Geometry

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by IRENE FISCHER and DUNSTAN HAYDEN

1967 ALLYN AND BACON, INC. Boston Rockleigh, N. J. Atlanta Dallas Belmont, Calif.

©

COPYRIGHT,

1965

By

ALLYN

AND

BACcon,

INC.

All rights reserved. No part of this book may be reproduced in any form, or by any means, without permission in writing from the publisher. Printed in the United States of America.

PREFACE This geometry course follows the recommendations for college preparatory mathematics made by the College Entrance Examination Board, in the Report of the Commission on Mathematics, New York, 1959. The selection and arrangement of the material was further influenced by the authors’ extensive experience in teaching and in practical research, and by their concern for better understanding and appreciation of modern ideas by the educated public as well as for thorough, intelligent training of future scientists. The presentation offers an entirely new approach to the teaching of high school geometry, characterized mainly by the following features: 1. Plane and solid geometry are treated simultaneously. 2. The ability to understand and visualize spatial relations is developed by drawing and reading orthogonal and oblique projections of simple three-dimensional objects. 3. Coordinate geometry in two and three dimensions is immediately introduced as a means of producing correct illustrative graphs. It is reintroduced formally in the second Unit as a useful theoretical method. 4. A serious attempt is made to convey to the student the significance of each postulate by discussing its logical content and the implications of an alternative choice. The postulates are of the traditional type, chosen and worded in a consistent manner so that they can be easily remembered. 5. The discussion of the peientifie use of geometry and abstract models which introduces Unit 1 is designed to arouse an interest for the “‘why”’ and “how.”’ A second reading of this material at the end of the course is suggested. 6. The conventional theorems are arranged under higher unifying aspects which make them appear as special applications of the great pervading ideas of similarity, congruence, symmetry, and rotation rather than as isolated facts. The important concepts of transformations and invariants under a given transformation are stressed, leading to an understanding of Euclidean geometry as the study of invariants under a certain group of transformations. Objectives: Although the appearance of this course is not conventional, the perceptive teacher will recognize the main content as the conventional subject matter traditionally taught for many years. A variety of topics are introduced just briefly in the text and in the Suggestions for Special Projects for the purpose of opening vistas. These departures from the main theme are intended to leave the student with the impression that there are many interesting things to do and, at the same time, provide sufficient material for experimentation.

iv

Preface

Organization of Text: The text consists of a Preparatory Chapter and six Units followed by supplements containing reviews of non-geometric topics, suggestions for special projects, and tables. The Preparatory Chapter has two purposes: (1) it offers an informal factual review of the geometric vocabulary, and (2) it provides a working knowledge of informal geometry through the visualizing and drawing of simple threedimensional objects. A summary at the end of each Unit is intended to help the student gain perspective of his progress by focusing his attention on the most important ideas of the Unit.

Lists of such items as theorems, postulates, corol-

laries, formulas, and words and phrases provide a point of easy reference. Cumulative tests covering the plane and solid geometry course are placed after Units 3 and 5. The exercises in these tests are arranged in groups according to topics to afford ease of selection by the teacher. The numerous exercises are graded by increasing difficulty, and are grouped according to topics to facilitate their assignment. Many more challenging exercises and topics, marked by asterisks (*), and invitations to try individual projects in the form of reports or term papers are located throughout the text.

Use of Text:

The text provides a three-track program:

Minimal Course: All sections not marked should be studied in some detail. The topics marked by a circle (©) may be discussed just briefly and informally. Sections marked | are departures from the main theme and can be omitted. Recommended Course: All topics marked © should be treated more thoroughly. Some, but not all, of the departures marked 1 should be covered in this track. The choice and thoroughness of study is left to the discretion of the teacher. Accelerated Course

(also for individual superior students):

All the ma-

terial offered should be covered to some degree.

It should be noted that the criterion is not necessarily the number of topics covered, but rather the depth of treatment of each topic. The topics in trigonometry or analytics may be omitted without affecting the rest of the geometry course. Should plane and solid geometry be offered independently this text can be used to great advantage in both courses. A Teachers’ Supplement containing commentaries on each Unit, and answers to all exercises and cumulative tests accompanies the text. Also included in this supplement is a chart suggesting a distribution of topics in each of the three tracks.

LF. D.H.

CONTENTS A PREPARATORY CHAPTER: INFORMAL GEOMETRY 1 2 3 4

THREE—-DIMENSIONAL GRAPHS HOW TO DRAW PICTORIAL VIEWS PICTORIAL VIEWS OF STRAIGHT-EDGED OBJECTS PLAN VIEW AND FRONT VIEW OF STRAIGHT-EDGED OBJECTS 5 CURVED SURFACES 6 THE SPHERE

1. A SYSTEM

OF POSTULATES

Part A

1 2 3 4

ON AXIOMS, THEOREMS, AND MODELS WHICH BASIC ASSUMPTIONS? BETWEEN CONNECTED AND SEPARATED The Concepts of Connected and Separated (4./) The Postulates of Separation (4.2)

5 INCIDENCE The Incidence Postulates (5.1) Some Properties of the Triangle: An Application of the Postulates (5.2)

65 65 7 Ue

6 PARALLEL Part

40 47 a2 57 7, 61

B

7 MEASUREMENT The Measure of Segments (7.1) The Measure of Angles (7.2)

86 87 94

8 EQUAL AND CONGRUENT Congruvent Segments (8.1) Congruent Angles (8,2)

99 99 103

9 PERPENDICULAR The Right Angle in the Plane (9,]) Perpendicular Lines and Planes (9.2) The Angle Between Two Planes (9.3)

115 115 121 127,

10 LARGER

AND

SMALLER

132

vi

Contents

CONGRUENCE,

2. SIMILARITY, 1 SIMILARITY

AND

SYMMETRY 146

CONGRUENCE

The Ratio of Similitude (1.1)

146

Congrvent Triangles (1,2) On Proofs (1.3)

153 164

The Quadrilateral: An Application of Congruent Triangles (1.4)

179

2 SIMILAR

TRIANGLES

189

Similarity Theorems (2.1) Dividing a Segment: An Application of Similar Triangles (2.2) Similarities in a Right Triangle (2.3)

189

200 204

Trigonometric Functions in a Right Triangle: An Application of Similar Triangles (2.4)

a i

Computing with Coordinates: An Application of Similar Triangles (2.5)

220

3 CONGRUENCE

AND

BILATERIAL

SYMMETRY

Direct and Inverse Congruence (3.1) The Symmetry Axis; The Symmetry Plane (3.2) Symmetric Triangles: Applications of Bilaterial Symmetry (3.3) The Isosceles Triangle (3.3.1) The Equilaterial Triangle (3.3.2) Circumscribed, Inscribed, Escribed Circles (3.4) Remarkable Points of a Triangle (3.5)

230

230 233 238 238 244 249 252

3. MAPPING 1 WHAT IS A 2 EXPANSION

MAPPING2

269

The Center of Similitudes (2.1)

PKS 2773

The Center of Symmetry (2.2)

27-0,

3 REFLECTION Trigonometric Functions of Any Angle: An Application of Reflection About an Axis (3.1) Polar Coordinates: An Application of Goniometric Functions (3.2)

280

4 PARALLEL

286

SHIFT

Vectors (4.1) Components of a Vector (4.2) 5 ROTATION

IN

THE

PLANE

Rotation Around a Point (5.1)

