Geometry Transformed: Euclidean Plane Geometry Based on Rigid Motions 1470463075, 9781470463076

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51

Geometry Transformed Euclidean Plane Geometry Based on Rigid Motions

James R. King

American Mathematical Society Institute for Advanced Study Park City Mathematics Institute

Geometry Transformed Euclidean Plane Geometry Based on Rigid Motions

UNDERGRADUATE

TEXTS

•

51

Geometry Transformed Euclidean Plane Geometry Based on Rigid Motions

James R. King

American Mathematical Society Institute for Advanced Study Park City Mathematics Institute

EDITORIAL COMMITTEE Giuliana Davidoff Maria Cristina Pereyra

Steven J. Miller James Pommersheim

Gerald B. Folland (Chair) The author gratefully acknowledges the support of the IAS/Park City Mathematics Institute during the preparation of this book. 2020 Mathematics Subject Classification. Primary 51-01, 51M04, 97B50, 51N10, 20H15.

For additional information and updates on this book, visit www.ams.org/bookpages/amstext-51

Library of Congress Cataloging-in-Publication Data Names: King, James R. (James Richard), 1943- author. Title: Geometry transformed : Euclidean plane geometry based on rigid motions / James R. King. Description: Providence, Rhode Island : American Mathematical Society, [2021] | Series: Pure and applied undergraduate texts, 1943-9334 ; volume 51 | Includes bibliographical references and index. Identifiers: LCCN 2020044394 | ISBN 9781470463076 (paperback) | ISBN 9781470464431 (ebook) Subjects: LCSH: Euclid’s Elements. | Geometry, Plane. | Rigidity (Geometry) | AMS: Geometry – Instructional exposition (textbooks, tutorial papers, etc.). | Geometry – Real and complex geometry – Elementary problems in Euclidean geometries. | Mathematics education – Educational policy and systems – Teacher education. | Geometry – Analytic and descriptive geometry – Affine analytic geometry. | Group theory and generalizations – Other groups of matrices – Other geometric groups, including crystallographic groups. Classification: LCC QA451 .K56 2021 | DDC 516.22–dc23 LC record available at https://lccn.loc.gov/2020044394

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established to ensure permanence and durability. Visit the AMS home page at https://www.ams.org/ 10 9 8 7 6 5 4 3 2 1

26 25 24 23 22 21

To Vicki, who has shared my life and made it so much richer

Contents

Introduction

xi

Advice for Students and Less Experienced Geometers Information for More Experienced Geometers

xiii xv

A Chapter Guide for Instructors and Others

xvii

Acknowledgments

xxi

Chapter 1.

Congruence and Rigid Motions

1

Rigid Motions

3

Informal Preview of a Problem Solution

4

Sameness Properties of Congruence

5

Exercises and Explorations

9

Chapter 2.

Axioms for the Plane

11

Incidence Axiom

12

Distance and the Ruler Axiom

12

Protractor Axiom and Angles

14

Plane Separation

17

Rigid Motions and Lines

19

The Other Axioms

20

Exercises and Explorations

21

Chapter 3.

Existence and Properties of Reflections

23

Deducing the Properties of Reflections

23

Isosceles Triangles and Kites

27

Circles and Lines

29

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Contents

Light, Angles, and Reflections

30

Paper Folding and Tools for Construction

31

Exercises and Explorations

32

Chapter 4.

Congruence of Triangles

35

Triangle Congruence Tests

36

Applications of Triangle Congruence

39

Properties of Rigid Motions

40

Midpoint Triangle and Angle Sum

41

Exercises and Explorations

42

Chapter 5.

Rotation and Orientation

45

Rotations and Double Reflections

46

Rotation-Reflection Relations

50

Symmetry at a Point

51

Orientation of a Plane

55

Orientation-Preserving and Orientation-Reversing Transformations

60

Exercises and Explorations

62

Chapter 6.

Half-turns and Inequalities in Triangles

67

Half-turn Properties

67

Inequalities with Angles

68

Circles and Distance to Lines

71

Reflections and the Triangle Inequality

75

Exercises and Explorations

79

Chapter 7.

Parallel Lines and Translations

83

The Euclidean Parallel Postulate

83

Transversals and Parallel Lines

85

Parallelograms

88

Rectangles

91

Midpoint Figures

93

Generalizing Parallelograms

96

Translations as Half-turn Products

98

Products of Translations

101

Direction from Translation

102

Direction and Rotation from Polar Angle

104

Vectors

107

Exercises and Explorations

108

Contents

Chapter 8.

ix

Dilations and Similarity

113

Similarity Theorems for Triangles

115

Right Triangles

117

The Regular Pentagon and Its Ratios

121

Ratios, Signed Ratios, and Scale Factors

123

Transversals of Parallels and Ratios

125

Parallel Segments and Centers of Dilation

128

Construction by Scaling Models

132

Harmonic Division

134

Composition of Dilations

136

Circles, Angles, and Ratios

139

Radical Axis, Intersections, and Triangle Existence

148

Centers of Dilation and the Midpoint Triangle

152

Exercises and Explorations

155

Chapter 9.

Area and Its Applications

161

Areas of Triangles and Parallelograms

161

Area Proofs of the Pythagorean Theorem

165

Area and Scaling

167

Area and the Circle

169

Affine Relationships and Area

172

Exercises and Explorations

175

Chapter 10.

Products and Patterns

177

Products of Rotations

178

Symmetry and 90-Degree Rotations

181

Triangles and 60 Degrees of Rotation

188

Translations and Symmetry

192

Tessellations and Symmetric Wallpaper Designs

194

Translations and Frieze Symmetry

200

Triple Line Reflection Products

208

Exercises and Explorations

211

Chapter 11.

Coordinate Geometry

217

Axes and Coordinates

217

Midpoints, Half-turns, and Translations

218

Lines, Dilations, and Equations

221

Euclidean Geometry and Cartesian Coordinates

223

Perpendicular Lines in the Coordinate Plane

225

x

Contents

Graphs and Transformations

228

Unit Circle and Rotation Formula

229

Complex Numbers and Transformations of the Plane

231

Barycentric Coordinates

233

Vectors and Affine Transformations

240

Axioms and Models

245

Conclusion

249

Exercises and Explorations

249

Bibliography

253

Index

255

Introduction

This book is an introduction to Euclidean plane geometry with axioms based on rigid motions. The initial spark for thinking about this topic was my reaction to reading the Common Core State Standards for Mathematics (CCSSM), especially this statement: Explain how the criteria for triangle congruence (ASA, SAS, and SSS) follow from the definition of congruence in terms of rigid motions.1 For almost all geometry textbooks, this statement would not make sense. Even in textbooks with a focus on geometrical transformations, such as Barker and Howe [1], the Side-Angle-Side criterion for triangle congruence (SAS) is taken as an axiom and is then used to prove the existence and properties of rigid motions. In contrast, a basic assumption of the CCSSM for plane geometry is the existence of rigid motions, transformations that preserve both distance and angle measure. The triangle congruence criteria are then proved as theorems. The rationale for this approach is that such a path into geometry is intuitive, easily modeled informally, and arrives at interesting theorems sooner than by other routes. Moreover, this approach gives students the valuable tools of geometrical transformations at an early stage along with all the traditional geometry tools. Hung-Hsi Wu [19] expounds this pedagogical point of view eloquently and in detail. So what would the rigorous mathematics of such an approach look like? What would the axioms be? How would the choice and flavor of topics be changed? And what would be the implications for teachers and students? I was presented a rare opportunity through my association with the IAS/Park City Mathematics Institute (PCMI)2 to ponder and respond to these questions. We in the PCMI program for teachers decided to develop a short online geometry course for teachers reflecting the Common Core approach to geometry. Working with my colleague and co-leader Gabriel Rosenberg, over two summers we collaborated with teachers developing ideas for such a course. Then, with Gabe as instructor, we offered the 1 2

CCSSM [14], standard ccss.math.content.hsg.co.b.8 PCMI is a program of the Institute for Advanced Study in Princeton, New Jersey.

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Introduction

course several times in a videoconferencing format. The five lessons for that course provided an initial outline for this book.

Transformations and Secondary Geometry For about a century, mathematicians and educators have recommended a greater role for transformations in the geometry curriculum, but for the most part, only small steps have taken place. As noted above, a difficulty has been that, with the most commonly used sets of axioms, the geometry required to define basic transformations, such as reflections, rotations, and translations, is developed fairly late in the course, when many opportunities for using these tools have already passed. This means that transformations appear more as an advanced topic or an enrichment topic than as a central part of geometry. The path urged by the Common Core makes these tools available at the very beginning in an accessible way. The key to this approach is to assume the existence of transformations that preserve both distance and angle, not just distance. This will be explained further, beginning in Chapter 1. Until recently, a second difficulty with introducing transformations into school geometry was that they seemed less visual and more abstract than figures. It is not clear how to draw a transformation. Pictures can show polygons more vividly than functions. This problem has been alleviated in recent years by the availability of dynamic geometry software that provides movement and interaction with transformations in a way that was impossible on paper. With such a dramatic change being proposed in the flow of the geometry curriculum, there’s a concern that teachers are being asked to teach in a new way while textbooks are still organized along an earlier model. There have been only a few resources that reflect the Common Core approach in useful detail. While this is not a high school textbook, I hope that anyone interested will see in the early chapters of this book a way to arrive at familiar territory of congruent triangles while also taking advantage of new tools. This is then followed up by topics such as parallel lines and similarity, intertwined with half-turns, translations, and dilations. I also hope that this book will be helpful to college and university instructors teaching geometry, especially when their students are future teachers who will want and need to understand the Common Core approach. Most of all, I hope that fellow lovers of geometry will find this an interesting path into their favorite subject. James King

Advice for Students and Less Experienced Geometers

The goal of this book is to present the beauty and richness of Euclidean geometry. The method chosen for exploring this mathematics is to begin by assuming a few simple axioms and then deducing the entire structure of interrelationships from these basic building blocks. And a key tool will be the transformations called rigid motions. If you have seen this kind of formal mathematical development before, you will now see how such reasoning can be used to develop geometry. If your experience with proofs is less extensive, you will have the opportunity to grow your toolkit for proofs as you work through theorems and learn more about geometry. Whatever your prior mathematical background, here are some suggestions for reading and working with this book. First, geometry is a visual subject. It is very important to look at figures and not just read words. Even better than looking is drawing. If you sketch figures or construct accurate drawings as you read, you will understand better and enjoy the geometry more. Drawing with any tools is valuable, but it will help enormously to include interactive geometry software in your repertoire. With such software you can draw a figure and then drag elements to create an unlimited number of examples. Since transformations play a big role in this book, the software should include the ability to reflect, rotate, and translate. Moving figures dynamically will really help you visualize these transformations. Second, while the book is mostly structured as a linear story, building block upon block, you have the freedom to read and think in ways that are most productive for you. You are not reading a mystery novel; at the beginning of each chapter, you can look ahead and see what the goals and big ideas seem to be. When you encounter a theorem, first really think (and draw) to make sure you understand what the theorem is saying. Then take a stab at an explanation of your

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Advice for Students and Less Experienced Geometers

own as to why the theorem is true. When you read the proof, does it reflect your ideas or does it go down a different path? Look ahead to see how the theorem is used. If you get stuck on something, spend a little time trying to figure out the point of difficulty, but then move on and return later. An idea or theorem or picture you encounter later may illuminate what came before. And if you sleep on it, the answer to the puzzle may become clear the next day; both experience and brain science confirm this. Finally, look at the Exercises and Explorations. Working on problems can provide valuable practice, to be sure. In addition, some interesting examples and special ideas are put there for working through or experimentation instead of being presented step by step in the main text.

Information for More Experienced Geometers

A primary goal of this book is to provide a reasonably rigorous development of plane Euclidean geometry from a set of axioms based on the existence of rigid motions of the plane. A secondary goal is to demonstrate, by example, how the presence of these tools changes the early experience of geometry. A third goal is to foster an appreciation of how the ways that transformations interact and affect the geometry of the plane. The geometry is based on six axioms. The first four axioms, about lines and angles, are essentially axioms of G. D. Birkhoff [2], often called the ruler and compass axioms. These axioms use the real numbers to arrive quickly at relationships of betweenness and separation at the price of restricting the number field to ℝ. The fifth axiom introduces rigid motions; for each line, it asserts the existence of a rigid motion that fixes the points of the line. In Chapter 3, this rigid motion is quickly proved to be the usual line reflection. Then line reflections are used to prove triangle congruence theorems, to define rotations, to study symmetry and also prove concurrence theorems and inequalities. The sixth axiom is a similarity axiom, asserting properties of dilations. This axiom is equivalent to the Euclidean Parallel Postulate. In Chapter 7, this axiom is used to prove the Euclidean Parallel Postulate as a theorem, at which point the whole of Euclidean plane geometry becomes available, including the theory of parallels and translations and then similarity. The last chapters of the book are devoted to an exploration of area, to considering the structure of symmetric patterns and tessellations on the whole plane, and finally to relating affine and Cartesian coordinates to the geometry that came before. Whether one is reading or teaching from the book, I would repeat the injunction in the Advice for Students section to include a lot of drawing, tracing, paper folding, and interactive geometry software to bring figures and transformations to life.

xv

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Information for More Experienced Geometers

As the titles of the chapters and sections suggest, the book is organized mostly around concepts such as parallel lines and similarity and the related transformations rather than around shapes, such as triangles or circles. Each chapter includes some basic or elementary theorems but also a few less common results or more advanced topics. Proofs feature transformations more than most texts. You may wish to look around to see where certain favorite topics are located and how things are put together (and what seems essential and what may be optional in a course). Non-Euclidean geometry is a topic important for some readers or instructors. The first five axioms in this book are axioms valid for the hyperbolic non-Euclidean plane. The book is structured so that all of the first six chapters, with a couple of isolated and easily avoidable exceptions, use only the first five axioms, so they can be used as an introduction to neutral geometry, valid for both the Euclidean and the non-Euclidean plane. (The principal exception is that Axiom 6 is used early to prove the angle sum theorem for triangles so that, after regular polygons are introduced, angles can be used in exercises; but this result is not used in the main text until the chapter on parallels.)

A Chapter Guide for Instructors and Others

This chapter outline is offered in the hope that it will complement the table of contents and provide helpful information for instructors and others in navigating the book and/or planning a course. The Basics: Chapters 1 to 4. The first four chapters begin with an introduction and statement of axioms and end with proofs of the triangle congruence theorems. An extract of this content could be adapted to a workshop or very short course for future or practicing teachers. Anyone using all or most of the book should not skip any sections of these short chapters. Chapter 1: This is a mostly descriptive introduction to the goals of the book. Rigid motions and congruence are defined and some properties are proved in a very general context. But except for a preview example, there is no actual plane geometry since there are as yet no axioms. For completeness, there is a formal proof of the equivalence properties of congruence, but students and instructors may find the justification by informal geometry adequate before moving on to Chapter 2. Chapter 2: This chapter states all the axioms and goes into the first four in detail. Ruler functions that measure distance on lines and protractor functions and associated polar angles that measure angles are key topics. Many basic geometry terms are defined. I believe that one could move briskly through this chapter and pick up a deeper understanding about rulers and protractors and terminology when encountering them in the geometrical proofs in Chapters 3 and 4. Chapter 3: This important chapter provides the first chance to use the rigid motion concept to prove basic but major theorems. Beginning with an existence axiom for a rigid motion that fixes points on a line, one proves that this rigid motion is the usual line reflection and then uses line reflection to prove theorems about perpendiculars, the properties of isosceles triangles (important for everything later), and other results. xvii

xviii

A Chapter Guide for Instructors and Others

Chapter 4: This is where all the triangle congruence theorems are proved using compositions of reflections. An important feature of these theorems is that the rigid motion defining congruence from △𝐴𝐵𝐶 to △𝐷𝐸𝐹 is unique. In addition, in the last section, the angle sum theorem for triangles is proved using the dilation axiom. If one wants to defer this axiom, equivalent to the Euclidean Parallel Postulate, until later, this result is only used for regular polygons until Chapter 7.

Rotations, Translations, and Dilations: Chapters 5 to 8. By the end of these chapters, one will have covered all or most of the topics about lines, angle, polygons, and circles that may be thought of as the essential core of an introduction to Euclidean plane geometry plus experience working with rotations, translations, and dilations in solving problems and proving theorems. New geometrical theorems are proven as new rigid motions are introduced. Chapter 5: The first part of this chapter introduces rotations as double line reflections, an essential topic for the rest of the book. This theory is applied to rotational and dihedral symmetry groups at a point, leading to a definition of regular polygons, with some details left to the exercises. The last section proves that an orientation can be defined on the plane by the order of vertices of a single triangle. Orientation is important for signed angle measure and orientation-preserving rigid motions, but some details of the proofs may be less critical. Chapter 6: This chapter features half-turns, inequalities within triangles, and the triangle inequality. The concept of distance from a point to a line is applied to tangent constructions and incircles, and reflected light paths culminate in the solution to Fagnano’s problem of finding the triangle of minimum perimeter inscribed in a given triangle. Chapter 7: Beginning with a proof of the Euclidean Parallel Postulate as a theorem using the dilation axiom, this chapter develops a number of theorems about parallelograms and rectangles, with half-turns as a main tool. This culminates in the very important definition of translations as products of half-turns and also as products of reflections in parallel lines, with proofs of their properties. At the end of the chapter, translations are applied to define orientation and polar angles on the whole plane, concluding with a brief introduction to vectors and a (nonstandard) vector notation. Many parts of this chapter are important for later developments. Chapter 8: Similarity is the topic of this long chapter. The early parts of the chapter contain basic material about similarity of triangles and other polygons (including the Pythagorean Theorem and the Golden Ratio) as well as ratios and signed ratios from transversals. The middle of the chapter applies dilations to solve problems about triangles and circles. The last part of the chapter is devoted to theorems concerning circles, including inscribed angles. Some topics near the end, such as Apollonian circles and the radical axis of circles, are interesting geometry topics not found in every beginning course.

A Chapter Guide for Instructors and Others

xix

Further Topics in the Plane: Chapters 9 to 11. The last chapters include one on area, one on products of rigid motions applied to symmetry, and one introducing coordinates. One could put together a short course (e.g., a one-quarter course) that includes the topics essential to a basic geometry course by covering most of the first eight chapters and then, as time permits, including topics from these chapters. With more time, each chapter offers a number of valuable geometric ideas. Chapter 9: This short chapter about area includes the usual elementary area formulas, a discussion of scaling and area, area proofs of the Pythagorean Theorem, and the formulas for circle area and perimeter (without explicitly estimating 𝜋). The chapter points out the effect of similarity transformations on area, ending with affine theorems about ratios and area and the introduction of shear transformations. The introduction of area requires adding some assumptions about the properties of area that were not part of the six axioms on which all the other proofs are based. Chapter 10: This long chapter is organized around the topic of composition of rigid motions and how the relationships from composition restrict possible symmetries. It begins by showing how to compute the composition of two rotations with different centers, or of a rotation and a translation, and applying this to a detailed examination of the symmetries of some wallpaper patterns, starting with a leisurely and detailed examination of a tile pattern with 90-degree rotational symmetry. Wallpaper groups are introduced and some other examples are given, but there is no attempt to look at every crystallographic group. The composition rules are then applied to prove Napoleon’s Theorem, the existence of the Fermat Point and the properties of several interesting tessellations. A complete proof that there are seven types of frieze symmetry is carried out in great detail, with lots of pictures. Then a final section fills in some remaining facts about glide reflections, completing the picture of the four kinds of rigid motions. Chapter 11: This chapter begins by constructing affine coordinates on the plane, proves formulas for half-turns and translations, and derives familiar equations for lines from dilations. Next comes the case of Cartesian coordinates and metric formulas: the distance formulas and dot products. There is a short section on why the (𝑥, 𝑦)-plane for graphs is not the Euclidean plane. There are rotation and reflection formulas and an introduction to the use of complex numbers to express these formulas. The latter part of the chapter introduces barycentric coordinates, proves Ceva’s Theorem, and explores affine transformations and the interpretation of determinant as signed area. The last topic returns to the very beginning of the book and proves that the Cartesian plane is a mathematical model satisfying the six axioms in Chapter 2.

Acknowledgments

I have many people to thank for help with this book and also for deepening my appreciation of geometry and supporting my interest in sharing it with others. This book was written as an outgrowth of the Teacher Leadership Program at the Park City Mathematics Institute (PCMI), which provided the opportunity to develop this material as part of its overall activities. For their support and encouragement, I wish to thank PCMI and its institutional sponsor, the Institute for Advanced Study, and also the other public and private financial sponsors of PCMI, including Math for America. I have benefited greatly from my decades-long association with the PCMI program for teachers. I have learned much and have had a great time, starting in 1991 with the visionary leadership of Herb Clemens and Naomi Fisher, when our new program was officially a geometry institute. My years of collaboration with Carol Hattan and Gail Burrill is something I treasure. I also want to mention the support of Brian Hopkins, whom I lured to PCMI as a grad student, and to express my appreciation for our current leaders, Rafe Mazzeo, Peg Cagle, and Monica Tienda. Other valuable geometrical experiences include years of teaching geometry for future teachers at the University of Washington and many summers of geometry in Seattle with our local PCMI offshoot called Northwest Mathematics Interaction (NWMI). Among the many NWMI people to thank, special mention goes to Joyce and Joe Frost, Art Mabbott, and Clint Chan, who persist in keeping our group going. The inspiration for this book was a desire to expand upon ideas presented in a PCMI online geometry course developed with PCMI colleague Gabriel Rosenberg, aided by two working groups of PCMI teachers, including Irene Espritu, Alice Hsaio, Tamuka Hwami, Robert Janes, Wendy Menard, Jennifer Tate, Robert Garber, Jen Katz, Daniel Kerns, Elizabeth McGrath, Vince Muccioli, and Brian Shay.

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Acknowledgments

For recognizing the benefits of installing rigid motions into the foundations of geometry, I offer a tribute to Hung-Hsi Wu. His persuasive advocacy for introducing rigid motions early in geometry instruction led to the adoption of this approach by the Common Core. For the book itself, I want to express my great appreciation to Bill Barker, who generously volunteered to read several chapters and then made suggestions that resulted in significant improvements. I also want to thank and recognize the editors and technical staff of the AMS, who suddenly had to adapt to working remotely in a pandemic year. Eriko Hironaka was very helpful and supportive throughout the development of this book, and copy editor Arlene O’Sean has been sharp-eyed in making corrections and generous and helpful while getting the book into print. And most of all, I want to thank my wife, Vicki, for so many things over the years but, in particular, for proofreading every page of this book even though she claims she did not understand any of it.

Chapter 1

Congruence and Rigid Motions

Congruence is a fundamental concept in geometry. However, it is surprisingly difficult to concoct a mathematically sound general definition in a textbook or in a course based on the most common axioms. The necessary ingredient for a good definition — a transformation called a rigid motion — typically only appears late in the logical development of the subject, rather than early when it is needed. In this book we modify the axiom set for plane geometry so that rigid motions are available at the outset. This provides a powerful new toolkit for geometrical reasoning in the early stages of geometry. Defining Congruence. Textbooks typically focus on congruence of triangles, which is explained reasonably well. But the general concept of congruence is often not defined. In classrooms one often hears the informal mantra of “same size, same shape”. This may be helpful as a suggestion of the intended meaning, but it is not a definition. For example, somewhat as a joke, we could propose that these first two figures are congruent. They are both rectangles (same shape) and have the same area (same size). No one would really suggest that these are congruent, but it does point out why a real definition should be based on unambiguous words that have clear mathematical meaning.

Figure 1. Same Size, Same Shape?

1

2

1. Congruence and Rigid Motions

If one tries to be more precise by using side length and angle measure in a definition, such as “corresponding side lengths and angle measures are equal”, the definition still does not explain why the polygons in Figure 2 and the polygonal paths in Figure 3 are not congruent.

Figure 2. Same Side Lengths, Same Angles

One can refine the definition of congruence by introducing internal angles and oriented angles to cover these examples, or one can specify that all distances between corresponding pairs of vertices are the same. But this will not apply to geometrical objects that are less polygonal and so are harder to deal with. What does one say about congruence when a figure includes a circular arc or two or more disconnected objects, not to mention an unbounded shape such as a line (or ray or parabola) as in Figure 4? To try to cover each kind of figure on a case-by-case basis would become a game of geometrical whack-a-mole, not a clear definition for a basic concept.

Figure 3. Same Side Lengths, Same Angles

Figure 4. A Disconnected Figure Containing an Arc, Isolated Points, and an Entire Line

There is a simple and intuitive, real-world answer to this problem if the two figures are drawn or printed on paper. One simply picks one up and lays it on the other to see whether they can be matched. This is the correspondence needed in the abstract definition as well. But rather than specify the correspondence as point 𝐴 to 𝐷, point 𝐵 to 𝐸, point 𝐶 to 𝐹, as we do for some triangles △𝐴𝐵𝐶 and △𝐷𝐸𝐹, this correspondence moves the whole plane (or the paper in the physical case). This is superposition. The principle of superposition says that two figures are congruent if one can be superimposed on the other. But how can one convert this physical movement into a mathematical definition?

Rigid Motions

3

When the Greeks developed their deductive approach to geometry, they lacked some language and concepts that are central in mathematics today. They did not have the system of real numbers, and they did not have the general concept of function. They clearly had superposition in their thoughts, because it occasionally appears without acknowledgment in the reasoning of Euclid. But they treated congruence as a formal equivalence without featuring the actual correspondence, the superposition, as an object that is part of the picture. Mathematics now does have the language for superposition well worked out. This is the language of transformations. We can use transformations explicitly in our axioms and our reasoning. Moreover, transformations conform as well or better to the idea of a geometrical tool in our modern world of computer-aided design and rotating photos on phones than do the older tools of straightedge and compass. In addition, transformations become geometrical objects that can be profitably studied in their own right, adding a new layer or richness to the geometry.

Rigid Motions The transformations that will model superposition will be called rigid motions. Then two figures will be congruent if there is a rigid motion that maps one to the other. In order to define rigid motions, we need some properties of the plane and the definition of transformation. We will soon introduce axioms for the plane, but at this point we can say the following: • The plane is a set whose elements are called points. • There is a distance measure: for any two points 𝐴 and 𝐵, the distance ‖𝐴𝐵‖ is a nonnegative real number. • There is also an angle measure: for any three points, 𝐴, 𝐵, 𝐶, with both 𝐵 and 𝐶 distinct from 𝐴, there is measure 𝑚∠𝐵𝐴𝐶 that is a real number between 0 and 180. A transformation of the plane is a function from the plane into itself that has an inverse. To be more precise, a function 𝑇 from the plane to itself is a transformation if there is another function 𝑇 −1 from the plane to itself so that for any point 𝐴, 𝑇 −1 (𝑇(𝐴)) = 𝐴 and 𝑇(𝑇 −1 (𝐴)) = 𝐴. In other words, each composition 𝑇𝑇 −1 = 𝑇 −1 𝑇 is the identity function 𝐼 that maps each point 𝐴 to itself. Definition 1.1. A rigid motion of the plane is a transformation of the plane that preserves distance and angle measure. This means a rigid motion 𝑇 is a transformation with these properties: • For any points 𝐴 and 𝐵, ‖𝐴𝐵‖ = ‖𝑇(𝐴)𝑇(𝐵)‖. • For any three points, 𝐴, 𝐵, 𝐶, with both 𝐵 and 𝐶 distinct from 𝐴, 𝑚∠𝑇(𝐴)𝑇(𝐵)𝑇(𝐶) = 𝑚∠𝐴𝐵𝐶. Now that we have defined rigid motions, we can state a general definition of congruence.

4

1. Congruence and Rigid Motions

Definition 1.2. A set 𝑈 in the plane is congruent to a set 𝑉 if there is a rigid motion 𝑇 such that 𝑇(𝑈) = 𝑉. Note. For a set 𝑈, the notation 𝑇(𝑈) means the image of 𝑈, the set of all the points that are images 𝑇(𝐴) for some point 𝐴 in 𝑈. Rigid Motion vs Isometry. This is a technical point that may be puzzling some readers. Readers who are already familiar with geometrical transformations may at this point be thinking something like this: “Oh, a rigid motion preserves distance. A transformation that preserves distance is called an isometry. But I thought an isometry always preserves angles. Why do we need this extra condition?” Readers applying real-world physical intuition may also be thinking that if distances between points are preserved, then angles are not distorted. There is enough truth here to be confusing. An isometry is a transformation that preserves distance. Nothing is said about angles. But in an axiomatic development of geometry that assumes the triangle congruence criterion Side-Angle-Side as an axiom, it is true — after a certain number of theorems have been proved — that one can prove that an isometry also preserves angle measure and, therefore, is a rigid motion. But in this book we are not assuming Side-Angle-Side as an axiom; instead we are going to assume the existence of certain rigid motions at the outset and then prove SideAngle-Side as a theorem. If we only assumed the existence of isometries as an axiom, this would not be possible, since we could not reason about angles. It is important that superposition preserve angle measure. So, by definition, before any theorems are proved, every rigid motion is an isometry, but there is no reason to assume that every isometry is a rigid motion. However, to complete the confusion, with our rigid motion axiom, we will eventually prove all the triangle congruence theorems and hence prove that distance preservation implies angle preservation; so an isometry will be proved to be a rigid motion. But for an axiom that one assumes without proof to serve as a basis for geometry, we need both conditions in the definition of a rigid motion. And this confirms the physical intuition about distance preservation and angle. In an axiomatic development of an abstract model of the plane, there are a lot of things that are true in the real world that also eventually turn out to be true in the abstract model. But these facts do not get built and proved in the abstract model until after some development of the theory.

Informal Preview of a Problem Solution Here is an example of how this concept of congruence can be used for a short, clear proof that reflects one’s geometric intuition. Since this is a preview, we will take the liberty of assuming in advance that we have already proved rigid motions map line segments to line segments. A segment with endpoints 𝐴 and 𝐵 is denoted 𝐴𝐵. Example 1.3. Suppose triangle △𝐴𝐵𝐶 is congruent to △𝐴′ 𝐵 ′ 𝐶 ′ . Let 𝐷 be the midpoint of 𝐵𝐶 and let 𝐷′ be the midpoint of 𝐵′ 𝐶 ′ . For points 𝐸 on 𝐴𝐶 and 𝐸 ′ on 𝐴′ 𝐶 ′ , assume ‖𝐶𝐸‖ = ‖𝐶 ′ 𝐸′ ‖. Prove that the quadrilateral 𝐷𝐶𝐸𝐹 is congruent to 𝐷′ 𝐶 ′ 𝐸 ′ 𝐹 ′ .

Sameness Properties of Congruence

5

A' C E F

D E'

A

B

C'

F'

B'

D'

Figure 5. Quadrilaterals in Congruent Triangles

Proof. A student steeped in triangle congruence might set off to find equal angles and lengths by proving congruence of some of the five subtriangles in each figure. After quite a number of steps, this will provide a proof that the corresponding sides and angles of the quadrilateral are congruent. But we are given a rigid motion 𝑇 that maps △𝐴𝐵𝐶 to △𝐴′ 𝐵 ′ 𝐶 ′ since the triangles are congruent. Once we show 𝑇 also maps 𝐷𝐶𝐸𝐹 to 𝐷′ 𝐶 ′ 𝐸 ′ 𝐹 ′ , this will prove the quadrilaterals are congruent. This follows from preservation of distance, since the distances defining 𝐷 and 𝐸 are the same as the distances defining 𝐷′ and 𝐸 ′ . In particular, 𝐷, which is the point on 𝐵𝐶 equidistant from 𝐵 and 𝐶, maps to the point on 𝐵 ′ 𝐶 ′ equidistant from 𝐵′ and 𝐶 ′ . Likewise, the image of 𝐸 is 𝐸 ′ , the point on segment 𝐴′ 𝐶 ′ at distance ‖𝐶𝐸‖ from 𝐶 ′ . Finally, the intersection 𝐹 of segments 𝐴𝐷 and 𝐵𝐸 must map by 𝑇 to the intersection of the segments 𝐴′ 𝐷′ and 𝐵 ′ 𝐸 ′ . □ The key idea is to exploit the explicit presence of a rigid motion 𝑇 whenever two figures are known to be congruent. Notice that we did not have to know anything about the behavior of 𝑇 beyond that it maps △𝐴𝐵𝐶 to △𝐴′ 𝐵′ 𝐶 ′ . Note. This proof used an important and convenient notational convention about congruence. When we write that △𝐴𝐵𝐶 is congruent to △𝐷𝐸𝐹, this means more than that there is a rigid motion 𝑇 mapping one triangle to the other. It also means that 𝑇(𝐴) = 𝐷, 𝑇(𝐵) = 𝐸, 𝑇(𝐶) = 𝐹. So the statement that △𝐴𝐵𝐶 is congruent to △𝐸𝐹𝐷 is a different statement. If we want merely to say the triangles are congruent without this convention, we can name 𝑡1 = △𝐴𝐵𝐶 and 𝑡2 = △𝐷𝐸𝐹 and then say that 𝑡1 is congruent to 𝑡2 . This convention carries over to segments 𝐴𝐵, quadrilaterals 𝐴𝐵𝐶𝐷, and to any figure that is defined in terms of a list of points.

Sameness Properties of Congruence In the first examples in this chapter, we have seen that the word “same” in our ordinary language has such rich and varied meanings that it is not precise enough for mathematical use. However, a concept of sameness or equivalence is an important one that we need in mathematics. The following theorem says that congruence has some key

6

1. Congruence and Rigid Motions

properties that we expect from our intuition about sameness. We write 𝑈 ≅ 𝑉 to denote that 𝑈 is congruent to 𝑉. Theorem 1.4. Congruence has the following three properties: (1) For any set 𝑈 in the plane, 𝑈 ≅ 𝑈. (2) If 𝑈 ≅ 𝑉, then 𝑉 ≅ 𝑈. (3) If 𝑈 ≅ 𝑉 and 𝑉 ≅ 𝑊, then 𝑈 ≅ 𝑊. Informally, if we think of congruence in terms of superposition, this is easy to see. If we draw the figures on paper, then not moving 𝑈 at all shows the first item. If we use a motion 𝑇 to move 𝑈 to 𝑉, then the reverse motion 𝑇 −1 moves 𝑉 to 𝑈, so congruence is symmetric. Finally, if we move 𝑈 to 𝑉 by a motion 𝑇 and then move 𝑉 to 𝑊 by the motion 𝑆, the combined motion 𝑆𝑇 of first 𝑇, then 𝑆 moves 𝑈 to 𝑊. This shows that congruence is transitive. The formal proof is pretty much the same, except there are surprisingly many steps to check off since we will need to prove some basic properties of rigid motions along the way. The main things to prove are that inverses and products of rigid motions are rigid motions (properties that we casually assumed with our informal motions but need to check to see that our formal definition really works as expected). We need to prove these properties of rigid motions that correspond to the three points of Theorem 1.4. Theorem 1.5. These statements about rigid motions are true: (1) The identity map 𝐼 is a rigid motion. (2) If 𝑇 is a rigid motion of the plane, then 𝑇 −1 is also a rigid motion. (3) If 𝑆 and 𝑇 are rigid motions of the plane, then the composition 𝑆𝑇 is also a rigid motion. Note. We write composition of transformations as a product, without the little circle that is sometimes used. Before proving this theorem, we first show how it is used to prove Theorem 1.4. Each item in Theorem 1.5 corresponds to the same-numbered item in Theorem 1.4. This proof is a formal version of our informal superposition reasoning above. Proof of Theorem 1.4. In each case, we prove 𝐸 ≅ 𝐹 by finding a rigid motion that maps 𝐸 to 𝐹. (1) The identity map 𝐼 is a rigid motion, and 𝐼(𝑈) = 𝑈, so 𝑈 ≅ 𝑈. (2) If 𝑈 ≅ 𝑉, there is a rigid motion 𝑇 with 𝑇(𝑈) = 𝑉. Then 𝑇 −1 (𝑉) = 𝑈. Since 𝑇 −1 is a rigid motion, 𝑉 ≅ 𝑈. (3) If 𝑈 ≅ 𝑉, there is a rigid motion 𝑇 so that 𝑇(𝑈) = 𝑉. If 𝑉 ≅ 𝑊, there is a rigid motion 𝑆 so that 𝑆(𝑉) = 𝑊. Then 𝑆𝑇(𝑈) = 𝑊, so 𝑈 ≅ 𝑊 since 𝑆𝑇 is a rigid motion. □

Sameness Properties of Congruence

7

Theorem 1.5 about rigid motions is very important. But we will not really use it until Chapter 3, so if you would prefer to read and understand the statement of the theorem now but defer going through the formal proof until we have spent some time with distance and angle measure, that is fine. (In fact, once you get the idea of how repetitious the proof of this theorem is, you may be satisfied without reading every step!) Proof of Theorem 1.5. First, we prove that distance is preserved in each case. Then afterwards, we will take up angle measure. We are not assuming any properties of distance except that it is a real-valued function 𝑑(𝐴, 𝐵) of two points. But instead of the usual function notation, we will write ‖𝐴𝐵‖ in place of 𝑑(𝐴, 𝐵). You will see that this reasoning is very general and quite formal, without regard for any special properties of 𝑑. By Definition 1.1, whenever we know a transformation 𝑇 is a rigid motion, we know that for any 𝐴 and 𝐵, ‖𝑇(𝐴)𝑇(𝐵)‖ = ‖𝐴𝐵‖. (1) The identity map is 𝐼. By definition, for any points 𝐴 and 𝐵, 𝐼(𝐴) = 𝐴 and 𝐼(𝐵) = 𝐵. Therefore, ‖𝐼(𝐴)𝐼(𝐵)‖ = ‖𝐴𝐵‖ by substitution. (2) We need to prove that for any points 𝐴 and 𝐵, ‖𝑇 −1 (𝐴)𝑇 −1 (𝐵)‖ = ‖𝐴𝐵‖. Since 𝑇 is a rigid motion, we apply it to the image points of 𝑇 −1 to get two things: ‖𝑇(𝑇 −1 (𝐴))𝑇(𝑇 −1 (𝐵))‖ = ‖𝑇 −1 (𝐴)𝑇 −1 (𝐵)‖ because 𝑇 is a rigid motion. But 𝑇(𝑇 −1 (𝐴)) = 𝐴 and 𝑇(𝑇 −1 (𝐵)) = 𝐵 since 𝑇𝑇 −1 = 𝐼, so ‖𝑇(𝑇 −1 (𝐴))𝑇(𝑇 −1 (𝐵))‖ = ‖𝐴𝐵‖ by substitution. (3) For any points 𝐴 and 𝐵, we need to prove ‖𝑆𝑇(𝐴)𝑆𝑇(𝐵)‖ = ‖𝐴𝐵‖. By definition, 𝑆𝑇(𝐴) = 𝑆(𝑇(𝐴)) and 𝑆𝑇(𝐵) = 𝑆(𝑇(𝐵)). Since 𝑆 is a rigid motion, ‖𝑆(𝑇(𝐴))𝑆(𝑇(𝐵))‖ = ‖𝑇(𝐴)𝑇(𝐵)‖ and this equals ‖𝐴𝐵‖ because 𝑇 is a rigid motion. The proof of angle preservation follows the identical pattern, except that we have a function of three points, 𝜇(𝐴, 𝐵, 𝐶), that is defined for any three points 𝐴, 𝐵, 𝐶 with 𝐵 ≠ 𝐴 and 𝐶 ≠ 𝐴. Once we have defined angles, we will denote 𝜇(𝐴, 𝐵, 𝐶) by 𝑚∠𝐴𝐵𝐶, but for now, we will stick with the usual functional notation because it seems a bit more readable for this proof. In every case below, we will assume the three points satisfy 𝐵 ≠ 𝐴 and 𝐶 ≠ 𝐴 so that 𝜇 is defined. For any such points, if 𝑇 is a rigid motion, then by Definition 1.1, 𝜇(𝑇(𝐴), 𝑇(𝐵), 𝑇(𝐶)) = 𝜇(𝐴, 𝐵, 𝐶). (1) The identity map is 𝐼. By definition, for any points 𝐴, 𝐵, and 𝐶, 𝐼(𝐴) = 𝐴, 𝐼(𝐵) = 𝐵, 𝐼(𝐶) = 𝐶. Therefore, 𝜇(𝐼(𝐴), 𝐼(𝐵), 𝐼(𝐶)) = 𝜇(𝐴, 𝐵, 𝐶) by substitution. (2) We need to prove that for any points 𝐴, 𝐵, and 𝐶, 𝜇(𝑇 −1 (𝐴), 𝑇 −1 (𝐵), 𝑇 −1 (𝐶)) = 𝜇(𝐴, 𝐵, 𝐶). We apply 𝑇 to the image points of 𝑇 −1 to get 𝜇(𝑇(𝑇 −1 (𝐴)), 𝑇(𝑇 −1 (𝐵)), 𝑇(𝑇 −1 (𝐶))). Since 𝑇 is a rigid motion this equals 𝜇(𝑇 −1 (𝐴), 𝑇 −1 (𝐵), 𝑇 −1 (𝐶)). But 𝑇𝑇 −1 (𝐴) = 𝐴, 𝑇𝑇 −1 (𝐵) = 𝐵, and 𝑇𝑇 −1 (𝐶) = 𝐶 so the expression equals 𝜇(𝐴, 𝐵, 𝐶) by substitution.

8

1. Congruence and Rigid Motions

(3) For any points 𝐴 and 𝐵, we need to prove 𝜇(𝑆𝑇(𝐴), 𝑆𝑇(𝐵), 𝑆𝑇(𝐶)) = 𝜇(𝐴, 𝐵, 𝐶). By definition, 𝑆𝑇(𝐴) = 𝑆(𝑇(𝐴)), 𝑆𝑇(𝐵) = 𝑆(𝑇(𝐵)), and 𝑆𝑇(𝐶) = 𝑆(𝑇(𝐶)). Since 𝑆 is a rigid motion, 𝜇(𝑆(𝑇(𝐴)), 𝑆(𝑇(𝐵)), 𝑆(𝑇(𝐶))) = 𝜇(𝑇(𝐴), 𝑇(𝐵), 𝑇(𝐶)). and this equals 𝜇(𝐴, 𝐵, 𝐶) because 𝑇 is a rigid motion. □ C'

A'

C

C''

S T

A

B'

B''

B

A''

Figure 6. Arrows and Primed Points Showing Transformations

A Few Words about Functional Notation. You may have noticed how extensively functional notation was used in the proofs above. From one point of view, this seems appropriate since transformations are functions. But despite its precision, functional notation can seem dense and hard to follow on the page, not to mention on a blackboard or a whiteboard. An alternative notation might be to write something like this, with the reasoning a bit less explicit: In Figure 6, let the 𝑇-images of 𝐴, 𝐵, 𝐶 be 𝐴′ , 𝐵 ′ , 𝐶 ′ , and let the 𝑆-images of 𝐴′ , 𝐵 ′ , 𝐶 ′ be 𝐴″ , 𝐵″ , 𝐶 ″ . Then if 𝑇 and 𝑆 are rigid motions, 𝑚∠𝐴𝐵𝐶 = 𝑚∠𝐴′ 𝐵 ′ 𝐶 ′ = 𝑚∠𝐴″ 𝐵 ″ 𝐶 ″ . This can be a more readable form of notation in the right context. But it does require an introductory explanation of what the primed images are, and it can be confusing when there are complicated things going on. If one is writing on a board or a document camera, drawing a diagram with a few well-placed arrows to go with 𝑆 and 𝑇 can be very helpful, as in Figure 6. In the coming chapters, we will sometimes use functional notation and sometimes we will use other ways to express what is going on, always aiming for clarity and readability. Circles. At this stage, with almost zero information about the plane, we cannot say much about any objects or shapes. But we can define one geometrical object just by having a distance function on the plane. Definition 1.6 (Circle). If 𝑂 is a point and 𝑟 is a positive real number, the circle with center point 𝑂 and radius 𝑟 is the set of points 𝑃 with distance ‖𝑂𝑃‖ = 𝑟. This is a familiar definition, but we will see in the exercises that, with an unfamiliar distance, the shape of the circle is not what we think of as circular.

Exercises and Explorations

9

Exercises and Explorations 1. (Taxicab Circles). This problem calls on your informal background knowledge and not anything we have proved so far. On a sheet of graph paper with (𝑥, 𝑦)coordinates, define the distance 𝑑(𝑃, 𝑄) to be |𝑥1 − 𝑥2 | + |𝑦1 − 𝑦2 | for 𝑃 = (𝑥1 , 𝑦1 ) and 𝑄 = (𝑥2 , 𝑦2 ). This is called the taxicab distance because it is the distance to go from 𝑃 to 𝑄 by traveling on streets, if the graph paper is viewed as a grid of city streets parallel to the axes. Pick a point 𝐴 on the graph paper and draw a circle of radius 5 using the taxicab distance in the circle definition. Draw some other circles and see how they are related. What transformations will map such circles to each other? 2. (Rigid Motions and Circles). Prove that a rigid motion of the plane maps a circle to a circle. (Actually an isometry also maps a circle to a circle.) 3. (Informal Exploration). Draw by tracing around a jar lid or other method to obtain a pair of identical shapes like the one on the left and another pair like the one on the right in Figure 7. Experiment with each pair. Superimpose one of the shapes onto the other as many ways as you can. What number of rigid motions did you find? How would you describe those rigid motions? How do the numbers and nature of the rigid motions for the two pairs of shapes compare?

Figure 7. Two Figures for Experiments

4. (Geometry of the Real Line). The real numbers ℝ have a geometry, with the distance between 𝑎 and 𝑏 defined as |𝑏 − 𝑎|. A transformation 𝐺 of the real numbers is an isometry if it preserves distance; in other words, |𝐺(𝑎) − 𝐺(𝑏)| = |𝑎 − 𝑏| for any 𝑎 and 𝑏. (a) For 𝐺, an isometry of ℝ, suppose 𝐺(1) = 8 and 𝐺(3) = 10. What is 𝐺(2)? What is 𝐺(9)? What is 𝐺(𝑡) for any 𝑡? (b) For 𝐻, an isometry of ℝ, suppose 𝐻(1) = 8 and 𝐻(3) = 6. What is 𝐻(9)? What is 𝐻(9/2)? What is 𝐻(𝑡) for any 𝑡? (c) Show that any transformation 𝐹 of ℝ that preserves distance must have one of these two formulas: 𝐹(𝑡) = 𝑡 + 𝑘 or 𝐹(𝑡) = −𝑡 + 𝑘, for some constant 𝑘. (d) One of the forms of 𝐹 above always has a fixed point 𝑐 so that 𝐹(𝑐) = 𝑐. For which form is this true, and what number is 𝑐?

10

1. Congruence and Rigid Motions

5. (Product Inverse). If 𝑆 and 𝑇 are transformations, show that the inverse of 𝑇𝑆 is 𝑆 −1 𝑇 −1 . Prove this with function notation and/or any other way you feel is an effective way to convince others. (Hint: If you put on socks and shoes, in what order do you take them off?) 6. (Proof Notation). If you feel it would be helpful, write one of the proofs above using some notation (function or primed) that was not used above. For example, prove that if 𝑇 is a rigid motion, then 𝑇 −1 preserves angle measure. 7. (Congruence Sets). Suppose 𝐹 and 𝐺 are two figures in the plane. Let 𝐶𝐹 be the set of figures congruent to 𝐹 and let 𝐶𝐺 be the set of figures congruent to 𝐺. Prove that if 𝐹 ≅ 𝐺, then 𝐶𝐹 = 𝐶𝐺 , but if 𝐹 is not congruent to 𝐺, then 𝐶𝐹 and 𝐶𝐺 have no elements in common.

Chapter 2

Axioms for the Plane

It is now time to insert a little geometry into our plane. The plane is intended to be an abstract model of the geometry of a flat surface such as a table or a piece of paper, with the difference that our plane will extend without end. So far we have points and some measures (with no details). We have also defined rigid motions, but except for the identity mapping, we have not yet claimed that any rigid motions actually exist. So there are things to do. In this chapter, we state the six axioms for plane geometry that will be used in this book. The first four are closely related to the well-known Ruler and Compass axioms of G. D. Birkhoff [2], [3]: • Incidence Axiom for points and lines • Ruler Axiom concerning distance measure • Protractor Axiom concerning angle measure • Plane Separation Axiom concerning boundaries and orientation We will explore consequences of these first four axioms in this chapter. From these axioms one can define some elementary but important geometrical concepts and prove a few key theorems. At the end of the chapter, we complete the set of axioms by stating the remaining two axioms: • Reflection Axiom • Dilation Axiom The Reflection Axiom will add rigid motions to our geometry. Significant geometrical connections between distance and angle measure will appear in Chapter 3 as consequences of the Reflection Axiom. This will lead to triangle congruence theorems in Chapter 4. In later chapters the importance of the Dilation Axiom will become evident. 11

12

2. Axioms for the Plane

Figure 1. A Ruler Measuring Distance in the Plane

Incidence Axiom The first axiom is very simple. The informal version is “two points determine a line”. Axiom 1 (Incidence). The plane consists of a nonempty set of elements called points and a nonempty set of subsets called lines. For any two distinct points, there is exactly one line containing the two points; it will be denoted 𝐴𝐵. As a consequence of this, two distinct lines cannot intersect in more than one point. This already rules out our model looking like the sphere. We might think of this as the unmarked straightedge axiom: for any two points on a paper, we can draw a line with a straightedge through the two points.

Distance and the Ruler Axiom The Ruler Axiom is an existence axiom for a distance measure and for “rulers” that provide distance measurement on a line by means of real numbers. Such a ruler is an abstraction of a marked straightedge. Axiom 2 (Ruler). The plane has a distance function. • Distance is a function that maps any two points 𝐴 and 𝐵 in the plane to a nonnegative real number ‖𝐴𝐵‖. • Every line 𝑚 in the plane has a ruler, a mapping of the real numbers ℝ onto 𝑚 so that distances are preserved. Another way of describing a ruler is that it is an isometry from ℝ to 𝑚. Explicitly, if 𝜌 is a ruler with 𝜌(𝑎) = 𝐴 and 𝜌(𝑏) = 𝐵, then the distance ‖𝐴𝐵‖ equals the standard distance between two real numbers, the absolute value |𝑎 − 𝑏|.

Distance and the Ruler Axiom

13

Remark 2.1. This axiom implies a number of important properties of points, lines, and distance. • For any 𝐴, ‖𝐴𝐴‖ = 0. This is true since, for 𝐴 on a line 𝑚 with ruler value 𝜌(𝑎) = 𝐴, the distance ‖𝐴𝐴‖ = |𝑎 − 𝑎| = 0. • A ruler is a one-to-one mapping, for if 𝜌 is a ruler on a line 𝑚 with 𝜌(𝑎) = 𝜌(𝑏) = 𝐴, then |𝑎 − 𝑏| = ‖𝐴𝐴‖ = 0, so 𝑎 = 𝑏. • For any two distinct points 𝐴 and 𝐵, the distance ‖𝐴𝐵‖ > 0. For a ruler 𝜌 on 𝐴𝐵, with 𝜌(𝑎) = 𝐴 and 𝜌(𝑏) = 𝐵 for distinct 𝑎 and 𝑏, the distance ‖𝐴𝐵‖ = |𝑎 − 𝑏| > 0. • There are an infinite number of points on each line, and the distances on the line are unbounded. • If 𝜌(𝑥) is a ruler, then so is 𝜌(𝑥 + 𝑘) or 𝜌(−𝑥 + 𝑘) for any real 𝑘. Therefore, any point on the line can have coordinate 0. The last item is true because, if we apply any distance-preserving transformation of the real numbers before mapping them by the ruler, this composition is still a ruler since the distance relations are still preserved. For example, if 𝜌 is a ruler for 𝑚, let the function 𝜌5 be defined as 𝜌5 (𝑥) = 𝜌(𝑥 + 5). Then 𝜌5 is also a ruler for 𝑚, since for any real numbers 𝑎 and 𝑏, ‖𝜌5 (𝑎)𝜌5 (𝑏)‖ = ‖𝜌(𝑎 + 5)𝜌(𝑏 + 5)‖, and this equals |(𝑎 + 5) − (𝑏 + 5)| = |𝑎 − 𝑏| since 𝜌 is a ruler. Exercise 1 explores these transformations of the real line. The real numbers assigned to each point of 𝑚 by the inverse map are the coordinates defined by the ruler. This inverse map will occasionally be designated as a function from points to numbers such as 𝑎 = 𝑥(𝐴) and 𝑏 = 𝑥(𝐵), but usually it will simply be stated that the real numbers 𝑎 and 𝑏 correspond to the points 𝐴 and 𝐵 in 𝑚. These real numbers can be used to prove things about points on a line 𝑚. For example, given a point 𝐴 on 𝑚 and any distance 𝑑, there are two points 𝐵 and 𝐶 on 𝑚 at distance 𝑑, with ‖𝐴𝐵‖ = ‖𝐴𝐶‖ = 𝑑, since for any real number 𝑎, there are two numbers 𝑏 = 𝑎 + 𝑑 and 𝑐 = 𝑎 − 𝑑 with |𝑏 − 𝑎| = |𝑐 − 𝑎| = 𝑑. Segments. From the Ruler Axiom, for distinct points 𝐴 and 𝐵 we can define 𝐴𝐵, the segment with endpoints 𝐴 and 𝐵. For the next few paragraphs, let the numbers 𝑎, 𝑏, 𝑐 correspond to the points 𝐴, 𝐵, 𝐶 on 𝐴𝐵. Segment Definition by Ruler: The segment 𝐴𝐵 is the set of points 𝐶 of 𝐴𝐵 such that the number 𝑐 belongs to the interval with endpoints 𝑎 and 𝑏: either 𝑎 ≤ 𝑐 ≤ 𝑏 or 𝑏 ≤ 𝑐 ≤ 𝑎. Equivalently, |𝑐 − 𝑎| + |𝑏 − 𝑐| = |𝑏 − 𝑎|. Segment Definition by Distance: From the absolute value equation above, one sees that the segment 𝐴𝐵 is the set of points 𝐶 of 𝐴𝐵 with ‖𝐴𝐶‖ + ‖𝐵𝐶‖ = ‖𝐴𝐵‖. Segment Interior Point: A point 𝐶 is an interior point of 𝐴𝐵, or is between 𝐴 and 𝐵, if it is in 𝐴𝐵 but is not one of the endpoints. Midpoint: The midpoint 𝑀 of 𝐴𝐵 is the point on the segment equidistant from 𝐴 and 𝐵. Hence, 2‖𝐴𝑀‖ = ‖𝐴𝐵‖.

14

2. Axioms for the Plane

Convexity: A set 𝑆 is convex if, for any points 𝐴 and 𝐵 in 𝑆, the segment 𝐴𝐵 is also contained in 𝑆. ⃗ with endpoint 𝐴 in the direcRays. For distinct points 𝐴 and 𝐵, there is a ray 𝐴𝐵 tion of 𝐵. We continue with the assumption that the numbers 𝑎, 𝑏, 𝑐 correspond to the points 𝐴, 𝐵, 𝐶 on 𝐴𝐵. ⃗ if 𝐴 is not between 𝐵 and 𝐶. In other Ray Definition: A point 𝐶 on 𝐴𝐵 is in 𝐴𝐵 words, 𝑐 ≥ 𝑎 if 𝑏 > 𝑎 and 𝑐 ≤ 𝑎 if 𝑏 < 𝑎. Ray Interior Point: The set of all points except the endpoint is the set of interior points of the ray. ⃗ is the ray consisting of endpoint 𝐴 and the Opposite Ray: The opposite ray of 𝐴𝐵 ⃗ points 𝐶 on 𝐴𝐵 not in 𝐴𝐵. In other words, 𝑐 ≥ 𝑎 if 𝑏 < 𝑎 and 𝑐 ≤ 𝑎 if 𝑏 > 𝑎. Directed Line: A choice of ray determines a direction or orientation of the line containing it. Two rays in a line define the same direction if their intersection is a ray; this occurs if one ray is contained in the other, so a line has two possible directions. Polygons. Now we can define polygons. Definition 2.2. A polygon 𝑃1 𝑃2 . . . 𝑃𝑛 is the figure consisting of 𝑛 distinct points, the vertices 𝑃𝑖 , and 𝑛 sides, the segments 𝑃1 𝑃2 , 𝑃2 𝑃3 , . . . , 𝑃𝑛−1 𝑃𝑛 , 𝑃𝑛 𝑃1 , such that (1) no interior point of a side is an intersection point of two sides and (2) any three consecutive vertices are noncollinear. The sequences of three consecutive vertices include 𝑃𝑛−1 , 𝑃𝑛 , 𝑃1 and 𝑃𝑛 , 𝑃1 , 𝑃2 . This definition excludes self-intersecting polygonal figures and nonclosed polygonal paths, which can also be interesting but are not included as polygons. The most familiar polygons have three sides (triangles), four sides (quadrilaterals), five sides (pentagons), and six sides (hexagons).

Protractor Axiom and Angles Next, we will consider how to measure angles. Alas, even in informal geometry this is a bit more complicated than measuring along a line. ⃗ ⃗ and 𝐴 Definition 2.3. An angle with vertex 𝐴 consists of two rays 𝐴𝐵 𝐶, denoted ∠𝐵𝐴𝐶, where 𝐴, 𝐵, 𝐶 are not collinear. ⃗ ⃗ and 𝐴 Two opposite rays 𝐴𝐵 𝐶 are said to form a straight angle and two equal ⃗ ⃗ rays 𝐴𝐵 and 𝐴𝐶 are said to form a zero angle even though these are not angles by the definition. This terminology is so well established that we will use it. But to be clear, if we say “angle” or refer to ∠𝐵𝐴𝐶 without explictly including zero or straight angles, this means an angle by the definition and does not include zero or straight angles. ⃗ Axiom 3 (Protractor). The plane has an angle measure that maps any two rays 𝐴𝐵 ⃗ and 𝐴 𝐶 with common endpoint to a real number 𝑚∠𝐵𝐴𝐶.

Protractor Axiom and Angles

15

Figure 2. A Wooden Protractor Measuring from 0 to 180

• The angle measure maps any angle ∠𝐵𝐴𝐶 to a positive number 𝑚∠𝐵𝐴𝐶, with 0 < 𝑚∠𝐵𝐴𝐶 < 180. ⃗ ⃗ =𝐴 • In addition, when 𝐴𝐵 𝐶, the angle measure of this zero angle is 𝑚∠𝐵𝐴𝐶 = 0. ⃗ ⃗ and 𝐴 If the rays 𝐴𝐵 𝐶 are opposite rays, the angle measure of this straight angle is 𝑚∠𝐵𝐴𝐶 = 180. • For any point A, there exists a protractor at A. This is a mapping 𝛿 from the real numbers onto the rays with endpoint 𝐴, such that 𝛿(𝑡) = 𝛿(𝑡 + 360) for every 𝑡, with ⃗ ⃗ and 𝛿(𝑐) = 𝐴 the property that 𝑚∠𝐵𝐴𝐶 = |𝑏 − 𝑐| when 𝛿(𝑏) = 𝐴𝐵 𝐶, provided that |𝑏 − 𝑐| ≤ 180. Note. A protractor is also called a polar angle parametrization at 𝐴. The degree is the unit of angle measure used here, since this is more convenient for geometry than radians. However, the number 360 could be replaced by another positive number to provide a different choice of unit. Example 2.4. Let’s try to connect this angle measurement with our geometric intuition. Imagine that you are holding a compass (the magnetic directional kind, not the kind that draws circles). The directions on the compass face are numbered from 0 to 360, with the directions precisely at 0 and 360 being the same direction. Suppose this 0 direction points rightward (eastward). If you are standing at point 𝐴, then the rays with endpoint 𝐴 indicate the directions ⃗ is numbered 40, sort of a northeast directhat you can be facing. Suppose direction 𝐴𝐵 ⃗ tion, and a direction 𝐴𝐶 is numbered 250, sort of southwesterly. Then we could try to compute the angle measure of ∠𝐵𝐴𝐶 by taking the difference between the numbers: |250 − 40| = 210. But this number is greater than 180, so it will not work as an angle ⃗ measure. However, the number 250 − 360 = −110 also corresponds to the ray 𝐴 𝐶, so we can try again with this number: | − 110 − 40| = 150. This is between 0 and 180, so 150 is the measure of ∠𝐵𝐴𝐶. Polar Angles. Using protractors directly is somewhat awkward; the inverse map from rays to real numbers is often more convenient. For a given protractor 𝛿 at 𝐴, we ⃗ has polar angle 𝑡 if 𝑡 is one of the real numbers for will say the point 𝐵 (or the ray 𝐴𝐵) ⃗ which 𝛿(𝑡) = 𝐴𝐵.

16

2. Axioms for the Plane

Two distinct rays never have the same polar angle. By Axiom 3, if the polar angles were the same, the angle measure of the rays would be zero and the rays would form a zero angle. ⃗ ⃗ ⃗ and 𝐴 ⃗ − 𝜃(𝐴 If two values of the polar angles of 𝐴𝐵 𝐶 satisfy |𝜃(𝐴𝐵) 𝐶)| < 180, we say the polar angles measure ∠𝐵𝐴𝐶. ⃗ will be denoted by 𝜃(𝐴𝐵), ⃗ even though the value A choice of polar angle for 𝐴𝐵 of this angle is ambiguous without context, since adding 360 to this number produces another polar angle. To get around this problem in a given situation and to choose a polar angle unambiguously, we can restrict the protractor 𝛿 to a half-open segment in the real numbers of length 360. For example, given the half-open interval [0, 360), which is the set of real numbers 𝑡, with 0 ≤ 𝑡 < 360, any protractor at 𝐴 will define a one-to-one correspondence be⃗ can be defined as tween this interval and the rays with endpoint 𝐴. Therefore, 𝜃(𝐴𝐵) ⃗ will be the the inverse of the restriction of 𝛿 to this interval, or in other words, 𝜃(𝐴𝐵) ⃗ < 360. polar angle of 𝐵 with 0 ≤ 𝜃(𝐴𝐵) Another choice of interval that is sometimes convenient is (−180, +180], the real numbers −180 < 𝑡 ≤ 180. More generally, one can choose unambiguous polar angles 𝜃 either in the interval 𝑎 ≤ 𝜃 < 𝑎 + 360 or the interval 𝑎 − 180 ≤ 𝜃 < 𝑎 + 180, where 𝑎 is any chosen real number. If 𝛿(𝑡) is a protractor function at 𝐴, so are 𝛿(𝑡+𝑘) and 𝛿(−𝑡+𝑘), since these changes of real numbers do not affect the absolute value of the differences. This provides a ⃗ convenience for computation: one may assign polar angle 0 to any convenient ray 𝐴𝐵. Some Angle Terminology. There are a number of terms that specify a relationship between two angles or describe the size of an angle. Supplementary Angles: Two angles are supplementary if the sum of their angle measures is 180. The vertices of the two angles do not have to be the same, but if ⃗ ⃗ and 𝐴 𝐴𝐵 𝐶 are opposite rays and 𝐷 is a point not on 𝐴𝐵, then ∠𝐵𝐴𝐷 and ∠𝐶𝐴𝐷 are supplementary angles. ⃗′ is the Vertical Angles: Two angles ∠𝐵𝐴𝐶 and ∠𝐵 ′ 𝐴𝐶 ′ are vertical angles if 𝐴𝐵 ′ ⃗ is the opposite ray of 𝐴 ⃗ ⃗ and 𝐴𝐶 opposite ray of 𝐴𝐵 𝐶. These vertical angles have the same angle measure. Right Angle: A right angle is an angle with measure 90. Two lines 𝐴𝐵 and 𝐴𝐶 are perpendicular if ∠𝐵𝐴𝐶 is a right angle. Acute Angle: An acute angle is an angle with measure less than 90, the measure of a right angle. Obtuse Angle: An obtuse angle is an angle with measure greater than 90, the measure of a right angle. Notice that a supplementary angle of a right angle is also a right angle. Therefore, two perpendicular lines form four right angles.

Plane Separation

17

Definition 2.5. The perpendicular bisector of a segment 𝐴𝐵 is the line through the midpoint of 𝐴𝐵 perpendicular to 𝐴𝐵. The perpendicular bisector of a segment always exists, since the midpoint 𝑀 exists by the Ruler Axiom and a right angle ∠𝐷𝑀𝐴 exists by the Protractor Axiom.

Plane Separation There is one more axiom that is important for relationships of order and separation. It ensures that our model really acts like a two-sided two-dimensional space. Axiom 4 (Plane Separation). For any line 𝑚, any point not on the line belongs to one of two disjoint sets called the half-planes of the line. • Points 𝑃 and 𝑄 are in different half-planes when 𝑃𝑄 intersects 𝑚. They are in the same half-plane if 𝑃𝑄 does not intersect 𝑚. ⃗ = 0, the points 𝑃 for which • Let 𝑚 = 𝐴𝐵. If a protractor is chosen so that 𝜃(𝐴𝐵) ⃗ ⃗ 0 < 𝜃(𝐴𝑃) < 180 lie in one half-plane, and the 𝑄 for which −180 < 𝜃(𝐴 𝑄) < 0 are in the other half-plane.

Polar angle = +125

P

A

B Q

Polar angle = -140

R

Polar angle = -15

Figure 3. Point 𝑃 in the Half-plane Opposite the Half-plane of 𝑄 and 𝑅

Note. We refer to the two half-planes as opposite one another. If two points are in different half-planes, we say they are on opposite sides of the line. Definition 2.6. Given an angle ∠𝐵𝐴𝐶, the interior of the angle is the intersection of the half-plane of 𝐴𝐶 containing 𝐵 and the half-plane of 𝐴𝐵 containing 𝐶. For △𝐵𝐴𝐶, the interior of the triangle is the intersection of the interiors of the three angles ∠𝐵𝐴𝐶, ∠𝐴𝐶𝐵, ∠𝐶𝐵𝐴. Definition 2.7. A polygon 𝑃1 𝑃2 . . . 𝑃𝑛 is a convex polygon if, for every line 𝑃𝑘 𝑃𝑘+1 that is a side extended, all the other vertices are in one half-plane of the line. The interior of a convex polygon is the intersection of all these half-planes It follows that in a convex polygon, for any vertex angle, the interior of the polygon is contained in the interior of the angle. Since half-planes are convex, the interior of an angle or the interior of a triangle or other convex polygon is convex also.

18

2. Axioms for the Plane

The definition of the interior of a nonconvex polygon in general is surprisingly technical and will not be attempted here, since it would be an inappropriate digression in an introductory work. For a polygon with only a few sides, such as a quadrilateral, it is not hard to dissect the figure into triangles or other convex polygons and then take the interior of the convex polygons, plus common edges, to be the interior. But in general one must account for polygons with hundreds of thousands of sides. And this is not just some sort of “abstract mathematics” problem, for such triangles really appear in computer graphics and animation, and they can twist into amazingly complicated shapes. One algorithm for determining which points are inside a polygon is to intersect the polygon with a line passing through sides but not vertices. Then the points of intersection of sides with the line divide the line into alternating segments of outside points and inside points. But just to prove that the number of intersection points is even, so that this algorithm works, is complicated and to show that the interior is well-defined is even more so.

C

A

D B

Figure 4. Intersection of 𝐶𝐷 with Ray Through Interior Point 𝐷

Theorem 2.8 (Addition of Angle Measure). A point 𝐷 is in the interior of ∠𝐵𝐴𝐶 if ⃗ ⃗ ⃗ and 𝐴 and only if some polar angle of 𝐴 𝐷 is between the polar angles of 𝐴𝐵 𝐶, provided those angles are chosen so that they measure ∠𝐵𝐴𝐶. In this case, 𝑚∠𝐵𝐴𝐷 + 𝑚∠𝐷𝐴𝐶 = 𝑚∠𝐵𝐴𝐶. ⃗ ⃗ = 𝑏, 𝜃(𝐴 ⃗ Proof. Let 𝜃(𝐴𝐵) 𝐶) = 𝑐, 𝜃(𝐴 𝐷) = 𝑑, with |𝑏 − 𝑐| < 180. Then we need to prove that 𝐷 is an interior point if and only if some choice of 𝑑 is between 𝑏 and 𝑐. Assume without loss of generality that 𝑏 is the smaller of 𝑏 and 𝑐, so 𝑏 < 𝑐 < 𝑏+180. 𝐷 is in the half-plane of 𝐴𝐵 containing 𝐶 when for some choice of 𝑑, 𝑏 < 𝑑 < 𝑏 + 180. 𝐷 is also in the half-plane of 𝐴𝐶 containing 𝐵 when 𝑑 < 𝑐. Thus, 𝑏 < 𝑑 < 𝑐 if and only if 𝐷 is in the interior of the angle. The angle measures are 𝑚∠𝐵𝐴𝐷 = 𝑑 − 𝑏, 𝑚∠𝐷𝐴𝐶 = 𝑐 − 𝑑, and 𝑚∠𝐵𝐴𝐶 = 𝑐 − 𝑏. This proves the final statement. □ ⃗ ⃗ ⃗ and 𝐴 If 𝐷 is in the interior of ∠𝐵𝐴𝐶, one also says 𝐴 𝐷 is between 𝐴𝐵 𝐶. All the points of this ray have polar angle 𝑑, so they are interior to ∠𝐵𝐴𝐶. ⃗ Definition 2.9. The angle bisector of ∠𝐵𝐴𝐶 is the ray 𝐴 𝐷 for which 𝑚∠𝐵𝐴𝐷 = 𝑚∠𝐷𝐴𝐶 = (1/2)𝑚∠𝐵𝐴𝐶. Using the notation from the proof of the theorem, the angle bisector exists and has polar angle 𝑑 = (𝑏 + 𝑐)/2.

Rigid Motions and Lines

19

Theorem 2.10 (Crossbar). Given any angle ∠𝐵𝐴𝐶: (a) The interior points of 𝐵𝐶 are interior to ∠𝐵𝐴𝐶. ⃗ (b) For any point 𝐷 interior to ∠𝐵𝐴𝐶, the ray 𝐴 𝐷 intersects the segment 𝐵𝐶. ⃗ are contained Proof. Statement (a) is true since 𝐵 is on 𝐴𝐵, so the interior points of 𝐵𝐶 ⃗ are contained in the half-plane of 𝐴𝐵 containing 𝐶. Likewise, the interior points of 𝐶𝐵 in the half-plane of 𝐴𝐶 containing 𝐵. The set of interior points of 𝐵𝐶 is the intersection of the interior points of these rays. To show (b), continue to use the polar angles 𝑏, 𝑐, and 𝑑 as above. Since 𝑏 < 𝑑 < 𝑐, we see that 𝐵 and 𝐶 are on opposite sides of 𝐴𝐷. Thus, 𝐵𝐶 intersects 𝐴𝐷 at a point in ⃗ the interior of ∠𝐵𝐴𝐶, therefore, at a point of 𝐴 𝐷. □

Rigid Motions and Lines We will now use our ruler and protractor axioms to prove some essential properties for doing geometry with rigid motions. Theorem 2.11 (Line Images). If T is a rigid motion, the T-image of a line, segment, or ray is, respectively, a line, segment, or ray. Proof. If 𝐴, 𝐵, 𝐶 are distinct collinear points on line 𝑛, then 𝑚∠𝐵𝐴𝐶 = 0 or 𝑚∠𝐵𝐴𝐶 = 180. Then 𝑚∠𝑇(𝐵)𝑇(𝐴)𝑇(𝐶) also equals 0 or 180. This proves that the image of a line is a line. Segments and rays are subsets of lines defined by distance relations. A point 𝐶 on 𝑛 is in segment 𝐴𝐵 if and only if ‖𝐴𝐶‖ + ‖𝐵𝐶‖ = ‖𝐴𝐵‖. This equation still holds true for the image points, so 𝑇(𝐶) is in the segment 𝑇(𝐴)𝑇(𝐵). Similar reasoning for rays ⃗ is ⃖⃖⃖⃖⃖⃖⃖⃖⃗ shows that the image of 𝐴𝐵 𝑇(𝐴)𝑇(𝐵). □

A

B

Figure 5. 𝐴 and 𝐵 fixed — All Line Points Fixed

This next theorem will play a key role in classifying rigid motions. Theorem 2.12 (Two Fixed Points). If a rigid motion T fixes two distinct points 𝐴 and 𝐵, then it fixes all points of 𝐴𝐵. Proof. To say 𝑇 fixes a point 𝐴 means that 𝑇(𝐴) = 𝐴. For a point 𝐶 on 𝐴𝐵, let 𝐷 = 𝑇(𝐶). If 𝐷 is distinct from 𝐶, then ‖𝐷𝐴‖ = ‖𝑇(𝐶)𝑇(𝐴)‖ = ‖𝐶𝐴‖, so 𝐴 is the midpoint of 𝐶𝐷. The same reasoning shows that 𝐵 is also the midpoint of 𝐶𝐷, so 𝐴 = 𝐵, a contradiction. So 𝐷 must be 𝐶. □

20

2. Axioms for the Plane

The Other Axioms The last two axioms are stated here. What follows from assuming these axioms will be investigated in coming chapters. The Reflection Axiom is an existence axiom for a very particular kind of rigid motion, not described in much detail in the axiom. The geometry of this rigid motion will be described in Chapter 3. Then by composing the transformations for different lines, in coming chapters we will prove many theorems of geometry and will generate familiar transformations such as rotations and translations. Already in Chapter 3 we will begin to see consequences for the geometry of triangles and other figures. Axiom 5 (Reflection). For any line 𝑚, there exists a rigid motion distinct from the identity transformation that fixes the points of 𝑚. We note by Theorem 2.12 that it is actually sufficient that this rigid motion fixes two points. The last of the six axioms concerns the properties of dilations. These transformations are not rigid motions but are transformations that will be defined later as similitudes. Definition 2.13. For point 𝐴 and a real number 𝑘 > 0, the dilation centered at 𝐴 with dilation ratio 𝑘, denoted 𝒟𝐴,𝑘 , is a transformation of the plane defined as follows: ⃗ (a) 𝒟𝐴,𝑘 (𝐴) = 𝐴; (b) if 𝑋 ≠ 𝐴, 𝑋 ′ = 𝒟𝐴,𝑘 (𝑋) is the point on 𝐴 𝑋 with ‖𝐴𝑋 ′ ‖ = 𝑘‖𝐴𝑋‖. Axiom 6 (Dilation). For every point 𝐴 and every 𝑘 > 0, the dilation 𝒟𝐴,𝑘 preserves angle measure and scales every distance by 𝑘: for all 𝑃′ = 𝒟𝐴,𝑘 (𝑃) and 𝑄′ = 𝒟𝐴,𝑘 (𝑄), ‖𝑃 ′ 𝑄′ ‖ = 𝑘‖𝑃𝑄‖. Note the difference in the nature of these two axioms. The Reflection Axiom is an existence axiom while the Dilation Axiom asserts properties of an existing transformation that is already defined. We will see in Chapter 7 that the Dilation Axiom implies the famous Euclidean Parallel Postulate as a theorem. In contrast, everything proved using only the first five axioms is true not only in Euclidean geometry, but in hyperbolic non-Euclidean geometry as well. For many readers, this will not be important, but anyone thinking about proving theorems that are true in both Euclidean and non-Euclidean geometry will note that, except for a very few well-identified and isolated exceptions, only the first five axioms are used in proofs until the end of Chapter 6. In Chapter 8, dilations will be central in the study of similarity. At that time, the definition of dilation will be extended to negative 𝑘.

Exercises and Explorations

21

Exercises and Explorations 1. (Rulers and Protractors). Let 𝜌(𝑥) be a ruler for a line 𝑚. It was claimed in Remark 2.1 that for a constant real number 𝑘, both 𝜌(𝑥 + 𝑘) and 𝜌(−𝑥 + 𝑘) are also rulers. Later it was similarly noted that if 𝛿(𝑡) is a protractor function at 𝐴, so are 𝛿(𝑡 + 𝑘) and 𝛿(−𝑡 + 𝑘). (a) Show that 𝜌(𝑥 + 𝑘) satisfies the definition of a ruler. Next show that 𝜌(−𝑥) is a ruler. Combine the two results to prove the statement in the remark. You may wish to supplement words and symbols with drawings. (b) Show how almost identical reasoning justifies the statement about protractors. 2. (Midpoints). Let points 𝐴 and 𝐵 correspond by a ruler on line 𝐴𝐵 to 𝑎 and 𝑏. (a) What real number 𝑚 corresponds to the midpoint 𝑀 of 𝐴𝐵? (b) If 𝐷 is a point corresponding to real number 𝑑 and if 𝐵 is the midpoint of 𝐴𝐷, what number is 𝑑? 3. (Angle Bisectors). From Definition 2.9, an angle bisector is a ray that divides an angle into two angles of equal measure. (a) For a protractor at 𝐴, let the polar angles of 𝐵, 𝐶, and 𝐷 be 𝑏, 𝑐, 𝑑. Assuming ⃗ |𝑏 − 𝑐| < 180 and 𝐷 is interior to ∠𝐵𝐴𝐶, for what value of 𝑑 will 𝐴 𝐷 bisect this angle? In other words, for what 𝐷 is 𝑚∠𝐵𝐴𝐷 = 𝑚∠𝐷𝐴𝐶? (b) Two lines 𝐴𝐵 and 𝐴𝐶 form four angles at 𝐴. Explain how the four angle bisectors are related. 4. (Relationships on a Line). Let points 𝐴 and 𝐵 correspond by a ruler on line 𝐴𝐵 to 𝑎 and 𝑏. (a) If 𝑎 = 3 and 𝑏 = 5, what number 𝑐 corresponds to the point 𝐶 on the line with ‖𝐴𝐶‖ = 7 and ‖𝐵𝐶‖ = 9? (b) If points 𝐴, 𝐵 on line 𝑚 correspond by a ruler to real numbers 3 and 5, what number 𝑐 corresponds to the point 𝐶 = 𝒟𝐴,5 (𝐵)? (c) What is a general formula for 𝑐 in terms of 𝑎 and 𝑏 if 𝐶 = 𝒟𝐴,𝑘 (𝐵)? 5. (Fixed points on ℝ). Suppose that 𝑓 is a map from the real numbers ℝ to ℝ that preserves distance. (a) If 𝑓(3) = 3, what are the possible values of 𝑓(7)? (b) If 𝑓(3) = 3, what are the possible values of 𝑓(𝑡), for any 𝑡? (c) If a point 𝑎 is a fixed point of 𝑓 (i.e., 𝑎 = 𝑓(𝑎)), what are the possible values of 𝑓(𝑡) for any other number 𝑡? (d) Based on your answers, if 𝑓 has two fixed points, what can you say about 𝑓(𝑡)? 6. (Plane Separation and Triangles). The plane separation axiom allows one to prove some facts that are basic and visually very believable but rather technical. One may choose to believe them without going through the proofs. However, they do provide good practice in reasoning with the Plane Separation Axiom. (a) In △𝐴𝐵𝐶, let 𝐷 be an interior point of 𝐵𝐶 and let 𝐸 be an interior point of 𝐶𝐴. Prove that 𝐴𝐷 and 𝐵𝐸 intersect at a point 𝐺, an interior point of the triangle.

22

2. Axioms for the Plane

(b) (Pasch Axiom). There are other axioms of plane separation that are used in place of Axiom 4. One is called the Pasch Axiom: if a line 𝑚 intersects side 𝐴𝐵 of △𝐴𝐵𝐶 at one point 𝐷, an interior point of side 𝐴𝐵, then 𝑚 also intersects either 𝐵𝐶 or 𝐶𝐴. Prove this statement as a theorem using the axioms in this chapter. (c) (Quadrilateral Shapes). Sketch a few quadrilaterals, convex and nonconvex. In your examples, is there always at least one diagonal that separates the other two vertices into opposite half-planes of the diagonal? If this is always true, how can one prove it from the axioms? If there are two such diagonals, what does this imply about the convexity or nonconvexity of the quadrilateral? Suggestion: For each statement, begin by drawing possible cases and explain with informal reasoning what is possible before undertaking a more formal explanation.

Chapter 3

Existence and Properties of Reflections

In the previous chapters, we have been describing properties of rigid motions, but without the Reflection Axiom, there is nothing that tells us that any rigid motions actually exist (except for the identity)! This axiom states the existence of some rigid motions that will turn out to be line reflections, with the usual geometric description. However, this description is not stated in the axiom. A bare minimum of existence is promised; in fact it is not even asserted that there is only one rigid motion whose existence is declared. But using the strong restrictions on behavior mandated by preservation of distance and angle measure, we will see that these vaguely described rigid motions must behave in exactly one way. Recall the statement of the axiom: Axiom 5 (Reflection). For any line 𝑚, there exists a rigid motion distinct from the identity transformation that fixes the points of 𝑚.

Deducing the Properties of Reflections At this point, we do not have a geometrical description of this rigid motion. The proof that this rigid motion is the usual reflection in a line with the usual geometrical description will come from the examination of a key figure. Lemma 3.1 (Y Figure). Given a line 𝐴𝐵, take any point 𝐶 not on the line. Then there is exactly one other point 𝐷 in the plane so that ‖𝐴𝐶‖ = ‖𝐴𝐷‖ and 𝑚∠𝐵𝐴𝐶 = 𝑚∠𝐵𝐴𝐷. The points 𝐶 and 𝐷 are in opposite half-planes of 𝐴𝐵. ⃗ = Proof. Choose the protractor 𝛿 for rays with endpoint 𝐴 so that the polar angle 𝜃(𝐴𝐵) 0. In the following, we choose polar angles in the interval −180 < 𝜃 < +180. ⃗ for which 𝑚∠𝐵𝐴𝐸 = 𝑡 have polar angle Let 𝑚∠𝐵𝐴𝐶 = 𝑡. Then the only rays 𝐴𝐸 ⃗ ⃗ +𝑡 or −𝑡. There are only two such rays. One is 𝐴 𝐶 itself. Call the other one 𝐴𝐹. 23

24

3. Existence and Properties of Reflections

F D A

B

M C

E

Figure 1. Only One Match to the Angle and Distance of 𝐶

Let ‖𝐴𝐶‖ = 𝑘. There is exactly one point on each ray at distance 𝑘 from 𝐴. One of ⃗ be 𝐷. these points is 𝐶. Let the point on 𝐴𝐹 Then ‖𝐴𝐶‖ = ‖𝐴𝐷‖ = 𝑘, and 𝑚∠𝐵𝐴𝐶 = 𝑚∠𝐵𝐴𝐷 = 𝑡. From the Plane Separation Axiom, these points lie in opposite half-planes of 𝐴𝐵, so segment 𝐶𝐷 intersects 𝐴𝐵. □ Lemma 3.2 (Reflection and Perpendicular Bisectors). For any 𝐴𝐵, let 𝑇 be a rigid motion that fixes the points of 𝐴𝐵. If 𝐶 is not a fixed point of 𝑇, then 𝐶 is not on 𝐴𝐵 and 𝑇(𝐶) is the point 𝐷 constructed in the Y Figure Lemma. Moreover, the segment 𝐶𝐷 intersects 𝐴𝐵 at 𝑀, the midpoint of 𝐶𝐷, and 𝐴𝐵 is the perpendicular bisector of 𝐶𝐷. Proof. Let 𝐷 = 𝑇(𝐶), a point distinct from 𝐶. 𝑇 is a rigid motion that preserves angle measure and distance. Therefore, 𝑚∠𝐵𝐴𝐶 = 𝑚∠𝐵𝐴𝐷 and ‖𝐴𝐶|‖ = ‖𝐴𝐷‖. So from the Y Figure Lemma, 𝐷 is uniquely determined and on the opposite side of 𝐴𝐵 from 𝐶. Let 𝑀 be the intersection of 𝐶𝐷 and 𝐴𝐵. (If 𝑀 = 𝐴, interchange the roles of 𝐴 and 𝐵 in the proof.) Since 𝑀 is on 𝐴𝐵, 𝑇(𝑀) = 𝑀. Since 𝑇 is a rigid motion, ‖𝐶𝑀‖ = ‖𝐷𝑀‖. Therefore 𝑀 is the midpoint of 𝐶𝐷. In addition 𝑚∠𝐶𝑀𝐴 = 𝑚∠𝐷𝑀𝐴 because 𝑇 maps one angle to the other. Since the sum of these angle measures is 180, each of these angles is a right angle. This means that 𝐴𝐵 is the perpendicular bisector of 𝐶𝐷. □ Theorem 3.3 (Three Fixed Points). If 𝑇 is a rigid motion with three noncollinear fixed points, then 𝑇 is the identify transformation. Furthermore, if 𝑆 and 𝑇 are two rigid motions that agree at three noncollinear points, then the transformations are the same: 𝑆 = 𝑇. Proof. Let 𝐴, 𝐵, 𝐶 be noncollinear fixed points of 𝑇. Suppose there is a point 𝐺 that is not a fixed point, so that 𝑇(𝐺) = 𝐻 is distinct from 𝐺. Then by the Two Fixed Points Theorem, Theorem 2.12, for each of the three lines 𝐴𝐵, 𝐵𝐶, 𝐶𝐴, all the points are fixed. Then by Lemma 3.2, 𝐴𝐵 is the perpendicular bisector of 𝐺𝐻 and 𝐴𝐶 is also the perpendicular bisector of 𝐺𝐻, as shown in Figure 2.

