular to a given line intersect in two antipodal points. Otherwise, the same argument applies. 4.5.9. Mimic the proof of Theorem 3.5.2. 4.5.10. (a) Half of a sphere, 2πr 2 .

Section 5.1 5.1.1. Circles with centers on y = 0, the line of reflection. 5.1.2. Solve x = y + 2 and y = 2 − x to find the only solution. All points on these circles are the same distance from the fixed point. No line is stable. ρ does not switch orientation. 5.1.3. Mirror reflection over y = 2 − x. No, it is a reflection over y = x − 2. 5.1.5. (a) Both formulas give functions. To show one-to-one and onto, solve y = α(x) and y = β(x) for x to see that the choice of x in each is unique. (b) (c) (d) (e) (g)

(2x − 1)3 , 2x 3 − 1. For α, 0, ±1; for√β, 0.5. For α, 1; for β, 3 0.5. √ 3 x, 0.5x + 0.5. √ √ 3 0.5x + 0.5, 0.5 3 x + 0.5.

Answers to Selected Exercises

529

5.1.7. (a) Rotation of 180◦ around (2, 3). (b) Solve (4 − x, 6 − y) = (a, b) to find the unique solution. (c) inverse is θ , fixed point is (2, 3) and stable lines are y = mx + 3 − 2m and x = 2. 5.1.9. (a) (b) (c) (d)

Rotation of 180◦ around (1, 2). Mirror reflection over y = x + 1. Dilation by a factor of 2 about (0, 0). Dilation by a factor of 0.5 about (2, −2).

Section 5.2 5.2.1. Isometries preserve distance and a circle is the set of points a fixed distance from its center. 5.2.2. Each point of k is fixed, so the entire line is fixed and hence stable. If m ⊥ k, each point on m goes to another point on m, so m is stable. 5.2.3. The identity is a translation of length 0 and a rotation of an angle of 0. 5.2.4. If the translation part of the glide reflection is the identity, we have a mirror reflection. Points on the line of reflection slide along the line and points off of this line switch sides. So no point is fixed by a glide reflection that isn’t a mirror reflection. The sliding aspect of a glide reflection means only lines parallel to the line of reflection have a chance of being stable. But other lines parallel to this line switch sides from the reflection part. 5.2.5. (a) (b) (c) (d)

(0, 1), (−3, 4), (−y − 1, x + 2), rotation of 90◦ around (−1.5, 0.5). (−3, 1), (0, 4), (x − 2, y + 2), translation of 2 to the left and 2 up. (−2, 0), (1, 3), (y − 1, x + 1), mirror reflection over y = x + 1. (−2, 2), (1, −1), (y − 1, −x + 1), rotation of 270◦ around (0, 1).

5.2.7. (a) (b) (c) (d)

(0, 2), (0, 3), (1, 4). (4, 2), (5, 2), (6, 3). τ ◦µ=µ◦τ composition is the translation τ ◦ τ taking (x, y) to (x + 4, y + 4).

5.2.9. (a) µx (x, y) = (x, −y), µ y (x, y) = (−x, y), µ3 (x, y) = (−y + 1, −x + 1). (b) µ y ◦ µx (x, y) = (−x, −y), rotation of 180◦ around (0, 0). (c) µ3 ◦ µx (x, y) = (y + 1, −x + 1), a rotation of 270◦ around (1, 0). µx ◦ µ3 (x, y) = (−y + 1, x − 1), a rotation of 90◦ around (1, 0). (d) µ3 ◦ µ y ◦ µx (x, y) = (y + 1, x + 1), glide reflection along y = x.

5.2.11. (a) µm fixes points on m, namely A and B. △AC D ∼ = △µk ◦ µm ( A)C D so Aµk ◦ µm ( A) = 2AC = 2d. Similarly for B. (b) Use SSS. Similar to (a). (c) Compose them. 5.2.13. Note that d(P, Q) + d(Q, R) = d(P, R) just when Q is between P and R and that isometries preserve distance.

530

Answers to Selected Exercises

5.2.15. τ 1 ◦ τ 2 = τ 2 ◦ τ 1 . Let P and Q be any points. Then Pτ 2 (P)∥Qτ 2 (Q), τ 2 (P)τ 1 ◦ τ 2 (P)∥τ 2 (Q)τ 1 ◦ τ 2 (Q), and ∠τ 1 ◦ τ 2 (P)τ 2 (P)P ∼ = ∠τ 1 ◦ τ 2 (Q)τ 2 (Q)Q. Because τ 1 and τ 2 are translations, corresponding sides are congruent. So by SAS Pτ 1 ◦ τ 2 (P) = Qτ 1 ◦ τ 2 (Q). Thus P Qτ 1 ◦ τ 2 (Q)τ 1 ◦ τ 2 (P) is a parallelogram and τ 1 ◦ τ 2 is a translation. 5.2.19. (a) V consists of translations, 180◦ rotations, vertical and horizontal mirror and glide reflections. Proof. Let l be a vertical line. ι(l) = l is vertical, so ι ∈ V. If α, β ∈ V, β(l) and so α(β(l)) are vertical, so α ◦ β ∈ V. Suppose α ∈ V and k the vertical line so that α(k) = l. Then α −1 (l) = k, so α −1 ∈ V.

Section 5.3 3 a 5.3.3. The matrix c 5.3.8. (a)

(b)

(c)

(d)

(e)

5.3.9. (a)

(b)

⎡

⎡ a b becomes ⎣ c d 0 4

⎤ 1 0 b ⎣0 1 0⎦. 0 0 1 ⎡ ⎤ 1 0 0 ⎣0 −1 0⎦. 0 0 1 ⎡ ⎤ 1 0 b ⎣0 −1 0⎦. 0 0 1 ⎡ ⎤ 0 1 0 ⎣1 0 0⎦. 0 0 1 ⎡ ⎤ 0 1 b ⎣1 0 b⎦. 0 0 1 ⎡ cos(30) − sin(30) ⎣ sin(30) cos(30) 0 0 ⎡ cos(30) − sin(30) ⎣ sin(30) cos(30) 0 0

b d 0

⎤ 0 0⎦.

1

⎤ 0 0⎦.

1 √ ⎤ 3.5 −√ 3 2 − 3 3/2⎦. 1

5.3.10. A is a rotation, B is a translation, C is a mirror reflection, D is a glide reflection, E is a rotation. 5.3.12. (b) c = 2 f . ⎡ ⎤ −1 0 0 5.3.16. (a) ⎣ 0 −1 0⎦ . 0 0 1

531

Answers to Selected Exercises

(b) (c) (d) (e)

⎡

⎤ −1 0 2u ⎣ 0 −1 2v ⎦ . 0 0 1 A direct isometry, a rotation of 180◦ . Lines through (u, v, 1). Translations twice the distance between the centers. You get the inverse translation.

5.3.20. a = ±1, b = 0, c free, d = 0, e = ±1, f = 0.

Section 5.4 5.4.2. The inverse of a contraction mapping is not a contraction mapping. 5.4.3. (a) (−2, −4, 1), [m, −1, 2m − 4], and [1, 0, 2]. (b) (2, 2, 1), no stable lines. ⎡ ⎤ ⎡ ⎤ 0 −4 −11 0 4 13 5.4.4. direct: ⎣4 0 −7 ⎦ and indirect: ⎣4 0 −7⎦. 0

0

1

0 0

1

5.4.5. (a) (b) (c) (d)

(1, 0, 1), (0, 2, 1), (−4, 0, 1), and (0, −8, 1). Spiral. (1, 1, 1), (−2, 2, 1), and (−4, −4, 1). Yes. √ M is a rotation by θ and a scaling by r , S is a rotation by θ/2 and a scaling by r . √ For C use a rotation by θ /3 and a scaling by 3 r and for N use a rotation by θ /n √ and a scaling by n r . ⎡ ⎤ ⎡ ⎤ 2 2 −2 2 −2 −2 5.4.7. (a) ⎣−2 2 4 ⎦ and ⎣−2 −2 4 ⎦. 0 0 1 0 0 1 ⎡ ⎤ ⎡ ⎤ −3 −4 10 −3 4 2 (b) ⎣−4 −3 11⎦ and ⎣−4 3 5 ⎦. 0 0 1 0 0 1

5.4.13. Follow Exercise 5.3.17 with the following substitutions.

(a) a = r cos θ, e = −r cos θ, and b = d = r sin θ . (c) and (d) In (a) set a = e = r cos θ , b = −r sin θ, and d = r sin θ . 5.4.15. (a) fixed points: (x, x, 1), stable line: [1, −1, 0]. (b) (0.5, 0.5, 1), [−1, −1, 1]. 5.4.20 (c) This IFS fractal is the part of Figure 5.32 on the x-axis. 5.4.21. The 4 corners of the square, (0, 0, 1), (1, 0, 1), (0, 1, 1), and (1, 1, 1), must go to points ⎡ ⎤ a b c in that square. For ⎣d e f ⎦, the following must be between 0 and 1: c, a + c, b + c, 0 0 1 a + b + c, f , d + f , e + f , and d + e + f .