284 286 291 294

294

Contents Regular Polygons (5.2) The Circle as a Limit (5.3) 6 ROTATION

IN

297 305

SPACE

309

Rotation Around an Axis (6.1) Regular Polyhedra (6.2) 7 GROUPS

OF

309 314

TRANSFORMATIONS

C2

Combining Transformations (7.1)

321

Repetitions (7.2)

325

4. CURVED JHE

SURFACES SGIRGLE

346

The Chord (1.1) The Arc and the Central Angle (1.2) The Inscribed Angle (1.3) Secants and Tangents (1.4) Power with Respect to a Circle (1.5) Two Circles (1.6)

346 351 355 364 B37 377

Common Tangents of Two Circles (1.7)

383

2 SURFACES

REVOLUTION

389

Axial and Cross Sections (2.1)

OF

389

Spheres (2.2)

393

Cones and Cylinders (2.3)

397

3 PLANE

SECTIONS

OF

CYLINDERS

Planes Containing a Generator (3.1)

5. AREA

vii

AND

CONES

401

401

The Ellipse (3.2)

403

The Hyperbola (3.3)

408

The Parabola (3.4)

415

The Conics (3.5)

A417

AND

VOLUME

1 UNITS OF MEASUREMENT 2 THE AREAS OF PLANE FIGURES

Quadrilaterals (2.1)

Triangles (2.2)

431 434

434

438

Similar Figures (2.3)

445

Polygons (2.4)

450

The Circle (2.5)

453

viii

Contents

A458 458 462 466

3 SURFACES Developable Surfaces (3.1) The Surface of the Sphere (3.2) Surfaces of Revolution

(3.3)

469 469 473 477 481

4 VOLUME Prisms and Cylinders (4.1) Pyramids and Cones

(4.2)

Polyhedra and Spheres (4.3) Solids of Revolution (4.4)

6. GEOMETRY 1 2 3 4 5 6

ON

THE SURFACE

THE SPHERICAL DISTANCE POLE AND POLAR SPHERICAL ANGLES THE SPHERICAL TRIANGLE POLAR TRIANGLES THE AREA OF THE SPHERICAL

Supplement Supplement Supplement Supplement Supplement

OF THE

SPHERE 496 501 505 508 ys) 518

TRIANGLE

|: Sets Il: Number Systems Ill: Equations in a Nutsheil IV: All About Proportions V: Numerical Computations

525 52? 232 539 544

Suggestions for Special Projects

546

Table |: Values of Trigonometric Functions Table Il: Powers and Roots

975 576

Index

577

Mathematical Symbols

segment AB ae ,

B,

es ray AB

line through

Wed.

carrier AB

right angle perpendicular(s)

S ILS V Vv

parallel(s)

|v |

angle(s)

parallelogram triangle(s) circle(s) arc with endpoints A, B similar equal not equal congruent greater than, larger than smaller than greater or equal to, not smaller than

Vi Ni = > >a Axel Ss S U ‘al

smaller or equal to, not

A

therefore quod erat demonstrandum total surface lateral surface volume vector absolute value, length of vector x-component of V y-component of V

is mapped into One-to-one correspondence between increases without bound set ofallx,suchthat... is (is not) an element of is a proper subset of union (one or the other) intersection (both) complement of A

greater than

Greek Alphabet A

alpha

B beta [ gamma A delta E epsilon Zozeta H eta SS YMLRDWR zt © theta

uc

I

iota

k K kappa dh uw v £5 o aw

A M N wiesl, O Il

lambda mu nu 9. omicron pi

rho sigma (s) tau upsilon phi chi psi KOK Deomega 21985 e266 e AMD

““Mnéels ayewmeTpntos eioiTw “LET

NO

ONE

UNTRAINED

IN

99

GEOMETRY

ENTER”

Inscription over the entrance of Plato’s Academy, center of higher learning in Athens, 4th century B. C.

A Preparatory

Chapter

The word geometry comes from the Greek words Ge (or Gaea), the name of the ancient Greek goddess of the Earth, and metrein,

which means “to measure.” Both words are also used in other combinations, such as geography (description of the earth), geomorphology (study of the structural formations of the earth), geology (science of the rocks of the earth), geodesy (measurement of the earth), meter, speedometer, metronome, and others.

Why do we study geometry? — Why do we study mathematics or theoretical physics? Apart from the interest in learning about the great ideas of our culture, which should appeal to most of you, there is a very practical reason:

these mathematical

sciences, of

which geometry is a branch, are very powerful tools for exploring the physical world around us. Those of you who may become engineers or experimental scientists will be using this tool to solve physical problems. The mathematician and the theoretical scientist will be more interested in the tool itself: its consistency, efficiency, and the possibility of its further development. In this basic course, you will not only acquire an efficient tool, but you will study this tool for its reliability. Some of the concepts that you are going to study are probably familiar to you, but in a general non-systematic way. In this Chapter you will find short summaries of geometric terms which should refresh your memory, and should serve as helpful reminders throughout the course. At the same time you will learn to look at these familiar bits of geometric knowledge in a more scientific manner than was possible in the lower grades. You will learn to use technical language, and to think more precisely. Also, you will learn to represent three-dimensional objects by clear drawings, which should help you to visualize geometric problems in three dimensions.

2

Preparatory Chapter

Let us see how much you remember. Name some geometrical figures. Name some geometrical objects. To describe these figures and objects, you may say various things about them. For example, you may say that they are big or small, round or with corners, green or blue, made of wood or of paper, etc. All such properties may be of interest in some connection, but not all are studied in geometry. What an object is made of and what color it has are not studied in geometry; but its shape and how it compares in size with another object are geometric properties. Think

of two cubes (dice, boxes) and two spheres (balloons,

tennis balls).

Make a variety of statements about pairs of these

four objects (two cubes, one cube and one sphere, two spheres)

and indicate which of these statements seem to express geometric properties and which do not. We shall talk about geometric properties in more detail as we proceed with this course. If you do not have a cube at hand when you want to describe it, then it is quite helpful to have at least a picture of acube. Can you make a sketch of a cube, or of a table, or of plane figures such as a Square or a triangle? Why is it easier, in general, to draw a sketch of a plane (two-dimensional)

figure than a three-dimensional ob-

ject? What are some of the ways in which we represent a threedimensional object on two-dimensional paper? (Think of photographs, maps, paintings, etc.) Figure Ia is a pictorial view (of what?), so called because it gives an over-all picture; but angles and sides are not necessarily represented in their true shape. If we wish to show the true shape, we can look at the object from the top and front and sides, and draw two or more of these views, as in Figure Ib. As

seen

front

from

Thus, for differthe

left

right

Informal Geometry

3

ent purposes, we may choose different ways of representation. In this Chapter you will learn how to make simple drawings of the geometric objects we are going to study.

1 THREE-DIMENSIONAL

GRAPHS

In algebra you made graphs as pictorial representations of sets! of ordered number pairs selected by some given relation from the set R of real numbers.”

For instance, the set of all pairs (x,y),

such that y is twice x, can be represented as in Figure I. In making such a graph, we first establish a one-to-one correspondence? between R and a line of abscissas * (x-axis), on which we represent x (the domain of the relation); then a one-to-one correspondence between R and another line through the zero point, called the line

of ordinates* (y-axis), on which we represent y (the range of the relation); and finally, a one-to-one correspondence between the set of number pairs (x,y) and the points of the plane. The abscissas and ordinates together are called coordinates.* Usually, but not necessarily, the coordinate axes are straight lines perpendicular to each other. In that case they are called rectangular Cartesian* coordinate axes, after the great philosopher and mathematician Descartes (17th century). The point O, common to both coordinate axes and_ representing the number pair (0,0), is called the origin; it separates the positive part of each axis (corresponding | 1 Words connected with sets and used in this paragraph are familiar to some of you. brief explanation of these is given in Supplement I. 2 See Supplement II on number systems. 3 These

words

come

from

Latin:

ordinare (to arrange in a certain order);

abscissa — abscindere

(to cut

off);

co — cum (together with).