Deducing the Properties of Reflections

25

H = T (G)

B

A

H = T (G )

G

C

Figure 2. Impossible Images of 𝐺 with 𝐴, 𝐵, 𝐶 Fixed

Thus 𝐴, 𝐵, 𝐶 are collinear with this assumption on 𝐺. Therefore, no such 𝐺 exists and 𝐺 = 𝑇(𝐺) for every 𝐺. In other words, 𝑇 is the identity transformation 𝐼. To prove the rest of the claim, let 𝑆(𝐴) = 𝑇(𝐴), 𝑆(𝐵) = 𝑇(𝐵), and 𝑆(𝐶) = 𝑇(𝐶). Then 𝑆 −1 𝑇 has 𝐴, 𝐵, 𝐶 as fixed points, so 𝑆 −1 𝑇 = 𝐼. Multiplying on the left by 𝑆 yields 𝑇 = (𝑆𝑆)−1 𝑇 = 𝑆(𝑆 −1 𝑇) = 𝑆𝐼 = 𝑆. □ This is a very powerful result. One can model it physically by tacking a piece of paper to a bulletin board. One tack allows a lot of swinging movement; two tacks allow no movement unless you remove the tacks, flip the paper over, and retack in the same holes. Three tacks allow no movement at all. Next is the theorem that completes the geometric description of reflection for every point. Theorem 3.4 (Line Reflection). Given any line 𝑚, there exists a unique rigid motion 𝑅 distinct from the identity that fixes the points of line 𝑚. For any point 𝐸 not on 𝑚, the image 𝐸 ′ = 𝑅(𝐸) is the point such that 𝑚 is the perpendicular bisector of 𝐸𝐸 ′ . We name this unique rigid motion reflection in line 𝑚, denoted ℛ𝑚 . Reflection in 𝐴𝐵 is denoted ℛ𝐴𝐵 .

E

C B' B

C'

E' Figure 3. Reflection in the Perpendicular Bisector of 𝐸𝐸 ′

Proof. By Theorem 3.3, an 𝑅 given by the Reflection Axiom fixes the points of 𝑚 but does not fix 𝐸 (or else it would be the identity). Therefore by Lemma 3.2, this description of 𝑅(𝐸) is correct for every 𝐸 not on 𝑚. □ Corollary. If line 𝑝 is perpendicular to line 𝑚, then ℛ𝑚 (𝑝) = 𝑝.

26

3. Existence and Properties of Reflections

Corollary. For any line 𝑚, the inverse of ℛ𝑚 is ℛ𝑚 . Proof. The perpendicular bisector of 𝐷𝐶 is the same as the perpendicular bisector of 𝐶𝐷. □ The geometric description of line reflection also tells us some useful properties concerning intersection and half-planes. Theorem 3.5 (Line Reflection and Half-Planes). Let 𝑚 be a line. (1) The reflection ℛ𝑚 maps each half-plane of 𝑚 to the opposite one. (2) If line 𝑛 is perpendicular to 𝑚, ℛ𝑚 maps each half-plane of 𝑛 to itself. Proof. By the Line Reflection Theorem, Theorem 3.4, any point 𝐴 not on 𝑚 is mapped to the opposite half-plane of 𝑚. This proves the first statement. For the second statement, let 𝐴 be a point of 𝑚 not on 𝑛 and let 𝐵 be a point in the same half-plane of 𝑛 as 𝐴, so 𝐴𝐵 does not intersect 𝑛. Since ℛ𝑚 (𝑛) = 𝑛, the reflection image of the segment 𝐴𝐵 cannot intersect 𝑛, so ℛ𝑚 (𝐴) = 𝐴 and ℛ𝑚 (𝐵) must be in the same half-plane. Since 𝐴 did not move, 𝐵 remains in its original half-plane of 𝑛 also. □ Theorem 3.6 (Perpendicular Parallels). Let 𝑚 be a line, and let 𝑝 and 𝑞 be distinct lines, each perpendicular to 𝑚. The lines 𝑝 and 𝑞 are parallel.

m

q p

Figure 4. Perpendicular to Line 𝑚 so Parallel

Proof. Suppose 𝑝 and 𝑞 intersect at a point 𝐴 not on 𝑚. Reflection in 𝑚 maps 𝑝 to 𝑝 and 𝑞 to 𝑞. Therefore, ℛ𝑚 (𝐴) is also a second, distinct intersection point of 𝑝 and 𝑞 in the opposite side of 𝑚 from 𝐴. This means 𝑝 and 𝑞 have two points in common, so they must be the same line. This contradicts the assumption that 𝑝 and 𝑞 intersect 𝑚 at distinct points. □ Theorem 3.7 (Dropping a Perpendicular). Let 𝑚 be a line. If 𝑃 is any point, there is a unique line 𝑝 through 𝑃 that is perpendicular to 𝑚. Proof. If 𝑃 is on 𝑚, then there is a unique line through 𝑃 that is perpendicular to 𝑚 by the Protractor Axiom. If 𝑃 is not on 𝑚, let 𝑃 ′ = ℛ𝑚 (𝑃). The line 𝑃𝑃 ′ is perpendicular to 𝑚 by the Line Reflection Theorem. There is only one such line, for any two lines perpendicular to 𝑚 are parallel so cannot both pass through 𝑃. □

Isosceles Triangles and Kites

27

The traditional name for construction of the perpendicular line to 𝑚 through 𝑃 is called Dropping a Perpendicular from a Point. Presumably one is dropping a vertical line like a plumb bob on a string to intersect a horizontal line. The intersection point of the perpendicular line with 𝑚 is called the foot of the perpendicular from 𝑃 to 𝑚.

Isosceles Triangles and Kites With the properties of line reflection established, we can prove all the usual relationships in any isosceles triangle. Recall from Definition 2.2 that a triangle is a threesided polygon. To spell this out: △𝐴𝐵𝐶 has three noncollinear vertices 𝐴, 𝐵, 𝐶 and three sides 𝐴𝐵, 𝐵𝐶, and 𝐶𝐴. The (interior) angles are the angles ∠𝐶𝐴𝐵, ∠𝐴𝐵𝐶, ∠𝐵𝐶𝐴, often referred to as “angle 𝐴”, etc. Definition 3.8 (Isosceles Triangle). An isosceles triangle is a triangle with two sides of equal length. If we label the two equal sides as 𝐴𝐵 and 𝐴𝐶, then the side 𝐵𝐶 is called the base of the triangle. The angles at 𝐵 and 𝐶 are called base angles. The angle at 𝐴 is called the apex angle. A triangle with three equal sides is called an equilateral triangle. This theorem points out a lot of congruent parts in an isosceles triangle. The theorem is labeled as a theorem about symmetry, since it shows that an isosceles triangle has a line of symmetry, which means that reflection in the line maps the figure onto itself. Theorem 3.9 (Isosceles Triangle Symmetry). Given an isosceles triangle 𝐴𝐵𝐶 with ‖𝐴𝐵‖ = ‖𝐴𝐶‖, the perpendicular bisector of the base 𝐵𝐶 passes through 𝐴 and bisects ∠𝐵𝐴𝐶. Also, the base angles ∠𝐵𝐶𝐴 and ∠𝐶𝐵𝐴 are congruent.

A C

M B Figure 5. Symmetry and Many Congruences in an Isosceles Triangle

⃗ as the angle bisector of ∠𝐵𝐴𝐶. Then the Y Figure Lemma, Proof. Construct 𝐴𝑃 Lemma 3.1, applies, and 𝐵 and 𝐶 are the mirror points in the lemma with respect to 𝐴𝑃. This means 𝐶 is the reflection of 𝐵 in 𝐴𝑃 and 𝐴𝑃 intersects 𝐵𝐶 at the midpoint 𝑀. Therefore the line 𝐴𝑃 = 𝐴𝑀 is the perpendicular bisector of the base 𝐵𝐶. The base angles are congruent since ℛ𝐴𝑀 maps one to the other. □

28

3. Existence and Properties of Reflections

The traditional word for a set of points satisfying a geometrical condition is locus. This next theorem says that the locus of points equidistant from 𝐴 and 𝐵 is the perpendicular bisector of 𝐴𝐵. Theorem 3.10 (Perpendicular Bisector as Locus). For any distinct points 𝐵 and 𝐶, the perpendicular bisector of 𝐵𝐶 is the set of points 𝑃 for which ‖𝑃𝐵‖ = ‖𝑃𝐶‖. Proof. To prove set equality, one must prove containment in two directions. If ‖𝑃𝐵‖ = ‖𝑃𝐶‖, then the isosceles theorem proves that 𝑃 is on 𝑚, the perpendicular bisector of 𝐵𝐶, except in the case when ∠𝐵𝑃𝐶 is a straight angle. But in this case, 𝑃 is the midpoint of 𝐵𝐶, so it is on 𝑚. For the converse, let 𝑄 be any point on 𝑚. Then ℛ𝑚 (𝐵) = 𝐶 and ℛ𝑚 (𝑄) = 𝑄, so ‖𝑄𝐵‖ = ‖𝑄𝐶‖ since ℛ𝑚 is a rigid motion. □ A quadrilateral was defined in Definition 2.2 as a polygon with four vertices. A quadrilateral 𝐴𝐵𝐶𝐷 has four vertex points 𝐴, 𝐵, 𝐶, 𝐷, no three of which are collinear, and four sides 𝐴𝐵, 𝐵𝐶, 𝐶𝐷, 𝐷𝐴, with no sides intersecting at interior points. A kite is a quadrilateral with two pairs of equal adjacent sides, e.g., ‖𝐴𝐵‖ = ‖𝐴𝐷‖ and ‖𝐶𝐵‖ = ‖𝐶𝐷‖. Theorem 3.11 (Kite Symmetry). If 𝐴𝐵𝐶𝐷 is a kite with ‖𝐴𝐵‖ = ‖𝐴𝐷‖ and ‖𝐶𝐵‖ = ‖𝐶𝐷‖, the diagonal 𝐴𝐶 is the perpendicular bisector of the segment 𝐵𝐷. D

D A C

B

C

A

B

Figure 6. Line Symmetry of Kites

Proof. Let 𝑀 be the midpoint of 𝐵𝐷. Apply the Isosceles Triangle Symmetry Theorem, Theorem 3.9, to △𝐴𝐵𝐷 and △𝐶𝐷𝐵. In △𝐴𝐵𝐷, the perpendicular bisector of the base 𝐵𝐷 contains the point 𝐴 and bisects ∠𝐵𝐴𝐷. Likewise, in △𝐶𝐷𝐵 the same perpendicular bisector of 𝐵𝐷 contains 𝐶 and bisects ∠𝐷𝐶𝐵. This says that the line 𝐴𝐶 is the perpendicular bisector of BD. This line also bisects the angles at 𝐴 and 𝐶. □ Notice that one of these kites in Figure 6 is not convex, but 𝐴𝐶 still is the perpendicular bisector of 𝐵𝐷. The proof of this theorem about equal base angles is left to Exercise 1. Theorem 3.12 (Equal Base Angles). For a triangle △𝐴𝐵𝐶, if 𝑚∠𝐴𝐵𝐶 = 𝑚∠𝐵𝐶𝐴, the triangle is isosceles.

Circles and Lines

29

Circles and Lines A circle with center 𝑃 was defined in Definition 1.6 as the set of points at distance 𝑟 from 𝑃, for some positive radius 𝑟. This means that if 𝐴 and 𝐵 are two points on the circle, either 𝑃 is the midpoint of 𝐴𝐵 or △𝑃𝐴𝐵 is isosceles with base 𝐴𝐵. Combining this observation with what we have learned about isosceles triangles can tell us about a number of other properties of circles. Some terminology will be used for a circle with center 𝑃: • A segment 𝑃𝐴 from the center to a point on the circle is called a radius segment, or just a radius if the meaning is clear. • A segment 𝐴𝐵 with 𝐴 and 𝐵 on the circle and with 𝑃 as midpoint is a diameter; both 𝑃𝐴 and 𝑃𝐵 are radius segments. A line through 𝑃 is a diameter line. Reflection in any line through 𝑃 maps the circle into itself, and for any reflection that maps the circle into itself, the reflection line passes through the center 𝑃. (The proof has been left as a problem in the exercises.) The intersections of a circle with a line can be characterized using what we know. Theorem 3.13 (Line-Circle Intersection). Let 𝑐 be a circle with center 𝑃 and radius 𝑟. If 𝑚 is a line, there are four ways in which the line can meet the circle. • The line and the circle may not intersect. • The line and the circle may intersect at one point 𝐴. Segment 𝑃𝐴 is perpendicular to 𝑚. • The line and the circle may intersect at two points 𝐴 and 𝐵, as a diameter line or not. Center 𝑃 is on the perpendicular bisector of 𝐴𝐵; it will be the midpoint if 𝐴𝐵 is a diameter. • The line cannot intersect the circle in more than two points. B A m B A

A

m m

Figure 7. Circle and Possible Line Intersections

Proof. If 𝐴 and 𝐵 are two distinct points of intersection of circle 𝑐 and line 𝑚, then by the Perpendicular Bisector as Locus Theorem, Theorem 3.10, the center 𝑃 is on the perpendicular bisector of 𝐴𝐵. This is the unique line 𝑛 through 𝑃 perpendicular to 𝑚.

30

3. Existence and Properties of Reflections

If 𝐴, 𝐵, and 𝐶 are three distinct points of intersection of 𝑐 and 𝑚, then 𝑛 must be the perpendicular bisector of both 𝐴𝐵 and 𝐴𝐶, which is impossible since this would mean 𝐵 = ℛ𝑛 (𝐴) = 𝐶. To prove the third case and then the second case, suppose 𝐴 is a point of intersection of 𝑐 and 𝑚. Then ℛ𝑛 must map 𝐴 to a point of intersection, since a reflection in a line through 𝑃 maps 𝑐 to itself and reflection in a perpendicular line maps 𝑚 to itself. If 𝐵 = ℛ𝑛 (𝐴) is distinct from 𝐴, then the third case holds. If 𝐵 = 𝐴, then 𝐴 must be on 𝑛, so 𝑛 = 𝑃𝐴. In this case 𝐵 = 𝐴 is the only point of intersection. If there were another intersection point 𝐶, then 𝐷 = ℛ𝑛 (𝐶) would be a third point of intersection, a case that has been ruled out above. Thus, in this case the second statement holds. Finally without trying to describe when this would happen, the first disjoint case may happen as well. □ The intersection of a circle and a line will be related to the distance of the line from the circle center in Theorem 6.8.

Light, Angles, and Reflections When speaking of the physics of reflection of light rays, one hears the statement “angle of incidence equals angle of reflection”. But when you look in a mirror, there is a copy of your face and your surroundings apparently just on the other side of the mirror, which seems to be a window into that scene. Some of this can be seen in our plane geometry of reflections if one views a line as a plane mirror from above. In Figure 8, points 𝐴 and 𝐵 are on one side of the mirror line and the line reflection 𝐵 ′ is on the other side. The observer at 𝐴 sees a reflection of ⃗′ , a ray that intersects the mirror at 𝐶. 𝐵 in the mirror by looking in the direction of 𝐴𝐵 There are congruent vertical angles in this figure and also an isosceles triangle 𝐵𝐶𝐵 ′ , with congruent angles at the apex 𝐶. So the angle of incidence does equal the angle of reflection.

B' C B

Figure 8. Line Reflection and Congruent Angles

A

Paper Folding and Tools for Construction

31

Paper Folding and Tools for Construction The proof of the geometric description of line reflection from the Reflection Axiom suggests why a paper fold through two points produces a nice, sharp straight line. The distance and angles on the paper are preserved by folding, so the destinations of all the points of the paper are determined, especially those of the points on the fold line. This connection offers many ways to experiment with line reflections by folding. Some of these are identified as experiments in the exercises at the end of the chapter. In addition to revealing mathematical relations, a paper fold and some scissors can produce all kinds of figures with line symmetry, from paper dolls to abstract shapes. Folding thin or translucent paper (such as patty paper) can also produce interesting traceable geometric figures, including ones that will appear in the next chapter. Actual glass or plastic mirrors are also tools for experimentation, though it is hard to actually draw the reflection behind a mirror. For this reason, semitransparent red plastic mirrors have been offered for classroom experiments since they allow one to see both the figure behind the mirror and the reflection of the figure in front of the mirror at the same time.

Figure 9. Line Symmetry from Folding and Cutting

The most versatile tool is dynamic geometry software that allows an abundance of examples with movable figures for experimentation. Compasses and Constructions. Euclid’s axioms for geometry were abstractions of drawing with an unmarked straightedge and a compass, so it was natural to use those tools when carrying out geometric constructions. The axioms in this book are abstractions of marked straightedges and protractors, so it may seem reasonable to use a marked ruler to measure distances when constructing geometric figures. However, compasses still are a more accurate way of constructing or copying lengths than using a physical ruler. The main problem with a physical marked ruler is that there are only a finite number of marks, so most distances have to be approximated. A second reason is the process of making a pencil mark at a point on the ruler depends on more hand and eye coordination than making a mark with a compass. Therefore, a physical marked ruler is not

32

3. Existence and Properties of Reflections

really a good model for the ruler functions in the Ruler Axiom. A better model is a line drawn on a computer screen with software, since the software will plot (as accurately as the screen allows) any distance on the line. In the exercises in the chapter, there are some directions for straightedge and compass constructions. These have the virtue of accuracy, but they also illustrate geometrical relationships between circles, lines, isosceles triangles and other figures. These methods complement the aforementioned paper folding, which is also a very accurate method for constructing certain lines.

Note about High School Texts. While transformations do not play a central role in most high school geometry textbooks, one exception is Geometry [6] from the University of Chicago School Mathematics Study Group. This book introduces line reflections early and uses them to prove such results as the symmetry properties of isosceles triangles. The set of axioms in that book assumes the parallel postulate before introducing line reflections and then assumes the basic properties of line reflections as postulates instead of proving them as theorems based on a general concept of rigid motion as is done here.

Exercises and Explorations 1. (Congruent Base Angles). Prove Theorem 3.12. Namely, prove for △𝐴𝐵𝐶 that if 𝑚∠𝐴𝐵𝐶 = 𝑚∠𝐴𝐶𝐵, then ‖𝐴𝐵‖ = ‖𝐴𝐶‖, so the triangle is isosceles. 2. (Rhombus Question). A rhombus is a quadrilateral 𝐴𝐵𝐶𝐷 with all four sides the same length. How are the diagonals of a rhombus related? Construct a rhombus by paper folding. 3. (Ruler Reflection on ℝ). Let 𝑛 be a line perpendicular to line 𝑚 at 𝐴. Reflection in 𝑛 maps 𝑚 to itself. Identifying line 𝑚 with ℝ by using a ruler; this defines a mapping of the real number line ℝ into itself. If the real number 𝑎 corresponds to 𝐴 and 𝑥 corresponds to a point 𝑋 on 𝑚, what number corresponds to ℛ𝑛 (𝑋)? 4. (Reflection of a Circle). Show that ℛ𝑚 maps a circle 𝑐 with center 𝑂 into itself, if and only if 𝑂 is on 𝑚. Notice that this implies that all lines through the center (the diameter lines) are lines of symmetry for 𝑐. 5. (Segment Image Experiment). Try this experiment either with geometry software or paper. Draw a segment and a line not intersecting the segment, preferably at no special angle with respect to the segment. Then construct the reflection of the segment (paper folding and tracing is a good option if you are working on paper). You now have four endpoints. Draw all the segments connecting the endpoints. Do the diagonals intersect on the fold line? Why? What happens if you repeat the experiment with a segment and a fold through the segment? What happens with special cases such as a segment parallel to or perpendicular to the fold line?

Exercises and Explorations

33

6. (X Symmetry Experiment). (a) Draw an X shape made of two lines (or as much as will fit on the paper) as in Figure 10. They need not be segments of equal length. Fold the paper so that the shape is folded onto itself. This fold is a line of symmetry of the figure consisting of two lines. Fold another line of symmetry. How are these lines related to the angles defined by the lines? How are they related to each other? (b) Fold a piece of paper; then fold it again so that the original fold is folded onto itself. What angle results? How is this related to the previous experiment?

Figure 10. Folds Are Symmetry Lines for This X

7. (Free Folds). Fold a piece of paper and then cut a shape; unfold and see what shape with line symmetry you have produced. It could be a geometric shape or a tree or a face or something else you design. 8. (Billiards). Draw your own example of a figure that shows where to look to see a figure in a mirror (or in what direction to aim a billiard ball off a cushion). This idea can be extended to reflections in two mirrors in a corner, as illustrated in Figure 11. Make your own mirror lines, place points 𝐴 and 𝐵, and find the billiard double cushion (or light double reflection) path between them. B''

B'

B A

Figure 11. Double Mirror Reflection Path from 𝐴 to 𝐵

9. (Symmetry Line Experiment). Find all the lines of symmetry for some familiar polygons: squares, rectangles, parallelograms, pentagons, hexagons (regular and otherwise). (Note: Some of these figures will not be “officially” defined until later chapters. This is informal experimentation at this stage.)

34

3. Existence and Properties of Reflections

10. (Perpendicular Bisector Construction). Draw a segment 𝐴𝐵 on a piece of paper. (a) Folding construction: Construct the perpendicular bisector of 𝐴𝐵 by folding the paper. How did you fold? (b) Compass and straightedge construction: Construct the perpendicular bisector by drawing two intersecting circles of the same radius, one centered at 𝐴 and the other at 𝐵. If there is one point of intersection not on line 𝐴𝐵, there will be two points of intersection. (Why?) Draw the line through the two points of intersection. Explain why this is the perpendicular bisector based on what was proved in this chapter. 11. (Angle Bisector Construction). Draw an angle ∠𝐵𝐴𝐶 on a piece of paper. (a) Folding construction: Construct the angle bisector of 𝐴𝐵 by folding the paper. (b) Compass and straightedge construction: Using the compass, construct points ⃗ ⃗ and 𝐸 on 𝐴 𝐷 on 𝐴𝐵 𝐶 equidistant from 𝐴. Then construct 𝐹 as in the previous problem to be an intersection point of two circles of the same radius with centers at 𝐷 and 𝐸. What shape is 𝐴𝐷𝐹𝐸 and what about its symmetry gives ⃗ is the angle bisector? a reason why the ray 𝐴𝐹

Chapter 4

Congruence of Triangles

Now that we have line reflections as a tool, we can establish congruence for simpler figures and build up to proving triangle congruence theorems. The first step may seem surprising: it is the congruence of points! Is this not obvious? Yes, it is obvious that if this were not true, our definition of congruence would be flawed. We do need to check that there is a rigid motion taking one point to the other. To say “same size, same shape” is not enough. But the proof is immediate. Theorem 4.1 (Congruence of Points). For any two distinct points 𝐴 and 𝐵, reflection in the perpendicular bisector of 𝐴𝐵 maps 𝐴 to 𝐵. Therefore, any two points are congruent. Proof. For distinct points, this follows immediately from the geometrical description in the Line Reflection Theorem, Theorem 3.4. If 𝐴 = 𝐵, let 𝑇 = 𝐼. □ That was fast. Now we can use this result for the next step, which is congruence of segments. Theorem 4.2 (Congruence of Segments). Two segments are congruent if and only if they have equal length.

m

B' C

n

B

D A

Figure 1. Isosceles Triangles △𝐶𝐷𝐵 ′ Formed by One Reflection

35

36

4. Congruence of Triangles

Proof. If we were using a different definition of congruence, this would not need a proof. But what we are really proving, again, is that there is a rigid motion that maps one segment to the other. One direction is immediate. If a rigid motion maps one segment to another, then the segments have equal length, so congruence implies equal length. To prove the converse, we will construct a rigid motion. Step 1: Let 𝐴𝐵 and 𝐶𝐷 be segments, with ‖𝐴𝐵‖ = ‖𝐶𝐷‖. For this first step, using the previous theorem for points, we choose a rigid motion 𝑇 that maps 𝐴 to 𝐶. If 𝐴 = 𝐶, let 𝑇 = 𝐼, the identity. Otherwise, let 𝑇 = ℛ𝑚 , reflection in the perpendicular bisector of 𝐴𝐶. We know 𝑇(𝐴) = 𝐶. Let 𝐵 ′ = 𝑇(𝐵). If 𝐵 ′ = 𝐷, then 𝑇 maps 𝐴𝐵 to 𝐶𝐷, and this proves that 𝐴𝐵 is congruent to 𝐶𝐷. If 𝐵 ′ is not 𝐷, we go to Step 2. Step 2: Since 𝑇 is a rigid motion, ‖𝐶𝐷‖ = ‖𝐶𝐵 ′ ‖. Therefore, △𝐶𝐷𝐵 ′ is an isosceles triangle with base 𝐷𝐵 ′ , or else 𝐶 is the midpoint of 𝐷𝐵 ′ . Let 𝑛 be the perpendicular bisector of 𝐷𝐵 ′ so that ℛ𝑛 (𝐵′ ) = 𝐷. By the Perpendicular Bisector as Locus Theorem, Theorem 3.10, 𝑛 passes through 𝐶 so ℛ𝑛 (𝐶) = 𝐶. Thus, ℛ𝑛 𝑇(𝐴) = 𝐶 and ℛ𝑛 𝑇(𝐵) = 𝐷. The rigid motion ℛ𝑛 𝑇 maps 𝐴𝐵 to 𝐶𝐷.

□

Note that the rigid motion from one segment to the other is either a line reflection or the product of two reflections. The rigid motion would be different if we began by mapping 𝐴 to 𝐷 instead of to 𝐶.

Triangle Congruence Tests Now we move on to the case of triangles, using our previous results. Theorem 4.3 (Side-Angle-Side, or SAS). Let △𝐴𝐵𝐶 and △𝐷𝐸𝐹 be two triangles such that ‖𝐴𝐵‖ = ‖𝐷𝐸‖, ‖𝐴𝐶‖ = ‖𝐷𝐹‖, and 𝑚∠𝐵𝐴𝐶 = 𝑚∠𝐸𝐷𝐹. Then △𝐴𝐵𝐶 is congruent to △𝐷𝐸𝐹. Moreover, the rigid motion of the congruence is unique.

D

C'

B A C

F

E

Figure 2. Isosceles △𝐷𝐹𝐶 ′ Formed by a Map from 𝐴𝐵 to 𝐷𝐸

Proof. Let 𝑈 be a rigid motion that takes 𝐴𝐵 to 𝐷𝐸; 𝑈 exists by the previous theorem. Let 𝐶 ′ = 𝑈(𝐶).

Triangle Congruence Tests

37

Triangles △𝐷𝐸𝐹 and △𝐷𝐸𝐶 ′ have a common side. Since 𝑚∠𝐸𝐷𝐶 ′ = 𝑚∠𝐸𝐷𝐹, ⃗′ is 𝐷 ⃗ 𝐹 or it is the unique ray with this angle measure on the opposite then either 𝐷𝐶 side of 𝐷𝐸. ⃗′ , then 𝐶 ′ = 𝐹 because ‖𝐷𝐶 ′ ‖ = ‖𝐷𝐹‖. Then the congruence is proved. ⃗ If 𝐷 𝐹 = 𝐷𝐶 If the rays are on opposite sides of 𝐷𝐸, forming equal angles ∠𝐸𝐷𝐹 and ∠𝐸𝐷𝐶 ′ , then 𝐷𝐸 bisects ∠𝐹𝐷𝐶 ′ except when the equal angles are right angles. Since 𝐷𝐹 and 𝐷𝐶 are congruent, △𝐷𝐹𝐶 ′ is isosceles with base 𝐹𝐶 ′ . The angle-bisecting line 𝐷𝐸 is also the perpendicular bisector of 𝐹𝐶 ′ , by Theorem 3.9. When the equal angles are right angles, 𝐷𝐸 is the perpendicular bisector of 𝐹𝐶 ′ by definition. This means that the reflection ℛ𝐷𝐸 fixes 𝐷 and 𝐸 but ℛ𝐷𝐸 (𝐶 ′ ) = 𝐹. Therefore, ℛ𝐷𝐸 𝑈 maps △𝐴𝐵𝐶 to △𝐷𝐸𝐹, and the two triangles are congruent. The uniqueness of the rigid motion from △𝐴𝐵𝐶 to △𝐷𝐸𝐹 follows from the Three Fixed Points Theorem, Theorem 3.3. It should be stressed that in this case the product transformation ℛ𝐷𝐸 𝑈 is unique, but not the factors. The transformation can be written as a product of rigid motions in many ways. □ This famous test for congruence is often abbreviated as SAS. As we have mentioned before, it is often taken as an axiom in other developments of geometry. The placement of the A between the two S’s indicates that the angle in the congruence test is between the two sides. Since the transformation 𝑈 that maps 𝐴𝐵 to 𝐷𝐸 is either the identity, a reflection, or a product of two reflections, the congruence ℛ𝐷𝐸 𝑈 between the triangles is either the identity or the product of one, two, or three reflections. The SAS criterion allows us a quick proof of angle congruence, which has not been addressed yet. Theorem 4.4 (Angle Congruence). Two angles ∠𝐴𝐵𝐶 and ∠𝐷𝐸𝐹 are congruent if and only if 𝑚∠𝐴𝐵𝐶 = 𝑚∠𝐷𝐸𝐹. Proof. By replacing the original points on the rays with different points on the same rays, we can assume that ‖𝐵𝐴‖ = ‖𝐵𝐶‖ = ‖𝐸𝐷‖ = ‖𝐸𝐹‖. Then we can apply SAS to show congruence of △𝐴𝐵𝐶 and △𝐷𝐸𝐹. The rigid motion of this congruence will map ∠𝐴𝐵𝐶 to ∠𝐷𝐸𝐹. □ There are two more triangle congruence tests: Side-Side-Side (SSS) and AngleSide-Angle (ASA). In each case we start with the same initial step to arrive at two triangles with a common side. Theorem 4.5 (Side-Side-Side, or SSS). Let △𝐴𝐵𝐶 and △𝐷𝐸𝐹 be two triangles such that segments 𝐴𝐵, 𝐵𝐶, and 𝐶𝐴 are congruent, respectively, to 𝐷𝐸, 𝐸𝐹, 𝐹𝐷. Then △𝐴𝐵𝐶 is congruent to △𝐷𝐸𝐹. Moreover, the rigid motion of the congruence is unique. Proof. Begin as before. Let 𝑈 map 𝐴𝐵 to 𝐷𝐸, with 𝑈(𝐶) = 𝐶 ′ . Again, if 𝐶 ′ = 𝐹, the theorem is proved. So we assume that 𝐶 ′ and 𝐹 are distinct.

38

4. Congruence of Triangles

C' B D A

F E

C Figure 3. Kite 𝐷𝐹𝐸𝐶 ′ Formed by a Map from 𝐴𝐵 to 𝐷𝐸

Because of the equal distances ‖𝐶 ′ 𝐷‖ = ‖𝐹𝐷‖ and ‖𝐶 ′ 𝐸‖ = ‖𝐹𝐸‖, both 𝐷 and 𝐸 are on the perpendicular bisector of 𝐶 ′ 𝐹, since the perpendicular bisector is the locus of points equidistant from 𝐶 and 𝐹 by Theorem 3.10. Therefore, 𝐷𝐸 is the perpendicular bisector of 𝐶 ′ 𝐹. This implies that 𝐶 ′ and 𝐹 are on opposite sides of 𝐷𝐸 and ℛ𝐷𝐸 (𝐶 ′ ) = 𝐹. Thus ℛ𝐷𝐸 maps △𝐷𝐸𝐶 ′ to △𝐷𝐸𝐹; and ℛ𝐷𝐸 𝑈 maps △𝐴𝐵𝐶 to △𝐷𝐸𝐹, so the triangles are congruent. The uniqueness of the rigid motion from △𝐴𝐵𝐶 to △𝐷𝐸𝐹 follows from the Three Fixed Points Theorem, Theorem 3.3, as in the proof of SAS. Again, this does not mean that the same rigid motion cannot be written as a product in many different ways. Notice that if neither angle 𝐷 nor angle 𝐸 is a right angle in △𝐷𝐸𝐹, then 𝐷𝐹𝐸𝐶 ′ □ is a kite, and 𝐷𝐸 is the diagonal that is a line of symmetry. Theorem 4.6 (Angle-Side-Angle, or ASA). Let △𝐴𝐵𝐶 and △𝐷𝐸𝐹 be two triangles such that segment 𝐴𝐵 is congruent to 𝐷𝐸, ∠𝐵𝐴𝐶 is congruent to ∠𝐸𝐷𝐹, and ∠𝐴𝐵𝐶 is congruent to ∠𝐷𝐸𝐹. Then △𝐴𝐵𝐶 is congruent to △𝐷𝐸𝐹. Moreover, the rigid motion of the congruence is unique. Proof. Begin as before: 𝑈 maps 𝐴𝐵 to 𝐷𝐸, with 𝐶 ′ = 𝑈(𝐶). . B

C' D A E

C F

Figure 4. Line of Symmetry 𝐷𝐸 after Map from 𝐴𝐵 to 𝐷𝐸

Applications of Triangle Congruence

39

⃗′ = 𝐷 ⃗ If 𝐶 ′ and 𝐹 are on the same side of 𝐷𝐸, then by the angle congruence, 𝐷𝐶 𝐹 ⃗′ = 𝐸𝐹. ⃗ But 𝐶 ′ and 𝐹 are each the intersection point of these same two rays, so and 𝐸𝐶 𝐶 ′ = 𝐹 and 𝑈 maps △𝐴𝐵𝐶 to △𝐷𝐸𝐹. If 𝐶 ′ and 𝐹 are on opposite sides of 𝐷𝐸, then reflection in 𝐷𝐸 fixes 𝐷 and 𝐸. Let 𝐶 = ℛ𝐷𝐸 (𝐶 ′ ). ″

Then 𝐶 ″ and 𝐹 are on the same side of 𝐷𝐸 with the same congruent angle relations as in the same-side case for 𝐶 ′ and 𝐹. Thus by the previous reasoning, 𝐶 ″ = 𝐹, and ℛ𝐷𝐸 𝑈 maps △𝐴𝐵𝐶 to △𝐷𝐸𝐹. Therefore, △𝐴𝐵𝐶 ≅ △𝐷𝐸𝐹. The uniqueness follows from the Three Fixed Points Theorem, Theorem 3.3, as in the proofs of SAS and SSS. □ This is a congruence theorem that only applies to right triangles. The proof is left to the exercises, where there is a brief discussion. Theorem 4.7 (Hypotenuse-Leg). Suppose △𝐴𝐵𝐶 and △𝐷𝐸𝐹 are right triangles with right angles at 𝐶 and 𝐹. If ‖𝐴𝐵‖ = ‖𝐷𝐸‖ (the hypotenuses) and ‖𝐴𝐶‖ = ‖𝐷𝐹‖ (legs), then the triangles are congruent.

Applications of Triangle Congruence These versions of SAS, SSS, and ASA say more than the traditional versions that just imply that certain lengths and angle measures are equal. The criteria here come with rigid motions attached, and this changes how the theorems can be applied. Here is an example, illustrated in Figure 5. Theorem 4.8 (Congruence for Convex Quadrilaterals). Let 𝐴𝐵𝐶𝐷 and 𝐸𝐹𝐺𝐻 be convex quadrilaterals with the angles at 𝐴, 𝐵, 𝐶, congruent to the angles at 𝐸, 𝐹, 𝐺, respectively, and also 𝐴𝐵 ≅ 𝐸𝐹 and 𝐵𝐶 ≅ 𝐹𝐺. Then 𝐴𝐵𝐶𝐷 is congruent to 𝐸𝐹𝐺𝐻.

A

B

H

D

E

G F C

Figure 5. Two Quadrilaterals with Congruent Angles and Sides as Marked

Proof. By SAS, △𝐴𝐵𝐶 is congruent to △𝐸𝐹𝐺, so there is a rigid motion 𝑇 that takes △𝐴𝐵𝐶 to △𝐸𝐹𝐺. Then 𝑇(∠𝐵𝐴𝐷) ≅ ∠𝐹𝐸𝐻 since this 𝑇 is a rigid motion. There⃗ ⃗ fore, 𝑇 must map 𝐴 𝐷 to 𝐸 𝐻, since both rays are in the same half-plane of 𝐸𝐹 as 𝐺 by ⃗ ⃗ convexity of 𝐸𝐹𝐺𝐻 and 𝑇(𝐴𝐵𝐶𝐷). Likewise, 𝑇 maps 𝐶 𝐷 to 𝐺 𝐻.

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4. Congruence of Triangles

⃗ ⃗ Since 𝐷 is the intersection of 𝐴 𝐷 and 𝐶 𝐷, the image of 𝐷 must be the intersection of ⃗ ⃗ the images of rays 𝐸 𝐻 and 𝐺 𝐻, which is 𝐻. Therefore, the 𝑇-image of 𝐴𝐵𝐶𝐷 is 𝐸𝐹𝐺𝐻 and the quadrilaterals are congruent. □

Properties of Rigid Motions The triangle congruence theorems in this chapter imply important facts about the nature of rigid motions. One is that every rigid motion is the product of three or fewer line reflections. Theorem 4.9 (Three Reflections). If 𝑇 is a rigid motion, then 𝑇 is the identity, a line reflection, or the product of two or three line reflections. Proof. For any triangle △𝐴𝐵𝐶, let 𝑇(𝐴) = 𝐷, 𝑇(𝐵) = 𝐸, 𝑇(𝐶) = 𝐹. Then △𝐴𝐵𝐶 ≅ △𝐷𝐸𝐹. Thus by SSS (or SAS or ASA), there is a product 𝑆 of no more than three line reflections that maps △𝐴𝐵𝐶 to △𝐷𝐸𝐹. By the Three Fixed Points Theorem, Theorem 3.3, 𝑆 = 𝑇. □ In Chapter 1, it was explained why an axiom was needed for the existence of rigid motions, not just isometries, because isometries could not be proved to be rigid motions without further assumptions. But at this point, the existence of rigid motions and the consequent proof of triangle congruence theorems make it possible to prove the promised theorem about isometries. Theorem 4.10 (Isometries Are Rigid). If 𝑇 is an isometry, then 𝑇 is a rigid motion. Proof. Recall that a transformation 𝑇 is an isometry if it preserves distance. To prove that 𝑇 is a rigid motion, one needs to show that it preserves angle measure as well. Consider any noncollinear points 𝐴, 𝐵, 𝐶. Apply 𝑇 to △𝐴𝐵𝐶. The image △𝑇(𝐴)𝑇(𝐵)𝑇(𝐶) has sides of the same length as the original. Applying SSS, we conclude △𝐴𝐵𝐶 ≅ △𝑇(𝐴)𝑇(𝐵)𝑇(𝐶). But if the triangles are congruent, then so are the angles: ∠𝐴𝐵𝐶 ≅ ∠𝑇(𝐴)𝑇(𝐵)𝑇(𝐶).

Figure 6. Objects Defined by Distance and Angle Measure Transported by Congruences

□

Midpoint Triangle and Angle Sum

41

A Comment about Triangle Congruence. Related to every triangle are a number of geometrical objects such as medians, altitudes, angle bisectors, perpendicular bisectors, incircles and excircles, among others, as illustrated in Figure 6. Since all of these objects are defined by some combination of distance, angle, and intersection, the same definitions apply to the image of a triangle if it is mapped by a rigid motion. This means that the image of each of these objects is the corresponding object of the image. This can be spelled out in particular cases, but the argument is always the same. State the definition, map the points, and observe that the same relations hold in the image. This is a powerful feature of the rigid motion attached to a congruence. With other definitions of congruence, these relations must be proved individually, or (quite commonly) assumed without proof.

Midpoint Triangle and Angle Sum In this section, we will apply the Dilation Axiom for the first time to set up the midpoint triangle as an application for the triangle congruence theorems of this chapter. The main reason for doing this now is to prove the angle sum of a triangle, so that angle sums can be used when exploring regular polygons in the next chapter. This result will not be used to prove other results until Chapter 7, where the Euclidean Parallel Postulate is introduced. Therefore, any readers who are interested in neutral geometry1 may wish to defer this theorem until Chapter 7, where it is proved again. Given △𝐴𝐵𝐶, the midpoint triangle of this triangle is the triangle whose vertices are the midpoints of the sides of △𝐴𝐵𝐶. Theorem 4.11 (Midpoint Triangle Theorem). Given △𝐴𝐵𝐶 and the midpoints 𝐷, 𝐸, 𝐹 of the sides 𝐵𝐶, 𝐶𝐴, 𝐴𝐵, respectively, the sides of △𝐷𝐸𝐹 dissect △𝐴𝐵𝐶 into four congruent subtriangles. A E

C

F

D

B

Figure 7. Triangles and Midpoints

Proof. As one sees in Figure 7, there are three corner triangles and one triangle in the middle. 1 Neutral geometry is the geometry that is true of both Euclidean and non-Euclidean plane geometry. All our axioms except for the Dilation Axiom belong to neutral geometry, so the results proved without this axiom are valid in neutral geometry.

42

4. Congruence of Triangles

The first step in the proof is to apply the dilation 𝑇 = 𝒟𝐴,1/2 to △𝐴𝐵𝐶. Then by the definition of dilation, 𝑇(𝐴) = 𝐴, 𝑇(𝐵) = 𝐹, and 𝑇(𝐶) = 𝐸, so the image is △𝐴𝐹𝐸. From the definition of dilation, ‖𝐴𝐹‖ = (1/2)‖𝐴𝐵‖ and ‖𝐴𝐸‖ = (1/2)‖𝐴𝐶‖. Also, ∠𝐹𝐴𝐸 = ∠𝐵𝐴𝐶. From the Dilation Axiom, ‖𝐹𝐸‖ = (1/2)‖𝐵𝐶‖, ∠𝐸𝐹𝐴 ≅ ∠𝐶𝐵𝐴, and ∠𝐴𝐸𝐹 ≅ ∠𝐴𝐶𝐵. In short, for this corner triangle, the angles are the same as the corresponding angles in the big triangle and the sides are half as long. The corresponding results can be obtained for the other two corner triangles using dilations centered at 𝐵 and 𝐶. With this information, we can conclude that the three corner triangles are congruent, using SAS, SSS, or ASA. This leaves the center triangle △𝐷𝐸𝐹. But we know one side length of this triangle from the shared side with a corner triangle. So from the corner triangle with vertex 𝐴, ‖𝐸𝐹‖ = (1/2)‖𝐵𝐶‖. From vertex 𝐵, ‖𝐹𝐷‖ = (1/2)‖𝐶𝐴‖, and from 𝐶, ‖𝐷𝐸‖ = (1/2)‖𝐴𝐵‖. But these are the same side lengths as for △𝐴𝐹𝐸, so these triangles are congruent by SSS. □

Thus, all four triangles are congruent. Looking at the angles at a midpoint, one gets the following corollary.

Theorem 4.12 (Angle Sum). For any triangle 𝐴𝐵𝐶, the sum of the measures of the angles at the vertices is 180 degrees. Proof. Let the measure of the angles of △𝐴𝐵𝐶 be 𝑎 = 𝑚∠𝐶𝐴𝐵, 𝑏 = 𝑚∠𝐴𝐵𝐶, 𝑐 = 𝑚∠𝐵𝐶𝐴. There are three angles with vertex 𝐷 in the figure. Their measures can be added and their sum is 180, the measure of a straight angle. But the angles at 𝐷 are angles from the four subtriangles, each of which has angles congruent to the angles of △𝐴𝐵𝐶. To be precise, 𝑏 = 𝑚∠𝐸𝐷𝐶, 𝑎 = 𝑚∠𝐸𝐷𝐹, and 𝑐 = 𝑚∠𝐹𝐷𝐵, so 𝑎 + 𝑏 + 𝑐 = 180.

□

Exercises and Explorations 1. (Comparing Reflection Sequences). Draw two congruent segments 𝐴𝐵 and 𝐶𝐷; then carry out a sequence of two line reflections to map one segment to the other with these differences. (a) Construct the sequence of line reflections, starting with a reflection from 𝐴 to 𝐶, that will map 𝐴𝐵 to 𝐶𝐷. (b) Then construct a second sequence starting with a reflection from 𝐵 to 𝐷. (c) How are the pairs of lines from the first construction related to those of the second construction? Are there any special points? Any relationship of angles?

Exercises and Explorations

43

(d) If a point 𝐸 is added to the figure to form △𝐴𝐵𝐸, will the image of this triangle be the same for both sequences of reflections? 2. (Symmetries of a Segment). Let 𝑗 be the segment with endpoints 𝐴 and 𝐵. The goal of this problem is to list and draw all the rigid motions that map 𝑗 to itself. Since the image of 𝑗 will be drawn on top of itself each time, as a visual aid, draw any nearby point 𝐶. Then draw the images of both 𝑗 and 𝐶 for each rigid motion that maps 𝑗 to itself. You may wish to use paper folding to model these rigid motions. (Note: The statement does not say that 𝐴 maps to 𝐴. Hint: The number of items in the list should be an even number that is not prime.) 3. (Two-sided Experiments). In each case, cut out a figure from cardboard and trace it on paper as suggested. (Alternatively, use software.) (a) Cut out a cardboard triangle of general shape. Color or decorate each side differently from the other. Repeat this experiment several times: trace the triangle on a sheet of paper and then trace a second copy at some other location on the paper, either turning the triangle over or not but noting the decoration each time. Construct lines of reflection by folding or drawing so that the product of the reflections will map the original triangle to the second. Based on the decorations you see, can you predict whether the number of reflections is odd or even? Explain. (b) Cut out a cardboard figure that resembles the letter F. Again, trace two copies on a sheet of paper and find a sequence of reflections that will take one shape to the other. Do this several times, for different positions of the F, including with the shape flipped over one or more times. Can you tell at a glance whether it will require one, two, or three reflections to move one F onto the other? What do you notice that makes this possible? 4. (Pentagon). For a circle with center 𝑂 and radius 𝑟, let 𝑃1 , 𝑃2 , 𝑃3 , 𝑃4 , 𝑃5 be points on ⃗𝑘 ) = 72𝑘 for each 𝑘 = 1, 2, 3, 4, 5. the circle with polar angle 𝜃(𝑂𝑃 (a) Use triangle congruence theorems to prove that the sides of the pentagon 𝑃1 𝑃2 𝑃3 𝑃4 𝑃5 are congruent. (b) Use triangle congruence theorems to prove that the sides of the star pentagon 𝑃1 𝑃3 𝑃5 𝑃2 𝑃4 are congruent, i.e. 𝑃1 𝑃3 ≅ 𝑃3 𝑃5 ≅ 𝑃5 𝑃2 ≅ 𝑃2 𝑃4 ≅ 𝑃4 𝑃1 . (A star is not a polygon according to our definition.) (c) Find rigid motions that prove these congruences directly without the triangle congruence theorems. Justify your claim. 5. (Hypotenuse-Leg). This is the congruence criterion for right triangles stated in Theorem 4.7. In a right triangle the side opposite the right angle is the hypotenuse and the other sides are called legs. Hypotenuse-Leg: Suppose △𝐴𝐵𝐶 and △𝐷𝐸𝐹 are right triangles with right angles at 𝐶 and 𝐹. If ‖𝐴𝐵‖ = ‖𝐷𝐸‖ (the hypotenuses) and ‖𝐴𝐶‖ = ‖𝐷𝐹‖ (legs), then the triangles are congruent. If we had already proved the Pythagorean Theorem, this would follow from SSS. But there is a nice proof using only the tools from this chapter. Write such a proof.

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4. Congruence of Triangles

6. (Midpoint Right Triangles). For an example of a right triangle △𝐴𝐵𝐶, with right angle at 𝐶, draw the midpoint triangle and compare with the subdivision of the triangle in Figure 7. What relations are in the right triangle figure that are not true in the general case? If 𝑀 is the midpoint of the hypotenuse 𝐴𝐵, what can you say about the distances from 𝑀 to the vertices of △𝐴𝐵𝐶?

Chapter 5

Rotation and Orientation

For geometrical foundations we have focused on line reflections. In this chapter we will expand our repertoire to rotations. We will see that certain products of reflections that we have already encountered are actually rotations. Before getting started with definitions and theorems, let’s look at a couple of examples that illustrate some ideas we will meet in this chapter. Figure 1 shows a triangle 𝑡 with a smaller triangle 𝑠 inside. Each triangle is reflected twice, and they arrive together at the same location, appearing to have been rotated about point 𝑂. But the two triangles took very different routes on their journey. Triangle 𝑡 was reflected first in 𝑂𝐴 and then reflected in 𝑂𝐵, so 𝑡′ = ℛ𝑂𝐴 (𝑡) and 𝑡″ = ℛ𝑂𝐵 (𝑡′ ) = ℛ𝑂𝐵 ℛ𝑂𝐴 (𝑡). In contrast, triangle 𝑠 was reflected in 𝑂𝐶 and then in 𝑂𝐷, with 𝑠″ = ℛ𝑂𝐷 (𝑠′ ) = ℛ𝑂𝐷 ℛ𝑂𝐶 (𝑠). But the final rigid motions mapping 𝑡 to 𝑡″ and 𝑠 to 𝑠″ appear to be the same transformation: ℛ𝑂𝐵 ℛ𝑂𝐴 = ℛ𝑂𝐷 ℛ𝑂𝐶 . We will see why this is true.

t'

B t''

A

s'' t s O

D

s' C

Figure 1. Two Different Double Reflections

45

46

5. Rotation and Orientation

Figure 2 is a shape that might result from looking into a kaleidoscope or from folding a paper three times into a triangle and then cutting it so that it unfolds into a “snowflake” shape. The figure has rotational symmetries and reflection symmetries. This illustrates how reflections and rotations around a central point 𝑂 interact to produce certain symmetry patterns.

Figure 2. Kaleidoscopic “Snowflake”

Rotations and Double Reflections Line reflections play the role of atoms from which we can build all the other rigid motions. In this chapter we will begin the study of products of two reflections. The short statement of the relationship is this: the composition of two reflections is a rotation if the lines intersect and it is a translation if they are parallel. This chapter will look at rotations. In the next chapter, before studying translations, we will see that we need to add another axiom, something that implies the Euclidean Parallel Postulate. Definition 5.1. A rotation is a rigid motion 𝑇 with a fixed point 𝑂 such that for every ⃗ ray 𝑂 𝐴, the angle measure 𝑚∠𝐴𝑂𝑇(𝐴) is the same. The point 𝑂 is called the center of the rotation. Note. The identity mapping 𝐼 is a rotation with 0 = 𝑚∠𝐴𝑂𝐼(𝐴). From the definition, the center 𝑂 is the unique fixed point for every rotation except 𝐼. Next come two theorems to say which rigid motions satisfy this descriptive definition. The first says that every rotation is a double reflection in lines intersecting at the center. The second says that every such double reflection is a rotation and gives more details about the behavior of rotations. Theorem 5.2 (One Fixed Point). If 𝑇 is a rigid motion with a single fixed point 𝑂, then 𝑇 is the product of two reflections ℛ𝑚 ℛ𝑛 , where 𝑚 and 𝑛 are both lines through 𝑂. Therefore, every rotation is the product ℛ𝑚 ℛ𝑛 , where 𝑚 and 𝑛 intersect at 𝑂. Proof. Take any point 𝐴 distinct from 𝑂 and let 𝐴′ = 𝑇(𝐴). Since ‖𝑂𝐴‖ = ‖𝑂𝐴′ ‖, 𝑂 is on the perpendicular bisector 𝑚 of 𝐴𝐴′ by the Locus Theorem, Theorem 3.10. Then ℛ𝑚 𝑇(𝑂) = ℛ𝑚 (𝑂) = 𝑂 and ℛ𝑚 𝑇(𝐴) = ℛ𝑚 (𝐴′ ) = 𝐴, so ℛ𝑚 𝑇 has two fixed points and Theorem 2.12 has 𝑂𝐴 as a line of fixed points. According to the Line Reflection Theorem, Theorem 3.4, either 𝑇 = 𝐼 or 𝑇 = ℛ𝑂𝐴 .

Rotations and Double Reflections

47

A' m O A

n

Figure 3. Rotation from 𝐴 to 𝐴′ as Two Reflections

If ℛ𝑚 𝑇 = 𝐼, then 𝑇 = ℛ𝑚 , which is not possible since 𝑇 has only one fixed point. So ℛ𝑚 𝑇 must be ℛ𝑛 , the reflection in 𝑛 = 𝑂𝐴, and 𝑇 = ℛ𝑚 ℛ𝑚 𝑇 = ℛ𝑚 ℛ𝑛 , the composition of reflections in two lines intersecting at 𝑂. This shows that every rotation except 𝐼 is the composition of reflections in two lines intersecting at the center 𝑂. In the case of 𝐼, there is not a unique center 𝑂, but 𝐼 = ℛ𝑚 ℛ𝑚 for any line through 𝑂. So every rotation is a double line reflection. □ Before stating the next theorem, we recall that if two lines intersect, they form two pairs of congruent vertical angles. So there are two angle measures, at least one of which is less than or equal to 90 degrees. We call this number the measure of the angle between the lines. Theorem 5.3 (Rotation). Let ℛ𝑚 and ℛ𝑛 be reflections in lines 𝑚 and 𝑛 that intersect at point 𝑂. The composition 𝑇 = ℛ𝑛 ℛ𝑚 is a rotation with center 𝑂. • The constant 𝑘 = 𝑚∠𝑃𝑂𝑇(𝑃) is twice the measure of the angle between the lines. ⃗ is increased by 𝑘 or else every • Given a protractor at 𝑂, the polar angle of every 𝑂𝑃 polar angle is decreased by 𝑘.

B X''

C

X'

A X

O

D Y

Figure 4. Isosceles Triangles from Double Reflections

Before the full proof, let us look at a picture that suggests why this is true. In Figure 4, there are two examples of double line reflection. The point 𝑋 ′ = ℛ𝑂𝐴 (𝑋) and 𝑋 ″ = ℛ𝑂𝐵 (𝑋 ′ ). These points form isosceles triangles △𝑋𝑂𝑋 ′ and △𝑋 ′ 𝑂𝑋 ″ . The

48

5. Rotation and Orientation

reflection lines are bisectors of the vertex angles at 𝑂, so that ∠𝐴𝑂𝐵 is formed from two halves of the isosceles angles at 𝑂. Thus, it is clear that 𝑚∠𝑋𝑂𝑋 ″ = 𝑚∠𝑋𝑂𝑋 ′ + 𝑚∠𝑋 ′ 𝑂𝑋 ″ = 2(𝑚∠𝐴𝑂𝑋 ′ + 𝑚∠𝑋 ′ 𝑂𝐵) = 2𝑚∠𝐴𝑂𝐵. This is rather convincing, but the second example in the figure shows 𝑌 = ℛ𝑂𝐶 (𝑋) and then 𝑋 ″ = ℛ𝑂𝐷 (𝑌 ). While it appears plausible that 𝑚∠𝐶𝑂𝐷 = 𝑚∠𝐴𝑂𝐵, it is much less clear what form of subtraction of the measures of the angles ∠𝑋𝑂𝑌 and ∠𝑌 𝑂𝑋 ″ will produce the desired rotation angle. This can be done, but it seems that one needs to deal with separate cases for different positions of 𝑋. Instead, the proof below uses a single calculation with the polar angles that provides additional information. ⃗ ⃗ Proof. Let 𝑂 𝐴 and 𝑂 𝐵 be rays in lines 𝑚 and 𝑛 with polar angles 𝑎 and 𝑏, respectively. For any point 𝑋, let 𝑋 ′ = ℛ𝑂𝐴 (𝑋) and let 𝑋 ″ = ℛ𝑂𝐵 (𝑋 ′ ). If 𝑥 is a polar angle of 𝑋, then by preservation of angle measure the number 𝑥′ , satisfying (𝑥′ − 𝑎) = −(𝑥 − 𝑎), is a polar angle of 𝑋 ′ . So 𝑥′ = 2𝑎 − 𝑥. By the same reasoning, 𝑥″ = 2𝑏 − 𝑥′ = 2(𝑏 − 𝑎) + 𝑥 is a polar angle of 𝑋 ″ . Thus the composition 𝑇 = ℛ𝑂𝐵 ℛ𝑂𝐴 adds the constant 2(𝑏 − 𝑎) to every direction angle. If we make the opposite choice of ray in either line of reflection, we change 𝑎 or 𝑏 ⃗′ and by 180 and so change 2(𝑏 − 𝑎) by an integer multiple of 360. Consequently 𝑂𝑋 ⃗ 𝑂 𝑋 ″ are unchanged. If we choose rays and polar angles 𝑎 and 𝑏 so that |𝑏 − 𝑎| ≤ 90, then the measure of the angle between the lines is 𝑚∠𝐴𝑂𝐵 = |𝑏 − 𝑎|. Thus 2|𝑏 − 𝑎| = 2𝑚∠𝐴𝑂𝐵 ≤ 180. Let 𝑘 = 2|𝑏 − 𝑎|. The sign of (𝑏 − 𝑎) reflects whether the direction 𝑥″ is obtained from direction 𝑥 by adding or subtracting the nonnegative number 𝑘. Since 𝑘 = |𝑥″ − 𝑥| ≤ 180, then 𝑘 = 𝑚∠𝑋𝑂𝑋 ″ . This completes the proof that ℛ𝑂𝐵 ℛ𝑂𝐴 is a rotation.

□

⃗ ⃗ ⃗ ⃗ Corollary. Let the polar angles of the rays 𝑂 𝐴, 𝑂 𝐵, 𝑂 𝐶, 𝑂 𝐷 be denoted by 𝑎, 𝑏, 𝑐, 𝑑, respectively. (1) If there is a real number ℎ so that 𝑐 = 𝑎 + ℎ and 𝑑 = 𝑏 + ℎ, then these double reflections are the same rotation with center 𝑂: ℛ𝑂𝐷 ℛ𝑂𝐶 = ℛ𝑂𝐵 ℛ𝑂𝐴 . (2) If any three of the lines are given, a unique fourth can be found so that ℛ𝑂𝐷 ℛ𝑂𝐶 = ℛ𝑂𝐵 ℛ𝑂𝐴 . Proof. For statement (1), if 𝑐 = 𝑎 + ℎ and 𝑑 = 𝑏 + ℎ, then 𝑑 − 𝑐 = 𝑏 − 𝑎. Then it is clear from the proof of the theorem that the two reflection products are the same. This equation can also be written as 𝑐 − 𝑎 = 𝑑 − 𝑏. Let ℎ = 𝑐 − 𝑎 = 𝑑 − 𝑏. For part (2), if 𝑎, 𝑏, 𝑐 are known, define ℎ = 𝑐 − 𝑎 and set 𝑑 = 𝑏 + ℎ. Similar reasoning applies to any of the other three choices of lines. □ Here is an example of such a double reflection to make the polar angle choices a bit more concrete. ⃗ ⃗ Example 5.4. Suppose that 𝑂 𝐴 has a polar angle of 0 and 𝑂 𝐵 has a polar angle of 115. Then 𝑘 = 230. We can consider 𝑇 to be a rotation by 230 degrees, or we can consider it

Rotations and Double Reflections

49

to be a rotation by −130 degrees. We would have obtained the −130 number if we had chosen the other ray for 𝑛, which would have a polar angle of −65. This example illustrates an important point: Note. A rotation is not a movie; it is two stills. While rotating a wheel counterclockwise by 230 degrees feels very different from rotating clockwise by 130 degrees, the rigid motion does not include the history. It only takes a figure at a starting position and moves it to a final position, without regard for intermediate steps. Therefore, the rotations by +230 or −130 degrees are exactly the same rigid motion; one does not rotate “farther” than the other.

Rotation by 230°

B

A

Rotation by –130°

Figure 5. Same Rotation by 230 or −130 Degrees

A movie showing the rotating of the wheel can be treated mathematically, but the model is not one rigid motion that is a rotation but rather a continuously varying family of rotations. Angle of Rotation and Signed Angle Measure. In the proof of the rotation theorem, the polar angles were changed in a positive direction 𝑘 (a counterclockwise rotation by 𝑘) or a negative direction −𝑘 (a clockwise rotation by 𝑘). But this distinction depended on the protractor, which was fixed during the course of the proof. With a physical protractor and paper, the counterclockwise direction can be reversed by flipping the protractor over or by flipping the paper over, in either case reversing the orientation of the plane. If at point 𝑂, the protractor 𝛿(𝑡) was chosen for the proof, then the protractor 𝛿(𝑡 + 𝑒), for any constant 𝑒, will have the same orientation. The rotation will continue to have the same positive or negative direction since the addition of a constant to the polar angles does not change the difference 𝑏 − 𝑎 = (𝑏 − 𝑒) − (𝑎 − 𝑒). But if the new protractor is 𝛿(−𝑡 + 𝑒), then the sign of 𝑏 − 𝑎 will be reversed and the counterclockwise direction will now be clockwise.

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Therefore, in the example above, a statement that the rotation is by an angle of −130 or +130 depends on the orientation of the plane at 𝑂 determined by the choice of protractor. We will have more to say about orientation in an upcoming section. Definition 5.5. For a rotation 𝑇 with center 𝑂, if a protractor is chosen, the angle of ⃗ rotation is the real number 𝑘 added to the polar angle of each ray 𝑂 𝐴 by the rotation, so that 𝜃(𝑇(𝐴)) = 𝜃(𝐴) + 𝑘. This rotation is denoted 𝑂 𝑘 . The angle of rotation is unique only up to the addition of an integer multiple of 360, since adding 360 does not change the rotation. The angle of rotation provides a signed measure of angle that takes direction of rotation into account. ⃗ ⃗ and 𝐴 Definition 5.6. Given a protractor at 𝐴 and two rays 𝐴𝐵 𝐶 defining ∠𝐵𝐴𝐶, the + ⃗ ⃗ =𝐴 signed angle measure 𝑚 ∠𝐵𝐴𝐶 is the angle of rotation 𝑘 such that 𝐴𝑘 (𝐴𝐵) 𝐶, with −180 < 𝑘 < 180. ⃗ ⃗ the difference of two polar angles From the definition, 𝑚+ ∠𝐵𝐴𝐶 = 𝜃(𝐴 𝐶)−𝜃(𝐴𝐵), is chosen so that the angle of rotation lies between −180 and +180. The sign of the measure depends on the positive direction defined by the protractor. Importantly, the mea⃗ ⃗ sure depends on the order of the rays, for 𝑚+ ∠𝐵𝐴𝐶 = −𝑚+ ∠𝐶𝐴𝐵 since 𝐴−𝑘 (𝐴 𝐶) = 𝐴𝐵 ⃗ ⃗ when 𝐴𝑘 (𝐴𝐵) = 𝐴𝐶.

Rotation-Reflection Relations For the next section about symmetry, it will be helpful to collect some facts about reflection products. Theorem 5.7 (Reflection Products). In each case below, 𝑆 and 𝑇 are rotations with center 𝑂. (1) For any line 𝑝 through 𝑂, there is a line 𝑞 through 𝑂 so that 𝑇 = ℛ𝑞 ℛ𝑝 ; there is also a line 𝑟 so that 𝑇 = ℛ𝑝 ℛ𝑟 . (2) For any three lines 𝑚, 𝑛, 𝑝 through 𝑂, ℛ𝑚 ℛ𝑛 ℛ𝑝 = ℛ𝑝 ℛ𝑛 ℛ𝑚 = ℛ𝑞 for some line 𝑞 through 𝑂. Consequently, ℛ𝑚 𝑇 and 𝑇ℛ𝑚 are line reflections. (3) The composition 𝑆𝑇 is a rotation with center 𝑂 (possibly the identity). The angle of rotation of 𝑆𝑇 is the sum of the angles of rotation of 𝑆 and 𝑇. (4) If M is the product of 𝑘 reflections in lines through 𝑂, then 𝑀 is a line reflection if 𝑘 is odd, and 𝑀 is a rotation if 𝑘 is even. Proof. Statement (1) is just a restatement in different notation of one point in the corollary to the Rotation Theorem, Theorem 5.3. In the case when 𝑈 is the identity, take 𝑞 = 𝑝.

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The proof of (2) uses (1). Let 𝑈 = ℛ𝑚 ℛ𝑛 , which is a rotation. By (1) there is a line 𝑞 so that 𝑈 = ℛ𝑞 ℛ𝑝 . Multiplying on the right by ℛ𝑝 , this becomes (ℛ𝑚 ℛ𝑛 )ℛ𝑝 = −1 (ℛ𝑞 ℛ𝑝 )ℛ𝑝 = ℛ𝑞 (ℛ𝑝 ℛ𝑝 ) = ℛ𝑞 . Since (ℛ𝑚 ℛ𝑛 ℛ𝑝 )−1 = ℛ𝑝 ℛ𝑛 ℛ𝑚 and ℛ𝑞 = ℛ𝑞 , the second product is also ℛ𝑞 . The proof of (3) follows by writing a product of four reflections as a product of a triple reflection with a single reflection, then applying (2). Since each rotation is a product of two line reflections, for some four lines through 𝑂, 𝑆𝑇 = (ℛ𝑚 ℛ𝑛 )(ℛ𝑝 ℛ𝑡 ) = (ℛ𝑚 ℛ𝑛 ℛ𝑝 )ℛ𝑡 = ℛ𝑞 ℛ𝑡 , by (2). This product of two reflections is a rotation. The proof of (4) is an inductive extension of (3). The details are left to Exercise 2. □

Symmetry at a Point It was mentioned earlier that certain figures, such as isosceles triangles, have line symmetry. This means that for some line 𝑚, there is a reflection ℛ𝑚 that maps the figure to itself. This is a special case of the definition of symmetry used in geometry. Definition 5.8. If 𝐹 is a set of points in the plane, a symmetry of 𝐹 is a rigid motion 𝑆 such that 𝑆(𝐹) = 𝐹. The set of all symmetries of 𝐹 is called the symmetry group of 𝐹. Example 5.9. An isosceles triangle has two symmetries: reflection in the perpendicular bisector of the base and the identity. A kite also has both a reflection and the identity as symmetries. But a rhombus has two reflections and two rotations: the identity and a rotation by 180 degrees (the product of the reflections). Rotational Symmetry. A very important property of symmetries is that the product of two symmetries of 𝐹 is also a symmetry of 𝐹. This is immediate: 𝑆𝑇(𝐹) = 𝑆(𝐹) = 𝐹 if 𝑆 and 𝑇 are symmetries. Also the inverse of a symmetry is a symmetry. And the identity is always a symmetry. This is what leads to the symmetry “group” terminology, since a symmetry group is an example of a transformation group. Definition 5.10. A transformation group is a nonempty set of transformations that is closed under multiplication (if 𝑆 and 𝑇 are in the group, so is 𝑆𝑇) and closed under inverses (if 𝑆 is in the group so is 𝑆 −1 ). Figure 6 gives two examples that have rotations centered at 𝑂 as symmetries. The figure on the left in Figure 6 has as symmetries the rotations with center 𝑂 and 1 2 3 4 5 rotation angles 72, 144, 216, 288, 0. These angles are multiples of 360 by 5 , 5 , 5 , 5 , 5 . Since the symmetries are closed under products, once we know 𝑂72 is a symmetry, so are all the powers 𝑂𝑘72 and these are exactly the rotations that are symmetries. We note that the 5/5 case is 𝑂360 = 𝑂0 = 𝐼. The figure on the right has seven rotational symmetries. Again, all the symmetries are rotations involving fractional multiples of 360; this time fractions with denominator 7. If one begins with 𝑇 = 𝑂360/7 , then the powers 𝑇 𝑘 for 𝑘 = 1, 2, 3, 4, 5, 6, 7 provide seven rotations of the form 𝑂360𝑘/7 , But if one begins with 𝑆 = 𝑂 (2/7)360 , then the

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Figure 6. Two Figures with Rotational Symmetry

powers 𝑆 𝑘 produce the same rotations but in this ordering of the fractional multiples 2 4 6 1 3 5 7 of 360: 7 , 7 , 7 , 7 , 7 , 7 , 7 . These figures and the symmetries play a dual role. As printed on the page, the lists of symmetries are the ones given. In the creation of the figures, the method was to create one small piece and then to apply all of the rotations in the list to produce the symmetric figure. So a group of transformations can generate a symmetric figure. But symmetries may occur in a figure that was created in other ways. One special feature of the two examples so far is the fact that the number of rotations is a prime number. If one picks an integer that is not prime as the denominator, then things get more complicated in terms of common factors. Reducing a fraction to its lowest terms will tell how many distinct powers the rotation has. Also, it should be noted that if an angle of rotation is not a rational multiple of 360, then all the powers of the rotation are distinct rigid motions and there are an infinite number of such powers. So if the rotational symmetries are a finite set, this excludes these irrational rotation angles.

Figure 7. Figures with 𝑘/12-fold Rotational Symmetry

In Figure 7 one shape was rotated by all the powers of a rotation 𝑂 (𝑘/12)360 , where 𝑘 = 1, 2, 3, 4, 5, 6. If one reduces these fractions to lowest terms, it becomes clear what the group of symmetries will be. Here is an important theorem about such symmetries. Theorem 5.11 (Generator of Rotational Symmetries). If the symmetries of a figure 𝐹 that are rotations with center 𝑂 are a finite set, they form a group made up of powers of a single rotation.

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Proof. The rotation 𝑆 with the smallest nonzero angle of rotation will be a generator; all the other rotations will be powers of 𝑆. Suppose that the rotation angle of 𝑆 is 𝑎. Pick a point 𝑃 distinct from 𝑂 and apply powers of 𝑆 to produce all the 𝑆 𝑘 (𝑃) (a finite number). If there is a rotation 𝑇 so that 𝑇(𝑃) is not among the 𝑃𝑘 , then it must be in the central angle between two points, say between 𝑆 3 (𝑃) and 𝑆 4 (𝑃). Then 𝑇𝑆 −3 has a smaller nonzero rotation angle than 𝑆, which is a contradiction. □ Regular Polygons. A group that is only made up of powers of one rotation is called a cyclic group of rotations. If the group contains 𝑛 rotations, then 𝑂360/𝑛 must generate the group. (As in the seven-sided star in Figure 6, other rotations may also generate the group.) If one takes a point 𝑃1 and defines the next point 𝑃2 = 𝑂360/𝑛 (𝑃1 ) and continues with 𝑃𝑘+1 = 𝑂360/𝑛 (𝑃𝑘 ), then 𝑃𝑛+1 = 𝑃1 and all later points repeat as well. This defines a set of 𝑛 distinct points 𝑃1 , 𝑃2 , . . . , 𝑃𝑛 . These points are all at the same distance from 𝑂, so they lie on a circle with center 𝑂. The powers of 𝑂360/𝑛 are symmetries of this set of points. The 𝑛 segments 𝑃1 𝑃2 , 𝑃2 𝑃3 , . . . , 𝑃𝑛 𝑃1 are congruent. Definition 5.12. A regular polygon with 𝑛 sides (a regular 𝑛-gon) is a polygon consisting of a set of 𝑛 vertices 𝑃1 , 𝑃2 , . . . , 𝑃𝑛 with 𝑛 rotation symmetries with a common center and 𝑛 congruent sides 𝑃1 𝑃2 , 𝑃2 𝑃3 , . . . , 𝑃𝑛 𝑃1 . There are other equivalent definitions of regular polygons that do not explicitly mention rotations. For example, they can be defined as convex polygons with all sides and all angles congruent. But then one needs a proof that the vertices are on a circle. Exercise 6 shows how to construct the rotations starting with equal angles and side lengths.

Figure 8. Polygons: Some Regular, Some Not

Dihedral Symmetry. The snowflake figure in Figure 2 has mirror symmetry as well as rotational symmetry. Since the product of two reflections is not a reflection but a rotation, the reflection symmetries of a figure are not a symmetry group. Instead the symmetries will be made up of both reflections and rotations. If we use what we have learned in Theorem 5.7 about the relationship between reflections and rotations, we get this important result. Theorem 5.13 (Reflection-Rotation Pairing). If for a line 𝑚 through 𝑂, the reflection ℛ𝑚 is a symmetry of a figure 𝐹, then the set of symmetries of 𝐹 that are rotations with center

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𝑂 is in one-to-one correspondence with the lines of symmetry of 𝐹 that pass through 𝑂. Such a one-to-one correspondence is defined by mapping each such rotation 𝑆 to 𝑆ℛ𝑚 . Proof. Multiply each rotation by the reflection and get a different reflection each time, exactly matching the number of rotations, not forgetting the identity among the rotations. □ Corollary. For a figure 𝐹 with a finite set of symmetries, all of which are either rotations centered at 𝑂 or reflections in lines through 𝑂, there are two possibilities: • The set of symmetries is a cyclic group of 𝑛 rotations. • The set of symmetries consists of a cyclic group of 𝑛 rotations and an equal number 𝑛 of line reflections. This is called a dihedral group. An example is the figure on the right of Figure 6 with seven rotational symmetries. There is a line of reflection through every one of the seven vertices. In Figure 8, the irregular hexagon at center left has the same symmetries as the regular 3-gon (equilateral triangle), three rotations of 0, 120, and 240 degrees, and three reflections. The regular hexagon has six rotations that are powers of 𝑂60 . There are only three lines of reflection through the vertices, since each such line contains two vertices. But there are three more reflection lines that are perpendicular bisectors of the sides. Similar reasoning applies to the regular 4-gon, also known as the square. A regular 𝑛-gon has 𝑛 reflection and 𝑛 rotation symmetries. If a figure has two mirror reflection symmetries, then the product is a rotation symmetry with rotation angle twice the angle between the two mirror lines. All the powers of the rotation are rotation symmetries, which are the products of an even number of reflections. The odd products of reflections are all reflection symmetries. This includes the symmetry of the two mirrors of a kaleidoscope, hence the name dihedral symmetry. If two mirrors are placed facing each other to form a wedge, then looking into the mirrors shows reflections of reflections. If the angles are chosen carefully, the scene appears to be a nonoverlapping design with dihedral symmetry as in Figures 2 and 9. Angles and Regular Polygons. The central rotation angle of a figure with 𝑛fold rotation symmetry is 360/𝑛. Using the Angle Sum Theorem, Theorem 4.12, one can compute from this the angles at each vertex of a regular polygon and other related figures. A regular 𝑛-gon can be divided into 𝑛 congruent isosceles triangles with the apex angle measure 360/𝑛. Then, since the sum of the angle measures is 180, each base angle of the isosceles triangles has measure (1/2)(180 − 360/𝑛). But the vertex angle of the 𝑛-gon has twice this measure, which is 180(𝑛 − 2)/𝑛. An example of this is the regular pentagon on the left of Figure 10, where the vertex angle is 180(3/5) = 108. The figure also shows on the right a regular star polygon, or pentagram, which is not a polygon by our definition since the sides intersect, but it does have the same dihedral symmetry and the same vertices. The angles in this figure can be deduced from the regular pentagon and from the isosceles triangles in the figure.

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Figure 9. Kaleidoscopic Image

Figure 10. Regular Pentagon and Star Pentagon

The vertices of star polygons satisfy the definition of a regular polygon, except that they are not polygons since they have self-intersections. Figure 6 shows a star polygon with seven vertices with a rotation of (2/7)360 mapping consecutive vertices. With a given set of five vertices, all the star pentagons are congruent. But there are two kinds of star 7-gons, the other obtained by rotating by (3/7)360. A few of these interesting figures will be explored in Exercise 5.

Orientation of a Plane For figures on a piece of paper or in a book, the intuitive sense of right and left or clockwise and counterclockwise rests on knowing which side of the paper is the top side and which is the bottom side. But for other planes, for example a plane in space that may be the face of a polyhedron or a tangent plane to a sphere, there may be no obvious top or bottom. In such cases it is common to indicate an orientation by specifying a triangle with an order sequence of vertices to indicate a clockwise or counterclockwise direction: △𝐴𝐵𝐶, △𝐵𝐶𝐴, and △𝐶𝐴𝐵 define one orientation and △𝐶𝐵𝐴, △𝐵𝐴𝐶, and △𝐴𝐶𝐵 define the opposite one. This section will connect the earlier discussion of rotation angle with this way of orienting the plane. For some readers, this may feel a bit technical, so we should mention that there is a shorter method for orienting the plane using translations as a

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consequence of Theorem 7.35, though that discussion does not include defining orientation from a triangle and it requires the Euclidean Parallel Postulate. Whether you dive deeply into this section or not, be sure to read the following short but important section on orientation-preserving and orientation-reversing transformations. B

A C Figure 11. Plane Orientation from a Triangle with Ordered Vertices

Orientation and Positive Half-Planes. We have already seen in Definition 5.5 for rotation angle that orientation at a point is equivalent to choosing one of two classes of protractors. A more visual way of expressing this choice is to indicate the positive half-plane associated with a ray or directed line. ⃗ the positive halfDefinition 5.14. Given a protractor 𝛿 at a point 𝑃, for any ray 𝑃𝐴 plane of this ray is the half-plane of 𝑃𝐴 consisting of points 𝐵 with the signed measure 𝑚+ ∠𝐴𝑃𝐵 > 0. The other half-plane is the negative half-plane. Remark 5.15. A choice of protractor at 𝑃 makes a choice of positive half-plane for every ray at 𝑃. A choice of positive half-plane for a ray at 𝑃 does not choose a specific protractor but rather one of the two sets of protractors that define orientation at 𝑃. Here are some things that can be deduced from the definition: ⃗ with polar angle 𝜃(𝑃𝐴) ⃗ = 𝑎 is the set of points 𝐵 • The positive half-plane of 𝑃𝐴 ⃗ with 𝑎 < 𝜃(𝑃𝐵) < 𝑎 + 180. • Protractors 𝛿(𝑡) and 𝛿(𝑡 + 𝑘) determine the same positive half-planes while 𝛿(−𝑡 + 𝑘) determines the opposite half-planes. ⃗ is the negative half-plane of 𝑃𝐴. ⃗ • The positive half-plane of the opposite ray of 𝑃𝐴 At this point we recall from page 14 of Chapter 2 that a directed line 𝐴𝐵 consists ⃗ and the other rays on the line with of the line along with the direction defined by 𝐴𝐵 the same direction. In the same way, a directed segment 𝐴𝐵 is a segment with the ⃗ on its containing line. direction of 𝐴𝐵 We will not introduce new notation for directed lines but will just use the words “directed line” in this manner to indicate that a direction is assigned. Therefore, the directed line 𝐵𝐴 is the same line as directed line 𝐴𝐵 but with opposite direction.

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Definition 5.16. An orientation of a line is the choice of a half-plane of the line as the positive half-plane for each direction of the line, with opposite directions having opposite positive half-planes. At every point 𝑃 of a line 𝐴𝐵, for either direction of the line there is a unique ray ⃗ in that direction. 𝑃𝑄 Definition 5.17. If a point 𝑃 with an orientation is on a line with an orientation, the orientations match if for a given direction both the positive half-plane of 𝑃 and the positive half-plane of the line are the same. The orientations at two points 𝐴 and 𝐵 are matching if the orientations at both 𝐴 and 𝐵 match the same orientation of the line 𝐴𝐵. In less formal terms, a directed line defines a ray or arrow at every point. If you are facing in a direction along the line, the line orientation tells you which side of the line is on your left and so does the orientation of the point where you are standing. The two match if what is on the left is the same. The following theorem is key to orienting the whole plane. Theorem 5.18 (Triangle Orientation). Let 𝐴, 𝐵, 𝐶 be a sequence of noncollinear points. This fixes an ordering of the vertices of △𝐴𝐵𝐶. For △𝐴𝐵𝐶, there is one set of matching orientations of the vertices and the sides so that the intersection of the positive hyperplanes for the directed lines 𝐴𝐵, 𝐵𝐶, and 𝐶𝐴 is the interior of △𝐴𝐵𝐶. This is called the counterclockwise orientation for the ordered triangle. In this case the signed angle measures 𝑚+ ∠𝐴𝐶𝐵, 𝑚+ ∠𝐶𝐵𝐴, 𝑚+ ∠𝐵𝐴𝐶 are all positive. There is also a second set of matching orientations at the vertices so that none of these positive hyperplanes contains the interior of △𝐴𝐵𝐶. The signed measures of the angles above are all negative. This is the clockwise orientation for the ordered triangle. For this theorem we distinguish the order of the vertices of the triangle. Since the three ordered triangles △𝐴𝐵𝐶, △𝐵𝐶𝐴, and △𝐶𝐴𝐵 have the same directed sides, the orientations and the positive half-planes and the counterclockwise orientation are the same for each of these triangles. For the triangles in the other order, △𝐶𝐵𝐴, △𝐵𝐴𝐶, and △𝐴𝐶𝐵, the directed sides in the theorem are 𝐶𝐵, 𝐵𝐴, 𝐴𝐶, so the counterclockwise orientation for these triangles is the clockwise orientation for the other ordered triangles The proof follows immediately from this next theorem, the In-Out Lemma. Lemma 5.19 (In-Out). For ∠𝐵𝐴𝐶, the signed measure 𝑚+ ∠𝐵𝐴𝐶 > 0 for an orientation ⃗ at 𝐴 if and only if the intersection of the positive half-planes at 𝐴 for directions 𝐶 𝐴 and ⃗ is the interior of ∠𝐵𝐴𝐶. 𝐴𝐵 The reason this is called the In-Out Lemma is that the two possible orientations at 𝐴 describe two in-out paths with the condition that the intersection of the positive half-planes of the two directed lines at 𝐴 be the interior of ∠𝐵𝐴𝐶.

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C

B

A Figure 12. In towards 𝐴 and Out from 𝐴 with Positive Half-planes on Left

In the case stated in the lemma, illustrated in Figure 12, one walks from 𝐶 to 𝐴 and then from 𝐴 to 𝐵 with the interior of △𝐶𝐴𝐵 always on the left. This agrees with our sense of a counterclockwise orientation of the triangle. If the orientation at 𝐴 is reversed and now 𝑚+ ∠𝐶𝐴𝐵 > 0, then one walks from 𝐵 to 𝐴 to 𝐶 with the triangle always on the right (since right is now the positive side). Proof. By definition, if the signed measure of ∠𝐵𝐴𝐶 is positive, 𝐶 is in the positive half-plane of the directed line 𝐴𝐵. The lemma is proved if we can show that 𝐵 is in ⃗ the positive half-plane of 𝐶𝐴 at 𝐴 with direction 𝐶 𝐴. Then the half-plane contains the interior of the angle. ⃗ = 𝑏. The point 𝐶 is in the positive half-plane of directed line 𝐴𝐵 if, for Let 𝜃(𝐴𝐵) ⃗ some choice of 𝑐 = 𝜃(𝐴 𝐶), 𝑏 < 𝑐 < 𝑏 + 180. A direction for 𝐶𝐴 is 𝑐 − 180, a polar angle opposite 𝑐. The condition for 𝐵 to be in the positive half-plane of directed 𝐶𝐴 is 𝑐 − 180 < 𝑏 < 𝑐, which is equivalent to the previous inequality. The converse follows from this also. When the sign of the angle measure is negative, the intersection of the negative half-planes is the angle interior. □ Now that the In-Out Lemma is proved, we can prove the Triangle Orientation Theorem, Theorem 5.18 above. Proof of the Triangle Orientation Theorem. Given an orientation at 𝐴, there is only one orientation at each of 𝐵 and 𝐶 that is matched with the orientation at 𝐴. The most important thing to prove is that these orientations at 𝐵 and 𝐶 also match. Then we must also show the intersection of the positive half-planes is the triangle interior. Consider △𝐴𝐵𝐶 and the directed lines 𝐴𝐵, 𝐵𝐶, 𝐶𝐴. Choose the orientation at 𝐴 so that 𝑚+ ∠𝐵𝐴𝐶 > 0 and then orient the points 𝐵 and 𝐶 to match this orientation at 𝐴. By the In-Out Lemma, the positive half-plane of the directed line 𝐴𝐵 contains the interior of ∠𝐵𝐴𝐶. This direction is the “in” direction for the orientation at 𝐵. Therefore, the “out” direction is directed line 𝐵𝐶 with positive half-plane containing the interior of ∠𝐶𝐵𝐴 and so containing the interior of the triangle.

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Also, the positive half-plane of the directed line 𝐶𝐴 contains the interior of ∠𝐵𝐴𝐶. This is the “out” direction at 𝐶, so the inward directed line 𝐵𝐶 must have its positive half-plane contain the interior of ∠𝐴𝐶𝐵, so containing the interior of the triangle. This shows that the orientations at 𝐵 and 𝐶 match this orientation of 𝐵𝐶. The In-Out Lemma shows that the oriented angle measures are positive for the three angles named in the theorem. We have also shown that the intersection of the three half-planes is the interior of the triangle. With these orientations and with these directions of the sides, the positive half-plane for each side is the half-plane containing the opposite vertex. For the last part of the proof, if the orientation at 𝐴 is reversed, then 𝑚+ ∠𝐶𝐴𝐵 > 0. The proof can be repeated for the triangle with vertices in order 𝐴𝐶𝐵, In this case the positive half-planes of the directed lines 𝐴𝐶, 𝐶𝐵, 𝐵𝐴 will also have intersection the interior of the triangle. But if one uses the opposite directions of these lines, then the positive half-planes are on the opposite side of the lines and so do not intersect the triangle interior. □ Orientation of the Whole Plane. Before stating the theorem, this is one more bit of useful language that we will define. Definition 5.20. A triangle △𝐴𝐵𝐶 with cyclicly ordered vertices is a triangle with the vertices taken in one of these three orders: 𝐴𝐵𝐶, 𝐶𝐴𝐵, 𝐵𝐶𝐴. The opposite cyclic order of the vertices would take the vertices in order 𝐶𝐵𝐴, 𝐴𝐶𝐵, 𝐵𝐴𝐶. Theorem 5.21 (Orientation of the Plane by a Cyclicly Ordered Triangle). To orient the entire plane: (1) Given an orientation at a point 𝐴, there is a unique orientation at each point of the plane so that for every pair of points 𝑃 and 𝑄, the orientations at these points match. This set of orientations is independent of the choice of the initial point 𝐴. If the orientation at the initial point is reversed, so is the orientation at every point. (2) For △𝐴𝐵𝐶 with cyclicly ordered vertices, the counterclockwise orientation of this triangle determines an orientation at 𝐴 and therefore a unique orientation at every point of the plane. (3) Given such an orientation at every point of the plane, for any △𝐷𝐸𝐹, there is a cyclic order of the vertices that will give the triangle a counterclockwise orientation. Proof. Part (1): Given an orientation at 𝐴, for any point 𝑃, there is a unique orientation at 𝑃 matching the orientation at 𝐴 along the line 𝐴𝑃. We must show that these orientations at any pair of points 𝑃 and 𝑄 match. This is clear if the three points are collinear. If they are not collinear, consider the ordered triangle △𝐴𝑃𝑄. By the Triangle Orientation Theorem, Theorem 5.18, there are two choices of matching orientations at the vertices, one for counterclockwise orientation and one for clockwise. Pick the choice that agrees with the initial orientation at 𝐴. Then by the theorem, the orientations at 𝑃 and 𝑄 match the one at 𝐴 and also match each other.