532

Answers to Selected Exercises

Section 5.5 5.5.3. No essential change for translation from Section 5.2. For glide reflection, replace “line k” with “plane P.” 5.5.5. No essential change for translation from Section 5.2. An (n + 1) × (n + 1) matrix with 1s on the main diagonal, any numbers in the far right column (except for a 1 in the lower right corner) and 0s elsewhere. 5.5.7. (a) cos(180◦ ) = −1 gives the upper two diagonal entries. Note that (0, 0, z, 1) is mapped to ⎡ itself. ⎤ ⎡ ⎤ 1 0 0 0 −1 0 0 0 ⎢0 −1 0 0⎥ ⎢ 0 1 0 0⎥ ◦ ⎥ ⎢ ⎥ (b) ⎢ ⎣0 0 −1 0⎦, ⎣ 0 0 −1 0⎦, rotation of 180 around the z-axis. 0 0 0 1 0 0 0 1 The other of these two matrices. (d) Rotations of 120◦ around opposite vertices of the cube. ⎡ ⎤ 1 0 0 0 ⎢0 0 −1 0⎥ ◦ ⎥ (e) For example, ⎢ ⎣0 1 0 0⎦ is a 90 rotation around the x-axis and 0 0 0 1 ⎡ ⎤ 0 1 0 0 ⎢0 0 1 0⎥ ◦ ⎢ ⎥ ⎣1 0 0 0⎦ is a 120 rotation around the axis through (1, 1, 1, 1) and (f)

5.5.9. (a)

(b) (c)

0 0 0 1 (−1, −1, −1, 1). ⎡ ⎤ 1 0 0 0 ⎢0 cos θ − sin θ 0⎥ ⎢ ⎥. ⎣0 sin θ cos θ 0⎦ 0 0 0 1 ⎡ ⎤ cos θ 0 − sin θ 0 ⎢ 0 1 0 k⎥ ⎢ ⎥ ⎣ sin θ 0 cos θ 0⎦ (or with the − on the other sin θ, depending on the 0 0 0 1 orientation of the axes). Yes. These rotation and translation matrices commute. a translation of 6 in the y-direction.

5.5.11. (a) (x, y, −x + 2, 1) which are points on the plane x + z − 2 = 0 or [1, 0, 1, −2], the fixed plane and the family [1, b, −1, d], mirror reflection. (c) (x, x, x, 1), which are points on a line, family of parallel planes [1, 1, 1, d], rotation of 120◦ . 5.5.13. (a) A 4 × 4 affine matrix where the upper left 3 × 3 submatrix is r times an orthogonal matrix. (b) The dilation affects only the upper left 3 × 3 submatrix, multiplying each of its entries by r .

Answers to Selected Exercises

533

(c) r 3 , the ratio of the volumes of the image of an object and the origin. (d) (0, 0, −2, 1), [1, −1, 1, 2].

Section 5.6 5.6.1. Circle x 2 + y 2 = r 2 inverts to x 2 + y 2 = 1/r 2 . 5.6.2. d(O, P) · d(O, ν c (P)) = d(O, ν c (P)) · d(O, P). 5.6.3. Case 1. r ̸= 0 gives q = ps/r and f (z) = p. Case 2. r = 0. Either p = 0 or s = 0. For p = 0, f (z) = q/s. For s = 0, f is undefined. As z → s/r , the denominator of f (z) goes to 0. As z → ∞ the limit makes the constant terms irrelevant. To convert the formulas of Theorem 5.6.4, first ignore the complex conjugate to rewrite r 2 /(z − w) + w as (r 2 + wz − w2 )/(z − w) = (wz + r 2 − w2 )/(z − w), which is a M¨obius transformation. Now apply complex conjugates. 5.6.4. ν( A) = (0, −2), ν(B) = (1, 1), ν(C) = C, ν(D) = (2, 4). 5.6.5. (a) ν D (1, 0) = (1, 0), ν C (ν D (1, 0)) = (4, 0), ν C (1, 0) = (4, 0) and ν D (ν C (1, 0)) = (0.25, 0). No. (b) ν D (0, 3) = (0, 1/3), ν C (ν D (0, 3)) = (0, 12), ν C (0, 3) = (0, 4/3) and ν D (ν C (0, 3)) = (0, 3/4). √ 5.6.7. (a) Center (0, 0) and radius 2, center (−1, 0) and radius 10. (b) Center must be on the line through (4, 0) and (1, 0). All x-axis except points between x =√ 1 and x = 4. (c) r = c2 − 5c + 4, where c is the x-coordinate of the center. 5.6.9. (a) x = 1.6 and x = 2.5. 5.6.12. (a) Translation. (b) If r > 0, a dilation by a ratio of r ; if r < 0, a dilation and a rotation of 180◦ . A rotation. A composition of a dilation and a rotation. (c) A mirror reflection over the real axis. (d) Use parts (a), (b), and (c) and composition. 5.6.13. (a) (b) (c) (e)

2/(z + 2i) − 2i. 4/(z + 3i) − 3i. E has center −1.5i and radius 0.5. −0.6i.

5.6.15. (b) The equator. Mirror reflection over the equator.

Section 6.1 6.1.1. Jelly fish drift in any (horizontal) direction, so there is no natural front/back axis. Single cellular organisms suspended in water can move in all directions, so spherical symmetry may be advantageous. 6.1.2. Rotations of multiples of 60◦ and mirror reflections over 6 lines through the center.

534

Answers to Selected Exercises

6.1.3. (a) No symmetry: F, G, J, L, P, Q, R; Vertical symmetry: A, M, T, U, V, W, Y; Horizontal symmetry: B, C, D, E, K; Rotational symmetry: N, S, Z; All of the above symmetries: H, I, O, X. (b) Some examples: Vertical (written vertically): TOMATO; Horizontal: CHOICE, BEDECK; Rotation: MOW. (c) A familiar example: MADAM, I’M ADAM. In a palindrome we switch the order of the letters, but we don’t reflect the individual letters. 6.1.5. Iranian: rotations of 0◦ , 120◦ , and 240◦ and 3 mirror reflections. Byzantine: rotations of 0◦ , 90◦ , 180◦ , and 270◦ and 4 mirror reflections. Afghani: rotations of 0◦ , 60◦ , 120◦ , 180◦ , 240◦ , and 300◦ . 6.1.7. Left figure: color preserving include rotations of multiples of 120◦ and 3 mirror reflections, including vertical; color switching include rotations of 60◦ , 180◦ and 300◦ and 3 mirror reflections, including horizontal. 6.1.9. (a) Mexican: translations, 180◦ rotations, vertical mirror reflections, and horizontal glide reflections. Chinese: translations, 180◦ rotations, vertical and horizontal mirror reflections, and horizontal glide reflections. 6.1.11. (a) 25%, 25%. Use rotations. (b) 33.3%, 33.3%.

Section 6.2 6.2.1. D6 , C4 . 6.2.2. Dn . 6.2.3. (a) Gothic: C3 , Islamic: D10 , Gothic: C2 . (b) equilateral: D3 , isosceles: D1 , scalene: C1 . (c) General: C1 , parallelogram: C2 , kite and isosceles trapezoid: D1 , rectangle and rhombus: D2 , and square: D4 . 6.2.5. (a) Rotations of 0◦ , 90◦ , 180◦ , and 270◦ . Yes, C4 . (b) Rotations of 45◦ , 135◦ , 225◦ , and 315◦ . No, closure and identity fail. (c) Yes, C8 . 6.2.6. (a) Triangular prism: 12; square prism: 16. (b) 4n. (c) Tetrahedron: 24, cube: 48, octahedron: 48, dodecahedron: 120, and icosahedron: 120.

Section 6.3 6.3.2. If τ is the smallest translation to the right, all other translations are of the form τ n , for some n. If n > 0, τ n represents a translation of n to the right. If n < 0, it is a translation of n to the left. τ 0 is the identity.

Answers to Selected Exercises

6.3.3. (a) (b) (c) (d) (e) (f)

p2mm. p211. p11g. p2mg. p11m. p1m1.

6.3.5. (a) (b) (c) (d) (e) (f)

pg. p2. p3. cm. p6m. cmm.

535

6.3.9. Use a sequence of numbers to denote the sequence of n-gons around any vertex. Then the options are (a) {6, 6, 6}, {4, 4, 4, 4}, and {3, 3, 3, 3, 3, 3}. 6.3.11. p4g, pgg. 6.3.15. (a) τ n (x, y) = (x + n, y). (b) ν(x, y) = (x, −y). (d) µd (x, y) = (2d − x, y).

Section 6.4 6.4.1. A cube is a special type of a square prism, which is a special type of a rectangular box. A shape has all the symmetries of a more general shape. 8,16,48. Rotations of 180◦ around the x-, y-, and z-axes, mirror reflections over the x y-, x z-, and yz-planes, the identity and the central symmetry. 6.4.3. 4n. Dnh . All symmetries of these prisms are symmetries of a prism with a regular 2n-gon as a base. ⎡ ⎤ −1 0 0 6.4.5. ζ = ⎣ 0 −1 0 ⎦. 0 0 −1

6.4.6. (a) Identity, 9 rotations of 90◦ , 180◦ , and 270◦ around the centers of opposite faces, 8 rotations of 120◦ and 240◦ around opposite vertices, and 6 rotations of 180◦ around centers of opposite edges; mirror reflections over 6 planes through opposite edges and 3 planes between opposite faces. Switch roles of faces and vertices to match cube and octahedron symmetries. 15 rotatory reflections. 6.4.11. The groups T, W, and P each have one of the Archimedean solids. The groups W and P each have 5 of them.