4 Descartes used the Latinized form Cartesius as his name on his books.

A

ordinate —

4

Preparatory Chapter

to the positive members of R) from the negative part. The coordinate system consisting of the origin and the two coordinate axes separates the plane into four quadrants. The first quadrant (1) corresponds

to those (x,y) whose

elements

are formed

from the

positive numbers only. The third quadrant (III) corresponds to those (x,y) whose elements are formed from negative numbers only. The second (II) and fourth (IV) correspond to number pairs with mixed elements: the points of II have negative abscissas and positive ordinates, while the points of IV have positive abscissas and negative ordinates. In representing geometric objects we shall also use a system of Cartesian coordinates, but in three dimensions. Objects are often described by stating how long, how wide, and how high they are. Length, width, and height are geometric properties. Itis very convenient to set up a rectangular coordinate system such that length, width, and height will be represented on x-, y-, z-axes which are

perpendicular to each other and pass through a common point, the origin. For example, the length of a rectangular room may be measured along one wall on the line where it meets the floor, and the width

along one of the adjacent walls also on the floor; the height is the distance between the floor and the ceiling, and can be measured along one of the lines where two adjacent walls meet. If we choose these three measuring lines such that they start from one corner of the room, then they can be considered as the x-, y-, z-axes of a three-dimensional coordinate system with the corner as the origin. On each of the three axes we choose a point to represent the number one, while the origin represents

the number zero.

The dis-

tance between these two points is the measuring unit (e.g. one foot) in the direction of this axis. As in the two-dimensional graphs a one-to-one correspondence can now be established between the points on each axis and the real numbers (how many feet each point is from the origin). Any point in the room can then be identified by three numbers; P(x =7, y= 4, z= 6) can be found by measuring 7 units along the x-axis, from there 4 units parallel to the y-axis, and from there 6 units straight up. More simply we may write P(7,4,6) and agree that the first number of such an ordered

number triple always refers to x, the second always to y, and the third always to z. Where do you think the point Q(4,6,7) is?

Informal Geometry Pm

5

Any point P in space corresponds to an ordered number triple (x,y,z), and any ordered number triple corresponds to a point in space.

Figure Ila is a picture of the corner of the room which we used as a coordinate system. The floor is indicated by the plane IT, containing the x- and y-axes. The two walls II, and II, meet in the z-axis. The wavy lines mean that only parts of the floor and the walls are shown. If [1,, I1,, and II; are considered as endless planes, then we have

Figure IIb. Figure Ib.

You will recognize Figure Ila as part of this larger

(a)

(b)

Can you see three planes J],, II,, IH, intersecting each other in the x-, y-, z-axes? Each of these planes contains two of the three

axes and is perpendicular to the third. These planes are called coordinate planes. The wavy contours indicate that these planes do not end here but extend through space. Can you see that these three endless planes divide space into eight parts? These parts are called octants. (Octo means eight in Latin.) Any system of three such planes can be considered as a coordinate system. To avoid using all eight octants, we can choose a convenient coordinate system such that a given object will be in only one octant,

preferably the one in which all coordinates are positive numbers. Example: Suppose you want to describe a table by giving the coordinates of its corners. You could again use the

6

Preparatory Chapter

room as a coordinate system. You could also use another coordinate system: e.g. one leg of the table is the z-axis, its end on the floor is the origin, the perpendicular lines on the floor from the origin to the next two legs are the x- and yaxes. To complete the coordinate system, we must choose a unit on each axis. Suppose John wants to use one inch as a unit, but Margaret wants to use one foot. Is there any real reason why one choice should be correct but not the other? Let both systems be used. After John and Mar-

garet finish measuring the x-, y-, z-coordinates for each of the four corners A, B, C, D of the table top, their number

triples will look different.

Suppose John’s answer for A

is (0,0,30); could you guess Margaret’s answer for the same

point?

Obviously, since 12 inches equal | foot, Margaret

will measure 24 ft where John measured 30 in.

Thus the

number triple (0,0,30) in the language of inches (one coordinate system) is equivalent to the number triple (0,0,23) in the language of feet (another coordinate system) in describing the position of the upper end of the table leg OA. The length of the leg is given by the z-coordinate. The answer in one coordinate system can be transformed into an equivalent answer in the other coordinate system by the transformation rule: 12 in.=1 ft. The leg itself does not become longer or shorter when it is described by the number 30 or 24; the length of the leg is independent of the coordinate system used to describe it. If a geometric property (such as the length of the table leg OA, or the distance between the points O and A) can be described by equivalent answers in different coordinate systems, it is called an invariant formation .-

(unchanging) under coordinate trans-

> The distance between two points is an invariant under various coordinate transformations. special cases of distances.

Length, width, height, depth are

EXERCISES 1. a. Choose a coordinate system in the room, calling the floor the xy-plane, and two adjacent walls accordingly the xzand yz-planes. Which are the x-, y-, z-axes? Where is the origin?

Informal Geometry b. Describe the position of a corner of your desk top (called point C) by measuring the three distances x, y, z. Starting at C measure backwards first z, then y, then x, until you reach the origin. How high is C above the floor? If it is 33 ft, then z= 3.5. The point on the floor directly below C is called C’. Now measure from C’ on the floor in a direction parallel to the y-axis until you hit the x-axis at point Cx Suppose you find that it is 4 ft from C’ to C,; then y= 4 ft. Now measure the distance from C, to the origin. Suppose it is 2 ft; thenx=2. The position of C is given by the number triple (2,4,3.5). Is the order in which these numbers are given important? c. Find a point D described by (3,5,3.5). Now you will measure in the opposite direction from b, starting at the origin. From there along the x-axis to D,, it is 1 ft longer than to C,. From D,, in the direction parallel to the y-axis to D’ it is also 1 ft longer than from C, to C’.. Thus D’ is a point on the floor farther away from the origin than C’ by 1 ft in the x-direction and 1 ft in the y-direction. Now measure 3.5 ft straight up (parallel to the z-axis) from D’ toreachD. We find D on the table top away from C by 1 ft in the x-direction and 1 ft in the y-direction. . a. Repeat

b. c. d. e.

Exercise

1c for E(4,5,3.5),

F(5,3,3.5),

G(any x,

any y,z=3.5). Ifall these points have z=3.5, is it correct to say that they all lie on the desk top (or its extended plane), just as C and D of Exercise 1? If z= 4.5 instead of 3.5, where would all the points of 2a lie? Where are the points if z= 0? Explain: All points that have the same z-coordinate lie in a plane parallel to the xy-plane (parallel to the floor). Explain: All points of the xy-plane (all points on the floor) are characterized by having z = 0.

. Repeat Exercise 1a, b for another point in the room, e.g. a point on the lamp. . Find the points (4,0,0), (3,0,0), (6,0,0). Where are all points characterized by having y=0 and z= 0? . Find the points (4,2,0), (3,2,0), (6,2,0). points characterized by z =0 and y= 2?

Where

are all the

. How many points are there that have z=0 and y=5? Where are they? Could one such point be on the x-axis? On the y-axis?