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Finally, to see that set of orientations does not depend on the choice of 𝐴, note that once one has orientations so that every pair matches, then the choice of orientation at any one point determines all of the orientations. And there are only two choices for this system. If one begins with a choice at 𝐴, then if one instead starts at 𝐵 with the orientation that matches the one at 𝐴, it must be the same set of orientations. Part (2): From Theorem 5.18, there is one choice of orientation that will make this cyclicly ordered triangle counterclockwise. So by (1) this determines orientations at every point of the plane uniquely. Part (3): Again by Theorem 5.18, there are two ways that △𝐷𝐸𝐹 can match orientations at the vertices. One of these must agree with the orientations on the whole plane. By re-ordering the sequence of vertices, one can make this clockwise or counterclockwise. □

Orientation-Preserving and Orientation-Reversing Transformations A rigid motion maps directed lines to directed lines and half-planes to half-planes, so such a transformation will either preserve or reverse the orientation of the plane at a point. And since the orientation at every point of the plane is defined by the orientation at a single point, a transformation will either preserve or reverse the orientation of the whole plane. Line reflections reverse orientation. In the proof of the Rotation Theorem, Theorem 5.3, the formula for the effect of a line reflection on polar angles was shown to be 𝑥′ = 2𝑎 − 𝑥, a mapping that reverses the sign of angles at the point. Clearly a product of two transformations that preserve orientation continues to preserve orientation. But a product of two transformations that reverse orientation also preserves orientation; since there are only two choices of orientation, two reversals arrive at the original orientation. On the other hand, a product of an orientationpreserving and an orientation-reversing transformation is orientation-reversing. Since it was shown earlier that every rigid motion is the product of one, two, or three line reflections, this separates the products of two reflections as orientationpreserving and the ones and threes as orientation-reversing. Theorem 5.22 (Odd and Even). Every rigid motion either preserves or reverses orientation. (1) If a rigid motion is the product of an even number of line reflections, it is orientationpreserving. If it is a line reflection or a product of an odd number of reflections, it is orientation-reversing. (2) If two segments 𝐴𝐵 and 𝐶𝐷 have the same length, there is exactly one orientationpreserving rigid motion and one orientation-reversing rigid motion that maps 𝐴𝐵 to 𝐶𝐷 and also one of each kind that maps 𝐴𝐵 to 𝐷𝐶. For convenience, we will sometimes use the terms “even” or “even parity” when referring to orientation-preserving rigid motions; the orientation-reversing ones have opposite parity and are called “odd” rigid motions or have “odd parity”.

Orientation-Preserving and Orientation-Reversing Transformations

B

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Odd Image of P

P

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Even Image of P

C

Figure 13. Odd and Even Images of a Segment and a Point

Proof. The first point is proved in the discussion before the statement of the theorem. For the second point, in the proof of the Theorem 4.2 on congruence of segments, a rigid motion 𝑆 was constructed that maps 𝐴𝐵 to 𝐶𝐷. This 𝑆 is either a line reflection ℛ𝑚 or the product of two reflections ℛ𝑛 ℛ𝑚 . The product 𝑇 = 𝑆ℛ𝐴𝐵 also maps 𝐴𝐵 to 𝐶𝐷: 𝑇 is even if 𝑆 is odd or vice versa. Therefore, 𝑆 and 𝑇 are orientation-preserving and orientation-reversing rigid motions from 𝐴𝐵 to 𝐶𝐷. To show uniqueness, pick the even transformation. We can suppose it is the one labeled 𝑆. Let 𝑈 be any rigid motion mapping 𝐴𝐵 to 𝐶𝐷. Then 𝑆 −1 𝑈(𝐴) = 𝐴 and 𝑆 −1 𝑈(𝐵) = 𝐵, so the line 𝐴𝐵 is fixed. This means that either 𝑆 −1 𝑈 = 𝐼 or 𝑆 −1 𝑈 = ℛ𝐴𝐵 by the Line Reflection Theorem, Theorem 3.4. If 𝑈 is even, then 𝑆 −1 𝑈 is also even, so the product must be 𝐼, and 𝑈 = 𝑆. If 𝑈 is odd, then 𝑆 −1 𝑈 = ℛ𝐴𝐵 so 𝑈 = 𝑆ℛ𝐴𝐵 = 𝑇. This shows that there is only one odd rigid motion and one even rigid motion mapping 𝐴𝐵 to 𝐶𝐷, namely 𝑆 and 𝑇. Of course, the same applies to maps from 𝐴𝐵 to 𝐷𝐶, with a different pair of maps. □ Example 5.23 (Equal Chords). Segments with endpoints on a circle are called chords of the circle. Here is a simple example of how knowledge about fixed points and orientation can be used to extend what is proved from triangle congruence. Let 𝐴, 𝐵, 𝐶, 𝐷 be points on a circle 𝑐 with center 𝑂. If 𝐴𝐵 ≅ 𝐶𝐷 (i.e., they are equal chords), then △𝐴𝑂𝐵 ≅ △𝐶𝑂𝐷 by SSS, since two sides of each triangle are radii. The congruence transformation 𝑇 maps 𝐴𝐵 to 𝐶𝐷 and 𝑂 to itself.

Figure 14. Congruent Chords

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5. Rotation and Orientation

Since 𝑂 is a fixed point of 𝑇, 𝑇 is either a line reflection or a rotation. Because 𝑂 is on the perpendicular bisector 𝑚 of 𝐴𝐵, ℛ𝑚 𝑇 is a rigid motion mapping 𝐴𝐵 to 𝐷𝐶 that also has 𝑂 as a fixed point. This rigid motion has opposite parity from 𝑇 so it is a rotation if 𝑇 is a reflection and a reflection if 𝑇 is a rotation. Thus, there exists a rotation that takes one chord to the other and also a reflection that does so (for some ordering of endpoints). This tells how the segments are positioned relative to one another; the four points — in some order — are the vertices of an isosceles trapezoid.1

Exercises and Explorations ⃗ ⃗ ⃗ 1. (Polar Angles and Reflections). At a point 𝑂, let the rays 𝑂 𝐴, 𝑂 𝐵, 𝑂 𝐶 have polar angles 𝑎 = 10, 𝑏 = 40, 𝑐 = 90. ⃗ (a) What is the polar angle 𝑑 of the ray 𝑂 𝐷 so that these rotations are equal: ℛ𝑂𝐴 ℛ𝑂𝐵 = ℛ𝑂𝐶 ℛ𝑂𝐷 ? Show your reasoning. ⃗ (b) What is the polar angle 𝑒 of the ray 𝑂 𝐸 so that these reflections are equal: ℛ𝑂𝐴 ℛ𝑂𝐵 ℛ𝑂𝐶 = ℛ𝑂𝐸 ? Show your reasoning. (c) Draw one or more careful figures to show that your answers match the drawing. 2. (Odd and Even). Complete the proof of statement (4) in Theorem 5.7 with a convincing informal or formal argument. 3. (Twelve-fold Symmetry). Figure 7 shows six figures with symmetry groups relating to the number 12. (a) What will the figure look like if the series continues for 𝑘 = 7, 8, 9, 10, 11, 12? If the shapes are numbered from 1 to 5 from left to right, are any of these shapes repeated? Are any not? Explain your reasoning. (b) Assume a figure has the rotations 𝑂90 and 𝑂150 in its symmetry group. What other rotational symmetries are products involving these two rotations? What is a generator of this group of symmetries? (c) Describe all the star polygons with 12 vertices. How is your answer related to the figures in Figure 7? 4. (Kaleidoscope). Consider four lines through a point 𝑂, equally spaced as in Figure 9. Number the lines consecutively 1, 2, 3, 4, and let the reflections in the lines be ℛ1 , ℛ2 , ℛ3 , ℛ4 . To answer these questions, you may use algebra or arithmetic, but following mappings from point to point by drawing arrows in the figures also is a good and very geometrical method. (a) Each of these products is a line reflection. Tell the name of each one: ℛ1 𝑂90 , ℛ2 𝑂90 , ℛ3 𝑂90 , ℛ4 𝑂90 . (b) Each of these products is a line reflection. Tell the name of each one: ℛ1 𝑂180 , ℛ2 𝑂180 , ℛ3 𝑂180 , ℛ4 𝑂180 . 1

A quadrilateral 𝐸𝐹𝐺𝐻 with the line through the midpoints of 𝐸𝐹 and 𝐺𝐻 being a line of symmetry.

Exercises and Explorations

63

5. (Angles of Regular 𝑁-gons). Use angle sum to investigate. (a) Find the vertex angles of some common regular 𝑛-gons: equilateral triangle, square, pentagon, hexagon, octagon. What is the general formula for 𝑛 sides? Also, find the angles in the pentagram. (b) For each of the previous regular polygons, find all the star polygons (up to congruence) that have the same vertices. Some will have no star polygons, some will have one or more, and some will have figures that seem to follow the rules for a star polygon but will have too few sides. Draw the examples, and then say something about how the nature of the number 𝑛 determines the number of star polygons. 6. (Defining Regular 𝑛-gons). The goal of this problem is to find a rotation symmetry in an alternate definition of regular polygon. (a) Consider a figure as in Figure 15 made up of points 𝐴, 𝐵, 𝐶, 𝐷, with 𝐴𝐵 ≅ 𝐵𝐶 ≅ 𝐶𝐷 and angles ∠𝐴𝐵𝐶 ≅ ∠𝐵𝐶𝐷, with 𝐴 and 𝐷 on the same side of 𝐵𝐶. Let ℛ1 be reflection in the angle bisector of ∠𝐴𝐵𝐶 and let ℛ2 be reflection in the perpendicular bisector of 𝐵𝐶. If 𝑂 is the intersection of these two reflection lines, show that ℛ2 ℛ1 is a rotation that maps △𝐴𝐵𝐶 to △𝐵𝐶𝐷.2 (b) If a polygon 𝑃1 𝑃2 . . . 𝑃𝑛 has every consecutive four points (including wraparound 𝑅𝑛 𝑅1 as consecutive) matching the description of 𝐴, 𝐵, 𝐶, 𝐷 above, show why this polygon is a regular polygon. (c) Show that a figure 𝑃1 𝑃2 . . . 𝑃𝑛 is a star polygon if it is as described in the previous item, except that sides may intersect.

B

C

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O

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Figure 15. Rotation of a Vertex Angle

2 The existence of the intersection point 𝑂, which is also the intersection of two perpendicular bisectors, requires Theorem 7.6. It is not true always without the Dilation Axiom or the equivalent Euclidean Parallel Postulate.

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5. Rotation and Orientation

7. (Rhombi in a Symmetric Figure). Use your knowledge of angles to find all the angles in the symmetric figure in Figure 16. You should assume any part of the figure that looks like a rhombus is a rhombus.

Figure 16. Symmetric Figure with Rhombi

8. (Angles in Symmetric Figures). Use your knowledge of angles to find all the angles in the symmetric figure in Figure 17.

Figure 17. Figures with 90-degree Rotational Symmetry

9. Let 𝑇 be a rigid motion that maps 𝐴𝐵 to 𝐵𝐴. Experiment with a segment 𝐴𝐵 on paper and convince yourself that 𝑇 must be one of only two rigid motions. Tell what they are and explain convincingly why this is so. D

C

A

B

Figure 18. Square with Design

Exercises and Explorations

65

10. (90-degree Rotations Experiment). Label and cut out a square 𝐴𝐵𝐶𝐷. Draw a simple triangular design on it that will be easy to trace, as in Figure 18, for example. (a) Trace your square. Then apply 𝐴90 to the square several times and trace the images to get a rotationally symmetric pattern. Then apply 𝐶90 to these four squares several times and trace to get a bigger symmetric figure. (b) Find an example of a triangle and its image by 𝐶90 𝐴90 . From what you see, what transformation is this product? (c) Next, find an example of a triangle and its image by 𝐶−90 𝐴−90 . Again, what rigid motion is the product? 11. (Transformation Groups). Transformation groups are defined in Definition 5.10. Tell which of the following sets are transformation groups and explain your answer. (Note: The identity map is considered to be a rotation; it is the rotation 𝐴0 for any 𝐴.) (a) Given a point 𝐴, the set of all rotations 𝐴𝜃 . (b) Given a point 𝐴, the set of all reflections ℛ𝑚 , for lines 𝑚 through 𝐴. (c) Given a point 𝐴, the set of all rigid motions that fix 𝐴. (d) The set of all orientation-preserving rigid motions. (e) The set of all orientation-reversing rigid motions.

Chapter 6

Half-turns and Inequalities in Triangles

The product of reflections in two lines perpendicular at 𝐴 is a rotation with center 𝐴 and angle of rotation 180 degrees. This is one case where orientation does not matter, since rotation by +180 or −180 is the same rotation. Such a rotation is called the halfturn at 𝐴 and is denoted by ℋ𝐴 . This transformation has some special properties.

Half-turn Properties Theorem 6.1. Let the half-turn ℋ𝐴 be rotation by 180 degrees with center 𝐴. (1) For any point 𝐵 distinct from 𝐴, ℋ𝐴 (𝐵) is the point on 𝐴𝐵 so that 𝐴 is the midpoint of 𝐵ℋ𝐴 (𝐵). (2) ℋ𝐴 ℋ𝐴 = 𝐼, so ℋ𝐴−1 = ℋ𝐴 . (3) For a line 𝑚, ℋ𝐴 (𝑚) = 𝑚 if 𝐴 is on 𝑚. If 𝑚 does not pass through 𝐴, then ℋ𝐴 (𝑚) is a line 𝑚′ parallel to 𝑚.

m A

m'

Figure 1. Half-turn Image of a Line

67

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6. Half-turns and Inequalities in Triangles

Proof. All these assertions except the last follow easily from the properties of rotations. It is also clear that if 𝑚 passes through 𝐴, then ℋ𝐴 maps the line to itself, though unlike the case of line reflection, the only point fixed on the line is 𝐴. If 𝑚 does not pass through 𝐴, it cannot intersect the line 𝑚′ = ℋ𝐴 (𝑚). If the lines intersect at point 𝑃, then their images ℋ𝐴 (𝑚) and ℋ𝐴 (𝑚′ ) would intersect at ℋ𝐴 (𝑃). But the image lines are the same as the originals: 𝑚′ = ℋ𝐴 (𝑚) and 𝑚 = ℋ𝐴 (𝑚′ ). Since distinct lines cannot intersect at two points 𝑃 and ℋ𝐴 (𝑃), then 𝑚′ = 𝑚. But if 𝑃 □ and ℋ𝐴 (𝑃) are on 𝑚, then so is their midpoint 𝐴, contrary to assumption. Some of the properties of half-turns may remind you of line reflections. In fact, a half-turn with center 𝐴 can be considered to be a reflection in point 𝐴; the statement (1) in the theorem is then a definition of point reflection. Half-turns are not only a kind of reflection; they are also dilations with dilation ratio −1. This will be explored further when we take up dilations with negative ratio.

Inequalities with Angles Let △𝐴𝐵𝐶 be a triangle; the (interior) angles at the 𝐴, 𝐵, and 𝐶 are the angles ∠𝐶𝐴𝐵, ∠𝐴𝐵𝐶, and ∠𝐵𝐶𝐴. At each vertex there are also two exterior angles that are supplementary to the interior angles. Definition 6.2 (Exterior Angle). In △𝐴𝐵𝐶, there are two exterior angles at 𝐴. One is ⃗ ⃗ ⃗ and the opposite ray of 𝐴 the angle formed by 𝐴𝐵 𝐶 and the other is formed by 𝐴 𝐶 and ⃗ the opposite ray of 𝐴𝐵. In Figure 2, ∠𝐵𝐴𝐷 is an exterior angle at 𝐴.

C'

B M C

A

D

Figure 2. Proof of Exterior Angle Theorem

Theorem 6.3 (Exterior Angle Theorem). In any △𝐴𝐵𝐶, an exterior angle at a vertex such as 𝐴 has greater measure than either of the two interior angles at the other vertices, 𝐵 or 𝐶. Consequently, the sum of the measures of any two angles in the triangle is less than 180. Since we know from Theorem 4.12 that the sum of all the angles equals 180, this inequality follows from addition and the fact that the exterior angle is supplementary to a vertex angle. But we include this proof partly because it shows the inequality more visually and partly because it does not rely on the Dilation Axiom, so it is valid in neutral geometry, a point that will be of interest to some readers. The proof of Euclid

Inequalities with Angles

69

also has some historical interest. It was criticized because his axioms could not prove the location of the point labeled 𝐶 ′ since a postulate like the Plane Separation Axiom was lacking. See Greenberg [8, pp. 118–120]. ⃗ ⃗ Proof. As in Figure 2, let 𝐷 be a point on 𝐴𝐶 so that 𝐴 𝐷 is the opposite ray of 𝐴 𝐶. Apply ℋ𝑀 to △𝐴𝐵𝐶, where 𝑀 is the midpoint of 𝐴𝐵. The points 𝐴 and 𝐵 map to 𝐵 and 𝐴, respectively, and the image of 𝐶 is a point 𝐶 ′ on the same side of 𝐴𝐶 as 𝑀 since 𝑀𝐶 ′ does not intersect this line. And 𝐶 ′ is on the opposite side of 𝐴𝐵 from 𝐶 and on the same side as 𝐷, since 𝐶𝐶 ′ intersects 𝐴𝐵 at 𝑀. ⃗′ is interior to ∠𝐵𝐴𝐷. This means that 𝑚∠𝐵𝐴𝐶 ′ < 𝑚∠𝐵𝐴𝐷. But Therefore, 𝐴𝐶 ′ ∠𝐵𝐴𝐶 is the half-turn image of ∠𝐴𝐵𝐶, the interior angle of △𝐴𝐵𝐶, at 𝐵. This shows that the sum of the measures of the angles at 𝐴 and 𝐵 is less than 180. The same construction and reasoning can be applied again, with the roles of 𝐴, 𝐵, and 𝐶 permuted, to show the same results for any of the other vertices. □ Theorem 6.4 (Greater Angle Theorem). In a triangle, if one side is longer than another, then the angle opposite the longer side is greater than the angle opposite the shorter side. Conversely, if one angle is greater than another, then the side opposite the greater angle is longer than the side opposite the smaller angle. A D

B

C

Figure 3. Greater Side and Greater Angle

Proof. Let the triangle be △𝐴𝐵𝐶. Suppose ‖𝐴𝐵‖ > ‖𝐴𝐶‖. Then let 𝐷 be the point on 𝐴𝐵 so that ‖𝐴𝐷‖ = ‖𝐴𝐶‖. ⃗ Then we have two inequalities of angle measure. First, since 𝐶 𝐷 is interior to ∠𝐴𝐶𝐵, the measure 𝑚∠𝐴𝐶𝐷 < 𝑚∠𝐴𝐶𝐵. But since △𝐴𝐶𝐷 is isosceles, ∠𝐴𝐶𝐷 ≅ ∠𝐴𝐷𝐶. Since ∠𝐴𝐷𝐶 is an exterior angle of △𝐷𝐶𝐵, 𝑚∠𝐷𝐵𝐶 < 𝑚∠𝐴𝐷𝐶. But ∠𝐷𝐵𝐶 = ∠𝐴𝐵𝐶; so we conclude that 𝑚∠𝐴𝐵𝐶 < 𝑚∠𝐴𝐶𝐵. Since ∠𝐴𝐶𝐵 is opposite the longer side 𝐴𝐵, this proves the first part of the theorem. The proof of the converse is an application of what was just proved. Suppose 𝑚∠𝐴𝐵𝐶 < 𝑚∠𝐴𝐶𝐵. The opposite sides are 𝐴𝐶 and 𝐴𝐵, respectively. If ‖𝐴𝐶‖ = ‖𝐴𝐵‖, the triangle is isosceles and the angles must be equal. If ‖𝐴𝐶‖ > ‖𝐴𝐵‖, then by what we just proved, 𝑚∠𝐴𝐵𝐶 > 𝑚∠𝐴𝐶𝐵. Therefore, it must be true that ‖𝐴𝐶‖ < ‖𝐴𝐵‖. □

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6. Half-turns and Inequalities in Triangles

The Greater Angle Theorem can be used to prove the well-known “shortest distance” inequality. Theorem 6.5 (Triangle Inequality). For any triangle, the sum of the lengths of any two sides is greater than the length of the third side. B

D

C

A Figure 4. Congruent Segments 𝐵𝐷 and 𝐴𝐵

Proof. Let the triangle be △𝐴𝐵𝐶. We will prove that ‖𝐴𝐵‖ + ‖𝐵𝐶‖ > ‖𝐴𝐶‖. This ⃗ to create an is the general case, since the labels are arbitrary. Extend side 𝐶𝐵 to 𝐶𝐵 external angle as in Figure 4. Let 𝐷 be the point on 𝐶𝐵 so that ‖𝐵𝐷‖ = ‖𝐵𝐴‖ and 𝐵 is between 𝐶 and 𝐷. Then △𝐵𝐴𝐷 is isosceles, so ∠𝐵𝐷𝐴 ≅ ∠𝐵𝐴𝐷. But ∠𝐵𝐴𝐷 is interior to ∠𝐶𝐴𝐷, so 𝑚∠𝐵𝐷𝐴 = 𝑚∠𝐵𝐴𝐷 < 𝑚∠𝐶𝐴𝐷. In △𝐶𝐷𝐴, ∠𝐶𝐴𝐷 is the angle at vertex 𝐴 and ∠𝐵𝐷𝐴 is the same angle as ∠𝐶𝐷𝐴, which is the angle at vertex 𝐷. By the inequality for these angles, the opposite sides in △𝐶𝐷𝐴 satisfy this inequality: ‖𝐴𝐶‖ < ‖𝐶𝐷‖. But ‖𝐶𝐷‖ = ‖𝐴𝐵‖ + ‖𝐵𝐶‖ by construction of 𝐷. □ Corollary. For any triangle, the difference of the lengths of any two sides is less than the length of the third side. Proof. For △𝐴𝐵𝐶, ‖𝐴𝐵‖ + ‖𝐵𝐶‖ > ‖𝐴𝐶‖, so ‖𝐴𝐵‖ > ‖𝐴𝐶‖ − ‖𝐵𝐶‖. Also ‖𝐴𝐵‖ + ‖𝐴𝐶‖ > ‖𝐵𝐶‖ implies ‖𝐴𝐵‖ > −‖𝐴𝐶‖ + ‖𝐵𝐶‖. Therefore, □

‖𝐴𝐵‖ > |‖𝐴𝐶‖ − ‖𝐵𝐶‖| .

B D C A Figure 5. 𝐶 Closer to 𝐴 than 𝐵

Theorem 6.6 (Half-plane Distance Inequality). For any segment 𝐴𝐵, a point 𝐶 is in the half-plane of the perpendicular bisector opposite 𝐵 if and only if ‖𝐴𝐶‖ < ‖𝐵𝐶‖.

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Proof. If 𝐵 and 𝐶 are on opposite sides of 𝑚, the perpendicular bisector of 𝐴𝐵, the segment 𝐵𝐶 intersects 𝑚 at a point 𝐷. Since 𝐷 is on 𝑚, ‖𝐵𝐷‖ = ‖𝐴𝐷‖ by Theorem 3.10. Therefore, ‖𝐵𝐶‖ = ‖𝐵𝐷‖ + ‖𝐷𝐶‖ = ‖𝐴𝐷‖ + ‖𝐷𝐶‖ > ‖𝐴𝐶‖ by the triangle inequality. Conversely, the inequality ‖𝐵𝐶‖ > ‖𝐴𝐶‖ implies that 𝐶 must be on the opposite side of 𝑚 from 𝐵. Since the distances are unequal, 𝐶 cannot be on 𝑚. If 𝐶 is on the same side of 𝑚 as 𝐵, then ‖𝐵𝐶‖ < ‖𝐴𝐶‖, so 𝐶 must be on the opposite side from 𝐵. □

Circles and Distance to Lines These inequalities in triangles can be used to prove some basic relationships about distance. We recall from Theorem 3.7 that a point 𝑄 on a line 𝑚 is called the foot of the perpendicular from 𝑃 to 𝑚 if 𝑄 is the intersection of 𝑚 with the perpendicular to 𝑚 through 𝑃. Theorem 6.7 (Distance from Point to Line). If 𝑚 is a line and 𝑃 is a point, let 𝑄 be the foot of the perpendicular from 𝑃 to 𝑚. (1) The point 𝑄 is closer to 𝑃 than any other point on 𝑚. Consequently, ‖𝑃𝑄‖ is defined to be the distance from the point 𝑃 to the line 𝑚. (2) If 𝐾 is a point on 𝑚, the distance ‖𝑃𝐾‖ is a strictly increasing function of ‖𝑄𝐾‖.

P

Q

K

L

Figure 6. Distances from 𝑃 to Points on a Line

Proof. To prove (1) let 𝑄 be the foot of the perpendicular from 𝑃 to 𝑚. Let 𝐾 be any other point on 𝑚. Then △𝑃𝐾𝑄 is a right triangle. The right angle at 𝑄 is greater than the angle at 𝐾, so the opposite sides satisfy ‖𝑃𝑄‖ < ‖𝑃𝐾‖. ⃗ To prove (2) consider two points 𝐾 and 𝐿 on 𝑚, with 𝐿 on 𝑄 𝐾. If ‖𝑄𝐿‖ > ‖𝑄𝐾‖, then 𝐾 is between 𝑄 and 𝐿. By the Exterior Angle Theorem, the sum of the measures of any two angles in a triangle is less than 180, so in a right triangle, two angles must be acute. Therefore, in △𝑃𝐾𝑄, the acute angle ∠𝑃𝐾𝑄 is an exterior angle of △𝑃𝐿𝐾 at 𝐾. So the supplementary angle ∠𝑃𝐾𝐿 is obtuse and so greater than ∠𝑃𝐿𝐾. Thus ⃗ ‖𝑃𝐾‖ < ‖𝑃𝐿‖. This proves the result for points on 𝑄 𝐾. If the figure is reflected in 𝑃𝑄, the points on the opposite ray have the same distance relations by symmetry. □

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6. Half-turns and Inequalities in Triangles

Intersections of Lines and Circles. Theorem 3.13 described the possible intersections of lines and circles but did not relate the intersection behavior to distance. Using the inequalities just proved, we can now prove this relationship. The proofs of the first two points are immediate. A quicker proof of the third point, with a precise value for ‖𝑄𝐴‖, follows from the Pythagorean Theorem, Theorem 8.11; but the proof here shows how the triangle inequality can be used to estimate distance and construct a point as a limit. Theorem 6.8 (Circle and Line Distance). Let 𝑐 be a circle with center 𝑃 and radius 𝑟. Let 𝑚 be a line, with 𝑑 the distance from 𝑃 to 𝑚. (1) If 𝑑 > 𝑟, 𝑚 does not intersect 𝑐. (2) If 𝑑 = 𝑟, 𝑚 intersects 𝑐 at one point 𝑄 so that 𝑃𝑄 is perpendicular to 𝑚 and 𝑃𝑄 = 𝑑. The line 𝑚 is called the tangent to the circle at 𝑄. (3) If 𝑑 < 𝑟, 𝑚 intersects 𝑐 at two points 𝐴 and 𝐵; the interior points of 𝐴𝐵 are inside 𝑐. Proof. Let 𝑄 be the foot of the perpendicular from 𝑃 to 𝑚. In case (1) all points on 𝑚 are at a distance greater than 𝑟 from 𝑃, so there is no intersection. In case (2), since ‖𝑃𝑄‖ = 𝑟, 𝑄 is on 𝑐. All other points of 𝑚 are farther from 𝑃. For case (3) we have 𝑄 inside the circle since ‖𝑃𝑄‖ < 𝑟. There is also a point 𝑆 on 𝑚 outside the circle. Pick any 𝑆 on 𝑚 with ‖𝑄𝑆‖ > 𝑟 + ‖𝑃𝑄‖. Then by the triangle inequality ‖𝑃𝑆‖ > ‖𝑄𝑆‖ − ‖𝑃𝑄‖ > 𝑟.

P L 1= L 2 Q = L0

A

U2

S = U0 = U1

Figure 7. Converging to Intersection 𝐴 within all 𝐿𝑛 𝑈𝑛

Since 𝑃𝑄 is a line of symmetry for both 𝑐 and 𝑚, any point of intersection on one side of this line has a reflected point of intersection on the other side. So it is sufficient ⃗ to find exactly one point of intersection of 𝑐 with 𝑄𝑆. By Theorem 6.7 the function ‖𝑃𝐾‖ on 𝑚 is a strictly increasing function of ‖𝑄𝐾‖. ⃗ with ‖𝑃𝐴‖ = 𝑟. To see that there is such a So there can be at most one point 𝐴 on 𝑄𝑆 point, we can find 𝐴 by a limiting process. We have upper and lower estimates for ‖𝑄𝐴‖: 𝐿0 = 𝑄 has ‖𝑃𝑄‖ < 𝑟 and 𝑈0 = 𝑆 has ‖𝑃𝑆‖ > 𝑟, so 𝐴 must be in 𝐿0 𝑈0 . Let 𝑀1 be the midpoint of 𝐿0 𝑈0 . Then either ‖𝑃𝑀1 ‖ < 𝑟 or ‖𝑃𝑀1 ‖ > 𝑟 (unless the distance equals 𝑟, in which case 𝑀1 = 𝐴). In the first case, let 𝐿1 = 𝑀1 and 𝑈1 = 𝑈0 and in the second case let 𝐿1 = 𝐿0 and 𝑈1 = 𝑀1 .

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Now 𝐴 must be in 𝐿1 𝑈1 , an interval with half the length of the previous one as shown in Figure 7. Continue inductively this way, defining a new 𝐿𝑛 and 𝑈𝑛 by this method. Each interval 𝐿𝑛 𝑈𝑛 contains 𝐴 and is half as long as the previous one. By carrying out this process, we can find as many values for the decimal expansion of ‖𝑄𝐴‖ as we wish. So a real number ‖𝑄𝐴‖ is defined in the limit and so determines the location of point 𝐴 on 𝑚. ⃗ the distance to 𝑃 is increasing, for every 𝑛 We need to check ‖𝑃𝐴‖ = 𝑟. Since on 𝑄𝑆 we know that ‖𝑃𝐿𝑛 ‖ ≤ ‖𝑃𝐴‖ ≤ ‖𝑃𝑈𝑛 ‖. Also, from being lower and upper estimates, we know that ‖𝑃𝐿𝑛 ‖ < 𝑟 < ‖𝑃𝑈𝑛 ‖. So both ‖𝑃𝐴‖ and 𝑟 are in the same real interval. The length of this interval is |‖𝑃𝑈𝑛 ‖ − ‖𝑃𝐿𝑛 ‖| < ‖𝐿𝑛 𝑈𝑛 ‖ by the corollary to the Triangle Inequality Theorem. Since this length is arbitrarily small, it must be true that ‖𝑃𝐴‖ = 𝑟. □ Angle Bisectors and Distance. With the definition of distance from a point to a line, it is possible to compare distances from a point to two distinct lines. ⃗ Theorem 6.9 (Angle Bisector Locus). For ∠𝐵𝐴𝐶, the angle bisector 𝐴 𝐷 is the locus of points equidistant from the sides of the angle. The locus of points equidistant from the two lines 𝐴𝐵 and 𝐴𝐶 consists of the two lines that contain the angle bisectors of the four angles defined by the intersecting lines.

Q B A

C

P

D

Q'

Figure 8. Bisector Points Equidistant from Angle Sides

Proof. To prove the angle bisector is this locus, one needs to prove that any point on ⃗ ⃗ and 𝐴 this ray is equidistant from the rays 𝐴𝐵 𝐶, and also that any equidistant point is on the bisector. ⃗ Let 𝑃 be a point on 𝐴 𝐷 and let 𝑄 be the foot of the perpendicular from 𝑃 to 𝐴𝐵. ⃗ for if it were on the opposite ray, then the right triangle The point 𝑄 must be on 𝐴𝐵, △𝑃𝐴𝑄 would have the acute ∠𝑃𝐴𝐵 as an exterior angle. The reflection ℛ𝐴𝐷 maps 𝐴𝐵 to 𝐴𝐶. Therefore, the image 𝑄′ = ℛ𝐴𝐷 (𝑄) must be the foot of the perpendicular from 𝑃 to 𝐴𝐶. Since ‖𝑃𝑄‖ = ‖𝑃𝑄′ ‖, 𝑃 is equidistant from both lines, with closest points on the rays that are the sides of the angle. Now suppose that a point 𝐹 is a distance 𝑑 from each of the lines 𝐴𝐵 and 𝐴𝐶. If 𝑑 > 0, let 𝐺 and 𝐻, respectively, be the feet of the perpendiculars from 𝐹 to these lines. Then △𝐴𝐹𝐺 and △𝐴𝐹𝐻 are right triangles with 𝐹𝐺 ≅ 𝐹𝐻 and 𝐴𝐹 ≅ 𝐴𝐹.

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6. Half-turns and Inequalities in Triangles

Therefore, the triangles are congruent by the Hypotenuse-Leg Theorem, Theorem 4.7. ⃗ is the angle bisector of ∠𝐺𝐴𝐻. This implies ∠𝐺𝐴𝐹 ≅ ∠𝐻𝐴𝐹, so 𝐴𝐹 The rays of the two intersecting lines 𝐴𝐵 and 𝐴𝐶 form four angles and hence four rays that are angle bisectors. From the first part of this proof, if a point is on one of the angle bisectors, the feet of the perpendiculars are on the sides of the corresponding angle. Therefore, the 𝐹 above is on one of these angle bisectors. If 𝑑 = 0, then 𝐹 = 𝐴 is on both bisecting lines. □ A consequence of this theorem is that any interior point on the angle bisector of an angle is the center of a circle inscribed in the angle. This means that both sides of the angle are tangent to the circle at points on the sides of the angle. This leads to one of the well-known theorems about concurrence of special lines in triangles.

C

E

D J

A'

A F

B

Figure 9. Inscribed Circle in a Triangle

Theorem 6.10 (Inscribed Circle of a Triangle). For any △𝐴𝐵𝐶, the three angle bisectors are concurrent at a point inside the triangle; they all pass through a point 𝐽 that is the center of a circle tangent to all three sides of the triangle. Proof. By the Crossbar Theorem, Theorem 2.10, the angle bisector of ∠𝐶𝐴𝐵 intersects 𝐵𝐶 at a point 𝐴′ . Then the angle bisector of ∠𝐴𝐵𝐶 = ∠𝐴𝐵𝐴′ intersects 𝐴𝐴′ at a point 𝐽 interior to both angles. Let 𝐷, 𝐸, 𝐹 be the feet of the perpendiculars from 𝐽 to the lines 𝐵𝐶, 𝐶𝐴, 𝐴𝐵. Then since 𝐽 is on the bisector of ∠𝐶𝐴𝐵, 𝐽𝐸 ≅ 𝐽𝐹. Also, since 𝐽 is on the bisector of ∠𝐴𝐵𝐶, 𝐽𝐷 ≅ 𝐽𝐹. Therefore, 𝐽𝐷 ≅ 𝐽𝐸 and 𝐽 is on the angle bisector of ∠𝐵𝐶𝐴. All this is true by Theorem 6.9. Let 𝑐 be the circle with center 𝐽 and radius ‖𝐽𝐷‖. This circle is tangent to each side at the points 𝐷, 𝐸, 𝐹. □ There is a quite different proof of this theorem that uses Theorem 5.7 instead of distance. Using the notation of Figure 9, let 𝑅 = ℛ𝐴𝐽 ℛ𝐹𝐽 ℛ𝐵𝐽 . Since all the lines pass through 𝐽, 𝑅 = ℛ𝑘 for some line through 𝐽. But ℛ𝑘 maps 𝐶𝐵 to 𝐶𝐴, so 𝑘 bisects an angle with vertex 𝐶 formed by these two lines. Therefore, 𝐶 is on 𝑘, so 𝑘 = 𝐶𝐽 bisects ∠𝐴𝐶𝐵, Thus the three angle bisectors are concurrent. This proof and others by means of reflection products can be found in Dodge [7].

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There is one more theorem about tangents that will be proved here. Instead of starting with the tangent lines 𝐴𝐵 and 𝐴𝐶 and constructing a circle, how can tangent lines be constructed given a circle and a point 𝐴 external to the circle? This is stated as a theorem here, and the construction is in the proof. E1 Q E2 O

A

Figure 10. Construction of Tangents Through External Point 𝐴

Theorem 6.11 (External Tangent Construction). Given a circle 𝑐 and a point 𝐴 outside the circle, there are two lines through 𝐴 that are tangent to the circle. Proof. Given a circle 𝑐 with center 𝑂 and radius 𝑟, ‖𝑂𝐴‖ = 𝑑 > 𝑟. Let 𝑎 be the circle with center 𝑂 through 𝐴. The radius of 𝑎 is 𝑑. For any point 𝑄 on the circle, the line 𝑚 perpendicular to 𝑂𝑄 at 𝑄 is tangent to 𝑐. The line 𝑚 intersects 𝑎 at two points, 𝐸1 and 𝐸2 . Let 𝑇1 and 𝑇2 be rotations with center 𝑂 so that 𝑇1 maps 𝐸1 to 𝐴, and 𝑇2 maps 𝐸2 to 𝐴. Then 𝑇1 (𝑚) and 𝑇2 (𝑚) are each tangent lines to 𝑐 through 𝐴 since the rotations map 𝑐 to itself and map tangent lines to tangent lines. The points of tangency are 𝑇1 (𝑄) and 𝑇2 (𝑄). These are all the tangent lines, since if there is any other tangent line through 𝐴, a rotation will map it to 𝑚. □ There is another very pretty construction of these tangent lines using a circle with diameter 𝑂𝐴. This construction is in Exercise 3.

Reflections and the Triangle Inequality The Triangle Inequality Theorem, Theorem 6.5, can be “folded” by line reflection to solve some problems about shortest paths.

B

A Figure 11. Seeking the Shortest Path

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6. Half-turns and Inequalities in Triangles

The Burning Tent. Here is a distance problem that is fun to explore with geometry software: given a line 𝑚 and points 𝐴 and 𝐵 on the same side of 𝑚, what is the shortest path from 𝐴 to some point 𝐶 on 𝑚 and then on to 𝐵? Using dynamic geometry software, place a sliding point 𝐶 on 𝑚 and then watch how the sum of the lengths of the segments 𝐴𝐶 and 𝐶𝐵 vary as one slides the point 𝐶. Spoiler warning: The solution will be presented in a couple of paragraphs, so stop here if you want to experiment first. This problem has some real-world applications and also some amusing but notso-real situations that it models. For example, if the line is a river, one might want to locate a pumping station on the river’s edge so that the amount of water pipe needed is minimal. One popular scenario assumes again that the line is a body of water with a person with a bucket at 𝐴 and a tent at 𝐵. The tent happens to be on fire and the person wishes to scoop up some water and pour it on the fire. To what point on the water’s edge should the person run? In fact this problem is widely known as the Burning Tent Problem! The triangle inequality offers a solution. In fact the method resembles the proof of the triangle inequality in that the secret is to straighten out the broken path to a segment between two points. Reflect one of the points, say 𝐵, in the line to produce an image 𝐵 ′ . For any point 𝐶 on the line, the path ‖𝐴𝐶‖ + ‖𝐶𝐵‖ = ‖𝐴𝐶‖ + ‖𝐶𝐵 ′ ‖ since △𝐵𝐶𝐵 ′ is an isosceles triangle. But ‖𝐴𝐶‖ + ‖𝐶𝐵 ′ ‖ ≥ ‖𝐴𝐵 ′ ‖ by the triangle inequality, with equality only when 𝐶 is the intersection 𝐷 of 𝐴𝐵′ with the line. B' C B

D

A

Figure 12. The Shortest Path by Reflection

The triangle inequality offers a new way to look at reflected paths followed by billiard balls or light rays. The Fermat Principle says that light will follow the shortest (and hence the fastest) path. Since this is not a physics book, we do not presume to explain the motivation of light rays, but this is the way to construct the reflected path of light in a mirror. For this path, the angle of incidence equals the angle of reflection. In thinking in terms of mirrors and light (as opposed to rivers and tents), if one wants to reflect in a mirror a laser or other beam of light onto an object, it seems natural just to aim the beam at the image of the object in the mirror. This construction can be extended to solve the light reflection problem for two or more mirrors by reflecting the reflected images. Fagnano’s Problem. Another shortest path problem is Fagnano’s problem. Given △𝐴𝐵𝐶, construct the inscribed triangle △𝐷𝐸𝐹 that has the shortest perimeter.

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This is a shortest path problem, but what makes it challenging is that one is not given the starting point. There are several approaches for solving this problem, and one solution uses rotation results from Chapter 5. A R R'

P2

Q' Q

P1 B

P

C

Figure 13. An Inscribed Triangle Unfolded by Reflection

Start with any inscribed triangle △𝑃𝑄𝑅, with point 𝑃 on 𝐵𝐶. Inspired by the Burning Tent Problem, we make the path shorter by reflecting △𝐴𝐵𝐶 and 𝑃 in side 𝐴𝐵 and side 𝐴𝐶 to get images 𝑃1 and 𝑃2 of 𝑃 as in Figure 13. One can envision unfolding the original triangle with two flaps. From now on, we will assume that △𝐴𝐵𝐶 has only acute angles. With this assumption, the segment 𝑃1 𝑃2 intersects side 𝐴𝐵 at 𝑄′ and side 𝐴𝐶 at 𝑅′ . By the triangle inequality (applied twice) the length of the segment 𝑃1 𝑃2 is less than or equal to the length of the path from 𝑃1 to 𝑄 to 𝑅 to 𝑃2 for any choice of 𝑄 and 𝑅. The perimeter of the △𝑃𝑄′ 𝑅′ equals ‖𝑃1 𝑃2 ‖, so this triangle has the shortest possible perimeter — provided that one vertex is 𝑃. This is a start at a solution but there is a different solution with different perimeter for each starting point 𝑃. What is the optimal starting point for getting the shortest path? Consider the relationship between 𝑃1 and 𝑃2 . These points are related by a double reflection in lines intersecting at 𝐴, so ℛ𝐴𝐶 ℛ𝐴𝐵 (𝑃1 ) = 𝑃2 . But ℛ𝐴𝐶 ℛ𝐴𝐵 is a rotation with center 𝐴 and rotation angle 2𝑚∠𝐵𝐴𝐶. The key point is that this is a constant angle not depending on 𝑃. M

N

L Figure 14. Nested Isosceles Triangles

Whatever 𝑃 is, the triangle △𝑃1 𝐴𝑃2 is an isosceles triangle with the same angle at 𝐴, even though the locations and lengths of the sides of the triangle are different for

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6. Half-turns and Inequalities in Triangles

different 𝑃. To compare the base lengths ‖𝑃1 𝑃2 ‖ of these triangles, we can move each one to a fixed angle ∠𝐿𝑀𝑁 congruent to ∠𝑃1 𝐴𝑃2 , moving 𝐴 to 𝑀 and moving the other vertices, one to each side of the angle. This gives a family of nested isosceles triangles as shown in Figure 14. The triangle with the shortest side length ‖𝐴𝑃1 ‖ = ‖𝐴𝑃2 ‖ = ‖𝐴𝑃‖ will have the shortest base length ‖𝑃1 𝑃2 ‖. This follows immediately if we invoke the Dilation Axiom, since the isosceles triangles are all dilations of one another and the base lengths are proportional to the side lengths. There is also a longer proof that does not require the Dilation Axiom; this proof is outlined in Exercise 7, with the justifications of steps left as a problem. This solves the Fagnano problem. Let 𝐷 be the foot of the perpendicular from 𝐴 to 𝐵𝐶. Then ‖𝐴𝐷‖ < ‖𝐴𝑃‖ for any other 𝑃 distinct from 𝐷 by Theorem 6.7. If 𝐷1 and 𝐷2 are the reflections of 𝐷 in the sides 𝐴𝐵 and 𝐴𝐶, then ‖𝐷1 𝐷2 ‖ is the shortest base length among the isosceles triangles. If 𝐸 and 𝐹 are the intersections of 𝐷1 𝐷2 with the sides 𝐴𝐶 and 𝐴𝐵, respectively, then 𝐷𝐸𝐹 is the triangle with the shortest perimeter. A E F

B

D

C

Figure 15. Fagnano Solution: Orthic Triangle 𝐷𝐸𝐹

The line 𝐴𝐷 is called the altitude of the triangle through 𝐴, and 𝐷 is the foot of the altitude. Each vertex of a triangle has an altitude. If the angles of the triangle are acute, the foot of each altitude is on a side of the triangle rather than on an extended side. In fact, for the Fagnano solution triangle, 𝐸 is the foot of the altitude through 𝐵 and 𝐹 is the foot of the altitude through 𝐶. Here is the reason: if we solve the Fagnano problem a second time, but starting with a point 𝑃 on a different side, such as 𝐴𝐵, then we must get the same solution triangle 𝐷𝐸𝐹 since this is the unique triangle with smallest perimeter. But again, the solution will require that the point on the side 𝐴𝐵 is the foot of the altitude through 𝐶, so 𝐶𝐹 must be an altitude! Thus, the vertices of △𝐷𝐸𝐹 are the feet of the three altitudes of △𝐴𝐵𝐶. This is sometimes called the orthic triangle (orthogonal means perpendicular). This solution is shown in Figures 15 and 16. Since the sides of the orthic triangle are reflected paths, as in the Burning Tent Problem, the sides of △𝐴𝐵𝐶 are bisectors of the exterior angles of △𝐷𝐸𝐹. But the

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perpendicular lines to the exterior bisectors are the bisectors of the interior angles. Therefore the altitudes of △𝐴𝐵𝐶 are the angle bisector lines of △𝐷𝐸𝐹, so they are concurrent at a point 𝐽, by Theorem 6.10.

A F

E J

B D

C

Figure 16. Orthic Triangle of 𝐴𝐵𝐶 or 𝐽𝐵𝐶 or 𝐴𝐽𝐶 or . . .

We have been assuming all the angles of the triangle are acute. If a triangle has an angle that is not acute, then it is no longer true that all the feet of the altitudes lie on the interior points of the segments forming the sides of the triangle. In that case, the picture of an inscribed triangle no longer holds, but something surprising happens. In Figure 16, the point 𝐽 is the point of concurrence of the altitudes of △𝐴𝐵𝐶. The angle at 𝐽 of △𝐽𝐵𝐶 is obtuse if ∠𝐵𝐴𝐶 is acute. But consider the altitudes of △𝐽𝐵𝐶. These are still the lines 𝐵𝐸, 𝐶𝐹, and 𝐽𝐷 = 𝐴𝐷. Therefore, the feet of the altitudes are still the vertices of △𝐷𝐸𝐹, which is the orthic triangle of △𝐽𝐵𝐶. But now the point of concurrence of the altitudes is 𝐴! Now consider also △𝐽𝐴𝐵 and △𝐽𝐴𝐶, so that we examine every triangle with vertices chosen from the four points 𝐴, 𝐵, 𝐶, 𝐽. What are the altitudes? Another proof of the concurrence of the altitudes — from a completely different perspective — will be given in Theorem 8.51. For a surprising connection to glide reflections, see Figure 10.42 and the accompanying discussion.

Exercises and Explorations 1. (Point Reflections). A point reflection ℛ𝐴 with center 𝐴 can be defined in any dimension in this way: ℛ𝐴 (𝑃) = 𝑃 ′ , where 𝐴 is the midpoint of 𝑃𝑃 ′ . (a) Check that in the plane, ℛ𝐴 is the same as ℋ𝐴 and that it is orientationpreserving. (b) In dimension 1, if your space is a line, what does ℛ𝐴 look like? Is it orientationpreserving? (c) In 3-space, imagine a point reflection at the center of a globe. What happens to the map of each country on the globe? Is ℛ𝐴 orientation-preserving in this case?

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(d) How is point reflection related to a 180-degree rotation about the axis 𝑁𝑆 through the north and south poles of the globe? Can this rotation in 3-space be considered a line reflection? What is the perpendicular bisector of a point and its image? 2. (Kites and Circles). Prove that a circle can be inscribed in any kite. Then construct an example. 3. (Straightedge and Compass Tangent Line Construction). Given a circle 𝑐 with center 𝑂 and a point 𝐴 outside the circle, a more traditional construction of tangent lines to 𝑐 through 𝐴 intersects 𝑐 with a circle with diameter 𝑂𝐴. (a) Refer to Theorem 7.20 for a relationship between semicircles and right angles, then use this relationship to explain why the construction works. (b) Carry out an example of the construction to see how it works in practice. 4. (Angle Bisectors and Perpendiculars). Draw a triangle △𝐴𝐵𝐶 and construct the angle bisector rays 𝑎, 𝑏, 𝑐 at each vertex 𝐴, 𝐵, 𝐶. Also, construct the line at each vertex perpendicular to the angle bisector. (a) Explain why each perpendicular line bisects the two exterior angles at its vertex. (b) Prove that any intersection of two external bisecting lines lies on one of the interior angle bisectors. (c) Label the three intersection points1 of the external bisectors as 𝑋, 𝑌 , 𝑍. What special lines in △𝑋𝑌 𝑍 are the lines containing the rays 𝑎, 𝑏, 𝑐? (d) How is this figure related to the solution of Fagnano’s problem? (e) Construct circles with centers 𝑋, 𝑌 , 𝑍 that are tangent to three extended sides of the triangle 5. (Double Reflection Experiments). Given two mirrors, the broken path in Figure 17 is the path of a light ray (or a billiard ball) from point 𝐴 to 𝐵 if it is reflected from both mirrors. (a) Draw several such figures with mirrors at different angles and construct the path from 𝐴 to 𝐵. Are there cases where there is no such doubly reflected path? (b) Instead of drawing in the plane, observe the double reflection of yourself or some other object in two glass mirrors meeting in a common edge. Will you see a “mirror image” of your face or the object? Explain. What will be the difference between the view when the mirrors are perpendicular and when the mirrors meet at another angle? B A

Figure 17. Double Reflection Path from 𝐴 to 𝐵 1 The proof of the existence of the intersection points in every case depends on Theorem 7.7 in the next chapter. For this theorem, one needs the Euclidean Parallel Property, as explained there.

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6. (Regions of Closeness Experiment). Draw three noncollinear points 𝐴, 𝐵, 𝐶 on a sheet of paper. (a) Fold the paper so that point 𝐴 is folded to point 𝐵; then unfold and fold again taking 𝐴 to 𝐶. Finally, unfold and fold again taking 𝐵 to 𝐶. If the first two folds intersect on your paper, how is the third fold related to them? (If the folds don’t intersect, start again with new points.) (b) On your paper, consider which points are equidistant from 𝐴 and 𝐵 and which are closer to 𝐴 than to 𝐵. Consider the same relations for 𝐴 and 𝐶. Shade those points that are closer to 𝐴 than to either 𝐵 or 𝐶. Then shade in a differer color or pattern the points that are closer to 𝐵 than either 𝐴 or 𝐶. What is the third unshaded region? (c) Is there a point in your figure that is the same distance from 𝐴, 𝐵, and 𝐶? How do you know this? Why must this point be present when two of the folds intersect? (d) Draw four points and color or label the four regions of points closer to one point than any of the others, either by folding or by drawing appropriate lines. Note: The reasoning to answer the questions in (c) will be repeated in the proof of Theorem 7.8. However, that theorem is not needed to answer this problem, which assumes a point of intersection as a starting point. Such regions of closeness for any number of points are called Voronoi regions. 7. (Comparing Isosceles Triangles). In comparing isosceles triangles with a common apex angle in the Fagnano solution, the distance relationships follow very precisely from the Dilation Axiom. But as an exercise, an inequality can be proved using only the other axioms. The suggested steps are shown in Figure 18. Since isosceles triangles can be divided into two right triangles, we prove an inequality for right triangles. Specifically, in the figure on the left side, we prove 𝐷𝐸 is shorter than 𝐵𝐶. (a) In the figure on the left, △𝐴𝐵𝐶 is a right triangle with point 𝐷 on the hypotenuse 𝐴𝐵. If 𝐷𝐸 is perpendicular to 𝐴𝐶, show that 𝐸 is an interior point of segment 𝐴𝐶 and so ‖𝐴𝐸‖ < ‖𝐴𝐶‖. (b) In the figure on the right (△𝐴𝐵𝐶 is the same but shown separately for clarity) let 𝐹 be the point on 𝐴𝐶 with 𝐹𝐶 ≅ 𝐴𝐸. Explain why 𝑚∠𝐵𝐹𝐶 > 𝑚∠𝐷𝐴𝐸. (c) Construct 𝐺 on 𝐵𝐶 with ∠𝐺𝐹𝐶 ≅ ∠𝐷𝐴𝐸 using the Crossbar Theorem. (d) By ASA, △𝐺𝐹𝐶 ≅ △𝐷𝐴𝐸 with ‖𝐶𝐺‖ < ‖𝐶𝐵‖. (e) Conclude: For right triangles △𝐴𝐵𝐶 and △𝐴′ 𝐵 ′ 𝐶 ′ , if ∠𝐶𝐴𝐵 ≅ ∠𝐶 ′ 𝐴′ 𝐵 ′ and the hypotenuse length ‖𝐴𝐵‖ > ‖𝐴′ 𝐵 ′ ‖, then ‖𝐵𝐶‖ > ‖𝐵 ′ 𝐶 ′ ‖.

B

B

D

A

E

G

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A

F

Figure 18. Comparing Right Triangles

C

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8. (Four Points and Altitudes). The several cases of acute vs obtuse angles illustrated in Figure 16 may seem confusing. Consider the following scenario and see how it fits the various cases in the figure: let 𝑃1 , 𝑃2 , 𝑃3 , 𝑃4 be four points so that for any pair of points the line through that pair is perpendicular to the line through the other pair. How many lines are there through these pairs of points? How many points of intersection? Does this scenario appear in Figure 16?

Chapter 7

Parallel Lines and Translations

This chapter marks an important point in the development of the geometry of the Euclidean plane. With the exception of a limited use of Triangle Angle Sum, proved by the Dilation Axiom, almost everything proved so far has used only the first five of our axioms. These axioms are also valid for the non-Euclidean plane geometry discovered by Lobachevsky and others in the nineteenth century, so theorems proved only with those axioms are true in both geometries. From now on, with the extensive use of the Euclidean Parallel Property, our text will be hewing exclusively to the Euclidean path.

The Euclidean Parallel Postulate While we have defined parallel lines as lines that do not intersect, we have proved very little about parallel lines so far. This is because we need something additional to make our geometry conform to the intuition we have about rectangles, about lined notebook paper, and many other flat things. The key to making our geometry Euclidean is an axiom that is equivalent to the Euclidean Parallel Postulate.

Figure 1. Playfair’s Version of the Parallel Postulate

There have been many statements equivalent to the Parallel Postulate proposed as axioms over the centuries. The axiom that we have chosen in this book is Axiom 6, the Dilation Axiom. This axiom will be used to prove a theorem about parallel lines that is equivalent to Euclid’s statement of the Parallel Postulate. The one we state and prove 83

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below is called the axiom of Playfair, named for its originator, the Scottish mathematician John Playfair. Euclid stated the Parallel Postulate as an axiom, but since we are assuming an equivalent axiom, the Dilation Axiom, it will be proved here as a theorem using dilation. Theorem 7.1 (Euclidean Parallel Postulate (EPP)). Let m be a line. For any point 𝑃 not on 𝑚, there is only one line through 𝑃 parallel to 𝑚. q

h P

p n

m

Q N

M

Q'

Figure 2. Dilation Proof of Parallel Postulate

Proof. Let ℎ be the line through 𝑃 perpendicular to 𝑚 and let 𝑝 be the line through 𝑃 perpendicular to ℎ. Let 𝑀 be the intersection of ℎ and 𝑚. Since both 𝑚 and 𝑝 are perpendicular to ℎ, 𝑝 is a line through 𝑃 parallel to 𝑚 by Theorem 3.6. Now suppose that 𝑞 is another line through 𝑃 distinct from 𝑝. Since this line is not perpendicular to ℎ, there is a point 𝑄 on the line with 𝑚∠𝑀𝑃𝑄 < 90. Construct the line 𝑛 through 𝑄 perpendicular to ℎ, with 𝑁 as the intersection of 𝑛 and ℎ. (All these perpendiculars can be constructed by line reflection.) These lines and points are shown in Figure 2. Now we have a triangle △𝑃𝑁𝑄, with a right angle ∠𝑃𝑁𝑄. The strategy is to dilate this triangle so that the image of 𝑄 is the intersection of 𝑞 and 𝑚. Let 𝐷𝑃,𝑘 be the dilation with center 𝑃 that maps 𝑁 to 𝑀. The value of 𝑘 is ‖𝑃𝑀‖/‖𝑃𝑁‖. Since this dilation maps lines through 𝑃 to themselves, the images of lines 𝑞 and ℎ are 𝑞 and ℎ. By the Dilation Axiom, which says angle measure is preserved by dilation, the right angle ∠𝑃𝑁𝑄 maps to a right angle ∠𝑃𝑀𝑄′ . But 𝑀𝑄′ = 𝑚, since both are lines through 𝑀 perpendicular to ℎ. This means that 𝑄′ is on both 𝑞 and 𝑚. So these lines intersect, and 𝑞 is not parallel to 𝑚. □ As an example of how one can use the EPP, here is a second proof of the Angle Sum Theorem, Theorem 4.12. If the EPP had been assumed as an axiom, this is the proof that would have come first. It shares some ideas with the proof in Chapter 4 but does not use dilation explicitly.

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Theorem 7.2 (Angle Sum). For any triangle △𝐴𝐵𝐶, the images of △𝐴𝐵𝐶 by halfturns centered at the midpoints of the sides form a triangle △𝐴′ 𝐵′ 𝐶 ′ , with △𝐴𝐵𝐶 as the midpoints. Therefore, the sum of the measures of the angles of △𝐴𝐵𝐶 is 180 degrees. Proof. Denote by 𝐹 and 𝐸 the midpoints of 𝐴𝐵 and 𝐴𝐶, respectively. If we apply ℋ𝐹 to △𝐴𝐵𝐶, ℋ𝐹 (𝐴) = 𝐵, ℋ𝐹 (𝐵) = 𝐴, and ℋ𝐹 (𝐶) = 𝐶 ′ . The halfturn is a rigid motion, so ∠𝐵𝐴𝐶 ′ ≅ ∠𝐴𝐵𝐶. Since by Theorem 6.1 half-turns map any line to a line parallel to it, line 𝐴𝐶 ′ is parallel to 𝐵𝐶. A

B'

C' E

F

B

C

D

A'

Figure 3. Angle Sum

Now apply ℋ𝐸 to △𝐴𝐵𝐶 and get the image △𝐶𝐵 ′ 𝐴. Again we see 𝐴𝐵′ is parallel to 𝐵𝐶, and ∠𝐶𝐴𝐵 ′ ≅ ∠𝐴𝐶𝐵. But since lines 𝐴𝐶 ′ and 𝐴𝐵′ are lines through 𝐴 parallel to 𝐵𝐶, they must be the same line, so 𝐵 ′ , 𝐴, and 𝐶 ′ are collinear. And 𝐴 is the midpoint of 𝐵 ′ 𝐶 ′ . The same reasoning applies to the midpoints of the other two sides of △𝐴′ 𝐵 ′ 𝐶 ′ . This means that the sum of the measures of the three angles ∠𝐵𝐴𝐶 ′ , ∠𝐵𝐴𝐶, and ∠𝐶𝐴𝐵 ′ must be 180 degrees. Since these angles are congruent to the angles of △𝐴𝐵𝐶, this proves the result. □ Corollary. The measure of an exterior angle of a triangle is equal to the sum of the measures of the interior angles at the other two vertices.

Transversals and Parallel Lines Definition 7.3. A line that intersects each of a pair of lines at two distinct points is called a transversal of the lines. The Parallel Postulate says something important about transversals: Theorem 7.4 (Transversal of Parallels). Suppose 𝑚 and 𝑛 are parallel lines. If 𝑝 is a line distinct from 𝑚 that intersects 𝑚, then 𝑝 also intersects 𝑛, so 𝑝 is a transversal. Proof. If 𝑝 intersects 𝑚 at 𝑀 and does not intersect 𝑛, then 𝑝 and 𝑚 are two lines through 𝑀 that are parallel to 𝑛. By the Parallel Postulate, the lines cannot be distinct. Therefore, 𝑝 must intersect 𝑛. □

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In the next theorem, a half-turn is used to establish important properties of transversals.

C

E

A M B

C'

D

Figure 4. Transversal with Half-turn at Midpoint

Theorem 7.5 (Transversal Theorem). Suppose 𝐴𝐵 is a transversal of 𝑐 and 𝑑, with 𝐴 on 𝑐 and 𝐵 on 𝑑. Let ℋ𝑀 be the half-turn with center 𝑀, the midpoint of 𝐴𝐵. (1) The line 𝑑 = ℋ𝑀 (𝑐) if and only if 𝑐 and 𝑑 are parallel. ⃗ ⃗′ is a parallel ray on the (2) If 𝐶 is not on 𝐴𝐵, let 𝐶 ′ = ℋ𝑀 (𝐶). Then ℋ𝑀 (𝐴 𝐶) = 𝐵𝐶 ′ opposite side of 𝐴𝐵. Also 𝐶𝐵 is parallel to 𝐶 𝐴. (3) If 𝐶 and 𝐷 are on opposite sides of 𝐴𝐵, then 𝑚∠𝐵𝐴𝐶 = 𝑚∠𝐴𝐵𝐷 if and only if 𝐶𝐴 is parallel to 𝐵𝐷. Proof. The line ℋ𝑀 (𝑐) is parallel to 𝑐 by Theorem 6.1. The Parallel Postulate says that if 𝑑 is also a line through 𝐵 parallel to 𝑐, then it must be the same line. This proves the first statement. If 𝐶 ′ = ℋ𝑀 (𝐶), then 𝐶𝐶 ′ intersects 𝐴𝐵 at 𝑀. So if 𝐶 is not on 𝐴𝐵, then 𝐶 and 𝐶 ′ are in opposite half-planes. Since 𝐴𝐶 ′ = ℋ𝑀 (𝐵𝐶), the lines 𝐶𝐵 and 𝐶 ′ 𝐴 are parallel. Also, since ℋ𝑀 is a rigid motion, 𝑚∠𝐵𝐴𝐶 = 𝑚∠𝐴𝐵𝐶 ′ . If 𝑚∠𝐵𝐴𝐶 = 𝑚∠𝐴𝐵𝐶 ′ = 𝑚∠𝐴𝐵𝐷, with 𝐶 and 𝐷 on opposite sides of 𝐴𝐵, then ⃗′ . Therefore 𝐶𝐴 is parallel to 𝐵𝐷 = 𝐵𝐶 ′ . Conversely, if the lines are parallel ⃗ 𝐵 𝐷 = 𝐵𝐶 ⃗′ since both are on the unique ⃗ and 𝐷 is on the opposite side of 𝐴𝐵, then 𝐵 𝐷 = 𝐵𝐶 parallel to 𝐶𝐴 through 𝐵. □ This theorem can be used to prove the version of the Parallel Postulate stated in Euclid’s Elements. It tells more about the location of the intersection of two nonparallel lines. Theorem 7.6 (Euclid’s Version of the Parallel Postulate). Given two points 𝐹 and 𝐷 on ⃗ and 𝐵 ⃗ the same side of 𝐴𝐵, if 𝑚∠𝐹𝐴𝐵 + 𝑚∠𝐷𝐵𝐴 < 180, then the rays 𝐴𝐹 𝐷 intersect at a point 𝐺. ⃗ Proof. Let 𝐸 be a point on the same side of 𝐴𝐵 as 𝐷, with 𝐴𝐸 parallel to 𝐵𝐷 and 𝐴 𝐶 ⃗ the opposite ray of 𝐴𝐸, as pictured in Figure 4. Let 𝑑 = 𝑚∠𝐴𝐵𝐷 and 𝑓 = 𝑚∠𝐵𝐴𝐹. Then 𝑚∠𝐵𝐴𝐶 = 𝑑 and 𝑚∠𝐵𝐴𝐸 = 180 − 𝑑 > 𝑓, by Theorem 7.5. Therefore, 𝐴𝐹 is not the unique parallel to 𝐵𝐷 through 𝐴, so it intersects 𝐵𝐷 at a point 𝐺.

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⃗ and 𝐵 ⃗ What remains to be proved is that 𝐺 is on the rays 𝐴𝐹 𝐷. To see this, we prove the points 𝐹 and 𝐷 are in the interior of ∠𝐵𝐴𝐸. From the Addition of Angle Measure Theorem, Theorem 2.8, the interior points of ⃗ are in the interior of ∠𝐸𝐴𝐵. To show 𝐷 is also in the interior we must show it is on 𝐴𝐹 the same side of 𝐴𝐵 as 𝐸 (true by assumption) and on the same side of 𝐴𝐸 as 𝐵 (true since the entire parallel line 𝐵𝐷 is on that side). ⃗ because that ray is on the opposite side Now 𝐺 cannot be on the opposite ray of 𝐴𝐹 ⃗ and so must be in the interior of 𝐴𝐸 from 𝐵𝐷, so the intersection point must be on 𝐴𝐹 of ∠𝐸𝐴𝐵. But this means that 𝐺 is on the same side of 𝐴𝐵 as 𝐷 and 𝐸, so it must be on ⃗ 𝐵 𝐷 rather than its opposite ray. □ The angles ∠𝐵𝐴𝐶 and ∠𝐴𝐵𝐷 are called the alternate interior angles defined by the transversal. As illustrated in Figure 5, two lines are parallel if and only if the alternate interior angles of a transversal are congruent. This includes the Important special case of a right angle, when the transversal is perpendicular to both lines. Theorem 7.7 (Perpendiculars to Intersecting Lines). Let 𝑚 and 𝑛 be distinct intersecting lines. If line 𝑚′ is perpendicular to 𝑚 and line 𝑛′ is perpendicular to 𝑛, then 𝑚′ and 𝑛′ are distinct intersecting lines as well. Proof. The lines 𝑚′ and 𝑛′ cannot be parallel, for if they were, both 𝑚 and 𝑛 would be perpendicular transversals of 𝑚′ and 𝑛′ and so would be parallel or equal. And 𝑚′ and 𝑛′ cannot be the same line for again 𝑚 and 𝑛 would be perpendicular to 𝑚′ and so parallel or equal. □

Figure 5. Transversals of Parallel Lines with Alternate Interior Angles

This theorem has a number of applications. It was already needed in Exercise 4 of Chapter 6 to prove that two bisectors of adjacent exterior angles of a triangle always intersect, because they are each perpendicular to an internal angle bisector. Another application is the next theorem, which concerns circumscribing a circle about any triangle. Theorem 7.8 (Triangle Circumcircle). For any △𝐴𝐵𝐶, there is a unique circle that passes through the points 𝐴, 𝐵, and 𝐶. The center 𝑂 of the circle is the point of concurrence of the perpendicular bisectors of the sides of the triangle.

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d

C D E

e

O

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A f

Figure 6. Circumcircle and Circumcenter of △𝐴𝐵𝐶

This circle is called the circumcircle of the triangle, and the point of concurrence 𝑂 is called the circumcenter of the triangle. Proof. Consider the bisectors of two of the sides, say the lines 𝑒 and 𝑓 that are the perpendicular bisectors of sides 𝐴𝐶 and 𝐴𝐵, as shown in Figure 6. By Theorem 7.7 these two lines intersect at a point 𝑂, since they are each perpendicular to a side of the triangle, and these sides are not collinear. Since the perpendicular bisector of a segment is the locus of points equidistant from the endpoints by Theorem 3.10, the points of 𝑒 are equidistant from 𝐴 and 𝐶 and the points of 𝑓 are equidistant from 𝐴 and 𝐶. Together this says that 𝑂 satisfies ‖𝑂𝐶‖ = ‖𝑂𝐴‖ = ‖𝑂𝐵‖. If we call this common distance 𝑟, the circle with radius 𝑟 and center 𝑂 passes through all three vertex points of △𝐴𝐵𝐶. For the final part of the proof, since ‖𝑂𝐶‖ = ‖𝑂𝐵‖, the point 𝑂 is equidistant from 𝐵 and 𝐶 and so is on 𝑑, the perpendicular bisector of 𝐵𝐶. Thus, 𝑂 is on all three perpendicular bisectors. □

Parallelograms The study of parallel lines leads to parallelograms. Note. We will say that two segments or rays are parallel if the lines they belong to are parallel. Definition 7.9. A quadrilateral 𝐴𝐵𝐶𝐷 is a parallelogram if its opposite sides are parallel, i.e., 𝐴𝐵 is parallel to 𝐶𝐷 and 𝐷𝐴 is parallel to 𝐵𝐶. Theorem 7.10 (Parallelogram Symmetry). For a parallelogram 𝐴𝐵𝐶𝐷, the opposite sides are congruent and the opposite angles are congruent. Also, the two diagonals 𝐴𝐶 and 𝐵𝐷 intersect at a point 𝑀 that is the midpoint of each diagonal. The half-turn ℋ𝑀 is a symmetry of the parallelogram. Proof. Let 𝑀 be the midpoint of the diagonal 𝐴𝐶. Then ℋ𝑀 (𝐴) = 𝐶 and ℋ𝑀 (𝐶) = 𝐴. The line 𝐴𝐶 is a transversal of both pairs of opposite sides. Therefore, by the Transversal Theorem, Theorem 7.5, ℋ𝑀 (𝐴𝐵) = 𝐶𝐷 and ℋ𝑀 (𝐴𝐷) = 𝐶𝐵.

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A Figure 7. Parallelogram with Diagonals

Point 𝐵 is the intersection of 𝐴𝐵 and 𝐶𝐵, so its image is the intersection of the image lines 𝐶𝐷 and 𝐷𝐴, namely 𝐷. So ℋ𝑀 (𝐷) = 𝐵. Therefore, since ℋ𝑀 (𝐵) = 𝐷 also, ℋ𝑀 is a symmetry of the parallelogram. This means that ℋ𝑀 maps 𝐴𝐵 to 𝐶𝐷 and 𝐵𝐶 to 𝐷𝐴, so the opposite sides are congruent. The half-turn with center 𝑀 also maps each angle of 𝐴𝐵𝐶𝐷 to its opposite angle, so the opposite angles are congruent. Finally, the midpoint of 𝐵𝐷 is also 𝑀, since ℋ𝑀 (𝐵) = 𝐷.

□

Corollary. A quadrilateral is a parallelogram if and only if it has point symmetry. Proof. The theorem has been proved in one direction. If 𝐴𝐵𝐶𝐷 has a half-turn ℋ𝑀 that is a symmetry, then ℋ𝑀 (𝐴𝐵) = 𝐶𝐷 and ℋ𝑀 (𝐵𝐶) = 𝐷𝐴, so the opposite sides are parallel, by the properties of half-turns. □ Earlier, the meaning of “direction” was defined for two rays or two segments on the same line. With these properties of parallels proven, it is possible to extend this definition to parallel rays and segments. ⃗ ⃗ and 𝐶 Definition 7.11. Two rays on parallel lines, 𝐴𝐵 𝐷, have the same direction if both 𝐵 and 𝐷 are on the same side of 𝐴𝐶. They have opposite direction if 𝐵 and 𝐷 are on opposite sides. The same definition applies for the same direction of segments 𝐴𝐵 and 𝐶𝐷 on parallel lines. Two directed lines that are parallel either have directing rays in the the same direction or the opposite direction. There are several properties of a quadrilateral 𝐴𝐵𝐶𝐷 that ensure it is a parallelogram. Theorem 7.12 (Equal Parallels). If two segments 𝐴𝐵 and 𝐶𝐷 are parallel and ‖𝐴𝐵‖ = ‖𝐶𝐷‖, then either • 𝐴𝐵 and 𝐶𝐷 have opposite direction and 𝐴𝐵𝐶𝐷 is a parallelogram with 𝐴𝐶 and 𝐵𝐷 as diagonals or • 𝐴𝐵 and 𝐶𝐷 have the same direction and 𝐴𝐵𝐷𝐶 is a parallelogram with 𝐴𝐷 and 𝐵𝐶 as diagonals. Therefore, if 𝐴𝐵𝐶𝐷 is a quadrilateral with two opposite sides parallel and congruent, then 𝐴𝐵𝐶𝐷 is a parallelogram.

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Figure 8. Congruent Parallel Segments

Proof. If 𝐵 and 𝐷 are on opposite sides of 𝐴𝐶, then 𝐴𝐵𝐶𝐷 is a parallelogram. Let 𝑀 be the midpoint of 𝐴𝐶. Then by the Transversal Theorem, Theorem 7.5, 𝐴𝐷 is parallel to 𝐶𝐵. Therefore 𝐴𝐵𝐶𝐷 is a parallelogram. Since 𝐴𝐶 and 𝐵𝐷 are diagonals that intersect at their common midpoint, 𝐴𝐵𝐷𝐶 is not a quadrilateral. As illustrated in Figure 8, if 𝐵 and 𝐷 are on the same side of 𝐴𝐶, then 𝐷 is an interior point of ∠𝐵𝐴𝐶. First, 𝐵 and 𝐷 are in the same half-plane of 𝐴𝐶. Second, 𝐷 and 𝐶 are on the same side of 𝐴𝐵 because 𝐷𝐶 is parallel to 𝐴𝐵. Putting these together, 𝐷 is interior to ∠𝐵𝐴𝐶. The Crossbar Theorem, Theorem 2.10, implies that 𝐶 and 𝐵 are on opposite sides of 𝐴𝐷. Then by the same reasoning as before 𝐴𝐵𝐷𝐶 is a parallelogram and 𝐴𝐷 and 𝐵𝐶 are diagonals that intersect at their common midpoint, so 𝐴𝐵𝐶𝐷 is not a quadrilateral. This implies that the last statement is true. □ A corollary of this theorem is that the midpoints of a parallelogram define more parallelograms and parallel lines. C

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Figure 9. Midpoints and Midsegments in a Parallelogram

Theorem 7.13 (Parallelogram Midsegments). If 𝐴𝐵𝐶𝐷 is a parallelogram, a segment 𝐸𝑊 connecting the midpoints of two opposite sides is parallel to the other two sides and forms a parallelogram with either of those sides. The two such segments intersect at 𝑀, the intersection point of the diagonals. Proof. Let the midpoints be labeled as in Figure 9. Then 𝐸 and 𝑊 are the midpoints of the opposite sides 𝐵𝐶 and 𝐷𝐴. The quadrilateral 𝐵𝐸𝑊𝐴 has two sides 𝐵𝐸 and 𝑊𝐴, parallel and of the same length, so it is a parallelogram by Theorem 7.12. Therefore,

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𝐸𝑊 is parallel and congruent to 𝐴𝐵. By the same reasoning, 𝑁𝑆, the other segment connecting midpoints of opposite sides, is parallel to 𝐷𝐴. Since the symmetry ℋ𝑀 maps each midpoint of a side of 𝐴𝐵𝐶𝐷 to the midpoint on the opposite side, 𝑀 is the midpoint of both segments 𝐸𝑊 and 𝑁𝑆. □ By symmetry 𝐸𝑁𝑊𝑆 is a parallelogram with the segments 𝐸𝑊 and 𝑁𝑆 as diagonals. Here is a second criterion for a quadrilateral to be a parallelogram. Theorem 7.14 (Congruent Opposite Sides). A quadrilateral 𝐴𝐵𝐶𝐷 with opposite sides of equal length is a parallelogram. Proof. By assumption ‖𝐴𝐵‖ = ‖𝐶𝐷‖ and ‖𝐵𝐶‖ = ‖𝐷𝐴‖. The diagonal 𝐴𝐶 is thus the common side of two triangles △𝐴𝐵𝐶 and △𝐶𝐷𝐴 that are congruent by SSS. Let 𝑇 be the rigid motion of the congruence defined by SSS. m B

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Figure 10. SSS Rigid Motion from △𝐴𝐵𝐶 to △𝐶𝐷𝐴

This rigid motion 𝑇 maps 𝐴𝐶 to 𝐶𝐴, so the midpoint 𝑀 of 𝐴𝐶 is a fixed point. Therefore, 𝑇 is either the half-turn ℋ𝑀 or the reflection ℛ𝑚 , where 𝑚 is the perpendicular bisector of 𝐴𝐶. If 𝑇 = ℋ𝑀 , the quadrilateral has a point symmetry and 𝐴𝐵𝐶𝐷 is a parallelogram. So what remains is to rule out the other case. Since 𝑇(𝐵) = 𝐷, if 𝑇 = ℛ𝑚 , then 𝐵 and 𝐷 must be on the same side of 𝐶𝐴, since the reflection maps each half-plane of 𝐶𝐴 into itself. It cannot be the case that 𝐵 is on 𝑚, for then 𝐵 = 𝐷 and 𝐴𝐵𝐶𝐷 is not a quadrilateral. Therefore, 𝐵 is in the 𝑚-half-plane of either 𝐴 or 𝐶. Suppose it is in the half-plane of 𝐴, as in Figure 10. The segment 𝐶𝐵 intersects 𝑚 at a point 𝑃, since 𝐵 and 𝐶 are in opposite half-planes. Applying ℛ𝑚 to 𝐶𝐵, the image 𝐴𝐷 also intersects 𝑚 at 𝑃. But this means that the sides 𝐶𝐵 and 𝐴𝐷 intersect at 𝑃, so 𝐴𝐵𝐶𝐷 is not a quadrilateral. □ This is an example of how having a rigid motion included in a triangle congruence test like SSS can help in a proof.

Rectangles Now we define a very familiar polygon that is a special parallelogram. Definition 7.15. A rectangle is a quadrilateral 𝐴𝐵𝐶𝐷 with each vertex angle a right angle.

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Theorem 7.16 (Rectangle Symmetry). If 𝐴𝐵𝐶𝐷 is a rectangle, then it is a parallelogram with the lines through the midpoints of opposite sides as lines of symmetry. Proof. Lines 𝐴𝐵 and 𝐶𝐷 are perpendicular to 𝐴𝐷, so they must be parallel. Likewise, 𝐵𝐶 and 𝐴𝐷 are parallel. This proves 𝐴𝐵𝐶𝐷 is a parallelogram. Let the midpoints be labeled as in Figure 11. It was proved in Theorem 7.13 that the midpoint segment 𝐸𝑊 is parallel to 𝐴𝐵 and 𝐷𝐶. So in a rectangle, 𝐸𝑊 is perpendicular to the transversals 𝐷𝐴 and 𝐵𝐶. Thus, the reflection ℛ𝐸𝑊 maps 𝐴𝐵 to 𝐷𝐶. The other midpoint line is also a line of symmetry. □

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Figure 11. Rectangle with Lines of Symmetry

Note. Because of the line symmetry, the two diagonals of a rectangle are congruent, since a line reflection maps one to the other. The converse of this statement is also true. In addition, if one angle of a parallelogram is a right angle, then all are right angles and the figure is a rectangle. The proofs are problems in Exercise 12. Every parallelogram, including rectangles, has a point symmetry. The product of the reflections in the two perpendicular midlines of a rectangle is a 180-degree rotation centered at the point of intersection. This is the half-turn ℋ𝑀 that defines the point symmetry. Thus, in general, parallelograms have two rotational symmetries, the identity and ℋ𝑀 , but rectangles also have, in addition, two line symmetries. This is consistent with the discussion of dihedral symmetry in Chapter 5. This agrees with a visual sense that rectangles are more symmetric than a general parallelogram. A square is a special rectangle that is yet more symmetric. It is a regular 4-gon and has four rotations and four reflections as symmetries. Distance Between Parallel Lines. One can use rectangles to define the distance between two parallel lines and also to define the unique parallel line of symmetry of a figure consisting of two parallel lines. Theorem 7.17 (Midline Theorem). Let 𝑝 and 𝑞 be parallel lines. (1) Every segment 𝑃𝑄 perpendicular to 𝑝 at 𝑃 and to 𝑞 at 𝑄 has the same length. (2) The perpendicular bisector of every such perpendicular 𝑃𝑄 is the same line 𝑚.

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(3) This line 𝑚 is parallel to 𝑝 and 𝑞 and is a line of symmetry for 𝑝 and 𝑞, i.e., ℛ𝑚 (𝑝) = 𝑞. (4) If a transversal intersects 𝑝 and 𝑞 at 𝐴 and 𝐵, the midpoint of 𝐴𝐵 is on 𝑚. P

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Figure 12. Midline with Transversals

Proof. Most of these statements are immediate consequences of properties of rectangles. If 𝑃 and 𝑅 are on 𝑝 and 𝑄 and 𝑆 are on 𝑞, then if 𝑃𝑄 and 𝑅𝑆 are perpendicular to 𝑝 and 𝑞, then 𝑃𝑄𝑆𝑅 is a rectangle and the opposite sides 𝑃𝑄 and 𝑅𝑆 have the same length. Also the perpendicular bisector 𝑚 of either 𝑃𝑄 or 𝑅𝑆 is the unique line of symmetry of the rectangle 𝑃𝑄𝑆𝑅 parallel to 𝑝 and 𝑞, so it is a well-defined line 𝑚 with ℛ𝑚 (𝑝) = 𝑞. If 𝐴𝐵 is a segment with 𝐴 on 𝑝 and 𝐵 on 𝑞 with midpoint 𝐶, then ℋ𝐶 (𝑝) = 𝑞 by Theorem 7.5. Therefore, ℋ𝐶 (𝑚) = 𝑚 and 𝐶 is on 𝑚. Another way to see this is to construct perpendicular segments 𝐴𝑇 and 𝑈𝐵 to form a rectangle as in Figure 12. □ Definition 7.18. For two parallel lines 𝑝 and 𝑞, the distance between 𝑝 and 𝑞 is the length of any segment 𝑃𝑄 perpendicular to 𝑝 at 𝑃 and to 𝑞 at 𝑄. The midline of the two parallel lines is the perpendicular bisector of any such 𝑃𝑄. This distance and this line are well-defined, according to the theorem. Consider the symmetries of a pair of parallel lines 𝑝 and 𝑞. By the Transversal Theorem, Theorem 7.5, and the Midline Theorem, every point 𝑀 on the midline is the center of a point symmetry ℋ𝑀 . Also, the midline is a line of symmetry, the only one parallel to 𝑝 and 𝑞. But every perpendicular transversal 𝑐 is a line of symmetry, for ℛ𝑐 will map 𝑝 to 𝑝 and 𝑞 to 𝑞 (and also the midline to itself as well). It is also true that the products of these symmetries are also symmetries. We will see some of these products later in the chapter.

Midpoint Figures For any triangle △𝐴𝐵𝐶, we encountered in Theorem 4.12 the triangle △𝐷𝐸𝐹 formed by the midpoints of the sides. More recently, in the proof of Theorem 7.2, we have seen that the images of △𝐴𝐵𝐶 by the half-turns centered at these midpoints create a triangle △𝐴′ 𝐵′ 𝐶 ′ that has the original △𝐴𝐵𝐶 as its midpoint triangle. All these triangles are shown in Figure 13. By Theorem 7.17, a midpoint line such as 𝐷𝐸 is the midline of the parallel lines 𝐴𝐵 and 𝐴′ 𝐵 ′ .

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Figure 13. Midpoint Triangles

Such structures of triangles within triangles can be built up to create arbitrarily large triangles tiled with congruent triangles, such as the one in Figure 14 with eight triangles to a side. Within △𝐴𝐵𝐶 in this figure, if one takes a subtriangle such as △𝐴𝐷𝐸, one can see that the ratio of the length of 𝐷𝐸 to the length of 𝐵𝐶 is the same as 𝐴𝐷/𝐴𝐵 and 𝐴𝐸/𝐴𝐶. Also one sees parallel lines and congruent angles such as ∠𝐴𝐷𝐸 ≅ ∠𝐴𝐵𝐶 in these triangles. A E C

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B Figure 14. Triangle Tiled with Congruent Triangles

This suggests how, if one assumed the Euclidean Parallel Postulate as an axiom instead of our Axiom 6 , one could prove the Dilation Axiom as a theorem for rational ratios and then extend to real ratios using a limit argument similar to that used in the proof of Theorem 6.8 about the intersection of a line with a circle. There are many other interesting features in this figure: parallelograms of different shapes and rigid motions that are products of half-turns. These rigid motions will be taken up in a few pages from now. Midpoint Quadrilaterals. A quadrilateral 𝐴𝐵𝐶𝐷 also has a midpoint figure 𝐽𝐾𝐿𝑀. Unlike the case of triangles, this figure is not similar to the original quadrilateral. Instead, it is always a parallelogram. Theorem 7.19 (Midpoint Parallelogram). For any quadrilateral 𝐴𝐵𝐶𝐷, the midpoints of the sides are the vertices of a parallelogram.