Section 6.5 6.5.1. p6m. 6.5.2. 180◦ rotation through a vertical or horizontal axis going midway between adjacent atoms.

536

Answers to Selected Exercises

6.5.4. (a) D2 . (b) D1 , D1 , and C2 . 6.5.7. (a) D6 . (b) D1 , D1 , and D2 ( for cyclic arrangement ClHHClHH). (c) D2 /D1 (for HHHClClCl), D1 /C1 (for HHClHClCl), and D6 /D3 (for HClHClHCl). 6.5.9. (a) p31m. (b) p6m. 6.5.12. (a) 0.9945. (b) 0.2273.

Section 6.6 6.6.1. The matrices have a scaling factor of r = 1/3, which corresponds with Koch’s method. We need four matrices because there are four smaller copies of the original. 6.6.6. (a) (b) (c) (d)

ln 2/ ln 2 = 1. ln 3/ ln 2 ≈ 1.585. ln 2/ ln 3 ≈ 0.631. ln 5/ ln 3 ≈ 1.465.

6.6.8. (a) ln 6/ ln 2 ≈ 2.585. (b) ln 13/ ln 3 ≈ 2.335. (c) ln 26/ ln 3 ≈ 2.966. 6.6.9. (a) 2n−1 /(3n ). (b) 1/3 + 2/9 + 4/27 + · · · = 1. Length left is 0. (d) 2n−1 /(k n ), 1/k + 2/k 2 + 4/k 3 + · · · = 1/(k − 2). Length left is (k − 3)/(k − 2). 6.6.12. (a) Estimates may range from d ≈ 1.38 to d ≈ 1.51. (The number of segments of various lengths can easily vary, significantly altering the estimates of d.)

Section 7.1 7.1.2. Larger. ←→ ←→ 7.1.5. (a) Let T be the intersection of P P ′ and Q Q ′ . 7.1.8. 0.6. Different constructions should give approximately the same point. 7.1.10. (a) In the definition of a harmonic set of points, interchange point and line and replace quadrangle with quadrilateral. (b) m⊥l, m is the other bisector of j and k. 7.1.11. (a) T1 = (0, a), T2 = ( a2 , a2 ). (d) y = (1 − 2a)x/(2 − a) + a/(2 − a).

Answers to Selected Exercises

537

7.1.14. (a) V , E, and D. 7.1.15. (a) y = −w. (b) midpoint. at infinity

Section 7.2 7.2.1. Define X −n to be the point such that H (X X 0 , X n X −n ). 7.2.3. Two distinct lines have exactly one point on them. There are at least four lines with no three on the same point. Every two distinct points have at least one line on both points. 7.2.4. Dual of 7.2.1: Two distinct points have exactly one line on both points. Every point has at least four distinct lines on it. If H ( pq, r s), then H ( pq, sr ). 7.2.2: If pq//r s, then qp//r s, qp//sr , r s//qp, sr// pq, and sr//qp. If a, b, c, and d are distinct concurrent lines, then exactly one of the following holds: ab//cd, ac//bd, or ad//bc. 7.2.3: If x p and xq are determined and p ̸= q, then x p and xq are distinct. 7.2.5. Each individual perspectivity preserves these properties, so their composition does. 7.2.6. A perspectivity with respect to a line k is a mapping of the lines u i on one point to the lines v i on another point so that v i is the image of u i if and only if u i , v i and k are concurrent. 7.2.6: A perspectivity preserves harmonic sets of lines and the relation of separation. That is, if a perspectivity from k maps the concurrent lines p, q, r , and s to the concurrent lines p ′ , q ′ , r ′ , and s ′ and H ( pq, r s), then H ( p ′ q ′ , r ′ s ′ ). Similarly, if pq//r s, then p ′ q ′ //r ′ s ′ . 7.2.7: A projectivity of the lines on a point is completely determined by three lines on the original point and their images. 7.2.7. (a) X a is between X b and X c . (b) X a is the midpoint of X b and X c . (c) j and k form two pairs of vertical angles. l is in one pair and m is in the other pair. 7.2.8. (b) m = −1. (e) Yes. a ), draw parallel lines to these through 7.2.12. (a) Given the construction for H (0 1, a 2a−1 ak . By Theorem 1.4.1 the corresponding the corresponding points 0, k, ak and 2a−1 ak . triangles are similar. Hence the sixth new line must go through 2a−1 1 n+1 (b) In part (a) let k = n+1 and a = n .

7.2.15. (a) s < 0 or 1 < s. 0 < t < r < 1. Yes because t satisfies the conditions r satisfies. 7.2.17. (b) For example, (vii) becomes If pq//r s, then p, q, r , and s are distinct, concurrent lines, pq//sr and r s// pq. 7.2.18. (b) In a complete quadrilateral with lines t1 , t2 , t3 , and t4 , the diagonal lines are the three ←−−−−−−→ ←−−−−−−−−→ ←−−−−−−−→ lines on the “opposite” points: (t1 · t2 )(t3 · t4 ), (t1 · t3 ) · (t2 · t4 ), and (t1 · t4 )(t2 · t3 ).

538

Answers to Selected Exercises

Section 7.3 7.3.1. (x, 2x, 0), [0, 0, c]. 7.3.2. Two points are two Euclidean lines through the origin O, which determine a unique Euclidean plane through O, i.e. a line. Consider the points of the x-axis, the y-axis, the z-axis, and the line through O and (1, 1, 1). Any two Euclidean planes through O intersect in a Euclidean line. 7.3.3. Note that three of the four terms in R(a, b, c, d) are negative. 7.3.6. y = x 2 and y = 1/x are functions. [1, 0 − b] intersects x 2 − yz = 0 in (b, b2 , 1) and (0, 1, 0). It intersects x y − z 2 = 0 in (b, 1/b, 1) and (0, 1, 0). (0, 1, 0). [0, 1, 0]. (1, 0, 0). √ 7.3.7. (±1, 1) and (± 2, 2). 7.3.8. (a) (1, m, 0). Parallel lines “meet” at infinity. (b) [m, −1, −mp + q]. This is the line through ( p, q, 1) with slope m. 7.3.10. (b) { A, C} separate {B, E} and {D, E}; {A, D} separate {B, C} and {B, E}; {D, E} separate {B, C}. (c) H ( AC, D E). (d) P = B, Q = C, R = E, S = D, T = A. 7.3.12. (a) 1 < x < 2, x = 1.6. 7.3.14. (a) (4, 2, 1), (−12, 0, 1), (12, 12, 1). (b) For A = (4, 2, 1) and B = (−12, 0, 1), k = [1, −8, 12], C = (24/7, 24/7, 1), and D = (7.2, 2.4, 1). 7.3.15. (a) S = (10, −9, 1), T = (1, 3, 1), U = (2.5, 1, 1), line: [−4, −3, 13]. 7.3.20. (a) −x z + x y + yz = 0, y − z = 0, and x + z = 0. (1, 0, 0) and (0, 1, 0). 7.3.22. (b) In nonhomogeneous coordinates: y = 1, y = −1, x = 1, x = −1, y = x, and y = −x. √ (c) Euclidean circle with center (0, 0, 1) and radius 2. (e) Euclidean hyperbola. 7.3.23. (a) −x 2 + yz = 0. (b) (x0 , x02 , 1), y = 2x0 x − x02 , [2x0 , −1, −x02 ].

Section 7.4

3 4 3 4 3 4 a b a 0 a b 7.4.2. (a) , where a ̸= 0; , where d ̸= 0; and , where a + b = c + d. 0 d c d c d ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ a b c a 0 c a b 0 (b) ⎣0 e f ⎦, where a ̸= 0; ⎣d e f ⎦, where e ̸= 0; ⎣d e 0⎦, where i ̸= 0; 0 h i g 0 i g h i ⎡ ⎤ a b c and ⎣d e f ⎦, where a + b + c = d + e + f = g + h + i. g h i

Answers to Selected Exercises

7.4.3. (a)

7.4.4. (a)

7.4.7. (a) (b) 7.4.10. (a) (c) 7.4.12. (a) (b) (d) (f)

539

3

4 3 0 2 , goes to 34 and H (0 1, 32 34 ). 1 2 3 ⎡ ⎤ 3 −1 0 ⎣4 4 2⎦. 1 1 2 ⎡ ⎤ a 0 0 ⎣0 a 0⎦. Dilation fixing the origin. 0 0 i The corresponding sides of △ABC and △A′ B ′ C ′ are parallel. ⎡ ⎤ 2 0 0 ⎣0 −0.5 1 ⎦, 4x 2 − 2yz = 0. 0 −0.5 −1 y = 2x 2 , a parabola. ⎡ ⎤ ⎡ ⎤ 1 0 0 1 0 −w ⎣ 0 1 0⎦, ⎣0 1 0 ⎦, and (1 + 2w)x 2 + y 2 − 2x z = 0. −w 0 1 0 0 1 2 2 For w = 0.5, 2x + y − 2x = 0. For w > −0.5 and w ̸= 0, the image is a Euclidean ellipse. For w = 0.5, [2, 0, −2] and [0.5, 1, −1].