On the z-axis?

7

8

Preparatory Chapter 7. How many points can you find that have z=0 and x= 5? Where are they? What can you say about their y-coordinate? . Pick some point on the wall called the yz-plane and write the number triple of its coordinates. Pick some more points on the same wall and write their coordinate triples. What do these triples have in common? Do the triples of all points on this wall have something in common? How must a coordinate triple look that represents some point on the yz-plane? . Repeat Exercise 8 for the xz-plane. How must a coordinate triple look that represents some point on the xz-plane?

10. Measure the length, width, and height of a bookshelf, using inches as units. Choose a coordinate system with origin at the lower left rear corner of the bookshelf, and write down

the number triples representing the other corners. a second coordinate system with its origin at the the room to the left of the bookshelf and write number triples representing the corners of the in the second system. Can you find a rule by triples of the one system can be transformed into of the other system?

Then take corner of down the bookshelf which the the triples

11. Choose one particular book on the bookshelf and describe its position with number triples within the two coordinate systems of Exercise 10. 12. Make a plan of your room at home. Describe the arrangement of the furniture in it by using a coordinate system with the origin in one corner of the room. Then estimate the height of that corner above street level and use that ground point as the origin of a second coordinate system with axes parallel to the axes of the first system. Describe the position of the furniture in this second coordinate system. Do you have to measure your furniture again for this second description? 13. What is the shape of an object that has 8 corners with the following coordinates:

A (0,0,0) (0,0,0) (0,0,0) (0,0,0) (2,0,0) ©= (2,0;0) & ac. d. -

B (3,0,0) (3,0,0) (3,0,0) (3,0,0) (5,0,0) (,0;0)

G (3,3,0) (3,3,0) (3,4,0) (3,4,0) (5,4,0) (5,3,0)

D (0,3,0) (0,3,0) (0,4,0) (0,4,0) (2,4,0) (2,3,0)

ah (0,0,3) (0,0,6) (0,0,5) (0,0,9) (2,0,9) (2,0,3)

E G (3,0,3) (3,3,3) (3,0,6) (3,3,6) (3,0,5) (3,4,5) (3,0,9) (3,4,9) (5,0,9) (5,4,9) G:0;3)'6,3,3)

H (0,3,3) (0,3,6) (0,4,5) (0,4,9) (2,4,9) 23.8)

Informal Geometry

9

14. Which points correspond to the set of (0,0,z) where z may be any real number?

15. Which points correspond to the set of (0,y,0) and the set of (x,0,0), where y and x may be any real number? 16. What is the set of all points with

z= 0?

17. What is the set of all points with y= 0? 18. What is the set of all points with x = 0? *19, Explain what each of the following means:

a. {(x,y,z) | x=0 and

y=0}

d. {(x,y,z) | z=0}

and z =0}

e. {(x,y,z) |y=0}

c. {(x,y,z) | y =0 and z=0}

f. {@,y,z) |x=0}

b. {(%,y,z) |x=0

2

HOW

TO DRAW

PICTORIAL VIEWS

In the previous Section we worked with an x-, y-, z-system in the

room, which helped us to describe three-dimensional objects. When studying three-dimensional geometry (solid geometry) we cannot always have three-dimensional models at hand; we must learn to use a drawing instead. In this Section we will learn to draw an x-, y-, z-system on paper and to locate points in it. In the next Section we shall apply this method to draw some of the basic geometric objects. To make a sketch of a three-dimensional object on two-dimensional paper, we may use a pictorial view as in Figure Ia (page 2) and Figure Ila (page 5). We first choose three straight lines through one point, representing the three coordinate axes x, y, z and the origin O of the system. Then we choose a unit on each axis to establish a one-to-one correspondence between the points of that axis and the set R of real numbers. Theoretically, any such system will enable us to complete the drawing. Practically, for good looks, it is advisable to draw the z-axis always upright; the y-axis horizontal and perpendicular to the z-axis; and the x-axis inclined, forming an angle w of 30° to 45° with the negative part of the y-axis.

(See Figure I.)

This method makes it seem as if the

x-axis and with it the whole xy-plane were sticking out of the paper, but seen sideways. Angle w represents a right angle in space; but it is not a right angle on the paper. w@ is called the angle of distortion. It is further advisable to make the units on the z-axis and y-axis equally long, but the unit on the inclined x-axis only 3 to 4

10

Preparatory Chapter

as long.

We may express such a choice as the ratio of distortion,

q = + for instance, or as the ratio of the three units, x:y:z = 3: 1:1.

P(3,2,4)

P(3,-2,4)

> +y

+X

+y

+x right -handed

left

-handed

The two coordinate systems in Figures I and II differ by the positive direction of the y-axis only. Try to imitate each system with three fingers of one hand held at right angles to each other (palm up and thumb towards you), using the thumb for x, the index finger for y, and the middle finger for z. Hold the middle finger up as the z-axis should be; then the palm together with the thumb and index finger will represent the horizontal xy-plane. For the one system you will need your right hand, for the other the left hand. While both systems can be used and are being used, it is more customary to use a right-handed system, as we shall do in this course. The distinction is important for correct interpretation of an ordered number triple (x,y,z). The point (3,2,4) referred to a right-handed system would be called (3,—2,4) in a left-handed

system, and vice versa. After the coordinate system is drawn, how do we locate a given point? In Figure III the point (2,3,4) is found by going from the origin O(0,0,0) two units in the x-direction to reach the point (2,0,0); from there three units in the y-direction to reach the point (2,3,0);

finally, from there four units up in the z-direction. In the sketch we do all this by drawing a parallel to the y-axis through (2,0,0), a parallel to the x-axis through (0,3,0), and a parallel to the z-axis through the intersection (2,3,0) of the former two parallels.

This

Informal Geometry

11

vertical line is intersected by a line through (0,0,4) which parallel to the line connecting O and (2,3,0).

itt

is

IV

Note that (2,3,4) gave us a route by which to reach P from the

origin. From Figure IV you can see that point P could have been reached also through other routes. Always starting from the origin O, we could first go to P,, from there to P’’, and from there

to P; or from O to P,, to P’’’ to P. Describe some other routes and make a drawing for each. All these different routes are represented by the same ordered number triple. The eight points O, P, P,, eel, P', P'', P'"’ are the corners

of a rectangular solid, which helps you to find and draw point P. The point P’(2,3,0) is in the xy-plane, directly under P; that means P’ has the same x- and y-coordinate as P but zero as z-coordinate. P’ is called the projection of P on the xy-plane. P’’ is the projection of P on the yz-plane. P’’’ is the projection of P on the xz-plane. As indicated by the superscript these projections are also called the first, second, and third projection of P.

Accord-

ingly, the planes containing them are called the first (xy), second (yz), and third (xz) projection planes. Projection planes are not always the same as coordinate planes, as they are here. In general, any plane may become a projection plane if we are interested in the view of an object in a certain direction. When you take a snapshot you hold the camera in a certain position hoping to get the view you are interested in. The exposed film is the projection plane, and the snapshot is the projection of the object on that projection plane.

12

Preparatory Chapter

The rectangular solid in Figure IV is seen in a certain direction, showing the front, top, and right side. It is viewed from above, right. The invisible rear edges are indicated by dotted lines. By changing angle w you can change the viewing direction. (See Figure V.) Try to visualize the direction in which the cube in Figure V is seen in each case. Look at a real cube from various directions until you see the same three sides as suggested by any one sketch. Of the eight vertices of the cube, six lie on the contour (outline).