Midpoint Figures

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Figure 15. Midpoint Quadrilaterals

Proof. Assume that 𝐵 and 𝐷 are on opposite sides of 𝐴𝐶. Using the labels in Figure 15 for the midpoints, the segments 𝐽𝐾 and 𝑀𝐿 are segments parallel to and in the same direction as the diagonal 𝐴𝐶. The lengths of these segments are both equal to ‖𝐴𝐶‖/2. Therefore, these segments are the opposite sides of a parallelogram if they are not collinear, but they cannot be collinear if 𝐵 and 𝐷 are on opposite sides of the diagonal. Thus, it only remains to check that for at least one of the diagonals, the other vertices are on opposite sides. Given quadrilateral 𝐴𝐵𝐶𝐷, if 𝐵 and 𝐷 are one the same side of 𝐴𝐶, then since the sides of the quadrilateral do not intersect, either 𝐷 is in the interior of △𝐴𝐵𝐶 or 𝐵 is in the interior of △𝐴𝐷𝐶. (This is to be proved in Exercise 7.) In either case, 𝐴 and 𝐶 are on opposite sides of 𝐵𝐷 so we should choose 𝐵𝐷 as the diagonal in the proof. □ For some cases, the midpoint quadrilateral is a special kind of parallelogram. There is a question about this in the exercises. Also, in the exercises is a method for proving the concurrence of medians of a triangle using midpoint quadrilaterals. Circumcircles of Right Triangles. Right triangles provide an interesting special case of midpoint triangles, as illustrated in Figure 16. Two of the midlines, 𝐷𝐹 and 𝐸𝐹, are perpendicular bisectors of legs. The parallelogram 𝐶𝐸𝐹𝐷 is a rectangle. From Theorem 7.8 we know the 𝐹, the intersection of these perpendicular bisectors, is the circumcenter of the triangle and is equidistant from the three vertices. This means that the segment 𝐴𝐵 is a diameter of the circumcircle. C

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This leads to an important relationship between circles and right angles. Theorem 7.20 (Right Angle Locus). If △𝐴𝐵𝐶 has a right angle at 𝐶, then the midpoint 𝐹 of 𝐴𝐵 is equidistant from all three vertices of the triangle. Therefore, 𝐶 is a point on the circle with diameter 𝐴𝐵. Conversely, if a segment 𝐴𝐵 is the diameter of a circle 𝑐, then for any point 𝐶 on 𝑐 distinct from 𝐴 and 𝐵, ∠𝐴𝐶𝐵 is a right angle. In addition, 𝑚∠𝐵𝐹𝐶 = 2𝑚∠𝐵𝐴𝐶, where 𝐹 is the midpoint of 𝐴𝐵 and the center of the circle. C E

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Figure 17. Congruent Angles in an Inscribed Right Triangle

Proof. We continue to use the same names for the points as in Figures 16 and 17. The first statement was already proved in the comments above, using properties of rectangles. To prove the second part of the theorem, take a triangle △𝐴𝐵𝐶 inscribed in a circle with diameter 𝐴𝐵. This triangle is divided by 𝐶𝐹 into two isosceles triangles, which in turn can be divided into two right triangles. In Figure 17, congruent angles are marked. Since the triangles △𝐴𝐹𝐶 and △𝐶𝐹𝐵 are isosceles, in each triangle the base angles are congruent. Let 𝑎 = 𝑚∠𝐹𝐴𝐶 = 𝑚∠𝐹𝐶𝐴 and 𝑏 = 𝑚∠𝐹𝐵𝐶 = 𝑚∠𝐹𝐶𝐵 be the measures of these base angles. In △𝐴𝐵𝐶, the measure of the angle at 𝐶 is the sum of two base angles, so 𝑚∠𝐴𝐶𝐵 = 𝑎 + 𝑏. But 𝑚∠𝐵𝐴𝐶 = 𝑎 and 𝑚∠𝐴𝐵𝐶 = 𝑏, so the sum of the angle measures of △𝐴𝐵𝐶 is 𝑎 + 𝑏 + (𝑎 + 𝑏) = 2𝑎 + 2𝑏 = 180, so that 90 = 𝑎 + 𝑏 = 𝑚∠𝐴𝐶𝐵. Therefore, the angle at 𝐶 is a right angle. To prove the final equation of angles, ∠𝐶𝐴𝐵 has measure 𝑎. The base angles in △𝐵𝐹𝐶 have measure 𝑏, so the central angle ∠𝐵𝐹𝐶 has measure 180 − 2𝑏 = 2𝑎. □

Generalizing Parallelograms One might ask if there is an “SSSS” test for parallelograms analogous to the SSS test for triangles. As many a do-it-yourselfer can attest, the answer is no. If one starts a bookcase by nailing four boards together to form a rectangle, it will be prone to collapse,

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for there are an infinite number of parallelograms with the same pair of side lengths. One needs some kind of cross-brace, such as a diagonal, to hold the shape.

Figure 18. Collapsing Parallelograms with the Same Side Lengths

This can also be seen with delivery boxes. If one breaks out the bottom, the four sides form a parallelogram that collapses flat. When we study translations in the next section, we will need such flattened parallelograms as well as ordinary parallelograms. So we extend our parallelogram definition with this additional one. Definition 7.21. A linear parallelogram 𝐴𝐵𝐶𝐷 consists of four collinear points 𝐴, 𝐵, 𝐶, 𝐷 with the midpoint 𝑀 of 𝐴𝐶 being also the midpoint of 𝐵𝐷. In other words, the figure has a point symmetry, with 𝑀 as the center. Theorem 7.22 (Linear Parallelogram). Let four collinear points 𝐴, 𝐵, 𝐶, 𝐷 correspond to real numbers 𝑎, 𝑏, 𝑐, 𝑑, 𝑚 by a ruler. Then 𝐴𝐵𝐶𝐷 is a linear parallelogram if 𝑎+𝑐 = 𝑏+𝑑, or equivalently if 𝐴𝐵 and 𝐷𝐶 have the same direction and length, meaning 𝑏 − 𝑎 = 𝑐 − 𝑑. In that case, 𝐴𝐷 and 𝐵𝐶 also have the same direction and length, since 𝑑 − 𝑎 = 𝑐 − 𝑏.

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D

Figure 19. Two Linear Parallelograms 𝐴𝐵𝐶𝐷

Proof. Working on the real line 𝑚 using a ruler, the midpoint of 𝐴𝐶 corresponds to (𝑎 + 𝑐)/2 and the midpoint of 𝐵𝐷 corresponds to (𝑏 + 𝑑)/2. If these midpoints are the same point, then 𝑎 + 𝑐 = 𝑏 + 𝑑, which implies the other equations. □ Such linear parallelograms can arise as limiting cases of well-known figures. For example, for a self-intersecting four-sided figure such at the ones in Figure 20, the midpoint figure may be a parallelogram or it may be a linear parallelogram. Some hints for a proof are in Exercise 8.

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7. Parallel Lines and Translations

D M A

L

J

B

D S

R

B

P

Q C

A

K

C

Figure 20. Midpoint Quadrilaterals for More General 𝐴𝐵𝐶𝐷

The next theorem combines a uniqueness property of parallelograms with a test for congruence. Theorem 7.23 (Fourth Parallelogram Point). Given three noncollinear points 𝐴, 𝐵, 𝐶, there is a unique point 𝐷 so that 𝐴𝐵𝐶𝐷 is a parallelogram. If 𝐴, 𝐵, 𝐶 are collinear, there is a unique point 𝐷 so that 𝐴𝐵𝐶𝐷 is a linear parallelogram. Proof. Let 𝑀 be the midpoint of 𝐴𝐶 and construct 𝐷 = ℋ𝑀 (𝐵). In the noncollinear case, this is a parallelogram since ℋ𝑀 is a symmetry. It is unique because for any parallelogram 𝐴𝐵𝐶𝐷, it is true that 𝐷 = ℋ𝑀 (𝐵). The same reasoning applies to the collinear case. In this case, using a ruler and real numbers, the formula for ℋ𝑀 (𝐵) is 2(𝑎 + 𝑐)/2 − 𝑏 = 𝑎 + 𝑐 − 𝑏. □ Corollary. Two parallelograms 𝐴𝐵𝐶𝐷 and 𝐸𝐹𝐺𝐻 are congruent if two pairs of adjacent sides and one pair of included angles are congruent; e.g., 𝐴𝐵 ≅ 𝐸𝐹, 𝐵𝐶 ≅ 𝐹𝐺, and ∠𝐴𝐵𝐶 ≅ ∠𝐸𝐹𝐺. This applies to linear parallelograms as well. Proof. By SAS for triangles, there is a unique rigid motion 𝑇 that maps △𝐴𝐵𝐶 to △𝐸𝐹𝐺. Let 𝐷 be the unique point so that 𝐴𝐵𝐶𝐷 is a parallelogram. Then the 𝑇-image of 𝐴𝐵𝐶𝐷 is a parallelogram 𝐸𝐹𝐺𝑇(𝐷). But 𝐻 is the unique point so that 𝐸𝐹𝐺𝐻 is a parallelogram, so 𝑇(𝐷) = 𝐻. For linear parallelograms, the angle will be either a zero angle or a straight angle. □ C

D

M B

A

Figure 21. Constructing a Parallelogram from Two Sides and an Angle

Translations as Half-turn Products In the study of parallels, we have used half-turns extensively to map one parallel segment to another. This was very efficient and effective for these purposes, but now we will use half-turns to expand our repertoire by defining the last well-known rigid motion, the translation.

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99

Definition 7.24. A translation is a rigid motion that is the product of two half-turns. An example of such a translation can be found in the parallelogram with midpoints in Figure 9. The rigid motion 𝑇 = ℋ𝑀 ℋ𝑆 maps 𝐴𝐵 to 𝐷𝐶, since ℋ𝑀 ℋ𝑆 (𝐴) = ℋ𝑀 (𝐵) = 𝐷 and ℋ𝑀 ℋ𝑆 (𝐵) = ℋ𝑀 (𝐴) = 𝐶. In addition the rigid motions ℋ𝐶 ℋ𝐸 and ℋ𝑁 ℋ𝑀 also map 𝐴𝐵 to 𝐷𝐶. We will see that these are all the same translation. To get a general idea of how such a product of half-turns will behave, let 𝐴 and 𝐵 be distinct points and let 𝑃 be any point. Then let 𝑃 ′ = ℋ𝐴 (𝑃) and 𝑃 ″ = ℋ𝐵 (𝑃′ ). P''

P B

A

P'

Figure 22. Translation of a Point as the Product of Half-turns

It is clear in Figure 22 that in △𝑃𝑃 ′ 𝑃″ , the segment 𝐴𝐵 is a midsegment parallel to 𝑃𝑃 ″ , in the same direction and half its length. This shows that translations have a property not shared by rotations and reflections: the distance from any point to its image is always the same. Before proceeding further, there is one observation about this translation that is already apparent. The line 𝐴𝐵 and all lines parallel to it are invariant lines of the translation ℋ𝐵 ℋ𝐴 . These are not lines of fixed points, but each point of such a line is mapped to another point of the same line. As we will see below, any other line is mapped into a line distinct from it but parallel. Now consider the relationship between the images of 𝑃 and those of another point 𝑄. For now, assume a general case when no three of the points 𝐴, 𝐵, 𝑃, 𝑄 are collinear and 𝑃𝑄 is not parallel to 𝐴𝐵. These special cases will be addressed afterwards. P''

P Q' B

A

Q''

Q P'

Figure 23. Translations of Two Points as the Product of Half-turns

Again, we observe that the segments 𝑃𝑃 ″ and 𝑄𝑄″ are parallel and have the same length. Also, since half-turns map segments to parallel segments, the segments 𝑃𝑄, 𝑃 ′ 𝑄′ , and 𝑃 ″ 𝑄″ are parallel, so 𝑃𝑃 ″ 𝑄″ 𝑄 is a parallelogram.

100

7. Parallel Lines and Translations

It seems reasonable to expect that the special cases also result in a parallelogram or a linear parallelogram, and this is so. It can be checked using our definition of a translation as a product of half-turns, but instead we can use an important alternative definition of a translation as a product of line reflections. HA ( geo)

P

n

m P' A

p

P'' B T (geo)

Rm ( geo)

geo

Figure 24. Translations of a Figure Named “geo” in Two Ways

Theorem 7.25 (Translations as Double Line Reflections). For distinct points 𝐴 and 𝐵, the translation ℋ𝐵 ℋ𝐴 = ℛ𝑏 ℛ𝑎 , where 𝑎 and 𝑏 are the lines perpendicular to 𝐴𝐵 at points 𝐴 and 𝐵. Proof. The half-turn at 𝐴 is the product of reflections in any two perpendicular lines at 𝐴, so one choice is ℋ𝐴 = ℛ𝐴𝐵 ℛ𝑎 . And at 𝐵, ℋ𝐵 = ℛ𝑏 ℛ𝐴𝐵 . Then ℋ𝐵 ℋ𝐴 = ℛ𝑏 ℛ𝐴𝐵 ℛ𝐴𝐵 ℛ𝑎 = ℛ𝑏 ℛ𝑎 . □ This gives a new perspective on the behavior of translations and a new tool. For any pair of parallel lines, the product ℛ𝑏 ℛ𝑎 is a translation whose invariant lines are the perpendicular transversals of the parallel lines 𝑎 and 𝑏. Figure 24 shows the translation of a figure both by a product of half-turns and by the corresponding line reflections. They both terminate at the image of 𝑇, but the intermediate steps are quite different. In addition, in this figure there are the same points 𝑃, 𝑃 ′ , and 𝑃″ as before, but this time on the line 𝐴𝐵. One can see how the lengths add up to ‖𝑃𝑃 ″ ‖ = 2‖𝐴𝐵‖. With this definition it is not difficult to take care of the missing pieces. Theorem 7.26 (Translation of a Segment). If 𝑇 is a translation distinct from 𝐼, then for any points 𝑃 and 𝑄, 𝑃𝑄 and 𝑇(𝑃)𝑇(𝑄) are opposite sides of a parallelogram or linear parallelogram. Proof. Let 𝑇 = ℋ𝐵 ℋ𝐴 . When needed, 𝑇 can also be written as 𝑇 = ℋ𝐷 ℋ𝐶 , where 𝐴𝐵𝐷𝐶 is a rectangle, since 𝑇 = ℛ𝐵𝐷 ℛ𝐴𝐶 . We continue to use the notation in Figure 23: 𝑃″ = 𝑇(𝑃) and 𝑄″ = 𝑇(𝑄). First, assume that 𝑃𝑄 is not parallel to or collinear with 𝐴𝐵. If neither 𝑃 nor 𝑄 is on 𝐴𝐵, then from the discussion of Figure 23, it follows that 𝑃″ 𝑄″ is parallel to 𝑃𝑄, so 𝑃𝑄𝑄″ 𝑃″ is a parallelogram. In case either 𝑃 or 𝑄 is on 𝐴𝐵, the proof can be carried

Products of Translations

101

out by a computation with rulers, but a simpler method is to use 𝑇 = ℋ𝐷 ℋ𝐶 , where neither 𝑃 nor 𝑄 is on 𝐶𝐷 and 𝐴𝐵𝐷𝐶 is a rectangle. In the case when 𝑃𝑄 is parallel to or collinear with 𝐴𝐵, let 𝑎 and 𝑏 be lines perpendicular to 𝐴𝐵 so that 𝑇 = ℛ𝑏 ℛ𝑎 . Then each line reflection reverses direction on 𝑃𝑄 and the segment 𝑃 ″ 𝑄″ is collinear with 𝑃𝑄 and has the same direction and length. Therefore, by Theorem 7.22, 𝑃𝑄𝑄″ 𝑃 ″ is a linear parallelogram. □ This theorem has a very powerful corollary. Corollary (One Point Determines Translation). If 𝑇 is a translation and if 𝐸 and 𝐹 = 𝑇(𝐸) are given, then 𝑇 is completely determined. For any other point 𝐺, 𝑇(𝐺) is the unique point such that 𝐺𝐸𝐹𝑇(𝐺) is a parallelogram or linear parallelogram. Notation. The unique translation that maps 𝐴 to 𝐵 will be denoted 𝒯𝐴𝐵 . The 𝐴𝐵 in this notation is far from uniquely determined since for any point 𝑃 and its translation image 𝑄 = 𝑇(𝑃), the translation is 𝒯𝑃𝑄 .

Products of Translations This section is devoted to two properties of the products of translations: closure and commutativity. A M N

C = V (B ) = VU (A)

B = U (A) Figure 25. Translation 𝑈𝑉

Theorem 7.27 (Closure). The product of two translations is a translation. The product of a translation and a half-turn is a half-turn. In other words, the set of all translations of the plane is closed under composition. Also, the set of all half-turns and translations together is also closed under composition. Each of these sets has the group property. This fact will play a role in the symmetric patterns studied in Chapter 10. Proof. Let 𝑈 and 𝑉 be translations. The task is to prove 𝑉𝑈 is also a translation. Choose any point 𝐴. Then 𝑈 = 𝒯𝐴𝐵 for 𝐵 = 𝑈(𝐴) and 𝑉 = 𝒯𝐵𝐶 for 𝐶 = 𝑉(𝐵). Let 𝑀 be the midpoint of 𝐴𝐵 and let 𝑁 be the midpoint of 𝐵𝐶. Then 𝑈 = ℋ𝐵 ℋ𝑀 since ℋ𝐵 ℋ𝑀 is a translation with ℋ𝐵 ℋ𝑀 (𝐴) = ℋ𝐵 (𝐵) = 𝐵, so ℋ𝐵 ℋ𝑀 = 𝒯𝐴𝐵 . By similar reasoning 𝑉 = ℋ𝑁 ℋ𝐵 , since ℋ𝑁 ℋ𝐵 (𝐵) = 𝐶. Multiplying these translations, 𝑉𝑈 = ℋ𝑁 ℋ𝐵 ℋ𝐵 ℋ𝑀 = ℋ𝑁 ℋ𝑀 . This is a translation that maps 𝐴 to 𝐶, so 𝑈𝑉 = 𝒯𝐴𝐶 .

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7. Parallel Lines and Translations

To prove that the product of a translation and a half-turn is a translation, one uses the same ideas. Let the half-turn be ℋ𝐵 and the translation be 𝒯𝐴𝐵 . Then continuing with the notation above, ℋ𝐵 𝒯𝐴𝐵 = ℋ𝐵 ℋ𝐵 ℋ𝑀 = ℋ𝑀 . Likewise, 𝒯𝐵𝐶 ℋ𝐵 = ℋ𝑁 ℋ𝐵 ℋ𝐵 = ℋ 𝑁 . □ The second theorem is a product property quite different from that for line reflections and most other products of rigid motions. Theorem 7.28 (Commutativity). Products of translations are commutative. For any translations 𝑆 and 𝑇, 𝑆𝑇 = 𝑇𝑆. C =T (S (A))

C = S (T (A)) D =T (A)

D = T (A) B = S (A) A

Apply T to AB.

B = S (A) A

Apply S to AD.

Figure 26. Commutative 𝑆 and 𝑇

Proof. If either 𝑆 or 𝑇 is the identity, the theorem is immediate. For the rest of the proof, we assume neither translation is the identity. This proof works for linear parallelograms also. Take any point 𝐴 and let 𝐵 = 𝑆(𝐴) and 𝐷 = 𝑇(𝐴). Apply 𝑇 to 𝐴𝐵 to get image 𝐷𝑇(𝐵). Since 𝐷𝑇(𝐵) is the image of 𝐴𝐵 by translation 𝑇, 𝐷𝐴𝐵𝑇(𝐵) is a parallelogram (or linear parallelogram), by Theorem 7.26. This is illustrated on the left side of Figure 26. Now compare with a second construction of a parallelogram. Beginning with 𝐴 and 𝐷, apply 𝑆 to 𝐴𝐷 with image 𝐵𝑆(𝐷). Then we have 𝐷𝐴𝐵𝑆(𝐷) is a parallelogram or linear parallelogram. This is illustrated on the right side of Figure 26. Now we have two parallelograms with three points in common. So the fourth points 𝑇(𝐵) and 𝑆(𝐷) must be the same point 𝐶 by Theorem 7.23. Therefore, 𝐶 = 𝑇(𝐵) = 𝑇𝑆(𝐴) and 𝐶 = 𝑇(𝐷) = 𝑆𝑇(𝐴). So 𝑆𝑇 = 𝒯𝐴𝐶 = 𝑇𝑆. □

Direction from Translation Now that the theory of translations is well developed, one can use translations to rephrase or reframe some earlier ideas in a simpler and more flexible form. ⃗ ⃗ and 𝐶 Previously, the concept of same direction for two parallel rays 𝐴𝐵 𝐷 was defined in Definition 7.11 as the rays being on the same side of 𝐴𝐶. For collinear rays, the definition on page 14 was that one ray was contained in the other. In either case, for ⃗ ⃗ These definitions can endpoint 𝐶 there is a unique ray 𝐶 𝐷 in the same direction as 𝐴𝐵. be combined into one case using translations. With this definition, some equivalence properties of the direction relation for rays become clear.

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103

⃗ ⃗ and 𝐶 Theorem 7.29 (Direction by Translation). The rays 𝐴𝐵 𝐷 have the same direction ⃗ ⃗ to 𝐶 if and only if 𝒯𝐴𝐶 maps 𝐴𝐵 𝐷. They have opposite direction if the half-turn ℋ𝑀 maps ⃗ ⃗ to 𝐶 𝐴𝐵 𝐷, where 𝑀 is the midpoint of 𝐴𝐶.

-- TAC --> B A

D

M

B' C

B'' Figure 27. Rays with the Same Direction Related by Translation

Proof. Let 𝑀 be the midpoint of 𝐴𝐶. Then 𝒯𝐴𝐶 = ℋ𝐶 ℋ𝑀 . Let 𝐵 ′ = 𝒯𝐴𝐶 (𝐵). ⃗ is a ray with endpoint Consider when 𝐴, 𝐵, 𝐶 are not collinear. The image 𝒯𝐴𝐶 (𝐴𝐵) ′ ′ ⃗ 𝐶 parallel to 𝐴𝐵. Since 𝐴𝐶𝐵 𝐵 is a parallelogram, 𝐵 and 𝐵 are on the same side of 𝐴𝐶. ⃗′ is the unique ray with endpoint 𝐶 in the same direction as 𝐴𝐵. ⃗ Therefore, 𝐶𝐵 ⃗′ is 𝐶𝐵 ⃗″ = ℋ𝐶 (𝐶𝐵 ⃗′ ) = ℋ𝑀 (𝐴𝐵), ⃗ where 𝐵 ″ = ℋ𝑀 (𝐵). The ray opposite 𝐶𝐵 The collinear case follows from a ruler computation since a translation adds a constant to each number and a half-turn reverses direction. □ Recall that the description of direction for rays carries over to directed segments ⃗ and and lines as well. Two segments 𝐴𝐵 and 𝐶𝐷 have the same direction if the rays 𝐴𝐵 ⃗ 𝐶 𝐷 have the same direction. The properties of translation make it straightforward to prove some informally evident properties of direction in this corollary. Corollary. These properties of direction follow from the theorem. ⃗ ⃗ ⃗ then 𝐴𝐵 ⃗ has the same direction as 𝐶 (1) If 𝐶 𝐷 has the same direction as 𝐴𝐵, 𝐷. ⃗ ⃗ ⃗ and 𝐸𝐹 ⃗ has the same direction as 𝐶 ⃗ (2) If 𝐶 𝐷 has the same direction as 𝐴𝐵 𝐷, then 𝐸𝐹 ⃗ has the same direction as 𝐴𝐵. ⃗ Here is a proof of the second item in the corollary. The assumption is that 𝐶 𝐷= ⃗ ⃗ and 𝐸𝐹 ⃗ = ⃗ = 𝒯𝐶𝐸 (𝐶 ⃗ = 𝒯𝐶𝐸 𝒯𝐴𝐶 (𝐴𝐵) 𝒯𝐴𝐶 (𝐴𝐵) 𝐷). Then since 𝒯𝐶𝐸 𝒯𝐴𝐶 = 𝒯𝐴𝐸 , 𝐸𝐹 ⃗ The rest is left to the reader. 𝒯𝐴𝐸 (𝐴𝐵).

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7. Parallel Lines and Translations

More Parallelograms. This next proposition unifies with translation language some earlier statements about parallel segments proved in Theorems 7.12 and 7.22. Theorem 7.30 (Direction and Congruent Segments). Let 𝐴𝐵 and 𝐶𝐷 be congruent segments. If they have the same direction, then 𝒯𝐴𝐶 maps 𝐴𝐵 to 𝐶𝐷 and 𝐴𝐵𝐷𝐶 is a parallelogram or linear parallelogram. If they have opposite direction, then ℋ𝑀 maps 𝐴𝐵 to 𝐶𝐷, where 𝑀 is the midpoint of 𝐴𝐶. In this case 𝐴𝐵𝐶𝐷 is a parallelogram or linear parallelogram. As an application of this theorem, we prove a theorem about an adjoining pair of parallelograms. E

F

C

D A

B

Figure 28. Two Linked Parallelograms Imply a Third

Theorem 7.31 (Third Parallelogram). Given figures 𝐴𝐵𝐶𝐷 and 𝐴𝐵𝐸𝐹, each a parallelogram or linear parallelogram, the figure 𝐷𝐶𝐸𝐹 is a parallelogram if the lines 𝐶𝐷 and 𝐸𝐹 are distinct and it is a linear parallelogram if they are equal. Proof. Since 𝐴𝐵𝐶𝐷 is a parallelogram or linear parallelogram, translation 𝒯𝐴𝐵 maps 𝐷 to 𝐶. By the same reasoning for 𝐴𝐵𝐸𝐹, 𝒯𝐴𝐵 maps 𝐹 to 𝐸. By Theorem 7.30, this means that 𝐶𝐷𝐹𝐸 is a parallelogram or a linear parallelogram. □ In addition to parallelograms formed by segments and their images, the following theorem is also true. Theorem 7.32 (Four Half-turn Centers). The translation ℋ𝐹 ℋ𝐸 = ℋ𝐻 ℋ𝐺 if and only if 𝐸𝐹𝐻𝐺 is a parallelogram or linear parallelogram. Proof. From Theorem 7.25, it follows that 𝐸𝐹 and 𝐺𝐻 are either parallel or collinear. The hypothesis is equivalent to ℋ𝐺 ℋ𝐻 ℋ𝐹 ℋ𝐸 = 𝐼 and this is equivalent to ℋ𝐻 ℋ𝐹 = ℋ𝐺 ℋ𝐸 . Therefore, if the points are not collinear, then 𝐹𝐻 is parallel to 𝐸𝐺, so 𝐸𝐹𝐻𝐺 is a parallelogram. The linear case follows from a coordinate computation. □

Direction and Rotation from Polar Angle In Chapter 2 protractors and polar angles at a point were introduced. Later, in Definition 5.6, the choice of a protractor at a point 𝐴 determined signed angle measure 𝑚+ ∠𝐵𝐴𝐶 as a difference of polar angles. But there was no way to relate polar angles between any two points of the plane. But now, using translations, these angles and measures can be extended uniformly to the whole plane.

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Definition 7.33. A global protractor 𝛿∗ on the plane is a set of protractors 𝛿𝑃 at each point 𝑃 such that for any real 𝑡 and any points 𝐴 and 𝐶, 𝛿 𝐶 (𝑡) = 𝒯𝐴𝐶 (𝛿𝐴 (𝑡)). ⃗ denote the polar angle of 𝐴𝐵 ⃗ correspondGiven such a global protractor, let 𝜃(𝐴𝐵) ⃗ ⃗ ⃗ then 𝜃(𝐶 ⃗ ing to 𝛿𝐴 , the protractor at 𝐴. For any point 𝐶, if 𝐶 𝐷 = 𝒯𝐴𝐶 (𝐴𝐵), 𝐷) = 𝜃(𝐴𝐵) ⃗ ⃗ ⃗ ⃗ because 𝛿 𝐶 (𝜃(𝐴𝐵)) = 𝒯𝐴𝐶 𝛿𝐴 (𝜃(𝐴𝐵)) = 𝒯𝐴𝐶 (𝐴𝐵) = 𝐶 𝐷. Conversely, if these polar ⃗ ⃗ angles are equal, by the same relationship 𝐶 𝐷 = 𝒯𝐴𝐶 (𝐴𝐵). = 25

= 25

B D

A C

Figure 29. Rays with the Same Direction and Polar Angle

This observation proves the following theorem. Theorem 7.34 (Direction and Global Polar Angle). Given a global protractor 𝛿∗ , two ⃗ ⃗ and 𝐶 rays 𝐴𝐵 𝐷 have the same direction if and only if they have the same polar angles ⃗ ⃗ 𝜃(𝐴𝐵) and 𝜃(𝐶 𝐷). ⃗ ⃗ and 𝜃(𝐶 This implies that two lines 𝐴𝐵 and 𝐶𝐷 are parallel if and only if 𝜃(𝐴𝐵) 𝐷) are equal or differ by 180 degrees, depending on whether the two rays have the same or opposite direction. This relationship between polar angles and translation can be used to construct a global protractor from a protractor at a single point. Theorem 7.35 (Existence of Global Protractors). For a protractor 𝛿𝐴 at any point 𝐴, there is a global protractor 𝛿∗ defined by 𝛿 𝐶 (𝑡) = 𝒯𝐴𝐶 (𝛿𝐴 (𝑡)) for all points 𝐶. Proof. The theorem defines a function 𝛿 𝐶 for every 𝐶. It must be shown that this ⃗ and 𝐶 ⃗ function is in fact a protractor at 𝐶. For any rays 𝐶𝑃 𝑄, choose values of 𝜃 so ⃗ − 𝜃(𝐶 ⃗ that |𝜃(𝐶𝑃) 𝑄)| ≤ 180. For 𝛿 𝐶 to be a protractor, this absolute value must equal 𝑚∠𝑃𝐶𝑄. ⃗′ be the images of these rays by translation 𝒯𝐶𝐴 . The polar angles ⃗′ and 𝐴𝑄 Let 𝐴𝑃 ⃗ ⃗ ⃗′ ). Since 𝛿𝐴 is known to be a pro⃗′ ) and 𝜃(𝐴𝑄 𝜃(𝐶𝑃) and 𝜃(𝐶 𝑄) are the same as 𝜃(𝐴𝑃 ′ ′ ′ ′ ⃗ − 𝜃(𝐶 ⃗ ⃗ )| = |𝜃(𝐶𝑃) ⃗ ) − 𝜃(𝐴𝑄 tractor, 𝑚∠𝑃 𝐴𝑄 = |𝜃(𝐴𝑃 𝑄)|. Because translations are rigid motions, 𝑚∠𝑃 ′ 𝐴𝑄′ = 𝑚∠𝑃𝐶𝑄. Thus 𝛿 𝐶 has the properties of a protractor. What remains to be proved is that any two points satisfy the global protractor relation. If 𝐸 is another point, then 𝛿𝐸 (𝑡) = 𝒯𝐶𝐸 (𝛿 𝐶 (𝑡)) for every 𝑡. By the definition in −1 the theorem statement, 𝛿 𝐶 (𝑡) = 𝒯𝐴𝐶 (𝛿𝐴 (𝑡)), so 𝛿𝐴 (𝑡) = 𝒯𝐶𝐴 (𝛿 𝐶 (𝑡)), since 𝒯𝐶𝐴 = 𝒯𝐴𝐶 . Then 𝛿𝐸 (𝑡) = 𝒯𝐴𝐸 (𝛿𝐴 (𝑡)) = 𝒯𝐴𝐸 𝒯𝐶𝐴 (𝛿 𝐶 (𝑡)) = 𝒯𝐶𝐸 (𝛿 𝐶 (𝑡)), since 𝒯𝐴𝐸 𝒯𝐶𝐴 = 𝒯𝐶𝐸 . □

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By choosing a global protractor on the plane, one can simplify relationships. For example, in Chapter 5, to define an orientation on the plane without translations, one had to compare protractors along various line segments. With a global protractor, the positive (counterclockwise) rotation direction is determined at every point by the protractor at the point. Another example, concerning rotations of rays, is given in the next paragraphs. Direction and Rotation. With a protractor one can measure both unsigned and signed angles at a point 𝑃. Assume 𝑃𝑘 is a rotation with rotation angle 𝑘, with 𝑘 cho⃗ if sen so that |𝑘| ≤ 180. We know from Definitions 5.5 and 5.6 that for any ray 𝑃𝑄, ′ ′ + ′ 𝑄 = 𝑃𝑘 (𝑄), then 𝑚∠𝑄𝑃𝑄 = |𝑘| and 𝑚 ∠𝑄𝑃𝑄 = 𝑘. With a global protractor, the same result is true for rays with endpoints different from 𝑃. Theorem 7.36 (Polar Angle Rotation). Given a global protractor 𝛿∗ let a rotation 𝑃𝑘 ′ 𝐵 ′ . Then 𝜃(𝐴 ′ 𝐵 ′ ) = 𝜃(𝐴𝐵) ⃗ to 𝐴 ⃗ ⃗ ⃗ + 𝑘. If 0 < |𝑘| < 180, the lines 𝐴𝐵 and 𝐴′ 𝐵′ map a ray 𝐴𝐵 intersect at a point 𝐶 forming an angle of measure |𝑘|. More specifically, the lines form ′ 𝐵 ′ ), with 𝑚∠𝐷𝐶𝐸 = |𝑘| and ⃗ ⃗ ⃗ and 𝐶 ⃗ an angle ∠𝐷𝐶𝐸, where 𝐶 𝐷 = 𝒯𝐴𝐶 (𝐴𝐵) 𝐸 = 𝒯𝐴′ 𝐶 (𝐴 𝑚+ ∠𝐷𝐶𝐸 = 𝑘.

A' B

B' Q

Q'

P

A

Figure 30. Polar Angle Change by a Ray Rotated about Center 𝑃

⃗ The rotation maps these rays ⃗ has the same direction as a ray 𝑃𝑄. Proof. The ray 𝐴𝐵 ′ ′ ′ ⃗ ⃗ to a pair of rays 𝐴 𝐵 and 𝑃𝑄 that also have the same direction, as illustrated in Figure ′ 𝐵 ′ ) = 𝜃(𝑃𝑄 ⃗′ ) = 𝜃(𝑃𝑄) ⃗ + 𝑘 = 𝜃(𝐴𝐵) ⃗ ⃗ + 𝑘. 30. Therefore, 𝜃(𝐴 If 0 < |𝑘| < 180, by Theorem 7.34, the lines 𝐴𝐵 and 𝐴′ 𝐵 ′ are not parallel so they intersect at a point 𝐶. ⃗ ⃗ ⃗ = 𝒯𝑃𝐶 (𝑃𝑄) ⃗ and 𝐶 ⃗′ ), so the translation 𝒯𝑃𝐶 The ray 𝐶 𝐷 = 𝒯𝐴𝐶 𝒯𝑃𝐴 (𝑃𝑄) 𝐸 = 𝒯𝐴𝐶 (𝑃𝑄 takes ∠𝑄𝑃𝑄′ to ∠𝐷𝐶𝐸 so the angles have the same measure and signed measure. □ This theorem allows one to deduce the nature of products of even (orientationpreserving) rigid motions. An even rigid motion is a rotation with rotation angle 𝑘 or else a translation, which can be assigned rotation number 0 since translations do not change polar angles. Let 𝑇1 and 𝑇2 be even rigid motions with rotation angles 𝑘1 and

Vectors

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𝑘2 . The product 𝑇2 𝑇1 changes the polar angle of a ray by adding 𝑘 = 𝑘1 + 𝑘2 . If 𝑘 = 0 (mod 360), the product is a translation. Otherwise, it is a rotation with rotation angle 𝑘 but center unknown from this information. Such products will be studied, and the centers of rotation located, in Theorem 10.2. Symmetries of a Pair of Parallel Lines. Some symmetries of a pair of parallel lines 𝑝 and 𝑞 were listed after the Midline Theorem, Theorem 7.17. One line of symmetry was the midline 𝑚. The others were reflections in perpendicular transversals and also half-turns with centers on 𝑚. It was also mentioned that products of these symmetries are symmetries. Now we know that products of reflections in two perpendicular transversals are translations in a direction parallel to 𝑝 and 𝑞. This is also true of the products of half-turns centered on 𝑚. So any translation with direction parallel to these lines is a symmetry. There is one other reflection product that has not been considered yet. This is the glide reflection. A glide reflection is defined to be the product of a line reflection and a translation parallel to the line of reflection. In the case of parallel lines, every glide reflection symmetry is the product of a translation symmetry and ℛ𝑚 . Such a product will translate the lines and then interchange 𝑝 and 𝑞. It is also true that a product of a half-turn symmetry and one of the perpendicular line reflections is also a glide reflection, a different way of writing the same symmetries. Glide reflections will be discussed in detail in Chapter 10. All these rigid motions are also the symmetries for a single line 𝑝, provided one takes 𝑚 = 𝑝.

Vectors When a translation can be written two different ways, as 𝒯𝐴𝐵 = 𝒯𝐶𝐷 , this means that 𝐴𝐵 and 𝐶𝐷 are congruent segments with the same direction. This is another case when a relationship is an equivalence relation, and this one is extremely useful in math and science and engineering. Definition 7.37. Two segments 𝐴𝐵 and 𝐶𝐷 represent the same vector if 𝒯𝐴𝐵 = 𝒯𝐶𝐷 . The notation for this relation will be [𝐴𝐵] = [𝐶𝐷]. Our use of [𝐴𝐵] to denote a vector is not standard notation. The vector [𝐴𝐵] is ⃗ but we are using this notation for rays. usually denoted by an arrow on top, as 𝐴𝐵, A'

[AE] = 3[AB] + (1/2)[CD] E

G

F A

B

C

Figure 31. Vector Addition and Multiplication by Real Numbers

D

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It is easy to see from this definition that the vector relation has these equivalence properties. • For all vectors, [𝐴𝐵] = [𝐴𝐵]. • If [𝐴𝐵] = [𝐶𝐷], then [𝐶𝐷] = [𝐴𝐵]. • If [𝐴𝐵] = [𝐶𝐷] and [𝐶𝐷] = [𝐸𝐹], then [𝐴𝐵] = [𝐸𝐹]. For the last item, given 𝐶𝐷 = 𝒯𝐴𝐶 (𝐴𝐵) and 𝐸𝐹 = 𝒯𝐶𝐸 (𝐶𝐷), then 𝐸𝐹 = 𝒯𝐶𝐸 𝒯𝐴𝐶 (𝐴𝐵) = 𝒯𝐴𝐸 (𝐴𝐵). The vector, the set of equivalent segments, is viewed as an object. Geometrically, the vector can be represented as an arrow with tail at 𝐴 and head at 𝐵. But it is also represented by an infinite number of other arrows, each one with the same length and the same direction. All such arrows are translations of each other. It is not yet clear just from this definition what value is gained by this concept, especially since the set of vectors is in one-to-one correspondence with the set of translations. We will see vectors in action in later chapters. There are some operations with vectors that we can define now. The first is addition. If 𝒯𝐶𝐷 𝒯𝐴𝐵 = 𝒯𝐸𝐹 , then [𝐴𝐵] + [𝐶𝐷] = [𝐸𝐹]. In particular, since 𝒯𝐵𝐶 𝒯𝐴𝐵 = 𝒯𝐴𝐶 , then [𝐴𝐵] + [𝐵𝐶] = [𝐴𝐶], as pictured in Figure 25. Also, in Figure 26, [𝐴𝐵] + [𝐴𝐷] = [𝐴𝐶]. This shows what is called the parallelogram law of addition. The vector [𝐴𝐴] acts as the zero for this addition; it can just be written as 0. Reversing the segment order is the additive inverse: [𝐴𝐵] + [𝐵𝐴] = [𝐴𝐴] = 0. Vector addition is commutative by Theorem 7.28. The other important vector operation is the multiplication of a vector by a real number. If 𝑟 > 0, the vector 𝑟[𝐴𝐵] is the vector 𝐶𝐷, where 𝐴𝐵 and 𝐶𝐷 have the same direction, but ‖𝐶𝐷‖ = 𝑟‖𝐴𝐵‖. If 𝑟 < 0, the direction is reversed and 𝑟[𝐴𝐵] = |𝑟|[𝐵𝐴]. Also, 0[𝐴𝐵] = 0. These operations are illustrated in Figure 31. The vector [𝐴𝐴′ ] = [𝐶𝐷], [𝐴𝐺] = (1/2)[𝐶𝐷], [𝐴𝐹] = 3[𝐴𝐵], and [𝐴𝐸] = 3[𝐴𝐵] + (1/2)[𝐶𝐷].

Exercises and Explorations 1. (Regular Polygon Angle). Derive the formula for the angles of a regular 𝑛-gon. 2. (Quadrilateral with Lines of Symmetry). Show that a quadrilateral with two perpendicular lines of symmetry is either a rectangle or a rhombus. 3. (Intersecting Railroad Tracks). Suppose 𝑚 and 𝑛 are parallel lines with distance 𝑑 between them, and suppose 𝑝 and 𝑞 are also parallel lines with the same distance 𝑑 between them. If the parallel pairs intersect, tell what kind of quadrilateral their intersection points define and prove your claim.

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4. (Circles and Parallelograms). Two problems: (a) Suppose parallelogram 𝐴𝐵𝐶𝐷 is inscribed in circle 𝑐. What properties must it have? Must it be a special kind of parallelogram? If so, what kind? (b) Suppose a circle 𝑐 is inscribed in a parallelogram 𝐴𝐵𝐶𝐷. What properties must 𝐴𝐵𝐶𝐷 have? Must it be a special parallelogram? If so, what kind? 5. (Midpoint Quadrilaterals). Theorem 7.19 proves that the midpoint quadrilateral 𝐽𝐾𝐿𝑀 of any quadrilateral 𝐴𝐵𝐶𝐷 is a parallelogram. Some of the figures below have midpoint quadrilaterals that are special kinds of parallelograms. Tell what they are and justify your answer. (a) A kite (b) A rhombus (c) A parallelogram (d) A rectangle (e) A square 6. (Midpoint Quadrilateral Symmetry). For the quadrilaterals in the previous problem, describe the set of all symmetries of the midpoint figure. Then answer these questions. (a) Is every symmetry of 𝐴𝐵𝐶𝐷 also a symmetry of the midpoint quadrilateral 𝐽𝐾𝐿𝑀? (b) Is every symmetry of the midpoint quadrilateral 𝐽𝐾𝐿𝑀 also a symmetry of 𝐴𝐵𝐶𝐷? (c) In either case, if yes, why? If not, is there any other relationship between the groups of symmetries? 7. (Nonconvex Quadrilaterals). In the proof of Theorem 7.19 it was asserted for a quadrilateral 𝐴𝐵𝐶𝐷 that if 𝐵 and 𝐷 are on the same side of 𝐴𝐶, then either 𝐷 is in the interior of △𝐴𝐵𝐶 or 𝐵 is in the interior of △𝐴𝐷𝐶 because the sides of the quadrilateral do not intersect except at the vertices. Prove this. Hint: Consider the regions of the plane created by the extended sides of △𝐴𝐵𝐶. Where can 𝐷 be? 8. (A More General Midpoint Quad). In Figure 32 there is an example on the right of an 𝐴𝐵𝐶𝐷 that is not a parallelogram but for which the midpoint figure 𝐽𝐾𝐿𝑀 is still a parallelogram. Use what we have learned about translations to prove that if 𝐴𝐵𝐶𝐷 is not assumed to be a quadrilateral but is just four distinct points, then

A

M

J

D L

C

A

M

D

B

D L

J

J

K B

B

K

Figure 32. Quadrilateral Midpoints

L

K

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the midpoint figure 𝐽𝐾𝐿𝑀 is still either a parallelogram or a linear parallelogram. Hint: Prove the translations ℋ𝐾 ℋ𝐽 and ℋ𝐿 ℋ𝑀 are equal because they agree at a point. What does this say about 𝐽𝐾𝐿𝑀? 9. (Translation Experiments). Here are some translation experiments using the midpoint triangle in Figure 13. You may need to sketch your own copy. Answer with points already labeled if possible, but describe and label some new points if needed. (a) What point is 𝒯𝐶𝐹 (𝐹)? How do you know? (b) Describe the image 𝒯𝐶𝐹 (△𝐷𝐸𝐹). (c) Tell some points 𝑃 and 𝑄 for which 𝒯𝑃𝑄 = ℋ𝐸 ℋ𝐷 . (d) Tell some points 𝑃 and 𝑄 for which 𝒯𝑃𝑄 = 𝒯𝐵𝐴 𝒯𝐷𝐹 . (e) For what point 𝑃 is ℋ𝑃 = ℋ𝐴 ℋ𝐸 ℋ𝐷 ? 10. (Fourth Parallelogram Point). Given a parallelogram 𝐴𝐵𝐶𝐷, we know from Theorem 7.23 that 𝐷 is uniquely determined when given 𝐴, 𝐵, 𝐶. But the parallelogram 𝐴𝐶𝐵𝐸, with the same three points in different order, is a different quadrilateral. Choose any noncollinear points 𝐴, 𝐵, 𝐶. Then draw all the parallelograms with these three points as vertices. 11. (Tiled Triangle). The triangle tiled with triangles in Figure 14 was created assuming the Euclidean Parallel Postulate as a theorem. By subdividing further with repeated midpoint triangles and taking limits, this tiled triangle suggests how the Dilation Axiom could be proved by assuming EPP as a postulate instead of the reverse, as we have done. (a) Check that the 𝐴𝐷/𝐴𝐵 and 𝐴𝐸/𝐴𝐶 equal ‖𝐷𝐸‖/‖𝐵𝐶‖. Check that ∠𝐴𝐵𝐶 ≅ ∠𝐴𝐷𝐸 and ∠𝐴𝐶𝐵 ≅ ∠𝐴𝐸𝐷. Thus, for this example, ratios and angles are preserved. (b) The ratios involved are also related to area, or equivalently, to the number of triangles in various figures. What is the total number of small triangles in △𝐴𝐵𝐶? What is the number in △𝐴𝐷𝐸? How is this related to the ratios of lengths? (c) Let 𝐴′ = 𝒯𝐸𝐶 (𝐴) and 𝐷′ = 𝒯𝐸𝐶 (𝐷). Tell how many small triangles are in each of these polygons: △𝐷𝐵𝐷′ , △𝐴′ 𝐷′ 𝐶, quadrilateral 𝐴𝐷𝐷′ 𝐴′ . How are these numbers related to ratios of length? 12. (Parallelograms that Are Rectangles). It was observed that rectangles have congruent diagonals and right vertex angles. Prove these converses. (a) Any parallelogram with congruent diagonals is a rectangle. (b) Any parallelogram with one vertex angle that is a right angle is a rectangle. 13. (Concurrence of Medians). A median of a triangle is a segment from a vertex to the midpoint of the opposite side. This is the first of several proofs of a theorem about medians of a triangle. (a) In a triangle △𝐴𝐵𝐶, let the medians be 𝐴𝐷, 𝐵𝐸, and 𝐶𝐹. Let 𝐺 be the intersection of 𝐵𝐸 and 𝐶𝐹. Consider the midpoint quadrilateral of 𝐴𝐵𝐺𝐶 and use it to find these ratios: 𝐵𝐺/𝐵𝐸 and 𝐶𝐺/𝐶𝐹. (b) Let 𝐻 be the intersection of 𝐵𝐸 and 𝐴𝐷. Find the ratios 𝐵𝐻/𝐵𝐸 and 𝐴𝐻/𝐴𝐷. (c) Explain how these two results prove that the three medians are concurrent at 𝐺 = 𝐻.

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111

14. (Translations as Products of Line Reflections). In Figure 33 the large triangle is mapped by one product of two line reflections and the small triangle by another, but the different products are the same translation. Prove that products of reflections in parallel lines ℛ𝑛 ℛ𝑚 and ℛ𝑛′ ℛ𝑚′ are equal if and only if there is a translation 𝑇 such that 𝑇(𝑚) = 𝑚′ and 𝑇(𝑛) = 𝑛′ . m

n

m'

n'

Figure 33. A Translation Shown as Two Different Double Line Reflections

⃗ ⃗ and 𝐶 15. (Rotating One Ray to Another). Let 𝐴𝐵 𝐷 be two rays that do not have the ⃗ ⃗ to 𝐶 same direction. Prove that there is a unique rotation 𝑃𝑘 that maps 𝐴𝐵 𝐷. Hint: How did we prove congruence of segments? 16. (Invariant Lines). A line 𝑚 is an invariant line for a rigid motion 𝑇 if 𝑇(𝑚) = 𝑚. If the points of the line are all fixed points, the line is invariant, but in general the points of the line will not all be fixed. (a) For a reflection ℛ𝑛 , find all the invariant lines, in addition to 𝑛. Explain your reasoning. (b) For a half-turn ℋ𝐴 , find all the invariant lines. Explain your reasoning. (c) For a translation 𝒯𝐴𝐵 , find all the invariant lines. Explain your reasoning. (d) For a glide reflection 𝒢𝐴𝐵 , find all the invariant lines. Explain your reasoning. (e) For a rotation 𝐴60 , find all the invariant lines? Explain your reasoning. 17. (Transformation Groups). Transformation groups were defined in Definition 5.10. Tell which of the following sets are transformation groups and justify your answer: (a) the set of all translations, (b) the set of all half-turns, (c) the set of all transformations that are either half-turns or translations, (d) for a given line 𝑚, the set of all translations that map this line into itself, ⃗ the set of all translations that map this ray into itself. (e) for a given ray 𝐴𝐵,

Chapter 8

Dilations and Similarity

In this chapter, we will begin the study of similar figures. Similarity is a relationship among figures much like congruence, except that instead of “same size, same shape,” it formalizes an idea of “enlarged or reduced size, same shape”. As with congruence, certain transformations will be used to define this relation; these transformations are called similitudes. Definition 8.1. A similitude with scale factor 𝑟 > 0 is a transformation that scales distances by the factor 𝑟. To be specific, if 𝐴 and 𝐵 are any points and 𝑆 is a similitude, then ‖𝑆(𝐴)𝑆(𝐵)‖ = 𝑟‖𝐴𝐵‖. If 𝑆 is a similitude, 𝑆 −1 is a similitude with scale factor 1/𝑟. To see ‖𝑆−1 (𝐴)𝑆 −1 (𝐵)‖ = (1/𝑟)‖𝐴𝐵‖ for all 𝐴 and 𝐵, let 𝐶 = 𝑆 −1 (𝐴) and 𝐷 = 𝑆 −1 (𝐵). Then this equation becomes ‖𝐶𝐷‖ = (1/𝑟)‖𝑆(𝐶)𝑆(𝐷)‖, which is true for all 𝐶 and 𝐷. By the Dilation Axiom, any dilation 𝒟𝑃,𝑟 is a similitude with scale factor 𝑟. From its definition, a rigid motion is a similitude with scale factor 1. Conversely, a similitude with scale factor 𝑟 = 1 is an isometry and so by Theorem 4.10 is a rigid motion. If 𝐹 and 𝐺 are similitudes with scale factors 𝑟 and 𝑠, then 𝐹𝐺 is a similitude with scale factor 𝑟𝑠, since ‖𝐹𝐺(𝐴)𝐹𝐺(𝐵)‖ = 𝑟‖𝐺(𝐴)𝐺(𝐵)‖ = 𝑟𝑠‖𝐴𝐵‖ for any points 𝐴 and 𝐵. Similitudes preserve angle measure. This is not part of the definition, but we can prove this as a theorem. Theorem 8.2 (Similitudes and Angles). Every similitude preserves angle measure. Consequently, since angle measures of 0 and 180 are preserved, a similitude maps lines to lines, segments to segments, and rays to rays. Proof. Let 𝑆 be a similitude with scale factor 𝑟. For any point 𝑃, 𝒟𝑃,1/𝑟 is a similitude with scale factor 1/𝑟. Then, as observed above for 𝐹𝐺, the product 𝑇 = 𝒟𝑃,1/𝑟 𝑆 is a similitude with scale factor 𝑟(1/𝑟) = 1, so it is an isometry and hence is a rigid motion by Theorem 4.10. So 𝑇 preserves angle measure. Since 𝑆 = 𝒟𝑃,𝑟 𝑇, the similitude 𝑆 is the product of two transformations that preserve angle measure, so it also preserves angle measure. 113

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To prove the second statement, let 𝐴, 𝐵, 𝐶 be three collinear points and let 𝐴′ , 𝐵′ , 𝐶 be the images of these points by 𝑆. Then 𝑚∠𝐵𝐴𝐶 is 0 or 180, so 𝑚∠𝐵 ′ 𝐴′ 𝐶 ′ is also 0 or 180, so the image points are collinear. To see that the image of a segment is a segment, consider the segment 𝐴𝐵. The point 𝐶 is on 𝐴𝐵 if and only if ‖𝐴𝐶‖+‖𝐶𝐵‖ = ‖𝐴𝐵‖. But the image points then satisfy ‖𝐴′ 𝐶 ′ ‖ + ‖𝐶 ′ 𝐵 ′ ‖ = 𝑟‖𝐴𝐶‖ + 𝑟‖𝐶𝐵‖ = 𝑟‖𝐴𝐵‖ = ‖𝐴′ 𝐵′ ‖, ′ 𝐵 ′ = 𝑆(𝐴𝐵). ⃗ ⃗ so 𝐶 ′ is on 𝐴′ 𝐵′ . Such reasoning also implies that 𝐴 □ ′

Definition 8.3. If 𝑈 and 𝑉 are two sets in the plane, 𝑈 is similar to 𝑉 if there is a similitude 𝑆 such that 𝑉 = 𝑆(𝑈). This can be written 𝑈 ∼ 𝑉. Similar figures have corresponding angles congruent but corresponding distances are proportional by ratio 𝑟. The relationship of similarity resembles that of congruence in many respects. This theorem echoes the statements of Theorems 1.4 and 1.5. Theorem 8.4 (Similarity Properties). The similitudes have these three group properties: (a) The identity 𝐼 is a similitude. (b) If 𝐹 is a similitude, so is 𝐹 −1 . (c) If 𝐹 and 𝐺 are similitudes, so is 𝐹𝐺. Therefore, simiilarity is an equivalence relation; namely for any sets: (a) 𝑈 ∼ 𝑈. (b) If 𝑈 ∼ 𝑉, then 𝑉 ∼ 𝑈. (c) If 𝑈 ∼ 𝑉 and 𝑉 ∼ 𝑊, then 𝑈 ∼ 𝑊. Proof. The stated properties of similitudes were proved in the remarks following Definition 8.1. The proof of the three properties of similarity follow exactly the proof of Theorem 1.4, with rigid motions replaced by similitudes. □ Dilations with Negative Ratio. Dilations with positive dilation ratio were defined in Definition 2.13, just before the statement of the Dilation Axiom. This axiom says something very special about the plane. Transformations somewhat like dilations can be defined on many spaces. For example, on a sphere, think of a center at the North Pole with points nearby being mapped by a ratio along north-south “rays” that are circles of longitude. Distances between points on the same “ray” are scaled by a ratio 𝑟 > 0, but distances between points on different circles of longitude are not, as is clear from looking carefully at a globe. Thus, the Dilation Axiom that asserts that angles are preserved and that distances are scaled by 𝑘 is a distinguishing property of the flat space that is the plane. The axiom asserts that a dilation with a positive dilation factor is a similitude. In Chapter 7 we have seen that this property is equivalent to the Euclidean Parallel Postulate. As promised earlier, we now extend the definition of a dilation to include negative dilation ratios. The definition is illustrated in Figure 1. Definition 8.5 (Dilation with real scale ratio). For point 𝑃 and a real number 𝑘 ≠ 0, the dilation centered at 𝑃 with a dilation ratio 𝑘, denoted 𝒟𝑃,𝑘 , is a transformation of the plane defined as follows: 𝒟𝑃,𝑘 (𝑃) = 𝑃. For 𝑋 ≠ 𝑃, there are two cases: ⃗ with ‖𝑃𝑋 ′ ‖ = 𝑘‖𝑃𝑋‖. • If 𝑘 > 0, 𝑋 ′ = 𝒟𝑃,𝑘 (𝑋) is the point on 𝑃𝑋 ⃗ and satisfies ‖𝑃𝑋 ′ ‖ = |𝑘|‖𝑃𝑋‖. • If 𝑘 < 0, 𝑋 ′ = 𝒟𝑃,𝑘 (𝑋) is on the opposite ray of 𝑃𝑋 With this extended definition, a dilation is still a similitude. If 𝑘 < 0, then 𝒟𝐴,𝑘 = ℋ𝐴 𝒟𝐴,|𝑘| , the product of a dilation with positive scale factor and a half-turn, therefore the product of two similitudes.

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115

C' C

A'' B''

C''

B'

B P

A

A'

Figure 1. Dilations of △𝐴𝐵𝐶 by Ratios 7/4 and −2/3

In fact, a half-turn is a dilation with dilation ratio −1 as well as being a rotation. The other dilation with |𝑘| = 1 is the identity map, 𝒟𝐴,1 = 𝐼. Like the half-turn, the general dilation maps lines to parallel lines, or to themselves. Theorem 8.6 (Dilation Line Image). A dilation 𝒟𝑃,𝑘 maps any line 𝑚 to a line 𝑚′ , where 𝑚′ = 𝑚 or 𝑚′ is parallel to 𝑚. Proof. There are three cases. • If 𝑘 = 1, then the dilation is the identity and 𝑚′ = 𝑚. • If 𝑃 is on 𝑚, then 𝑚′ = 𝑚 by definition. • If 𝑃 is not on 𝑚 and 𝑘 ≠ 1, then let 𝐴 be a point of intersection of 𝑚 and 𝑚′ . Then 𝑚′ also contains 𝒟𝑃,𝑘 (𝐴) = 𝐴′ . The points 𝐴 and 𝐴′ are distinct, since 𝑘 ≠ 1. Since both points are on 𝑚′ , this implies 𝑚′ = 𝐴𝐴′ . But 𝑃 is on 𝐴𝐴′ , so this would imply that 𝑃 is on 𝑚′ and so also on 𝑚, which is not true. Therefore, there is no such 𝐴, and the lines are parallel. □ Dilations also have properties with respect to direction and orientation. • A dilation with positive dilation ratio preserves direction. • A dilation with negative dilation ratio reverses direction. • All dilations are orientation-preserving. The direction properties are clear from Definition 7.11. If, as in Figure 1, the segment 𝐴′ 𝐵 ′ is the 𝒟𝑃,𝑘 image of 𝐴𝐵, then 𝐵 and 𝐵′ are in the same half-plane of 𝑃𝐴 if 𝑘 > 0 and in opposite half-planes if 𝑘 < 0. A dilation is orientation-preserving since a triangle such as △𝑃𝐴𝐶 in Figure 1 is mapped to △𝑃𝐴′ 𝐶 ′ with the same orientation.

Similarity Theorems for Triangles We begin our study of similarity with some quick proofs of similarity criteria for triangles. For each of the triangle congruence theorems SAS, ASA, and SSS, there is a generalization to similar triangles that we will also call SAS, ASA, SSS. The congruence theorems will now be special cases when the scaling factor = 1. There is one exception to the naming: AA will often be used instead of ASA. The reason will become clear below.

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One can prove many things by applying a dilation and then referring to a congruence theorem. For example, to prove any two segments 𝐴𝐵 and 𝐶𝐷 are similar, let 𝑘 = ‖𝐶𝐷‖/‖𝐴𝐵‖. Then 𝒟𝐴,𝑘 will map 𝐴𝐵 to a segment 𝐴𝐵 ′ that has the same length as 𝐶𝐷. Since there is a congruence 𝑇 mapping 𝐴𝐵′ to 𝐶𝐷, then 𝑇𝒟𝐴,𝑘 is a similitude mapping 𝐴𝐵 to 𝐶𝐷. The same general method proves similarity tests for triangles, starting with a generalization for SAS. Theorem 8.7 (SAS [Similarity Version]). Let △𝐴𝐵𝐶 and △𝐷𝐸𝐹 be two triangles such that angles ∠𝐵𝐴𝐶 and ∠𝐸𝐷𝐹 are congruent and the adjacent sides are proportional by the same ratio 𝑘 > 0: ‖𝐷𝐸‖ = 𝑘‖𝐴𝐵‖, and ‖𝐷𝐹‖ = 𝑘‖𝐴𝐶‖. Then △𝐴𝐵𝐶 is similar to △𝐷𝐸𝐹 with scale factor 𝑘. Proof. Let △𝐴′ 𝐵 ′ 𝐶 ′ be the image by 𝒟𝐴,𝑘 of △𝐴𝐵𝐶. Then ‖𝐴′ 𝐵 ′ ‖ = 𝑘‖𝐴𝐵‖, ‖𝐴′ 𝐶 ′ ‖ = 𝑘‖𝐴𝐶‖, and ∠𝐵 ′ 𝐴′ 𝐶 ′ ≅ ∠𝐵𝐴𝐶. Therefore, △𝐴′ 𝐵 ′ 𝐶 ′ ≅ △𝐷𝐸𝐹 by SAS. Let 𝑇 be the rigid motion of this congruence. Then 𝑆 = 𝑇𝒟𝐴,𝑘 is a similitude such that 𝑆(△𝐴𝐵𝐶) = △𝐷𝐸𝐹, so the two triangles are similar. □ The same argument, scaling one figure by a dilation and then composing with a rigid motion, also proves generalizations of ASA and SSS. Theorem 8.8 (AA or ASA [Similarity Version]). Let △𝐴𝐵𝐶 and △𝐷𝐸𝐹 be two triangles such that angles ∠𝐵𝐴𝐶 ≅ ∠𝐸𝐷𝐹 and ∠𝐴𝐵𝐶 ≅ ∠𝐷𝐸𝐹. Then △𝐴𝐵𝐶 is similar to △𝐷𝐸𝐹 with scale factor 𝑘 = ||𝐷𝐸||/||𝐴𝐵||. Proof. Let 𝑘 = ‖𝐷𝐸‖/‖𝐴𝐵‖. Choose the dilation 𝒟𝐴,𝑘 just as in the previous theorem, with △𝐴′ 𝐵 ′ 𝐶 ′ being the image of △𝐴𝐵𝐶. Then there is a rigid motion 𝑇 as before, but this time by ASA, since ‖𝐴′ 𝐵 ′ ‖ = ‖𝐷𝐸‖. Then 𝑇𝒟𝐴,𝑘 is a similitude mapping △𝐴𝐵𝐶 □ to △𝐷𝐸𝐹. Note: Since the existence of the scale factor 𝑘 is automatic, the congruence of the two angles is sufficient to prove similarity for some scale factor. This accounts for dropping the S in ASA to get AA. Theorem 8.9 (SSS [Similarity Version]). Let △𝐴𝐵𝐶 and △𝐷𝐸𝐹 be triangles with corresponding sides proportional: 𝑘 = ‖𝐷𝐸‖/‖𝐴𝐵‖ = ‖𝐸𝐹‖/‖𝐵𝐶‖ = ‖𝐹𝐷‖/‖𝐶𝐴‖. Then △𝐴𝐵𝐶 is similar to △𝐷𝐸𝐹 with scale factor 𝑘. The proof proceeds just as for the others, except that the existence of 𝑇 follows from SSS. Here is one more proposition that follows from applying a rigid motion theorem. Theorem 8.10 (Similitude Uniqueness). If △𝐴𝐵𝐶 ∼ △𝐷𝐸𝐹, there is a unique similitude that maps 𝐴, 𝐵, and 𝐶 to 𝐷, 𝐸, and 𝐹. For any two segments 𝐸𝐹 and 𝐺𝐻, there is a unique orientation-preserving similitude that maps 𝐸𝐹 to 𝐺𝐻. Proof. Suppose that 𝑆 1 and 𝑆 2 are similitudes that map △𝐴𝐵𝐶 to △𝐷𝐸𝐹. Then 𝑆 −1 2 𝑆1 is a rigid motion mapping △𝐴𝐵𝐶 to △𝐴𝐵𝐶. This must be the identity map 𝐼 by Theorem 3.3, so 𝑆 1 = 𝑆 2 .

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117

For the segments, let 𝐷 = 𝒟𝐸,𝑟 , where 𝑟 = ‖𝐺𝐻‖/‖𝐸𝐹‖. If 𝐹 ′ = 𝐷(𝐹), then 𝐸𝐹 ≅ 𝐺𝐻. By Theorem 5.22, there is a unique orientation-preserving rigid motion 𝑇 that maps 𝐸𝐹 ′ to 𝐺𝐻. Let 𝑆 = 𝑇𝐷. If 𝑈 is any other such similitude, then it must be true that 𝑈𝐷−1 = 𝑇, by uniqueness of this rigid motion. □ ′

Right Triangles These theorems can be used to prove results without explicit reference to similitudes. For example, here is a proof of a famous theorem. Theorem 8.11 (Pythagorean Theorem). Let △𝐴𝐵𝐶 be a right triangle, with right angle at 𝐶. Then ‖𝐴𝐵‖2 = ‖𝐴𝐶‖2 + ‖𝐵𝐶‖2 . C a

b A

D

B c D

D bb/c A

aa/c

ab/c

ab/c C

C

B a

b

Figure 2. Similar Triangles for the Pythagorean Theorem

Proof. Let 𝐷 be the foot of the altitude through 𝐶. The segment 𝐶𝐷 is perpendicular to 𝐴𝐵, and the point 𝐷 is within the segment 𝐴𝐵 since the angles at 𝐴 and 𝐵 are acute. Also, let 𝑐 = ‖𝐴𝐵‖, the hypotenuse, and 𝑏 = ‖𝐴𝐶‖ and 𝑎 = ‖𝐵𝐶‖. The triangles △𝐴𝐶𝐷 and △𝐶𝐵𝐷 are similar to △𝐴𝐵𝐶 by AA, since each one has a right angle and also shares one of the angles with △𝐴𝐵𝐶. The side lengths of △𝐴𝐵𝐶 opposite the angles at 𝐴, 𝐵, and 𝐶 are (‖𝐵𝐶‖, ‖𝐴𝐶‖, ‖𝐴𝐵‖) = (𝑎, 𝑏, 𝑐). The scale factor from △𝐴𝐵𝐶 to △𝐴𝐶𝐷 is 𝑏/𝑐 since 𝑐 is the hypotenuse length of the larger triangle and 𝑏 is the hypotenuse length of the smaller. The corresponding sides of △𝐴𝐶𝐷 opposite the angles at 𝐴, 𝐶, 𝐷 are (‖𝐶𝐷‖, ‖𝐴𝐷‖, ‖𝐴𝐶‖) = (

𝑎𝑏 𝑏2 , , 𝑏) . 𝑐 𝑐

The scale factor from △𝐴𝐵𝐶 to △𝐶𝐵𝐷 is 𝑎/𝑐 since 𝑐 is the hypotenuse length of the larger triangle and 𝑎 is the hypotenuse length of the smaller. The corresponding

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8. Dilations and Similarity

sides of △𝐶𝐵𝐷 opposite the angles at 𝐶, 𝐵, 𝐷 are (‖𝐷𝐵‖, ‖𝐶𝐷‖, ‖𝐵𝐶‖) = (

𝑎2 𝑎𝑏 , , 𝑎) . 𝑐 𝑐

Since 𝐷 is on 𝐴𝐵, the length 𝑐 = ‖𝐴𝐵‖ = ‖𝐴𝐷‖ + ‖𝐷𝐵‖ =

𝑏2 𝑎2 + . 𝑐 𝑐 □

Multiplying by 𝑐, this becomes 𝑐2 = 𝑎2 + 𝑏2 .

Notice that there is no dilation that maps △𝐴𝐶𝐷 to △𝐴𝐵𝐶. However, if △𝐴𝐶𝐷 is first reflected in the angle bisector of ∠𝐶𝐴𝐵 to a triangle △𝐴𝐶 ′ 𝐷′ , then the segments 𝐵𝐶 and 𝐶 ′ 𝐷′ are parallel and there is a dilation that maps one to the other. One application of the Pythagorean Theorem is to find the lengths of special lines such as the diagonals of squares or the altitudes of equilateral triangles. Some examples are found in the exercises. Trigonometric Functions. In applications of geometry to architecture, mechanics, surveying, or navigation, one needs to find unknown lengths from some known measurements. The fundamental shape in this practical activity is the right triangle. Because knowing the dimensions of right triangles has been of fundamental importance in many practical endeavors, calculations were made and put into tables for quicker practical use. C C ||AC|| = sec

||AC|| = 1 ||BC|| = sin A

||AB|| = cos

B

A

||AB|| = 1

||BC|| = tan B

Figure 3. Right Triangles for Trig Tables

For right triangles, one right angle is known; then given one acute angle 𝜃 and one length equal to 1, by ASA, the other measures of the triangle are determined. If the measured length of the side corresponding to the unit length in the table is 𝐿, then 𝐿 is the scale factor that can be used to multiply the sides in the table to compute the other lengths in the desired right triangle. For some applications, it may be more convenient to make the hypotenuse the side with the unit length, or it may be more convenient to make one of the legs a unit length. The other side lengths associated with either of these choices are functions of 𝜃. They are called trigonometric functions and are given names corresponding to the choice of unit side — sine and cosine for unit hypotenuse and tangent and secant for unit leg — as shown in Figure 3.

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119

Perhaps this is a good moment to recognize that while actual tables of these numbers (and even their logarithms!) were on bookshelves not so many decades ago, now our “tables” are in our computers, calculators, or even our phones. Intersections and Constructions from the Pythagorean Theorem. Given a circle of radius 𝑟 and center 𝑃 and a line 𝑚 at distance 𝑑 < 𝑟 from 𝑃, in Theorem 6.8 it was proved by a limit process that the line intersects the circle at two points 𝐴 and 𝐵. What this theorem did not prove was the length of 𝐴𝐵. In fact, since this theorem was proved without the Dilation Axiom, there was more than one possible length1 . Using the Dilation Axiom and the Pythagorean Theorem, it is easy to find this length in the Euclidean plane. Theorem 8.12. Given a circle of radius 𝑟 and center 𝑃, a line 𝑚 at distance 𝑑 < 𝑟 from 𝑃 intersects 𝑐 in two points 𝐴 and 𝐵, with ‖𝐴𝐵‖ = 2√𝑟2 − 𝑑 2 . Proof. Let 𝑄 be the foot of the perpendicular from 𝑃 to 𝑚. Then ‖𝑃𝑄‖ = 𝑑. If we assume Theorem 6.8, the points 𝐴 and 𝐵 exist and △𝑃𝑄𝐴 and △𝑃𝑄𝐵 are congruent right triangles. The Pythagorean Theorem says that 𝑑 2 + ‖𝑄𝐴‖2 = 𝑟2 . Solving this equation, ‖𝐴𝐵‖ = 2‖𝑄𝐴‖ = 2√𝑟2 − 𝑑 2 . In fact, we can prove this theorem without assuming Theorem 6.8 at all. Just let 𝐴 and 𝐵 be the two points on 𝑚 at equal distance √𝑟2 − 𝑑 2 from 𝑄. Then in the right triangle △𝑃𝑄𝐴, we can compute ‖𝑃𝐴‖ = 𝑟 from ‖𝑃𝐴‖2 = ‖𝑃𝑄‖2 + (√𝑟2 − 𝑑 2 )2 = 𝑟2 to prove that the points 𝐴 and 𝐵 are on the circle. □ The Pythagorean Theorem can also answer a triangle existence question: given three side lengths, does there exist a triangle with those side lengths? Consider that we have proved three congruence theorems for triangles — SAS, ASA, SSS. In each case, it was assumed that two triangles already exist and that they should be compared. But one can also ask whether given the same data about side lengths and angles, does such a triangle exist? The answer for SAS is immediate. Given positive real numbers 𝑎 and 𝑏 and a real number 0 < 𝜙 < 180, one can construct ∠𝐵𝐶𝐴 with 𝑚∠𝐵𝐶𝐴 = 𝜙, ‖𝐵𝐶‖ = 𝑎, and ‖𝐶𝐴‖ = 𝑏 to create △𝐴𝐵𝐶 with the given side lengths and angle measure. For ASA, two angle measures 𝜙 and 𝜃 and a side length 𝑐, let 𝐴𝐵 have length 𝑐. Then construct angles with these measures at the endpoints, using rays on the same side of 𝐴𝐵. If 𝜙 + 𝜃 < 180, the rays intersect at a point 𝐶 by Theorem 7.6, Euclid’s version of the Parallel Postulate. Given three side lengths, 𝑎, 𝑏, 𝑐, for there to be a triangle △𝐴𝐵𝐶 with these side lengths, Theorem 6.5 says that it is necessary that the lengths satisfy the triangle inequality for each sum of length pairs: 𝑎 < 𝑏 + 𝑐, 𝑏 < 𝑐 + 𝑎, 𝑐 < 𝑎 + 𝑏. The existence question asks whether this is a sufficient condition for a triangle to exist with these side lengths. 1

The length would be different in the non-Euclidean plane.

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8. Dilations and Similarity

To construct such such a triangle, one could start with a segment 𝐴𝐵 of length 𝑐 and then draw circle 𝑐𝐴 with center 𝐴 and radius 𝑏 and circle 𝑐 𝐵 with center 𝐵 of radius 𝑎. If the circles intersect at a point 𝐶, as in Figure 4, then △𝐴𝐵𝐶 has the required side lengths. So will the circles intersect?

C

cB

cA B

A B'

D

A'

Figure 4. Existance of a Triangle with Three Given Side Lengths

Theorem 8.13 (Triangle from Three Side Lengths). Given three positive numbers 𝑎, 𝑏, 𝑐 with each number strictly less than the sum of the others, there exists a triangle with these side lengths. Proof. Assume that 𝑐 is greater than or equal to either of the others; then 𝑎 + 𝑏 > 𝑐 is sufficient to satisfy the triangle inequality, since the other two conditions are automatic when 𝑐 ≥ 𝑎 and 𝑐 ≥ 𝑏. We will try to construct △𝐴𝐵𝐶 with 𝑎 = ‖𝐵𝐶‖, 𝑏 = ‖𝐶𝐴‖, and 𝑐 = ‖𝐴𝐵‖. Let 𝑐𝐴 be the circle with radius 𝑏 and let 𝑐 𝐵 be the circle with radius 𝑎 . As a first step, we consider how these circles intersect the segment 𝐴𝐵. Since both 𝑎 ≤ 𝑐 and 𝑏 ≤ 𝑐, each circle will intersect 𝐴𝐵 in one point. Let 𝐴′ and 𝐵 be these points on 𝐴𝐵, with ‖𝐴𝐴′ ‖ = 𝑏 and ‖𝐵𝐵′ ‖ = 𝑎. Since 𝑐 < 𝑎 + 𝑏, then ‖𝐴𝐵 ′ ‖ = 𝑐 − 𝑎 < 𝑏, so 𝐵′ is closer to 𝐴 than 𝐴′ , and 𝐵′ is an interior point of 𝑐𝐴 . Similar reasoning shows that 𝐴′ is an interior point of 𝑐 𝐵 . As shown in Figure 4, the points on the segment will be in this order: 𝐴, 𝐵 ′ , 𝐴′ , 𝐵, possibly with 𝐴 = 𝐵 ′ or 𝐵 = 𝐴′ . The interior points of segment 𝐵 ′ 𝐴′ are inside both circles. ′

Suppose the two circles do intersect at two points 𝐶 and 𝐶 ′ . Let 𝐷 be the midpoint of 𝐶𝐶 ′ . The point 𝐷 is also the foot of the perpendicular from 𝐶 to 𝐴𝐵. According to Theorem 8.12, if we start with a 𝐷 on 𝐴𝐵 satisfying 𝑏2 − ‖𝐷𝐴‖2 = 𝑎2 − ‖𝐷𝐵‖2 > 0, then points of intersection of each circle with the perpendicular line to 𝐴𝐵 through 𝐷 are the same and so these points are also the points of intersection of the two circles. For points 𝑋 on 𝐴𝐵, let Δ be the difference of these two expressions that should be equal: Δ(𝑋) = (‖𝑋𝐴‖2 − 𝑏2 ) − (‖𝑋𝐵‖2 − 𝑎2 ). The 𝐷 we seek will be a zero of this function.

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121

In order to find such a 𝐷, choose a ruler on 𝐴𝐵 with the numbers 0 and 𝑐 corresponding to 𝐴 and 𝐵 and with coordinate 𝑥 for a point 𝑋 on the line 𝐴𝐵. Using these coordinates, Δ(𝑋) = 𝑥2 − 𝑏2 − ((𝑐 − 𝑥)2 − 𝑎2 ) = 2𝑐𝑥 − 𝑐2 − 𝑏2 + 𝑎2 . This is a nonconstant linear function of the real variable 𝑥, so it has only one zero: 𝑥0 = (𝑐2 + 𝑏2 − 𝑎2 )/2𝑐. Let 𝐷 be the point corresponding to this solution 𝑥0 . Now it remains to check that 𝐷 is inside both circles. Since Δ is a linear function, we will show that 𝐷 is in 𝐵 ′ 𝐴′ by observing that the values of Δ have different signs at 𝐵 ′ and 𝐴′ , so the point 𝐷 with value 0 is between them. Since 𝐵 ′ is on 𝑐 𝐵 , then 0 = 𝑎2 − ‖𝐵 ′ 𝐵‖. In addition, ‖𝐵′ 𝐴‖ < 𝑏, so 𝑏2 − ‖𝐵 ′ 𝐴‖2 > 0. Therefore, Δ(𝐵′ ) < 0. Similar reasoning shows Δ(𝐴′ ) > 0. Therefore, 𝐷 is in 𝐵 ′ 𝐴′ and so is interior to both circles, with ‖𝐴𝐷‖ < 𝑏 and ‖𝐵𝐷‖ < 𝑎. Thus, the quantities in the displayed equation are equal and positive. The intersection points of either circle with the line through 𝐷 perpendicular to 𝐴𝐵 are the points 𝐶 on the line with ‖𝐶𝐷‖ = √𝑏2 − ‖𝐷𝐴‖2 = √𝑎2 − ‖𝐷𝐵‖2 . Therefore, each of these points is on both circles. □ Corollary. A circle 𝑐 1 with center 𝑂1 and radius 𝑟1 and a circle 𝑐 2 with center 𝑂2 and radius 𝑟2 intersect at a point 𝐶 if the three distances 𝑟1 , 𝑟2 , and ‖𝑂1 𝑂2 ‖ satisfy the triangle inequality. Proof. Construct △𝑂1 𝑂2 𝐶 with side lengths ‖𝑂1 𝐶‖ = 𝑟1 and ‖𝑂2 𝐶‖ = 𝑟2 .

□

This proof is closely related to the proof that will be given for Theorem 8.47; this theorem has a condition for intersection of circles without an explicit reference to lengths.

The Regular Pentagon and Its Ratios Another application of the triangle similarity theorems is the study of the regular and star pentagons. In Figure 5, the star pentagon, or pentagram, is inscribed in the regular pentagon, and some interesting polygons inside are shaded. To get started, let us look at the angles in these figures.

108° - 108° - 108° - 108° - 108°

36° - 108° - 36°

72° - 36° - 72°

72° - 108° - 72° - 108°

Figure 5. Pentagon and Star Pentagon with Internal Shapes and Angles

The regular pentagon is made up of isosceles triangles with the central vertex angles of 360/5 = 72 degrees (shown in Figure 5.10). The base angles of these central triangles have measure (180 − 72)/2 = 54 degrees. The vertex angles of the pentagon are twice this, so they have measure 108.

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8. Dilations and Similarity

The first shaded triangle in Figure 5 is an isosceles triangle with apex angle 108 degrees, so its base angles are 36 degrees. For the purpose of this discussion, we will call this the “wide isosceles triangle”. The next shaded triangle is the “narrow isosceles triangle”. A key thing to notice is that the two unshaded regions on the right and left of the narrow triangle are wide isosceles triangles. Thus, the angle at a vertex of the pentagon is divided into three angles, two of which are known to have measure 36 degrees. Therefore, the third angle, the apex angle of the narrow isosceles triangle, also has measure 108 − 2 ∗ 36 = 36 degrees, and so these three angles are congruent. Since the apex angle of the narrow isosceles triangle is 36 degrees, the base angles are 72 degrees. The last shaded shape is a parallelogram that is a rhombus. One pair of opposite angles is 108 degrees, so the other pair is 2 ∗ 36 = 72 degrees. Now that we know the angles in Figure 5, what are the lengths? There are two key lengths to start with. Let 𝑝 be the length of the side of the pentagon and let 𝑠 be the longer length of a side of the star pentagon with the same vertices. C D p

pp/s p

p

A s B Figure 6. Golden Isosceles Triangle Extracted from Pentagram

Figure 6 shows the narrow isosceles triangle △𝐴𝐵𝐶. From what we have shown about angles in Figure 5, we know that 𝑚∠𝐶𝐵𝐷 = 𝑚∠𝐷𝐵𝐴 = 𝑚∠𝐶𝐴𝐵 = 36. Therefore, △𝐵𝐶𝐷 is similar to △𝐴𝐵𝐶 by AA, with scale factor 𝑝/𝑠. But △𝐷𝐴𝐵 is also isosceles, since it has two base angles of 36 degrees. In fact, this triangle is similar to the wide isosceles triangle. This figure has three segments with length equal to the side of the pentagon: ‖𝐴𝐷‖ = ‖𝐷𝐵‖ = ‖𝐵𝐶‖ = 𝑝. Both segments 𝐴𝐵 and 𝐴𝐶 have length 𝑠, and from of the scale factor, ‖𝐶𝐷‖ = 𝑝2 /𝑠. From this, it is possible to determine the ratio 𝑠/𝑝. Since 𝑠 = 𝑝 + 𝑝2 /𝑠, then 𝑠/𝑝 = 1 + (𝑠/𝑝)−1 . Let 𝜙 = 𝑠/𝑝; then 𝜙2 = 𝜙 + 1, or 𝜙2 − 𝜙 − 1 = 0. Using the quadratic formula and choosing the root that is greater than 1, we conclude 𝜙=

1 + √5 . 2

This is the famous Golden Ratio. If one looks at the inscribed pentagram in the pentagon, it is full of ratios that are golden.

Ratios, Signed Ratios, and Scale Factors

123

Ratios, Signed Ratios, and Scale Factors Ratios are important, and a lot of ratios show up when reasoning with similarity. This section points out some important distinctions when working with them. Every similitude 𝑆 has a scale factor 𝑟. The ratios between distances in a figure such as a quadrilateral 𝐴𝐵𝐶𝐷 and the corresponding distances in the image 𝑆(𝐴𝐵𝐶𝐷) = 𝐴′ 𝐵 ′ 𝐶 ′ 𝐷′ all equal 𝑟; e.g., ‖𝐴′ 𝐵 ′ ‖/‖𝐴𝐵‖ = 𝑟, ‖𝐶 ′ 𝐷′ ‖/‖𝐶𝐷‖ = 𝑟, etc. D'

D

A' A

C' B'

C

B

Figure 7. Similar Figures with Scale Factor and Internal Ratios

An example is shown in Figure 7, where the scale factor is 3/5, so ‖𝐴′ 𝐵 ′ ‖ ‖𝐵′ 𝐶 ′ ‖ ‖𝐶 ′ 𝐷′ ‖ ‖𝐷′ 𝐴′ ‖ = = = = .6. ‖𝐴𝐵‖ ‖𝐵𝐶‖ ‖𝐶𝐷‖ ‖𝐷𝐴‖ But there are also internal ratios in these geometrical figures. A ratio of distances within 𝐴𝐵𝐶𝐷 is equal to the ratio of corresponding distances within 𝐴′ 𝐵 ′ 𝐶 ′ 𝐷′ , but there is no reason for such internal ratios within a single figure such as 𝐴𝐵𝐶𝐷 to be equal to each other. Here are two examples from Figure 7: ‖𝐶𝐷‖ ‖𝐶 ′ 𝐷′ ‖ ‖𝐴𝐵‖ ‖𝐴′ 𝐵 ′ ‖ = = 2; = = .75. ‖𝐷𝐴‖ ‖𝐷′ 𝐴′ ‖ ‖𝐵𝐶‖ ‖𝐵′ 𝐶 ′ ‖ The reason that the corresponding internal ratios are equal follows from the scaling, as in this example: 𝑟‖𝐶𝐷‖ ‖𝐶𝐷‖ ‖𝐶 ′ 𝐷′ ‖ = = . ′ ′ ‖𝐷 𝐴 ‖ 𝑟‖𝐷𝐴‖ ‖𝐷𝐴‖ Theorem 8.14. Similitudes preserve internal ratios. More specifically, if 𝑆 is a similitude for any 𝐴, 𝐵, 𝐶, 𝐷 with 𝑆-images 𝐴′ , 𝐵 ′ 𝐶 ′ , 𝐷′ , these ratios are equal: ‖𝐴𝐵‖/‖𝐶𝐷‖ = ‖𝐴′ 𝐵′ ‖/‖𝐶 ′ 𝐷′ ‖. Proof. The proof is essentially contained in the equation displayed before the theorem. □ The choice of two kinds of equal ratios can sometimes lead to confusion. The important point is that the scale factor is a ratio that is different from the others in that it controls the change in size for all distances. The other internal ratios represent relationships within the figures that are preserved, unrelated to the scale factor 𝑟. To emphasize this distinction, we will try to use the term scale “factor” rather than scale “ratio”.

124

8. Dilations and Similarity

Signed Ratios. A third ratio that plays an important role with similitudes is the signed ratio of points on a line. This ratio is only defined when the points are collinear. Definition 8.15. If 𝐴, 𝐵, 𝐶, 𝐷 are points on a line 𝑚 with 𝐶 ≠ 𝐷, the signed ratio 𝐴𝐵/𝐶𝐷 is a real number with |𝐴𝐵/𝐶𝐷| = ‖𝐴𝐵‖/‖𝐶𝐷‖. The ratio is positive if the rays ⃗ ⃗ and 𝐶 𝐴𝐵 𝐷 have the same direction, and it is negative if the rays have the opposite direction. Signed ratios are preserved by similitudes. The absolute value is preserved by scaling. Since rays are mapped to rays and since two rays on a line have the same direction if one is contained in the other, the direction relationship is also preserved by a similitude. D

C

E

A

B

Figure 8. Signed Ratios 𝐶𝐷/𝐴𝐵 = −2 and 𝐶𝐸/𝐴𝐵 = 2

Example 8.16. Consider the collinear points in Figure 8 with ‖𝐴𝐵‖ = 3, ‖𝐶𝐷‖ = ‖𝐶𝐸‖ = 6. Then ‖𝐶𝐷‖/‖𝐴𝐵‖ = ‖𝐶𝐸‖/‖𝐴𝐵‖ = 2. But 𝐶𝐷/𝐴𝐵 = −2 and 𝐶𝐸/𝐴𝐵 = 2, since 𝐴𝐵 and 𝐶𝐸 have the same direction while 𝐶𝐷 has the opposite direction from 𝐴𝐵. Suppose we wish to locate an unknown point 𝑃. If we know that ‖𝐶𝑃‖/‖𝐴𝐵‖ = 2, we know that 𝑃 is either 𝐷 or 𝐸, since the ratio tells us that ‖𝐶𝑃‖ = 6. But if we know 𝐶𝑃/𝐴𝐵 = 2, we know that 𝑃 = 𝐸, but if 𝐶𝑃/𝐴𝐵 = −2, then 𝑃 = 𝐷. The signed ratio locates the point 𝑃 uniquely. Signed Ratio Definition of Dilation. One important place where this relationship appears is in the definition of a dilation, which has separate cases for positive and negative dilation ratios. It is possible to unite this definition using signed ratios. If we consider the definition for the dilation 𝒟𝑃,𝑘 with center 𝑃 and real dilation ratio 𝑘, the dilation takes a point 𝐴 to the point 𝐴′ on 𝑃𝐴 with 𝑃𝐴′ /𝑃𝐴 = 𝑘. This is an alternate definition of 𝒟𝑃,𝑘 that just has one case. k = -2.5

k = -1

k =0

k =1

P

A

k =2

k =4

k =7

Figure 9. Dilations of Point 𝐴 with Center 𝑃 and Various Ratios 𝑘

Real Number Signed Ratio Formula. Since the points in a signed ratio are on a line, it is useful to see how the ratio 𝐴𝐵/𝐶𝐷 can be expressed using real numbers 𝑎, 𝑏, 𝑐, 𝑑 assigned by a ruler. The formula is 𝑏−𝑎 𝐴𝐵 = . 𝐶𝐷 𝑑−𝑐 This clearly has the correct absolute value. The ratio is positive if both 𝑐 < 𝑑 and 𝑎 < 𝑏 or if both 𝑐 > 𝑑 and 𝑎 > 𝑏. It is negative otherwise. This agrees with the direction relationships in the definition.