7.4.13. (a) [−4, −3, 5] and [0, −1, −1]. 7.4.14. (a) tangent at A: [1, −1, 0], at B : [−1, −1, 0], where A = (1, 1, 1), B = (−1, 1, 1), k = [0, −1, 1]. (b) S = (0, 0.5, 1), T = (0, 1, 0), Q = (0, 1, 1).

Section 7.5 7.5.1. Note that X x · X −x = (1 − x 2 )I . 7.5.2. Compare h-inner product with the regular inner product. 7.5.3. Consider the bottom row. 7.5.6. (a) S = (−0.8, 0.6, 1), T = (0.8, 0.6, 1), d H ((0, 0, 1), (0.5, 0, 1)) = |log(1/3)| ≈ 0.477, and d H ((0, 0.6, 1), (0.5, 0.6, 1)) = |log(3/13)| ≈ 0.637. 7.5.8. (d) The x-axis. Rotations of 180◦ . ⎡√ ⎤ 1 − b2 0 0 7.5.9. Yb = ⎣ 0 1 b⎦. Transposes of each other. 0 b 1

540

Answers to Selected Exercises

Section 7.6 7.6.1. h-inner product of two vectors ( p, q, r, s) and (t, u, v, w) is ( p, q, r, s) ·h (t, u, v, w) = pt + qu + r v − sw. They are h-orthogonal if this product is 0. The h-length of ( p, q, r, s) is ( p, q, r, s) ·h ( p, q, r, s). 7.6.2. (a) [0, 0, 0, 1]. (b) [1, 1, 1, −5]. 7.6.3. (a) Four distinct points (a, b, c, d), (e, f, g, h), (i, 0 0a e 0 0b f coplanar if and only if the determinant 00 0c g 0d h ⎡ ⎤ 2 0 0 0 ⎢4 −1 0 0⎥ ⎥ 7.6.4. (a) ⎢ ⎣0 −2 2 0⎦. 0

0

j, k, l), and (m, n, o, p), in P3 are 0 i m 00 j n 00 = 0. k o 00 l p0

4 1

7.6.5. (a) By a quadric surface we mean a symmetric invertible 3 × 3 matrix. Two such matrices represent the same conic if and only if one is the multiple of the other by some real number λ ̸= 0. A point P is on a quadric surface S if and only if P T S P = 0. ⎡ ⎤ 1 0 0 0 ⎢0 1 0 0 ⎥ ⎥ (b) x 2 + y 2 + z 2 − t 2 = 0, ⎢ ⎣0 0 1 0 ⎦. 0 0 0 −1

7.6.7. (a) n ≤ 3. (b) n ≤ 4.

7.6.8. (a) The change to −0.25 brings the vanishing points closer to the origin, moving the apparent position of the viewer closer to the cube. Vx = (2, 1.73), Vy = (−3.46, 1), and Vz = (0, −3.46). ⎡ ⎤ 1 0 0 7.6.12. (a) ⎣0 1 0 ⎦. 2 2 −1 (b) The image of (x, 0, 1) is (x, 0, 2x − 1) = (x/(2x − 1), 0, 1), which is on the x-axis on either side of [0, 1]. Note that for 0 < x < 1/2, 2x − 1 < 0. The diagonal is pointwise fixed.

Section 8.1 A1 B2 C3 8.1.1. (a) C2 A3 B1. B3 C1 A2 8.1.2. (a) Shift two to the right and one down. Shift one right and two down.

541

Answers to Selected Exercises

8.1.3. 1 2 3, 4 5 6, 7 8 9; 1 6 8, 2 4 9, 3 5 7;

1 4 7, 2 5 8, 3 6 9; and 1 5 9, 2 6 7, 3 4 8. Four days.

8.1.5. (b) 4k + 1. 8.1.8. (a) A pentagon, 5 points.

Section 8.2 8.2.3. All lines have infinitely many points on them and each point is on infinitely many lines. 8.2.4. In spherical geometry two lines intersect in two points, whereas in single elliptic geometry and projective geometry two lines intersect in exactly one point. 8.2.6. (b) The points of the plane are the girls. Each day corresponds to a family of parallel lines with points on one of those lines representing girls in the same row for that day. 8.2.8. (b) Suppose k∥l and m intersects k. If m didn’t intersect l, there would be two parallels to l through the intersection of k and m, contradicting axiom (iii). 8.2.10. (a) 4, 4. 8.2.16. (a) (ii) (c) The dual of Theorem 7.2.1 (i) is weaker than axiom (i). Parts (ii) and (iii) of Theorem 7.2.1 are duals of each other. The dual of part (iv) of Theorem 7.2.1 is weaker than part (v). 8.2.17. (a) A triangle.

Section 8.3 8.3.1. Since k = 3, each variety must be on a block with two other varieties, giving an odd number of varieties. 8.3.3. (a) 5 varieties with each pair forming a block. b = 10, r = 4. 8.3.4. (a) b = v(v − 1)/6, r = (v − 1)/2. 8.3.5. (a) r ≥ k. (b) r = k = n + 1, v = b = k 2 − k + 1 = n 2 + n + 1. 8.3.6. (a) For k = 3 and λ = 1, the first equation of Theorem 7.3.1 gives v = 2r + 1, so v is odd. 8.3.7. Rotate {1, 2, 5, 7} to obtain the lines.

542

Answers to Selected Exercises

8.3.13. (a) 1 0 0 1 0 0 0 1 1 1 0

0 1 0 0 1 0 0 0 1 1 1

1 0 1 0 0 1 0 0 0 1 1

1 1 0 1 0 0 1 0 0 0 1

1 1 1 0 1 0 0 1 0 0 0

0 1 1 1 0 1 0 0 1 0 0

0 0 1 1 1 0 1 0 0 1 0

0 0 0 1 1 1 0 1 0 0 1

1 0 0 0 1 1 1 0 1 0 0

0 1 0 0 0 1 1 1 0 1 0

0 0 1 0 0 0 1 1 1 0 1

. 5 and 2.

Section 8.4 8.4.1. There is a nonzero constant k so that ka = a ′ and kb = b′ . [0, 1, 0] has points of the form (x, 0, 1) and [0, 1, 2] has points of the form (x, 3, 1). No point is on both. 8.4.2. There are only 4 points on a line in PZ23 . 8.4.3. (0, 0, 1), (1, 1, 1), (2, 4, 1), (3, 4, 1), and (4, 1, 1). Additional projective point: (0, 1, 0). 8.4.4. (a) (b) (c) (d)

[1, 2, 1]. (2, 0, 1). [2, 2, 1]. [−2, −1, 4], yes. (5, −6, 1), yes. [2, 2, −8], yes.

8.4.6. (a) Switches x- and y-coordinates. Similar to mirror reflection. (b) Rotation of 90◦ around (1, 1, 1) for both planes. 8.4.8. (a) For x = b, the tangent is y = 2bx − b2 or [2b, −1, −b2 ]. (b) Each line is tangent. For example, for x = 0, [0, 4, 0] at (0, 0, 1) and for x = 3, [1, 4, 1] at (3, 4, 1). 8.4.9. Many answers are possible. (a) [2, 2, 2] has 8 points. (b) Consider [2, 1, 0] and [0, 1, 0]. 8.4.10. (a) (2, 0, 1). (b) [3, 1, 4]. (d) (3, 0, 1) is the only choice for S. 8.4.12. (a) [3, 4, 2], [1, 4, 2], and (0, 3, 4). 8.4.16. (a) Fixed points of the form (x, x, z), which are on the line [1, 4, 0]. The other stable lines are [4, 4, c] for any c. This matches the mirror reflection over y = x. 8.4.18. (a) (1, 0, 1), (2, 0, 1), (0, 1, 1), and (0, 2, 1).

Answers to Selected Exercises

543

8.4.20. x 2 + y 2 = 1 has (0, 1, 1), (0, 4, 1), (1, 0, 1), and (4, 0, 1). x 2 + 4y = 0 has (1, 1, 1), (4, 1, 1), (2, 4, 1), (3, 4, 1), and (0, 0, 1). x 2 + 3y 2 = 1 has (1, 0, 1), (4, 0, 1), (2, 2, 1), (2, 3, 1), (3, 2, 1), and (3, 3, 1). 8.4.21. (a) [1, 1, 1, 4]. 8.4.22. (a) 775, 806. 8.4.23. (a) (2, 2, 1, 2).

Section 9.1 9.1.1. (a) So the bending can match, x = 0, x 2 + (y − b)2 = b2 , or x 2 + y 2 − 2by = 0 for b ≥ 0. √ (b) y = b − b2 − x 2 , b = 0.5 is the radius. 9.1.2. (a) x 2 + (y − b)2 = (b − 1)2 for b ≤ 1, b = 0, radius = 1. 9.1.6. (a) A and B are on a line of longitude (x z-plane) at the same latitude determined by c, where c = 0 gives the equator and c = π/2 gives the north pole. (b) 1. (c) cos(c). (d) d(c) = π cos(c), a half circle. (e) D(c) = π − 2c. (f) At c = 0, both give a half circle of radius 1. At c = π /2, A = B, so the distance is 0.