The contour is, of course, always visible, and therefore

it is always drawn as a heavy line.

Of the two vertices within the

contour, one is in reality in front and visible; the other is in the rear and invisible, and therefore connected with three dotted lines

to its neighbors. Note that by changing the dotted lines to full ones and the inner full ones to dotted ones, you also change the viewing direction, although w is not changed.

(g)-150°

(h)-150°

V @ is measured

counterclockwise

from the negative side of the y-axis;

therefore @ = +30° in (a) and w =—30° in (e).

Note that parallel edges stay parallel in all views. Since the y- and z-axes were not changed (only w was changed), all edges parallel to them stay horizontal and vertical. The right angles be-

Informal Geometry

13

tween them stay right angles, but right angles formed with the x-axis do not. Study Figure V by making a quick freehand copy of each case, after deciding from which side it is seen. Then close the book and make freehand sketches for a given direction.

Agreements on Labeling 1. Points are labeled with capital letters, lines with lowercase letters, angles with arabic numerals or lower-case

Greek letters (a> B> y» 6> +++» w), planes with capital

Greek letters (il 2,.0): 2. AB means the segment between points A and B. Tips for Drawing Quickly and Accurately You should have two draftsman’s right triangles (preferably one

with 30° and 60° angles, the other with 45°

angles) or one such triangle (one with 30° and 60° angles) and a short straightedge. 1. To draw parallel lines a, b quickly and accurately, use the long edge of one of the triangles, say triangle A, to draw or line up with the first line a, while the left short

edge of A rests against the long edge of the second triangle B (or against the straightedge), which is held securely in place with the left hand. (See Figure VIa.) Now move the first triangleA with your right hand along the supporting second triangle B in the direction of the arrow until it is in position to draw the parallel line b (Figure VIb).

2. To draw a line c perpendicular to line a through point P, start out as before (Figure Vla), then turn triangle A

as the arrow in Figure VIc indicates, until its other short edge rests against the supporting triangle B. Then move triangle A as indicated until its long edge reaches point P (Figure VId). If more parallels or perpendiculars are needed,

those parallels which are in the same

direction should be drawn in the same operation. 3. To transfer distances along the axis or from one line a to another line a’, use a folded strip of paper, the fold

14

Preparatory Chapter

Vi

being a perfect straight line (Figure VII). On the strip, mark the given distance AB which is on line a, then line the strip up with a’ such that the mark for A is at A’; the mark for B will then give you B’ such that AB = AB". a'

A

8B

a B'

A

Vil

4. Your 30°-60° right triangle will help in drawing the x-axis with a distortion angle of 30°. Line up the long

Informal Geometry

15

edge of the triangle A against the y-axis and move it parallel as shown in Figure VIIla (always supported by the fixed second triangle), until its appropriate short edge passes through the origin. If the distortion angle

is 45°, the 45° triangle may be used in the same manner.

Vill

The right angle of your triangle may be used in the same way to draw a perpendicular line through a given point P. (Figure VIIIb.)

EXERCISES =

. Make a pictorial

2. oe). 4.

SH

*6.

view

of the following

points:

A(3,5,0),

B(1,2,0), C(4,3,0), D(4,0,0), £(0,5,0), F(0,4,3), G(0,2,4), H(0,0,5), J(5,0,2), K(3,0,4), L(2,0,3), M(3,4,5) Where are the points with z= 0? with x=0? with y= 0? Where are the points with z= 4? with x=3? with y=2? Make a pictorial view of the following points: A(1,1,2), BOD). C25). D440), BG.4.2), FA52;3)..GO33;2), H(3,4,2), J(2,3,4), K(2,1,4) Make a neat drawing of Figure V (not freehand), but with all edges twice as long. Be sure to know which faces are visible. Write underneath each case whether it is seen from above or from below, right or left. Where are all points that have x= y?

*7, Where a. x=y b. x=y c. x=y

are all points that have: and also z= 3? and alsoz > 3? and also 0 Sz 3 3?

16

Preparatory Chapter

*8. Make a pictorial view of the set of points (x,y,z) such that (Since z=0, the points arenn the y=2, z=0. 22x25, xy-plane; all points in the xy-plane with the same y-value are on a straight line parallel to the x-axis at what distance? Now mark the limits of x on

the x-axis;

all points between these limits in x would lie in a band parallel to the y-axis. We want only those that lie also on the line y= 2.)

+9: Make a pictorial view of the set of points (x,y,z) such that SS 4 = ye

ce

*10. Make pictorial views of a cube, with g = sand a—45-s135— —4S5°, —135°. Which faces must appear as true squares?

3

PICTORIAL VIEWS

OF STRAIGHT-EDGED

OBJECTS

A. Do you know the names and shapes of the objects in Figure I? Identify as many as you can before reading further. The objects in Figure I may be described as follows: (a) A cube has six equal squares as faces, therefore it is also called

a hexahedron.

(In Greek

hex means

six, hedra

means seat or side.) How many corners does it have? How many edges? A segment joining two vertices (corners) that do not lie in the same face is called a diagonal of the cube.

(b)

A rectangular solid has six rectangles as faces;

but, in

general, only the opposite faces are congruent (have the same

shape

and size).

The cube is a special case

of a

rectangular solid. A segment joining two vertices that do not lie in the same face is called a diagonal of the rectangular solid. (c) A parallelepiped has six parallelograms as faces; opposite faces are congruent. Since the square and the rectangle are special cases of a parallelogram, the cube and the rec-

ieee

Informal Geometry

F

(f)

(g)

C

(i)

17

18

Preparatory Chapter

(d)

(f)

tangular solid are special cases of a parallelepiped. (In Greek, epipedon means base.) and (e) are prisms. According to the form of the base, (d) is called a hexagonal prism and (e) a triangular prism. If the lateral edges are perpendicular to the base as in (d), it is a right prism; otherwise, it is an oblique prism as in (e). A regular prism is a right prism whose base is a regular polygon. and (g) are pyramids.

If the base is a square as in (f), it

is called a square pyramid; (g) is called a triangular pyramid. The perpendicular distance from the vertex V to the plane of the base is the height of the pyramid. If the base is a regular polygon and the foot F of the altitude is in the center of the base, we have a regular pyramid. (h A frustum of a pyramid is the lower part of a pyramid which was cut by a plane parallel to the base. The upper and lower base are similar figures (same shape, different —S

size).

9)

(Frustum means piece in Latin.)

A regular tetrahedron has four congruent equilateral triangles as faces. (In Greek tetra means four.) How many corners and how many edges are there? A regular octahedron has eight congruent equilateral triangles as faces.

(Octo means eight in Greek and in Latin.)

It looks like two regular square pyramids put together, but their heights have a certain relation to their common base: their heights together are as long as the diagonal of their common square base. The perpendicular to the square ABCD isthe axis EF. There are three equal axes, EF = AC = BD; they are perpendicular to each other as are the axes of a Cartesian coordinate system with origin in center O. Each of these axes is perpendicular to a square, the common base of another pair of square pyramids. Can you see these three squares? How many vertices and edges does an octahedron have? Note that in these pictorial views: 1. parallel lines appear as parallel lines. 2. right angles are sometimes distorted. 3. hidden edges are indicated by dotted lines. - Do you know the names and shapes of figures that could be faces of geometric objects?