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125

In thinking about this ratio, an important special case is the ratio 𝑃𝑋/𝑃𝐴, a ratio with one point repeated, such as the ratio in the definition of dilation above, as illustrated in Figure 9. In this case, if the values 𝑝, 𝑎, 𝑥 are assigned to 𝑃, 𝐴, 𝑋, then 𝑃𝑋/𝑃𝐴 = (𝑥 − 𝑝)/(𝑎 − 𝑝), a linear function of 𝑥 that equals 0 at 𝑝 and equals 1 at 𝑎. Thus, the ratio is a coordinate function for 𝑋 with 𝑃 at the origin and unit point at 𝐴.

Transversals of Parallels and Ratios In this section we will prove Theorem 8.17 about parallel lines, transversals, and ratios. There will be one immediate application to angle bisectors of a triangle, but the theorem is more significant than that. It underlies the foundations of coordinates as well as being the basis of graphs of linear equations. A'

A B

B' C'

C

D

D'

Figure 10. Notebook Paper with Transversals

Before stating the theorem to provide some intuition as to why Theorem 8.17 is true, let us consider an experiment with ruled notebook paper. If one draws a transversal across the lines on a piece of notebook paper, the segments between consecutive lines of the paper all have the same length. One can compute signed ratios such as 𝐴𝐵/𝐶𝐷 by counting steps rather than measuring, provided that the points are intersection points of the transversal and the lines of the paper as in Figure 10. Then if one draws a second transversal, not parallel to the first, the lengths of the segments between consecutive lines will be different from those of the first transversal. Nonetheless, if one labels points 𝐴′ , 𝐵 ′ , 𝐶 ′ , 𝐷′ on this new transversal and on the same lines as before, then 𝐴′ 𝐵 ′ /𝐶 ′ 𝐷′ = 𝐴𝐵/𝐶𝐷 since the step counts are the same even if the lengths of the segments in the ratios are different. If this figure reminds you of graphs of linear equations, this is not a coincidence, as we will see in Chapter 11. By the way, one can actually use notebook paper in this way to mark the edge of another paper into eighths, or some other fraction. This theorem generalizes the notebook example, as illustrated in Figure 11. Theorem 8.17 (Transversal Ratios). Given parallel lines 𝑎, 𝑏, 𝑐, 𝑑 and two transversals 𝑚 and 𝑛, let the intersections of 𝑚 with 𝑎, 𝑏, 𝑐, 𝑑 be 𝐴, 𝐵, 𝐶, 𝐷 and the intersections with 𝑛 be 𝐴′ , 𝐵 ′ , 𝐶 ′ , 𝐷′ . Then these signed ratios are equal: 𝐴𝐵 𝐴′ 𝐵 ′ = ′ ′. 𝐶𝐷 𝐶𝐷

126

8. Dilations and Similarity

It is not required that the lines and points be distinct, except for 𝑐 and 𝑑. If 𝑚 and 𝑛 intersect at a point 𝑃, then 𝑃 can be one of the four points. a

d

b

c C'

n

B' A'

D P

A

B

D'

C

m

Figure 11. Parallels with Transversals

Proof. First, we will prove the theorem when 𝑚 and 𝑛 are not parallel. Let 𝑃 be the intersection point of these two transversals. Assume that 𝑃 is not one of the four designated points; that case will be covered at the end of the proof. Consider a ruler on the line 𝑚 with 0 as the coordinate of 𝑃 and 𝑥𝐴 , 𝑥𝐵 , 𝑥𝐶 , 𝑥𝐷 the coordinates of 𝐴, 𝐵, 𝐶, 𝐷. There are dilations with center 𝑃 that take 𝐴 to each of 𝐴, 𝐵, 𝐶, 𝐷: 𝐴 = 𝒟𝑃,𝑘𝐴𝐴 (𝐴), 𝐵 = 𝒟𝑃,𝑘𝐴𝐵 (𝐴), 𝐶 = 𝒟𝑃,𝑘𝐴𝐶 (𝐴), 𝐷 = 𝒟𝑃,𝑘𝐴𝐷 (𝐴). The ratios of dilation are 𝑘𝐴𝐴 = 1, 𝑘𝐴𝐵 = 𝑥𝐵 /𝑥𝐴 , 𝑘𝐴𝐶 = 𝑥𝐶 /𝑥𝐴 , 𝑘𝐴𝐷 = 𝑥𝐷 /𝑥𝐴 . Therefore, the signed ratio 𝐴𝐵/𝐶𝐷 can be written in terms of these ratios of dilation: 𝑥 − 𝑥𝐴 (𝑘 − 𝑘𝐴𝐴 )𝑥𝐴 (𝑘 − 𝑘𝐴𝐴 ) 𝐴𝐵 = 𝐵 = 𝐴𝐵 = 𝐴𝐵 . 𝐶𝐷 𝑥𝐷 − 𝑥𝐶 (𝑘𝐴𝐷 − 𝑘𝐴𝐶 )𝑥𝐴 (𝑘𝐴𝐷 − 𝑘𝐴𝐶 ) The same analysis can be performed on the line 𝑛. However, the dilation 𝒟𝑃,𝑘𝐴𝐵 that takes 𝐴 to 𝐵 also takes the line 𝑎 to the parallel line 𝑏. So it takes the intersection of 𝑎 with 𝑛 to the intersection of 𝑏 with 𝑛; namely it maps 𝐴′ to 𝐵 ′ . All the other dilations that take 𝐴 to each of 𝐴, 𝐵, 𝐶, 𝐷 also take 𝐴′ to 𝐴′ , 𝐵 ′ , 𝐶 ′ , 𝐷′ , so the dilation ratios are exactly the same. This means that (𝑘 − 𝑘𝐴𝐴 ) 𝐴𝐵 𝐴′ 𝐵 ′ = 𝐴𝐵 = ′ ′. 𝐶𝐷 𝐶𝐷 (𝑘𝐴𝐷 − 𝑘𝐴𝐶 ) This proves the case when 𝑚 and 𝑛 intersect. In fact, this proves the similarity of the two figures 𝐴𝐵𝐶𝐷 and 𝐴′ 𝐵 ′ 𝐶 ′ 𝐷′ , each figure consisting of these collinear points. When 𝑚 and 𝑛 are parallel, the figure is full of parallelograms. The translation 𝒯𝐴𝐴′ maps 𝑚 to the parallel line through 𝐴′ , which is 𝑛. It also maps each of the parallel lines

Transversals of Parallels and Ratios

127

𝑎, 𝑏, 𝑐, 𝑑 into itself, so 𝒯𝐴𝐴′ defines a congruence from 𝐴𝐵𝐶𝐷 to 𝐴′ 𝐵 ′ 𝐶 ′ 𝐷′ . Therefore, all signed ratios must be the same. For the special case when the intersection point 𝑃 is on one of the four parallel lines, add a new transversal 𝑞 that intersects the four parallel lines at 𝐴″ , 𝐵 ″ , 𝐶 ″ , 𝐷″ , none of which is on 𝑚 or 𝑛. Then by the proof above, 𝐴𝐵/𝐶𝐷 = 𝐴″ 𝐵 ″ /𝐶 ″ 𝐷″ = 𝐴′ 𝐵 ′ /𝐶 ′ 𝐷′ . The proof did not make any assumptions about the relative positions or distinctness of the points beyond what is needed for the ratios to be defined. For example 𝐴𝐶/𝐴𝐵 = 𝐴′ 𝐶 ′ /𝐴′ 𝐵 ′ . □ One possible transversal through 𝑃 in the situation above is a transversal perpendicular to all the parallel lines. In this case the ratios are ratios of distances between parallel lines. As an application of this theorem, here is a proposition relating angle bisectors to ratios. Theorem 8.18 (Angle Bisector Ratios). In a triangle △𝐴𝐵𝐶 (except for the isosceles case when 𝐴𝐵 ≅ 𝐴𝐶) the bisectors of the interior and exterior angles at a vertex 𝐴 divide the opposite side 𝐵𝐶 internally and externally in the ratio of 𝑘 = ‖𝐴𝐵‖/‖𝐴𝐶‖. To be more specific, if the internal angle bisector intersects 𝐵𝐶 at 𝑃 and the external angle bisector intersects the extended side 𝐵𝐶 at 𝑄, then 𝑃𝐵/𝑃𝐶 = −𝑘 and 𝑄𝐵/𝑄𝐶 = 𝑘. When 𝐴𝐵 ≅ 𝐴𝐶, 𝑃 is the midpoint of 𝐵𝐶 and the external angle bisector is parallel to 𝐵𝐶.

C' A C'' Q B

P

C

Figure 12. Angle Bisector Ratios

Proof. The proof is illustrated in Figure 12. The internal and external angle bisectors at 𝐴 are perpendicular, since together they bisect supplementary angles whose sum is a straight angle. The intersection 𝑃 of the internal angle bisector with 𝐵𝐶 always exists. Assuming ⃗ is not perpendicular to 𝐵𝐶, the external angle bisector is not the angle bisector 𝐴𝑃 parallel to 𝐵𝐶 and intersects that line at a point 𝑄.

128

8. Dilations and Similarity

Let 𝐶 ′ = ℛ𝐴𝑄 (𝐶). This point is on 𝐵𝐴 and 𝐴𝐶 ≅ 𝐴𝐶 ′ . The lines 𝐶𝐶 ′ and 𝐴𝑃 are parallel since both are perpendicular to 𝐴𝑄. Therefore, the signed ratios 𝐴𝐵/𝐴𝐶 ′ = 𝑃𝐵/𝑃𝐶 by Theorem 8.17. (Or one can view this as a dilation of a line with center 𝐵.) Since 𝐴 is between 𝐵 and 𝐶 ′ , the signed ratio 𝐴𝐵/𝐴𝐶 ′ = −‖𝐴𝐵‖/‖𝐴𝐶‖ = −𝑘. For the other ratio, construct point 𝐶 ″ on 𝐵𝐴 by reflecting 𝐶 in 𝐴𝑃. Then 𝐶𝐶 ″ and 𝐴𝑄 are parallel, since both are perpendicular to 𝐴𝑃. Applying the same reasoning, ‖𝐴𝐵‖/‖𝐴𝐶 ″ ‖ = 𝐴𝐵/𝐴𝐶 ″ = 𝑄𝐵/𝑄𝐶 = 𝑘. This time the ratio is positive because 𝐴 is not on 𝐵𝐶 ″ . In the exceptional case of an angle bisector of angle 𝐴 perpendicular to the side 𝐵𝐶, the line 𝐴𝑃 is a line of symmetry of △𝐴𝐵𝐶, which is isosceles with 𝐴𝐵 ≅ 𝐴𝐶. In this case 𝑃 is the midpoint of 𝐵𝐶 and 𝑃𝐵/𝑃𝐶 = −1. In this case the bisector of the exterior angle is parallel to line 𝐵𝐶 and the intersection 𝑄 does not exist. However, if the sides are very close to being of equal length, the ratio 𝑄𝐵/𝑄𝐶 is very close to 1. □

Parallel Segments and Centers of Dilation This is a key theorem about the unique orientation-preserving similitude that maps a segment to a parallel segment. Theorem 8.19 (Parallel Segment Centers). If 𝐴𝐵 and 𝐶𝐷 are two parallel or collinear segments with the same direction, the following statements are true. (1) If ‖𝐴𝐵‖ ≠ ‖𝐶𝐷‖, there is a dilation 𝒟𝑃,𝑘 , with 𝑘 > 0, which maps 𝐴 to 𝐶 and 𝐵 to 𝐷. There is another dilation 𝒟𝑄,−𝑘 which maps 𝐴 to 𝐷 and 𝐵 to 𝐶. (2) If ‖𝐴𝐵‖ = ‖𝐶𝐷‖, the translation 𝒯𝐴𝐶 maps 𝐴 to 𝐶 and 𝐵 to 𝐷, and ℋ𝑄 = 𝒟𝑄,−1 maps 𝐴 to 𝐷 and 𝐵 to 𝐶, where 𝑄 is the intersection of 𝐴𝐷 and 𝐵𝐶. (3) The unique orientation-preserving similitudes that map 𝐴𝐵 to 𝐶𝐷 and to 𝐷𝐶 are the dilations or translation defined above. D B N C

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Figure 13. The Two Dilation Centers for Parallel Segments

Proof. Statement (2) was proved in Theorem 7.30. To prove the first statement, begin by assuming that 𝐴𝐵 and 𝐶𝐷 are parallel but not collinear. The lines 𝐴𝐶 and 𝐵𝐷 are distinct transversals; let 𝑃 be the intersection point

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of these lines. The intersection exists, given the assumption that ‖𝐴𝐵‖ ≠ ‖𝐶𝐷‖. For if these lines were parallel, then 𝐴𝐵𝐷𝐶 would be a parallelogram with ‖𝐴𝐵‖ = ‖𝐶𝐷‖. Let 𝒟𝑃,𝑘 be the dilation that takes 𝐴 to 𝐶. Then the line 𝑎 = 𝐴𝐵 is mapped to a line 𝑎′ through 𝐶 parallel to 𝑎. Then 𝑎′ = 𝐶𝐷, by the uniqueness of parallel lines from the Euclidean Parallel Postulate, Theorem 7.1. By the definition of dilation, 𝒟𝑃,𝑘 (𝐵) is the intersection of 𝑎′ and 𝑃𝐵 = 𝐵𝐷, so 𝒟𝑃,𝑘 (𝐵) = 𝐷. The center 𝑄 of a second dilation is the intersection of 𝐴𝐷 and 𝐵𝐶. The reasoning is the same but the ratio is now −𝑘 because the direction is reversed. Thus, the figure is arranged as shown in Figure 13. For the case when 𝐴𝐵 and 𝐶𝐷 are collinear, we will give an algebraic proof using a ruler on the line. A more geometrically interesting proof follows from Theorem 8.21, in which the centers of dilation are located by considering the segments to be diameters of circles. If the points on the line are represented by real numbers 𝑎, 𝑏, 𝑐, 𝑑, then 𝑘 = (𝑑 − 𝑐)/(𝑏 − 𝑎). For a dilation mapping 𝐴 to 𝐶, the center of dilation 𝒟𝑃,𝑘 is represented by real number 𝑝 satisfying (𝑐 − 𝑝)/(𝑎 − 𝑝) = 𝑘, an equation with solution 𝑝 = (𝑐 − 𝑘𝑎)/(1 − 𝑘). Since the dilation maps 𝐵 to 𝐷, it must also be true that 𝑝 = (𝑑 − 𝑘𝑏)/(1 − 𝑘). These two equations have the same solution 𝑝 for the given value of 𝑘. A similar computation using −𝑘 produces a real number 𝑞 = (𝑑 + 𝑘𝑎)/(1 + 𝑘) for the other center 𝑄. This completes the proof of the first assertion. For statement (3), Theorem 8.10 proves that the similitudes listed are unique, and the other points of this proof demonstrate precisely what the maps are. □ Corollary. Let 𝐴, 𝐵, 𝐶 be points on line 𝑚 and let 𝐷, 𝐸, 𝐹 be points on a parallel or equal line 𝑛 such that 𝐴𝐶/𝐴𝐵 = 𝐷𝐹/𝐷𝐸. If 𝑆 is the unique dilation or translation that maps 𝐴𝐵 to 𝐷𝐸, then 𝑆(𝐶) = 𝐹. In particular, if the signed ratio is 1/2, then 𝐶 is the midpoint of 𝐴𝐵 and is mapped to 𝐹, the midpoint of 𝐷. Proof. Signed ratios are preserved by similitudes, and these signed ratios determine 𝐶 and 𝐹 uniquely. □ Centers of Dilation. A point 𝐻 is called a center of dilation of two figures if there is a dilation 𝒟𝐻,𝑘 that takes one figure to the other. The theorem says that unequal parallel segments have two centers of dilation if orientation is not taken into account. A single center of dilation relates two triangles with corresponding sides parallel. Theorem 8.20 (Triangles with Corresponding Sides Parallel). Suppose △𝐴𝐵𝐶 and △𝐷𝐸𝐹 have corresponding sides parallel or collinear. Then the triangles are similar. If they are congruent, the similitude that takes △𝐴𝐵𝐶 to △𝐷𝐸𝐹 is either a translation or a half-turn. Otherwise the similitude is a dilation 𝒟𝑃,𝑘 . Proof. Suppose one pair of corresponding sides is not congruent; e.g., ‖𝐴𝐵‖ ≠ ‖𝐷𝐸‖. The segments 𝐴𝐵 and 𝐷𝐸 are parallel or collinear, so by Theorem 8.19, there is a dilation 𝒟𝑃,𝑘 that takes 𝐴 to 𝐷 and 𝐵 to 𝐸. Then 𝒟𝑃,𝑘 maps 𝐴𝐶 to the parallel line 𝐷𝐹

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Figure 14. Triangles with Corresponding Sides Parallel

and also maps 𝐵𝐶 to 𝐸𝐹. Therefore, it maps the intersection point 𝐶 of these lines in △𝐴𝐵𝐶 to the intersection point 𝐹 of these lines in △𝐷𝐸𝐹. This completes the proof for this case. If the segments 𝐴𝐵 and 𝐷𝐸 have the same length, then by Theorem 7.30 there is either a translation or a half-turn that maps 𝐴 to 𝐷 and 𝐵 to 𝐸. Then the same reasoning as before shows that this rigid motion maps △𝐴𝐵𝐶 to △𝐷𝐸𝐹. □ Theorem 8.19 about centers of dilation for segments can also be applied to finding centers of dilation for circles. Theorem 8.21 (Circle Centers of Dilation). Let 𝐴𝐵 and 𝐶𝐷 be parallel or collinear segments with midpoints 𝑀 and 𝑁, respectively. If 𝑐 𝑀 and 𝑐 𝑁 are circles with these segments as diameters, then any center of dilation for these segments is also a center of dilation for the two circles. It is also a center of dilation for any other pair of parallel or collinear diameters of these circles. Proof. If 𝒟𝑃,𝑘 is a dilation that maps 𝐴𝐵 and 𝐶𝐷, then it must map midpoint to midpoint by distance scaling so the center of 𝑐 𝑀 is mapped to the center of 𝑐 𝑁 . Since 𝐴 is mapped to 𝐶, the points at distance ‖𝐴𝑀‖ from 𝑀 are mapped to the points at distance ‖𝐶𝑁‖ from 𝑁. Therefore, 𝑃 is a center of dilation of the circles. If the dilation maps 𝐴𝐵 to 𝐷𝐶, the reasoning is the same. If 𝐴′ 𝐵 ′ is any other diameter of 𝑐 𝑀 , then the segment is mapped to a parallel segment or collinear segment with endpoints on 𝑐 𝑁 and midpoint at 𝑁. □

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Figure 15. Constructing Centers of Dilation of Two Circles

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This gives a practical method of constructing the centers of dilation of two circles with distinct centers, as shown in Figure 15. Let 𝐴𝐵 and 𝐶𝐷 be parallel diameters of the circles. An easy choice would be the diameters perpendicular to the line of centers 𝑀𝑁. Then construct the centers 𝑃 and 𝑄 by intersecting 𝐴𝐶 and 𝐵𝐷 and also intersecting 𝐴𝐷 and 𝐵𝐶 as before. This also gives a graphical method of constructing centers of dilation of two collinear segments. Construct two circles with the segments as diameters and then construct the centers 𝑃 and 𝑄 from parallel but not collinear diameters by intersecting lines.

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Figure 16. Centers of Similitude on or Inside the Circles

Figure 15 shows circles that do not intersect. In such a case, the centers of dilation will both be outside the circles. But this method applies to all circles. Some examples are shown in Figure 16, where a center 𝑃 or 𝑄 may be inside both circles or on both circles. It is not possible for a center such as 𝑃 to be inside or on one circle and not the other, for the figure consisting of the first circle along with 𝑃 is similar to the figure of the second circle along with 𝑃. There are some interesting special cases, when the circles are congruent or concentric. This will be taken up in the exercises. Dilations, Inscribed Circles, and Common Tangent Construction. If a circle is inscribed in an angle, a dilation centered at the vertex of the angle maps the circle to another circle inscribed in an angle formed by the same lines. Figure 17 shows both a dilation with center 𝑃 with a positive ratio and a dilation with center 𝑄 with a negative ratio. Looking at this a different way, if a dilation center for two circles is 𝑃, then any tangents to one circle through 𝑃 will be dilated to themselves, so the dilated image circle will also have the same tangents. Thus, to construct common tangents to two circles, first find any centers of dilation that are outside or on the circles. If a dilation center is outside a circle, construct the two tangents to the circle; these will be common tangents to both circles. If the dilation center is on a circle, it will be on both circles and the tangent to the circle at that point is a common tangent. Among other things, this implies that a center of dilation is either outside both circles, on both circles, or inside both circles.

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Figure 17. Inscribed Circles as Dilation Images

Example 8.22 (Constructing Tangents Common to Two Circles). In Figure 18 in each circle the diameter perpendicular to the line of centers was constructed. The two dilation centers of the circles were then constructed by intersecting the four lines through the endpoints of the diameters with the line of centers. One of these dilation centers is inside the circles, so no tangent lines will pass through it. The other dilation center is outside the circles. Two tangent lines to the larger circle through this dilation center were constructed by the method of Theorem 6.11. These tangent lines are also tangent to the smaller circle. The tangent lines are shown as blue lines of medium width. The lines through the endpoints of the diameters are thin red lines. Caution! For these diameters, the lines through the endpoints are close enough to the tangent lines that a hasty person might think these were the tangent lines.

Figure 18. Common Tangents and Dilation Centers

Construction by Scaling Models Here is a construction problem: given an angle ∠𝐴𝐵𝐶 and a point 𝐸 in the interior of the angle, construct a circle passing through 𝐸 that is inscribed in the angle. ⃗ it would not be difficult to construct a circle through 𝐸 If the point 𝐸 were on 𝐵𝐴, that is inscribed in the angle, for the center of the circle is the intersection of the angle ⃗ at 𝐸. bisector with the perpendicular to 𝐵𝐴

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Figure 19. Construction: Two Inscribed Circles Through Point 𝐸

But what if the point 𝐸 is in the interior of the angle? The solution seems not so straightforward. But we will see that it is easy to construct a figure that is similar to the desired inscribed circle. Then this scale model of the solution can be dilated to construct a true solution. In Figure 19, ∠𝐴𝐵𝐶 and point 𝐸 are given. Construct any circle inscribed in ∠𝐴𝐵𝐶. Then intersect this circle with 𝐵𝐸. There will be two points of intersection, 𝐹 and 𝐺. The circle, along with either of these points, is a dilation image of one of the two solutions to the desired circle construction. The dilation with center 𝐵 that maps 𝐹 to 𝐸 will be one solution of the construction problem and the dilation that takes 𝐺 to 𝐸 will be the other. A W

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Figure 20. Inscribed Square Problem

A second construction problem is a classic problem offered as an example of problem-solving by Polya [15]. It is the problem of constructing a square inscribed in a triangle, with one square side contained in a side of the triangle, as shown on the left side of Figure 20. The problem is to construct in △𝐴𝐵𝐶 a square 𝑊𝑋𝑌 𝑍 with 𝑋𝑌 in 𝐵𝐶 and with 𝑊 on 𝐴𝐵 and 𝑍 on 𝐴𝐶. The method of solution is on the right. Take any convenient point 𝐸 on 𝐴𝐵 and construct the small square 𝐸𝐹𝐺𝐻 where three of the four vertices satisfy the required conditions, but 𝐻 does not. If one constructs a line through 𝐻 parallel to 𝐴𝐶, then this figure is a dilated model of the solution. The point 𝑍 of the solution is the intersection ⃗ of 𝐵 𝐻 with 𝐴𝐶. The distance from 𝑍 to 𝐵𝐶 determines the length of the side of the square, which then can be constructed. In particular, 𝒟𝐵,‖𝐵𝑍‖/‖𝐵𝐻‖ maps 𝐸𝐹𝐺𝐻 to 𝑊𝑋𝑌 𝑍. Note: The parallel line through 𝐻 is not necessary for the construction; it was just an aid in visualizing a scale model of the solution.

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Figure 21. Inscribed Triangle Problem

A more elaborate construction problem with a similar solution idea is the problem of inscribing one triangle in another. For example, in Figure 21 there are shown two triangles, △𝐴𝐵𝐶 on the left and △𝐷𝐸𝐹 on the right (with a shaded blue interior). The problem is to construct triangle △𝐷′ 𝐸 ′ 𝐹 ′ (with shaded green interior) similar to △𝐷𝐸𝐹, and with vertices 𝐷′ , 𝐸 ′ , 𝐹 ′ on the sides opposite 𝐴, 𝐵, 𝐶 in △𝐴𝐵𝐶. A solution is to notice that the related problem of circumscribing a triangle similar to △𝐴𝐵𝐶 around △𝐷𝐸𝐹 is not so difficult. Just construct three parallel lines through the vertices of △𝐷𝐸𝐹, each parallel to a side of △𝐴𝐵𝐶. The resulting triangle △𝐴′ 𝐵 ′ 𝐶 ′ formed by the parallels is similar to △𝐴𝐵𝐶 by Theorem 8.20. The unique dilation that maps this triangle to △𝐴𝐵𝐶 also maps △𝐷𝐸𝐹 to the desired triangle inscribed in △𝐴𝐵𝐶. Remark 8.23. These constructions of inscribed squares and triangles will in some situations result in figures with vertices on the extended sides of △𝐴𝐵𝐶. In special cases, the parallel lines through 𝐷, 𝐸, 𝐹 will be concurrent, so they will not form a △𝐴′ 𝐵′ 𝐶 ′ . If one rotates △𝐷𝐸𝐹, the construction in Figure 21 will produce a different triangle inscribed in △𝐴𝐵𝐶 and so will be an additional solution to the inscription problem. The Miquel Theorem states further interesting properties of a figure with one triangle inscribed in another (see Exercise 11).

Harmonic Division We have seen that ratios relate the midpoints 𝑀 and 𝑁 of two unequal parallel segments to the centers of dilation 𝑃 and 𝑄 of the segments. Specifically, 𝑃𝑁/𝑃𝑀 = 𝑘 = −𝑄𝑁/𝑄𝑀. One can say that 𝑃 and 𝑄 divide 𝑀𝑁 internally and externally by a constant 𝑘 ≠ 1. This relationship among four points occurs so often in geometry that it is given a name, harmonic division. Definition 8.24 (Harmonic Division). One says that 𝑃 and 𝑄 divide 𝑀𝑁 harmonically, or that 𝑃𝑄 divides 𝑀𝑁 harmonically, if the four points are collinear and 𝑃𝑁 𝑄𝑀 𝑃𝑁 𝑄𝑀 = −1 = . 𝑃𝑀 𝑄𝑁 𝑄𝑁 𝑃𝑀 Remark 8.25. The fact that both products of ratios equal −1 implies that when 𝑃𝑄 divides 𝑀𝑁 harmonically, then 𝑀𝑁 also divides 𝑃𝑄 harmonically, although the above

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ratio 𝑘 for 𝑃𝑄 is not the corresponding ratio for 𝑀𝑁. Also, one should note that the order within a pair (i.e., 𝑃𝑄 or 𝑄𝑃) does not affect the harmonic relationship. This symmetrical relationship between pairs of points on a line appears in many guises in geometry. It has previously shown up with angle bisector ratios in Theorem 8.18. It also appears in the geometry of perspective drawing and in the geometry of circles. Here is an example with circles. Theorem 8.26 (Circles of Apollonius). Given distinct points 𝐴 and 𝐵 and a positive constant 𝑘 ≠ 1, the set of points 𝑃 satisfying ‖𝑃𝐴‖/‖𝑃𝐵‖ = 𝑘 is a circle 𝑐 𝑘 . This circle intersects 𝐴𝐵 in points 𝐶1 and 𝐶2 that divide 𝐴𝐵 harmonically. When 𝑘 = 1, the set is the perpendicular bisector of 𝐴𝐵.

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Figure 22. Points 𝑃 and 𝑄 on a Circle of Apollonius

Proof. For some 𝑘 > 0, let ‖𝑃𝐴‖/‖𝑃𝐵‖ = 𝑘. For △𝑃𝐴𝐵, let 𝐶1 be the intersection of 𝐴𝐵 with the internal angle bisector at 𝑃 and let 𝐶2 be the intersection of the line with the external angle bisector. By Theorem 8.18, 𝐶1 and 𝐶2 divide 𝐴𝐵 harmonically, with 𝐶1 𝐴/𝐶1 𝐵 = −𝑘 and 𝐶2 𝐴/𝐶2 𝐵 = 𝑘. In addition, ∠𝐶1 𝑃𝐶2 is a right angle, so by the Right Angle Locus Theorem, Theorem 7.20, the point 𝑃 is on the circle 𝑐 𝑘 with diameter 𝐶1 𝐶2 , as illustrated in Figure 22. Suppose 𝑄 is another point with ‖𝑄𝐴‖/‖𝑄𝐵‖ = 𝑘. Repeating the same construction for △𝑄𝐴𝐵, since the 𝑘 is the same, the points 𝐶1 and 𝐶2 are the same, since their locations are determined by the signed ratios. Thus 𝑄 is on the same circle. In order to see that every point 𝑄 on 𝑐 also satisfies ‖𝑄𝐴‖/‖𝑄𝐵‖ = 𝑘, consider that this ratio must equal some positive number 𝑘′ . Then, repeating the reasoning above, 𝑄 is on a circle 𝑐′ with a diameter that divides 𝐴𝐵 harmonically, with the endpoints of the diameter 𝐶1′ 𝐶2′ satisfying the equations 𝐶1′ 𝐴/𝐶1′ 𝐵 = −𝑘′ and 𝐶2′ 𝐴/𝐶2′ 𝐵 = 𝑘′ . If one looks carefully at the ratios, one sees that if 𝑘′ > 𝑘, then the segment 𝐶1′ 𝐶2′ is contained in the interior of 𝐶1 𝐶2 , so the circles with these two diameters do not intersect. Therefore a point 𝑄 on 𝑐 much have 𝑘′ = 𝑘. So the set of points 𝑃 with ‖𝑃𝐴‖/‖𝑃𝐵‖ = 𝑘 is a circle 𝑐 𝑘 as claimed. □ Definition 8.27. Given two points 𝐴 and 𝐵, the circle of Apollonius of 𝐴 and 𝐵 with ratio 𝑘 is the set of points 𝑃 so that ‖𝑃𝐴‖/‖𝑃𝐵‖ = 𝑘.

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Figure 23. Circles of Apollonius

Figure 23 shows several circles of Apollonius based on 𝐴 and 𝐵. From the definition of these circles, it is clear that the perpendicular bisector of 𝐴𝐵 is a line of symmetry. Reflection in the line reflects the circle 𝑐 𝑘 to the circle 𝑐 1/𝑘 . As 𝑘 approaches 1, the circles get very large and approach the perpendicular bisector, which as part of this family can be thought of as a circle of infinite radius intersecting 𝐴𝐵 at the midpoint of 𝐴𝐵 and also at a point at infinity.

Composition of Dilations By Theorem 8.19, if 𝐷1 and 𝐷2 are each either a dilation or a translation, then for any segment 𝐴𝐵, the segment 𝐴′ 𝐵 ′ = 𝐷2 𝐷1 (𝐴𝐵) is a segment parallel to or collinear with 𝐴𝐵. And by this theorem, this implies that 𝑆 = 𝐷2 𝐷1 is either a dilation or a translation. If 𝑆 is a dilation, the sign of the dilation ratio is positive if 𝐴𝐵 and 𝐴′ 𝐵′ have the same direction and it is negative if the directions are opposite. The following theorem adds some details. Theorem 8.28 (Product of Dilations). For dilations 𝒟𝐸,𝑒 and 𝒟𝐹,𝑓 , if 𝑒𝑓 ≠ 1, then 𝒟𝐹,𝑓 𝒟𝐸,𝑒 = 𝒟𝐺,𝑔 , where 𝑔 = 𝑒𝑓 and 𝐸, 𝐹, 𝐺 are collinear. If 𝑒𝑓 = 1, the product 𝒟𝐹,𝑓 𝒟𝐸,𝑒 = 𝑇, where 𝑇 is a translation with direction parallel to 𝐸𝐹. Note. This implies that while the set of all translations is a transformation group, the set of all dilations is not. But the dilations and the translations together are a transformation group. Proof. As a special case, if 𝐸 = 𝐹, the product is already known to be 𝒟𝐸,𝑒𝑓 from the definition of dilation, so from now on we assume 𝐸 ≠ 𝐹. As explained already, the product 𝑆 = 𝒟𝐹,𝑓 𝒟𝐸,𝑒 must be either a dilation or a translation. Since the scale factor = |𝑒𝑓|, it must be a dilation if |𝑒𝑓| ≠ 1. Since dilations with negative ratio reverse direction, the product 𝑆 must have ratio 𝑒𝑓, a number which is positive if there are either no reversals or two reversals and negative otherwise. Thus 𝒟𝐹,𝑓 𝒟𝐸,𝑒 = 𝒟𝐺,𝑒𝑓 for some 𝐺. If 𝑒𝑓 = −1, then 𝑆 is a half-turn.

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From the definition, the center of a dilation is collinear with any point 𝐴 and its image. Since 𝐸 ′ = 𝑆(𝐸) = 𝒟𝐹,𝑓 (𝐸), the image 𝐸 ′ is on 𝐸𝐹. But also, since 𝑆 = 𝒟𝐺,𝑒𝑓 , the center 𝐺 is on the line 𝐸𝐸 ′ = 𝐸𝐹. The remaining case is 𝑒𝑓 = 1. Since 𝐸 and 𝐸 ′ are distinct, 𝑆 is not the identity, so it must be the translation 𝒯𝐸𝐸 ′ , which is a translation with direction parallel to 𝐸𝐹. □ A E

F

G

C

B

Figure 24. Menelaus Theorem as Dilation Product

A geometrical construction of 𝐺 is shown in Figure 24. Take any point 𝐵 not on 𝐸𝐹; then 𝐴 = 𝒟𝐸,𝑒 (𝐵) and 𝐶 = 𝒟𝐹,𝑓 (𝐴), so 𝐶 = 𝒟𝐺,𝑔 (𝐵). Then 𝐺 is both on 𝐸𝐹 and also on 𝐵𝐶 so it is the intersection of the two lines. The example in the figure has negative 𝑒 and 𝑓. When 𝑒𝑓 = 1 and the product is a translation, a figure like Figure 24 would show 𝐸𝐹 parallel to 𝐵𝐶, with the translation being 𝒯𝐵𝐶 . As a numerical example, suppose 𝑒 = −1/2 and 𝑓 = −2. Then 𝐵𝐶 has 3 times the length of 𝐸𝐹 and the same direction, since the segments 𝐵𝐴 and 𝐶𝐴 are divided internally in the ratio 2 ∶ 1. Figure 24 shows a product of dilations but also illustrates a theorem about triangles. Given three points on three lines, the Theorem of Menelaus tells how ratios on each line can determine when the points are collinear. Theorem 8.29 (Theorem of Menelaus). For a triangle △𝐴𝐵𝐶, let 𝐸, 𝐹, 𝐺 be points on the sides 𝐶𝐴, 𝐴𝐵, 𝐵𝐶 extended. These points are collinear if and only if 𝐸𝐴 𝐹𝐶 𝐺𝐵 = 1. 𝐸𝐵 𝐹𝐴 𝐺𝐶 Proof. This follows from the previous theorem, stated as −1 𝒟𝐹,𝑓 𝒟𝐸,𝑒 = 𝐼. 𝒟𝐺,𝑔

The dilation centers are 𝐸, 𝐹, 𝐺 and the ratios are 𝑒=

𝐸𝐴 , 𝐸𝐵

𝑓=

𝐹𝐶 , 𝐹𝐴

𝑔−1 =

𝐺𝐵 . 𝐺𝐶

□

We can see the same Menelaus relationship in a figure with three circles of differing size, as in Figure 25. Part of what is in this rich figure is captured by the following theorem. But there is even more!

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P1

a3 Q2

P2 Q1

Q3 a1

P3 a2

Figure 25. Dilation Centers Among Three Circles

Theorem 8.30 (Three Circle Theorem). Given three circles with different radii, each pair of circles is related by a dilation with positive ratio centered at a point 𝑃 and a dilation with negative ratio with center at a point 𝑄. These six dilation centers lie on four lines, with three centers per line. More precisely, the three 𝑃 centers lie on one line, and the other lines each have two 𝑄 points and one 𝑃. Proof. The proof consists of organizing the information and applying Theorem 8.28. Let the circles be 𝑐 1 , 𝑐 2 , 𝑐 3 . Denote by 𝐸𝑖𝑗 the dilation with positive ratio that maps 𝑐𝑗 to 𝑐 𝑖 and denote by 𝐹𝑖𝑗 the dilation with negative ratio that maps 𝑐𝑗 to 𝑐 𝑖 . Then, for each permutation (𝑖, 𝑗, 𝑘) of (1, 2, 3), the product 𝐸 𝑘𝑗 𝐸𝑗𝑖 = 𝐸 𝑘𝑖 , since this map is a dilation from 𝑐 𝑖 to 𝑐 𝑘 with positive ratio. Also, 𝐹 𝑘𝑗 𝐹𝑗𝑖 = 𝐸 𝑘𝑖 for the same reason. This implies that the centers of the three dilations in each equation are collinear. −1 Since 𝐸𝑗𝑖 = 𝐸𝑖𝑗 , these two dilations have the same center. Rather than have double subscripts, we can label the center of 𝐸𝑖𝑗 with the number that is neither 𝑖 nor 𝑗. So 𝑃1 , 𝑃2 , 𝑃3 are the centers of 𝐸23 , 𝐸31 , 𝐸12 and 𝑄1 , 𝑄2 , 𝑄3 are the centers of 𝐹23 , 𝐹31 , 𝐹12 .

This implies that these sets of points are collinear: {𝑃1 , 𝑃2 , 𝑃3 }, {𝑃1 , 𝑄2 , 𝑄3 }, {𝑄1 , 𝑃2 , 𝑄3 }, {𝑄1 , 𝑄2 , 𝑃3 }. □ In Figure 25, the four lines of dilation centers cited by the theorem are the thicker blue lines. The extended triangle △𝑂1 𝑂2 𝑂3 of centers of the circles consists of three solid red lines of thinner weight (the centers are not labeled in the figure to keep the total labeling simpler). There are also three concurrent dashed black lines: 𝑂1 𝑄1 , 𝑂2 𝑄2 , 𝑂3 𝑄3 . The explanation of why these lines are concurrent is given by Ceva’s Theorem, Theorem 11.14, which states that if the product in the Menelaus Theorem equals −1 instead of +1, the three points on the extended sides of the triangle lie on three concurrent lines. In the picture, the points 𝑄3 and 𝑃3 divide 𝑂1 𝑂2 harmonically, since they are dilation centers with the same ratios, except for sign.

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Although the proof of Ceva’s Theorem wlll come later, we cannot resist pointing out that with this theorem and the Menelaus Theorem, one can begin with collinear points 𝐵, 𝐶, 𝐺, then construct with straightedge the point 𝐻 so that 𝐺𝐻 divides 𝐵𝐶 harmonically, as shown in Figure 26. A E

F J

B

H

G

C

Figure 26. Harmonic Division of 𝐵𝐶 by 𝐺𝐻 by Straightedge

Returning to the circles, the dilation centers in this figure are also the intersection points of common tangents to pairs of circles, so drawing the tangents would be another way to represent this relationship. If this were not complicated enough, the Three Circles Theorem applies to arrangements of intertwined circles that are more challenging to visualize than the disjoint arrangement of circles in Figure 25.

Circles, Angles, and Ratios An inscribed angle in a circle is an angle ∠𝐵𝐴𝐶, where all three points 𝐴, 𝐵, 𝐶 are on the circle. For this section we will need an important theorem about the measure of inscribed angles. Already in the Right Angle Locus Theorem, Theorem 7.20, it was proved that if 𝐴𝐷 is a diameter of a circle 𝑐 with center 𝑂, then for any point 𝐵 on 𝑐, the measure of the inscribed angle 𝑚∠𝐵𝐴𝐷 = (1/2)𝑚∠𝐵𝑂𝐷. B

O

A

D

C

Figure 27. Inscribed and Central Angles in a Circle

We would like to have this angle relation apply to all inscribed angles ∠𝐵𝐴𝐶, but there is a problem illustrated by the example in Figure 27. In this figure, 𝑚∠𝐵𝐴𝐶 = 125, with 𝑚∠𝐵𝐴𝐷 = 65 and 𝑚∠𝐷𝐴𝐶 = 60. By Theorem 7.20, 𝑚∠𝐵𝑂𝐷 = 130 and 𝑚∠𝐷𝑂𝐶 = 120, but 𝑚∠𝐵𝑂𝐶 = 110 and cannot equal

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250 = 130 + 120, since angle measure can be no greater than 180. So we need a way of measuring angles larger than a straight angle. A solution can be found in the figure: one can measure the angle of an arc. The points of the circle that are interior to ∠𝐵𝑂𝐶 form an arc from 𝐵 to 𝐶 containing 𝐴 with angle measure 110. The points on the circle outside the angle form a larger arc from 𝐶 to 𝐵 containing 𝐷 that will be defined to have measure 360 − 110 = 250. The first is the minor arc and the second is the major arc determined by 𝐵 and 𝐶. Definition 8.31. An arc of a circle 𝑐 with endpoints 𝐵 and 𝐶 on the circle is the intersection of 𝑐 with one of the half-planes of 𝐵𝐶, plus the endpoints. Two points 𝐵 and 𝐶 on a circle define two arcs, one for each half-plane. Remark 8.32. For any noncollinear points 𝐴, 𝐵, 𝐶, there is a unique arc with endpoints 𝐵 and 𝐶 that contains 𝐴. The arc must be an arc of the only circle that contains the three points, the circumcircle of △𝐴𝐵𝐶. Then the required arc is the arc of the circumcircle with endpoints 𝐵 and 𝐶 that contains 𝐴. The measure of an arc with endpoints 𝐵 and 𝐶 is defined by extending angle measure. If 𝐵𝐶 is not a diameter, the arc defined by the half-plane containing the center 𝑂 of the circle is called the major arc; the other is the minor arc. The measure of the minor arc equals 𝑚∠𝐵𝑂𝐶 and the measure of the major arc is 360 − 𝑚∠𝐵𝑂𝐶. If 𝐵𝐶 is a diameter, the arcs are congruent, and each has measure 180. A

C

B

ˆ ˆ = 210, 𝑚𝐵𝐶 ˆ = 150, and 𝑚∠𝐵𝐴𝐶 = 75 Figure 28. Measures: 𝑚𝐵 𝐴𝐶 = 𝑚𝐶𝐵

Notation for an Arc. Given points 𝐵 and 𝐶 on a circle 𝑐, there are two arcs with ˆ is ambiguous. More inforthese endpoints. Therefore, just writing an arc symbol 𝐵𝐶 mation is needed to say which arc is meant. ˆ is an unambiguous designation for the If a point 𝑃 on the arc is known, then 𝐵𝑃𝐶 arc. But a natural choice of 𝑃 is not always present. ˆ unambiguous, we take an orientation on the plane to desTo make the symbol 𝐸𝐹 ignate the positive, or counterclockwise, direction for polar angles. Then choose polar angles 𝑒 = 𝜃(𝑂𝐸) and 𝑓 = 𝜃(𝑂𝐹), with 0 < 𝑓 − 𝑒 < 360. The (counterclockwise) ˆ on a circle with center 𝑂 is defined to be the set of points 𝑃 on the circle with arc 𝐸𝐹 ˆ = 𝑓 − 𝑒. 𝑒 ≤ 𝜃(𝑃) ≤ 𝑓. The measure of the arc is 𝑚𝐸𝐹

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141

This notation is very helpful in proving the next proposition about arc measure addition and subtraction. ˆ are arcs on a circle 𝑐 with 𝑚𝐸𝐹 ˆ and 𝐹𝐺 ˆ+ Theorem 8.33 (Arc Measure Addition). If 𝐸𝐹 ˆ ˆ ˆ ˆ 𝑚𝐹𝐺 < 360, then 𝑚𝐸𝐺 = 𝑚𝐸𝐹 + 𝑚𝐹𝐺. Proof. Let 𝑓 = 𝜃(𝐹); then values of the polar angles 𝑒 = 𝜃(𝐸) and 𝑔 = 𝜃(𝐺) can be ˆ is well-defined and 𝑔 − 𝑒 = 𝑚𝐸𝐺. ˆ chosen with 𝑒 < 𝑓 < 𝑔. If 𝑔 − 𝑒 < 360, the arc 𝐸𝐺 ˆ ˆ Since 𝑔 − 𝑒 = (𝑔 − 𝑓) + (𝑓 − 𝑒) = 𝑚𝐸𝐹 + 𝑚𝐹𝐺, the addition formula holds. □ From this addition formula, it is also true that subtraction formulas hold; e.g., ˆ − 𝑚𝐸𝐹 ˆ ˆ = 𝑚𝐹𝐺. 𝑚𝐸𝐺 Before stating the theorem about inscribed angles, we introduce one more way to ˆ will denote designate an arc. Given a circle 𝑐 and an angle ∠𝐵𝐴𝐶, the symbol ∠𝐵𝐴𝐶 the points of 𝑐 interior to, or on, ∠𝐵𝐴𝐶. This set is an arc in each of these three cases: • 𝐵, 𝐶, and 𝐴 are on 𝑐, so ∠𝐵𝐴𝐶 is an inscribed angle. • 𝐵 and 𝐶 are on 𝑐 and 𝐴 is interior to 𝑐. ⃗ • 𝐵 and 𝐴 are on 𝑐 and 𝐴 𝐶 is tangent to 𝑐 at 𝐴. In each case, the arc is the intersection of 𝑐 with a half-plane of 𝐵𝐶. If 𝐴 is exterior to 𝑐, with 𝐵 and 𝐶 on 𝑐, then the set will consist of two arcs. Theorem 8.34 (Inscribed Angle Theorem). If ∠𝐵𝐴𝐶 is an inscribed angle in circle 𝑐, 1 ˆ In other words, the measure of the inscribed angle is one-half then 𝑚∠𝐵𝐴𝐶 = 2 𝑚∠𝐵𝐴𝐶. the measure of the arc included in the angle. Proof. Let 𝐷 be the point so that 𝐴𝐷 is a diameter as in Figure 27. From Theorem 7.20, 1 ˆ Then by subtracting angle measure ˆ and 𝑚∠𝐷𝐴𝐶 = 1 𝑚∠𝐷𝐴𝐶. 𝑚∠𝐵𝐴𝐷 = 2 𝑚∠𝐵𝐴𝐷 2 if both 𝐵 and 𝐶 are in the same half-plane of 𝐴𝐷 and adding angle measure if they are 1 ˆ on opposite sides, one gets 𝑚∠𝐵𝐴𝐶 = 2 𝑚∠𝐵𝐴𝐶. □

Figure 29. Inscribed Angles in Circumscribed Regular Polygons and Star Polygons

ˆ =𝐵 ˆ Returning to Figure 27, as an example, 𝑚∠𝐵𝐴𝐶 = 125 and ∠𝐵𝐴𝐶 𝐷𝐶, with ˆ = 250. Another example of inscribed angles is provided by the pentagon and 𝑚∠𝐵𝐴𝐶 pentagram in Figure 29. The vertices of the pentagon divide the circle into equal arcs

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8. Dilations and Similarity

of measure 360/5 = 72. The included arc in any vertex angle of the pentagon has arc measure 3 ∗ 72 = 216, so the vertex angle has measure 108. But for the pentagram, the included arc has measure 72, so the vertex angle has measure 36. The same reasoning can be applied to the nonagon on the right side of the figure. The circle is divided into nine congruent arcs with measure 40 = 360/9. A vertex angle of the nonagon has an included arc of measure 7 ∗ 40 = 280, so the vertex angle of the nonagon has measure 140. Then we see that the measure of the vertex angle of the star nonagon shown is 20. What are the measures of the two angles formed at a vertex by a combination of a side segment from the regular nonagon and one of the sides of the star nonagon?

B

C

Figure 30. Locus of Points 𝐴 on Either Side of 𝐵𝐶 with 𝑚∠𝐵𝐴𝐶 = 120

Corollary (Angle Arc Locus). Given a segment 𝐵𝐶, the locus of points 𝐴 in a given halfplane with a given angle measure 𝑚∠𝐵𝐴𝐶 is an arc. Therefore, if 𝐷 is on the same side ˆ ˆ of 𝐵𝐶 as 𝐴 with 𝑚∠𝐵𝐷𝐶 = 𝑚∠𝐵𝐴𝐶, then 𝐵 𝐷𝐶 = 𝐵 𝐴𝐶. Proof. It is sufficient to show that, given 𝐵𝐶, knowing the measure 𝑡 = 𝑚∠𝐵𝐴𝐶 and ˆ knowing which half-plane of 𝐵𝐶 contains 𝐴 determines 𝐵 𝐴𝐶 uniquely with no further information about 𝐴. For any 𝐴 not on 𝐵𝐶, let 𝑂 be the center of the circumcircle of △𝐵𝐴𝐶. If 𝑡 < 90, ˆ the measure of the included arc of ∠𝐵𝐴𝐶 is 2𝑡, and 2𝑡 = 𝑚∠𝐵𝑂𝐶. Since 𝐵 𝐴𝐶 is the major arc, it is on the same side of 𝐵𝐶 as 𝑂. If 𝑡 > 90, then 𝑚∠𝐵𝑂𝐶 = 360 − 2𝑡. In this ˆ case 𝐵 𝐴𝐶 is the minor arc on the opposite side of 𝐵𝐶 from 𝑂. ˆ In either case, if the location of 𝑂 is known, this is sufficient to determine 𝐵 𝐴𝐶. But △𝐵𝑂𝐶 is an isosceles triangle with base 𝐵𝐶, with angle measure at 𝑂 and base angle measure all known from 𝑡. Since the half-plane containing 𝑂 is also specified, ˆ this is sufficient to determine 𝑂 uniquely and also 𝐵 𝐴𝐶. ˆ If 𝑡 = 90, 𝑂 is the midpoint of 𝐵𝐶 and 𝐵 𝐴𝐶 is the semicircle on the same side of □

𝐵𝐶 as 𝐴.

Corollary (Tangent-Chord). Given a circle 𝑐, let 𝐴 and 𝐵 be points on 𝑐 and let 𝐴𝐸 be 1 ˆ The arc is the set of points interior to tangent to 𝑐 at 𝐴. Then 𝑚∠𝐸𝐴𝐵 = 2 𝑚∠𝐸𝐴𝐵. ∠𝐸𝐴𝐵, plus the endpoints 𝐵 and 𝐴. Proof. Let 𝐷 be the point so that 𝐴𝐷 is a diameter of 𝑐. Then the inscribed angle ∠𝐴𝐵𝐷 is a right angle, so 𝑚∠𝐴𝐷𝐵 + 𝑚∠𝐷𝐴𝐵 = 90. Also, 𝑚∠𝐸𝐴𝐵 + 𝑚∠𝐷𝐴𝐵 = 90, so 1 ˆ since ∠𝐸𝐴𝐵 ˆ and ∠𝐴𝐷𝐵 ˆ are the same arc. 𝑚∠𝐸𝐴𝐵 = 𝑚∠𝐴𝐷𝐵 = 𝑚∠𝐸𝐴𝐵 □ 2

Circles, Angles, and Ratios

143

A

E B

O

D Figure 31. Tangent-Chord

Example 8.35. The inscribed angle theorem reveals some surprising circle relations. Here is one. Take three circles 𝑎, 𝑏, 𝑐 with centers 𝐴, 𝐵, 𝐶, all passing through a point 𝑃. Let the other points of intersection be 𝑋, 𝑌 , 𝑍 as shown in Figure 32. Then for any general point 𝑃𝐴 on circle 𝑎, draw a 𝑃𝐴 𝑍 through 𝑍, the point of intersection of the circles 𝑎 and 𝑏. Let the other intersection of this line with 𝑏 be 𝑃𝐵 . Then repeat. Let line 𝑃𝐵 𝑋 intersect the circle 𝑐 at 𝑃𝐶 . The next line 𝑃𝐶 𝑌 does not create a new point; it passes through the original 𝑃𝐴 . If 𝑃𝐴 moves around circle 𝑎, the triangle 𝑃𝐴 𝑃𝐵 𝑃𝐶 follows around in an animation.

PA

PB

B Z

P

X

A C Y

PC

Figure 32. Circle Intersection Path

Why does this work? Here is a proof for circles and point 𝑃𝐴 as in Figure 32. The answer can be found by looking at two circles at a time. In the figure, there are three congruent angles 𝜃 > 0 marked near the centers of the circles. We will prove that ˆ ˆ 𝜃 = 𝑚𝑃ˆ 𝐴 𝑍𝑃 = 𝑚𝑃𝐵 𝑋𝑃 = 𝑚𝑃𝐶 𝑌 𝑃. If this is true, then it will explain why the step from 𝑃𝐶 to 𝑃𝐴 constructs the original 𝑃𝐴 since it is the unique point with 𝜃 = 𝑚𝑃ˆ 𝐴 𝑍𝑃. ˆ Start in circle 𝑎 by defining 𝜃 = 𝑚𝑃ˆ 𝐴 𝑍𝑃. Then 𝑚∠𝑃𝐴 𝑍𝑃 = 360 − 𝜃. The inscribed angle theorem says that 𝑚∠𝑃𝑍𝑃𝐴 = 180 − 𝜃/2. Since 𝑃𝐴 , 𝑍, 𝑃𝐵 are collinear, the angles ∠𝑃𝑍𝑃𝐴 and ∠𝑃𝑍𝑃𝐵 are supplementary. Therefore, 𝑚∠𝑃𝑍𝑃𝐵 = 𝜃/2. Applying the ˆ inscribed angle theorem again in circle 𝑏 shows that 𝑚𝑃 𝐵 𝑋𝑃 = 𝜃.

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Continuing with the same reasoning applied to the next pair of circles leads from ˆ 𝑃𝐵 to 𝑃𝐶 , with 𝜃 = 𝑚𝑃 𝐶 𝑌 𝑃. Then the next step leads to 𝑃𝐴 , the only point on 𝑎 satisfying 𝜃 = 𝑚𝑃ˆ 𝑍𝑃. 𝐴 In fact, nothing restricts this argument to three circles. One can apply this to four or five circles, or with any number one can keep track of. Note. In this proof we relied heavily on the arrangement of elements in the figure. The three circles can be positioned quite differently, and the angle reasoning is not the same when 𝑃𝐴 is on the other side of 𝐴𝐵. The result is still true, but one must take more care with signed angle measure to prove the general case. One way to address this is to write the rotations 𝐴𝜃 and 𝐵𝜃 as products of line reflections. Let 𝑚 and 𝑛 be the diameters of circles 𝑎 and 𝑏 perpendicular to 𝑃𝐴 𝑍. Then 𝑍 = ℛ𝑚 (𝑃𝐴 ) = ℛ𝑛 (𝑃𝐵 ), so 𝑃 = ℛ𝐴𝐵 ℛ𝑚 (𝑃𝐴 ) = ℛ𝐴𝐵 ℛ𝑛 (𝑃𝐵 ). The double line reflections are rotations with center 𝐴 and center 𝐵, respectively. Since the translation 𝒯𝐴𝐵 maps the pair of lines 𝑚, 𝐴𝐵 to 𝑛, 𝐴𝐵, the rotation angles are the same angle 𝜃, so 𝐴𝜃 (𝑃𝐴 ) = 𝐵𝜃 (𝑃𝐵 ) = 𝑃. This figure is related to the Miquel Theorem (see Exercise 11). More Lines, Circles, and Angles. So far we have proved two theorems about angles formed when two lines intersect a circle. In both cases, the lines intersect at a point on the circle. Now we will consider angles and arcs when the lines intersect outside the circle and inside the circle. Figure 33 shows examples of each case. We will begin when the point of intersection 𝑃 is outside a circle 𝑐. A line through 𝑃 intersecting the circle in two points is called a secant line. Theorem 8.36 (Angle and Arcs from Secants). Let 𝑐 be a circle and 𝑃 a point outside the circle. Let 𝐴, 𝐵, 𝐶, 𝐷 be distinct points on the circle such that 𝐴 is on 𝑃𝐶 and 𝐵 is on 𝑃𝐷. Let 𝛼 ˆ be the arc with endpoints 𝐶 and 𝐷 contained in ∠𝐶𝑃𝐷, and let 𝛽ˆ be the arc ˆ with endpoints 𝐵 and 𝐴 contained in ∠𝐶𝑃𝐷. Then 𝑚∠𝐶𝑃𝐷 = (1/2)(𝑚ˆ 𝛼 − 𝑚𝛽). Remark 8.37. If the points 𝐴, 𝐵, 𝐶, 𝐷 are arranged in counterclockwise order, as in ˆ and 𝛽ˆ = 𝐵𝐴. ˆ Recall the definition Figure 33, then simpler notation is available: 𝛼 ˆ = 𝐶𝐷 ˆ of ∠𝐶𝑃𝐷, the set of points on 𝑐 within the angle ∠𝐶𝑃𝐷. This set consists of this pair of arcs. Proof. We will follow the notation in Figure 33. Considering △𝑃𝐶𝐵, the exterior angle at 𝐵 has measure equal to the sum of the measures of the angles at 𝑃 and 𝐶; therefore, 𝑚∠𝐶𝑃𝐵 = 𝑚∠𝐶𝐵𝐷−𝑚∠𝑃𝐶𝐵. But the angles on the right side of the equation are ˆ Second, 𝑚∠𝑃𝐶𝐵 = (1/2)𝑚𝐵𝐴. ˆ inscribed angles in 𝑐. First, 𝑚∠𝐶𝐵𝐷 = (1/2)𝑚𝐶𝐷. □ In the remaining case, 𝑃 is inside the circle 𝑐 as shown in Figure 33. Rather than secant lines, this theorem is about an angle defined by intersecting chords. Theorem 8.38. Let 𝑐 be a circle and 𝑃 a point inside the circle. Let 𝐴, 𝐵, 𝐶, 𝐷 be distinct points on the circle such that 𝑃 is the intersection of 𝐴𝐶 and 𝐵𝐷. Then 𝑚∠𝐶𝑃𝐷 = ˆ + 𝑚∠𝐴𝑃𝐵). ˆ (1/2)(𝑚∠𝐶𝑃𝐷

Circles, Angles, and Ratios

145

D C

A

P

P

C

A B D

B Figure 33. Lines Intersecting a Circle

Proof. In △𝑃𝐷𝐴, ∠𝐶𝑃𝐷 is an external angle and thus has measure equal to the sum 𝑚∠𝑃𝐴𝐷 + 𝑚∠𝑃𝐷𝐴. But each of these is an inscribed angle with measure equal to ˆ + 𝑚∠𝐴𝑃𝐵). ˆ one-half the included arc measure. Thus 𝑚∠𝐶𝑃𝐷 = (1/2)(𝑚∠𝐶𝑃𝐷 □ Secants, Chords, and the Power of a Point Function. We continue with the same figures to show that certain products of lengths are equal. The statement of the theorem will be using signed products, which we define before proceeding. Definition 8.39. Let 𝐴, 𝐵, 𝐶, 𝐷 be collinear points, some of which may be the same. The signed product 𝐴𝐵 ⋅ 𝐶𝐷 is a real number 𝑘 with |𝑘| = ‖𝐴𝐵‖ ‖𝐶𝐷‖. If this product is nonzero, 𝑘 has positive sign when 𝐴𝐵 and 𝐶𝐷 have the same direction and it has negative sign for opposite direction. Theorem 8.40 (Products from Secants and Chords). Let two lines through 𝑃 intersect a circle 𝑐 in points 𝐴 and 𝐶 for one line and points 𝐵 and 𝐷 for the other line. These signed products of lengths are equal: 𝑃𝐴 ⋅ 𝑃𝐶 = 𝑃𝐵 ⋅ 𝑃𝐷. If the radius of the circle is 𝑟 and the center is 𝑂, then each of these signed products equals ‖𝑂𝑃‖2 − 𝑟2 . Proof. We will continue to use the notation in the proofs of Theorems 8.36 and 8.38, with labels as in Figure 33. The plan of the proof is to show △𝑃𝐷𝐴 ∼ △𝑃𝐶𝐵 by AA. First, the angles at 𝑃 are congruent, ∠𝐴𝑃𝐷 ≅ ∠𝐵𝑃𝐶. When 𝑃 is outside 𝑐, the angles are shared angles. When 𝑃 is inside 𝑐, the angles at 𝑃 are vertical angles, so they are congruent in both cases. Second, ∠𝑃𝐷𝐴 ≅ ∠𝑃𝐶𝐵 because both are inscribed angles including the same ˆ when 𝑃 is inside. Therefore, arc, namely 𝛽ˆ when 𝑃 is outside the circle and ∠𝐴𝐶𝐵 △𝑃𝐷𝐴 ∼ △𝑃𝐶𝐵 by AA for either location of 𝑃. Ratios of corresponding sides are equal, so 𝑃𝐶/𝑃𝐷 = 𝑃𝐵/𝑃𝐴 and 𝑃𝐴⋅𝑃𝐶 = 𝑃𝐵 ⋅𝑃𝐷 The signed product is positive if 𝑃 is outside the circle and negative if 𝑃 is inside the circle. If 𝑃 is on the circle, the product is 0. If we pick the line through 𝑃 to be 𝑂𝑃, then the formula follows. Let the line 𝑂𝑃 intersect 𝑐 in points 𝐸 and 𝐹. Since this 𝐸𝐹 is a diameter, ‖𝐸𝐹‖ = 2𝑟. Let 𝑑 = ‖𝑂𝑃‖. When 𝑃 is outside the circle, one of ‖𝑃𝐸‖ and ‖𝑃𝐹‖ equals 𝑑 + 𝑟 and the other is 𝑑 − 𝑟; therefore 𝑃𝐸 ⋅ 𝑃𝐹 = (𝑑 + 𝑟)(𝑑 − 𝑟) = 𝑑 2 − 𝑟2 . When 𝑃 is inside the circle, the two distances are 𝑟 − 𝑑 and 𝑟 + 𝑑. The signed product has a minus sign since the segments have different directions, so 𝑃𝐸 ⋅ 𝑃𝐹 = −(𝑟 + 𝑑)(𝑟 − 𝑑) = 𝑑 2 − 𝑟2 . Finally, if 𝑃 is on the circle, the formula equals 0. □

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Theorem 8.40 leads to this definition. Definition 8.41. Given a point 𝑃 and a circle 𝑐 with center 𝑂 and radius 𝑟, the power of 𝑃 with respect to the circle equals ‖𝑂𝑃‖2 − 𝑟2 . Remark 8.42. If 𝑃 is fixed and outside 𝑐 and point 𝐴 on circle 𝑐 approaches a point of tangency 𝑇, so does the other intersection 𝐵. Therefore the constant product 𝑃𝐴 ⋅ 𝑃𝐵 approaches ‖𝑃𝑇‖2 , so this must also equal the power of 𝑃. A geometrical proof of this and its consequences will be found in the next theorem. Power of a Point, Orthogonal Circles, and Harmonic Division. In general, two curves in the plane are said to be orthogonal at a point of intersection if the tangent lines to each curve at the point are perpendicular. If the curve is a circle, the tangent line at a point 𝑇 is perpendicular to the diameter through the point, so orthogonal circles look like the circles in Figure 34. B T

c1 P1

P2

A

c2 S

Figure 34. Orthogonal Circles

Theorem 8.43. Suppose circles 𝑐 1 and 𝑐 2 have centers 𝑃1 and 𝑃2 . The circles are orthogonal at 𝑇 if and only if 𝑃1 𝑇 and 𝑃2 𝑇 are perpendicular. This occurs if and only if the power of 𝑃1 with respect to circle 𝑐 2 is ‖𝑃1 𝑇‖2 . Let a secant through 𝑃1 intersect 𝑐 2 at 𝐴 and 𝐵. If 𝑐 1 and 𝑐 2 are orthogonal, then 𝑐 1 is also orthogonal to any other circle passing through 𝐴 and 𝐵. Proof. The line 𝑃1 𝑇 is tangent to 𝑐 2 at 𝑇 if and only if it is perpendicular to 𝑃2 𝑇. But then △𝑃1 𝑇𝑃2 is a right triangle with leg lengths ‖𝑃𝑇‖ and 𝑟2 . By the Pythagorean Theorem, ‖𝑃𝑇‖2 = ‖𝑃1 𝑃2 ‖2 − 𝑟22 , the power of 𝑃1 with respect to 𝑐 2 . The circles are orthogonal if 𝑟1 = ‖𝑃𝑇‖, so 𝑟12 equals the power of 𝑃 with respect to 𝑐 2 . To prove the second part of the theorem, note that for any circle 𝑐 3 through points 𝐴 and 𝐵, the power of the point 𝑃 with respect to 𝑐 3 is the same as for 𝑐 2 by Theorem 8.40 . Therefore, 𝑟1 is the radius such that a circle with center 𝑃1 is orthogonal to 𝑐 3 . □ The reflection of △𝑃1 𝑇𝑃2 in the line of centers 𝑃1 𝑃2 is another right triangle, △𝑃1 𝑆𝑃2 with 𝑃𝑆 tangent to 𝑐 2 , and the circles are also orthogonal at 𝑆. There is a connection between this figure and harmonic division. To see this, we will rewrite the equation for harmonic division of 𝐶𝐷 by 𝐴𝐵 in a different form.

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147

A

B C

M

D

Figure 35. Harmonic Division of 𝐶𝐷 by 𝐴𝐵 with Midpoint 𝑀 of 𝐶𝐷

In Definition 8.24, the equation for harmonic division was written with a minus sign. Instead, we can reverse the orientation of one segment in the equation to get this form: 𝐴𝐶/𝐴𝐷 = 𝐶𝐵/𝐵𝐷. In this expression all the segments will be oriented in the same direction, so it will be easier to work with. Let 𝑀 be the midpoint of 𝐶𝐷 with 𝑟 = ‖𝐶𝑀‖ = ‖𝑀𝐷‖, as illustrated in Figure 35. Then the equation becomes ‖𝑀𝐴‖ − 𝑟 𝑟 − ‖𝑀𝐵‖ = . ‖𝑀𝐴‖ + 𝑟 ‖𝑀𝐵‖ + 𝑟 Cross-multiplying and canceling like terms, this becomes ‖𝑀𝐴‖‖𝑀𝐵‖ = 𝑟2 . Now we relate this to orthogonal circles. Let 𝑐 be the circle with diameter 𝐶𝐷 and center 𝑀 and let 𝑑 be any circle containing 𝐴 and 𝐵. Then the expression ‖𝑀𝐴‖‖𝑀𝐵‖ is the power of the point 𝑀 with respect to 𝑑 and 𝑟 is the radius of a circle centered at 𝑀 that is orthogonal to 𝑑, namely the circle 𝑐. This proves the following relationship between these circles.

A

B

Figure 36. Circles of Apollonius of 𝐴 and 𝐵 with Orthogonal Circles

Theorem 8.44. If the interval 𝐶𝐷 divides 𝐴𝐵 harmonically, then the circle with diameter 𝐶𝐷 is orthogonal to every circle through 𝐴 and 𝐵. Therefore, every circle through 𝐴 and 𝐵 is orthogonal to every circle of Apollonius of 𝐴 and 𝐵. Proof. The proof of the first statement is given in the paragraph before the theorem. The proof of the second statement is immediate from Definition 8.27. □

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Radical Axis, Intersections, and Triangle Existence We saw in the previous section how power of a point is related to orthogonal circles. What if, instead of constructing a circle 𝑒 orthogonal to a single circle 𝑐, one wants to construct a circle 𝑒 orthogonal to two circles 𝑐 1 and 𝑐 2 ? From Theorem 8.43, we see that the center of such a circle must have the same power with respect to both 𝑐 1 and 𝑐 2 , since the radius of 𝑒 is the square root of the power. This leads to a definition of a set. Definition 8.45. The radical axis of two circles 𝑐 1 and 𝑐 2 is the set of points 𝑃 so that the power of 𝑃 with respect to each of the circles is the same. If 𝑐 1 and 𝑐 2 are the circles with centers 𝑂1 and 𝑂2 and radii 𝑟1 and 𝑟2 , the radical axis is the set of points satisfying ‖𝑃𝑂1 ‖2 − 𝑟12 = ‖𝑃𝑂2 ‖2 − 𝑟22 . From the equation we see that for distinct concentric circles, the radical axis is empty. Theorem 8.46. The radical axis of two circles with distinct centers is a line perpendicular to the line of centers. Proof. We continue with the notation from the displayed equation above. Let the function Δ(𝑃) be the difference between the two power functions: Δ(𝑃) = (‖𝑃𝑂1 ‖2 − 𝑟12 ) − (‖𝑃𝑂2 ‖2 − 𝑟22 ). The radical axis is the set of points for which Δ(𝑃) = 0. For any point 𝑃, let 𝑄 be the foot of the perpendicular from 𝑃 to 𝑂1 𝑂2 . By the Pythagorean Theorem, ‖𝑄𝑂1 ‖2 = ‖𝑃𝑂1 ‖2 − ‖𝑃𝑄‖2 and ‖𝑄𝑂2 ‖2 = ‖𝑃𝑂2 ‖2 − ‖𝑃𝑄‖2 . Therefore, substituting this into the definition of Δ, we see that Δ(𝑃) = Δ(𝑄). In particular, 𝑃 is on the radical axis of the two circles if and only if 𝑄 is also. Now to see whether there are points 𝑄 on the line of centers that are on the radical axis, we introduce a ruler on 𝑂1 𝑂2 , with real numbers 𝑎1 , 𝑎2 , 𝑥 corresponding to 𝑂1 , 𝑂2 , 𝑄. Using these real numbers, we see that Δ(𝑄) = (𝑥 − 𝑎1 )2 − 𝑟12 − (𝑥 − 𝑎2 )2 + 𝑟22 = 2(𝑎2 − 𝑎1 )𝑥 + 𝑟22 − 𝑟12 . This is a linear function with nonzero coefficient of 𝑥, so there is exactly one point 𝑆 on the line with Δ(𝑆) = 0. This point 𝑆 is the only point of the radical axis on 𝑂1 𝑂2 . Let 𝑠 be the line through 𝑆 perpendicular to 𝑂1 𝑂2 . For any point 𝑃 on this line, 𝑆 is the foot of the perpendicular from 𝑃, so Δ(𝑃) = Δ(𝑆) = 0, so 𝑃 is on the radical axis. Moreover, if a point 𝑃 is on the radical axis, 𝑆 must be the foot of the perpendicular, since Δ(𝑃) = 0. This proves that the line 𝑠 is the radical axis. □ Now that we know that the radical axis is a line (thus justifying “axis” in the name), where is it located? Since the radical axis contains all the centers of circles orthogonal to 𝑐 1 and 𝑐 2 , Figure 36 gives us a good idea. For any two of the Apollonian circles of 𝐴𝐵,

Radical Axis, Intersections, and Triangle Existence

P1

A

c1

B

P2

P1 c2

149

P2 C

P1

D

c1

P2 F

E c2

c2

Figure 37. Three Cases of Radical Axes of Circles

the perpendicular bisector of 𝐴𝐵 is the radical axis. And for any two circles through 𝐴 and 𝐵, 𝐴𝐵 must be the radical axis. The next theorem completes this picture. Theorem 8.47. Given two circles 𝑐 1 and 𝑐 2 with distinct centers 𝑂1 , 𝑂2 and radii 𝑟1 , 𝑟2 , let the segment 𝑑𝑖 be the diameter of 𝑐 𝑖 contained in 𝑂1 𝑂2 . Then four cases describe the location of the radical axis. (1) Disjoint circles external to each other: the radical axis intersects 𝑂1 𝑂2 at a point between the circles. (2) Interlocked circles: One endpoint of each diameter 𝑑𝑖 is inside the other circle and the other endpoint is outside. The radical axis intersects 𝑂1 𝑂2 at a point within both circles. Moreover, the circles intersect at two points on the radical axis. (3) Nested circles: One circle is completely inside the other one. The radical axis lies outside both circles. (4) Tangent circles: The circles are tangent either internally or externally. The radical axis is the common tangent line. Proof. We will continue with the function Δ from the preceding proof. Let Π𝑖 (𝑃) be the power of 𝑃 with respect to 𝑐 𝑖 . The key to the proof is shown in Figure 37, which shows the first three cases. In each case, a segment is picked out: 𝐴𝐵, 𝐶𝐷, and 𝐸𝐹. For segments 𝐴𝐵 and 𝐶𝐷, the sign of Δ is different at the two endpoints, so Δ, a linear function on 𝑂1 𝑂2 , must have its single zero interior to the segment. In the third case, Δ has the same sign at both endpoints of 𝐸𝐹, so the zero cannot be in 𝐸𝐹. To make this clear in the figure, the endpoints at which Δ has positive sign are colored blue and the negative endpoints are colored red. We give more details for case (2), the interlocked circles, and leave the rest for cases (1) and (3) to the reader.2 In this case 𝐷 is an endpoint of a diameter of 𝑐 1 that is interior to 𝑐 2 . Therefore, Π1 (𝐷) = 0 and Π2 (𝐷) < 0, so Δ(𝐷) > 0. The point 𝐶 is an endpoint of a diameter of 𝑐 2 that is interior to 𝑐 1 , so Π2 (𝐸) = 0 and Π1 (𝐸) < 0. Therefore, Δ(𝐸) < 0. This means the 2

This proof is the same as that of Theorem 8.13 but in a broader context.

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point 𝑆 with Δ(𝑆) = 0 is interior to 𝐶𝐷. The line 𝑠 through 𝑆 perpendicular to 𝑂1 𝑂2 is the radical axis by Theorem 8.46. The point 𝑆 is interior to both circles, so the distance from 𝑠 to 𝑂1 is less than 𝑟1 . We know by Theorem 6.8 that such a line must intersect 𝑐 1 in two points. But any point of intersection of the radical axis with 𝑐 1 is a point where Π1 is 0 since the point is on 𝑐 1 , but also Π2 is 0 at this point because Π1 = Π2 at points of the radical axis. Therefore, this point of intersection must also be on 𝑐 2 . This implies that in case (2) the two circles must intersect in two points. For the last case, if two circles are tangent at a point 𝑇, then 𝑇 is a point on the line of centers with Π1 (𝑇) = Π2 (𝑇) = 0, so 𝑇 is on the radical axis, which is the common tangent line. □ In addition to the information about radical axes in general, this theorem implies an important fact about circle intersection stated differently from the corollary to Theorem 8.13. Corollary. Let 𝑐 1 and 𝑐 2 be two circles. If there is a point of 𝑐 1 that is interior to 𝑐 2 and another point of 𝑐 1 that is exterior to 𝑐 2 , then the circles intersect in two distinct points. Proof. Let Π𝑖 (𝑃) be the power of 𝑃 with respect to 𝑐 𝑖 and let Δ(𝑃) = Π1 (𝑃)−Π2 (𝑃). We have seen that for any 𝑘, the set of 𝑃 for which Δ(𝑃) = 𝑘 is a line parallel to the radical axis of 𝑐 1 and 𝑐 2 . The set of points for which Δ > 0 is one half-plane of the radical axis, and the set where Δ < 0 is the opposite one. For a point 𝑃 on 𝑐 1 , Δ(𝑃) = 0 − Π2 (𝑃). Let 𝐸 and 𝐹 be points on 𝑐 1 , with 𝐸 exterior to 𝑐 2 and with 𝐹 interior to 𝑐 2 . Then Δ(𝐸) < 0 and Δ(𝐹) > 0, so the points are on opposite sides of the radical axis. Therefore, this line, the radical axis, must intersect 𝑐 1 at two points. At these points Δ = −Π2 = 0, so these are two points of intersection of 𝑐 1 and 𝑐 2 . □ While we have a theorem about the location of the radical axis, for disjoint circles it is not clear how to construct the radical axis in a straightforward manner. An answer is given by the next theorem. Theorem 8.48 (Radical Center). Let 𝑐 1 , 𝑐 2 , 𝑐 3 be three circles with noncollinear centers. The three radical axes of the pairs of circles are concurrent at a point called the radical center of the three circles. The radical center is the center of the unique circle orthogonal to all three circles. Proof. This proof is almost identical to the proof of concurrence of perpendicular bisectors of a triangle. Let Π𝑖 be the power function of 𝑐 𝑖 . Since each radical axis of two circles is perpendicular to the line of centers, by Theorem 7.7, any two radical axes intersect, since the centers are not collinear. Let 𝐴 be the intersection of the radical axis 𝑡12 of 𝑐 1 and 𝑐 2 and the radical axis 𝑡23 of 𝑐 2 and 𝑐 3 . Then Π1 (𝐴) = Π2 (𝐴) since 𝐴 is on 𝑡12 , and also Π2 (𝐴) = Π3 (𝐴) since 𝐴 is on 𝑡23 . But since Π1 (𝐴) = Π3 (𝐴), then 𝐴 is also on the radical axis of 𝑐 1 and 𝑐 3 , so the axes are concurrent at 𝐴. If the circle 𝑎 is the circle with center 𝐴 and radius Π1 (𝐴), it is orthogonal to all three of the given circles. Since any such circle must be on all three radical axes, this is the only such circle. □

Radical Axis, Intersections, and Triangle Existence

151

A

c2

c1 e

Figure 38. Constructing Radical Axis of 𝑐 1 and 𝑐 2 by Radical Center 𝐴

Remark 8.49. One can use this theorem to construct the radical axis of two disjoint circles. Just draw a third circle 𝑒 that intersects each of the two given circles in two points so that the radical axes with the new circle are lines of intersection. These lines through the intersection points are radical axes intersecting at the radical center 𝐴 of the three circles. The line through 𝐴 perpendicular to the line of centers of the original circles is the desired radical axis. This is illiustrated in Figure 38. The original circles are thicker green circles and the radical axis is the thicker line through 𝐴. Orientation-Reversing Similitudes. The similar triangles in the proof of power of a point in Theorem 8.40 differed from those in theorems about triangles with parallel sides. The similitude relating them was orientation-reversing instead of orientationpreserving. If one of the triangles was dilated to a size so that the new triangle was congruent to the other, it would not be related by a rotation or translation but rather by orientation-reversing rigid motion. The simplest example of such an orientation-reversing transformation is the product of a dilation with a line reflection. A very commonly occurring instance is when the line of reflection passes through the dilation center. This can produce a sort of antiparallel triangular figure such as shown in Figure 39. Such figures occurred in the proof of the Pythagorean Theorem and in the Golden Triangle. B

D

B'

P A'

C

A

Figure 39. Product of a Dilation with Reflection in Angle Bisector

In this figure, the parallel segments coming from the dilation are related by signed ratios on each line through the center 𝑃: 𝑃𝐴′ /𝑃𝐴 = 𝑃𝐵 ′ /𝑃𝐵, while after the reflection, the points on each line are related by signed products: 𝑃𝐴 ⋅ 𝑃𝐶 = 𝑃𝐵 ⋅ 𝑃𝐷 (or equivalently, 𝑃𝐷/𝑃𝐴 = 𝑃𝐶/𝑃𝐵).

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One final transformation product that will be mentioned here for contrast is orientation-preserving; it is the dilation rotation product. The simplest case to visualize is when the centers of dilation and rotation are the same. In Figure 40, a point is transformed by a similitude that is the product of a dilation with ratio 1.05 and a rotation by 20 degrees. This transformation is applied to a point and then applied to its image and to the next image again and again to produce a spiral with 24 points.

Figure 40. Spiral from a Dilation Rotation

Centers of Dilation and the Midpoint Triangle Since the corresponding sides of a triangle △𝐴𝐵𝐶 and its midpoint triangle △𝐷𝐸𝐹 are parallel, we can apply Theorem 8.20 to find the center of the dilation mapping △𝐴𝐵𝐶 to △𝐷𝐸𝐹. Since the directions of the parallel segments are opposite, the dilation will have negative ratio −1/2 and the center will be located on the segments connecting corresponding points. In this case, these segments have a name. They are the medians, the segments connecting each vertex of △𝐴𝐵𝐶 to the midpoint of the opposite side. The existence of the center of dilation is a proof that the medians are concurrent at a point 𝐺. The point of concurrence is called the centroid of △𝐴𝐵𝐶. Thus, the dilation from △𝐴𝐵𝐶 to △𝐷𝐸𝐹 is 𝒟𝐺,−1/2 . This proves the following theorem. Another proof was in the Chapter 7 exercises. Theorem 8.50 (Concurrence of Medians). For any △𝐴𝐵𝐶 the medians 𝐴𝐷, 𝐵𝐸, and 𝐶𝐹 concur at a point 𝐺 such that 1/3 = 𝐷𝐺/𝐷𝐴 = 𝐸𝐺/𝐸𝐵 = 𝐹𝐺/𝐹𝐶. A

E

F

O

G

B C

D

B

Figure 41. Dilation Image of the Circumcircle

Centers of Dilation and the Midpoint Triangle

153

We know from Theorem 7.8 that every triangle has a circumcircle passing through its three vertices. The center of this circle, the circumcenter 𝑂, is the point of concurrence of the perpendicular bisectors of the three sides, as shown in Figure 41. Now when the dilation 𝒟𝐺,−1/2 is applied to △𝐴𝐵𝐶 and its circumcircle, the image of the circumcircle of △𝐴𝐵𝐶 is the circumcircle of △𝐷𝐸𝐹, a circle of radius 𝑟/2 with center 𝐵 = 𝒟𝐺,−1/2 (𝑂). And the images of the perpendicular bisectors of the larger triangle are the perpendicular bisectors of △𝐷𝐸𝐹. These points and circles are shown in Figure 41. But the perpendicular bisectors of △𝐴𝐵𝐶 are also special lines in the smaller triangle. For example, the perpendicular bisector of 𝐵𝐶 is the line through the midpoint 𝐷 of 𝐵𝐶 that is perpendicular to 𝐵𝐶. But since 𝐸𝐹 is parallel to 𝐵𝐶, this line is also perpendicular to 𝐸𝐹, and this line is an altitude of △𝐷𝐸𝐹 through 𝐷. Therefore, the three concurrent perpendicular bisectors of △𝐴𝐵𝐶 are also concurrent altitudes of △𝐷𝐸𝐹.

A E O

G B

F

C

H

D B Figure 42. Perpendicular Bisectors of △𝐴𝐵𝐶 Dilated to Altitudes of △𝐴𝐵𝐶

Since △𝐷𝐸𝐹 is similar to △𝐴𝐵𝐶 this means that the altitudes of △𝐴𝐵𝐶 are the dilation images of the altitudes of △𝐷𝐸𝐹 and so are also concurrent. The similitude that maps △𝐷𝐸𝐹 to △𝐴𝐵𝐶 is the inverse of the similitude in the other direction, so it is 𝒟𝐺,−2 . The dilation 𝒟𝐺,−2 maps perpendicular bisectors of △𝐴𝐵𝐶 to the altitudes of △𝐴𝐵𝐶 concurrent at the image of 𝑂, which is labeled 𝐻. The point 𝐻 is called the orthocenter of the triangle. Recall that the concurrence of altitudes was already encountered in the solution of the Fagnano problem as shown in Figure 6.16. This is shown in Figure 42; let us walk through the details of this figure.