Section 9.2

K K ! 9.2.1. x ′ (t) and y ′ (t) can’t both be 0 so Kc′ (t)K = x ′ (t)2 + y ′ (t)2 > 0.

9.2.2. (b) x = π /2 is the line of symmetry. (x − π /2)2 + y 2 = 1.

. 9.2.4. (a) r (x) = (1 + e2x )3/2 /e x√ (b) y = −x + 1, r (0) = 2 2 ≈ 2.828, center is (−2, 3), (x + 2)2 + (y − 3)2 = 8. (c) minimum at x = 0.5 ln(0.5) ≈ −0.3, which has a radius of approx. 2.598. 9.2.7. (a) Near x = ±1. (b) κ(0) = 0.125, κ(1) = 14/(51.5 ) ≈ 1.2522. 9.2.8. (a) At an inflection point we expect f ′′ (x) to be 0, and κ(x) can’t be smaller than 0. Radius is infinite. −−→ −−→ −−→ −−→ 9.2.9. (a) c′ (t) = (2t, 1 − 2t) and√c′′ (t) = (2, −2) give c′ (t) · c′′ (t) = 8t − 2. (b) κ(0) = 2, κ(0.25) = 4 2 ≈ 5.656. 9.2.10. (b) Maximum curvature of 2 at t = π /2 + wπ, minimum curvature of 0.25 at t = wπ. (c) Maximum at π /4 + wπ/2, minimum at wπ/2, crosses when t = π /2 + wπ.

544

Answers to Selected Exercises

→ 9.2.13. (a) Definitions A smooth space curve is a function − c from a real interval into R3 , writ−→ ten c(t) = (x(t), y(t), z(t)) so that x ′ , y ′ , z ′ , x ′′ , y ′′ ,and z ′′ exist and are continuous and in addition, x ′ (t), y ′ (t), and z ′ (t) are not simultaneously all 0. (b) circle of longitude. (c) κ(t) = 1.

Section 9.3 9.3.4. Nu = (− sin(u) cos(v), cos(u) cos(v), 0) and Nv = (− cos(u) sin(v), − sin(u) sin(v), cos(v)). 9.3.9. (a) These are all of the points at a distance of R from the z-axis. ! ! 9.3.10. (a) sx = (1, 0, ∓x/ R 2 − x 2 − y 2 ), s y = (0, 1, ∓y/ R 2 − x 2 − y 2 ), which are not orthogonal. 9.3.11. (a) At points with u = kπ and v = mπ, positive curvature. At points with u = π/2 + kπ and v = π /2 + mπ, negative curvature. 9.3.15. (a) s(u, v) = (cos(u) cos(v), sin(u) cos(v), a sin(v)). (b) su = (− sin(u) cos(v), cos(u) cos(v), 0) and sv = (− cos(u) sin(v), − sin(u) sin(v), a cos(v)). & ' (d) Nu =

cos(v) cos(v) √−a2 sin(u) , √ a2 cos(u) 2 2 2 2 a cos (v)+sin (v)

a cos (v)+sin (v)

a su . a 2 cos2 (v)+sin2 (v)

,0 = √

9.3.16. (a) su = (−(r cos(v) + R) sin(u), (r cos(v) + R) cos(u), 0) and sv = (−r cos(u) sin(v), −r sin(u) sin(v), r cos(v)). √ √ 9.3.18. (a) s(u, v) = (cos(u)√1 + v 2 , sin(u)√1 + v 2 , v). (c) s(u, v) = (cos(u) v 2 − 1, sin(u) v 2 − 1, v). Curvature is always positive.

Section 9.4 9.4.2. E = 1 = G, F = 0. For these values ds 2 = du 2 + dv 2 gives Euclidean distance from the Pythagorean theorem. 9.4.3. su = (− f (v) cos(u), f (v) sin(u), 0), sv = ( f ′ (v) cos(u), f ′ (v) sin(u), g ′ (v)), v) = f (v)2 , F(u, v) = 0, and G(u, v) = f ′ (v)2 + g ′ (v)2 = 1.

E(u,

9.4.5. For v 0 = 0, sin(v 0 ) = 0, so N (t) = (cos(t), sin(t), 0) = −T ′ (t).

Section 10.1 10.1.1. Vertices of a square, a rectangle, a parallelogram, {(0, 0), (1, 0), (0, 1), (2, 3)}, {(0, 0), (1, 0), (0, 2), (2, 3)}. The circle with center A going through B and the circle with center B going through A intersect in two points C and D. However, the distances AB and C D are unequal. 10.1.2. Yes.

Answers to Selected Exercises

545

10.1.3. Draw all the diagonals from one vertex. 10.1.4. Label the vertices from A to J going counterclockwise from the lower right. Place guards at A, B, and E. Extend C D, F E, G H , and J I to where they intersect AB at K , L, M, and N , respectively. There must be a guard in each of △BC K , L F G M, and △A J N in order to see the points C, G, and J . 10.1.5. Triangle, square, hexagon. The vertex angles must divide 360◦ . 10.1.6. Rotate △ABC 180◦ about the midpoint of AB to get △B AC ′ . Then AC BC ′ is a parallelogram and we can tile the plane as a slanted checkerboard with copies of AC BC ′ and hence with copies of △ABC. 10.1.7. Region A is between regions B and D and between regions C and E. Hence the perpendicular bisectors of A with each of these regions shrink the regions B, C, D and E more than the bisectors for B and D or for C and E. 10.1.8. The plane is infinite, so at least one of finitely many regions has to be infinite to cover it all. 10.1.9. (a) No: two equilateral △s back to back. 10.1.10. (a) (b) (c) (d)

3. 3. 3, 4. 3, 4, 5, 6.

10.1.11. (a) 3: {1, 2, 3, 4}, 4: {1, 2, 3, 5}, 5: {1, 2, 4, 7}, 6: {1, 2, 4, 8}. 10.1.13. D2 (5) = 2, D2 (6) = 3 = D2 (7), D2 (8) = 4. 10.1.15. (a) 2. (b) 5. 10.1.16. (a) 2. (b) 5. 10.1.17. (a) One of the diagonals is interior. Place a guard on this diagonal. (b) a hexagon. 10.1.18. (a) 2 guards at opposite vertices. (b) 3 guards at vertices that are not all adjacent. 10.1.20. All five give monohedral tilings. 10.1.21. The vertex angles of two octagons and a square add to 360◦ .

Section 10.2 10.2.1. Two well chosen guards at vertices or elsewhere suffice. 10.2.3. Suppose Dd (n) is given. Let V = {v 1 , . . . , v n , v n+1 } be any set of n + 1 vertices in d dimensions. Then {v 1 , . . . , v n } has at least Dd (n) distances, forcing V to have at

546

Answers to Selected Exercises

least that many. Hence Dd (n) ≤ Dd (n + 1). Now suppose W = {w1 , . . . , wn } is a set of n points in d dimensions with exactly Dd (n) distances. We can embed W in d + 1 dimensions and still have the same number of distances. So Dd+1 (n) ≤ Dd (n). 10.2.4. (a) n (b) Suppose we have n numbers 1, 2, . . . , k, where all adjacent differences are 1 or 2. Then we get all distances from 1 to k − 1. The largest k can be is 2n − 2, so we can get up to 2n − 3 different distances. 10.2.6. (a) 3. (b) 9. 10.2.7. (a) n(r ): 7, 19, 37, 61; d(r ): 3, 8, 15, 23. 10.2.9. (b) n(n − 1)/2, n. (Note: n(n − 1)/2 − n = n(n − 3)/2.) The polygon is convex so every segment except outside edges is interior and so a diagonal. 10.2.10. (a) 2, 3 or 4 diagonals. 10.2.11. (a) 2, 3, 4, 5. 10.2.12. (b) 1, 2, 3, 5. 10.2.13. (a) One guard must be in △AB H to see B and one must be in △E F G to see E. (b) “Square” the left peak: Make AB vertical and add in B ′ above C and level with B. 10.2.17. (b) 2, 3, 4, 5. 10.2.18. (a) 3 guards suffice. Note: There is just one corner of 270◦ , so essentially just one design. 10.2.19. (a) ⌈n/2⌉. Position the guards to cover the outside. Any one of them also covers the inside.

Section 10.3 10.3.1. Color the big squares with two colors as with a checkerboard. Use the third color for the small squares. Since two adjacent big squares also are adjacent to a small square, they must each be a different color. 10.3.2. Use each polygonal tile as the base of a prism with rectangular sides of, say, height h, perpendicular to the plane of the tiling. Stack layers of these prisms on top of one another to fill space. 10.3.4. No. Each angle is 120◦ so every vertex must have 3 polygons meeting at it. If a short edge of one polygon is matched with a long edge, there will be a 60◦ angle. So matching edges must be the same length. However at each vertex each polygon has a long and a short edge, so this is impossible. 10.3.6. (a) Suppose the measures of the angles are A, B, C, A, B, and C, consecutively. The angle sum is 720◦ so A + B + C = 360 and any the angle sum of any three consecutive angles is 360.