Informal Geometry

19

1. A triangle can be a right triangle (one right angle), acute (each angle smaller than a right angle), isosceles (two equal sides), equilateral (three equal sides), scalene (no two sides

equal). - A quadrilateral (four-sided figure) can be a square, a rectangle, a parallelogram, a rhombus (parallelogram with equal sides), a trapezoid (only one pair of parallel sides), a deltoid or kite (two pairs of adjacent sides equal), or irregular. . A polygon has many vertices. A hexagon has six vertices, a pentagon five, an octagon eight, adecagon ten. Triangles and quadrilaterals are also polygons. If all sides and angles are equal, we have aregular polygon. The square is a regular polygon. (In Greek penta=5, deca= 10, polys = many, gonia = angle.) - Do you know what types of angles may be found in these faces?

right

acute

obtuse

90%

~S=hile

>90°

:

reflex

>180°

A straight angle is 180°, the sum of two right angles. A round angle is 360°, the sum of four right angles. Two complementary angles add up to a right angle. Two supplementary angles add up to a straight angle.

EXERCISES Group I

1. Which quadrilaterals have: a. equal sides? c. two pairs of parallel sides? b. four right angles? dd. two pairs of adjacent equal sides? . Which regular polygons do you know by specific names?

. Which special prisms have specific names? *4, Make a Venn diagram (see Supplement I on Sets) showing the relation between the various kinds of parallelograms.

20

Preparatory Chapter

Group I 1. Draw a pictorial view of the objects described by their coordinates in Exercise 13a through f of Section 1. a pictorial view of a cube whose edge is 2 units long. te. Draw (Think of the coordinate system in the corner of a room and think of a cube lying in that corner. See Figure I on page 2.) . Draw a pictorial view of two cubes, one on top of the other, with edges each 3 units long. . Draw a pictorial view of a cube whose edge is 3 units. Then draw a second cube on top of the first and make its edge only

2 units long. The upper cube could have various positions on the top square of the lower cube; draw two different positions. . Draw a pictorial view of a cube (= 45°, g=35, the edges along the y- and Zz z-axes = 2 inches). Mark the midpoints of each edge. Cut the cube into two equal parts by a plane parallel to the xy-plane, a second plane parallel to the xz-plane, and a third plane parallel to the yz-plane. All three cutting planes together divide the cube into how many small cubes? Watch the heavy and dotted lines.

un

. If one

of the small

cubes in Exercise 5 were moved out, edges that had been invisible (dotted lines) would become visible (heavy lines). Make a drawing of the remaining object for each of the small cubes moved out. (Auxiliary lines are drawn as fine lines.)

Informal Geometry

. Repeat Exercise 6 for w = 120°,

g=3.

. What has been cut out of this cube?

w= 120°,

21

Repeat the drawing for

q=%. (See Figure below.)

. Draw various objects of your own design by cutting out parts from a cube. . Draw these steps for w = 60° and for w = 120°.

11. Draw a cube and mark the midpoint of each face by intersecting its diagonals. The six midpoints are vertices of a regular octahedron. Draw it. 12. Draw a cube and draw face diagonals.

the

tetrahedron

formed

by six

*Group III

Draw pictorial views with g =5 and w = 30° of the following objects: . A regular octahedron, 6 units high, stands on

the xy-plane in point F(3,4,0). (First find F(3,4,0). Then E is 6 units above F, therefore E=(3,4,6). The center O is halfway between E and F. The other two axes of the octahedron pass through O, and, since

22

Preparatory Chapter nothing else is said about them, can be chosen parallel to the x- and y-axes. This is simplest because we know the units in these directions: A and C are 3 units in the y-direction on either side of O; B and D are 3 units in the x-direction on either side of O. Complete the square ABCD, which appears as a parallelogram; draw the parallels neatly with the twotriangle method; connect E above and F below, drawing hidden edges as dotted lines.) 2. A regular octahedron, 8 units high, with center at the coordinate origin O. (Mark 4 units on each axis on either side of O, then connect these 8 points.)

3. A cube with side 3 units, having one vertex in O(0,0,0).

4. A regular square pyramid, edge of base 4 units long, height 6 units, one vertex of the base at O(0,0,0).

5. A right triangular prism, 5 units A(1,1,0), B(5,2,0), C(3,4,0).

o

4

high, with base

points

PLAN VIEW AND FRONT VIEW OF STRAIGHT-EDGED OBJECTS

A drawback in representing geometric objects by pictorial views is that the shapes of the objects appear distorted. For instance, we know that the faces of a cube are identical squares, yet in Figure V on page 12, not all of these faces appear as such. If it is important to see the true shape of an object, the representation in plan view and front view is preferable to the pictorial view.

The plan view (Figure Ia,b) is obtained by looking down at the object in the direction of the z-axis perpendicular to the xy-plane. The xy-plane is called the first projection plane I1,. If a cube is lying on II,, we would see only the top square EFGH. The bottom square ABCD is hidden by the top square; the vertical edges EA, FB, GC, HD are hidden by the vertices (points) E, F, G, and H. In Section 2 we called the point P’, lying directly under P on II,, the first projection of P. The straight line through P, perpen-

dicular to II, and intersecting H, at P’ is called a projector. Figure Ia.)

(See

If we pass projectors through all points of the object

and intersect them with II,, we obtain the first projection of the

object.

Figure Ib is the plan view or first projection of the cube.

Informal Geometry

23

Notice that all points on a vertical edge, AE for example, have the same projector and, therefore, the same first projection, A’. The plan view alone does not give a complete graphic description of the cube. A square could also be the plan view of a rectangular solid of any height. The front view will show which one it is. The plan view and front view together give the complete description of the object.

24

Preparatory Chapter

The front view or second projection (Figure Ic, d) is obtained by looking at the object in the direction of the x-axis perpendicular The yz-plane is called the second projection to the yz-plane. view is the set of points in I], which are interfront plane I1,. The sections of IT, with all second projectors through the object. Figure II shows the connection between the plan view and front view. A cube is lying on II, at a short distance in front of Il). The heavy red straight arrows show the direction of the projectors. The figure produced on I],is shown to the right as the square A'B"F"E". (Double prime, because it is a second projection.) The figure produced on IT, coincides with the bottom square ABCD and is seen sideways and distorted in Figure Ia, since it lies on a

plane that should be perpendicular to II,. In order to see the true shape, we imagine the plane II, turned down as indicated by the curved red arrows until it lies in the same plane as II,._ The re-

sulting projections are shown at the right.

The distance of H’G"

from the y-axis is the true, undistorted distance of the cube from

II,. The first projection, A'B'C’D’, is the plan view; the second projection, A”B"F"E”, is the front view.

Let us repeat the procedure of how to find the plan and front views of a vertex or of any point P in general: In Figure Ia consider point P and its projectors. PP’ is on the first projector and

can in turn be projected ontoII, as P"P..

PP” is on the second pro-

jector and can in turn be projected onto IT, as ae. If we imagine II, turned down, then the segments Bt P,, and eve, lie on the same line perpendicular to the y-axis and form a single segment 1d P'P, a (See Figure IIb.) This relationship helps us to line up the front and plan views as in Figure IIb. For example, if the plan view F’ of the vertex F is already drawn, we know that the corresponding front view F” must lie on a perpendicular to the y-axis through F’. > The plan and front views of any point P lie on a perpendicular to the y-axis.

The projectors are very helpful in drawing and in understanding the plan and front views. They are drawn as fine lines in order to distinguish them from the edges of the object, which should be drawn as heavy lines. Since the projectors are perpendicular to the y-axis, they are parallel to the x- and z-axes respectively. (See

Informal Geometry

Figure II.)

the views.