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The black triangle △𝐴𝐵𝐶 and the gray interior of the midpoint triangle △𝐷𝐸𝐹 are shown. In this example, the triangles are obtuse, so the blue perpendicular bisectors of △𝐴𝐵𝐶 are concurrent at a point 𝑂 outside of the triangle. Therefore, 𝑂 is also outside △𝐷𝐸𝐹. The image of these blue altitudes of △𝐷𝐸𝐹 when mapped by 𝒟𝐺,−2 is the set of red altitudes of △𝐴𝐵𝐶, with the point of concurrence 𝐻 exterior to △𝐴𝐵𝐶. This figure points out that concurrence of altitudes is true for all triangles and not just acute ones since the concurrence of perpendicular bisectors and the existence of the circumcircle of a triangle is true for any triangle. This discussion is a proof of this concurrence theorem for any triangle. Theorem 8.51 (Concurrence of Altitudes of a Triangle). For any triangle △𝐴𝐵𝐶 the altitudes are concurrent at a point 𝐻, called the orthocenter of the triangle. There is one more act to this drama. Figure 42 also shows the segment containing two circle centers and two points of concurrence. The relationship is shown in more detail in Figure 43. O

B G

H

Figure 43. Spacing of Points on the Euler Line

Consider the relationship among these points of concurrence. First, the center 𝐵 of the circumcircle of △𝐷𝐸𝐹 is 𝒟𝐺,−1/2 (𝑂). Therefore, 𝑂, 𝐺, and 𝐵 are collinear with signed ratio 𝐺𝐵/𝐺𝑂 = −1/2. Next, 𝐻 = 𝒟𝐺,−2 (𝑂) so 𝐺𝐻/𝐺𝑂 = −2. All four points are on a line, in the order 𝑂, 𝐺, 𝐵, 𝐻. The distances are ‖𝐺𝑂‖ = 2‖𝐺𝐵‖ and ‖𝐺𝐻‖ = 4‖𝐺𝐵‖, so ‖𝐻𝑂‖ = 6‖𝐺𝐵‖, since 𝑂 and 𝐻 are on opposite sides of 𝐺. This means the midpoint of 𝑂𝐻 is 𝐵, so 𝐻𝐵/𝐻𝑂 = 1/2. This says that 𝐻𝐺 divides 𝐵𝑂 harmonically! As we learned in Theorem 8.19, these are centers for dilations that map a circle of radius 𝑟 centered at 𝑂 to a circle of radius 𝑟/2 centered at 𝐵. We have already examined how dilation 𝒟𝐺,−1/2 maps the circumcircle of △𝐴𝐵𝐶 to the circumcircle of △𝐷𝐸𝐹. But now we see that 𝒟𝐻,1/2 also maps 𝑂 to 𝐵 with ratio 1/2, so it also maps the large circumcircle to the smaller one. This dilation highlights more special points on the circumcircle of △𝐷𝐸𝐹. Let △𝐴′ 𝐵 ′ 𝐶 ′ be the image of △𝐴𝐵𝐶 by 𝒟𝐻,1/2 . This new triangle has vertices on the circumcircle of △𝐷𝐸𝐹 and sides parallel to the corresponding sides of △𝐷𝐸𝐹 since they are related by dilations, but the directions of the segments are reversed. So the triangles are related by a half-turn, specifically, △𝐴′ 𝐵 ′ 𝐶 ′ = ℋ𝐵 (△𝐷𝐸𝐹). One can see this from the product of two dilations, since △𝐴′ 𝐵 ′ 𝐶 ′ is the image of △𝐷𝐸𝐹 by 𝒟𝐻,1/2 𝒟𝐺,−2 . This product is ℋ𝐵 since it is a dilation with ratio −1 that has 𝐵 as a fixed point. The locations of the points 𝐴′ , 𝐵 ′ , 𝐶 ′ are simple to describe: they are the midpoints of 𝐻𝐴, 𝐻𝐵, 𝐻𝐶.

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155

A A' C1

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B1

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C'

B

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O A1

B' D

B

Figure 44. Nine Points on the Circle

A final set of three notable points consists of the feet of the altitudes of △𝐴𝐵𝐶, labeled 𝐴1 , 𝐵1 , 𝐶1 in Figure 44. These are the vertices of the orthic triangle. To see that these points are also on the circumcircle of △𝐷𝐸𝐹, consider that each segment 𝐷𝐴′ , 𝐸𝐵 ′ , 𝐹𝐶 ′ is a diameter of this circle, since the endpoints are related by a half-turn with center at 𝐵. But also, ∠𝐷𝐴1 𝐴′ is a right angle, since 𝐴𝐴! is an altitude. This right angle has a diameter 𝐷𝐴′ as hypotenuse, so it is inscribed in the circle with this diameter by the Right Angle Locus Theorem, Theorem 7.20. The same reasoning applies to the other feet of altitudes. Pulling this together, there are nine special points 𝐷, 𝐸, 𝐹, 𝐴′ , 𝐵 ′ , 𝐶 ′ , and 𝐴1 , 𝐵1 , 𝐶1 on this midpoint circumcircle — the midpoints, the feet of altitudes, and the midpoints of segments from orthocenter 𝐻 to each vertex. For this reason the circumcenter of the midpoint triangle is sometimes called the Nine-Point Circle of △𝐴𝐵𝐶, though it is also called the Euler Circle. Finally, the line containing the centers 𝑂, 𝐺, 𝐵, 𝐻 is called the Euler Line, after the mathematician who figured out the relationship among them. In this section we have seen how one dilation leads to complex relationships among three kinds of concurrent lines in a triangle and special centers in a triangle.

Exercises and Explorations 1. (Lengths in Squares and Equilateral Triangles). Use the Pythagorean Theorem to find the length of a diagonal of a square with side 𝑠. Also find the length of an altitude of an equilateral triangle with side 𝑠.

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2. (Diagonal of a Box). Consider a rectangular box with base a rectangle of sides 3 and 4 inches. If the height of the box is 5 inches, what is the distance between any two opposite corners of the box. 3. (Similar Subrectangles). (a) Let 𝐴𝐵𝐶𝐷 be a rectangle with 𝑀 the midpoint of 𝐴𝐵 and 𝑁 the midpoint of 𝐶𝐷. If 𝐴𝑀𝑁𝐷 is similar to 𝐴𝐵𝐶𝐷, what is the ratio of length to width of 𝐴𝐵𝐶𝐷. (b) In the previous case, the rectangle was similar to a rectangle obtained by folding it in half. What is the ratio of length to width if the rectangle is divided into three congruent rectangles (or 𝑛 rectangles), each similar to 𝐴𝐵𝐶𝐷, as in Figure 45?

D

N

N'

C

A

M

M'

B

Figure 45. Subrectangles Similar to the Large Rectangle

4. (Golden Rectangle). In Figure 46 the rectangle 𝐴𝐵𝐶𝐷 is similar to the subrectangle 𝐵𝐶𝐹𝐸, where 𝐴𝐸𝐹𝐷 is a square. Prove that the ratio of length to width is the golden ratio 𝜙.

D

F

C

A

E

B

Figure 46. Golden Rectangle

5. (Ratios in a Triangle). (a) In Figure 47 the signed ratios 𝐴𝐵′ /𝐴𝐵 = 𝐴𝐶 ′ /𝐴𝐶 = 1/3. Let 𝐷 be the intersection of 𝐵𝐶 ′ and 𝐶𝐵 ′ . If 𝑀 and 𝑁 are the midpoints of 𝐵𝐶 and 𝐵 ′ 𝐶 ′ , tell why all the points 𝐴, 𝐷, 𝑁, and 𝑀 are on 𝐴𝑀. What is the ratio 𝐷𝑀/𝐴𝑀? (b) If instead of 1/3, the equal ratios 𝐴𝐵 ′ /𝐴𝐵 = 𝐴𝐶 ′ /𝐴𝐶 = 2/3, what is 𝐷𝑀/𝐴𝑀?

Exercises and Explorations

157

A N

B'

C'

D

B

M

C

Figure 47. Ratios in a Triangle

6. (Centers of Dilation of Two Circles). (a) Draw two disjoint circles of different radius and construct the centers of dilation and the common tangents for three cases: (a) intersecting circles, (b) tangent circles, (c) one circle contained within the other. (b) Draw two cases of two congruent circles (one case of disjoint circles and one case intersecting) and construct the center(s) of dilation. What is the dilation ratio? What are the common tangents in each case and how are they constructed? (c) Construct two concentric circles, one with twice the radius of the other. What are the center(s) of dilation? What are the dilation ratios? (d) Explain more fully what was meant by the remark concerning Figure 16 that for two circles there cannot be a center of dilation that is on or within one circle and not the other. 7. (Rectangle Dilation Centers). Let 𝐴𝐵𝐶𝐷 be a rectangle and 𝐴′ 𝐵′ 𝐶 ′ 𝐷′ be another rectangle with corresponding sides parallel as in Figure 48, with ‖𝐴𝐵‖/‖𝐴𝐷‖ = ‖𝐴′ 𝐵 ′ ‖/‖𝐴′ 𝐷′ ‖. Show that there are two points 𝑃 and 𝑄 that are the centers of dilations mapping 𝐴𝐵𝐶𝐷 to 𝐴′ 𝐵 ′ 𝐶 ′ 𝐷′ . (These points will be distinct except for a few special positions of the rectangles.) Explain why there is only one dilation center for triangles with respective sides parallel, but there are two for these rectangles. D

C D'

C'

A' A

B' B

Figure 48. Rectangles in Parallel

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8. Dilations and Similarity

A

B

B'

A' D

C'

C

Figure 49. A Subtriangle with Collinear Bases and Parallel Sides

8. (Dilated Triangle). In Figure 49 there are triangles △𝐴𝐵𝐶 and △𝐴′ 𝐵 ′ 𝐶 ′ with 𝐵′ 𝐶 ′ contained in 𝐵𝐶 and the other corresponding sides parallel. If ‖𝐵𝐵 ′ ‖ = 2, ‖𝐵′ 𝐶 ′ ‖ = 1, and ‖𝐵𝐶‖ = 7, find ‖𝐵𝐷‖, where 𝐷 is the intersection of 𝐴𝐴′ and 𝐵𝐶. 9. (Circles and Geometric Mean). Let 𝐴𝐵 be the diameter of a circle and 𝐶𝐷 be a chord perpendicular to 𝐴𝐵 at a point 𝑃. (a) If ‖𝐴𝑃‖ = 2 and ‖𝑃𝐵‖ = 7, what are the lengths ‖𝐶𝑃‖ and ‖𝐷𝑃‖? (b) The geometric mean of two positive real numbers 𝑎 and 𝑏 is √𝑎𝑏. Based on the previous example, describe a geometrical construction of the geometric mean of two lengths. (c) Carry out an example of this construction to find the geometric mean of 3 cm and 5 cm. (d) Explain how this construction proves that the geometric mean is always less than the arithmetic mean (𝑎 + 𝑏)/2 except when 𝑎 = 𝑏.. 10. (Cyclic Quadrilaterals). A quadrilateral is said to be cyclic if it can be inscribed in a circle. Prove that a quadrilateral is cyclic if and only if the opposite angles of the quadrilateral are supplementary. Then tell whether a quadrilateral must be cyclic if one pair of opposite angles is supplementary. 11. (Miquel Theorem). Given △𝐴𝐵𝐶, let 𝐴′ , 𝐵 ′ , 𝐶 ′ be points on sides 𝐵𝐶, 𝐶𝐴, 𝐴𝐵, respectively. Then the circumcircles of △𝐴𝐵 ′ 𝐶 ′ , △𝐵𝐶 ′ 𝐴′ , and △𝐶𝐴′ 𝐵 ′ are concurrent at a point 𝑃. (a) Prove this theorem (the Miquel Theorem). Hint: Let 𝑃 be an intersection point of two circumcircles and then look for a cyclic quadrilateral. (b) Discuss how Figure 32 is a figure for the Miquel Theorem except that it starts with circles and ends up with a triangle. (c) Consider the triangle formed by the three circumcircle centers in the Miquel Theorem figure. How are the angles of this triangle related to the angles of △𝐴𝐵𝐶? The Miquel Theorem is the center of a very rich collection of relationships between triangles and special points (e.g., see Coxeter-Greitzer [5, page 62]).

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159

12. (Parallel Chords). Let two circles intersect at points 𝑃 and 𝑄 and let one line through 𝑄 intersect the first circle at 𝐴 and the second circle at 𝐶. Let a second line through 𝑃 intersect the first circle at 𝐵 and the second circle at 𝐷. Prove that 𝐴𝐵 and 𝐶𝐷 are parallel. Two cases are shown in Figure 50. C

A Q

B

P

C

A

Q

P

D

D

B

Figure 50. Lines Through Circle Intersections and Chords

13. (Harmonic Division and Orthogonal Circles). (a) Draw and label a triangular figure with the same ratios as in Figure 47. (b) Explain why 𝐴𝐷 divides 𝑀𝑁 harmonically. (c) Construct a circle with diameter 𝑀𝑁 and a circle through 𝐴 and 𝐷. (d) Draw the radii from the centers of the circles to the points of intersection and check that they are perpendicular. (e) Explain why the circles are perpendicular. 14. (Orthogonal Circles). Draw two disjoint circles of different radius and construct the radical axis of the two circles. Then construct two circles orthogonal to both of the original circles. 15. (Radical Center). Draw three circles with noncollinear centers and construct a circle orthogonal to all three original circles. 16. (A Transformation Group). Let 𝐸 be the set of all similitudes that map each line 𝑚 to itself or to a parallel line. (a) Prove that 𝐸 is a transformation group. (b) Show that 𝐸 is the set of similitudes that are either dilations or translations.

Chapter 9

Area and Its Applications

Areas of Triangles and Parallelograms Area, a familiar and useful topic in the real world, is not only an interesting topic in geometry; it also provides an additional tool for understanding geometrical relationships. While it is simple in some respects, it is not so simple in others. The simple part includes the familiar formulas for areas of rectangles, triangles, parallelograms, and other basic polygons. The not-so-simple part is expressing a general definition of area and then proving it agrees with the simple formulas but also applies to shapes such as circles and other nonpolygonal shapes. To derive an area formula for circles, limits are involved. This is as far as we will be going in this book.1 Area should have these properties: • Congruent figures have the same area. • For a square 𝑆 of side length 1, the area 𝒜(𝑆) = 1. • If the set 𝑃1 contains 𝑃2 , then 𝒜(𝑃1 ) ≥ 𝒜(𝑃2 ). • If a set 𝑃 is the union of a finite number 𝑃1 , 𝑃2 , . . . , 𝑃𝑘 of convex polygons and their interiors such that the polygon interiors do not intersect, then 𝒜(𝑃) = 𝒜(𝑃1 ) + 𝒜(𝑃2 ) + ⋯ + 𝒜(𝑃𝑘 ). Squares and the Definition of Area. All our work with area will rely on the areas of squares. Consider how only one definition makes sense. A square of integer side 𝑚 can be divided into 𝑚2 unit squares of side 1, so the area should be 𝑚2 . For example, the square on the left of Figure 1 is 9, assuming the small squares have side 1. If the area of a square of side 1 is divided into squares of side length 1/𝑛, then since there are 𝑛2 such small squares, the area of the small square should be 1/𝑛2 . The area of 1 If one investigates really complicated sets in the plane, such as fractals, it turns out that a set may have no area, or it may have more than one value competing to be area.

161

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9. Area and Its Applications

Figure 1. Areas of Squares

a square of side length 𝑚/𝑛 should be (𝑚/𝑛)2 since there are 𝑚2 squares of side length 1/𝑛 and area 1/𝑛2 in this square. This reasoning shows that the area of a square of side 𝑠 should be 𝑠2 if 𝑠 is rational. This is illustrated for 𝑚/𝑛 = 6/10 = .6 by the square on the right in Figure 1, For irrational 𝑠, the number can be approximated arbitrarily closely by decimal fractions. For example, if a decimal approximation of 𝑠 is .628, then dividing the side of the unit square into squares with sides 1/10 will place a square with side 𝑠 between squares with sides .6 and 7, so the area of this square will be between .36 and 49, not a very close estimate. Dividing the unit square into squares with sides 1/100 puts a square with side 𝑠 between squares with sides .62 and .63, so the area is between .622 = .3844 and .632 = .3969, so this produces the area accurate to a couple of decimal places. For more decimal places, any part of the decimal expansion of the area of the square of side 𝑠 can thus be computed by squaring a sufficiently long decimal expression for 𝑠. This leaves 𝑠2 as the only possible area that conforms to the second property above. For computing the area of other shapes, there are two paths one can take. The most general path is to define area by filling up the shape as much as possible with small squares and adding up the areas of the small squares. Taking ever smaller squares gives a larger and more accurate approximation of the area, much as we did for squares with irrational sides above. Taking a limit will produce a value for the area, at least for familiar geometric shapes. The full development of this general approach requires technical arguments that would be distracting from the main thread of this book, so we will only touch on this approach a few times, assuming some statements without proof. Instead, we will take an older and geometrically more interesting approach, which is to cut shapes apart and re-assemble them into other shapes whose areas are known.

Area of Rectangles. Shapes that can be assembled from squares, like the rectangle on the left in Figure 2, will have an area equal to the sum of the area of the squares, in this case 3×4 = 12. If squares are subdivided into smaller squares, such as a division into 4 smaller congruent squares, then the small squares have an area of 1/4 for each small square. Since the shaded figure inside the rectangle on the right contains 24 such squares, it has an area of 6. If a rectangle has sides 𝑎 and 𝑏 that are rational, then by choosing common denominators, one can divide such rectangles into squares and get the formula that area equals 𝑎 × 𝑏.

Areas of Triangles and Parallelograms

163

Figure 2. Area of Rectangle and of Figure Built from Squares

For the general case, a limit argument is not hard, but there are direct proofs using dissection and the area formula for a square. In Figure 3, a rectangle 𝑅 with sides 𝑎 and 𝑏 is placed in a rotationally symmetric way into a square with sides 𝑎 + 𝑏. Then the sum of the areas of the four rectangles is the difference of the areas of squares with sides 𝑎 + 𝑏 and 𝑎 − 𝑏, so 4𝒜(𝑅) = (𝑎 + 𝑏)2 − (𝑎 − 𝑏)2 = 4𝑎𝑏. Therefore, 𝒜(𝑅) = 𝑎𝑏.

b a

Figure 3. Rectangle Area as the Difference of Square Areas

Area of Parallelograms. For a parallelogram 𝑃, the area formula of length of base times height, 𝒜(𝑃) = 𝑏ℎ, is usually justified with a figure like Figure 4. Here a triangle is cut off and translated to the other end of the parallelogram to form a rectangle with the same base and the same height, which is the distance between the parallel lines. This is certainly an adequate figure to get students to remember the formula and it covers many cases, but it does not really cover the general case. Consider the parallelogram in Figure 5 where neither perpendicular segment from the top side to the bottom meets the base. Since we are supposed to be able to use the formula for whichever side we choose as the base, it is not a solution to choose the longer side as base. There is, however, another reasonably simple solution as shown in the Figure 5. The entire parallelogram is enclosed in a rectangle of area 𝑑ℎ, as shown. Translate the triangle in the upper left corner to form a rectangle of area (𝑑 − 𝑏)ℎ. Then the difference in areas is 𝑏ℎ. In this proof, it is not clear how the parallelogram can be cut apart and assembled into the rectangle. To see such a dissection, there is another visualization (shown in Figure 6) that calls for as many translations of the parallelogram as needed so that the

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9. Area and Its Applications

D

A

C

E

B

F

Figure 4. Area of Parallelogram d F

D

C h

A

B

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h b

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c

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Figure 5. Area of a More General Parallelogram

rectangle formed by the perpendicular segments at the vertices on one parallel side are filled with pieces of the translated parallelograms. This idea of filling the space between two parallel lines provides a way for dissecting one shape into another. D C

A

A

B

D

B C

C'

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Figure 6. Assembling a Rectangle from a More General Parallelogram

Area of a Triangle. There are a number of dissections and assemblies that show the formula for the area of a triangle. A triangle can be doubled by a half-turn into a parallelogram (or a parallelogram can be halved by a diagonal into two congruent triangles). Any of the three sides of a triangle can be designated the base. Then the distance from the other vertex to the line of the base is the altitude of the triangle for this base. The area has the well-known formula 𝐴 = 𝑏ℎ/2. Figure 7 shows two triangles. In △𝐴𝐵𝐶 the foot 𝐷 of the altitude 𝐶𝐷 is on a side of the triangle. In △𝐸𝐹𝐺, the foot of the altitude 𝐺𝐻 is on line 𝐸𝐹 but outside the

Area Proofs of the Pythagorean Theorem

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Figure 7. Area of Triangle from Parallelogram Area

side 𝐸𝐹. In either case, the area is one-half the base times the altitude: 𝒜(△𝐴𝐵𝐶) = ‖𝐴𝐵‖‖𝐶𝐷‖/2 and 𝒜(△𝐸𝐹𝐺) = ‖𝐸𝐹‖‖𝐺𝐻‖/2. Instead of doubling the triangle, it can be cut apart and re-assembled as a rectangle, as shown in Figure 8.

Figure 8. Area Formula for a Triangle from Dissection to a Rectangle

The area of a kite can be deduced from the area of a triangle. This is posed as a problem in the exercises.

Area Proofs of the Pythagorean Theorem There are many area proofs of the Pythagorean Theorem. If 𝑇 is a right triangle with hypotenuse length 𝑐 and the other side lengths 𝑎 and 𝑏, the Pythagorean relationship 𝑐2 = 𝑎2 + 𝑏2 can be considered as an equation of areas. It is often represented by a picture like Figure 9, with the triangle 𝑇 and three squares attached to the sides. The area of the square on the hypotenuse equals the sum of the areas of the other two squares.

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Figure 9. Pythagorean Theorem Expressed as an Area Equation

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Figure 10. Dissection and Area Proof of the Pythagorean Theorem

Figure 10 shows a famous and venerable area proof. Let 𝑇 = △𝐴𝐵𝐶, a right triangle with hypotenuse 𝐴𝐵 of length 𝑐 and with leg lengths 𝑎 and 𝑏. Arrange four triangles congruent to 𝑇 in the corners of a square of side 𝑎 + 𝑏 in a rotationally symmetric manner. This is shown on the left side of the figure. (The triangles are arranged as the rectangles were arranged in Figure 3.) By rotational symmetry, the quadrilateral 𝐵𝐴𝐵 ′ 𝐴′ formed by the four hypotenuses is a square with side length 𝑐. The area of this square is the difference of the area of the big square less the four triangle areas. On the right of Figure 10 is the same square with the same four triangles repositioned so that the empty space consists of two squares, one of side length 𝑎 and the other 𝑏. Therefore, 𝑐2 = 𝒜(𝐵𝐴𝐵 ′ 𝐴′ ) = 𝑎2 + 𝑏2 . Notice that the geometrical figure on the right is an area model for the algebraic equation (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏2 . A3 B3

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Figure 11. Perigal’s Dissection Proof of the Pythagorean Theorem

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Finally, in Figure 11 there is a remarkable dissection area proof of the Pythagorean Theorem due to Perigal. This figure is full of relationships that can be explained by rigid motions. There are rotations within squares that establish congruences of segments and angles. Then there are translations that relate the distances in 𝑊𝑋𝑌 𝑍 to distances in the largest square and translations that map the four parts of the square on side 𝐴𝐶 to the congruent quadrilaterals in the square on the hypotenuse. In the exercises, there will be some suggestions about how to prove that this relationship of dissections is correct. Also, this decomposition will appear again in Figure 10.22 arising from a symmetric pattern and tessellation in the plane.

Area and Scaling Since congruent figures have the same area, mapping by a rigid motion does not change the area of a plane figure. But a similitude with scaling factor 𝑘 does change the area; in fact, it scales it by the ratio 𝑘2 . Again, starting with a square 𝑆 of side 𝑠, if 𝑈 is a similitude with scaling factor 𝑘, the image of the square will have side length 𝑘𝑐. So it will have area (𝑘𝑐)2 = 𝑘2 𝑐2 = 𝑘2 𝒜(𝑆). Likewise, the 𝑈 image of a rectangle with side lengths 𝑎 and 𝑏 will be a rectangle with side lengths 𝑘𝑎 and 𝑘𝑏, so the area of the image is 𝑘𝑎𝑘𝑏 = 𝑘2 𝑎𝑏, which is 𝑘2 times the area 𝑎𝑏 of the original rectangle. The formulas for the areas of parallelograms and triangles, being based on the formula for rectangles, also show the lengths, including the altitude scaling by 𝑘 and the area by 𝑘2 . One can also get a vivid picture of scaling for triangle area by examining triangles that are partitioned by similar triangles scaled by ratio 1/𝑛. Figure 12 shows examples of such triangles for 𝑛 = 2, 3, 4, 5. For these examples, count the number of triangles when the large triangle 𝑇 is scaled by ratio 1/𝑛. This indicates that the area of each small triangle is (1/𝑛)2 × 𝒜(𝑇).

Figure 12. Dissection of a Triangle by Similar Subtriangles

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The scaling of triangle area can be illustrated by a triangle dissection as shown in the Figure 13. The shaded triangle is dilated to the large triangle 𝑇 by a scale factor of 10/7. One can check by counting that the area of the shaded area is 49 small triangle areas and the area of 𝑇 is 100; so the area is scaled by the ratio (10/7)2 .

Figure 13. Dilation of the Shaded Triangle by Scale Factor 10/7

The area of a shape defined as the limit of sums of areas of squares contained within the shape also scales by 𝑘2 since the area of each square scales by this factor. This leads to other pictures illustrating the Pythagorean Theorem. If one takes any shape 𝑃 and attaches figures similar to 𝑃 on each side of a right triangle so that they are scaled proportionally to the length of the sides, then the sum of the areas of the figure on the legs equals the area of the figure on the hypotenuse. The shapes can be quite varied. Three are shown in Figure 14. In each case the area of the largest shape is equal to the sum of the other two areas. A frequent figure in the early days of computer graphics was a teapot, so one can find pictures of teapots on the sides of a right triangle.

Figure 14. Examples of Similar Shapes Attached to the Sides of a Right Triangle

While some shapes seem a bit whimsical, this idea leads to one of the shortest and clearest proofs of the Pythagorean Theorem. When a right triangle 𝑇 is divided by an altitude into two similar subtriangles, then these triangles can be reflected in the legs to attach them on the outside and the whole triangle 𝑇 can be reflected in the hypotenuse, as in Figure 15. The original triangle then has three similar triangles attached with scale factors proportional to the lengths of the sides. The triangle on the hypotenuse has area 𝒜(𝑇), and the other two similar triangles have areas (𝑎/𝑐)2 𝒜(𝑇)

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and (𝑏/𝑐)2 𝒜(𝑇). But the sum of the areas of the smaller triangles is the area of 𝑇 since 𝑇 is divided into triangles congruent to them. Thus, ((𝑎/𝑐)2 + (𝑏/𝑐)2 )𝒜(𝑇) = 𝒜(𝑇), or 𝑎2 + 𝑏2 = 𝑐2 . This proves the Pythagorean Theorem using the scaling of area of similar triangles.

Figure 15. An Area Proof of the Pythagorean Theorem by Paper Folding

One can cut out a paper model of this figure to illustrate this reasoning. The large triangle folds and covers 𝑇 and then the two smaller triangles fold and also cover 𝑇.

Area and the Circle So far all the polygonal shapes whose areas we have considered have had triangulations, meaning that the figure can be cut up into triangles that only overlap on the edges. Thus, their areas can be computed by a finite sum of triangle areas. Circles do not have this property; the area of the interior of a circle is the limit of a sequence of areas of polygons approximating the area of the circle.

Figure 16. Approximating a Circle with Regular Polygons

In Figure 16 a circle of radius 1 is approximated on the left by an inscribed and a circumscribed regular octagon. On the right, the number of sides is doubled and the inscribed and circumscribed polygons have 16 sides. The areas of the inscribed polygons are less than that of the circumscribed polygons. It is clear that the areas of the inscribed 2𝑛 -gons are increasing, since each one is

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contained in the next. Likewise the areas of the circumscribed polygons are decreasing since each one is contained in the previous one. Each inscribed polygon is a union of congruent isosceles triangles. Figure 17 shows the detail of one such triangle, along with the circumscribed polygon, as the number of sides is doubled. (The central angle has been made much bigger for visibility.) Suppose the polygon has 2𝑛 sides. In this figure, the segment 𝐴1 𝐴2 is a side of the inscribed polygon and 𝑃𝐴1 and 𝑃𝐴2 are each one-half of a side of the circumscribed polygon. The figure 𝑂𝐴1 𝑃𝐴2 is a kite with right angles at 𝐴1 and 𝐴2 . Therefore, the angles at 𝑂 and 𝑃 are supplementary. The area of the isosceles triangle △𝑃𝐴1 𝐴2 is 2−𝑛 of the difference in area between the inscribed and circumscribed polygons. A2 Q2 O

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Q1 A1 Figure 17. Details of Inscribed-Circumscribed Polygons

Let 2𝜃 = 𝑚∠𝐴1 𝑂𝐴2 . The base angles of △𝑃𝐴1 𝐴2 each have measure 𝜃 by angle sum. If the number of sides is doubled, then there is a vertex 𝐵 for the new inscribed polygon. Then a side of the new circumscribed polygon is 𝑄1 𝑄2 , which is perpendicular to 𝑂𝐵 and so parallel to 𝐴1 𝐴2 . Therefore, 𝑚∠𝑃𝑄1 𝐵 = 𝜃. This angle is an exterior angle of the isosceles triangle △𝐵𝐴1 𝑄1 with measure equal to the sum of the measures of its base angles. This means that 𝑚∠𝐵𝐴1 𝑄1 = 𝜃/2 and so 𝐴⃗ 1 𝐵 is the angle bisector of ∠𝑃𝐴1 𝑀. This proves that ‖𝑀𝐵‖ < ‖𝑃𝐵‖, since the ratio of these segments is the same as the ratio of ‖𝑀1 𝐴1 ‖/‖𝑃𝐴1 ‖ by Theorem 8.18. This, in turn, since 𝑀𝐴1 is parallel to 𝐵𝑄1 , shows that ‖𝐴1 𝑄1 ‖ < ‖𝐴1 𝑃‖/2. Putting this together, we can conclude that 𝒜(△𝐴1 𝑄1 𝐵) < (1/2)𝒜(△𝐴1 𝑃𝐵) < (1/2)𝒜(△𝐴1 𝑃𝑀). The first inequality is true because of the relation between the bases: ‖𝐴1 𝑄1 ‖ < ‖𝐴1 𝑃‖/2. The second inequality is true because one height is less than that other. Since △𝐴1 𝑃𝑀 has one-half the area of △𝐴1 𝑃𝐴2 , the sum of the areas of the shaded triangles is less than one-half the area of △𝐴1 𝑃𝐴2 . Therefore, the difference in areas between the inscribed polygon and the circumscribed polygon is diminished by more than one-half.

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Definition of 𝜋. The bounded and increasing sequence of inscribed areas must have a limit. This limit will be the area of the circle; the area of a circle of radius 1 is given the Greek letter 𝜋 as a name. As we have just seen, the decreasing sequence of circumscribed area also converges to the same limit. The numerical value of this limit can be computed approximately by the areas of these polygons, but there are much more efficient methods for finding vast numbers of digits of the decimal expansion using calculus methods. There are two important aspects of this picture that we will point out in this section. • Any circle of radius 𝑟 is similar to a circle of radius 1 with a scale factor of 𝑟. Therefore, the area is scaled by 𝑟2 . • Since 𝜋 is defined to be the area of the circle of radius 1, the area of a circle of radius 𝑟 must be 𝜋𝑟2 . The second observation concerns the circumference 𝐶 of the circle. This is the first instance of length that we have encountered that is not simply the sum of lengths of segments. The perimeter of the each polygon is simply the sum of the lengths of the sides of the polygon. In the sequence of inscribed and circumscribed regular polygons as in Figure 16 above, again we find that the sequence of inscribed perimeters is increasing with the number of sides and the perimeters of the circumscribed polygons are decreasing. In Figure 17, by the triangle inequality, 2‖𝐴1 𝐵‖ = ‖𝐴1 𝐵‖ + ‖𝐵𝐴2 ‖ > ‖𝐴1 𝐴2 ‖ and ‖𝑄1 𝑄2 ‖ < 2‖𝑃𝑄1 ‖, the latter number being one-half the side length for the 𝐻 circumscribed sides. Thus, the perimeters of the inscribed polygons converge to a number which is a length that we call the circumference of the circle. The circumscribed lengths converge to the same limit; we omit the proof since we do not need this result here. A complete proof is given in Barker and Howe [1]. This reference also considers the inscribed and circumscribed polygons with the number of sides not equal to some power of 2. For a circle of radius 𝑟, each inscribed regular 𝑛-gon is made up of 𝑛 congruent isosceles triangles, with the heights being numbers 𝑟𝑛 that converge to the radius 𝑟 as 𝑛 increases. The base of each triangle is 𝑠𝑛 , the side length of the 𝑛-gon. The area of the inscribed 𝑛-gon is approximately equal to the sum of the areas of these triangles, each of which is 𝑟𝑛 𝑠𝑛 /2, so the sum is (𝑟𝑛 /2)(𝑛𝑠𝑛 ). The limit of 𝑟𝑛 /2 is 𝑟/2 and 𝑛𝑠𝑛 is the perimeter, whose limit is the circumference 𝐶. Taking a limit gives a second area formula. • The area and circumference are related by 𝜋𝑟2 = 𝑟𝐶/2. • This implies 𝐶 = 2𝜋𝑟. Therefore, the 2 in the relation between the area and circumference formulas is the same 2 found the area formula for a triangle. If the only role of 𝜋 were to be the name of the area of a circle, that would not really say much. But the appearance of 𝜋 in the formula for circumference as well, connecting area and length, makes this a key formula introduces 𝜋 as an important constant.

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In addition to the formula for the entire circumference, the same reasoning provides the measure of length for any arc of the circle in terms of the central angle. This is simply based on the proportion of sides of the inscribed polygon that are also inscribed in the arc. If the central angle of the arc is 𝛼 degrees, then the length of the arc is 𝛼/360 times the entire circumference. So the formula for the arc length is (𝛼/360)2𝜋𝑟 = 𝛼𝜋𝑟/180, where the angle is measured in degrees. In working with arc length, the formula is simpler if the angle is measured in radians, for then the 180 degrees is replaced by 𝜋 radians, and the formula is simply 𝛼𝑟, where the angle is measured in radians. But for angles of polygons, degrees give simpler numbers to work with.

Affine Relationships and Area Triangle Ratios and Area. Since the formula for triangle area is one-half base times height, 𝑏ℎ/2, if either 𝑏 or ℎ is multiplied by a real number 𝑟, so is the area. Here is an example. In △𝐴𝐵𝐶 let 𝐷 be a point on 𝐵𝐶 with 𝐵𝐷/𝐵𝐶 = 1/3. Then area 𝒜(△𝐴𝐵𝐷) = (1/3)𝒜(△𝐴𝐵𝐶).

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Figure 18. Ratios of Areas of Triangles

This reasoning can be applied twice, changing which side is the base. Let 𝐸 be the midpoint of 𝐴𝐵. Then 𝒜(△𝐸𝐵𝐷) = (1/2)𝒜(△𝐴𝐵𝐷), taking 𝐸𝐵 and 𝐴𝐵 as the bases of the two triangles, since the heights are the same: the distance from 𝐷 to 𝐴𝐵. Putting this together with the earlier equation, we conclude 𝒜(△𝐸𝐵𝐷) = (1/6)𝒜(△𝐴𝐵𝐶). Theorem 9.1 (Affine Scaling and Area). Given a △𝐴𝐵𝐶, if point 𝐷 is on 𝐵𝐶 with 𝐵𝐷/𝐵𝐶 = ℎ and point 𝐸 is on 𝐵𝐴 with 𝐵𝐸/𝐵𝐴 = 𝑘 , then 𝒜(𝐸𝐵𝐷) = |ℎ𝑘|𝒜(△𝐴𝐵𝐶). Proof. The proof is essentially the same as the reasoning in the example. First 𝒜(△𝐴𝐵𝐷) = |ℎ|𝒜(△𝐴𝐵𝐶), since |ℎ| is the ratio of the bases, both segments on 𝐵𝐶. Then for triangles △𝐸𝐵𝐷 and △𝐴𝐵𝐷, consider the bases for the area formula to be the sides on 𝐴𝐵. Then again, 𝒜(△𝐸𝐵𝐷) = |𝑘|𝒜(△𝐴𝐵𝐷). Putting the relations together proves the theorem. □

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Affine Geometry. Most theorems in plane Euclidean geometry depend heavily on distance and angle measure, congruence and symmetry. This includes theorems involving perpendicular lines and isosceles triangles. But some theorems, such as Theorem 9.1 above, do not depend on the size or shape of △𝐴𝐵𝐶, only on incidence of points and lines and also signed ratios on lines. The theorems of plane geometry that depend only on incidence and signed ratios are called theorems of affine geometry. A familiar example is this theorem about medians: the medians of a triangle are concurrent at a point that divides each median in the ratio 2:1. This again is a theorem that does not depend on any special metric feature of the triangle. A B'

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Figure 19. Side Ratio 1/3 and Area Ratio 1/2

Another example is shown in Figure 19, a figure that is also in Exercise 5 of Chapter 8. If the ratios 𝐴𝐵′ /𝐴𝐵 = 𝐴𝐶 ′ /𝐴𝐶 = 1/3, then 𝐷, the intersection of 𝐵𝐶 ′ and 𝐶𝐵 ′ , is the midpoint of 𝐴𝑀. Therefore, 𝒜(△𝐷𝐵𝐶) = (1/2)𝒜(△𝐴𝐵𝐶). This is true for any triangle with any angle measures and any side length. The reason that 𝐷 is the midpoint is that 𝒟𝐴,1/3 and 𝒟𝐷,−1/3 are dilations that map 𝐵𝐶 to 𝐵′ 𝐶 ′ . Then 𝐴𝑁/𝐴𝑀 = 1/3 implies that 𝑁𝑀/𝐴𝑀 = 2/3, and 𝐷𝑁/𝐷𝑀 = −1/3 implies 𝐷𝑀/𝑁𝑀 = 3/4. So 𝐷𝑀/𝐴𝑀 = 1/2. Shears and Affine Transformations. Our description of affine geometry has been informal and imprecise so far. In fact, there is a definition of affine geometry based on transformations. These affine transformations preserve points and line incidence and also signed ratios on lines. We will study these transformations further in Chapter 11, but we will give one example here. A shear transformation is an affine transformation with a line of fixed points 𝑚 so that any line 𝑛 parallel to 𝑚 is translated to itself by a distance proportional to the distance of 𝑛 from 𝑚. A shear transformation 𝑆 can be defined by three points, 𝐴, 𝐵, and 𝐷, as illustrated in Figure 20. Let 𝑚, the set of fixed points, be the line through 𝐵 parallel to 𝐴𝐷, and let 𝑆(𝐴) = 𝐷. For any point 𝑃 on 𝐴𝐷, the image 𝑄 = 𝑆(𝑃) = 𝒯[𝐴𝐷] (𝑃). For any 𝑃 not on 𝐴𝐷 or 𝑚, 𝑄 = 𝑆(𝑃) is defined as follows: Let 𝐶 be the point of intersection of 𝑃𝐴 and 𝑚. The image 𝑄 is the point on 𝐶𝐷 such that 𝑃𝑄 is parallel to 𝑚. In other words, 𝐶𝑄/𝐶𝐷 = 𝑟 = 𝐶𝑃/𝐶𝐴, and 𝑆(𝑃) = 𝒯𝑟[𝐴𝐷] (𝑃).

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C Figure 20. The Shear Image 𝑄 of 𝑃

This shear transformation sends lines to lines, though it may not be obvious at this point. In Figure 20, if one begins with △𝐴𝐵𝐶 and △𝐷𝐵𝐶 defining 𝑆, areas within △𝐴𝐵𝐶 are preserved. For example, 𝒜(△𝐴𝐵𝑃) = 𝒜(△𝐷𝐵𝑄) since each is the same fraction of the area of the whole triangle that contains it. Areas within △𝐴𝐵𝐶 are preserved generally since a tessellation of this triangle by a large number of congruent triangles is mapped to such a tessellation of △𝐷𝐵𝐶. In fact, 𝑆 is an area-preserving map of the whole plane since the plane can be tessellated by triangles congruent to △𝐴𝐵𝐶. This property will be explored in more detail when coordinate formulas for shears and other affine transformations are introduced in Chapter 11, on page 241 and following pages. Marion Walters Theorem. A beautiful example of an affine theorem is called the Marion Walters Theorem after the mathematics educator who proved it and also made it an example for mathematical problem-solving. A

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Figure 21. Figure for the Marion Walters Theorem

Each side of the triangle in Figure 21 has two points that divide the side into three equal lengths. Each of the points is connected by a segment to the vertex opposite the side. These six segments form a hexagon in the center of the triangle. The question is this: what is the area of the hexagon compared to the area of the entire triangle? More precisely, what is the ratio of these two areas?

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The answer is left to the exercises. What is intriguing about this theorem (besides the surprise of the answer) is that it lends itself to so many strategies for finding areas of regions of the triangle that are not in the hexagon. And the areas do not depend on the shape of the triangle, for this is an affine relationship.

Exercises and Explorations 1. (Rectangle Area Formula). In Figure 22 there are squares of side 𝑎 and side 𝑏 inside a larger square. Explain how this figure can be used to derive the area formula for rectangles. b b

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Figure 22. Two Squares Inside a Square

2. (Trapezoid Area). Derive a formula for the area of a trapezoid, a quadrilateral with two opposite sides parallel. 3. (Kite Area). Let 𝐴𝐵𝐶𝐷 be a kite with 𝐴𝐵 ≅ 𝐴𝐷 and 𝐶𝐵 ≅ 𝐶𝐷. If ‖𝐴𝐶‖ = 𝑝 and ‖𝐵𝐷‖ = 𝑞, find and prove a formula for the area of the kite in terms of 𝑝 and 𝑞. 4. (Perigal Dissection). Answers to the first three parts of this problem suggest a proof of the correctness of the Perigal dissection in Figure 11. Let the lengths of the sides of the right triangle △𝐴𝐵𝐶 be 𝑎, 𝑏, 𝑐. The 𝑏 square is the square on the side opposite 𝐵, the 𝑎 square is the square opposite 𝐴, and the 𝑐 square is on the hypotenuse 𝐴𝐵. (a) The 𝑏 square is decomposed into four congruent quadrilaterals by two perpendicular segments through the center of the square. Find a parallelogram that explains why the length of each of these segments is 𝑐, hence why the four quadrilaterals fit neatly into the sides of the 𝑐 square. (b) Use symmetry to explain why the “hole” in the middle of the 𝑐 square must be itself a square of some size. (c) Consider how the image of 𝑋𝐴𝐵𝑍 by 𝒯𝑋𝑍 shows a relationship between 𝑎 and the side lengths of the quadrilaterals in the 𝑏 square. This relation can be used to show why the small square in the center of the 𝑐 square has side 𝑎. (d) Put this all together as a proof of the Perigal dissection.

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5. (Midpoint Quadrilateral Area). Let 𝐴𝐵𝐶𝐷 be a quadrilateral. By Theorem 7.19, the midpoint quadrilateral 𝐽𝐾𝐿𝑀 is a parallelogram. Prove that the area of 𝐽𝐾𝐿𝑀 is one-half the area of 𝐴𝐵𝐶𝐷. The proof should be valid for both convex and nonconvex quadrilaterals. 6. (Circle Area and Circumference Estimate). Given a circle of radius 1, inscribe and also circumscribe a hexagon. (a) Show the lower and upper estimates for the area of the circle provided by these two hexagons. (b) Show the lower and upper estimates for the circumference of the circle provided by these two hexagons. (c) If you were seeking a practical estimate for 𝜋 from this work, what working value would you assign to 𝜋 and why? 7. (Ratios of Areas). Figure 23 shows some shaded polygons from the Marion Walters figure, Figure 23. For each shaded polygon, find the ratio of the area of the polygon to the area of the whole triangle. Using some of your results, find the ratio of the area of the complement of the center hexagon to the area of the triangle and thus find the area ratio for the hexagon itself. A

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Figure 23. Polygons in the Marion Walters Figure

8. (Formula for a Shear). Although we have not introduced coordinates yet, the concept of the (𝑥, 𝑦)-plane is familiar to everyone. Let 𝐹(𝑥, 𝑦) = (𝑥 + 𝑑𝑦, 𝑦) for some 𝑑 ≠ 0. (a) Pick a value for 𝑑, such as 𝑑 = 3. On graph paper pick some points and draw arrows from each point to its image by 𝐹 to get an idea of the behavior of 𝐹. In particular, what are the images of the points on the 𝑥-axis and the 𝑦-axis and also the images of points on the lines with equations 𝑥 = 1, 𝑥 = 10, 𝑦 = 1, 𝑦 = 10, and 𝑥 = 𝑦. (b) Is 𝐹 a shear transformation? Explain why your answer is correct.

Chapter 10

Products and Patterns

For thousands of years people have been intrigued by symmetric, repeating patterns in their art and architecture — from Chinese porcelain to Roman tiles, from the complex designs of Islamic art to the repeating lizards of Maurits Escher. Interestingly, despite great differences in style, repeating patterns that cover the plane share an underlying structure. We have the tools in our knowledge of rigid motions to understand why such designs share a finite number of patterns — seventeen for patterns on the whole plane. The goal of this chapter is to introduce and explain how relationships created by composing symmetries shape but also limit such patterns of symmetry. We will begin by describing the remaining products of rigid motions, such as products of rotations with different centers. Then we will apply this knowledge to investigate symmetric patterns on the plane; these investigations will reveal how relationships between rigid motions and their products constrain possible shapes and patterns in the plane. A first example of this is in Figure 1. There are two labeled points, 𝐴 and 𝐵, in the upper-right region of the figure. The figure was generated from a single small square with a face by rotating it with center 𝐴 by angles, 0, 90, 180, and 270, creating the upper right-hand corner square made of four small squares. Then this 2 × 2 square was rotated by the same angles with center 𝐵 to create the 4 × 4 square in the figure. Each square in the figure was produced from 𝑛 𝑚 the small square in the upper right-hand corner by a rigid motion of the form 𝐵90 𝐴90 . Looking at the faces, one sees translations and half-turns and also rotations by multiples of 90 degrees in the lower-left corner. In fact, if one were not told that rotations at 𝐵 were employed, it would seem that the pattern was obtained by translating the original 2 × 2 square by two square side-lengths, to the left and then down. If this figure were rotated further by powers of 𝐴90 and then by powers of 𝐵90 , and on and on indefinitely, the shape would grow without bound. The repetition of this pattern could also be obtained by translation.

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Figure 1. Wallpaper Pattern from Two 90-degree Rotations

If the whole plane were covered in this way, the symmetries would include rotation symmetries with centers that are translates of 𝐴 (indicated by blue points) and also such centers at the translates of 𝐵 (indicated by red points), noting that 𝐴 and 𝐵 are not related by translation symmetries of the pattern. The other square corners are centers of half-turns. What makes all this happen using only products involving rotations about two points? We will begin by studying products of rotations with rotations and then also products with translations.

Products of Rotations Rotations are products of two reflections in intersecting lines. A rotation 𝐴𝛼 , with center 𝐴 and rotation angle 𝛼, can be written as a double reflection many different ways. Two key insights were used to prove Theorem 5.7 about products of rotations with a common center: • In writing 𝐴𝛼 as a double reflection, you can pick as either one of the reflection lines any line through 𝐴 you like, but then the second line is determined. To be more precise, according to Theorem 5.7, if one picks any line 𝑚 through 𝐴, there are lines 𝑝 and 𝑞 so that 𝐴𝛼 = ℛ𝑚 ℛ𝑝 = ℛ𝑞 ℛ𝑚 . • In computing the product 𝐴𝛽 𝐴𝛼 , one can show 𝐴𝛽 𝐴𝛼 = 𝐴𝛼+𝛽 by choosing the second mirror line of the first rotation and the first mirror line of the second to be the same line; they cancel! To analyze the product of rotations with different centers, we can use the same approach to analyze the product 𝐵𝛽 𝐴𝛼 . Again, we will adjust the mirrors so that two

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are identical. This time we must pick the canceling mirror line to be 𝑚 = 𝐴𝐵, since that is the only line that passes through both centers. Given this line 𝑚 as the starting point, then our choices of line 𝑛 through 𝐵 and line 𝑝 through 𝐴 are forced so that 𝐵𝛽 = ℛ𝑛 ℛ𝑚 and 𝐴𝛼 = ℛ𝑚 ℛ𝑝 . Then the product of four reflections is 𝐵𝛽 𝐴𝛼 = ℛ𝑛 ℛ𝑚 ℛ𝑚 ℛ𝑝 = ℛ𝑛 ℛ𝑝 . If the lines 𝑛 and 𝑝 intersect, this is a rotation with center at the point of intersections. If the lines are parallel, the double reflection is a translation. By looking at the geometry of these lines, we can determine precisely which rigid motion it is. Example 10.1. The product 𝐵90 𝐴90 = 𝐶180 , where 𝐶 is the right angle vertex of the isosceles right triangle 𝐴𝐵𝐶 shown in Figure 2. As explained above, we choose our shared mirror 𝑚 to be 𝐴𝐵. The mirrors for a 90degree rotation must meet at a 45-degree angle. Therefore, if line 𝑝 = 𝐴𝐶 and 𝑛 = 𝐵𝐶, the rotations 𝐴90 = ℛ𝑚 ℛ𝑝 and 𝐵90 = ℛ𝑛 ℛ𝑚 . Then compute the product: 𝐵90 𝐴90 = ℛ𝑛 ℛ𝑚 ℛ𝑚 ℛ𝑝 = ℛ𝑛 ℛ𝑝 = ℛ𝐵𝐶 ℛ𝐴𝐶 = 𝐶180 = ℋ𝐶 . The rotation angle at 𝐶 is 180 since the mirror lines at 𝐶 meet at a 90-degree angle and 2 ∗ 90 = 180. Since the sign of a rotation depends on orientation, the orientation of △𝐴𝐵𝐶 is important. In the example just shown, the triangle is oriented in a clockwise or negative direction. The 45-45-90 triangle △𝐴𝐵𝐷 in Figure 2 is oriented in a counterclockwise or positive direction. In this case, we can re-orient the triangle to be the clockwise △𝐵𝐴𝐶, with 𝐴90 𝐵90 = 𝐷180 = ℋ𝐷 . We can also find a translation 𝐵90 𝐴−90 in this figure. The rotation 𝐴−90 = ℛ𝐴𝐵 ℛ𝐴𝐷 , since the direction of the rotation is reversed. This time the product is a translation: 𝐵90 𝐴−90 = ℛ𝐵𝐶 ℛ𝐴𝐵 ℛ𝐴𝐵 ℛ𝐴𝐷 = ℛ𝐵𝐶 ℛ𝐴𝐷 . This translation is equal to ℋ𝐶 ℋ𝐴 = 𝒯𝐴𝐴′ , where 𝐴′ is the reflection of 𝐴 in 𝐵𝐶. This translation can also be denoted 𝒯2[𝐷𝐵] .

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The Rotation Product Theorem. We will now consider the general case. Theorem 10.2 (Product of Rotations). The product of rotations 𝐵𝛽 𝐴𝛼 is a rotation 𝐶𝛼+𝛽 if 𝛼 + 𝛽 ≠ 0 or ± 360. Otherwise, the product is a translation when 𝐵 ≠ 𝐴, or the identity if 𝐵 = 𝐴. The direction of the translation is not parallel to 𝐴𝐵 unless 𝛼 = ±180. Proof. This general proof resembles the example above, but with a choice of angles. Again, let 𝑚 = 𝐴𝐵 be the common line of reflection. Let 𝑝 be the line through 𝐴 so that 𝐴𝛼 = ℛ𝑚 ℛ𝑝 and let line 𝑛 through 𝐵 be the line with 𝐵𝛽 = ℛ𝑛 ℛ𝑚 , as in Figure 3. Then as before 𝐵𝛽 𝐴𝛼 = ℛ𝑛 ℛ𝑚 ℛ𝑚 ℛ𝑝 = ℛ𝑛 ℛ𝑝 . If the lines 𝑛 and 𝑝 intersect at a point 𝐶, then the product is a rotation with center 𝐶. If the lines are parallel, the product is a translation. When trying to visualize the location of 𝐶, the “kite of composition” 𝐴𝐶𝐵𝐶 ′ may help. In Figure 3, 𝐶 ′ is the reflection of 𝐶 in 𝑚, so it is also 𝐴𝛼 (𝐶). Moreover, 𝐵𝛽 (𝐶 ′ ) = 𝐶. This shows in another way that 𝐶 is the fixed point, hence the center, of 𝐵𝛽 𝐴𝛼 . It remains to consider how the angles 𝛼 and 𝛽 affect the result. Using translationinvariant global protractors and polar angles on the plane, Theorem 7.36 shows that a rotation such as 𝐴𝛼 adds 𝛼 to the polar angle of any ray, not just rays with endpoint 𝐴. Therefore, the image of a ray by the product 𝐵𝛽 𝐴𝛼 must have 𝛼 + 𝛽 added to its polar angle. • If 𝛼 + 𝛽 is not 0 or ±360, the product must be a rotation by this angle. • If 𝛼 + 𝛽 equals 0 or ±360, the direction of an image ray will stay the same so the product is a translation or the identity. The direction of this image ray will be the same as the original ray only when this angle sum equals 0 or ±360. The direction of an image ray will stay the same as the original if and only if the product is a translation or the identity. Therefore, in the other cases the product is a rotation 𝐶𝛼+𝛽 for some point 𝐶. Thus the lines 𝑛 and 𝑝 must intersect except when the sum of the rotation angles is 0 or ±360. In Figure 3, where both rotation angles are positive, the angles at 𝐴 and 𝐵 in △𝐴𝐵𝐶 have angle measure 𝛼/2 and 𝛽/2. The marked angle at 𝐶 is an exterior angle, so its C

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Figure 3. Rotation at 𝐶 as the Product of Rotations at 𝐴 and 𝐵

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measure is (𝛼 + 𝛽)/2, as it should be. To see the rotations 𝐶𝛼 and 𝐶 𝛽 in the figure, let 𝑚′ be the line through 𝐶 parallel to 𝑚. Then 𝐶𝛼 = ℛ𝑚′ ℛ𝑝 and 𝐶 𝛽 = ℛ𝑛 ℛ𝑚′ so this confirms a second way that ℛ𝑛 ℛ𝑝 = ℛ𝑛 ℛ𝑚′ ℛ𝑚′ ℛ𝑝 = 𝐶 𝛽 𝐶𝛼 = 𝐶𝛼+𝛽 . A

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Figure 4. Translation from 𝐴 to 𝐴′ as the Product of Rotations at 𝐴 and 𝐵

Next, we consider the direction of the translation when 𝑝 and 𝑛 are parallel. An example is in Figure 4 for rotations 𝐴𝛼 and 𝐵−𝛼 ; the marked mirror angles are congruent alternate interior angles. The translation is the product 𝑇 = ℛ𝑛 ℛ𝑝 . To find the direction, let 𝐴′ = 𝑇(𝐴) so that 𝑇 = 𝒯𝐴𝐴′ . But 𝑇(𝐴) = ℛ𝑛 ℛ𝑝 (𝐴) = ℛ𝑛 (𝐴). The only way that ℛ𝑛 (𝐴) can be on 𝑚 = 𝐴𝐵 is for 𝑛 to be perpendicular to 𝑚. But the parallel 𝑝 must also be perpendicular to 𝑚, which means 𝛼 = ±180. The two rotations are the half-turns ℋ𝐴 and ℋ𝐵 . One final special case not yet mentioned is when 𝛼 = 𝛽 = 0. In this case both rotations and the product are the identity, and one can say 𝐴 = 𝐵. □ For transformations 𝑆 and 𝑇, the conjugation of 𝑇 by 𝑆 is the product 𝑈 = 𝑆𝑇𝑆 −1 . This “transforms” 𝑇 by 𝑆, and 𝑇 = 𝑆 −1 𝑈𝑆 is the inverse conjugation. Corollary. The product of a translation 𝑇 and a rotation 𝐴𝛼 is a rotation 𝐶𝛼 for some 𝐶. The conjugation 𝑇𝐴𝛼 𝑇 −1 = 𝑇(𝐴)𝛼 . Proof. Since 𝑇 = 𝑄180 𝑃180 for some 𝑃 and 𝑄, the product 𝑇𝐴𝛼 is the product of rotations 𝑄180 and 𝑃180 𝐴𝛼 . This is a rotation by 180 + (180 + 𝛼) = 𝛼, The conjugate has 𝑇(𝐴) as a fixed point, so this rotation is 𝑇(𝐴)𝛼 . □

Symmetry and 90-Degree Rotations We have already computed the product of two 90-degree rotations in Example 10.1. What we will explore now is how the triangle in Figure 2 can be used as a modular building block for figures with 90-degree rotational symmetry. Consider that in a clockwise-oriented1 isosceles right triangle △𝐴𝐵𝐶 with right angle at 𝐶, these relations are true: • 𝐶180 = 𝐵90 𝐴90 , • 𝐴90 = 𝐵−90 𝐶180 , • 𝐵90 = 𝐶180 𝐴−90 . 1 Clockwise and counterclockwise depend on an orientation being chosen for the plane. Then an orientationpreserving similitude will take a clockwise triangle to a clockwise triangle.

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Figure 5. Centers of Products of 𝐴90 and 𝐵90

The second and third equations follow from the first. For example, to derive the second equation from the first: 𝐵90 𝐴90 = 𝐶180 implies 𝐵−90 𝐶180 = 𝐵−90 𝐵90 𝐴90 = 𝐴90 . If we have any such 45-45-90 isosceles right triangle, we have these equations. But from the equations, one can fill in a missing vertex. If we have two vertices of such a triangle with the two appropriate rotations located there, the missing 90- or 180-degree rotation at the third vertex is the product of rotations at the known vertices. Assume that 𝐴90 and 𝐵90 are symmetries of a figure 𝐹. Since products of symmetries are symmetries, all products formed by repeatedly multiplying these rotations centered at 𝐴 and 𝐵 will be symmetries. In Figure 5, we see other centers of rotations that must also be symmetries for this reason. We have already seen in Example 10.1 that 𝐶180 and 𝐷180 are products of 𝐴90 ′ and 𝐵90 . From the clockwise △𝐵 ′ 𝐴𝐶, we get 𝐵90 = 𝐴−90 𝐶180 . From △𝐵𝐴′ 𝐶 we get ′ 𝐴90 = 𝐶180 𝐵−90 . Continuing with more triangles, we find symmetries that are 90-degree rotations at 𝐵′ and 𝐵 ″ as well. Centers of 180-degree rotation, half-turns, are at 𝐶, and at 𝐷′ , 𝐷″ , 𝐷‴ . Translations are also products. In Example 10.1, we saw that the translation 𝐵90 𝐴−90 = ℋ𝐶 ℋ𝐴 = 𝒯𝐴𝐴′ is a symmetry of 𝐹. Also, 𝐴−90 𝐵90 = ℋ𝐶 ℋ𝐵 = 𝒯𝐵𝐵′ is a symmetry as well. A clear pattern is emerging of rotation centers and translations that must be symmetries of 𝐹 without knowing anything else about 𝐹. Note. If these computations start to seem complicated, now that we have a pattern for where the centers are located, we can see relations and check our work by following the path of a single point, especially if it will be a fixed point. For example, in Figure 5 ′ ′ ″ ″ we see that ℋ𝐴 (𝐵″ ) = 𝐵 and also 𝐵90 (𝐵) = 𝐵 ″ , so 𝐵90 𝐴180 = 𝐵270 = 𝐵−90 since 𝐵 ″ is fixed. The rotation angle is 270 = 90 + 180. Another example is 𝐴−90 𝐵90 . We see that 𝐵90 (𝐵) = 𝐵 and 𝐴−90 (𝐵) = 𝐵 ′ , so the translation is 𝒯𝐵𝐵′ . Example 10.3. Let’s refer back to our first picture, Figure 1. This figure was built from two 90-degree rotations. When looking for symmetries, remember that the symmetries will not be symmetries of the bounded figure on the page but a figure made by translating and repeating this pattern to cover the whole plane.

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The points colored red and blue are 90-degree centers, and the points colored yellow or green are 180-degree centers just as in Figure 5. See where in this figure the named points of Figure 5 are located. Looking at the relative position of the faces, you can recognize whether the corner of a square is a 90-degree center or a 180-degree center. Also look for the translations that are symmetries. Infinite Graph Paper and 𝑝4 Symmetry. Based on this emerging picture, we will construct an example of a figure with 𝐴90 and 𝐵90 as symmetries. Picture a square tiling of the whole plane or square graph paper of unbounded extent. We can construct such a figure on the plane.

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Figure 6. Infinite Square Tiling ℒ with Some Centers of Symmetry

Start with a square 𝑊𝑋𝑌 𝑍, like the one in Figure 6. Let 𝑇1 = 𝒯𝑊 𝑋 and 𝑇2 = 𝒯𝑊 𝑍 . Then let ℒ be made of two sets of equally spaced parallel lines, the translates 𝑇1𝑚 (𝑊𝐶) (vertical lines in the figure) and the 𝑇2𝑛 (𝑊𝑋) (horizontal lines) for all integers 𝑚 and 𝑛. The lines of ℒ define a tiling by congruent squares. The rotational symmetries of ℒ map lines to lines, so the rotations must have rotation angles ±90 or 180. Every point is located in some square tile, including any center 𝑃 of 90-degree rotation. The image of 𝑃90 of this square must also be a square tile, either the square itself or an adjoining square. In the first case, 𝑃 must be the center of the square. In the second case, it must be a vertex. All the centers of quarter-turn rotational symmetry will also be centers of half-turn symmetry, since if 𝑃90 is a symmetry, so is 𝑃180 . In addition, a half-turn centered at the midpoints of an edges of a square tile will map the tile to a neighbor, so these half-turns are also symmetries. The translation symmetries of ℒ are clearly the products formed from 𝑇1 and 𝑇2 . For any two squares, there is a translation that will map one to the other.

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Now place a point 𝐴 in the center of one square 𝑠0 of this tiling and denote by 𝐵 a vertex of the same square. Then if we begin with 𝐴90 and 𝐵90 and form all possible products of any length of these two rotations, the products will include translations and rotations with centers including the points from Figure 5. The point 𝐴′ will be the center of an adjoining square, The points 𝐵 ′ , 𝐵 ″ , 𝐵 ‴ will be the remaining vertices of the square 𝑠0 with center 𝐴. The points 𝐶 and 𝐷 will be midpoints of edges of 𝑠0 . This is shown in Figure 6, where the earlier Figure 5 is placed inside the square tiling. Now we will prove that the set of symmetries of ℒ and the set of rigid motions that are repeated products of 𝐴90 and 𝐵90 are the same. Theorem 10.4. The set of orientation-preserving symmetries of ℒ , the square tiling of the plane, is the set of rigid motions generated by a 90-degree rotation 𝐴90 with center the center point of a square tile and 𝐵90 , where 𝐵 is a vertex of the same square tile. Proof. We will sometimes refer to 𝐴90 as the 𝐴 generator and to 𝐵90 as the 𝐵 generator of the set of products. Let the set of these product rigid motions be denoted as 𝒫. First, note that both 𝐴90 and 𝐵90 are symmetries of ℒ. Clearly each one maps each set of parallel lines in ℒ to the perpendicular set of parallel lines in ℒ, since the distances to the lines are preserved. Therefore, since products and inverses of symmetries are symmetries, the set 𝒫 of products of the generators must be a subset of the set of symmetries of ℒ. So the only question is whether every symmetry is in 𝒫. The translation symmetries of ℒ are all products in 𝒫 because the generators of these translations, 𝑇1 and 𝑇2 , are in 𝒫. The translation 𝑇1 = 𝒯𝐴′ 𝐴 and 𝑇2 = 𝒯𝐵𝐵′ , each of which has been shown earlier to be a product of the 𝐴 and 𝐵 generators. Next suppose that 𝑃90 is a rotational symmetry of ℒ. The point 𝑃 is in some square tile 𝑠. The square with 𝐴 as center is 𝑠0 . If 𝑇 is the translation that maps 𝑠 to 𝑠0 , then 𝑇𝑃90 𝑇 −1 is a rotation 𝑄90 with center 𝑄 = 𝑇 −1 (𝑃) in the square 𝑠0 by the corollary on page 181. Therefore, 𝑄 must either be 𝐴 or a vertex of 𝑠0 , one of the points 𝐵, 𝐵 ′ , 𝐵 ″ 𝐵 ‴ . This is true because 𝑄 is a fixed point of this product rigid motion, and such a product of translations and a rotation is a 90-degree rotation by Theorem 10.2. But every such 𝑄90 in 𝑠0 is known to be a product of the generators, so it is in 𝒫. The translation 𝑇 is also in 𝒫. Therefore, 𝑃90 = 𝑇 −1 𝑄90 𝑇 is in 𝒫. This shows that every 90-degree rotational symmetry of ℒ is in 𝒫. Once we have these centers, the half-turn symmetries are products of the 90-degree rotations, as in Figure 2, so the theorem is proved. □ Definition 10.5. A plane figure 𝐹 has 𝑝4 rotational symmetry if the rotational symmetries of 𝐹 can be generated by 90-degree rotations with distinct centers. As we have seen, if a figure has 𝑝4 rotational symmetry, the figure comprised of centers of rotation of these symmetries must be a figure similar to the figure made of the centers of rotational symmetry of ℒ. Here are some things we learned. As before, let the 𝐴90 and 𝐵90 be the rotations that generate the rotational symmetries of 𝐹. • There is a tiling of the plane by squares so that 𝐴 is at the center of a square and 𝐵 is at a vertex. The centers of 90-degree rotational symmetry are the vertices and the centers of the squares.

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• The translation symmetries take any square tile to any other. • In terms of these translation symmetries, the set of translates of 𝐴 is the set of square tile centers and the translates of 𝐵 are the square vertices. No member of one of these sets can be translated by a symmetry to a member of the other. • The 180-degree centers at midpoints of sides of square are also divided into two disjoint sets by translation. This explains the color coding in the figures that we have seen so far in the chapter. The points of the same color are the ones that can be translated to each other by translation symmetries of the pattern. This description so far has a flaw: it distinguishes the role of 𝐴90 at the center from 𝐵90 on the corner. But if one looks again at ℒ in Figure 6, one sees that the 𝐴-translates are also arranged into the corners of a square tiling with the 𝐵-translates at the center. This new tiling is congruent to ℒ since the translation 𝒯𝐴𝐵 maps the new tiling to ℒ. If we combine these two sets of lines together into a new figure ℳ, we have a new model of a square tiling by squares of half the side length of the ℒ tiles. So now we have two examples of square tilings of the plane that have 𝑝4 rotational symmetry: • the model ℒ with one set of 90-degree rotations at the center of the squares and generating translations 𝑇1 and 𝑇2 with displacement distance equal to the side of a square, • the model ℳ with all 90-degree rotations at the vertices of the squares and generating translations 𝑇1 and 𝑇2 with displacement distance equal to twice the side of a square. Example 10.6. To make ℳ more concrete, return to Figure 1. Each square of this figure has 90-degree rotations at two corners and two 180-degree centers at the other corners, one from each of the separate sets of each type. We can see this easily if we look at the face designs in the tile. Using the notation of Figure 5, in any tile in which the face is right-side-up, the 𝐴 center is in the upper-right corner and the 𝐵 is in the lower left; the 𝐶 translate is in the upper left and the 𝐷 translate is in the lower right. The translations that are symmetries will only map a tile to another tile with the faces looking in the same direction, so these translations will not map a tile to an adjacent tile. But one can see the ℒ figure in this example also. Instead of thinking of a tile as being a square with a face, now let a tile be a 2 × 2 square arrangement of tiles with four faces, each looking in a different direction. Then a tiling by translating all these tiles is exactly model ℒ. Line Reflections and 𝑝4 Symmetry. Since the products formed from two 90degree rotations with distinct centers must include all the translations and rotations described above, any design on the plane with two 90-degree rotations with distinct centers as symmetries must include the same arrangement of centers and translations described above. In addition, there may be line reflection symmetries as well. One example is the figure ℒ itself. The lines of symmetry are shown in Figure 7. Reflection in any of the

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C

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Figure 7. Lines of Symmetry in 𝑝4 Pattern and 𝑝4 Centers for ℒ

lines making up the figure ℒ itself is a symmetry. But the parallel lines through the 𝐴 centers (the lines in ℳ) are also lines of reflection. And the lines that are diagonals of the tiles of ℒ are lines of symmetry as well. In fact, the lines of symmetry of each square tile are also lines of symmetry of ℒ. Example 10.7. A more visually interesting example of reflection symmetries in a 𝑝4 pattern of rotational symmetries is Figure 8. Find the rotation centers and the translation symmetries, and observe how the mirror lines intersect at the rotation centers (as they must, since the product of two reflection symmetries is a rotation symmetry).

Figure 8. A Pattern from a Cathedral

Again, to speak of 𝑝4 we imagine tiling the plane by translations of this figure. This pattern has the horizontal, vertical, and diagonal lines of symmetry just described. The rotation centers are the intersection points of these lines that meet at 45 or 90 degrees (as they must be, since the product of two reflection symmetries is a rotation symmetry).

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There are two repeating shapes in the design. Any shape can be mapped to a congruent shape by translations. The 𝐴-centers are at the centers of symmetry of one of these shapes and the 𝐵 centers are at the centers of symmetry of the others. The halfturn centers are the points where four shapes come together.

Figure 9. Some Tiles for the Cathedral Pattern

Naming Symmetry Patterns: 𝑝4, 𝑝4𝑚, 𝑝4𝑔. Can a figure with 𝑝4 rotational symmetry have more lines of symmetry than this? Can it have fewer?

Figure 10. Reflections Plus 𝑝4 Rotational Symmetry

Since the reflection of a tile in a line of symmetry must be another tile, a line of symmetry of the whole figure must either be a line of symmetry of square tiles that it intersects or it must reflect the tile to another tile. This rules out any lines beyond those already mentioned. As to whether there can be fewer lines of symmetry, that question is partially answered by the Reflection-Rotation Pairing Theorem, Theorem 5.13. If a line of symmetry passes through a center of rotation symmetry, the number of lines of symmetry must equal the number of rotational symmetries centered at the point. Thus, if a line of symmetry passes through 𝐵90 , then the product of the line reflection with the four distinct powers of 𝐵90 produces four line reflections. So if one of the edge lines or diagonal lines of a square passes through corner 𝐵, then both edge lines and both diagonal lines at 𝐵 will be lines of symmetry. But one of the diagonal lines passes through 𝐴, so there are now four lines of symmetry through 𝐴. So all the lines of symmetry must be present if any are, since the translation lines through any tile translate to lines of symmetry in any of the other tiles. Therefore, a pattern with 𝑝4 rotations and fewer lines of symmetry different from those of ℒ can only have lines that intersect at half-turn centers at the midpoints of the edges of the tiles. This is in fact possible. The example in Figure 11 has the same 𝑝4 pattern of rotation centers as ℒ but also reflection symmetries at lines through the midpoints of the tiles that do not pass through the center or vertices. In this figure the mirror lines are horizontal or vertical

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Figure 11. An Example of 𝑝4𝑔

and the square tiles are at a 45-degree angle from the horizontal. The points are colored by the same color scheme as in the previous figures. There are other figures with symmetries of this type that are not so complicated. One example is in Figure 10. Remember that the squares formed by the reflection lines are midpoint squares of the ℒ tiling. We will be leaving this lengthy exploration of the 𝑝4 pattern of rotational symmetry and moving on to other relationships. But before we do, we need to say something about names. The symmetry groups of such repeating patterns that cover the plane are called wallpaper groups or crystallographic groups. We will say more about them later in the chapter. But these groups have names (sometimes more than one name, unfortunately). We have talked about 𝑝4 rotational symmetry, but to be clear, here are the names of the symmetry groups we have seen. • Type 𝑝4: The symmetry group of a figure consisting of only rotations generated by two 90-degree rotations at two centers 𝐴 and 𝐵. • Type 𝑝4𝑚: The rotational symmetries of a figure are a 𝑝4 but there are also line reflection symmetries through all centers of rotation. • Type 𝑝4𝑔: If the rotational symmetries of a figure are a 𝑝4 but there are also line reflection symmetries that do not pass through the 90-degree centers of rotation.

Triangles and 60 Degrees of Rotation Another important special case of the product of two rotations is 𝐵60 𝐴60 = 𝑃120 . Figure 12 shows an equilateral triangle with its center and midpoints. The product 𝐵60 𝐴60 = ℛ𝐵𝐸 ℛ𝐴𝐵 ℛ𝐴𝐵 ℛ𝐴𝐷 = ℛ𝐵𝐸 ℛ𝐴𝐷 = 𝑃120 . Also, 𝐶60 𝐵60 = 𝑃120 , so 𝐶60 = 𝑃120 𝐵−60 .

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C

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Figure 12. Product of 𝐴60 and 𝐵60 is 𝑃120

Another product to note is 𝐶120 𝐵120 𝐴120 = 𝐼 for an equilateral triangle oriented counterclockwise as △𝐶𝐵𝐴. Since the angle sum is 360, this product is either a translation or the identity. It is easy to see that 𝐶 is a fixed point, so the product is 𝐼. Equivalently, starting from scratch with a segment 𝑍𝑌 , 𝑍120 𝑌120 = 𝑋−120 where the center 𝑋 is the third vertex of an equilateral triangle △𝑍𝑌 𝑋. If we tessellate the plane with congruent equilateral triangles by extending the pattern in Figure 13, the centers of 60-degree rotational symmetry are at the vertices of the triangles, and the centers of the triangles are centers of 120-degree rotation. Since 3 × 60 = 180, the vertices such as 𝐴 are also half-turn centers: 𝐴360 = ℋ𝐴 . In addition, the midpoints of the sides are centers of half-turn symmetry. For example, 𝑃120 𝐴60 = 𝐸180 = ℋ𝐸 .

Figure 13. Part of Triangular Tessellation with Symmetry Lines

Such a tessellation by triangles will also have reflection symmetries, with some lines of symmetry being the sides of triangles and others the perpendicular bisectors of sides. Translation symmetries will take any triangle vertex to any other; e.g., 𝒯𝐴𝐵 = ℋ𝐹 ℋ𝐴 . In contrast, the 120-degree centers in the middle of the triangles are not all translation symmetries of each other; the centers of upward pointing triangles are translates of each other, as are the centers of the downward pointing triangles. In Figure 13 the lines of symmetry through the triangle centers are dashed lines.

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In the triangular tessellation in Figure 13, if one glues together six triangles at a time, one can also have a tessellation by hexagons sharing edges. Or one can leave triangular spaces between hexagons and get a tessellation made of hexagons and triangles sharing edges as in Figure 46 in the exercises. Napoleon’s Theorem. One of the most striking examples showing the tight relationships among rotations and their products is a theorem attributed (probably not very accurately) to Napoleon Bonaparte. Theorem 10.8 (Napoleon’s Theorem). Let △𝐴𝐵𝐶 be any triangle. Construct on each side of the triangle an equilateral triangle that shares that side and label the centers of these triangles 𝐴′ , 𝐵 ′ , 𝐶 ′ . Then △𝐴′ 𝐵 ′ 𝐶 ′ is an equilateral triangle.

C

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Figure 14. Napoleon’s Theorem

The triangles for Napoleon’s Theorem are shown in Figure 14. The proof will use the labels there. Proof. Consider the central rotations in each equilateral triangle. Starting with 𝐴 in ′ the figure, 𝐵120 (𝐴) = 𝐶 and then 𝐴′120 (𝐶) = 𝐵. ′ The product 𝐴′120 𝐵120 = 𝑃240 = 𝑃−120 for some point 𝑃, with 𝑃−120 (𝐴) = 𝐵. As was noted above, this means 𝑃 is the third vertex for an equilateral triangle oriented as △𝐴′ 𝐵 ′ 𝑃. ′ ′ But 𝐶−120 also maps 𝐴 to 𝐵. We claim 𝐶−120 = 𝑃−120 so 𝐶 ′ = 𝑃. −1 ′ ′ This is true because 𝑃−120 𝐶−120 = 𝑃120 𝐶−120 is either the identity 𝐼 or a translation, since 120 − 120 = 0. But this transformation fixes 𝐴, so it cannot be any translation ′ except 𝐼. Therefore, 𝐶120 = 𝑃120 . □

Fermat Point. While we are examining the figure of a triangle with equilateral triangles on the sides, there is another feature of this figure that we should mention. It is illustrated in Figure 15. The segments 𝐴𝐴∗ , 𝐵𝐵 ∗ , 𝐶𝐶 ∗ from a vertex of △𝐴𝐵𝐶 to the farther vertex of the opposite equilateral triangle are concurrent at a point. This point is called the Fermat point of the triangle. It has a number of remarkable properties. But first, we must prove that these segments are concurrent.

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A*

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B* C* A Figure 15. Concurrence of Segments at the Fermat Point

Theorem 10.9. In any triangle △𝐴𝐵𝐶 with each angle measure less than 120, construct on each side of the triangle an equilateral triangle that shares that side. Label the three unnamed vertices 𝐴∗ , 𝐵∗ , 𝐶 ∗ on the sides opposite 𝐴, 𝐵, 𝐶, respectively. Then the segments 𝐴𝐴∗ , 𝐵𝐵 ∗ , and 𝐶𝐶 ∗ are congruent and are concurrent at a point 𝐹, each pair of segments meeting at an angle of 60 degrees. Proof. Orient the triangle as in Figure 16. Each segment such as 𝐴𝐴∗ is a diagonal of a convex quadrilateral formed from two triangles, in this case 𝐴𝐵𝐴∗ 𝐶. Therefore, 𝐴𝐴∗ intersects the other diagonal 𝐵𝐶 at an interior point. An application of the Crossbar Theorem shows that any of the segments 𝐴𝐴∗ , 𝐵𝐵 ∗ , 𝐶𝐶 ∗ intersects each of the others at a point interior to △𝐴𝐵𝐶. These three intersection points are not yet known to be the same. Consider the rotation 𝐴−60 . The vertex 𝐶 = 𝐴−60 (𝐵∗ ) and 𝐶 ∗ = 𝐴−60 (𝐵). Therefore, 𝐶𝐶 ∗ ≅ 𝐵𝐵 ∗ . Let 𝐹 be the intersection of 𝐶𝐶 ∗ and 𝐵𝐵 ∗ . The lines are not parallel because one is the rotation image of the other. In fact the lines form an angle of 60 degrees at 𝐹 by Theorem 7.36. Let 𝑐 1 , 𝑐 2 , 𝑐 3 be the circumcircles of the equilateral triangles, labeled as marked in Figure 16. Applying the Angle Arc Locus corollary to the Inscribed Angle Theorem, Theorem 8.34, we will see that 𝐹 is on each of these circles. Each side of an equilateral triangle divides its circumcircle into a major arc of measure 240 and a minor arc of measure 120. Since 𝑚∠𝐵 ∗ 𝐴𝐶 = 𝑚∠𝐵 ∗ 𝐹𝐶 = 60, the two points 𝐴 and 𝐹 are on the major arc of 𝑐 2 defined by 𝐵 ∗ 𝐶 since the two points are on the same side of 𝐵 ∗ 𝐶. Likewise, 𝐴 and 𝐹 are on the major arc of 𝑐 3 defined by 𝐵𝐶 ∗ . In addition, since 𝑚∠𝐵𝐹𝐶 = 120, the point 𝐹 is on the minor arc of 𝑐 1 defined by 𝐵𝐶. Thus 𝐹 is the point of concurrence of these three circles. But this shows the segments 𝐴𝐴∗ and 𝐵𝐵 ∗ also intersect at 𝐹. If we start afresh and repeat the same reasoning, we conclude that the intersection point of these segments is the point of concurrence of the same three circles and so must also be 𝐹. Thus 𝐹 is on all three segments and is a point of concurrence as claimed. □

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A* c1

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Figure 16. Concurrence of Circles at the Fermat Point

One consequence of this theorem is that the three segments 𝐹𝐴, 𝐹𝐵, and 𝐹𝐶 meet symmetrically at angles of 120 degrees. Such an equiangular picture is associated with minimization and in fact the point 𝐹 is the point that minimizes the sum of the distances to each vertex of △𝐴𝐵𝐶. This will be left to the exercises, with a figure that gives a big hint.

Translations and Symmetry In the examples with 90-degree and 60-degree rotations, we were led to consider unbounded symmetric repetitions of the pattern. Such repeating symmetric patterns over the whole plane are commonly called wallpaper patterns. Symmetric patterns that are unbounded but extend only in one direction, like the one in Figure 17, are called frieze patterns.

Figure 17. A Frieze Pattern

In either case, an essential ingredient in such a pattern is a collection of translations. When repeated, translations move points farther and farther in an unbounded way. This is not true of a single rotation or reflection. It is true of glide reflections, but since the square of a glide reflection is a translation, this means that any unbounded symmetric pattern will still include a group of translations among the symmetries. But there are restrictions on a set of translations for it to be in the symmetry group of such a pattern. To begin with, there must be a minimum displacement distance for the translations. If a figure has arbitrarily small translation symmetries, then the pattern will not be the repeat of a reasonable figure in a region but will be a dense set akin to all the irrational numbers sitting inside the real numbers.

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S (A ) T-2(A )

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Figure 18. 𝑇𝑣 as a Generator of Translations in One Direction

Theorem 10.10 (Frieze Translation Symmetries). Given a wallpaper or frieze pattern, suppose that 𝑉 is a translation among the symmetries with minimum displacement, 𝑑 = ‖𝐴 𝑉(𝐴)‖ > 0 for any 𝐴. Then any translation symmetry with direction parallel to 𝑉 is equal to 𝑉 𝑛 , a power of 𝑉. Since all of the powers 𝑉 𝑛 are symmetries, the set of all translation symmetries with direction parallel to 𝑉 is just the set of the 𝑉 𝑛 . For a frieze pattern, when all translation symmetries have the same direction, these powers are all the translation symmetries. Proof. To see this, suppose that 𝑆 is another translation with direction parallel to 𝑉. Then, taking any point 𝐴, the points 𝑉 𝑛 (𝐴) are equally spaced points in both directions on a line 𝑚𝑉 . This is shown in Figure 18, using the vector notation 𝑉 𝑚 = 𝒯𝑚𝑣 . If 𝑆 is not a power of 𝑉, then the point 𝑆(𝐴) will also be on this line between two of the points, 𝑉 𝑘 (𝐴) and 𝑉 𝑘+1 (𝐴), so 𝑆(𝐴) is closer than 𝑑 to 𝑉 𝑘 (𝐴). But then 𝑉 −𝑘 𝑆(𝐴) is a translation symmetry with smaller displacement than 𝑉, for ‖𝐴 𝑉 −𝑘 𝑆(𝐴)‖ = ‖𝑉 𝑘 (𝐴) 𝑆(𝐴)‖ < ‖𝑉 𝑘 (𝐴) 𝑉 𝑘+1 (𝐴)‖ = 𝑑. This contradicts the assumption that 𝑑 is the shortest displacement. Therefore, all the translation symmetries with the direction parallel to the direction of 𝑉 will be powers of 𝑉. In other words 𝑉 will generate the translations with direction parallel to 𝑚𝑉 . □

T- v +2w(A )

T-2v+2w(A ) T-2v+w (A ) T-2v(A ) T-2v- w (A ) T-2v-2w(A )

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Figure 19. A Lattice Defined by 𝑉 and 𝑊

If there are other translation symmetries which are not powers of 𝑉, let 𝑊 be any such translation. Then 𝑊 maps the line 𝑚𝑉 to a parallel line as shown in Figure 19. In this figure there are shown some images of a point 𝐴 by the powers 𝑉 𝑘 𝑊 ℎ (𝐴) using vector notation, with 𝑉 = 𝒯𝑣 and 𝑊 = 𝒯𝑤 . Then 𝑉 𝑘 𝑊 ℎ = 𝒯𝑘𝑣 𝒯ℎ𝑤 = 𝒯𝑘𝑣+ℎ𝑤 . Now among all the translations that do not map 𝑚𝑉 to itself, choose one 𝑊 so that the distance from 𝑊(𝑚𝑉 ) to 𝑚𝑉 is minimum. As one can see from the figure, 𝑊 is not a unique choice because any 𝑊𝑉 𝑘 = 𝒯𝑤+𝑘𝑣 will map 𝑚𝑉 to the same line.

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After choosing one 𝑊, then all the points 𝒯𝑘𝑣+ℎ𝑤 (𝐴) for integers 𝑘 and ℎ form what is called a lattice. The points form the vertices of a tessellation by parallelograms. Any other translation 𝑆 must be a product of powers of 𝑉 and 𝑊. If not, then 𝑆(𝐴) is not in the lattice defined by 𝑉 and 𝑊, so it must be interior to one of the parallelograms with 𝒯𝑘0 𝑣+ℎ0 𝑤 (𝐴) as a vertex. Then 𝑊 −ℎ0 𝑉 −𝑘0 𝑆(𝐴) must be in one of the four parallelograms with vertex 𝐴. And so 𝑊 −ℎ0 𝑉 −𝑘0 𝑆 maps 𝑚𝑉 to a line closer to 𝑚𝑉 than the image of 𝑊, which contradicts the choice of 𝑊. Each of these parallelograms is a fundamental parallelogram of the symmetry pattern. For a wallpaper pattern, the entire design can be obtained by covering the plane with translated copies of what is in a fundamental parallelogram. However, if there are rotations or other symmetries in the pattern, some choices of parallelogram may be preferred because they fit well with the other symmetries. This appears in the examples so far. In Figure 1 with the faces, any one of the 2 × 2 arrangements of 4 small squares would be a fundamental parallelogram. But this is not the only choice. A nonrectangular parallelogram with a horizontal bottom edge of 2 units and height of 2 units would also be a fundamental parallelogram. However, such a parallelogram would not interact well with the rotations. In this case a small square is a fundamental domain for the pattern since the rotation symmetries can map it to fill out a 2 × 2 fundamental parallelogram. The example in Figure 8 has the same 𝑝4𝑚 rotation symmetries and reflections. The small tile on the left in Figure 9 is a fundamental domain for the 𝑝4 rotation pattern while the large tiles are fundamental parallelograms. When reflections are included, a fundamental domain for 𝑝4𝑚 is a right triangle obtained by dividing the small tile by its line of symmetry. For the triangular example in Figures 12 and 13, a fundamental parallelogram that fits with the rotations would be a rhombus formed by joining two adjacent equilateral triangles at a common edge, since two generators of the lattice are translations from one triangle vertex to another, e.g., 𝒯𝐴𝐵 and 𝒯𝐴𝐶 .