Answers to Selected Exercises

547

10.3.8. (a) The angle sum is 1080◦ = x · 90◦ + (8 − x)270◦ and so x = 6. (b) By symmetry the 270◦ angles are opposite. 10.3.9. (b) Any two lengths will work. The polygon with the shorter edge is surrounded by polygons with the longer edge. 10.3.11. (a) 2 colors. (c) 3 colors. Use two alternating colors for the triangles and the third color for the hexagons. 10.3.12. (a) Connect midpoints of each edge to create four similar triangles to the bigger one. (One can do a similar construction with n 2 smaller triangles.)

Section 10.4 10.4.1. The boundaries are two parallel lines. Three parallel lines. 10.4.2. All but the ones with P−1,1 and P1,−1 . All but P1,1 and P−1,−1 . 10.4.3. Suppose part of the perpendicular bisector of the sites A and B forms part of the boundary of the region surrounding B. Then the point C so that B is the midpoint of A and C is another site and the perpendicular bisector of the sites C and B forms part of the boundary of the region surrounding B. Thus these sides are centrally symmetric with respect to B. This holds for all sides of the region of B and so all regions. 10.4.4. The tiling is made of squares. 10.4.5. The regions are regular hexagons. 10.4.7. Sites D, E, F, and G, where D is the intersection of the angle bisectors of △ABC and E, F, and G are the mirror reflections of D in the three sides of the triangle. 10.4.8. (a) Sites: vertices of equilateral triangle △ABC. Voronoi vertex, D, is the center of △ABC and the edges are rays from D perpendicular to the sides of △ABC. (b) Use A, B, C, and D from part (a) as sites. The Voronoi diagram has an equilateral triangle and three rays, giving 3 vertices and 6 edges. 10.4.10. (a) k + 1 sites with k at the vertices of a regular k-gon and the last one at the center of the k-gon. √ √ (b) For example when k = 3, 6 sites at (±1, 0), (±2, 3), and (±2, − 3). 10.4.14. (a) Yes, the sites are at the centers of the triangles and lie at the vertices of a tiling by regular hexagons and form part of a lattice. 10.4.15. (a) The ray of points (x, a+b ) for x < 0, the ray (x, b−a ) for x > a and the segment 2 2 a+b ). between (0, 2 ) and (a, b−a 2 10.4.17. (a) A great circle is the perpendicular bisector. (b) Three half circles through the poles at angles of 120◦ .

Acknowledgements We gratefully acknowledge the permissions we received to use the following. Figure 1.0, courtesy the Estate of R. Buckminster Fuller. Figure 1.51, from Galileo, Two New Sciences, copyright 1954 by Dover Publications, New York, reprinted with permission. Chapter 2. Example 1, from Dubnov, Mistakes in Geometric Proofs, copyright 1963 by D.C. Heath, Lexington, MA, reprinted with permission. Figure 3.0, St. Joseph Government Center, courtesy of Murray A. Mack, HMA Architects. Figures 4.0, 4.39, and 4.40, courtesy of Douglas Dunham. Figure 4.5, courtesy of Caren Diefenderfer. Figure 5.1, from Wade, Geometric Patterns and Borders, copyright 1982 by Nostrand Reinhold Co., New York, reprinted with permission. Figure 5.3, from Thompson, On Growth and Form, copyright 1942 by Cambridge University Press, New York, reprinted with permission. Figure 5.40, courtesy of Murray Mack. Figures 6.0, 6.1, 6.3, 6.6, 6.12, 6.16, 6.17, 6.18, 6.20, 6.23, 6.24, 6.25, 6.26, 6.28, 6.29, from Wade, Geometric Patterns and Borders, copyright 1982 by Nostrand Reinhold Co., New York, reprinted with permission. Figure 6.9, from Bentley and Humphrey, Snow Crystals, copyright 1962, by Dover Publications, New York, reprinted with permission. Figures 6.19 and 6.27, from Crowe and Washburn, “Groups and geometry in the ceramic art of San Ildefonso,” Algebra, Groups and Geometries, copyright 1985 by Hadronic Press, Palm Harbor, FL, reprinted with permission. Figure 6.31, courtesy of David Paul Lange, O.S.B. Figure 6.37, from Holden and Morrison, Crystals and Crystal Growing, copyright 1982 by MIT Press, Cambridge, MA, reprinted with permission.

549

550

Acknowledgements

Figure 6.39, from Peterson, The Mathematical Tourist: Snapshots of Modern Mathematics, p. 208. Copyright 1988 by Ivars Peterson. Used by permission of Henry Holt and Company, LLC. All rights reserved. Figure 6.48, from Mandelbrot, The Fractal Geometry of Nature, p. 265. Copyright 1977, 1982, 1983 by Benoit B. Mandelbrot. Used by permission of Henry Holt and Company, LLC. All rights reserved. Figures 6.51 and 6.53, courtesy of U.S. Geological Survey. Figure 6.52, from Moore and Persaud, The Developing Human: Clinically Oriented Embryology, p. 248. Copyright 2003 by Elsevier, reprinted with permission. Figures 7.0 and 7.1, from Art Resource, reprinted with permission. Figure 10.1, from Johnson, The Ghost Map, copyright 2006 by Penquin Books, New York, courtesy of UCLA Snow Site. Figure 11.1, courtesy of David Paul Lange, O.S.B. Appendix B, from School Mathematics Study Group, Geometry, copyright 1961 by Yale University Press, New Haven, reprinted with permission. Appendix C, from Hilbert, The Foundations of Geometry, 2nd ed., translated E. Townsend, copyright 1921 by Open Court, Peru, IL, reprinted with permission.

Index Terms AAS, 11, 493 (I-26) absolute conic, 355, 358 absolute quadric surface, 367 acceleration vector, 509 Achilles and the Tortoise, 4 affine geometry, 358 affine matrix, 215 plane, 215 three-dimensional, 235 n-dimensional, 241 affine plane, 379ff, 396 affine space, 397, 400 affine transformation, 215, 243, 358 alternate interior angles, 27, 494 alternate exterior angles, 31, 494 altitude of a triangle, 41 analysis, 11 analytic geometry, 98 analytic projective geometry, 337 analytic model, 98 angle, 3, 49, 70, 500 angle bisector, 13 angle defect for polyhedra, 46 angle of parallelism, 162 angle of two great circles, 49 angle-angle-side, 11, 493 (I-26) angle-side-angle, 11, 493 (I-26) angle sum, 3 antipodal points, 49 antiprism, 51, 291 apex, 42 arc length, 438 Archimedean axiom, 501 Archimedean solid, 63, 289 area, 36, 176 arithmetic mean, 7 art gallery problem, 449, 462

art gallery theorem, 463 ASA, 11, 493 (I-26) arithmetic mean asymptote, 108, 341 atoms, 294 axiom, 69 axiom of linear completeness, 501 axiomatic system, 68 axioms, projective geometry, 327ff axis of rotation, 237 Babylonian mathematics, 2 balanced incomplete block design, 386 barycentric coordinates, 118, 366 base of a pyramid, 42 base of a Saccheri quadrilateral, 169 B.C.E. (before the common era), xviii between, 366 Bezi´er curve, 126ff BIBD, 386 bilateral symmetry, 262 binormal, 433 bipyramid, 53 bisector, 12 block, 386 bonds, 294 Brianchon’s theorem, 336 Bruck-Ryser theorem, 381 CAD, 126ff, 241, 362 cases, 514 Catalan number, 461 Cavalieri’s principle, 56, 498 cell, 144 center of inversion, 246 central angle, 21 central symmetry, 221, 243

551

552 centroid of a triangle, 41 chaos, 489 characteristic axiom, hyperbolic geometry, 153 characteristic equation, 241 chemical structure, 294 circle, 79 circular points at infinity, 358 circumcenter of a triangle, 20 circumscribed circle, 20 closure, 200 code, 390 codeword, 390 collinear, 29, 328 collineation, 347, 362, 398 color group, 281 coloring, 473 color preserving group, 281 color preserving symmetry, 271, 282 color switching symmetry, 271, 282 color symmetry, 271, 282 combinations, 448 combinatorics, 375 commensurable, 3 Common Core State Standards (CCSS), xv, xvi, xx, 196, 208 complete (axiomatic system), 86 complete quadrangle, 320 complete quadrilateral, 321 complex conjugate, 103, 249 complex numbers, 103–104, 248ff composition of functions, 198 computer-aided design, 126ff, 241, 362 conclusion, 511 concurrent, 20, 328 conformal, 252 congruent, 3, 11, 76, 166 conic, 104ff, 341, 370 conic surface, 135 consistent, 83 constructible angle, 25 constructible number, 24 constructible polygon, 19 constructions, 12 continuity axiom (projective), 330 contraction mapping, 230 contradiction, 512 converse, 27 convex, 7, 366 convex hull, 464 convex set of points, 458 correlation, 371 corresponding angles, 31, 494 counterexample, 513