25

Hence there is no need to include the x- and z-axes in

The units on all three axes are equal.

Example 1; Draw the plan and front views of a one inch cube lying on II,, with one vertical edge in [],, and an adjacent lateral face inclined to II, at an angle of 60°.

(a)

Figure IIa is a pictorial view of the required cube. Draw the plan and front views of this cube, as you read these explanations: First draw the horizontal y-axis. The plan view of the cube is a square of 1 inch sides, with one corner on the y-axis and one side inclined to it at 60°. The other side must then be inclined at 30°. (Review “Tips for Drawing” (page 14) for this 60° angle; check the right angles and the parallel sides of the square by the twotriangle method.) Now draw fine lines representing the projectors through the vertices of the square perpendicular to the y-axis (draw them all at once by lining up the long edge of one triangle with the y-axis, then turning it through 90° with the support of the second triangle, then moving it to each vertex). The height of the cube is seen in the front view from the y-axis upward: it is equal to the side of the square, 1 inch. The bottom and top squares appear in front view as the parallel segments A"B"D"C” and E"F"H"G". Draw these segments and the vertical

26

Preparatory Chapter

edges as heavy lines, and indicate by a dotted heavy line that DH (front view D"H") is a rear edge. The bottom vertices 4,B,C,D lie in I], and are, therefore, iden-

tical with their first projections. (The first projectors have zero length in this case.) If you wish to indicate this fact, you may omit the prime marks

in the plan view.

Likewise,

the vertices

D and H lie in II, and are identical with their second projections. You may omit the double prime marks, if you wish. Example 2: Draw the plan and front views of a regular square pyramid of | inch height and | inch base edge. The base square has the same position as: (a) the base of the cube in Example 1, and the pyramid points upward; (b) the top square of the cube in Example 1, and the pyramid points downward

with vertex in II,; (c) the base of the cube in

Figure I; that is, a base edge is parallel to the y-axis. y!!

A"

iB!

op!

c!!

y!

A"'

fell

To draw Figure [Va start with the same square that was used in Figure IIb. The diagonals of this square are the first projections of the lateral edges, and their intersection must be the first

projection V' of the vertex V. V” must be on a projector through V’ and | unit above the y-axis (why?). The second projections of the base points lie on the y-axis as in Example 1; their connections with V” are the second projections of the lateral edges. V"D" is drawn as a dotted line because it is a rear edge. To draw Figure IVb start again with the square in the plan view as in Figure [Va. Since the pyramid points downward, the lateral

Informal Geometry

27

edges are hidden by the base when looking down at it. Therefore, the plan views of these edges are drawn in dotted lines. V” lies on a projector through V’ and also on the y-axis (why?). The front view of the base square appears as the segment A”B”D"C" parallel to the y-axis in 1 unit height (compare with Example 1). The front view of the lateral edges is again found by connecting each of A”,B",C",D" with V". And again, the edge DV is in the

rear and invisible.

Therefore D’V” is drawn as a dotted line.

Figure IVc differs from IVa and b only by the position of the pyramid in relation to II],. |Since two edges of the base square are perpendicular to the y-axis, two pairs of projectors through the vertices coincide. In this position the front view of the pyramid does not show much detail. A position where the projectors do not coincide is, therefore, preferable. Example 3: It is easy to draw a regular octahedron if one thinks of it as a double square pyramid of a certain height. (See Figure Ij on page 17, and also the exercises in Section 3.) This height, from one vertex to the opposite vertex, must be equal to the diagonal of the common base square (because this diagonal connects another pair of vertices).

The plan view is a square and its diagonals. The center of the square is the

first projection of both vertices E and

A

F. The diagonals are drawn as full lines because the edges of the upper part are visible, hiding those of the

lower part.

E”F”=A'C’.

Why are

F"D" and D"E" drawn in dotted lines? (The square in the plan view was turned against the y-axis so that all four pairs of lateral edges would have sepa-

rate segments as front views.)

N

B'

Example 4: Draw the plan and front views of an oblique pyramid which has a regular hexagon as base, and a height equal to the major diagonal of the base. How to construct a regular hexagon of side s: Use your compasses to draw a circle of radius s; mark off the same distance s between

28

Preparatory Chapter

the vertices of the hexagon on the circumference of the circle, as shown. Since the hexagon is symmetrical, there are several checks for

a neat drawing: sides 2,3 and

5,6 must be parallel to the major diagonal 1,4 and should be drawn as parallel lines. (Use the two-triangle method.)

The

same

holds

for

sides 1,2 and 4,5 and diag-

|

o4,

onal 3,6; and for sides 3,4 and 1,6 and diagonal 2,5. vy"!

To draw the pyramid let us assume that it stands on I],. The plan view of the base is a regular hexagon. (No prime marks are necessary, why?) Let us give it a position such that each vertex has a separate projector. The front view of the hexagon is the segment A"F"B"E"C"D" on the y-axis. The vertex V does not project into the center of the hexagon because the pyramid is oblique. We shall assume V" is outside the hexagon. V” is on a projector through V’ and is as much above the y-axis as the pyramid is

Informal Geometry

high — namely the length of the diameter of the base. connects

7

with A.B,G,D

esr.

29

Now we

and V" with’ A",B’.C’,D" EF" F’”.

How do we decide which edges are drawn in dotted lines? The contours are always full lines. From the plan view, we can tell that E and F are nearer to [I], than

B and C.

Therefore E and F

and the lateral edges through them are in the rear and hidden, and their front views are drawn in dotted lines. The plan view helped us to draw the front view. Likewise, the front view will help us to draw the plan view: the edges through D are nearer to [], than other parts of the object in the same line of sight when we look down at it. These edges are therefore hidden, and their first pro-

jections DV’ and DE and DC are drawn in dotted lines. EXERCISES Group I 1. II, is the floor of the room, II, is one wall, and the y-axis is the

line where the floor and this wall meet. Imagine the following objects lying on the floor, with one edge of the base in the y-axis. What is their plan view? (What do you see if you look straight down?) a.acube, b. a rectangular solid, c. a regular square pyramid, d. a triangular right prism. 2. Suppose the objects in Exercise 1 were lifted straight up to lie on a shelf at about your eye level. What is their front view? (What do you see if you look at them in a direction perpendicular to the wall II,?) 3. The two squares and their diagonals at the right are plan views of two regular square pyramids. One is pointing upward and one is pointing downward; which is which?

Group II

1. Draw the plan and front views of a 13 inch cube lying on IL,;

one vertical edge is in II, and an adjacent face is inclined against II, at an angle of 30°.

(See Example 1.)

30

Preparatory Chapter 2. Draw the plan and front views of a regular hexagonal prism standing on II,.

(See Example 4.)

3. Draw the plan and front views of a regular octahedron. Example 3.)

(See

4. Draw the plan and front views of a triangular oblique pyramid. (Look at the pictorial view in Figure Ig on page 17.)

*5, Study the plan and front views given below and draw a pictorial view of the objects which they represent.

©

5S CURVED

SURFACES

Do you know the names and shapes of the following objects? Vv

V

Pictorial Views

Informal Geometry

31

DOO Plan and Front Views

The cone differs from the pyramid in the shape of the base. Instead of a polygon there is a curved figure. If the base is a circle, then the cone is called a circular cone.