Tessellations and Symmetric Wallpaper Designs This section will offer a small sample of tessellations and symmetric wallpaper patterns, with an emphasis on how the product rules of rigid motions constrain the possible symmetric patterns that cover the plane in a periodic way, i.e., can be built from one basic tile that can be translated to cover the plane. Interested readers are encouraged to explore more in the exercises but also in the extensive literature on the subject. Grünbaum and Shephard [9] is the definitive work on tessellations (including much more general types than our wallpaper patterns). Schattschneider [16] combines a mathematic treatment of wallpaper and other patterns with an analysis of the mathematics in the art of Maurits Escher. Sykes [18] offers a collection of examples of design from architecture, especially medieval cathedrals. Stevens [17] also offers examples from design and architecture arranged by symmetry pattern. The fundamental underpinnings of these designs are based in a small number of tessellations by simple or regular objects. More elaborate schemes are constructed on

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these foundations, but the relationship among symmetries remains among a limited number of possibilities, in fact among 17 possibilities, for it has been proved that there are only 17 crystallographic groups that can be symmetry groups for such wallpaper patterns. Proofs and the classification of these symmetry patterns can be found in Barker and Howe [1, Chapter 8] and Martin [12, Chapter 11]. A key idea in the proof is that a rotation 𝐴360/𝑛 can be a symmetry of a wallpaper pattern only if 𝑛 = 2, 3, 4, 6. The reasons for this are explored in the exercises. We have already encountered tessellations by congruent triangles and parallelograms of any shape, as shown in Figure 7.14 and Figure 19 of this chapter. Various tessellations by rectangles are exhibited by brick walls. The tessellations by a single congruent regular polygon are very important.

Figure 20. Tessellations by Regular Polygons

There are only three: triangles, squares, and hexagons, as in Figure 20. That these are the only tessellations by regular polygons can be seen from their vertex angle measure. At least three polygons have to fit together at a common vertex; if the angles are equal, then each angle must be no greater than 360/3 = 120 degrees. This rules out all regular polygons with more than seven sides. The only other possibility would be regular pentagons, with a vertex angle of 108. However three times this angle is less than 360 and four times the angle is greater than 360; these do not fit together to form a tessellation. A Two-Square Tessellation. The tessellation in Figure 21 is frequently encountered in tile floors. Besides providing an attractive design, it also illustrates the Pythagorean Theorem!

Figure 21. Tessellation by Squares of Two Sizes

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If this pattern is repeated over the whole plane by translation, it seems clear that there are centers of 90-degree rotation at all the centers of the squares. But this means this symmetry pattern is the pattern called 𝑝4. We know then that one fundamental parallelogram is a square with corners at the centers of adjacent large squares and another is a square with vertices at the centers of the small squares. The midpoints of the sides of either set of fundamental squares are the same points; they are the centers of 180-degree rotational symmetry.

A B

C

Figure 22. Tessellation with Fundamental Parallelograms

Figure 22 shows squares with vertices at the centers of the larger squares. The translation symmetries of the tessellation that define this fundamental parallelogram are translations that map any large square tile to any other large square, at the same time mapping every small square tile to another small one. Centers are mapped to centers and vertices are mapped to vertices. One map of vertices is pointed out as the translation 𝒯𝐵𝐴 in the lower left part of the figure. The segment 𝐵𝐴 is part of a right triangle △𝐴𝐵𝐶. If we denote the side length of the larger squares in the tessellation as 𝑎 and the side length of the smaller ones as 𝑏, then ‖𝐴𝐵‖ = 𝑐 is the length of the hypotenuse. And the length of the hypotenuse is also the length of the side of a square that is a fundamental parallelogram. This seems to be another area proof of the Pythagorean Theorem. Since this tessellation is made up of repeating translation units made of two squares, of total area 𝑎2 + 𝑏2 , the area should equal the area 𝑐2 of the other set of units, the fundamental parallelograms. Indeed, in looking at a square of side 𝑐, it is dissected into one square of side 𝑎 and four congruent quadrilaterals. One can see that the quadrilaterals can be assembled to make a square of side 𝑏, proving the Pythagorean Theorem. In fact this is exactly the dissection for the area proof of Perigal in Figure 9.11! When the Perigal dissection was introduced, it seemed rather hard to imagine how one would think of cutting up a square in such a manner. Perhaps it still seems that way, but interesting dissections can result from taking a fundamental parallelogram of a tessellation made of multiple shapes. In fact, in Figure 21, the dissection that results from taking a fundamental parallelogram with corners at the centers of the small squares will give another dissection of a square of side 𝑐 that looks quite different.

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Napoleon Tessellation. Here is another tessellation that is connected to an earlier theorem. If we take the figure for the Napoleon Theorem, Theorem 10.8, and extend it by rotating shapes by 120 degrees with centers at the centers of the equilateral triangles, the result is a tessellation, part of which is shown in Figure 23 along with some of the underlying centers of rotation.

Figure 23. Napoleon Theorem Tessellation

There are several interesting things to note. The angles do, in fact, add up to 360 degrees at each vertex, half from the shaded triangle and half from three 60-degree contributions from the equilateral triangles. The 120-degree rotations centered at the centers of the three sizes of Napoleon equilateral triangles are symmetries of the tessellation. The product of two 120-degree rotations at adjacent centers is a 240-degree rotation at a third such center point, so these rotations are the only symmetries except for the translations that are products of a 120-degree rotation and a 120-degree rotation. There are no line reflection symmetries and no 60-degree or 180-degree rotational symmetries. Each translation symmetry for this 120-degree rotation pattern takes one of the Napoleon equilateral triangles to a congruent one with corresponding sides parallel. Looking at the figure to see parallel sides, we observe that these triangles are divided by parallelism into three distinct sets. The translations will map a center to another center in the same set. The same translation will take a shaded nonequilateral triangle to a triangle with corresponding sides parallel. The rotation centers form the vertices of a regular tessellation by equilateral triangles like the one in Figure 20. However, these vertices are not 60-degree centers, but 120-degree centers. Each triangle in this tessellation has one vertex from each of the three kinds of centers. Two triangles that share an edge will form a rhombus. The endpoints of the longer diagonal of the rhombus will be centers of the same kind, so there is a translation from one endpoint to the other. This figure was generated by taking the Napoleon figure and rotating it. Another way to arrive at this figure would be to start with three rotation centers at the vertices of an equilateral triangle △𝑋𝑌 𝑍.

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Pick any point 𝐴 in △𝑋𝑌 𝑍 and let 𝐵 = 𝑍120 (𝐴) and 𝐶 = 𝑌120 (𝐵). Then 𝑋120 (𝐶) = 𝐴, since 𝑋120 𝑌120 𝑍120 = 𝐼, by the remarks pertaining to Figure 12. This then produces the same family of tessellations starting from the rotation centers. Quadrilateral Tessellations. Unless one has seen it before, it is surprising to learn that every quadrilateral tessellates the plane. It is an interesting classroom or workshop activity to take cut-outs of a general quadrilateral (especially if it is not convex) and invite participants to cover part of a table with these shapes. It is even better to do it with dynamic geometry software so that one can vary the shape in real time. But setting up the software first requires understanding the relationships among the quadrilaterals in order to connect them. Figures 24 and 25 show two examples of quadrilateral tessellations. Based on our experience so far, it makes sense to look for rigid motions that are symmetries. One can see pretty clearly that there are translations present in each case. Why should they be there?

Figure 24. Convex Quadrilateral Tessellation

Figure 25. Nonconvex Quadrilateral Tessellation

In looking at the tessellations in Figures 24 and 25, one sees that each quadrilateral is attached to its neighbor with a shared side by a half-turn centered at the midpoint of the common side. Therefore, if one goes from a quadrilateral to a neighbor and then to a second neighbor, one has moved the quadrilateral by the composition of two half-turns which is a translation. So there are these translations in the tessellations.

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But this reasoning so far would also be true if one tried to tessellate by adjoining general pentagons in the same way. But the pentagons will overlap and not fit together. Why do the half-turns and translations for quadrilaterals create a lattice, when those for a pentagon would not? The answer is that the midpoint figure of a general quadrilateral is a parallelogram, as we learned in Theorem 7.19. A

M

J

D L

C

A

M

D L

J

K B

B

K

C

Figure 26. Quadrilateral Midpoints

Since the midpoints of a single quadrilateral 𝐴𝐵𝐶𝐷 form a parallelogram 𝐽𝐾𝐿𝑀, a half-turn centered at a vertex of the parallelogram will map the original midpoint parallelogram to a congruent midpoint parallelogram in a quadrilateral with a shared side with 𝐴𝐵𝐶𝐷. The half-turn products ℋ𝐾 ℋ𝐽 = 𝒯𝐴𝐶 and ℋ𝑀 ℋ𝐽 = 𝒯𝐵𝐷 will map a vertex of 𝐴𝐵𝐶𝐷 to the opposite vertex by a diagonal translation. Therefore, they will map the midpoint quadrilateral to the midpoint parallelogram of a quadrilateral with a shared vertex, not a side. This is visible in Figure 27, where the midpoint parallelograms are shaded. The gray parallelograms are related to the green parallelograms by half-turns. All the gray ones are related by translation symmetries, as are the green ones. The lattice of translations is generated by 𝑇𝑣 = 𝒯2[𝐽𝐾] = 𝒯𝐴𝐶 and 𝑇𝑤 = 𝒯2[𝐽𝑀] = 𝒯𝐵𝐷 as in Figure 19. A fundamental parallelogram contains four parallelograms, one gray parallelogram and one green one and also two unshaded parallelograms congruent to the midpoint parallelogram. A fundamental domain for the entire symmetry group of translations and

Figure 27. Midpoint Parallelograms in Convex Quadrilateral Tessellation

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half-turns could be one quadrilateral or could consist of one shaded and one unshaded parallelogram. A fundamental domain for the translation symmetries alone can be two quadrilaterals that share an edge. Because of the segment pattern inside, it is easy to see which unshaded parallelograms are related by translation and which by half-turns. In this convex case, the four triangles in the original quadrilateral that do not belong to the midpoint parallelogram can be assembled into one of the unshaded parallelograms. This concludes a brief taste of the vast subject of tessellations and wallpaper patterns. Some additional examples are in the exercises.

Translations and Frieze Symmetry Recall that a frieze pattern is a figure on the plane that has one translation and its powers as symmetries but no other translations. Such a pattern was illustrated in Figure 17, in a section where possible translation symmetries were discussed. As for the 17 crystallographic groups, the rules for forming products of rigid motions limit the number of possible symmetry groups for frieze patterns. For frieze patterns the number of possible symmetry groups is seven. In this section we will prove this and see how what we have learned about products allows us to analyze all these possible symmetries. Theorem 10.11 (Seven Frieze Patterns). Given a figure on the plane whose translation symmetries consist of the powers 𝑇 𝑛 of a single translation 𝑇, the symmetry group of the figure is one of seven groups listed in the cases below. The proof will be spelled out as the various possible cases are explored in detail in the next pages. As we have already seen in Theorem 10.10, if all the symmetry translations of a figure in the plane have parallel direction, they are all powers of a single generating translation 𝑉 = 𝒯𝑣 with the minimum displacement 𝑑 = ‖𝐴 𝑉(𝐴)‖. Based on what we have learned about products of rigid motions, we can now say what other kinds of symmetries may be present in a frieze pattern. Case 1 — Translations Only. The simplest pattern is one with no other symmetries besides translations. One example of such a pattern is shown in Figure 28. This is also true of Figure 29. Even though the individual stars that are design elements have rotational symmetry, these rotations are not symmetries of the whole design.

Figure 28. Frieze with Translation Symmetry Only

Note. In this figure and in all the figures illustrating frieze patterns, the frieze consists of the finite pattern shown extended infinitely in two directions by powers of the fundamental translation symmetry 𝑉.

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Figure 29. Another Frieze with Translation Symmetry Only

Symmetry Group for Case 1: • 𝑉 𝑛 , powers of translation 𝑉.

Rotations and Reflections. In order to consider what rotations and reflections may possibly be symmetries of a frieze pattern, one should note that such a symmetry will not be alone. For a pattern, assume that the translation 𝑉 = 𝒯𝑣 generates all the translation symmetries of the figure, with 𝑑 = ‖𝐴𝑉(𝐴)‖ for any 𝐴. Then 𝑉 relates other symmetries by conjugation, as in the corollary on page 181. • For a rotation symmetry 𝐴𝛼 , the product 𝑉𝐴𝛼 𝑉 −1 = 𝑉(𝐴)𝛼 is a second rotation symmetry with the same rotation angle. Therefore, the product of rotations 𝑉(𝐴)−𝛼 𝐴𝛼 is a translation symmetry. By Theorem 10.2, the direction of this translation is not parallel to 𝐴𝑉(𝐴) unless 𝛼 = ±180. In other words, 𝐴𝛼 = ℋ𝐴 is the only possible rotation symmetry at 𝐴. • For a reflection symmetry ℛ𝑎 , the product 𝑉ℛ𝑎 𝑉 −1 = ℛ𝑉 (𝑎) is a second reflection symmetry in a line parallel to or equal to the first. If the line 𝑎 is not parallel to the direction of 𝑉, the product of reflections ℛ𝑉 (𝑎) ℛ𝑎 is a translation symmetry with direction perpendicular to the two lines. For this translation to have the same direction as 𝑉, the line 𝑎 must be perpendicular to the direction of 𝑉. • There is one other possibility. If the line 𝑎 is parallel to the direction of 𝑉, then the line is mapped by 𝑉 to itself, and the product 𝑉ℛ𝑎 𝑉 −1 = ℛ𝑉 (𝑎) = ℛ𝑎 , so no new reflection is introduced. Thus, reflection in such a line can be a symmetry of the frieze pattern. However, there can be only one such line. If 𝑎 and 𝑏 are lines parallel to the direction of 𝑉, the product ℛ𝑏 ℛ𝑎 is a translation with direction perpendicular to the direction of 𝑉. Therefore, both these reflections cannot be symmetries of a frieze pattern.

Case 2 — Half-turns and Translations. The only possible rotation symmetries are half-turns. If ℋ𝐴 is a symmetry of a frieze pattern, then all the products 𝑉 𝑛 ℋ𝐴 𝑉 −𝑛 are also half-turn symmetries with centers at all 𝑉 𝑛 (𝐴), a set of equally spaced collinear points. But there is a second set of half-turns that must be symmetries as well. Let 𝐵 be the midpoint of 𝐴𝑉(𝐴). Then ℋ𝐵 ℋ𝐴 is a translation that maps 𝐴 to 𝑉(𝐴), so 𝑉 = ℋ𝐵 ℋ𝐴 . Therefore, the product ℋ𝐵 = 𝑉ℋ𝐴 must be a symmetry also. Then, as for 𝐴, all the translates 𝑉 𝑛 (𝐵) must also be centers of half-turn symmetry.

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So, starting with a single half-turn symmetry ℋ𝐴 and the translation 𝑉, we have a line containing an infinite set of half-turn centers spaced evenly in a pattern 𝐴𝐵𝐴𝐵𝐴𝐵, with the distance between centers equal to 𝑑/2. And the 𝐴 points are translates of each other by powers of 𝑉, as are the 𝐵 points. Some examples of frieze symmetry with only translations and rotations are shown in Figures 17 and 30. The centers of rotation are indicated in the latter figure; they are visibly spaced apart one-half the distance of the translation displacement. The figure has the same symmetry if those points are not visible.

Figure 30. Translation and Rotation Symmetry

Symmetry Group for Case 2: • 𝑉 𝑛 , powers of translation 𝑉 with displacement distance 𝑑. • Half-turns ℋ𝐴 , with centers spaced at distance 𝑑/2 on a line parallel to direction of 𝑉. Case 3 — Reflections in Lines Perpendicular to Translation Direction. Assume that ℛ𝑝 is a symmetry of a frieze pattern, with 𝑝 perpendicular to the direction of 𝑉. Then all the products 𝑉 𝑛 ℛ𝑝 𝑉 −𝑛 are line reflection symmetries with mirror lines 𝑉 𝑛 (𝑝), a set of equally spaced parallel lines. The story for these perpendicular mirrors is much like that for half-turns. If 𝐴 is a point on 𝑝, then 𝑉 = 𝒯𝐴𝑉 (𝐴) and also 𝑉 = ℛ𝑞 ℛ𝑝 , where 𝑞 is the perpendicular bisector of 𝐴𝑉(𝐴) and also the midline of 𝑝 and 𝑉(𝑝). So, again, ℛ𝑞 = 𝑉ℛ𝑝 is a symmetry, and the lines are spaced at distance 𝑑/2 apart, with a set of 𝑝 lines being translates by 𝑉 at distance 𝑑 apart and the same for the translates of 𝑞.

Figure 31. Translation and Perpendicular Mirror Symmetry

An example of frieze symmetry consisting of translations and perpendicular reflections is shown in Figure 31. This figure can be interpreted several ways. It can be viewed as two separate examples of this kind of frieze symmetry separated by a horizontal line. Or the entire figure can be taken as a single example of such symmetry.

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Symmetry Group for Case 3: • 𝑉 𝑛 , powers of translation 𝑉 with displacement distance 𝑑. • Reflections ℛ𝑝 in lines perpendicular to direction of 𝑉 with mirror lines spaced 𝑑/2 apart. Case 4 — Reflection in Line Parallel to Translation Direction. If a reflection symmetry ℛ𝑚 is parallel to the direction of 𝑉, there is only one such line. The frieze pattern looks like a pattern of translations alone but with the addition of a reflection of the whole pattern in a line parallel to the direction of 𝑉. Two such patterns are shown in Figures 32 and 33. This reflection also adds a new symmetry as well. The product 𝑉ℛ𝑚 is a glide reflection. The second figure differs from the first in that the reflection line is a line of symmetry for the arrow polygon, so the reflection maps the polygon into itself rather than to a second disjoint polygon.

Figure 32. Frieze with Parallel Mirror Symmetry

Figure 33. Frieze with Parallel Mirror Symmetry: Symmetric Motif

Symmetry Group for Case 4: • 𝑉 𝑛 , powers of translation 𝑉 with displacement distance 𝑑. • 𝐺 𝑛 , powers of glide reflection 𝐺 with 𝐺 2 = 𝑉 2 . • Line reflection ℛ𝑚 . Case 5 — Glide Reflection and Translation. These figures from Case 4, most visibly Figure 32, show a new rigid motion in the roster of possible symmetries: the glide reflection. Glide reflections were defined at the end of Chapter 7. Here we will repeat the definition with some useful notation.

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Figure 34. Glide Translation Symmetry as Footprints

Definition 10.12. The glide reflection 𝒢𝐴𝐵 is the product ℛ𝐴𝐵 𝒯𝐴𝐵 . This glide reflection can be written as a product in a couple of other useful ways. To see this, let 𝑚 = 𝐴𝐵. Then 𝒯𝐴𝐵 = ℛ𝑞 ℛ𝑝 , where 𝑝 and 𝑞 are lines perpendicular to 𝑚 at 𝑃 = 𝐴 and 𝑄 the midpoint of 𝐴𝐵. Since reflections in perpendicular lines commute, we can write the glide reflection as a triple reflection in several ways: 𝒢𝐴𝐵 = ℛ𝑚 (ℛ𝑞 ℛ𝑝 ) = (ℛ𝑚 ℛ𝑞 )ℛ𝑝 = ℛ𝑞 (ℛ𝑚 ℛ𝑝 ) = (ℛ𝑞 ℛ𝑝 )ℛ𝑚 . These grouped products show the glide reflection in a variety of guises: 𝒯𝐴𝐵 = ℛ𝐴𝐵 𝒯𝐴𝐵 = ℋ𝑄 ℛ𝑝 = ℛ𝑞 ℋ𝑃 = 𝒯𝐴𝐵 ℛ𝐴𝐵 . T (E )

E q

T(P ) = G(P )

Q

p E'

P m

G( E )

HP (E )

R m( E )

Figure 35. Glide Reflection 𝐺 as Product of Half-turn and Line Reflection

Figure 35 illustrates these three mirrors and how a glide reflection 𝐺 can be ℛ𝑞 ℋ𝑃 and can also be 𝑇ℛ𝑚 = ℛ𝑚 𝑇, where the translation 𝑇 = ℛ𝑞 ℛ𝑝 . For future use, we will capture some of this information in a theorem. Theorem 10.13 (Half-turn Reflection Product). If 𝑛 is a line and 𝐴 is a point, let 𝐵 = ℛ𝑛 (𝐴). If 𝐴 is not on 𝑛, these products are glide reflections: ℛ𝑛 ℋ𝐴 = 𝒢𝐴𝐵 and ℋ𝐴 ℛ𝑛 = 𝒢𝐵𝐴 . If 𝐴 is on 𝑛, then these products are reflections in the line through 𝐴 perpendicular to 𝑛. All this was proved in the formulas and discussion above. One important well-defined feature of a glide reflection is its invariant line. This is a line, 𝑚 in Figure 35, that is mapped into itself: 𝐺(𝑚) = 𝑚. For a translation, any line parallel to its direction is an invariant line, but a glide reflection has only one. The invariant line of 𝐺 is the line that contains the midpoint of the segment 𝐸𝐺(𝐸) for every point 𝐸. This midpoint is the point 𝐸′ in the figure. It is the midpoint of the diagonal of a rectangle, so it is clearly on line 𝑚. No other line can be an invariant line, for if 𝐸

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is not on 𝑚, then 𝐺(𝐸) is in the opposite half-plane yet 𝐸𝐺 2 (𝐸) is a segment parallel to 𝑚. These three points cannot be collinear, so 𝐸 cannot be on an invariant line. From the commutativity ℛ𝐴𝐵 𝒯𝐴𝐵 = 𝒯𝐴𝐵 ℛ𝐴𝐵 , it follows that the square of a glide 2 2 reflection is a translation: 𝒢𝐴𝐵 = 𝒯𝐴𝐵 ℛ𝐴𝐵 ℛ𝐴𝐵 𝒯𝐴𝐵 = 𝒯𝐴𝐵 = 𝒯2[𝐴𝐵] . And if the glide reflection is a symmetry of a frieze pattern, then this translation must be a power of 𝑉, so 2[𝐴𝐵] = 𝑘𝑣 for some integer 𝑘.

Figure 36. Glide Translation Symmetry Only

If 𝑘/2 is an integer, then 𝑉 −𝑘/2 𝒢𝐴𝐵 = ℛ𝐴𝐵 . So the reflection in the line 𝐴𝐵 parallel to the direction of 𝑉 is a symmetry. Thus, if 𝑘/2 is an integer, the symmetry group is the group of Case 4. The translation 𝒯𝐴𝐵 can equal 𝑉. This is the case in Figures 32 and 33. In each case, the product of 𝑉 with the reflection in line 𝑚 is a glide reflection, so the square of the glide reflection is 𝑉 2 . In Figure 32, if we number the faces from left to right, the glide reflection takes an even-numbered face on the top and moves it to an odd-numbered face on the bottom. At the same time, an even-numbered face on the bottom is mapped to an odd-numbered face on the top. In Figure 33, the symmetric arrows are reflected onto themselves. The glide reflection is present but harder to see. Therefore, the only new cases will be when 𝑘/2 = 𝑝 + (1/2) where 𝑝 is an integer. In this case, 𝑉 −𝑝 𝒢𝐴𝐵 = ℛ𝐴𝐵 𝒯𝑣/2 is a glide reflection symmetry with translation displacement 𝑑/2. The square of this glide reflection is 𝒯𝑣 = 𝑉. In Case 5, we assume that 𝑉 is the same translation symmetry and that there is a glide reflection 𝒢𝐴𝐵 with ‖𝐴𝐵‖ = 𝑑/2. Products of the glide reflection and the translations are either translations of the form 𝑉 𝑛 or glide reflections of the form 𝑉 𝑛 𝒢𝐴𝐵 . Two examples are the footprint track of Figure 34 and the faces of Figure 36. It should be mentioned here that if two glide reflections 𝐺 1 = 𝑇1 𝑅𝑛1 and 𝐺 2 = 𝑇2 𝑅𝑛2 are symmetries of a frieze pattern, then the two invariant lines must be the same: 𝑛1 = 𝑛2 . If not, then 𝐺 1 𝐺 2 = 𝑇1 𝑅𝑛1 𝑅𝑛2 𝑇2 = 𝑇1 𝑆𝑇2 , where 𝑆 is a translation in a direction perpendicular to the direction of 𝑉, which is not allowed as a symmetry. So the invariant lines are the same. With the same invariant line, all the glide reflections must be of the form 𝑉 𝑛 𝒢𝐴𝐵 . Symmetry Group for Case 5: • 𝑉 𝑛 , powers of translation 𝑉 with displacement distance 𝑑. • 𝐺 𝑛 , powers of glide reflection 𝐺 with 𝐺 2 = 𝑉.

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Case 6 — Half-turn and Reflection in Line Perpendicular to Translation Direction. Suppose the symmetries of a frieze figure include both half-turns and reflections in lines perpendicular to the direction of 𝑉 but no other reflections. What was proved in Cases 2 and 3 still holds; the half-turn centers are spaced at distance 𝑑/2 apart on a line 𝑚. The lines of reflection perpendicular to 𝑚 are also spaced at distance 𝑑/2 apart. If any reflection line 𝑝 contains a half-turn center 𝐴, then the product ℋ𝐴 ℛ𝑝 = ℛ𝑚 which is not a reflection allowed in this case. Therefore, given the distances of the spacings, between any adjacent half-turn centers 𝐴 and 𝐵, there must be a reflection line 𝑝. The line 𝑝 must be the perpendicular bisector of 𝐴𝐵. The product ℛ𝑝 ℋ𝐴 ℛ𝑝 = ℋ𝐴′ , where 𝐴′ = ℛ𝑝 (𝐴). This ℋ𝐴′ is a symmetry since it is the product of symmetries. But also, 𝐴′ is on the same side of 𝑝 as 𝐵. Since the distance from 𝐴 to 𝑝 is less than 𝑑/2, then ‖𝐴𝐴′ ‖ < 𝑑. The only possible center of half-turn symmetry at this distance is 𝐵, so 𝑝 is the perpendicular bisector of 𝐴𝐵, at a distance of 𝑑/4 from 𝐴 and from 𝐵. From Theorem 10.13, we know that ℛ𝑝 ℋ𝐴 = 𝒢𝐴𝐵 . So this glide reflection is a 2 symmetry, with the displacement of 𝒯𝐴𝐵 equal to 𝑑/2, so 𝒢𝐴𝐵 = 𝑉. Thus, the symmetries of Case 6, assuming half-turns and reflections in perpendicular lines, also include a glide reflection 𝐺 whose square is 𝑉. If at the outset we had assumed the existence of a glide reflection 𝐺 and a half-turn symmetry, a product of the two would be a reflection in a line perpendicular to the invariant line of 𝐺. Or if one assumed the existence of 𝐺 and a reflection in a line perpendicular to the invariant line of 𝐺, then a product of the two would be a half-turn symmetry. So given any two among half-turn symmetries, perpendicular line reflections or glide reflections, the third type of symmetry will also be present. All this can be seen in Figure 37. The frieze pattern is supplemented with red lines and points indicating the symmetries. The points are half-turn centers such as ℋ𝐴 and the vertical lines indicate the reflections such as ℛ𝑞 . The horizontal line is the invariant line 𝑚 of 𝐺. For this figure, reflection in the line 𝑚 is not a symmetry. In the figure, a face and its 𝐺-images and its 𝑉-images are looking in the same direction, but the half-turns and line reflections reverse the direction of a face. The glide reflection translates an image in the horizontal direction by the distance between

Figure 37. Face Frieze with Half-turns and Line Reflections

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half-turn centers or the distance between mirror lines, which is 𝑑/2. The translation 𝑉 moves an image twice as far.

Figure 38. Flag Frieze with Half-turns and Line Reflections

Figure 38 shows the same symmetry pattern without the centers and lines drawn in. This contrasts with Figure 30, which has half-turns but not the mirrors. Symmetry Group for Case 6: • 𝑉 𝑛 , powers of translation 𝑉 with displacement distance 𝑑. • 𝐺 𝑛 , powers of glide reflection 𝐺 with 𝐺 2 = 𝑉. • Half-turns ℋ𝐴 , with centers spaced at distance 𝑑/2 on a line parallel to direction of 𝑉. • Reflections ℛ𝑝 in mirror lines that are perpendicular bisectors of adjacent half-turn centers.

Case 7. The last pattern of the seven symmetry patterns or symmetry groups for friezes is obtained by adding reflection ℛ𝑚 in the invariant line 𝑚 to the previous set of symmetries. If ℛ𝑝 is a symmetry that is reflection in a line perpendicular to 𝑚 at 𝐴, then ℛ𝑝 ℛ𝑚 = ℋ𝐴 , so ℋ𝐴 is also a symmetry. By the same equation, if 𝐴 is a center of half-turn symmetry, then the line ℛ𝑝 is a symmetry. So unlike Case 6, in this case the centers of half-turn symmetry will be on the perpendicular lines of reflection symmetry. In fact, if one considers all the lines of reflection symmetry — the line 𝑚 and the lines perpendicular to 𝑚 spaced at distance 𝑑/2 apart — then all the other symmetries are products of reflections in these lines.

Figure 39. Face Frieze with All Possible Symmetries

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We already observed that the half-turns are products of the form ℛ𝑝 ℛ𝑚 and ℛ𝑞 ℛ𝑚 . The translation 𝑉 = ℛ𝑞 ℛ𝑝 , and the glide reflection 𝐺 = 𝑉ℛ𝑚 . Such a frieze pattern is illustrated in Figure 39, where the mirrors are very clearly included in the figure. This contrasts with Case 5, the other case that includes ℛ𝑚 as a symmetry. One sees in Figure 32 a string of translated figures and their reflection 𝑚, all facing the same direction. In this case the glide reflection symmetry is present but not half-turns or reflections in lines perpendicular to 𝑚.

Figure 40. Circle Pattern with All Possible Frieze Symmetries

In contrast to Figure 39, Figure 40 has the same reflection pattern, but the lines and points of symmetry are not shown. However, they can be located using intersection points of the circles. Symmetry Group for Case 7: • 𝑉 𝑛 , powers of translation 𝑉 with displacement distance 𝑑. • 𝐺 𝑛 , powers of glide reflection 𝐺 with 𝐺 2 = 𝑉 2 . • Line reflection ℛ𝑚 in 𝐺-invariant line 𝑚. • Half-turns ℋ𝐴 , with centers spaced at distance 𝑑/2 on 𝑚. • Reflections ℛ𝑝 in mirror lines that are lines through half-turn centers perpendicular to 𝑚. This completes the survey of the seven frieze symmetry patterns. These frieze symmetries were first encountered in Chapter 7 as the symmetries of a pair of parallel lines. But for a pair of parallel lines, no separation of half-turns or line reflections is required.

Triple Line Reflection Products Earlier in the chapter, all the possible products of orientation-preserving rigid motions were described. Since every rigid motion is the product of at most three line reflections, the orientation-reversing rigid motions are either line reflections or the product of three reflections. We have named two kinds of orientation-reversing rigid motions: the line reflections and the glide reflections. The main aim of this section is to prove that every such rigid motion is one of these two kinds. This will complete the work of classifying rigid motions: any rigid motion is a line reflection, a translation, a rotation, a line reflection, or a glide reflection.

Triple Line Reflection Products

209

Theorem 10.14 (General Triple Reflections). Let 𝑎, 𝑏, and 𝑐 be three lines, and let 𝑆 = ℛ𝑐 ℛ𝑏 ℛ𝑎 . (1) If 𝑎, 𝑏, and 𝑐 are concurrent at 𝑂, then 𝑆 is reflection in a line through 𝑂. (2) If each pair of 𝑎, 𝑏, and 𝑐 is either parallel or the same line, then 𝑆 is reflection in a line parallel to these lines (or equal to one of them). (3) In all other cases, 𝑆 is a glide reflection. Note: The glide reflection cases are these: the set of all intersections of pairs of lines consists of either three points (the vertices of a triangle) or two points (the intersection of a transversal with two parallel lines). Proof. We will prove that the triple intersection is the product of a half-turn and a line reflection; this is a glide reflection by Theorem 10.13. Either 𝑎 and 𝑏 intersect or 𝑏 and 𝑐 intersect in a point (or both). If 𝑎 and 𝑏 intersect at a point 𝑃, the rotation ℛ𝑏 ℛ𝑎 can be written as a different product ℛ𝑏′ ℛ𝑎′ , where 𝑏′ is perpendicular to 𝑐 at a point 𝐶. Then 𝑆 = ℋ𝐶 ℛ𝑎′ is a glide reflection since 𝐶 is not on 𝑎′ . If 𝑏 and 𝑐 intersect, by applying the same approach to the rotation ℛ𝑐 ℛ𝑏 , we can write 𝑆 = ℛ𝑐′ ℋ𝐴 , also a glide reflection. □ Abundance of Glide Reflections. There is implicit in this theorem an interesting point about what is common and what is rare in the world of rigid motions. The glide reflections are the least familiar and least transparent of the rigid motions. The Common Core decided not to mention them by name at all. But what may seem surprising is that almost all orientation-reversing rigid motions are glide reflections, in some probabilistic sense. If you pick three lines in a truly random way, your probability of picking concurrent or parallel lines is zero. So glide reflections are very common. Another way of seeing the same thing is to cut out an asymmetrical shape, like one resembling a letter F, and tracing it on a piece of paper. Then flip it over and toss it randomly onto the piece of paper. You will surely not get a line reflection connecting them. What is left is a glide reflection.

Figure 41. Midpoints from Shapes with Opposite Orientation

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Try connecting corresponding points with segments and mark the midpoints. They will be collinear. This is the invariant line of the glide reflection. This experiment is illustrated in Figure 41. Invariant Glide Reflection Lines in a Triangle. Glide reflections are not too hard to visualize if they are presented as translation in a direction with a reflection in a parallel line. But given a triangle with sides 𝑎, 𝑏, 𝑐, how does one visualize 𝐺 = ℛ𝑐 ℛ𝑏 ℛ𝑎 ? Suppose these lines are the extended sides of △𝐴𝐵𝐶, with 𝐴, 𝐵, 𝐶 being the vertices opposite 𝑎, 𝑏, 𝑐. In the proof of Theorem 10.14, the triple reflection was proved to be a glide reflection by writing it as the product of a half-turn and a line reflection. For example (with a change of label), 𝐺 = ℋ𝐹 ℛ𝑎′ , where 𝐹 is the foot of the perpendicular to 𝑐 through 𝐶. In other words, 𝐹 is the foot of the altitude of the triangle through 𝐶. In the same way, 𝐺 = ℛ𝑐′ ℋ𝐷 , where 𝐷 is the foot of the altitude through 𝐴. But this tells us two points on the invariant line of 𝐺. When a glide reflection is written as a product in the form ℋ𝑃 ℛ𝑚 or ℛ𝑛 ℋ𝑃 , the point 𝑃 is on the invariant line, as shown in Theorem 10.13 and illustrated in Figure 35. Therefore, we know that the invariant line is 𝐷𝐹.

k

A

E

C

E1 F B

D

E0

Figure 42. Invariant Line 𝐷𝐹 of Product of Three Reflections

It only remains to find the translation parallel to the invariant line. But this figure may start to look familiar. Since the feet of two altitudes play such an important role, the point 𝐸 on 𝑏, the foot of the third altitude, is a key to the translation. Let 𝐸0 = ℛ𝑎 (𝐸) and let 𝐸1 = ℛ𝑏 (𝐸). Then 𝐺(𝐸0 ) = ℛ𝑐 ℛ𝑏 ℛ𝑎 (𝐸0 ) = ℛ𝑐 ℛ𝑏 (𝐸) = ℛ𝑐 (𝐸) = 𝐸1 . This is what is needed to identify 𝐺 provided that 𝐸0 is on 𝐸𝐹. But we know this figure because we have seen it before. It is the unfolding of the orthic triangle that was the solution of Fagnano’s problem, shown in Figure 6.15 of Chapter 6! The orthic triangle is shaded in Figure 42. When two sides of the orthic triangle are unfolded by reflection in the sides of △𝐴𝐵𝐶, the images of the altitude feet lie on a line, the line known to be the invariant line of 𝐺. Therefore, 𝐺 = 𝒢𝐸0 𝐸1 .

Exercises and Explorations

211

It must be confessed that in Chapter 6 this unfolding was only justified for acute triangles, so we have not proved this is a way to find this translation vector for all triangles, though it is true. Another way is to notice that when we write 𝐺 = ℋ𝐹 ℛ𝑎′ , the line 𝑎′ is a line through 𝐶 perpendicular to the invariant line 𝐷𝐹. Therefore 𝐺 = 𝒢𝐹𝐹 ′ , where 𝐹 ′ = ℛ𝑎′ (𝐹). For the case when the three lines are two parallels and a transversal, similar reasoning applies.

Exercises and Explorations 1. (Direction of a Translation). For distinct points 𝐴 and 𝐵, the product of rotations 𝑇𝛼 = 𝐵−𝛼 𝐴𝛼 is a translation. The segment 𝐴𝑇𝛼 (𝐴) shows the displacement by this translation. (a) Draw 𝐴 and 𝐵 and the segments 𝐴𝑇𝛼 (𝐴) for 𝛼 a multiple of 60: 60, 120, 180, 240, 300. This shows directions of translations produced with 60-degree centers of rotation at 𝐴 and 𝐵. (b) Draw another figure with 𝐴 and 𝐵 and repeat the same exercise but for 𝛼 a multiple of 72: 72, 144, 216, 288. (c) Note the differences in these examples for rotation by 360/𝑛, with one example 𝑛 being even and the other with 𝑛 odd. State what differences you can see will exist between even and odd examples of 𝑛. 2. (Checkerboard). We have analyzed the symmetries of a square tiling with all tiles the same. What are the the symmetries of a chessboard made of alternating black and white squares when the symmetries do not send black squares to white squares. (If you are uneasy about color in geometry, you can put a point at the center of the black squares but not the white squares and then ignore color). Do we know a name for this symmetry group? 3. (Example of 𝑝4𝑔). Make a copy of Figure 10. Mark all the centers of rotational symmetry and the lines of reflection symmetry and indicate the translation symmetries. Also, draw a square tile that will cover the plane when translated by symmetries. 4. (Rotating a Rotation). Let 𝐴𝛼 and 𝑃𝜃 be rotations. The goal of this problem is to show in general that the product 𝑃𝜃 𝐴𝛼 𝑃−𝜃 = 𝑃𝜃 (𝐴)𝛼 . In other words, this premultiplication by 𝑃−𝜃 and post-multiplication by 𝑃𝜃 (an example of the conjugation introduced on page 181) relocates 𝐴𝛼 by “rotating” this rigid motion by 𝑃𝜃 .

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C

B'

B A'

C'

A P120

Figure 43. Rotating a Rotation

(a) Before tackling the general case, verify this relation for the example in Figure 43. In this case, 𝐶 = 𝐴45 (𝐵) and △𝐴′ 𝐵′ 𝐶 ′ is the 𝑃120 image of △𝐴𝐵𝐶. Compare the images of 𝐴′ 𝐵 ′ by the rigid motions 𝐴′60 and 𝑃120 𝐴60 𝑃−120 . (b) Prove the relationship in general. One knows the product is a rotation, and one knows its angle. So all one needs to do is find the center of the rotation. 5. (Reflecting a Rotation). A second example of conjugation would be to take the same rotation 𝐴𝛼 as in the previous problem and consider the product ℛ𝑚 𝐴𝛼 ℛ𝑚 . This is a rotation, but the relationship is not quite the same as in the previous problem. State what this rotation is and prove it. 6. (Rotations Not in Wallpaper Groups). One can use the results about conjugation in Exercise 4 to see why most rotations cannot belong to a wallpaper symmetry group. (a) In Figure 44, 𝐶60 and 𝐷60 are rotations. Use conjugation to show that all the other points of the hexagon are also centers of 60-degree rotation that are products of 𝐶60 and 𝐷60 . How do the lengths of the edges of the hexagon compare with the length of 𝐶𝐷? (b) In the same figure, 𝐴360/7 and 𝐵360/7 are rotations. Use conjugation to show that all the other points of the heptagon are also centers of 360/7-degree rotation that are products of 𝐴360/7 and 𝐵360/7 . How do the lengths of the edges of the heptagon compare with the length of 𝐴𝐵? (c) If a rotation 𝐴𝛼 is a symmetry of a wallpaper pattern, there must be a minimum distance between centers of rotation centers by the same angle. Suppose that 𝐴360/7 is such a symmetry and that 𝐴360/7 and 𝐵360/7 are chosen to be this minimum distance. Why does this lead to a contradiction? (d) Generalizing the case of 360/7, why is it not possible for a rotation by 360/𝑛 to be a symmetry of a wallpaper pattern for 𝑛 > 6?

A

B

C

D

Figure 44. Centers of Rotation Products of Two Rotations

Exercises and Explorations

213

A

B

Figure 45. 72-degree Rotation Centers

7. (Rotations by 72 Not in Wallpaper Groups). For 𝐴𝛼 to be a symmetry of a wallpaper pattern, there must be a minimum distance between all centers of rotation by 𝛼 that are also symmetries. In Exercise 4, it was shown that this excludes all rotations by 360/𝑛 for 𝑛 > 6. In this problem, the case 𝑛 = 5 will also be excluded. This leaves only 360/𝑛 for 𝑛 = 2, 3, 4, 6 as possible rotation symmetries for wallpaper patterns. (a) In Figure 45, the points 𝐴 and 𝐵 are centers of 72-degree rotation. Use the conjugation method of Exercise 4 to show that all the points in the figure are centers of 72-degree rotation. (b) Explain why Figure 45 shows that no 72-degree rotation can be a symmetry of a wallpaper pattern. 8. (Tessellation with triangles and hexagons). Find the centers of rotational symmetry and the lines of symmetry in Figure 46 and compare with the centers and lines in Figure 13. Can you draw a picture in which the two figures are overlaid with matching center patterns?

Figure 46. Tessellation by Hexagons and Triangles

9. (Parhexagon). A parhexagon is a hexagon with opposite sides that are parallel and congruent. (a) Prove that one can tile the plane with any parhexagon. (b) For a general parhexagon (i.e., no extra congruences or symmetries that are not present in every parhexagon), describe the group of symmetries of the tessellation. (c) What tessellation described in the text has symmetries most like those of a parhexagon tessellation? Is there a visible reason for this?

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10. Products and Patterns

10. (Frieze Symmetry Analysis 1). Tell what case of frieze symmetry applies to Figure 47. Draw your own version and indicate all the lines of symmetry, half-turn centers of symmetry, as well as the glide reflections and translations.

Figure 47. A Frieze XOXOXOX

11. (Frieze Symmetry Analysis 2). In Figure 48, two triangular friezes are shown, one of which has shaded triangular interiors. (a) What case number applies to the upper frieze? What is the displacement distance for the fundamental translation 𝑉? (b) Consider color symmetries of the lower frieze, meaning symmetries of the frieze that map shaded triangles to shaded triangles. What case corresponds to this group of symmetries? What is the shortest displacement distance of the translations that are color symmetries?

Figure 48. Triangular Friezes

12. (Design from a Right Triangle). Starting with an isosceles right triangle, the plane can be tessellated by reflecting the triangle in its sides and then continuing to reflect the images in their sides. If there is a design in the triangle, such as that in Figure 49, a wallpaper pattern is created. (a) Using computer software or a cut-out triangle, make an example of such a wallpaper pattern (with enough repeats to see the overall symmetry pattern). (b) What are the rotation centers and where are they located? What size square is a fundamental parallelogram?

Exercises and Explorations

215

Figure 49. Right Isosceles Triangle with Design

13. (Distance-Minimizing Fermat Point). Show that in Figure 50 the Fermat point minimizes the distance sum ‖𝑃𝐴‖ + ‖𝑃𝐵‖ + ‖𝑃𝐶‖ among all points 𝑃 in the acute triangle △𝐴𝐵𝐶. Big Hint: In the figure, 𝑃′ is 𝐴60 (𝑃). The key is to explain why the length of the broken path 𝐵 ∗ 𝑃 ′ 𝑃𝐵 equals ‖𝑃𝐴‖ + ‖𝑃𝐵‖ + ‖𝑃𝐶‖. That suggests why the distance sum would be smaller if 𝑃 and 𝑃′ were on 𝐵𝐵 ∗ . Once this is established, why does placing 𝑃 on 𝐵𝐵 ∗ in this way show that 𝑃 is the Fermat point? A*

C B P

B*

P' A

C*

Figure 50. Minimizing the Sum of the Distances to the Vertices

14. (Glide Reflection Midpoint Experiment). Cut out an asymmetrical shape and carry out the midpoint experiment pictured in Figure 41. Trace the shape, flip it over, and reposition it in no special way. Find the midpoints of several pairs of corresponding points and identify the glide reflection that maps the first trace to the second. Then use this information to construct the glide reflection image of some point not on either shape. Repeat the experiment a few times to develop more insight into the behavior of glide reflections.

Chapter 11

Coordinate Geometry

Coordinates attach numbers to points in such a way that algebra can be used to investigate geometry. In the 17th century, Fermat and Descartes developed the beginnings of the coordinate systems that are so familiar today. To express rigid motions and distance and angle relationships, we will want a Cartesian coordinate system with perpendicular axes and rulers on each axis. But there are applications that do not require such a restricted coordinate system, so we begin more simply with parallel lines and ratios to define what are called affine or linear coordinates. Then we develop formulas for half-turns and translations and equations for lines in affine coordinates. Afterwards, for the metric part of Euclidean geometry, we will restrict ourselves to Cartesian coordinates, which are coordinates with more constraints on angles and distances. The introduction of coordinates will connect the abstract model of the plane developed up to now with algebra and the (𝑥, 𝑦)-plane ℝ2 . This will provide some new algebraic tools for studying geometry, but the geometry will also provide insight into the algebra.

Axes and Coordinates In the plane, take a triangle △𝐸0 𝐸1 𝐸2 as the reference triangle for a system of coordinates based on parallel lines and signed ratios. On the line 𝐸0 𝐸1 , let 𝑥(𝑃) = 𝐸0 𝑃/𝐸0 𝐸1 . This ratio is a coordinate proportional to a ruler on the line, with 𝑥(𝐸0 ) = 0 and 𝑥(𝐸1 ) = 1. In the same way, on 𝐸0 𝐸2 , let 𝑦(𝑃) = 𝐸0 𝑃/𝐸0 𝐸2 . The first axis, or 𝑥-axis, of the coordinate system is 𝐸0 𝐸1 ; the second axis, or 𝑦-axis, is 𝐸0 𝐸2 . The origin of the system is the point 𝐸0 . For any point 𝑃 in the plane, let 𝑝1 be the line through 𝑃 parallel to or equal to the 𝑦-axis. This line intersects the 𝑥-axis at a point 𝑃1 . Define 𝑥(𝑃) = 𝑥(𝑃1 ). 217

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11. Coordinate Geometry

E 0P2 E 0E2 = 2.00

P (3,2)

P2

Q (-1, .5)

E2

E0

E1

P1

E 0P1 E 0E1 = 3.00

Figure 1. Axes, Parallel Lines, and Coordinates

Again, in the same way, let 𝑝2 be the line through 𝑃 parallel to the 𝑥-axis, intersecting the 𝑦-axis at 𝑃2 . Define 𝑦(𝑃) = 𝑦(𝑃2 ). The figure 𝐸0 𝑃1 𝑃𝑃2 is a parallelogram (or linear parallelogram). Define the affine coordinates of 𝑃 with respect to this reference triangle, or axis system, to be the pair (𝑥(𝑃), 𝑦(𝑃)). Figure 1 shows two points and their coordinates, the point 𝑃 with coordinates (3, 2) and 𝑄 with coordinates (−1, .5). The coordinates (𝑥, 𝑦) define a map of the plane to real 2-space ℝ2 . So long as the coordinate system remains unchanged, we will refer to points by their coordinates, either as 𝑃(2, 3) or even by saying 𝑃 = (2, 3). Equations for other lines will come soon, but from the definition of coordinates, it is clear that 𝑥 = 2 is the equation of a line parallel to the 𝑦-axis, and 𝑦 = 3 is the equation of a line parallel to the 𝑥-axis. Note (Terminology). We have called these coordinates affine coordinates, but such (𝑥, 𝑦)-coordinates may also be called linear coordinates. What is the difference? For affine coordinates we must specify three points, the origin and then the unit points on the axes. Linear coordinates have the same definition except that they occur in a space where the origin is already understood, such as ℝ2 , where (0, 0) is normally taken to be the origin, or the vectors in the plane, where the zero vector is understood to be the origin. However, this can seem to be a very fine distinction once an origin 𝐸0 has been settled on and designated as 𝑂.

Midpoints, Half-turns, and Translations For any two points 𝐴 and 𝐵 on the plane, one can find the coordinates of the midpoint of 𝐴𝐵 using what has been proved about parallelograms and parallels. The figure is full of overlapping parallelograms that can be used to prove this formula, but 𝐴𝐵 is also a transversal of two sets of parallel lines (if it is not parallel to an axis), so the Transversal Ratios Theorem, Theorem 8.17 can be applied.

Midpoints, Half-turns, and Translations

219

B ( b 1, b 2) B2

D( a1,b 2)

E2

M2 A( a1,a 2)

M C( b 1,a 2)

A2

A1

M1

E0

E1

B1

Figure 2. Parallelograms and Midpoints

Theorem 11.1 (Midpoint Formula). In linear coordinates, the formula for the midpoint 𝑀 of 𝐴𝐵 is 1 (𝑥(𝑀), 𝑦(𝑀)) = (𝑥(𝐴) + 𝑥(𝐵), 𝑦(𝐴) + 𝑦(𝐵)). 2 This can also be written more compactly as 1 (𝐴 + 𝐵) 2 where a sum of points here means a sum of number pairs. 𝑀=

Proof. Let 𝐴 = (𝑎1 , 𝑎2 ) and 𝐵 = (𝑏1 , 𝑏2 ). A point 𝑀 is the midpoint of 𝐴𝐵 when the signed ratio 𝐴𝑀/𝐵𝑀 = −1. If 𝐴𝐵 is not parallel to the 𝑦-axis, Theorem 8.17 says that ratios are preserved, so 𝐴1 𝑀1 /𝐵1 𝑀1 = −1 and 𝑀1 is also a midpoint. The midpoint of 𝐴1 𝐵1 has coordinate (1/2)(𝑎1 + 𝑏1 , 0) = (1/2)(𝑥(𝐴) + 𝑥(𝐵)), since the signed ratio 𝑏 − 𝑎1 (1/2)(𝑎1 + 𝑏1 ) − 𝑎1 = 1 = −1. 𝑎1 − 𝑏1 (1/2)(𝑎1 + 𝑏1 ) − 𝑏1 For the case when 𝐴𝐵 is parallel to the 𝑦-axis, then 𝑀1 = 𝐴1 = 𝐵1 , so the 𝑥coordinates are all equal and it remains true that 𝑥(𝑀) = (1/2)(𝑥(𝐴) + 𝑥(𝐵)). The same reasoning applies to the other axis so 𝑦(𝑀) = (1/2)(𝑦(𝐴) + 𝑦(𝐵)).

□

A formula for any half-turn is an immediate consequence of the midpoint formula. Theorem 11.2 (Half-turn Formula). For any half-turn ℋ𝑃 , the formula for the image of a point 𝐴 is ℋ𝑃 (𝐴) = 2𝑃 − 𝐴. Proof. This is the midpoint formula rearranged. If 𝐵 = ℋ𝑃 (𝐴), then 𝑃 is the midpoint of 𝐴𝐵, so 𝑃 = (1/2)(𝐴 + 𝐵). If 𝐴 and 𝑃 are known, this becomes a formula for the image of 𝐴: 𝐵 = 2𝑃 − 𝐴. □ Figure 3 shows examples of ℋ𝑃 for 𝑃(0, 2). If 𝐷 = (−1, 1), then ℋ𝑃 (𝐷) = 2(0, 2) − (−1, 1) = (1, 3) and ℋ𝑃 (𝐸1 ) = 2(0, 2) − (1, 0) = (−1, 4).

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C

HP(E 1 )

HP (D ) B

T(C)

P A

D

E2 E0

E1

T(B )

T (A )

Figure 3. Examples: A Half-turn and a Translation

Since translations are compositions of half-turns, this also provides a formula for translations. Theorem 11.3 (Translation Formula). The formula for translation 𝑇 = ℋ𝐵 ℋ𝐴 is 𝑇(𝑃) = 𝑃 + 2(𝐵 − 𝐴). If 𝑇 is written as 𝒯𝐶𝐷 , then 𝑇(𝑃) = 𝑃 + (𝐷 − 𝐶). Proof. The first formula is simply a substitution: ℋ𝐵 ℋ𝐴 (𝑃) = ℋ𝐵 (2𝐴 − 𝑃) = 𝑃 + 2(𝐵 − 𝐴). If 𝑇(𝐶) = 𝐷, then 𝐷 = 𝐶 + 2(𝐵 − 𝐴). □ The example in Figure 3 shows 𝑇(𝐴) − 𝐴 = (1, −1) − (−2, 1) = (3, −2). Then 𝑇(𝐵) = (1, 2) + (3, −2) = (4, 0) and 𝑇(𝐶) = (2, 4) + (3, −2) = (5, 2). The translation is represented by three directed segments that are translates of one another. The translation formula suggests a way to represent vectors by means of number pairs. Since 𝒯𝐴𝐵 = 𝒯𝐶𝐷 if and only if 𝐵 − 𝐴 = 𝐷 − 𝐶, then representing [𝐴𝐵] with the number pair 𝐵 − 𝐴 would be consistent with the vector concept that two vectors are equal if their associated translations are equal. In the example in Figure 3, 𝑇 = 𝒯𝑣 = 𝒯[3,−2] . This list of formulas has a lot of consequences, some of which we have seen before and some not. As a starting point, consider what the translation formula means for parallelograms. We learned in Theorem 7.30 that 𝐴𝐵𝐶𝐷 is a parallelogram (or linear parallelogram) if and only if the segment 𝐷𝐶 is a translate of 𝐴𝐵. This means that 𝐶 = 𝒯𝐴𝐷 (𝐵), or 𝐶 − 𝐵 = 𝐷 − 𝐴. This is consistent with the fact that the midpoints of the diagonals are the same: (1/2)(𝐴 + 𝐶) = (1/2)(𝐵 + 𝐷). Theorem 7.19 proved that the midpoint quadrilateral 𝐽𝐾𝐿𝑀 of any quadrilateral 𝐴𝐵𝐶𝐷 is a parallelogram. Using coordinates and the midpoint formula, the following computation shows 𝐽𝐾𝐿𝑀 is a parallelogram or linear parallelogram for any four points 𝐴, 𝐵, 𝐶, 𝐷 (with the caveat that a pair of sides may have length 0 if 𝐽 = 𝐾 or 𝐽 = 𝑀): 𝐽−𝐾 =

1 1 1 1 1 (𝐴 + 𝐵) − (𝐵 + 𝐶) = (𝐴 − 𝐶) = (𝐴 + 𝐷) − (𝐷 + 𝐶) = 𝑀 − 𝐿. 2 2 2 2 2

Lines, Dilations, and Equations

221

Lines, Dilations, and Equations How are lines represented in this coordinate system? Let’s begin with lines through the origin 𝐸0 . Also, since we will be referring to the origin a lot, to avoid so many subscripts, let’s use 𝑂 as another name for this point. In thinking about points on a line through the origin 𝑂, it is natural to think of dilations with center 𝑂. For every 𝑡, the dilations 𝒟𝑂,𝑡 (𝑃) lie on 𝑂𝑃; in fact, every point on this line is one of these dilation images. What are the coordinates of these points? Theorem 11.4 (A Dilation Formula). If 𝑃 = (𝑎, 𝑏), then 𝒟𝑂,𝑡 (𝑃) = (𝑡𝑎, 𝑡𝑏). In more compact notation, 𝒟𝑂,𝑡 (𝑃) = 𝑡𝑃. Proof. For 𝑃, consider the coordinate parallelogram 𝑂𝑃1 𝑃𝑃2 from Figure 1. The dilation not only maps 𝑃 to a point, it also maps the entire parellogram to a new coordinate parallelogram. For example, 𝑃1 = (𝑎, 0) is the point on 𝑂𝐸1 with 𝑂𝑃1 /𝑂𝐸1 = 𝑎. If 𝑃1′ denotes the dilation image of 𝑃1 and 𝐸1′ denotes the dilation image of 𝐸1 , then, since signed ratios are preserved by dilations, 𝑂𝑃1′ /𝑂𝐸1′ = 𝑎. But 𝑂𝐸1′ /𝑂𝐸1 = 𝑡, so 𝑂𝑃1′ /𝑂𝐸1 = (𝑂𝑃1′ /𝑂𝐸1′ )(𝑂𝐸1′ /𝑂𝐸1 ) = 𝑡𝑎, and 𝑃1′ = (𝑡𝑎, 0). In the same way the image of 𝑃2 , 𝑃2′ = (0, 𝑡𝑏), so 𝒟𝑂,𝑡 (𝑃) = (𝑡𝑎, 𝑡𝑏). □

P' = tP

P'2 =tP 2

P

P2 E2 E 0 =O

E1

P1

P'1 = tP 1

Figure 4. Dilation of 𝑃 by Ratio 𝑡

Thus, the function 𝐿(𝑡) = 𝑡𝑃 from ℝ to 𝑂𝑃 is a parametrization of this line through the origin. It is not a ruler, since the difference of the parameters does not measure the distance between points, but it is proportional to a ruler by a scale factor.

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Line Equation and Slope. When 𝑃 = (𝑎, 𝑏), according to this parametrization, for any point (𝑥, 𝑦) on 𝑂𝑃, there is a 𝑡 so that (𝑥, 𝑦) = 𝒟𝑂,𝑡 (𝑃) = (𝑡𝑎, 𝑡𝑏). The equation 𝑥 = 𝑡𝑎 says that 𝑡 = 𝑥/𝑎, provided 𝑎 ≠ 0, so the second coordinate satisfies the equation 𝑦 = 𝑏𝑡 = (𝑏/𝑎)𝑥. Conversely, any point (𝑥, (𝑏/𝑎)𝑥) is the dilation of 𝑃 by ratio 𝑥/𝑎. Therefore, the points (𝑥, 𝑦), where 𝑥 and 𝑦 are in the solution set of the equation 𝑦 = (𝑏/𝑎)𝑥, is the set of points on the line 𝑂𝑃. If 𝑎 = 0, the line is the 𝑦-axis, so the equation for the line is 𝑥 = 0 since the dilations of 𝑃 have coordinates (0, 𝑡𝑏). Here is another way to picture the constant 𝑏/𝑎. If a line through 𝑂 is not the 𝑦-axis, then the line is a transversal for all lines parallel to this axis and so intersects the parallel line through 𝐸1 with equation 𝑥 = 1. Let the intersection point be (1, 𝑚). Then 𝑚 = (𝑏/𝑎)1 = 𝑏/𝑎. So the 𝑦-coordinate of the intersection with the line 𝑥 = 1 is 𝑚 = 𝑏/𝑎. This constant 𝑚 is given the name the slope of the line, and the letter 𝑚 is often used for this number. Using this slope notation, the equation for a line through 𝑂 looks like this: 𝑦 = 𝑚𝑥. Note (Meaning of Slope). We have used the slope terminology here since that is what one rather universally calls 𝑚 in this equation. However, for affine coordinates, slope need not indicate steepness or even whether a line is rising from left to right on a computer screen or a piece of paper. The triangle defining the coordinates may be oriented in directions very different from graph paper. If the 𝐸1 is very far from 𝐸0 in a vaguely northwest direction and 𝐸2 is close to 𝐸0 in an easterly direction, the meaning of slope 𝑚 = 1 will be very different from what it means for more familiar coordinate systems. Line Parametrization. Now that we have a parametrization for lines through the origin, what is a parametrization for a general line 𝑞 = 𝑄1 𝑄2 ? Theorem 11.5 (Affine Parametrization of a Line). A parametrization of the line 𝑄1 𝑄2 is 𝑃(𝑡) = 𝑄1 + 𝑡(𝑄2 − 𝑄1 ) = (1 − 𝑡)𝑄1 + 𝑡𝑄2 . The signed ratio 𝑃(𝑡)𝑄1 /𝑄2 𝑄1 = 𝑡. Proof. The translation 𝒯𝑄1 𝑂 maps 𝑞 to the parallel line 𝑞′ though the origin 𝑂. The point 𝒯𝑄1 𝑂 (𝑄2 ) = 𝑄′2 = 𝑄2 − 𝑄1 is on 𝑞′ . By the parametrization for lines through 𝑂, all points 𝑃 on 𝑞′ are of the form 𝑡(𝑄2 − 𝑄1 ) for some 𝑡, where 𝑡 = 𝑃𝑂/𝑄′2 𝑂. The inverse translation 𝑇 −1 maps this point to the point 𝑄1 + 𝑡(𝑄2 − 𝑄1 ). Thus, we have a parametrization for the line 𝑄1 𝑄2 . The signed ratio is preserved by translation. □ Note: Since the sum of the real coefficients is 1, this is a weighted average of the two points. This property was hidden in the formula, 𝑡𝑃, which really should be written (1 − 𝑡)𝑂 + 𝑡𝑃. Equation for a Line Through Two Points. What is an equation satisfied by the points of 𝑞? Of course, if the line is parallel to the 𝑦-axis, the equation is 𝑥 = 𝑥1 and there is no slope 𝑚. So from now on, assume that 𝑞 is not such a line.

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223

Consider the slope of the parallel line 𝑞′ through 𝑂. If 𝑄1 = (𝑥1 , 𝑦1 ) and 𝑄2 = (𝑥2 , 𝑦2 ), then the point with coordinates 𝑄2 − 𝑄1 = (𝑥2 − 𝑥1 , 𝑦2 − 𝑦1 ) is on the line, so the equation 𝑦 = 𝑚𝑥 becomes 𝑦 = 𝑚𝑥 =

(𝑦2 − 𝑦1 ) 𝑥. (𝑥2 − 𝑥1 )

This slope 𝑚 can be computed from any two points 𝑄1 and 𝑄2 on the line, since the value of 𝑚 is determined by the slope of the parallel line through 𝑂. Now, to find an equation satisfied by every point 𝑄 = (𝑥, 𝑦) on the line, pick any point 𝑄0 on 𝑞 (which may be 𝑄1 or 𝑄2 but need not be). Then the translation 𝒯𝑄0 𝑂 maps 𝑞 to 𝑞′ , the parallel line through 𝑂, and maps 𝑄 to 𝑄 − 𝑄0 = (𝑥 − 𝑥0 , 𝑦 − 𝑦0 ). Since this point is on 𝑞′ , it satisfies this equation: 𝑦 − 𝑦0 = 𝑚(𝑥 − 𝑥0 ) =

(𝑦2 − 𝑦1 ) (𝑥 − 𝑥0 ). (𝑥2 − 𝑥1 )

This equation is the point-slope form of the equation of a line. If one picks 𝑄0 = 𝑄1 , then there are fewer symbols in the formula, for then 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ). Two General Line Equations. If the constant terms in this equation are collected together, then the equation above takes on a simpler form. This is an equation for a line not parallel to the 𝑦-axis: 𝑦 = 𝑚𝑥 + 𝑏. The intersection (0, 𝑏) of this line with the 𝑦-axis is obtained by setting 𝑥 = 0. This point is the 𝑦-intercept, and this is the slope-intercept form of the equation of a line. There is one other common form of the equation, for real constants 𝑎, 𝑏 𝑐, with (𝑎, 𝑏) ≠ (0, 0). This form includes lines parallel to the 𝑦-axis, since these lines have equation 𝑥 = 𝑐: 𝑎𝑥 + 𝑏𝑦 = 𝑐. This equation is not unique, since any multiple of it by a nonzero 𝑘 has the same solution set: 𝑘𝑎𝑥 + 𝑘𝑏𝑦 = 𝑘𝑐. But if the coefficients 𝑎 and 𝑏 are not changed, then each change of constant 𝑐 has as solution a different line parallel to 𝑚, since for 𝑐 1 ≠ 𝑐 2 , the equations 𝑎𝑥 + 𝑏𝑦 = 𝑐 1 and 𝑎𝑥 + 𝑏𝑦 = 𝑐 2 have no common solution. This completes the derivation of the equation of a line in affine coordinates.

Euclidean Geometry and Cartesian Coordinates Now we move on to the more familiar picture of perpendicular Cartesian coordinates. To work with distances and angles, we need a more restricted kind of coordinates. These Cartesian coordinates are simply affine coordinates where △𝐸0 𝐸1 𝐸2 , the reference triangle, is a right isosceles triangle, with right angle at 𝐸0 and sides of length 1. Again, we will write the origin 𝑂 = 𝐸0 . Then the coordinates on each axis are ruler coordinates that are signed distances, not just ratios. The coordinate parallelograms 𝑂𝑃1 𝑃𝑃2 are now rectangles.

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11. Coordinate Geometry

4

(0,mx )

P (x,y)

3

2

(0,m) 1

Q (1,m)

O – 2

(0,0)

(1,0)

2

(x ,0))

4

– 1

– 2

Figure 5. Slope of a Line Through the Origin

Everything that we proved for affine coordinates remains true in the Cartesian case, but some features look different enough that we will revisit them. For the equation 𝑦 = 𝑚𝑥, the slope 𝑚 is now a ratio of side lengths of a rectangle, as illustrated in Figure 5. It indicates steepness in a way that the general affine case did not. Also, the similar triangles, attached to a line to derive the point-slope form of the line equation, are now right triangles, all of which are similar and have the same internal ratio of side length, which is the slope. This is illustrated in Figure 6.

Figure 6. Similar Triangles Dangling from Parallel Lines

Perpendicular Lines in the Coordinate Plane

225

Distance Formula. Let 𝑃1 and 𝑃2 be points in the plane with coordinates (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ). Then if 𝑀 is the point with coordinates (𝑥2 , 𝑦1 ) and 𝑁 is the point with coordinates (𝑥1 , 𝑦2 ), then 𝑃1 𝑀𝑃2 𝑁 is a rectangle, with sides parallel to the axes, as shown in Figure 7. 2

2

2

By the Pythagorean Theorem, ‖𝑃1 𝑃2 ‖ = ‖𝑃1 𝑀‖ + ‖𝑀𝑃2 ‖ . Since 𝑃1 𝑀 is parallel to the 𝑥-axis, the coordinate 𝑥 is a ruler, so the length of this segment is |𝑥2 − 𝑥1 |. Likewise, the length of 𝑀𝑃2 is |𝑦2 − 𝑦1 |. This gives the formula for the distance in terms of coordinates: ‖𝑃1 𝑃2 ‖ = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 . Now that there is a way to measure distance, one can identify a few rigid motions without much calculation. Again, let’s denote 𝐸0 by 𝑂. The reflection ℛ𝑂𝐸1 (𝑥, 𝑦) = (𝑥, −𝑦). This clearly fixes the points of the 𝑥-axis and it also preserves distance, since the change of sign does not change the value of the square terms in the distance formula. The reflection in the 𝑦-axis maps (𝑥, 𝑦) to (−𝑥, 𝑦). Another simple reflection formula is reflection in the diagonal line 𝑠 with equation 𝑦 = 𝑥. The reflection ℛ𝑠 (𝑥, 𝑦) = (𝑦, 𝑥) fixes the points of this line and again does not change the value of the expressions in the distance formula. This reflection also maps each axis to the other. The general formula for reflections is more complicated. We will take it up when we consider the general formula for rotations.

(0,y1 )

N ( x2 , y1 )

(0,y2 )

P2 ( x2, y2 )

E2 O

E1

(x2 ,0)

P1 ( x1, y1)

M ( x1, y2 )

(x1,0)

Figure 7. Coordinates and Distance

Perpendicular Lines in the Coordinate Plane The formula for 𝑂90 , the 90-degree rotation centered at 𝑂, can be deduced in several ways. Given 𝑃 = (𝑎, 𝑏), the rotation will rotate the entire coordinate rectangle for 𝑃 to a new rectangle. The point (𝑎, 0) is rotated to the point (0, 𝑎) on the 𝑦-axis. The point (0, 𝑏) is rotated to (−𝑏, 0) on the 𝑥-axis. Therefore, this new rectangle is the coordinate rectangle for (−𝑏, 𝑎). This gives a formula 𝑂90 (𝑎, 𝑏) = (−𝑏, 𝑎) valid for all points in the plane.

226

11. Coordinate Geometry

A second way to see this is from a product formula. Let 𝑠 be the line 𝑥 = 𝑦 as in the previous section. We know that 𝑂90 = ℛ𝑠 ℛ𝑂𝐸1 , and ℛ𝑠 ℛ𝑂𝐸1 (𝑥, 𝑦) = ℛ𝑠 (𝑥, −𝑦) = (−𝑦, 𝑥). The rotation of this rectangle for a point 𝑃 with positive coordinates is pictured in Figure 8. From this, one can already see the coordinates of 𝑃 ′ . This formula is consistent with the formula for half-turns, for 𝑂290 (𝑎, 𝑏) = 𝑂90 (−𝑏, 𝑎) = (−𝑎, −𝑏), which is the formula for ℋ𝑂 . (0,b) (- b,a)

(a,b)

(0,a)

(- b,0) (0,0)

(a,o)

Figure 8. Rotation by 90 Degrees

This tells us something about the slope of perpendicular lines. The line 𝑂𝑃 has slope 𝑏/𝑎, but the rotated line 𝑂𝑃 ′ has slope −𝑎/𝑏. Since every line is parallel to a line through 𝑂, this proves this result: Theorem 11.6 (Perpendicular Line Equation). If the equation of a line is 𝑦 = 𝑚𝑥 + 𝑏, the equation for every line perpendicular to this line is 𝑦 = (−1/𝑚)𝑥 + 𝑐, for some 𝑐. Euclidean Geometry of the General Line Equation. All the equations for a line can be converted to the form 𝑎𝑥 + 𝑏𝑦 = 𝑐. If 𝑏 ≠ 0, this line has slope −𝑎/𝑏. Let 𝑚 be the line through 𝑂 with equation 𝑎𝑥 + 𝑏𝑦 = 0. If one plots the point 𝑁 = (𝑎, 𝑏), the line 𝑂𝑁 has slope 𝑏/𝑎, so it is perpendicular to 𝑚 by the previous theorem. Thus, the direction from 𝑂 towards 𝑁 is perpendicular to 𝑚; this is sometimes called the normal direction. The point 𝑁 satisfies the equation 𝑎𝑥+𝑏𝑦 = 𝑎2 +𝑏2 , so this equation is the equation of a line parallel to 𝑚 at distance 𝑑 = ‖𝑂𝑁‖ = √𝑎2 + 𝑏2 from 𝑚, since 𝑂𝑁 is a segment perpendicular to both lines. Any point 𝑡𝑁 satisfies 𝑎𝑥+𝑏𝑦 = (𝑎2 +𝑏2 )𝑡 = 𝑑 2 𝑡 and is on a parallel line at distance |𝑡|𝑑 from 𝑚. The point 𝑁𝑐 = (𝑐/(𝑎2 + 𝑏2 ))𝑁 is the intersection point of 𝑂𝑁 and the line with equation 𝑎𝑥 + 𝑏𝑦 = 𝑐. We can replace the original coefficient pair (𝑎, 𝑏) with a unit normal point 𝑈 on 𝑂𝑁 at distance 1 from 𝑂. Let 𝑈 = (𝑝, 𝑞) = (1/𝑑)𝑁 = (𝑎/𝑑, 𝑏/𝑑). This is the equation of a line parallel to 𝑚 at distance |𝑐| from 𝑚: 𝑝𝑥 + 𝑞𝑦 =

𝑎 √𝑎2

+

𝑏2

𝑥+

𝑏 √𝑎2

+ 𝑏2

𝑦 = 𝑐.

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227

The lines with 𝑐 > 0 are the lines on the same side of 𝑚 as (𝑎, 𝑏). The lines with 𝑐 = 0, 1, 2, 3, . . . are parallel lines spaced a unit distance apart. For any point 𝑀 = (𝑚, 𝑛) on the line 𝑎𝑥 + 𝑏𝑦 = 𝑐, 𝑐 ≥ 0, consider the right triangle △𝑀𝑂𝑁𝑐 . If 𝜃 = 𝑚∠𝑀𝑂𝑁𝑐 , then the distance |𝑐|/√𝑎2 + 𝑏2 = ‖𝑂𝑁𝑐 ‖ = ‖𝑂𝑀‖ cos 𝜃 = √𝑚2 + 𝑛2 cos 𝜃 or 𝑎𝑚 + 𝑏𝑛 = 𝑐 = √𝑎2 + 𝑏2 √𝑚2 + 𝑛2 cos 𝜃. Definition 11.7. The expression 𝑎𝑚 + 𝑏𝑛 is called the dot product 𝑁 ⋅ 𝑀, with 𝑁 = (𝑎, 𝑏) and 𝑀 = (𝑚, 𝑛). From our interpretation of the equation 𝑎𝑚 + 𝑏𝑛 = 𝑐, the value of the dot product 𝑐 is given by this equation: 𝑁 ⋅ 𝑀 = ‖𝑂𝑁‖‖𝑂𝑀‖ cos 𝜃. This formula is also true for a negative dot product when the angle 𝜃 will be obtuse. This case is left to the exercises. The dot product has been introduced here as an algebraic formula applied to number pairs. The geometric interpretation of 𝑁 ⋅ 𝑀 given in the equation does not come into play unless an origin 𝑂 has been designated. Therefore, the dot product is not really a function of two points, since it implicitly requires more information. The resolution of this awkward situation is that the dot product is really a pairing of vectors, not points! So the dot product of two vectors, [𝐴𝐵] ⋅ [𝐶𝐷] is the algebraic formula (𝐵 − 𝐴) ⋅ (𝐷 − 𝐶). For vectors, the formula with cos 𝜃 still holds if 𝜃 is interpreted as the difference between the polar angles of the two vectors. This agrees with what has gone before if one interprets 𝑁 ⋅ 𝑀 as [𝑂𝑁] ⋅ [𝑂𝑀]. The dot denoting this product is the same notation used earlier for signed products of collinear segments. This should not lead to confusion since for such segments the product 𝐴𝐵 ⋅ 𝐶𝐷 equals [𝐴𝐵] ⋅ [𝐶𝐷] because for such segments 𝜃 is 0 or 180, so cos 𝜃 = ±1. Algebraic properties of the dot product simplify calculations with number pairs. Here 𝑀, 𝑁, 𝑃, 𝑄 are any points. The demonstrations of these formulas are either substitution calculations or were already carried out in the paragraphs above. Commutative Property: 𝑁 ⋅ 𝑀 = 𝑀 ⋅ 𝑁. Distributive Property: (𝑎𝑀 + 𝑏𝑁) ⋅ 𝑃 = 𝑎𝑀 ⋅ 𝑃 + 𝑏𝑁 ⋅ 𝑃. Perpendicular Directions: (𝑀 −𝑃)⋅(𝑁 −𝑄) = 0 if and only if 𝑀𝑃 is perpendicular to 𝑁𝑄. Distance: ‖𝑃𝑄‖2 = (𝑃 − 𝑄) ⋅ (𝑃 − 𝑄).

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Graphs and Transformations If 𝑓 is a function from the real numbers to the real numbers, we graph this function by plotting the points (𝑥, 𝑦) such that 𝑦 = 𝑓(𝑥). This is closely related to the geometry of the plane; in fact, the equation of a line can be interpreted as the graph of the function 𝑓(𝑥) = 𝑚𝑥 + 𝑏. It is also true that certain transformations are performed on graphs, with the transformed graphs then interpreted as functions. Here are some examples. • The transformation 𝑅(𝑥, 𝑦) = (𝑦, 𝑥) maps the graph to its reflection in the line with equation 𝑥 = 𝑦; the image can be interpreted as the graph of 𝑓−1 . • The transformation 𝑆(𝑥, 𝑦) = (𝑥 + 𝑎, 𝑦 + 𝑏) maps the graph to the graph of the function 𝑓(𝑥 − 𝑎) + 𝑏. • The transformation 𝑇(𝑥, 𝑦) = (−𝑥, 𝑦) maps the graph to the graph of the function 𝑓(−𝑥). • The transformation 𝑈(𝑥, 𝑦) = (2𝑥, 2𝑦) maps the graph to the graph of the function 2𝑓(𝑥/2). • The transformation 𝑉(𝑥, 𝑦) = (2𝑥, 𝑦) maps the graph to the graph of the function 𝑓(𝑥/2). This looks a lot like the geometry of the Euclidean plane. Almost all of these transformations are ones we have seen. But there are several fundamental ways in which the space of graphing is not the plane that we have been studying. First, most of the transformations of the Euclidean plane are not allowable in the space of graphs. It does not make sense to rotate a graph by 30 degrees or reflect it in the line with equation 𝑦 = 2𝑥. Second, while some of the transformations applied to graphs are similitudes, others are not and do not preserve angles or scale distance. Transformation 𝑉 in the list above is an example. Also, in the graph world, the slope has units and can change if the units change from seconds to hours or from dollars to euros. There is a reason for this difference. The Euclidean plane is a homogeneous space. All locations and all directions are equivalent. Even when we place coordinates on the plane, if we are in Euclidean geometry, we still have no preferred directions, just simplified equations perhaps. However, the space of graphing is not a unified two-dimensional space; rather it is the joining of two one-dimensional spaces as a product. This is seen most clearly when the function being graphed is not a function of a distance to a distance but a function of time to money, or temperature to density. The quantities measured by the 𝑥 and the 𝑦 can be completely different, and the scales on the axes can be different. If one is plotting time versus billions of dollars, then the visual slope of the graph on the page depends on whether one unit on the graph paper is one million or ten billion. Just because two mathematical quantities are described by two real numbers, they need not be the same thing. Even points and vectors are not the same kind of objects. The space of graphing is not really ℝ2 but ℝ1 ×ℝ1 , a product. This means that the space

Unit Circle and Rotation Formula

229

consists of pairs, with the first element a number from one space, the 𝑥-space, and the second from a second space, the 𝑦-space. The first element may be time measured in years and the second may be dollars measured in megabucks. Just because both spaces are ℝ, there is no reason that the two are otherwise related. So, in short, one can learn something about graphs from geometry about slope and other properties, but one should be very cautious when applying geometrical relationships to the world of graphs.

Unit Circle and Rotation Formula Rotations 𝑂 𝜃 map each circle centered at 𝑂 onto itself. The unit circle, the circle of radius 1 with center 𝑂, will play an important role in the formula for rotations. E2 = (0,1)

U( )

V( ) E1 =(1,0) O

Figure 9. Points on Unit Circle

⃗1 , Let polar angles at 𝑂 be measured with a protractor that maps the angle 0 to 𝑂𝐸 ⃗2 . Then every point on the unit circle has an angle with 90 being the polar angle of 𝑂𝐸 that can be taken to be in the range from 0 to 360, or from −180 to 180. A point 𝑈(𝜃) corresponding to polar angle 𝜃 on the unit circle has coordinates (𝑐, 𝑠), with 𝑐2 + 𝑠2 = 1. We can extend the definition of trigonometric functions from acute angles to all of these angles by setting 𝑈(𝜃) = (cos 𝜃, sin 𝜃). One way of thinking about the point 𝑈(𝜃) is that it is a rotated point, namely 𝑂 𝜃 (𝐸1 ). Denote by 𝑉(𝜃) the rotation of 𝑈(𝜃) by 90 degrees; by the formula for this rotation, 𝑉(𝜃) = (− sin 𝜃, cos 𝜃) = 𝑂 𝜃 (𝐸2 ). Rotation Formula for Center 𝑂. One can rotate a point 𝑃 = (𝑥, 𝑦) by angle 𝜃 along with its coordinate rectangle 𝑂𝑃1 𝑃𝑃2 , with diagonal 𝑂𝑃 and vertices at (𝑥, 0) and (0, 𝑦) as shown in Figure 10. The rotation images by 𝑂 𝜃 of the points on the axes are multiples of 𝑈(𝜃) and 𝑉(𝜃): 𝑃1′ = 𝑂 𝜃 (𝑥, 0) = 𝑥𝑈(𝜃) and 𝑃2′ = 𝑂 𝜃 (0, 𝑦) = 𝑦𝑉(𝜃). This is clear from the direction and length of the segments 𝑂𝑃1′ and 𝑂𝑃2′ . Then 𝑃 ′ = 𝑂 𝜃 (𝑃) is a translate of 𝑂: 𝑃 ′ = 𝒯𝑂𝑃1′ 𝒯𝑃1′ 𝑃 ′ (𝑂), which equals 𝒯𝑂𝑃1′ 𝒯𝑂𝑃2′ (𝑂) since 𝑂𝑃1′ 𝑃 ′ 𝑃2′ is a rectangle. Thus, 𝑃′ = 𝑂 + 𝑃1′ + 𝑃2′ = 𝑥𝑈(𝜃) + 𝑦𝑉(𝜃)

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P' P 2 =(0,y )

P

P 1' P 2' P 1 =(x ,0) O

Figure 10. Rotation of 𝑃 and Its Coordinate Rectangle

and so 𝑂 𝜃 (𝑃) = 𝑂 𝜃 (𝑥, 𝑦) = 𝑥(cos 𝜃, sin 𝜃) + 𝑦(− sin 𝜃, cos 𝜃). For a formula for 𝐴𝜃 , centered at a point 𝐴 that is not 𝑂, simply translate by 𝒯𝐴𝑂 , then rotate with center 𝑂, then translate back: 𝒯𝑂𝐴 𝑂 𝜃 𝒯𝐴𝑂 (𝑃) = 𝐴 + 𝑂 𝜃 (𝑃 − 𝐴). From the Product of Rotations Theorem, Theorem 10.2, this is a rotation by angle 𝜃. Since 𝐴 is a fixed point, the product must be 𝐴𝜃 . The rotation formula implies the addition formulas for cosine and sine. First apply the formula: 𝑈(𝜃 + 𝜙) = 𝑂 𝜃 (𝑈(𝜙)) = 𝑂 𝜃 (cos 𝜙, sin 𝜙) = cos 𝜙(cos 𝜃, sin 𝜃) + sin 𝜙(− sin 𝜃, cos 𝜃). Then note the value of 𝑈(𝜃 + 𝜙): (cos(𝜃 + 𝜙), sin(𝜃 + 𝜙)) = (cos 𝜙 cos 𝜃 − sin 𝜙 sin 𝜃, cos 𝜙 sin 𝜃 + sin 𝜙 cos 𝜃). For the special case 𝜃 = 𝜙, this gives formulas for cos 2𝜃 and sin 2𝜃. Line Reflection Formula. A formula for reflection in a line 𝑚 through 𝑂 can be derived in the same way. Let 𝑚 = 𝑂𝑈(𝜃). P' P1' P2 = (0,y)

P

U( )

O

P2' P1 = (x,0)

E1

Figure 11. Reflection of 𝑃 and Its Coordinate Rectangle

• The reflection image 𝑃1′ = ℛ𝑚 (𝑃1 ) is also the rotation image 𝑂2𝜃 (𝑃1 ), since ∠𝑃1 𝑂𝑈(𝜃) ≅ 𝑃1′ 𝑂𝑈(𝜃) and 𝑂𝑃1 ≅ 𝑂𝑃1′ , with 𝜃 = 𝑚∠𝑃1 𝑂𝑈(𝜃). Therefore, 𝑃1′ = 𝑥𝑈(2𝜃).

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231

• The reflection image 𝑃2′ = ℛ𝑚 (𝑃2 ) is also the rotation image 𝑂2(𝜃−90) (𝑃2 ), since ∠𝑃2 𝑂𝑈(𝜃) ≅ 𝑃2′ 𝑂𝑈(𝜃) and 𝑂𝑃2 ≅ 𝑂𝑃2′ with |𝜃 − 90| = 𝑚∠𝑃2 𝑂𝑈(𝜃). Therefore, 𝑃2′ = 𝑦𝑉(2𝜃 − 180) = −𝑦𝑉(2𝜃). As for rotations, this leads to a formula for 𝑃′ = ℛ𝑚 𝑃 = 𝑃1′ + 𝑃2′ : ℛ𝑂𝑈(𝜃) (𝑃) = 𝑥𝑈(2𝜃) + 𝑦𝑉(2𝜃) = 𝑥(cos 2𝜃, sin 2𝜃) + 𝑦(sin 2𝜃, − cos 2𝜃). We relied on Figure 11 to derive this formula. But now that we have a formula, we can check directly that the function is the correct reflection for all 𝜃. Let 𝐹 be the function with the formula given in the displayed equation. One can check by a calculation that this 𝐹 preserves distance, so it is a rigid motion. Also one can calculate 𝐹(𝑂) = 𝑂, 𝐹(𝑈(𝜃)) = 𝑈(𝜃), and 𝐹(𝑉(𝜃)) = −𝑉(𝜃). Thus, 𝐹 agrees with the line reflection ℛ𝑂𝑈(𝜃) at these three points, so it must be the same rigid motion. Some details of this calculation will appear in the exercises.

Reflections in Lines Not Through 𝑂. Again, for a reflection in a line 𝑛 not through 𝑂, let 𝐴 be a point of 𝑛, and let 𝑛′ = 𝒯𝐴𝑂 (𝑛). Then 𝒯𝑂𝐴 ℛ𝑛′ 𝒯𝐴𝑂 is reflection in 𝑛, so ℛ𝑛′ (𝑃) = 𝐴 + ℛ𝑛′ (𝑃 − 𝐴). With this formula, we now have formulas for translation, dilation, rotation, and reflection. In the next two sections, we will show how these formulas can be expressed with different mathematical tools or in different mathematical language.