Index

cross polytope, 139 cross product, 503, 509 cross ratio, 339 cross section, 510 crystallographic group, 290 crystallographic restriction, 277 crystals, 290, 294 cube, 51 cuboctahedron, 52 curvature of a curve, 415 curvature of a surface, 154, 427 curve, 509 cyclic group, 267 cycloid, 115 decomposition of a figure, 17 defect of a hyperbolic triangle, 178 definition, 69 degenerate conic, 108, 341 derivative vector, 509 Desargues’ theorem, 322, 370 Descartes’ formula, 46 descriptive geometry, 136 design theory, 385ff determinant, 505 diagonal points, 335 diagrams in proofs, 70, 71 diameter of a circle, 79 diameter of an ellipse, 107 differential geometry, 409ff dihedral angle, 140 dihedral group, 267 dilation, 225 direct isometry, 206 direct proof, 511 directrix, 106 discrete geometry, 448ff discrete pattern, 273 distance, 99, 134, 214, 355 dodecahedron, 55 dot product, 503, 509 double elliptic geometry, 185 doubling the cube, 16 dual, 92, 144, 331, 362, 382 dual of a polyhedron, 53 duality, 331ff, 362, 382 dynamical systems, 201, 489 ear of a polygon, 462 edges, 19, 42 Egyptian mathematics, 2 eigenvalue, 219, 506 eigenvector, 219, 506

553

Index

Elements, The, 10, 491 ellipse, 105 ellipsoid, 135 equal (functions), 197 equal (in measure), 15 equivalence relation, 259, 380 equivalent polygons, 177 Erlanger programm, 209, 212 error-correcting code, 389 Euclidean geometry, 1ff, 358 Euclidean isometry, 202, 241 Euler’s formula, 44 example, 513 excess of a spherical triangle, 51 exterior of a conic, 353 externally tangent circles, 40 faces, 42 ff (“following” for page references), 51 field, 394 15-schoolgirl problem, 375 fifth postulate, 28, 492 finite geometry, 373ff first fundamental form, 437 first stellation, 55 fixed point, 197, 347 flow chart, wallpaper groups, 279 foci, focus, 106 fortress problem, 454 fortress theorem, 464 fractal, 228, 230, 304ff fractal curve, 309 fractal dimension, 307 fractal surface, 309 frequency, 47 frieze pattern, 273 fundamental theorem, projective geometry, 333 Gauss-Bonnet theorem, 439 Gaussian curvature of a surface, 427 general theory of relativity, 409, 440 generate, 274 generic, 451 geodesic, 155, 409, 433 geodesic dome, 47 geometric mean, 7 geometry (definition), 209 glide reflection, 206 global positioning system, 440 golden ratio, 8 GPS, 440 great circle, 48 Greek mathematics, 2ff

group, 200 guard point, 62 h-inner product, 357 h-length, 357 h-orthogonal, 357 half plane, 76 half plane model, hyperbolic geometry, 157, 253 Hamming distance, 390 harmonic (music), 326 harmonic set of lines, 325 harmonic set of points, 320 Hausdorff dimension, 305 helix, 116 Hilbert’s axioms, 77ff, 499 homogeneous coordinates, 337 homogeneous second degree equation, 340 horocycle, 191, 258 horolation, 258 hyperbola, 105 hyperbolic distance, 182, 355 hyperbolic geometry, 152ff, 247, 355 hyperbolic glide reflection, 258 hyperbolic isometry, 250, 355 hyperbolic translation, 258, 356 hyperboloid of one sheet, 136 hyperboloid of two sheets, 135 hypercube, 137 hypercycle, 258 hyperplane, 134, 241 hypothesis, 511 icosahedral group, 289 icosahedron, 54 ideal line, 318 ideal point, 318 identity function, 199 identity matrix, 505 if and only if, 514 IFS, 228, 230 image, 197 incenter, 21 incidence matrix, 390 incommensurable, 3 independent, 84 indirect isometry, 206 induction proof, 512 inscribed angle, 21 inscribed circle, 21 inscribed polygon, 19 interior angles on the same side, 31, 494 interior of a conic, 353, 355 inverse function, 199

554

Index

inverse matrix, 505 inversion, 246 inversive line, 246 inversive plane, 246 invertible matrix, 505 irrational, 3 isomer, 299 isometry, 202 Euclidean plane, 202 Euclidean n-dimensional, 241 spherical, 235, 241 isometric, 212 isomorphic, 87 isosceles, 17 iterated function system, 228, 230 iteration, 304

minor of a matrix, 505 mirror reflection, 203, 237 M¨obius transformation, 250 model, 82 mod (n), modulo, 395 monohedral tile, 450 motif, 264

Kirkman’s 15 schoolgirl problem, 375 kite, 32 Klein model, hyperbolic geometry, 157

oblate, 430 oblong number, 7 octahedral group, 289 octahedron, 53 omega point, 165, 355 omega triangle, 165 on, 99, 134, 216, 337, 341 one-point perspective, 323 operational definition, 69 opposite interior angles, 29 orbit of a point, 269 orbit-stabilizer theorem, 269 order, affine plane, 379 order, field, 395 order, projective plane, 383 orientable, 188, 488 orientation, 196 oriented point, 365 oriented projective geometry, 365 origin, 133 orthocenter, 41 orthogonal, 237, 503, 509 orthogonal circles, 154, 247 orthogonal matrix, 241 orthogonal polygon, 454 orthonormal basis, 237, 504 osculating circle, 414 oval, 399

latitude, 49, 424 law of cosines, 23 law of the lever, 26 lemniscate, 117 length, 3, 237, 503 length of a vector, 418, 509 lightlike (relativity), 303, 367 limiting curve, 191 line, 3, 99, 133, 216, 337, 362 line conic, 345 line segment, 3, 366 linear algebra, 503 linear combination, 504 linear transformation, 504 locus problem, 105 logically equivalent, 27 longitude, 49, 424 Lorentz transformation, 297, 368 lunes, 8, 49 manifold, 439 matrices, matrix, 504 matrix multiplication, 504 mean curvature, 445 median of a triangle, 41, 100 metamathematics, 83 Michelson-Morley experiment, 297 midline, 273 midpoint, 18 minimal surface, 445 Minkowski geometry, 367

Napoleon’s theorem, 145 National Council of Teachers of Mathematics (NCTM), xv, xx n-gon, 19 non-orientable surface, 188, 365 norm of a vector, 509 normal, 414 normal line, 415

Pappus’ theorem, 336 parabola, 106 paraboloid, 135 parallel, 26, 134, 379, 492 parallel postulate, 26, 492, 498 parallelogram, 29

555

Index

parametric equations, 114 partial derivative, 510 Pascal’s theorem, 336 Pasch’s axiom, 77, 499 Penrose tiling, 296 pentamino, 483 perpendicular, 12, 143 perpendicular bisector, 12 perspective (art), 318 perspective (computers), 362ff perspective from a line, 321 perspective from a point, 321 perspectivity, 320 Pick’s theorem, 145 plane, 134, 238, 362 plane curve, 509 plane tiling, 450 platonic solid, 44 Playfair’s axiom, 28, 498, 501 Poincar´e conjecture, 489 Poincar´e model, hyperbolic geometry, 157, 247 point, 99, 133, 214, 235, 337, 362 polar, 252, 326 polar coordinates, 116 pole, 252, 326 polygon, 19 polyhedra, polyhedron, 42 polytope, 137 postulate, 12, 28, 76 power of a point, 39 principle of mathematical induction, 512 prism, 43 product of matrices, 505 projection mapping, 242 projective geometry, 318ff projective plane, 382ff, 397 projective space, 362ff, 399 projective transformation, 346ff projectivity, 333, 347 prolate, 430 proof, 70 proof by contradiction, 512 properties of matrices, 506 proportional, 33 pseudosphere, 156, 431 pyramid, 42 Pythagorean theorem, 3, 495 (I-47) quadric surface, 368 quadrilateral, 29 quasicrystal, 296

radians, 50 radius, 79 radius of curvature, 415 ratio of proportionality, 33 rational line, 120 rational numbers, 120 rational point, 120 ray, 70, 500 real numbers, 82 rectangle, 30 rectangular box, 51 recursive, 461 regular polygon, 19 regular polyhedron, 44 regular polytope, 137 regular star figure, 62 regular star polygon, 61 regular tiling, 484 relatively consistent, 83 relativity theory special, 263, 297 Galilean, 297 general, 409, 440 reproducing an angle, 13 rep-tile, 476 rhombus, 22 Richardson’s equation (dimension), 308 right angle, 12 rotary reflection, 237 rotation, 203, 237 Saccheri quadrilateral, 169 SAS, 11, 492 (I-4), 498, 500 scalar, 503 scalar multiple, 133 scaling ratio, 223 Sch¨onflies’ notation, 289 screw motion, 240 segment, 3, 499 self-dual, 53, 362 self-similarity, 304 semi-regular tiling, 484 sensed parallel lines, 162 separation (projective geometry), 329 separation axiom (or postulate), 76, 77, 498, 500 separation axiom (projective geometry), 329 set, 401 shape operator, 427 shear, 227 side-angle-side, 11, 492 (I-4) side-side-side, 11, 492 (I-8)