Complex Numbers and Transformations of the Plane The complex numbers have the form 𝑧 = 𝑎 + 𝑖𝑏, where 𝑎 and 𝑏 are real numbers and 𝑖2 = −1. The numbers can be represented by points in the Cartesian coordinate plane by identifying this number with (𝑎, 𝑏). Addition agrees with coordinatewise addition of points in the plane: (𝑎 + 𝑖𝑏) + (𝑐 + 𝑖𝑑) = (𝑎 + 𝑐) + 𝑖(𝑏 + 𝑑). Multiplication is (𝑎 + 𝑖𝑏)(𝑐 + 𝑖𝑑) = 𝑎𝑐 + 𝑖(𝑎𝑑 + 𝑏𝑐) + 𝑖2 𝑏𝑑 = (𝑎𝑐 − 𝑏𝑑) + 𝑖(𝑎𝑑 + 𝑏𝑐). The complex conjugate 𝑧 of 𝑧 = 𝑎 + 𝑖𝑏 is 𝑎 − 𝑖𝑏. The product 𝑧𝑧 is a nonnegative real number: 𝑧𝑧 = (𝑎 + 𝑖𝑏)(𝑎 − 𝑖𝑏) = 𝑎2 + 𝑏2 . The square root of this number is the modulus, the distance to 0: |𝑧| = |𝑎 + 𝑖𝑏| = √𝑎2 + 𝑏2 . Since the product 𝑖(𝑎 + 𝑖𝑏) = −𝑏 + 𝑖𝑎, the function that maps a complex number 𝑧 to 𝑖𝑧 is the rotation 𝑂90 of the plane, since the formulas are the same. More generally, if |𝑤| = 1, then the complex number 𝑤 is on the unit circle and has the form 𝑤 = 𝑢(𝜃) = cos 𝜃 + 𝑖 sin 𝜃 for some 𝜃. (For the rest of this section, the temporary notation 𝑢(𝜃) will denote this complex number with modulus 1.) For 𝑧 = (𝑥 + 𝑖𝑦), the formula for this product is the formula for 𝑂 𝜃 (𝑧): 𝑢(𝜃)(𝑥 + 𝑖𝑦) = (cos 𝜃 + 𝑖 sin 𝜃)(𝑥 + 𝑖𝑦) = 𝑥(cos 𝜃 + 𝑖 sin 𝜃) + 𝑦(− sin 𝜃 + 𝑖 cos 𝜃).

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More generally, the function 𝑆𝛼 (𝑧) = 𝛼𝑧, where 𝛼 = 𝑎 + 𝑖𝑏 is a constant nonzero complex number, is a transformation of the plane. The inverse of the transformation is multiplication by 𝛼−1 = 𝛼/|𝛼|2 . 4

2Z

3

2Z + i Z

2

Z 1

–4

–2

2 + 1i

i

2

iZ

4

–1

–2

Figure 12. Geometry of Complex Product (2 + 1𝑖)𝑧

Any nonzero complex number 𝛼 = |𝛼|(𝛼/|𝛼|), where (𝛼/|𝛼|) = 𝑢(𝜃) has modulus 1. Multiplication by the real number |𝛼| is dilation with center 𝑂 and dilation ratio |𝛼|. Therefore, the transformation 𝑆𝛼 (𝑧) can be factored into 𝒟𝑂,|𝛼| 𝑂 𝜃 , the product of a dilation and a rotation. This is an orientation-preserving similitude. The example of complex multiplication in Figure 12 can be viewed two ways. It is a dilation rotation by 𝑧 of 2 + 1𝑖, scaling the coordinate rectangle of 2 + 1𝑖 by |𝑧| and rotating by 𝑧/|𝑧|. One can also view this as a dilation rotation of 𝑧 by 2 + 1𝑖 as a construction, starting with a triangle △0 2 (2 + 1𝑖) and then building a similar triangle with base segment 0 2𝑧 in place of segment 0 2 as the base, then adding a perpendicular segment 0 𝑖𝑧 of length |𝑧| instead of the side of length 1 in the original triangle. This second interpretation constructs a rectangle similar to the coordinate rectangle of 2 + 1𝑖 and shows how the angle of rotation and the length of (2 + 1𝑖)𝑧 emerge from the geometry. Translations have the form 𝑇(𝑧) = 𝑧 + 𝛽, for some complex number 𝛽. Therefore, 𝑈(𝑧) = 𝛼𝑧 + 𝛽 is an orientation-preserving similitude and is a rigid motion if |𝛼| = 1. In fact, every such similitude and rigid motion can be written in this form. A proof will be explored in the exercises. The complex conjugate map 𝐶(𝑧) = 𝑧 is reflection in the 𝑥-axis. Then 𝑆𝛼 𝐶𝑆𝛼−1 is reflection in line 0𝛼. The formula for this map is 𝑆𝛼 𝐶𝑆𝛼−1 (𝑧) = 𝑆𝛼 𝐶(𝛼𝑧/|𝛼|2 ) = 𝑆𝛼 (𝛼𝑧/|𝛼|2 ) = (𝛼/|𝛼|)2 𝑧 = 𝑢(𝜃)2 𝑧 = 𝑢(2𝜃)𝑧. This agrees with the formula derived in the previous section. The composition of this map with translations provides a complex expression for any line reflection.

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233

The flexibility provided by the algebra of complex numbers makes some geometrical theorems easier to prove. Theorem 11.8 (Similitude Fixed Point). Every orientation-preserving similitude that is not a translation has a unique fixed point. Proof. Such a transformation has the form 𝑈(𝑧) = 𝛼𝑧 + 𝛽, with 𝛼 ≠ 1. A fixed point 𝑧 satisfies 𝑧 = 𝛼𝑧 + 𝛽. A unique solution of this equation is found by elementary algebra: 𝑧0 = 𝛽/(1 − 𝛼). □ One can check that if |𝛼| < 1, then for any 𝑧, the sequence of images 𝑈 𝑘 (𝑧) approaches 𝑧0 as a limit. This is illustrated by the spiral in Figure 8.40. Functions of complex numbers can also be used to describe additional geometrical transformations of the plane. For example, 𝐽(𝑧) = 1/𝑧 is a transformation called reflection in the unit circle; this fixes the points on the unit circle but exchanges the interior with the exterior, much as line reflection does. Inversion leads to some beautiful geometry of circles that is beyond the scope of this book. Subsets of complex numbers are also used to model non-Euclidean plane geometry. References include Hahn [10].

Barycentric Coordinates In this section we will introduce barycentric coordinates, which locate points as barycenters, or centers of mass. In Theorem 11.5, we saw that any point on 𝑃𝑄 can be written as (1 − 𝑡)𝑃 + 𝑡𝑄 for some real 𝑡. This expression represents the same point whatever the coordinate system. Other expressions in the coordinates of 𝑃 and 𝑄 will not define points on the plane in a way that is independent of the coordinate system. For example, the expression 𝑃+𝑄 denotes the point 𝑁 such that 𝑃𝑂𝑄𝑁 is a parallelogram. If the origin 𝑂 is changed, so is 𝑁. In contrast, (1 − 𝑡)𝑃 + 𝑡𝑄 always denotes the point 𝐴 on 𝑃𝑄 with 𝐴𝑃/𝑄𝑃 = 𝑡. Center of Mass of a Segment. A more symmetric way of writing the parametric expression for a point 𝐺 on a line 𝑃1 𝑃2 is to say that 𝐺 = 𝑐 1 𝑃1 + 𝑐 2 𝑃2 , where 𝑐 1 + 𝑐 2 = 1. The point 𝐺 is the center of mass, the barycenter, of a system consisting of a mass 𝑐 1 at 𝑃1 and 𝑐 2 at 𝑃2 . This follows from the law of levers of Archimedes: at the barycenter, the products of mass times lever arm are equal and opposite. In this case, 𝑐 1 (𝑃1 − 𝐺) = −𝑐 2 (𝑃2 − 𝐺). If there are masses 𝑚1 and 𝑚2 at these points that do not sum to 1, there is still a center of mass. Let 𝑀 = 𝑚1 + 𝑚2 ; then set 𝐺 = (𝑚1 /𝑀)𝑃1 + (𝑚2 /𝑀)𝑃2 . For example, if one puts 2 grams at 𝑃1 and 5 grams at 𝑃2 , then the center of mass is (2/7)𝑃1 + (5/7)𝑃2 ; this is the point on the segment with 𝐺𝑃1 /𝐺𝑃2 = −5/2. When speaking of masses, the numbers 𝑐 𝑖 are nonnegative and the point 𝐺 will belong to the segment 𝑃1 𝑃2 . In the formula for 𝐺, negative real numbers are allowed, with 𝐺 possibly located at any point of 𝑃1 𝑃2 . But the physical analogy must be extended to springs or balloons that pull upward, opposite from gravity.

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G P1

P2

Figure 13. Barycenter of Masses 2 and 5

Affine Combinations and Barycenters. We will extend this idea of a center of mass to a system of more than two points. The next theorem will set the stage. Theorem 11.9 (Affine Combinations). Concerning 𝑐 1 𝑃1 + 𝑐 2 𝑃2 + ⋯ + 𝑐𝑛 𝑃𝑛 : • An expression 𝑐 1 𝑃1 + 𝑐 2 𝑃2 + ⋯ + 𝑐𝑛 𝑃𝑛 is called an affine combination of points if 𝑐 1 + 𝑐 2 + ⋯ + 𝑐𝑛 = 1. Such an expression equals a point 𝐺 independent of the coordinate system. • If the sum 𝑐 1 + 𝑐 2 + ⋯ + 𝑐𝑛 = 0, the expression 𝑐 1 𝑃1 + 𝑐 2 𝑃2 + ⋯ + 𝑐𝑛 𝑃𝑛 equals a vector independent of the coordinate system. • If 𝑐 1 + 𝑐 2 + ⋯ + 𝑐𝑛 = 1, the vector 𝑐 1 (𝑃1 − 𝐺) + 𝑐 2 (𝑃2 − 𝐺) + ⋯ + 𝑐𝑛 (𝑃𝑛 − 𝐺) is the zero vector. This is the sum of the masses times the lever arm at 𝐺. Proof. Here is a proof for three points, 𝐺 = 𝑐 1 𝑃1 + 𝑐 2 𝑃2 + 𝑐 3 𝑃3 , where 𝑐 1 + 𝑐 2 + 𝑐 3 = 1. If 𝑐 1 + 𝑐 2 = 0, then 𝑐 1 (𝑃1 − 𝑃2 ) = 𝑣 is a vector and 𝐺 = 𝒯𝑣 (𝑃3 ) is a point independent of coordinates. This completes the proof in this case. Now assuming that 𝑐 1 + 𝑐 2 ≠ 0, let 𝑄 = 1/(𝑐 1 + 𝑐 2 )(𝑐 1 𝑃1 + 𝑐 2 𝑃2 ), a point on 𝑃1 𝑃2 . Then 𝐺 = (1 − 𝑐 3 )𝑄 + 𝑐 3 𝑃3 since 𝑐 1 + 𝑐 2 = 1 − 𝑐 3 ; this 𝐺 is a point on 𝑄𝑃3 . Both 𝑄 and 𝐺 are independent of the coordinate system, by Theorem 11.5. Next, if 𝑐 1 +𝑐 2 +𝑐 3 = 0, let 𝑣 = 𝑐 1 𝑃1 +𝑐 2 𝑃2 +𝑐 3 𝑃3 be computed by adding coordinates in some coordinate system. Let 𝐸 = 𝑃1 . The point 𝐹 = 𝐸 + 𝑣 = (1 + 𝑐 1 )𝑃1 + 𝑐 2 𝑃2 + 𝑐 3 𝑃3 is a well-defined point independent of any coordinate system, since 1 + 𝑐 1 + 𝑐 2 + 𝑐 3 = 1. Therefore, 𝑣 = [𝐸𝐹] is a vector independent of the coordinate system. Lastly, if 𝑐 1 + 𝑐 2 + 𝑐 3 = 1, the vector 𝑐 1 (𝑃1 − 𝐺) + 𝑐 2 (𝑃2 − 𝐺) + 𝑐 3 (𝑃3 − 𝐺) = 𝑐 1 𝑃1 + 𝑐 2 𝑃2 + 𝑐 3 𝑃3 − (𝑐 1 + 𝑐 2 + 𝑐 3 )𝐺 = 𝐺 − 𝐺 = 0. This proves the theorem for 𝑛 = 3 by using the already-known result for 𝑛 = 2. This reasoning can be continued inductively, step by step in the same manner, for any number of summands, but we will leave the details to the reader. In what follows, the case 𝑛 = 3 is the case we will need. □ Example 11.10. For a parallelogram 𝐴𝐵𝐶𝐷, the center of symmetry 𝑀 can be written in an infinite number of ways: for example, (1/2)(𝐴 + 𝐶), (1/4)(𝐴 + 𝐵 + 𝐶 + 𝐷), or (1/2)(2𝐴 − 𝐵 + 2𝐶 − 𝐷). In Theorem 8.50 we learned that the medians of a triangle are concurrent at a point called the centroid of the triangle. The centroid of △𝐴𝐵𝐶 is 𝐺 = (1/3)(𝐴 + 𝐵 + 𝐶). To see this, let 𝑀 = (1/2)(𝐴 + 𝐵) be the midpoint of 𝐴𝐵. Then 𝐶𝑀 is a median, and

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A

C

(C + A)/2

G ( A+ B )/2

(B + C )/2

B Figure 14. Centroid 𝐺 = (1/3)(𝐴 + 𝐵 + 𝐶)

𝐺 = (2/3)𝑀 + (1/3)𝐶 is the point on the median 𝐶𝑀 with 𝐺𝑀/𝐶𝑀 = 1/3. By the same reasoning, this 𝐺 is on any median, so the medians are concurrent at 𝐺. Unlike the example of the parallelogram, there is no other affine combination of the vertices that equals 𝐺. Why this is unique will be shown in Theorem 11.11. Barycenter of Masses on a Triangle. Given a triangle △𝑃1 𝑃2 𝑃3 , a triple of numbers (𝑐 1 , 𝑐 2 , 𝑐 3 ), with 𝑐 1 + 𝑐 2 + 𝑐 3 = 1, defines a point 𝐺 = 𝑐 1 𝑃1 + 𝑐 2 𝑃2 + 𝑐 3 𝑃3 . This points can be interpreted as the center of mass for three masses located at 𝑃𝑖 when the 𝑐 𝑖 ≥ 0. This follows from the third point in Theorem 11.9 by the lever principle of Archimedes. As a physical example, imagine supporting and balancing a rigid triangle on the edge of a ruler that passes under 𝑃3 . With a mass at each vertex, to balance the remaining two masses, the ruler should pass under the center of mass 𝑄 of the side 𝑃1 𝑃2 . Then to balance the whole structure on a fingertip at a point of 𝑄𝑃3 , one can conceptually replace the segment 𝑃1 𝑃2 with masses at each endpoint by its center of mass 𝑄 with mass 𝑐 1 + 𝑐 2 . This is the reasoning used in the proof of Theorem 11.9 to find the balance point at 𝐺. This is the same balance principle used in a hanging mobile, where every pair of objects is hung from a stick attached by a thread at the center of mass of the two objects. Theorem 11.11 (Barycentric Coordinates). Given a triangle △𝑃1 𝑃2 𝑃3 , every point 𝑃 of the plane can be expressed in exactly one way as 𝑃 = 𝑐 1 𝑃1 +𝑐 2 𝑃2 +𝑐 3 𝑃3 , where 𝑐 1 +𝑐 2 +𝑐 3 = 1. This unique triple of numbers can be used as coordinates for 𝑃. These are called the normalized barycentric coordinates of 𝑃 with respect to this triangle. Proof. This is actually not new; these are affine coordinates in disguise. If we set 𝑃3 = 𝑂, then 𝑐 1 𝑃1 + 𝑐 2 𝑃2 + 𝑐 3 𝑃3 = (1 − 𝑐 1 − 𝑐 2 )𝑂 + 𝑐 1 𝑃1 + 𝑐 2 𝑃2 ). Then (𝑐 1 , 𝑐 2 ) are the affine coordinates of 𝑃 with respect to △𝑂𝑃1 𝑃2 .

□

One may wonder what is the point of repackaging affine coordinates. We will see that treating the three vertices of the coordinate triangle symmetrically has interesting consequences. In the case of masses at the vertices 𝑃𝑖 where the nonzero sum of the masses is not 1, the masses (𝑚1 , 𝑚2 , 𝑚3 ) still determine a point. Let 𝑀 = 𝑚1 + 𝑚2 + 𝑚3 and let 𝑐 𝑖 = 𝑚𝑖 /𝑀. Then 𝐺 = (1/𝑀)(𝑚1 𝑃1 + 𝑚2 𝑃2 + 𝑚3 𝑃3 )

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is a point that is a center of mass when the 𝑚𝑖 ≥ 0. The numbers (𝑚1 , 𝑚2 , 𝑚3 ) are barycentric coordinates of 𝐺. Unlike (𝑥, 𝑦)-coordinates, barycentric coordinates are homogeneous. The number triples (𝑘𝑚1 , 𝑘𝑚2 , 𝑘𝑚3 ) define barycentric coordinates that are considered the same for any 𝑘 ≠ 0, and they determine the same point. The one restriction is that 𝑀 = 𝑚1 + 𝑚2 + 𝑚3 ≠ 0. Example 11.12. On the left side of Figure 15 there is shown a △𝐴𝐵𝐶 with masses 1, 2, and 4, at the respective vertices. So (1, 2, 4) are barycentric coordinates for a point, with 𝑀 = 7. The barycenters of each pair of masses on the sides are 𝐷 = (2/6)𝐵 + (4/6)𝐶, 𝐸 = (4/5)𝐶 + (1/5)𝐴, 𝐹 = (1/3)𝐴 + (2/3)𝐵 with the barycenter for three masses 𝐺 = (1/7)𝐴 + (2/7)𝐵 + (4/7)𝐶. The ratios on the sides are 𝐷𝐵/𝐷𝐶 = −2, 𝐸𝐶/𝐸𝐴 = −1/4, 𝐹𝐴/𝐹𝐵 = −2. A, m = 1

G= (1/7)(1A + 2B + 4C) 2 GB

F

B, m = 2

G D

E

1 GA 4 GC

C, m = 4

Figure 15. Barycenter and Vector Sum

Segments representing the three vectors from G to the vertices are shown in different colors. On the right side of the figure there is plotted the sum of the three vector lever arms using the same colors: 1(𝐴 − 𝐺) + 2(𝐵 − 𝐺) + 4(𝐶 − 𝐺), which is 0 as it should be, since 0 = 𝑐 1 (𝑃1 − 𝐺) + 𝑐 2 (𝑃2 − 𝐺) + 𝑐 3 (𝑃3 − 𝐺). We can see in this example a way to show that the barycenter is inside the triangle if the masses are positive. On the segments 𝐴𝐷, 𝐵𝐸, and 𝐶𝐹, the negative signed ratios 𝐺𝐷/𝐺𝐴 = −1/6, 𝐺𝐸/𝐺𝐵 = −2/5, and 𝐺𝐹/𝐺𝐶 = −4/3 show that 𝐺 is an interior point of each segment. Cevians and Ceva’s Theorems. In Example 11.12, the segments 𝐴𝐷, 𝐵𝐸, and 𝐶𝐹 are called cevians. The signed ratios of the endpoint of the cevians on the sides satisfied this equation: (𝐷𝐵/𝐷𝐶) (𝐸𝐶/𝐸𝐴) (𝐹𝐴/𝐹𝐵) = (−2)(−1/4)(−2) = −1. We will now look at cevians and these ratios in general. Definition 11.13. A segment joining any vertex of △𝐴𝐵𝐶 to a point of the (extended) opposite side is called a cevian. A cevian line is a line containing a cevian. For a point 𝐺 with normalized barycentric coordinates (𝑐 1 , 𝑐 2 , 𝑐 3 ) with respect to △𝐴𝐵𝐶, the line 𝐴𝐺 will intersect 𝐵𝐶 at a point 𝐷 = (1 − 𝑡)𝐴 + 𝑡𝐺 where 𝑐 1 = 0.

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237

This means that the barycentric coordinates of 𝐷 will be (0, 𝑐 2 , 𝑐 3 ) and the normalized barycentric coordinates will be (0, 𝑐 2 /(1 − 𝑐 1 ), 𝑐 3 /(1 − 𝑐 1 )). The point of intersection will not exist, and 𝐴𝐺 will be parallel to 𝐵𝐶 when 𝑐 1 = 1. The signed ratio 𝐷𝐵/𝐷𝐶 will equal −𝑐 3 /𝑐 2 except when 𝑐 2 = 0 and the ratio is not defined. Now let point 𝐺 have coordinates (𝑐 1 , 𝑐 2 , 𝑐 3 ) with respect to △𝑃1 𝑃2 𝑃3 for which none of the 𝑐 𝑖 equals 0 or 1. Then the triangle has three cevian lines 𝑃𝑖 𝑄 𝑖 concurrent at 𝐺 and with this relationship among the ratios on the sides: −𝑐 3 −𝑐 1 −𝑐 2 𝑄1 𝑃2 𝑄2 𝑃3 𝑄3 𝑃1 = = −1. 𝑄1 𝑃3 𝑄2 𝑃1 𝑄3 𝑃2 𝑐2 𝑐3 𝑐1 If all the 𝑐 𝑖 > 0, then the cevian segments concur at 𝐺 at a point inside the triangle. The converse is an important triangle concurrence theorem. To avoid the special cases, the theorem is stated for cevian segments. Theorem 11.14 (Ceva’s Theorem). In △𝑃1 𝑃2 𝑃3 , let the three cevians 𝑃1 𝑄1 , 𝑃2 𝑄2 , 𝑃3 𝑄3 have endpoints 𝑄1 , 𝑄2 , 𝑄3 on the sides of the triangle, not at a vertex. Then the cevians are concurrent at a point 𝐺 if and only if 𝑄1 𝑃2 𝑄2 𝑃3 𝑄3 𝑃1 = −1. 𝑄1 𝑃3 𝑄2 𝑃1 𝑄3 𝑃2 Proof. The“only if” claim is all that is left to prove, given what has already been proved, The first step of the proof is to use two ratios on the sides, 𝑟2 and 𝑟3 , to find the barycentric coordinates of 𝐺, the point of concurrence of two of the cevians 𝑃2 𝑄2 and 𝑃3 𝑄3 . Assume that there is mass 𝑚1 = 1 at 𝑃1 ; then 𝑟2 = 𝑄2 𝑃3 /𝑄2 𝑃1 = −𝑚1 /𝑚3 , so 𝑚3 = −1/𝑟2 . Also, 𝑟3 = 𝑄3 𝑃1 /𝑄3 𝑃2 = −𝑚2 /𝑚1 , so 𝑚2 = −𝑟3 . Therefore, (1, −𝑟3 , −1/𝑟2 ) are barycentric coordinates for the intersection point. Then let 𝑀 = 1 − (1/𝑟2 ) − 𝑟3 . Since 𝑀 > 1, we can set 𝑐 1 = 1/𝑀, 𝑐 2 = −𝑟3 /𝑀, 𝑐 3 = −1/(𝑟2 𝑀) as the normalized barycentric coordinates for a point 𝐺 that is the intersection of the cevians through 𝑃2 and 𝑃3 . The last part of the proof is to show that the intersection 𝐽 of 𝑃1 𝐺 with 𝑃2 𝑃3 is the point 𝑄1 . Since we have the barycentric coordinates of 𝐺, we know 𝐽 is the barycenter of 𝑃2 𝑃3 with mass −𝑟3 at 𝑃2 and −1/𝑟2 at 𝑃3 . The point 𝐽 divides the segment in the ratio 𝐽𝑃2 /𝐽𝑃3 = −(1/𝑟3 )/𝑟2 = −1/𝑟2 𝑟3 . Since 𝑟1 𝑟2 𝑟3 = −1, this ratio is 𝑟1 and 𝐽 = 𝑄1 . □ Note. This proof also applies to cases when two of the 𝑄 𝑖 are on extended sides of the triangle, but for some cases the 𝑃𝑖 𝑄 𝑖 will be parallel and not concurrent. We have encountered a number of special lines in triangles. Perpendicular bisectors are not cevians. But medians, angle bisectors, and altitudes are all cevians, and this theorem applies to them. Figure 16 shows another barycentric coordinate example with one coordinate negative. The point 𝑆 has barycentric coordinates (4, −1, 3), or (2/3, −1/6, 1/2), relative to △𝐴𝐵𝐶. Since the second coordinate is negative, the vertices 𝐵 and 𝑆 are on opposite sides of 𝐴𝐶. The segment ratios on the sides and their product are shown in the figure.

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DB = 3.00 DC EC = –1.33 EA

F A

FA = 0.25 FB

S E

DB FA EC · · = –1.00 DC FB EA B

C

D

Figure 16. Cevian Lines and Ratios of Mixed Sign

Note. The Menelaus Theorem, Theorem 8.29, asserts that when the product of ratios is positive instead of negative, the three points 𝑄 𝑖 are collinear. This provides a straightedge construction of harmonic division, as shown in Figure 8.26. Areal Coordinates. The normalized barycentric coordinates of points in a triangle are sometimes called areal coordinates. The reason for this can be seen in the example in Figure 15. In this figure, the ratio 𝐷𝐺/𝐷𝐴 = 1/7. This means that the height of △𝐺𝐵𝐶 is 1/7 the height of △𝐴𝐵𝐶. Since the base of each triangle is 𝐵𝐶, the area 𝒜(△𝐺𝐵𝐶) = (1/7)𝒜(△𝐴𝐵𝐶). By the same reasoning, 𝒜(△𝐺𝐶𝐴) = (2/7)𝒜(△𝐴𝐵𝐶)

and

𝒜(△𝐺𝐴𝐵) = (4/7)𝒜(△𝐴𝐵𝐶).

Thus, the three numbers (1/7, 2/7, /4/7) are the normalized barycentric coordinates but also represent the fractions of the total area 𝒜(△𝐴𝐵𝐶) that are the areas of the three subtriangles with vertex 𝐺. And the sum of these coordinates is 1, since the sum of the three triangle areas is 𝒜(△𝐴𝐵𝐶). In Figure 16 the area ratios can be computed in the same way. Let 𝔸 = 𝒜(△𝐴𝐵𝐶). The area ratios are 𝒜(△𝑆𝐵𝐶)/𝔸 = 2/3, 𝒜(△𝑆𝐶𝐴)/𝔸 = 1/6, and 𝒜(△𝑆𝐴𝐵)/𝔸 = 1/2. This time 2/3 + 1/6 + 1/2 ≠ 1 since the three triangles do not subdivide the original triangle. Instead, the sum of two of their areas is the area of the quadrilateral 𝐴𝐵𝐶𝑆. However, the sum of the ratios will still be 1 if 𝒜(△𝑆𝐶𝐴) is considered negative. This makes sense since the orientation of the triangle is opposite that of all the other triangles. In the figure one sees that removing this area from the area of the quadrilateral 𝐴𝐵𝐶𝑆 does equal the area of △𝐴𝐵𝐶. And the calculation shows this also: 2/3 − 1/6 + 1/2 = 6/6 = 1. This idea of signed area, of setting area to be negative if a figure has negative orientation, is an idea that appears in many guises in geometry. Example 11.15. For △𝐴𝐵𝐶 with angle measures 𝛼, 𝛽, and 𝛾 at the vertices 𝐴, 𝐵, and 𝐶, the barycentric coordinates of the circumcenter can be deduced from the areas of △𝑂𝐴𝐵, △𝑂𝐵𝐶, and △𝑂𝐵𝐶 as shown in Figure 17. For instance, △𝑂𝐴𝐵 is an isosceles triangle with angle measure 2𝛾 at 𝑂, since ∠𝐵𝐴𝐶 is an inscribed angle. The equal sides of the triangle have length 𝑟, the radius of the circle. The area of this isosceles

Barycentric Coordinates

239

triangle is twice the area of the right triangle △𝐴𝑂𝐹. The legs of this triangle are 𝑟 sin 𝛾 and 𝑟 cos 𝛾, so the area is 𝑟2 sin 𝛾 cos 𝛾 = (𝑟2 /2) sin 2𝛾. Dividing each area by 𝑟2 , the barycentric coordinates of 𝑂 are (sin 2𝛼, sin 2𝛽, sin 2𝛾). This formula remains true when one of the angles is obtuse, though the area of the corresponding isosceles triangle will be negative. To get the areal or normalized barycentric coordinates, one must either divide by the sum of these numbers or use the area of △𝐴𝐵𝐶.

A E C

O

F

D B

Figure 17. The Arial Coordinates of the Circumcenter

Intersections from Barycentric Coordinates. Confronted by points 𝐸 and 𝐹 on the sides of a triangle, as in Figure 18, one can use barycentric coordinates to find the coordinates of the intersection 𝐺. If we know that 𝐴𝐸/𝐴𝐵 = 1/5 and 𝐴𝐹/𝐴𝐶 = 2/5, then each point can be viewed as a barycenter on a side. The point 𝐸 is the barycenter of a mass 4𝑚 at 𝐴 and 1𝑚 at 𝐵, while 𝐹 is the barycenter of mass 3𝑛 at 𝐴 and 2𝑛 at 𝐶. If we set 𝑚 = 3 and 𝑛 = 4, we get barycentric coordinates (12, 3, 8) with respect to △𝐴𝐵𝐶. Now it is also immediate to see that the intersection of 𝐴𝐺 with 𝐵𝐶 has coordinates (0, 3, 8). A

F E

G

C

B

Figure 18. Barycentric Coordinates of 𝐺 from Ratios 𝐴𝐸/𝐴𝐵 and 𝐴𝐹/𝐴𝐶

Of course this is not the only way to find these coordinates. A very sure way is to choose affine coordinates, then find the equations for the lines 𝐵𝐹 and 𝐶𝐸, and then solve the pair of equations. But for some simple problems the method with masses is more direct and is perhaps more insightful. This method can be used to find the barycentric coordinates of all the points in the Marion Walters figure, Figure 9.21. Then the arial coordinates can be used to find areas of regions in the figure to solve the problem. That figure exhibits affine symmetry in

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that many points have symmetric coordinates because the coordinate and area ratios are the same as they would be for the figure in an equilateral triangle. Another application of barycentric coordinates is to the intersection of sides of a triangle with a line parallel to a side. In Figure 19, we begin with the point 𝑃1 on 𝐴𝐵, then intersect 𝐴𝐶 with the line through 𝑃1 parallel to 𝐵𝐶 to get 𝑃2 . Then we repeat the same procedure to get point 𝑃3 on 𝐵𝐶 and then 𝑃4 on 𝐴𝐵. The point 𝑃4 will not be the same as 𝑃1 (unless 𝑃1 is the midpoint), but if we continue, the next time the point that should be 𝑃7 is equal to 𝑃1 . P2

A

P5

P1

C P6

P4 B

P3

Figure 19. Repeating Parallel Paths in a Triangle

It is not too difficult to see why this works without using barycentric coordinates, but it is very clear with them. If the barycentric coordinates of 𝑃1 with respect to △𝐴𝐵𝐶 are (𝑝, 𝑞, 0), then the parallel intersection at 𝑃2 has coordinates (𝑝, 0, 𝑞), since the mass at 𝐴 does not change and the mass at 𝐵 moves to 𝐶. Continuing, we get the repeating sequence (𝑝, 𝑞, 0), (𝑝, 0, 𝑞), (0, 𝑝, 𝑞), (𝑞, 𝑝, 0), (𝑞, 0, 𝑝), (0, 𝑞, 𝑝), (𝑝, 𝑞, 0). This repeated pattern also holds, with the same coordinate reasoning, for a point 𝑃1 on the extended side of the triangle, though the figure will now be a polygon. If one does the construction with dynamic geometry software, it can be animated,

Vectors and Affine Transformations In this section, we will use vectors to study some transformations more general than similitudes. We begin with a brief recapitulation of what we have said about vectors so far, beginning with the definition and notation in Chapter 7. Vectors and Coordinates. Vectors appear as objects with direction and magnitude in many areas of geometry. As we have seen, in the Euclidean plane there is a one-to-one correspondence between the set of vectors and the set of translations: an ordered segment 𝐴𝐵 represents both the vector [𝐴𝐵] and the translation 𝒯𝐴𝐵 . The coordinate formula for 𝒯𝐴𝐵 (𝑃) is 𝑃 + (𝐵 − 𝐴), so [𝐴𝐵] can also be represented by the number pair 𝐵 − 𝐴. Both the vector and the translation can be represented visually by arrows with tail at 𝐴 and head at 𝐵. But if 𝐶𝐷 is another ordered segment so that 𝐴𝐵𝐷𝐶 is a parallelogram or linear parallelogram, then the arrow from 𝐶 to 𝐷 also represents the same vector and translation: [𝐴𝐵] = [𝐶𝐷].

Vectors and Affine Transformations

241

Figure 20. Coordinates Expressed Using Vectors or Translations

Addition of vectors corresponds to composition of transformations, so 𝑢 = 𝑣 + 𝑤 when 𝒯ᵆ = 𝒯𝑣 𝒯𝑤 . In coordinates, this is just addition of number pairs: (𝐵 − 𝐴) + (𝐷 − 𝐶). Geometrically, starting at any point 𝐴, let 𝐵 = 𝒯𝑣 (𝐴) and 𝐷 = 𝒯𝑤 (𝐴); then 𝑢 = 𝑣 + 𝑤 = [𝐴𝐶], where 𝐶 is the unique point so that 𝐴𝐵𝐶𝐷 is a parallelogram or linear parallelogram. Multiplication by a real number 𝑘, e.g., 𝑘[𝐴𝐵], corresponds to translation 𝒯𝐴𝐶 , where 𝐶 = 𝒟𝐴,𝑘 (𝐵). For a linear coordinate system based on △𝐸0 𝐸1 𝐸2 , let vectors 𝑒 1 = [𝐸0 𝐸1 ] and 𝑒 2 = [𝐸0 𝐸2 ]. For a point 𝑃 with coordinates (𝑥, 𝑦), 𝑃 = 𝐸0 + 𝑥(𝐸1 − 𝐸0 ) + 𝑦(𝐸2 − 𝐸0 ). Or in vector terms, [𝐸0 𝑃] = 𝑥𝑒 1 + 𝑦𝑒 2 . The points within △𝐸0 𝐸1 𝐸2 have 0 ≤ 𝑥 + 𝑦 ≤ 1. This is the shaded triangle in Figure 20. Shear Transformations. Shear transformations were defined in Chapter 9 and depicted in Figure 9.20. Let us consider a shear transformation 𝑆 related to the linear coordinate system defined by △𝐸0 𝐸1 𝐸2 . The base will stay fixed, so 𝑆(𝐸0 ) = 𝐸0 and 𝑆(𝐸1 ) = 𝐸1 , but 𝑆(𝐸2 ) = 𝐸2′ , where 𝐸2 𝐸2′ is parallel to 𝐸0 𝐸1 . In other words, [𝐸2 𝐸2′ ] = 𝑘𝑒 1 for some real 𝑘, so [𝐸0 𝐸2′ ] = 𝑘𝑒 1 + 𝑒 2 . This transformation of the triangle vertices determines the images of the other points in the plane: 𝑆(𝐸0 + 𝑥𝑒 1 + 𝑦𝑒 2 ) = 𝐸0 + 𝑥𝑒 1 + 𝑦(𝑘𝑒 1 + 𝑒 2 ). This says 𝑆(𝑥, 𝑦) = (𝑥 + 𝑘𝑦, 𝑦) is the coordinate formula for 𝑆. Each line parallel to the 𝑥-axis with equation 𝑦 = 𝑐 is translated to itself by the vector 𝑐𝑘𝑒 1 . And any point (0, 𝑦) = 𝐸0 + 𝑦𝑒 2 on the 𝑦-axis maps to the point 𝐸0 + 𝑦(𝑘𝑒 1 + 𝑒 2 ) = (𝑘𝑦, 𝑦), as prescribed in Chapter 9. One way to picture this is to imagine the coordinate tessellation of the plane by △𝐸0 𝐸1 𝐸2 and triangles congruent to it, with vertices being all the points with integer coordinates. Then all the triangles are transformed consistently so that the tessellation of the plane is mapped to the coordinate tessellation defined by △𝐸0 𝐸1 𝐸2′ . In other words, a point with affine coordinates in the △𝐸0 𝐸1 𝐸2 system is mapped to the point with the same coordinates in the △𝐸0 𝐸1 𝐸2′ system. One important property of this shear transformation is that all areas are preserved. The area of a figure and its sheared image are the same. If one tessellates the

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E2

E0

E 2'

E1 Figure 21. Tessellation by Triangles and Sheared Images

plane with triangles congruent to △𝐸0 𝐸1 𝐸2 , then this tessellation is transformed to a tessellation by triangles congruent to 𝑆(△𝐸0 𝐸1 𝐸2 ) and so each has the same area. Moreover, the plane can be tessellated by arbitrarily small triangles similar to △𝐸0 𝐸1 𝐸2 by taking midpoint triangles and then midpoint triangles of those to get triangles smaller than any pre-determined size. Thus, for a figure like a circle or a regular pentagon, the area can be approximated by a sum of areas of small triangles; then the area of the image is approximated in the same way by the image triangles that have the same areas as the originals.

Figure 22. Sheared and Translated Circle and Regular Pentagon

The sheared images in Figure 22 have also been translated, not changing the area but making them easier to see. The image of the circle (which can be shown to be an ellipse) has the same area as the circle. And the irregular pentagon has the same area as the regular pentagon. Shear transformations in other directions can be defined with respect to any other triangles. And more general transformations can be defined as products of various shears and rigid motions. These products will also preserve area. Affine Transformations. As products of shears and rigid motions and other similitudes are formed, the analysis of their behavior will be more complex, but the products will always map collinear points to collinear points and preserve signed ratio on each line. This property will be used as a definition. Definition 11.16. An affine transformation of the plane is a transformation that maps lines to lines and preserves signed ratio.

Vectors and Affine Transformations

243

All rigid motions and other similitudes are affine transformations. It is not hard to show that products and inverses of affine transformations are also affine transformations. Theorem 11.17 (Affine Transformation Formula). Given affine coordinates, any transformation 𝑇 that has a coordinate formula 𝑇(𝑥, 𝑦) = (𝑎𝑥 + 𝑏𝑦, 𝑐𝑥 + 𝑑𝑦) + (𝑒, 𝑓) is an affine transformation, and conversely. Note: 𝑇 is assumed to be a transformation. A function with 𝑎 = 𝑏 = 𝑐 = 𝑑 = 1 will not have an inverse. The necessary condition on 𝑎, 𝑏, 𝑐, 𝑑 for 𝑇 to be a transformation will be considered below. Proof. It is sufficient to show that the image of a point 𝑠𝑃 + 𝑡𝑄 has the same form, where 𝑃 = (𝑝1 , 𝑝2 ), 𝑄 = (𝑞1 , 𝑞2 ), and 𝑠 + 𝑡 = 1. 𝑇(𝑠𝑃+𝑡𝑄) = (𝑎(𝑠𝑝1 +𝑡𝑞1 )+𝑏(𝑠𝑝2 +𝑡𝑞2 ), 𝑐(𝑠𝑝1 +𝑡𝑞1 )+𝑑(𝑠𝑝2 +𝑡𝑞2 ))+𝑠(𝑒, 𝑓)+𝑡(𝑒, 𝑓), for (𝑒, 𝑓) = 𝑠(𝑒, 𝑓) + 𝑡(𝑒, 𝑓) since 𝑠 + 𝑡 = 1. Collecting like terms, this equals 𝑠𝑇(𝑃) + 𝑡𝑇(𝑄). To see the converse, let 𝑇 be an affine transformation and let 𝐸0 , 𝐸1 , 𝐸2 be the vertices of the triangle defining the affine coordinates. If the coordinates of the images of these points are (𝑒, 𝑓), (𝑎 + 𝑒, 𝑐 + 𝑓), and (𝑏 + 𝑒, 𝑑 + 𝑓), then any point 𝑃 = (1 − 𝑥 − 𝑦)𝐸0 + 𝑥𝐸1 + 𝑦𝐸2 with coordinates (𝑥, 𝑦) has 𝑇-image with coordinates (1 − 𝑥 − 𝑦)(𝑒, 𝑓) + 𝑥(𝑎, 𝑐) + 𝑥(𝑒, 𝑓) + 𝑦(𝑏, 𝑑) + 𝑦(𝑒, 𝑓) = (𝑎𝑥 + 𝑏𝑦, 𝑐𝑥 + 𝑑𝑦) + (𝑒, 𝑓). □

Figure 23. Compression-Expansion Scaling (2𝑥, 𝑦/3)

One example of such an affine transformation is a scaling transformation 𝑇(𝑥, 𝑦) = (𝑎𝑥, 𝑏𝑦), where 𝑎𝑏 ≠ 0. This differs from a dilation in that the scaling in the 𝑥 and 𝑦 directions is not the same. In directions when 𝑎 or 𝑏 is greater than 1, this is an expansion in the direction of 𝑒 1 or 𝑒 2 . When less than one, it is compression. Linear Transformations. Affine transformations induce transformations on vectors. If we write PQ in coordinates as 𝑄 − 𝑃 = (𝑣 1 , 𝑣 2 ), then applying 𝑇 yields 𝑇(𝑄) − 𝑇(𝑃) = (𝑎𝑣 1 + 𝑏𝑣 2 , 𝑐𝑣 1 + 𝑑𝑣 2 ). The coefficients 𝑎, 𝑏, 𝑐, 𝑑 are the same, but the translation part of the formula disappears. This is consistent with the property of vectors that they have no fixed location. In fact, as we saw when we derived the rotation formula, it may be clearer to work with transformations that map the origin 𝑂 to itself and then add translations later.

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Definition 11.18. A linear transformation is an affine transformation that has the origin 𝑂 as a fixed point. Thus, any linear transformation has the form 𝑇(𝑥, 𝑦) = (𝑎𝑥 + 𝑏𝑦, 𝑐𝑥 + 𝑑𝑦). In a note above, it was observed that while we have shown that any linear transformation has this form, for some coefficients this formula is not a transformation. We will explore this by looking at the image of △𝐸0 𝐸1 𝐸2 . We see that 𝑇(𝐸0 ) = (0, 0), 𝑇(𝐸1 ) = (𝑎, 𝑐), and 𝑇(𝐸2 ) = (𝑏, 𝑑). We wish to find out more about this set of points by applying a couple of shear transformations. This is illustrated by Figure 24, with (𝑎, 𝑐) = (2, −1) and (𝑏, 𝑑) = (1, 1). We will assume that 𝑎 ≠ 0. If this is not true, we can interchange the role of 𝐸1 and 𝐸2 and then assume 𝑏 ≠ 0. If both 𝑎 = 0 and 𝑏 = 0, then the points all are on the 𝑦-axis 𝐸0 𝐸2 , so this function 𝑇 is not a transformation, since transformations map noncollinear point to noncollinear points.

ST ( E2)

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Figure 24. Shearing the Image of the Coordinate Triangle

Now assuming 𝑎 ≠ 0, let 𝑆 be the shear parallel to 𝐸0 𝑇(𝐸1 ) such that 𝑆-image of 𝑇(𝐸2 ) will be on the line through 𝑇(𝐸2 ) parallel to 𝐸0 𝑇(𝐸1 ). Chose the shear 𝑆 to be the shear that takes 𝑇(𝐸2 ) to a point on the 𝑦-axis. This point will have the form 𝑆(𝑏, 𝑑) = (𝑏, 𝑑) + 𝑘(𝑎, 𝑐) = (0, 𝑡), so 𝑘 = −𝑏/𝑎. This means that the image 𝑆𝑇(𝐸2 ) = 𝑆(𝑏, 𝑑) = (𝑏, 𝑑) − (𝑏/𝑎)(𝑎, 𝑐) = (0, (𝑎𝑑 − 𝑏𝑐)/𝑎). In Figure 24, this is the point 𝑆𝑇(𝐸2 ) = (1, 1) + (−1/2)(2, −1) = (0, 3/2). There is one exception: if this line through (𝑏, 𝑑) intersects the 𝑦-axis at (0, 0), then 𝑎𝑑 − 𝑏𝑐 = 0, the shear is not defined, and 𝑇 is not a transformation since the images of 𝐸0 , 𝐸1 , 𝐸2 are collinear. Henceforth, we assume 𝑎𝑑 − 𝑏𝑐 ≠ 0. In the example in Figure 24, this quantity equals 3. Next applying a shear 𝑆 ′ parallel to the 𝑦-axis that will map 𝑇(𝐸1 ) to a point on the 𝑥-axis will not change the points on the 𝑦-axis but will have 𝑆 ′ 𝑆𝑇(𝐸1 ) = (𝑎, 0). In Figure 24 this point is (2, 0).

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Let

𝑎𝑦 𝑥 𝐻(𝑥, 𝑦) = ( , ). 𝑎 𝑎𝑑 − 𝑏𝑐 Then 𝐻𝑆 ′ 𝑆𝑇 = 𝐼, the identity map, since it fixes the vertices of △𝐸0 𝐸1 𝐸2 . This means 𝑇 has an inverse 𝑆 −1 𝑆 ′−1 𝐻 −1 and so is a transformation when 𝑎𝑑 − 𝑏𝑐 ≠ 0. The Determinant of an Affine Transformation. The coefficients in the formula 𝑇(𝑥, 𝑦) = (𝑎𝑥 + 𝑏𝑦, 𝑐𝑥 + 𝑑𝑦) are commonly presented in the form of a matrix: 𝑎 𝑏 ( ). 𝑐 𝑑 The quantity 𝑎𝑑 − 𝑏𝑐 is called the determinant of the matrix or, in this case, the determinant of the transformation 𝑇. The determinant of an affine transformation is defined by the same formula as well; we will interpret this quantity in terms of area. Continuing with the notation above, the 𝑆𝑆 ′ 𝑇 image of △𝐸0 𝐸1 𝐸2 is a triangle with base 𝑎 and height |𝑎𝑑 − 𝑏𝑐|/𝑎, with area (1/2)|𝑎𝑑 − 𝑏𝑐|. Thus, the ratio of the area of the image to the original area is |𝑎𝑑 − 𝑏𝑐|, the absolute value of the determinant. By reasoning similar to that which showed shears do not change area, one can conclude that an affine transformation 𝑇 scales the area of any shape by the factor |𝑎𝑑 − 𝑏𝑐|. The sign of the determinant also has meaning. Since shears are orientation-preserving transformations of the plane, the product 𝑆𝑆 ′ 𝑇 is orientation-preserving if and only if 𝑇 is orientation-preserving and the triangles 𝑇(△𝐸0 𝐸1 𝐸2 ) and 𝑆𝑆 ′ 𝑇(△𝐸0 𝐸1 𝐸2 ) have the same orientation. But the latter triangle has vertices (0, 0), (𝑎, 0), and (0, (𝑎𝑑 − 𝑏𝑐)/𝑎). For a triangle 𝐴𝐵𝐶 with 𝐴 = 𝑂, 𝐵 = (𝑥0 , 0), and 𝐶 = (0, 𝑦0 ), the triangle has positive orientation when 𝑥0 𝑦0 > 0. In the case of 𝑇, this means that 𝑇 is orientationpreserving if (𝑎𝑑 − 𝑏𝑐) > 0 and it is orientation-reversing otherwise. If we take unit area to be the area of △𝐸0 𝐸1 𝐸2 , then 𝑇(△𝐸0 𝐸1 𝐸2 ) has signed area equal to the determinant 𝑎𝑑 − 𝑏𝑐. The absolute value |𝑎𝑑 − 𝑏𝑐| = 𝒜(𝑇(△𝐸0 𝐸1 𝐸2 ))/ 𝒜(△𝐸0 𝐸1 𝐸2 ) while the sign indicates whether the two triangles have the same orientation. If one thinks in terms of the unit square, the determinant is the signed area of the parallelogram that is the image of the square.

Axioms and Models To end the book, we would like to complete the circle that began by assuming a set of six axioms for Euclidean geometry and conclude by proving that there is a model for this geometry based on the real numbers. When one creates a set of axioms for a mathematical structure, one has to be concerned about whether there actually exist any examples of the structure, or whether from contradictions or excessive restrictions in the axioms one is proving theorems about nothing. An example of this genuine concern was the early period of discovery in non-Euclidean plane geometry. Mathematicians such as Lobachevsky assumed that

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there could be more than one parallel to a line through a given point and proved beautiful and surprising theorems. But some mathematicians were very concerned that this purported non-Euclidean plane had some as yet undiscovered contradiction that meant it did not really exist, even as a mathematical abstraction. It was not until some models for the non-Euclidean plane were constructed by Poincaré and Klein that the geometry was accepted to be on a sound footing. While the soundness of the concept of the Euclidean plane is not in question here, it is a satisfying exercise to see that one can define a mathematical model for the plane and prove our six axioms as theorems. Theorem 11.19 (Plane Model ℰ for the Six Axioms). Let a model ℰ for plane geometry be given by taking the points of ℝ2 as the points of the plane and with lines, distance, and angle measure as specified in the following definitions. Then this is a model for Euclidean geometry in which the six axioms of Chapter 2 are valid. The points of ℰ are number pairs such as 𝐴 = (𝑎1 , 𝑎2 ). We will add such points and multiply them by real numbers in our usual manner. We will also rely on computations with coordinates from earlier in the chapter to avoid needless repetition. The First Three Axioms. Definition 11.20 (Lines in ℰ). An ℰ-line, or simply line, is the set of solutions of an equation 𝑎𝑥 + 𝑏𝑦 = 𝑐, for real numbers 𝑎, 𝑏, 𝑐 with (𝑎, 𝑏) ≠ (0, 0). Proposition 11.21 (Axiom 1). ℰ consists of a nonempty set of elements called points and a nonempty set of subsets called lines. For distinct points 𝑃 and 𝑄, there is exactly one ℰ-line containing 𝑃 and 𝑄. Proof. The sets of points and lines are clearly not empty. For distinct 𝑃 = (𝑝1 , 𝑝2 ) and 𝑄 = (𝑞1 , 𝑞2 ), let us consider the set of points (1 − 𝑡)𝑃 + 𝑡𝑄, where 𝑡 is any real number. Then the points (𝑥, 𝑦) of this set are solutions of −(𝑞2 − 𝑝2 )𝑥 + (𝑞1 − 𝑝1 )𝑦 = −(𝑞2 − 𝑝2 )𝑝1 + (𝑞1 − 𝑝1 )𝑝2 . This equation, obtained by eliminating 𝑡 from the parametrization, can be expressed more transparently with a dot product. If we set the normal direction 𝑁 = (−(𝑞2 − 𝑝2 ), 𝑞1 − 𝑝1 ), then 𝑁 ⋅ (𝑄 − 𝑃) = 0, and the equation is 𝑁 ⋅ (𝑥, 𝑦) = 𝑁 ⋅ 𝑃. In the development of the equation of a line earlier in this chapter, it was shown by computations in ℝ2 that the solution set not only contains 𝑃𝑄 but equals this set. □ Definition 11.22. The norm of a pair 𝐴 = (𝑎1 , 𝑎2 ) is |𝐴| = √𝑎21 + 𝑎22 . The distance function in ℰ is ‖𝐴𝐵‖ = |𝐵 − 𝐴| = √(𝑏1 − 𝑎1 )2 + (𝑏2 − 𝑎2 )2 for 𝐴 and 𝐵 = (𝑏1 , 𝑏2 ). Proposition 11.23 (Axiom 2). The plane ℰ has a distance function. • Distance is a function that maps any two points 𝐴 and 𝐵 in the plane to a nonnegative real number ‖𝐴𝐵‖.

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• Every line 𝑚 in the plane has a ruler, a mapping of the real numbers ℝ onto 𝑚 so that distances are preserved. Proof. Distance was defined above. Let the line 𝑚 = 𝐴𝐵. If 𝑑 = ‖𝐴𝐵‖, then we claim 𝜌(𝑡) = (1 − (𝑡/𝑑))𝐴 + (𝑡/𝑑)𝐵 is a ruler. To see this for any real numbers 𝑢, 𝑣, 𝜌(𝑣) − 𝜌(𝑢) = ((𝑣 − 𝑢)/𝑑)(𝐵 − 𝐴). Therefore, ‖𝜌(𝑣), 𝜌(𝑢)‖ = (|𝑣 − 𝑢|/𝑑)|𝐵 − 𝐴| = |𝑣 − 𝑢|, proving distance preservation as required. □ Before moving on to the next axiom, we will introduce translations into the model with the usual formula. Definition 11.24. A translation is a transformation 𝑇 of ℝ2 of the form 𝑇(𝑋) = 𝑋 + 𝐶 for some 𝐶 in ℝ2 . Note. For any translation, |𝑇(𝑋)−𝑇(𝑌 )| = |𝑋 −𝑌 | so translations preserve ℰ-distance. As usual, 𝒯𝐴𝐵 (𝑋) = 𝑋 + 𝐵 − 𝐴. Definition 11.25. As before, for any real 𝜃, let 𝑈(𝜃) = (cos 𝜃, sin 𝜃) and let 𝑂 = (0, 0). ⃗ and the protractor 𝛿𝐴 at point Then the ℰ-protractor at 𝑂 is the mapping 𝛿(𝜃) = 𝑂𝑈(𝜃) 𝐴 is 𝒯𝑂𝐴 (𝛿(𝜃)) = ⃖⃖⃖⃖⃖⃖⃖⃖⃖⃖⃖⃗ 𝐴(𝐴 + 𝑈(𝜃)). Definition 11.26. Angle measure is defined in ℰ as follows: 𝑚∠𝐵𝐴𝐶 = |𝑏 − 𝑐| when ⃗ ⃗ and 𝛿𝐴 (𝑐) = 𝐴 𝛿𝐴 (𝑏) = 𝐴𝐵 𝐶, provided that |𝑏 − 𝑐| ≤ 180. Proposition 11.27 (Axiom 3). The ℰ-plane has an angle measure that maps any two ⃗ ⃗ and 𝐴 rays 𝐴𝐵 𝐶 with common endpoint to a real number 𝑚∠𝐵𝐴𝐶. • The angle measure maps any angle ∠𝐵𝐴𝐶 to a positive number 𝑚∠𝐵𝐴𝐶, with 0 < 𝑚∠𝐵𝐴𝐶 < 180. ⃗ ⃗ =𝐴 • In addition, when 𝐴𝐵 𝐶, the angle measure of this zero angle is 𝑚∠𝐵𝐴𝐶 = 0. ⃗ ⃗ If the rays 𝐴𝐵 and 𝐴𝐶 are opposite rays, the angle measure of this straight angle is 𝑚∠𝐵𝐴𝐶 = 180. • For any point A, there exists a protractor at A. This is a mapping 𝛿 from the real numbers onto the rays with endpoint 𝐴, such that 𝛿(𝑡) = 𝛿(𝑡 + 360) for every 𝑡, with ⃗ ⃗ and 𝛿(𝑐) = 𝐴 the property that 𝑚∠𝐵𝐴𝐶 = |𝑏 − 𝑐| when 𝛿(𝑏) = 𝐴𝐵 𝐶, provided that |𝑏 − 𝑐| ≤ 180. Proof. This all follows immediately from the definitions of protractor and angle mea⃖⃖⃖⃖⃖⃖⃖⃖⃖⃖⃖⃗ ⃗ sure, noting that 𝑂𝑈(𝜃 + 180) is the opposite ray of 𝑂𝑈(𝜃). □ This short proof for what is arguably the most complicated of the axioms may seem unsettling, but if one thinks about it, it is a setup. The candidate for protractor is defined. Then angle measure is defined so that this protractor precisely fulfills the requirements of the axiom. The only point that is left to well-known mathematics is that the mapping 𝑈(𝜃) from ℝ to the unit circle is periodic and onto. If one accepts this, there is nothing left to prove. But perhaps the nagging question is whether this is really the angle measure that we expect in the Cartesian plane. Yes, it is. For example, if 𝐴 = (1, 0) and 𝐵 = (3, 4),

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⃗ where 𝑈(𝜃) = (3/5, 4/5). Then ⃗ what is 𝑚∠𝐴𝑂𝐵? Since |(3, 4)| = 5, 𝑂 𝐵 = 𝑂𝑈(𝜃), 𝑚∠𝐴𝑂𝐵 = arccos(3/5), converted from radians to degrees. So there is really substance in the proof, but the substance is in 𝑈 and the trigonometric functions. Before considering the final three axioms, we will introduce rotations as rigid motions in ℰ. Definition 11.28. The rotation 𝑂 𝜃 is the transformation defined as 𝑂 𝜃 (𝑥, 𝑦) = 𝑥𝑈(𝜃)+ 𝑦𝑉(𝜃), where 𝑉(𝜃) = (− sin 𝜃, cos 𝜃). For any point 𝐴, 𝐴𝜃 = 𝒯𝑂𝐴 𝑂 𝜃 𝒯𝐴𝑂 . Proposition 11.29. Translations and rotations are both rigid motions of ℰ, meaning that they preserve distance and angle measure. Proof. It was already noted that translations preserve distance. They preserve angle measure because angle measure was defined as translation invariant. Rotations preserve distance by the same computation given earlier for the Cartesian plane. And they preserve measure since they just add a constant to each polar angle. □ The Final Three Axioms — In Part. For the final three axioms, we will follow a different strategy. We will prove each of Axioms 4 and 5 only for a line that is the 𝑥-axis and we will prove Axiom 6 only for the point 𝑂. Then we will transport the 𝑥-axis to any line or 𝑂 to any point by means of a rigid motion. Definition 11.30 (Half-planes of the 𝑥-axis). For the 𝑥-axis, the set of points (𝑥, 0), the upper half-plane is the set of points (𝑥, 𝑦) with 𝑦 > 0 and the lower half-plane is the set with 𝑦 < 0. Proposition 11.31 (Axiom 4 for the 𝑥-axis). Let 𝑚 denote the 𝑥-axis; then any point not on line 𝑚 belongs to one of two disjoint sets called the half-planes of the line. • Points 𝑃 and 𝑄 are in different half-planes when 𝑃𝑄 intersects 𝑚. They are in the same half-plane if 𝑃𝑄 does not intersect 𝑚. ⃗ = 0, the points 𝑃 for which • Let 𝑚 = 𝐴𝐵. If a protractor is chosen so that 𝜃(𝐴𝐵) ⃗ ⃗ < 180 lie in one half-plane, and the 𝑄 for which −180 < 𝜃(𝐴 0 < 𝜃(𝐴𝑃) 𝑄) < 0 are in the other half-plane. Proof. The segment 𝑃𝑄 intersects the 𝑥-axis when (1 − 𝑡)𝑃 + 𝑡𝑄 at 𝑡 = 𝑝2 /(𝑝2 − 𝑞2 ), a number between 0 and 1 when 𝑝2 and 𝑞2 have opposite sign. Also, 𝑈(𝜃) is in the upper half-plane when 0 < 𝜃 < 180 and in the lower half-plane when −180 < 𝜃 < 0. □ Proposition 11.32 (Axiom 5 for the 𝑥-axis). For the 𝑥-axis, there exists a rigid motion distinct from the identity transformation that fixes the points of the 𝑥-axis. Proof. Let 𝑅(𝑥, 𝑦) = (𝑥, −𝑦). This is clearly a rigid motion that fixes the points of the 𝑥-axis. □ Definition 11.33. The ℰ-dilation 𝒟𝑂,𝑘 is the mapping 𝒟𝑂,𝑘 (𝑥, 𝑦) = (𝑘𝑥, 𝑘𝑦) = 𝑘(𝑥, 𝑦). Proposition 11.34 (Axiom 6 for 𝑂). For every point 𝑂 and every 𝑘 > 0, the dilation 𝒟𝑂,𝑘 preserves angle measure and scales every distance by 𝑘: for all 𝑃 ′ = 𝒟𝑂,𝑘 (𝑃) and 𝑄′ = 𝒟𝑂,𝑘 (𝑄), ‖𝑃 ′ 𝑄′ ‖ = 𝑘‖𝑃𝑄‖.

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Proof. A direct substitution proves the scaling statement. 𝑃 ′ = 𝒟𝑂,𝑘 (𝑃) = 𝑘𝑃 and 𝑄′ = 𝒟𝑂,𝑘 (𝑄) = 𝑘𝑄, so ‖𝑃 ′ 𝑄′ ‖ = |𝑃 ′ − 𝑄′ | = 𝑘|𝑃 − 𝑄| = 𝑘‖𝑃𝑄‖. As for angles, 𝒟𝑂,𝑘 ′ 𝐵 ′ in the same direction. Therefore, 𝒟 ⃗ to a parallel ray 𝐴 ⃗ ⃗ maps a ray 𝐴𝐵 𝑂,𝑘 (𝐴𝐵) = ⃗ This means that the dilation image of an angle ∠𝐵𝐴𝐶 is also a translation 𝒯𝐴𝐴′ (𝐴𝐵). image of this angle, so the measures are the same. □ The Final Three Axioms. To complete the proof that the final three axioms hold in ℰ, we will use translations and rotations, which are known to be rigid motions in ℰ. To prove the Plane Separation Axiom, Axiom 4, for any line 𝑚, first rotate the 𝑥axis so that it is parallel to 𝑚; then translate the rotated line to 𝑚. Then the half-planes and all the relations are preserved by the rigid motion 𝑆 that is the composition of these two rigid motions. To prove the Reflection Axiom for a line 𝑚, again choose the same rigid motion 𝑆 that takes the 𝑥-axis to 𝑚. Then if 𝑅 is the reflection that we defined for the 𝑥-axis, the rigid motion 𝑆𝑅𝑆 −1 is a rigid motion that fixes the points of 𝑚. Finally, the dilation 𝒟𝐴,𝑘 with the usual definition equals 𝒯𝑂𝐴 𝒟𝑂,𝑘 𝒯𝐴𝑂 . Since translations are rigid motions, the scaling relation will remain true. Thus we have connected back to the beginning with a model that satisfies Axioms 1–6 of Chapter 2.

Conclusion While the geometry in this chapter is interesting in its own right, the concepts and tools introduced here extend to areas of geometry that would require several more books to pursue. In plane geometry, the complex numbers are essential for the geometry of circles and inversions, which can in turn be used to model non-Euclidean geometry. Barycentric coordinates can be extended to include a line at infinity with 𝑐 1 +𝑐 2 +𝑐 3 = 0, leading to projective geometry. Beyond the plane, affine and Cartesian coordinates can be extended to any dimension. Matrices and vectors in all dimensions are everywhere in mathematics and its applications. Our hope is that this introduction to these tools and concepts through plane Euclidean and affine geometry will provide background and motivation for further exploration, either in the deeper regions of Euclidean geometry itself or in other areas beyond.

Exercises and Explorations 1. (Pencils of Lines). For two functions 𝑓0 (𝑥, 𝑦) = 𝑎0 𝑥 + 𝑏0 𝑦 − 𝑐 0 and 𝑓1 (𝑥, 𝑦) = 𝑎1 𝑥 + 𝑏1 𝑦 − 𝑐 1 , the equations 𝑓0 (𝑥, 𝑦) = 0 and 𝑓1 (𝑥, 𝑦) = 0 are the equations of lines 𝑚0 and 𝑚1 . Let 𝑡 be any real number; then (1 − 𝑡)𝑓0 (𝑥, 𝑦) + 𝑡𝑓1 (𝑥, 𝑦) = 0 is also the equation of a line 𝑚𝑡 . (a) Prove that if 𝑚0 and 𝑚1 intersect at a point 𝑃, then all the 𝑚𝑡 pass through 𝑃.

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(b) Prove that if 𝑚0 and 𝑚1 are parallel, then all the 𝑚𝑡 are parallel (or equal) to these lines. (c) The set of all lines that pass through a point 𝑃 is called the pencil of lines through 𝑃. The set of lines parallel to or the same as a line 𝑚 is called the pencil of lines parallel to 𝑚. Explain why any two lines belong to a unique pencil and any distinct pencils either intersect in one line or no lines. (If the words line and pencil are interchanged in this sentence, do these words sound familiar?) (d) Let 𝑚 ∥ 𝑛 mean that 𝑚 and 𝑛 belong to the same parallel pencil. Show that this relation satisfies the three criteria in Theorem 1.4 when ≅ is replaced by ∥. Such a relation is called an equivalence relation. Why is 𝑚 ∥ 𝑛 an equivalence relation while the relation “𝑚 is parallel to 𝑛” is not? 2. (Ruler for a Line). For constants 𝑚 and 𝑏, the map 𝑃(𝑥) = (𝑥, 𝑚𝑥+𝑏) is a parametrization of a line 𝑝 but not necessarily a ruler. (a) For what values of 𝑚 and 𝑏 will this be a ruler? (b) Find rulers for some lines with equations such as 𝑥 + 𝑦 = 1 and 3𝑥 + 4𝑦 = 5. 3. (Negative Dot Product). If 𝑚∠𝑀𝑂𝑁 = 𝜃, show that the dot product 𝑁 ⋅ 𝑀 = ‖𝑂𝑁‖‖𝑂𝑀‖ cos 𝜃, even when the dot product is 0 or negative. 4. (Reflection Formula). Using the notation 𝑂 = (0, 0), 𝑈(𝜃) = (cos 𝜃, sin 𝜃), and 𝑉(𝜃) = (− sin 𝜃, cos 𝜃), let 𝐹 be the function with this coordinate formula: 𝐹(𝑥, 𝑦) = 𝑥𝑈(2𝜃) + 𝑦𝑉(2𝜃) = 𝑥(cos 2𝜃, sin 2𝜃) + 𝑦(sin 2𝜃, − cos 2𝜃). (a) Confirm that 𝐹(𝑂) = 𝑂, 𝐹(𝑈(𝜃)) = 𝑈(𝜃), and 𝐹(𝑉(𝜃)) = −𝑉(𝜃). (b) Show that 𝐹 preserves distance. (c) Explain why this proves that 𝐹 = ℛ𝑂𝑈(𝜃) . 5. (Area and Barycentric Coordinates). For the point 𝑆 in Figure 16, check by computing barycenters and ratios on each triangle side that the areas are what was stated in the text. 6. (Radical Axis). Given a circle 𝑐 with center 𝐴 and radius 𝑟, the power of a point 𝑃 with respect to 𝑐 is ‖𝐴𝑃‖2 − 𝑟2 . The radical axis of two circles 𝑐 1 and 𝑐 2 is the set of points 𝑃 for which the two powers are.equal. (a) Use coordinates to write a formula for the the radical axis of two circles. Show this set is a line perpendicular to the line of centers when the circle centers are distinct. (b) Compare this coordinate proof with the proof of Theorem 8.46. How are they similar, and how do they differ? 7. (Transformation Conjugation and Complex Numbers). Let 𝑆(𝑧) = 𝛼𝑧 and 𝑇(𝑧) = 𝑧 + 𝛽. Use complex numbers to compute two transformation conjugations from these transformations (not to be confused with complex conjugate). (a) Generalize the corollary on page 181 by computing the conjugation 𝑇𝑆𝑇 −1 and finding its fixed point using Theorem 11.8. Since 𝑇 is a translation and 𝑆 can be a rotation or a dilation, this generalizes the corollary. (b) Compute also the conjugate 𝑆𝑇𝑆 −1 . What kind of transformation is this? Draw a picture of this when 𝛼 = 2.

Exercises and Explorations

251

(c) Let 𝑅(𝑧) = 𝑧 (yes, complex conjugate!). Compute 𝑅𝑆𝑅−1 . What kind of transformation is this? How does this result differ from that of the corollary when |𝛼| = 1? 8. (Incenter Barycentric Coordinates). For a triangle △𝐴𝐵𝐶, let 𝑎, 𝑏, and 𝑐 be the lengths of the sides opposite 𝐴, 𝐵, 𝐶. (a) Show that (𝑎, 𝑏, 𝑐) are the barycentric coordinates of the incenter, the point of concurrence of the angle bisectors. (b) What are the barycentric coordinates for the three excenters, the points of concurrence of two exterior angle bisectors and one interior bisector? Hint: Theorem 8.18. 9. (Midpoints and Affine Transformations). 𝑇 is an affine transformation that maps segment 𝐴𝐵 to 𝐶𝐷. (a) If 𝑃(𝑡) = (1 − 𝑡)𝐴 + 𝑡𝐵, show that 𝑇(𝑃(𝑡)) = 𝑄(𝑡), where 𝑄(𝑡) = (1 − 𝑡)𝐶 + 𝑡𝐷, for 0 ≤ 𝑡 ≤ 1. (b) Prove that the set of midpoints of 𝑃(𝑡)𝑄(𝑡) is a segment or a point. (For an isometry 𝑇, this is called Hjelmslev’s Theorem, Coxeter [5, Section 3.6].) 10. (Affine Congruence). Prove that in affine geometry all triangles are alike (“affinecongruent”). In other words, for any two triangles, there is an affine transformation that takes one to the other. Hint: Rather than try to find a formula in one step, you can write the affine transformation as the product of simpler transformations that step by step transform one triangle to coincide with the other one. 11. (Uniqueness of Affine Transformations). Prove that if two affine transformations 𝑆 and 𝑇 map the vertices of a triangle △𝐴𝐵𝐶 to the same points, then 𝑆 = 𝑇. Hint: To simplify the problem, let △𝐴𝐵𝐶 define a coordinate system. 12. (Area and Determinants). In a Cartesian coordinate system, let 𝐴 = (𝑎1 , 𝑎2 ) and 𝐵 = (𝑏1 , 𝑏2 ). If 𝐶 = 𝐴 + 𝐵, prove that the area of a parallelogram 𝑂𝐴𝐶𝐵 is the 𝑎 𝑏1 absolute value of the determinant of ( 1 ), with the sign of the determinant 𝑎2 𝑏2 indicating the orientation of this parallelogram. 13. (Bézier Curve). For three points 𝐴, 𝐵, 𝐶, let 𝐹(𝑡) = (1 − 𝑡)𝐴 + 𝑡𝐵 and 𝐺(𝑡) = (1 − 𝑡)𝐵 + 𝑡𝐶. The image curve of the function 𝑃(𝑡) = (1 − 𝑡)𝐹(𝑡) + 𝑡𝐺(𝑡) is called the Bézier curve with 𝐴, 𝐵, 𝐶 as control points. The curve is used in computer graphics.

C A P (t )

F (t )

G (t ) B

Figure 25. Bézier Curve for 0 ≤ 𝑡 ≤ 1

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(a) Draw three points 𝐴, 𝐵, 𝐶, then several examples of 𝐹(𝑡), 𝐺(𝑡), and 𝑃(𝑡) for several values of 𝑡. It would be best if you could do this with dynamic geometry software and let 𝑡 vary, but it is otherwise helpful to see an example with several points plotted. (b) For the points 𝐴 = (−𝑐, 1), 𝐵 = (0, 0), and 𝐶 = (𝑐, 1), show that this curve satisfies the equation 𝑦 = 𝑎𝑥2 − 𝑏 for some 𝑎 and 𝑏. This is the equation for a parabola. Note: There is an affine transformation that will map these control points to any distinct and noncollinear 𝐴, 𝐵, 𝐶 and so will map this parabola to the corresponding Bézier curve. Therefore, all such Bézier curves are parabolas. 14. In Figure 26, one sees on the left how a one-dimensional object, a segment, is translated to another segment that is the opposite side of a parallelogram. In the middle figure, a parallelogram is translated to what looks like the opposite face of the plane projection of a three-dimensional object, such as a cube or a parallelepiped. This figure is repeated on the right, without the arrows. (a) In the figure on the right, show that the four “diagonals” indicated by the dashed lines are concurrent. Is this point a center of mass? (b) Draw a new figure in this sequence by translating the figure on the right to produce a figure with 16 vertices that appears to be the projection of a fourdimensional object such as a hypercube. (c) In this new figure, describe 8 diagonals that generalize the diagonals from the previous case; then show that they are concurrent.

Figure 26. Vectors and Pictures of Higher Dimensions

Bibliography

[1] William Barker and Roger Howe, Continuous symmetry: From Euclid to Klein, American Mathematical Society, Providence, RI, 2007. MR2362745 [2] George D. Birkhoff, A set of postulates for plane geometry, based on scale and protractor, Ann. of Math. (2) 33 (1932), no. 2, 329–345, DOI 10.2307/1968336. MR1503058 [3] George David Birkhoff and Ralph Beatley, Basic geometry, 3rd ed., Chelsea Publishing Company, New York, 1959. MR0117615 [4] Richard G. Brown, Transformational geometry, Dale Seymour, Palo Alto, 1973. [5] H. S. M. Coxeter, Introduction to geometry, 2nd ed., Wiley, New York, 1980. [6] Arthur Coxford, Zalman Usiskin, and Daniel Hirschhorn, The University of Chicago School Mathematics Project Geometry, Scott-Foresman, Glenview, IL, 1991. [7] Clayton W. Dodge, Euclidean geometry and transformations, revised reprint of the 1972 original, Dover Publications, Inc., Mineola, NY, 2004. MR2067883 [8] Marvin Jay Greenberg, Euclidean and non-Euclidean geometries: Development and history, 3rd ed., W. H. Freeman and Company, New York, 1993. MR1261866 [9] Branko Grünbaum and G. C. Shephard, Tilings and patterns, W. H. Freeman and Company, New York, 1987. MR857454 [10] Liang-shin Hahn, Complex numbers and geometry, MAA Spectrum, Mathematical Association of America, Washington, DC, 1994. MR1269108 [11] Euclid, Euclid’s Elements, all thirteen books complete in one volume, The Thomas L. Heath translation, edited by Dana Densmore, Green Lion Press, Santa Fe, NM, 2002. MR1932864 [12] George Edward Martin, Transformation geometry: An introduction to symmetry, Undergraduate Texts in Mathematics, Springer-Verlag, New York-Berlin, 1982. MR718119 [13] E. A. Maxwell, Geometry by transformations, Cambridge, Cambridge, 1975. [14] National Governors Association Center for Best Practices and Council of Chief State School Officers, Common Core State Standards for Mathematics: High School Geometry, 2010, www.corestandards.org/Math/Content/HSG/CO/B/ 7/. [15] G. Pólya, How to solve it: A new aspect of mathematical method, 2nd ed., Princeton Science Library, Princeton University Press, Princeton, NJ, 1988. MR1090087 [16] Doris Schattschneider, Visions of symmetry: Notebooks, periodic drawings, and related work of M. C. Escher, W. H. Freeman and Company, New York, 1990. MR1189799 [17] Peter S. Stevens, Handbook of regular patterns: An introduction to symmetry in two dimensions, MIT Press, Cambridge, Mass.-London, 1980. MR625845 [18] Mabel Sykes, A source book of problems for geometry: Based upon industrial design and architectural ornament [1912], Cornell University Library, Ithaca, NY, 2009. [19] Hung-Hsi Wu, Teaching geometry in grade 8 and high school according to the Common Core Standards, 2013, https: //math.berkeley.edu/~wu/CCSS-Geometry_1.pdf. [20] I. M. Yaglom, Geometric transformations I, AMS-MAA Press, Providence, 1962.

253

Index

affine geometry affine transformations, 240–245 shears, 241 area ratios, 172–175 linear transformations determinant, 245 shear transformation, 173 area, 173 vectors linear transformations, 243 angle acute, 16 addition, 18 bisector as locus, 73 definition ∠𝐴𝐵𝐶, 14 inscribed, 139 interior, 17 measure 𝑚∠𝐴𝐵𝐶, 3, 14 measure, signed 𝑚+ ∠𝐵𝐴𝐶, 50 obtuse, 16 ⃗ 15, 16 polar angle 𝜃(𝐴𝐵), protractor definition, 15 right, 16 straight, 14 supplementary, 16 vertical, 16 zero, 14 area affine ratios, 172–175 Marion Walters Theorem, 174 definition, 161 parallelogram, 163 rectangle, 162

scaling by similitude, 167–169 shear transformations, 173 shears, 241 signed area, 238 determinant, 245 triangle, 164 axiom dilation, 20 incidence, 12 plane separation, 17 protractor, 14 reflection, 20 ruler, 12 barycenter, 235 circle arc ˆ 140, 141 ∠𝐵𝐴𝐶, arc measure, 140 area, 169–172 𝜋, 171 centers of dilation common tangent construction, 131 three circles, 138 two circles, 130–131 chord definition, 61 equal chord theorem, 61 circles of Apollonius, 135 circumference, 171 definition, 8 external tangent construction, 75, 80 inscribed angle theorem, 141 intersection with lines

255

256

distance and intersection, 72 intersection triangle, 143 number of points, 29 secants, chords, power of point, 145 orthogonal circles, 146 radical axis, 148–151 radical center, 150 complex numbers rotation, 231 congruence definition, 4 equivalence relation, 5 of angle measure, 37 of convex quadrilaterals, 39 of points, 35 of segments, 35 of triangles angle-side-angle (ASA), 38 hypotenuse-leg (HL), 39 side-angle-side (SAS), 36 side-side-side (SSS), 37 superposition, 1–3 conjugation of transformation, 181 convex set, 14 coordinates affine, 217 dilation formula, 221 half-turn formula, 219 midpoint formula, 219 translation formula, 220 affine transformations, 240–245 barycentric coordinates, 233–240 areal coordinates, 238 Ceva’s Theorem, 237 Cartesian, 223 complex numbers, 231 distance formula, 225 dot product, 227 normal, 226 reflection formula, 230 rotation formula, 229 equations of lines, 221–224 graphing 𝑦 = 𝑓(𝑥), 228 reference triangle, 217 vectors, 240 crossbar theorem, 19 dilation axiom, 20 centers of dilation, 128 three circles, 138 two segments, 129

Index

two triangles, 129 composition Menelaus theorem, 136 construction by scaling, 132–134 definition, 20 negative ratio, 114 direction definition, 89 directed line, 56, 57 redefined by translation, 103 distance, 3 ‖𝐴𝐵‖, 12 Cartesian coordinates, 225 minimizing distance by reflection, 75 Fagnano problem, 76 to line, 71 point to line, 71 ruler, 12 Fermat Point, 191, 215 fixed point one fixed point, 46 three fixed points, 24 two fixed points, 19 glide reflection 𝒢𝐴𝐵 , 204 definition, 107 invariant line, 204 product half-turn and reflection, 204 three reflections, 209 group cyclic group, 53 symmetry group, 51 transformation group, 51 half-plane definition, 17 half-turn ℋ𝐴 , 67 definition, 67 line image, 67 harmonic division, 134 circles of Apollonius, 135 invariant lines, 111, 204 isometry isometry angle preservation, 40 vs rigid motion, 4 kite

Index

definition, 28 symmetry, 28 line 𝐴𝐵, 12 directed line, 14, 56 image by rigid motion, 19 locus definition, 28 equal distance from two points, 28 equal distance to two lines, 73 right angle vertex, 96 vertex of angles of equal measure, 142 midpoint, 13 Napoleon’s Theorem, 190 orientation at a point by a half-plane, 56 by protractor, 49 of the plane by a global protractor, 106 by a triangle, 57–60 parallel lines distance between, 92 Euclidean Parallel Postulate, 83, 94 midline, 93 perpendiculars to line, 26 transversals, 85–87 definition, 85 main theorem, 86 ratios, 125–128 parallelogram, 88–91 area, 163 congruent parallel segments, 89 definition, 88 fourth point, 98 linear, 97 midsegments, 90 symmetry, 88 perpendicular through a point, 26 perpendicular bisector definition, 16 distance inequality, 70 equidistance locus, 28 isosceles triangle, 27 polygon convex polygon, 17 definition, 14

257

regular angle, 54 definition, 53 regular pentagon golden ratio, 121 pentagram, 54 power of a point, 146 protractor axiom, 14 definition, 15 global protractor 𝛿∗ , 105 Pythagorean Theorem, 117 area proofs, 165–166 Perigal dissection, 167 quadrilateral convex congruence, 39 cyclic, 158 definition, 28 kite, 28 midpoint quadrilateral, 94, 109, 220 radical axis of circles, 148–151 radical center, 150 ray direction, 14 opposite ray, 14 rectangle, 91–92 area, 162 symmetry, 92 reflection ℛ𝑚 , ℛ𝐴𝐵 , 25 axiom, 20 definition, 25 fundamental theorem, 25 light rays, 30 Y Figure Lemma, 23 rigid motion definition, 3 glide reflection 𝒢𝐴𝐵 , 204 definition, 107 half-turn, 67 isometry angle preservation, 40 orientation-preserving or -reversing, 60 product of 1, 2, or 3 reflections, 40 products half-turn and reflection, 204 products with distinct centers 60-degree rotation, 188–190 90-degree rotations, 179, 181 general rotations, 180

258

rotation, 46 rotation 𝑂 𝑘 , 50 definition, 46 half-turn, 67 orientation and angle of rotation, 50 product of two reflections, 47–49 reflection product theorem, 50 symmetry, 51 ruler definition and axiom, 12 segment 𝐴𝐵, 13 definition, 13 midpoint, 13 similarity definition, 114 internal ratio, 123 scale factor, 123 signed ratio, 123 tests for triangles, 115–116 similitude area scaling, 167–169 definition, 113 preserves angle measure, 113 uniqueness, 116 symmetry definition, 51 dihedral, 53–54 frieze seven groups, 200–208 translation generator, 192 quarter-turn 𝑝4, 𝑝4𝑚, 𝑝4𝑔, 183–188 rotational, 51, 53 generator, 52 symmetry group, 51 wallpaper lattice, 194 translations, 192 tessellation Napoleon’s Theorem, 197 quadrilateral, 198 regular polygons, 195 two squares Pythagorean Theorem, 195 transformation, 3 orientation-preserving or -reversing, 60 transformation group, 51 translation 𝒯𝐴𝐵 , 101

Index

as half-turn product, 99 as line reflection product, 100 from image of one point, 101 product of translations is a translation, 101 is commutative, 102 segment image, 100 triangle, 14 altitude, 78 orthic triangle, 78, 155 angle bisector ratio, 127 angle sum, 42, 85 area, 164 Ceva’s Theorem, 237 circumcircle, 88 concurrence altitudes, 79, 153 angle bisectors, 74 medians, 110, 152 perpendicular bisectors, 87 corresponding parallel sides, 129 equilateral, 27 Euler line, 155 exterior angle, 68 Fagnano problem, 76–79 inequality exterior angle, 68, 85 greater angle–greater side, 69 triangle inequality, 70 inscribed circle, 74 interior as half-plane intersection, 17 isosceles base angles, 27 bisector, 27 definition, 27 symmetry, 27 midpoint triangle, 93–96 as dilation image, 152 fundamental theorem, 41 Miquel Theorem, 158 nine-point circle, 155 Pythagorean Theorem, 117 area proofs, 165–166 Perigal dissection, 167 trigonometric functions, 118 vector, 107 [AB], 107 addition, 108 dot product, 227 real number multiplication, 108

Selected Published Titles in This Series 51 49 48 47

James R. King, Geometry Transformed, 2021 Carl G. Wagner, A First Course in Enumerative Combinatorics, 2020 R´ obert Freud and Edit Gyarmati, Number Theory, 2020 Michael E. Taylor, Introduction to Analysis in One Variable, 2020

46 Michael E. Taylor, Introduction to Analysis in Several Variables, 2020 45 Michael E. Taylor, Linear Algebra, 2020 44 Alejandro Uribe A. and Daniel A. Visscher, Explorations in Analysis, Topology, and Dynamics, 2020 43 Allan Bickle, Fundamentals of Graph Theory, 2020 42 41 40 39

Steven H. Weintraub, Linear Algebra for the Young Mathematician, 2019 William J. Terrell, A Passage to Modern Analysis, 2019 Heiko Knospe, A Course in Cryptography, 2019 Andrew D. Hwang, Sets, Groups, and Mappings, 2019

38 Mark Bridger, Real Analysis, 2019 37 Mike Mesterton-Gibbons, An Introduction to Game-Theoretic Modelling, Third Edition, 2019 36 Cesar E. Silva, Invitation to Real Analysis, 2019 ´ 35 Alvaro Lozano-Robledo, Number Theory and Geometry, 2019 34 33 32 31

C. Herbert Clemens, Two-Dimensional Geometries, 2019 Brad G. Osgood, Lectures on the Fourier Transform and Its Applications, 2019 John M. Erdman, A Problems Based Course in Advanced Calculus, 2018 Benjamin Hutz, An Experimental Introduction to Number Theory, 2018

30 Steven J. Miller, Mathematics of Optimization: How to do Things Faster, 2017 29 Tom L. Lindstrøm, Spaces, 2017 28 Randall Pruim, Foundations and Applications of Statistics: An Introduction Using R, Second Edition, 2018 27 Shahriar Shahriari, Algebra in Action, 2017 26 25 24 23

Tamara J. Lakins, The Tools of Mathematical Reasoning, 2016 Hossein Hosseini Giv, Mathematical Analysis and Its Inherent Nature, 2016 Helene Shapiro, Linear Algebra and Matrices, 2015 Sergei Ovchinnikov, Number Systems, 2015

22 21 20 19

Hugh L. Montgomery, Early Fourier Analysis, 2014 John M. Lee, Axiomatic Geometry, 2013 Paul J. Sally, Jr., Fundamentals of Mathematical Analysis, 2013 R. Clark Robinson, An Introduction to Dynamical Systems: Continuous and Discrete, Second Edition, 2012

18 17 16 15

Joseph L. Taylor, Foundations of Analysis, 2012 Peter Duren, Invitation to Classical Analysis, 2012 Joseph L. Taylor, Complex Variables, 2011 Mark A. Pinsky, Partial Differential Equations and Boundary-Value Problems with Applications, Third Edition, 1998

14 13 12 11

Michael E. Taylor, Introduction to Differential Equations, 2011 Randall Pruim, Foundations and Applications of Statistics, 2011 John P. D’Angelo, An Introduction to Complex Analysis and Geometry, 2010 Mark R. Sepanski, Algebra, 2010

For a complete list of titles in this series, visit the AMS Bookstore at www.ams.org/bookstore/amstextseries/.

The only prerequisite for this book is a basic understanding of functions. Some previous experience with proofs may be helpful, but students can also learn about proofs by experiencing them in this book—in a context where they can draw and experiment. The eleven chapters are organized in a flexible way to suit a variety of curriculum goals. In addition to a geometrical core that includes finite symmetry groups, there are additional topics on circles and on crystallographic and frieze groups, and a final chapter on affine and Cartesian coordinates. The exercises are a mixture of routine problems, experiments, and proofs.

For additional information and updates on this book, visit www.ams.org/bookpages/amstext-51

AMSTEXT/51

This series was founded by the highly respected mathematician and educator, Paul J. Sally, Jr.

Photo courtesy of James R. King

Many paths lead into Euclidean plane geometry. Geometry Transformed offers an expeditious yet rigorous route using axioms based on rigid motions and dilations. Since transformations are available at the outset, interesting theorems can be proved sooner; and proofs can be connected to visual and tactile intuition about symmetry and motion. The reader thus gains valuable experience thinking with transformations, a skill that may be useful in other math courses or applications. For students interested in teaching mathematics at the secondary school level, this approach is particularly useful since geometry in the Common Core State Standards is based on rigid motions.