556 signed curvature, 427 similar, 34, 41 similarity, 223 simplex, simplices, 139 simply connected, 488 single elliptic geometry, 185ff, 358 site, 453 smooth plane curve, 415 smooth surface, 426 SMSG postulates, 76ff, 497 spacelike (relativity), 303, 368 sphere, 48, 424 spherical cap, 58 spherical excess, 51 spherical geometry, 185ff spherical isometry, 235 spherical triangle, 50 spheroid, 430 spiral of Archimedes, 418 spline, 130 square, 30 square matrix, 504 squaring the circle, 16 SSS, 11, 493 (I-8) stabilizer of a point, 268 stable, 197, 238, 347 standard basis, 504 statistical design theory, 374 statistical self-similarity, 305 Steiner quadruple system, 392 Steiner triple system, 388 stereographic projection, 254 straight angle, 12 straightedge and compass constructions, 12 subgeometry, 354ff, 367 subgroup, 274 subspace, 504 sum, 133 summit of Saccheri quadrilateral, 169 supplementary angles, 79 surface, 426ff, 509 surface of revolution, 428 symmetric design, 391 symmetric group, 291 symmetric matrix, 504 symmetries of a prism, 289 symmetry, 264 symmetry group, 264 synthetic, 11 tangent, 15, 341, 399 tangent plane, 510 tangram, 61

Index

taxicab geometry, 84, 125 tetrahedral group, 289 tetrahedron, 42 tetromino, 455 theorem, 70 36-officer problem, 374 three body problem, 489 three-point perspective, 323 tile, 450 tilings, 450ff timelike (relativity), 303, 368 topology, 488 torsion, 410, 420, 445 torus, 425 total angle defect, 46 tractrix, 431 transcendental, 17 transformation, 197 transformation group, 200 translation, 203 transpose, 237, 504 transversals, 26 trapezoid, 39 triangle, 3 triangle inequality, 493 (I-20) triangular number, 7 triangulation, 449, 459 trilinear plot, 118 trisecting an angle, 16 two-point perspective, 323 ultraparallel lines, 162 undefined term, 69 unit normal, 426, 510 unit tangent vector, 419, 509 vanishing point, 318 variety, 386 vector, 133, 503 vector function, 509 velocity vector, 509 vertex, vertices, 42, 203 vertical angles, 79, 493 (I-15) volume, 43 Voronoi diagram, 453 Voronoi edge, 478 Voronoi region, 453 Voronoi vertex, 478 wallpaper group flow chart, 279 wallpaper pattern, 273 Zeno’s paradoxes, 4

557

Index

People (biography pages in bold) Archimedes, 25 Aristotle, 5, 474 Barnsley, Michael, 228, 230 Beltrami, Eugenio, 156, 161, 424 Bernoulli, Jakob, 116, 410 Bernoulli, Johann, 377 ´ B´ezier, Pierre Etiene, 126 Bolyai, J´anos, 153, 169 Bolyai, W., 17 Bonnet, Pierre Ossian, 439 Bravais, Auguste, 263, 288 Brianchion, Charles Julien, 336 Cayley, Arthur, 318, 354, 361 Cavalieri, Bonaventura, 56 Chv´atal, V´aclav, 450 Clairaut, Alexis-Claude, 410 Coxeter, H. S. M., 180, 263, 288, 293 Crowe, Don, 280 da Vinci, Leonardo, 267, 318 Dehn, Max, 17 Desargues, Girard, 318, 322, 336 Descartes, Ren´e, 46, 98, 104, 423, 477, 487 Dirichlet, Peter Gustave Lejeune, 477 D¨urer, Albrecht, 263, 318, 319 Eddington, Sir Arthur, 409, 440 Einstein, Albert, 155, 297, 411, 444 Eratosthenes, 8 Erdos, ˝ Paul, 449, 457, 459, 466 Escher, M. C., 180, 288, 293 Euclid, 10ff, 318, 487 Eudoxus, 4 Euler, Leonhard, 46, 98, 114, 374, 377, 410 Fano, Gino, 374, 382 Fedorov, Vyatseglav, 278 Fermat, Pierre de, 98, 114, 487 Fisher, Sir Ronald A., 374, 385, 393 Fisk, Steve, 462 Francesca, Piero della, 318 Fuller, Buckminster, 1, 46, 59, 293 Galileo Galilei, 36, 297, 423 ´ Galois, Evariste, 263 Gauss, Carl Friedrich, 19, 153, 160, 410, 424, 437, 439 Gerwien, P., 17

G¨odel, Kurt, 83, 94 Gr¨unbaum, Branko, 477 Hamilton, William Rowan, 361 Hausdorff, Felix, 305 Helmholtz, Hermann, 235 Hessel, J. H. C., 263, 288, 290 Hilbert, David, 77, 81 Jordan, Camille, 263 Kant, Immanuel, 152 Kelvin, Lord, 378 Kepler, Johannes, 267, 283, 318, 423 Khayyam, Omar, 169, 174 Kirkman, Rev. Thomas, 375, 378, 388 Klein, Felix, 185, 196, 209, 212, 263, 278, 318, 346, 354 Koch, Helge von, 304 Lagrange, Joseph Louis, 263 Leibniz, Gottlieb Wilhelm, 409, 423 Lie, Sophus, 196, 234, 263 Lindemann, Ferdinand, 16 Lobachevsky, Nikolai, 153, 169 Lorentz, Hendrik, 263, 297, 489 Mandelbrot, Benoit, 304, 312, 488 Minkowski, Hermann, 299 M¨obius, Augustus, 118, 137, 195, 250, 254, 263, 318, 487 Monge, Gaspard, 136, 145, 318, 336 Nasir Eddin, 60 Newton, Sir Isaac, 297, 409, 423, 430 Oresme, Nicole, 98 Pappus, 336 Pascal, Blaise, 114, 318 Penrose, Sir Roger, 296 Perelman, Grigori, 489 Plato, 4 Pl¨ucker, Julius, 149, 254, 318, 346, 388 Poincar´e, Henri, 212, 247, 263, 297, 488 Poncelet, Jean Victor, 318, 336 Pythagoras, 2 Pythagoreans, 2 Richardson, Lewis, 308 Riemann, Georg, 154, 185 189, 411, 439 Saccheri, Giovanni, 152, 175

558 Schl¨afli, Ludwig, 137 Senechal, Margorie, 304 Snow, John, 447, 477 Steiner, Jacob, 346, 388 Tait, Peter Guthrie, 378 Thales, 33, 511 Theaetetus, 4

Index

⟨g1 , g2 , . . . , gn ⟩, the subgroup generated by g1 , g2 , . . . , gn , 274 ·h , h-inner product, 357 H (P Q, RS), harmonic set, 320, 339 H ( jk, lm), harmonic set, lines, 325 I , identity matrix, 505

Van Staudt, Karl, 318 Vi`ete, Franc¸ois, 98 Voronoi, Georgy, 477

k · l, intersection of lines, 328 κ( p, q), curvature of a surface, 427 κ(t), curvature of a curve, 416

Wantzel, Pierre, 16 Washburn, Dorothy, 280 Wiles, Andrew, 114

M, shape operator, 427 M −1 , inverse of M, 216, 505 M T , transpose of M, 504 m∠ABC, measure of ∠ABC, 21

Zeno, 4

Notation AB, length of AB, 3 ← → AB, line on A and B, 3 −→ AB, ray from A through B, 70 AB, segment between A and B, 3 ∠ABC, angle with vertex at B, 3 △ABC, triangle with vertices A, B, and C, 3 −−−−→ −−−−−→ (a, b, c) · (d, e, f ), dot product, 338, 503, 509 a + bi, complex conjugate, 103, 249 α −1 , inverse of α, 199 AF2 , affine plane over F, 396 AFd , affine space over F, 400 B(t0 ), binormal, 433 cm, etc. wallpaper groups, 279 C, complex numbers, 249 C# , extended complex numbers, 248 Cn , cyclic group, 267 −→ c(t), curve, 415, 509 −→ −→ ∥c(t)∥, norm (length) of c(t), 418, 503, 509 −− → −→ c′ (t), derivative (tangent) of c(t), 418, 509 −− → c′′ (t), acceleration vector, 509 Dd (n), minimum number of distances, 449 det(M), determinant of M, 505 Dn , dihedral group, 267, 289 Dnh , symmetry group of a prism, 289 ∂x , partial derivative, 510 ∂u F, field, 395

N, the natural numbers, 512 ⌊n⌋, the greatest integer less than or equal to n, 459 ⌈n⌉, the least integer greater than or equal to n, 467 n!, n Cn D factorial, 290 , combinations of n, k at a time, 448 k N (u, v), unit normal vector, 426, 510 P, icosahedral group, 289 = oriented point, 365 P, p111, etc. frieze pattern groups, 276 PF2 , projective plane over F, 397 PFd , projective space over F, 399 pg, etc. wallpaper groups, 279 P Q//RS, separation, 329, 339 Q, the rational numbers, 120 R, the real numbers, 82 R(P, S, U, W ), cross ratio, 339 r (x), radius of curvature, 416 Sn , symmetric group, 290 su , partial derivative of surface, 426, 510 s(u, v), surface, 426 T, tetrahedral group, 289 T (t), unit tangent vector, 419, 509 T (n), triangulations of a convex n-gon, 461 (v, k, λ), parameters of a BIBD, 385 V or (S), Voronoi diagram of S, 453 − → → v ×− w , cross product, 503, 509

559

Index

W, octahedral group, 289 X x , hyperbolic translation, 356 Zn , integers modulo n, 395 " end of a proof, xvii ♦ end of an example, xvii

* exercise answered in back of book, xvii ∼ =, congruent, 3 ⊥, perpendicular, 12 ∥, parallel, 26, 379 ∼, similar, 34 ⊕, velocity addition in relativity, 298 ∞, added point for inversive plane, 246