If the foot F of the

altitude is in the center of the circle, the cone is called a right circular cone; otherwise, it is called an oblique circular cone. The straight lines connecting the vertex V with each point of the base are the elements or the generators of the cone. The straight line from the vertex to the center of the base is the axis of the cone. The height of the cone is the distance from the vertex to

the plane of the base. If the cone is cut by a plane parallel to the base, the part of the cone between the plane and the base is called the frustum of the cone. The cylinder may be thought of as a special case of a cone. Imagine the height of a cone becoming larger and larger as the vertex moves further and further from the base. Then the elements of the cone will more and more approach parallel lines. If they are parallel then there is no vertex, and we have a cylinder. Usually we think of cylinders as cut off by parallel planes. There are general and circular cylinders, right and oblique cylinders. Figure I contains the pictorial views of various cones and cylinders. The circular base does not appear as a circle, but as an ellipse. The plan and front views in Figure II are more easily drawn because the circles appear in their true shape.

32

Preparatory Chapter

m An ellipse can be considered as the view of a circle seen from the side.

Look at a glass of water. You will see the circular opening as a circle only if you look straight down at it; otherwise it will appear as an ellipse. This fact is helpful in drawing an ellipse: Imagine the circle to be enclosed in a square and diameters drawn parallel to its sides. The endpoints 4,B,C,D

of these diameters

are the points where the sides of the square touch the circle. If the circle is seen as an ellipse, the square around it may be seen either as a rectangle (Figure IIIa), or as a parallelogram (IIIb). In either case the di-

ameters AB and CD are parallel to its sides and can easily be drawn through the midpoint. A,B,C,D

the ellipse touches

At the points

the sides of the “square.”

These

portions of the curve are sketched in first, and are then connected with a smooth curved line. Cc

@

Sel D

D (a)

(b)

Il EXERCISES 1. Draw plan and front views measurements as indicated.

of the following objects, with

Informal Geometry

33

2. Draw plan and front views of the frustum of a right circular cone and of an oblique circular cone; both cones were cut off in half their height. *3. Draw several ellipses enclosed in given rectangles. *4. Draw several ellipses enclosed in given parallelograms. *5. Draw pictorial views of several cones and cylinders.

o

6

THE SPHERE

A sphere is the set of all points in space that are at the same distance from a fixed point called the center. The globe: From your geography course you know that a globe is a representation of the earth as a sphere, giving the location of each place on the earth in a special type of coordinate system: Jongitude and latitude. The longitude of a place tells us on which meridian it lies. In Figure I, NS is the rotational axis of the earth connecting north and N

rien

60°W

(a)

30°W

(b)

south poles. The meridians are circles lying in planes passing through the rotational axis. Figure Ia is a pictorial view of some of the meridians.

(Think of the wedges in an orange.)

meridian the circumscribed

For one

square has been drawn to represent

34

Preparatory Chapter

the meridional plane. The square appears as a parallelogram and the inscribed circle as an inscribed ellipse. In the plan view (Figure Ib) these meridional planes appear as straight lines through the center and therefore are more easily drawn. The angle between any two of them appears undistorted, which obviously is a In the front view (Figure Ib) most meridians great advantage. appear as ellipses. For one of them the circumscribed square has again been drawn; it appears as a rectangle. After choosing one of the meridians arbitrarily as the 0° meridian (customarily the one through Greenwich, England) the others are labeled by the angle their planes form with the 0° plane. These longitude angles appear undistorted in the plan view of Figure Ib. The longitudes are counted in the east and west direction until they meet at 180°; or they are counted only in the east direction until 360° brings us back to Greenwich again. The meridians are counted as semicircles; two semicircles in the same meridional plane are 180° apart in the counting system. The word meridian comes from Latin and means noon. All places on the same meridian (same longitude) have noon time (the

time when the sun stands highest above the horizon) at the same moment.

As the earth rotates about its axis in one day, noon will

occur at one meridian after another.

When it is noon in Green-

wich, it will not be noon on the 90th meridian West (in what country?) until the earth has rotated through 90°. (Remember 9OZE

60°E SO°E

180°

0°

A definition is an agreement of how to use a new word.

A definition must use words already known, or else we would

not understand the agreement. The definition of a digit only makes sense to someone who can count. A baby would not understand it, nor would a person who did not know the language. The definition of a square only makes sense if we already know what a rectangle is and what is meant by equal sides. If we do not know these words, then they should be defined first in terms of other words that we do know. Where is the beginning of such a chain? How does a child learn new words? If you have a chance to watch a two-year-old pick up new words, you may notice that he first picks up the new sound, then tries it out in childish

sentences, watching the effect he makes on grown-ups. Eventually, he understands the meaning of the word from learning about its usage. Similarly, you learn a new game by studying its rules. In chess, for instance, each piece is defined by the rules of how it may be moved from one place on the chess board to another. It is not at all important whether a chess piece is artistically carved of ebony or only a piece of cardboard with a name written on it; it is not at all important what it is, but how it is supposed to be used. Suppose your chess king moves two spaces. Your opponent will tell you that you should correct your error, since the king can only move one space at a time according to chess rules. If you then say: “Nothing prevents me from moving this piece two spaces; let us play this way,” your opponent may say: “All right, but then this piece is not a chess king; we will be playing some other game, not chess.” The chess king is defined by the system of chess rules. Let us examine the second question: why if— then? Sometimes chess problems or bridge problems are published in a newspaper or a magazine posing the question: Is it possible at this stage of the game for player A to win? Suppose you see the picture of such a game in a foreign language paper that you cannot read. You ask yourself: “Is it chess or some other similarlooking game which is unknown to me? /f it is chess, then I can solve the problem. The answer is: A cannot possibly win. If it is not the chess game I know, then I cannot tell, because the rules

44

Unit1

of the game might be different, the meaning of the pieces might be different, and the problem would be a different one.” Your statement that player A is the loser, is comparable to the theorem about the quadrilateral. The chess theorem is true, if the rules of The geometric theorem is true, if the rules chess are observed. of geometry (basic statements or postulates) are accepted. Your tormentor who started all of this trouble told you that the theorem did not apply to real figures. He was playing another game. Your answer that the theorem applied to a different type of triangle was quite correct. Whether a theorem is true or not depends on the rules of the game. The rules of geometry are the basic statements or postulates. By considering geometry as a game with certain rules two ideas are suggested: (1) that the rules might be changed resulting in a different game; and (2) that it does not make much sense to say these rules are true or not true, but rather that we accept them or that we do not accept them. These ideas are reflected in calling the basic statements basic assumptions. Once these assumptions have been agreed upon, a collection of theorems may be deduced from them (derived by logical reasoning), forming a deductive system (assumptions and theorems together). If real triangles have different angle sums each time you measure them, they cannot form a deductive system. This explains why the deductive system of geometry must deal with ideal triangles rather than with real ones. But how can this deductive system, which deals with ideal figures, help us in studying practical problems, which deal with real figures? (This is the third question.) Astronomers, for instance, study the movements

of the planets,

including the earth, around the sun. Do you think it is very importait in such problems to consider that the surface of the earth with its mountains and valleys is very irregular, or that in your neighborhood this surface is being changed by a bulldozer leveling a hill to make way for a new shopping center? Obviously, the problem of the path (orbit) of a planet around the sun will be clearer, if we omit all confusing detail and assume the earth to be a sphere. On the other hand, the city planner who is concerned with the new shopping center has studied several different problems: Where are new houses planned? Will these be big apartment houses with many people or small one-family homes? Will there be need for a large or a small shopping center? Do you think it is

A System of Postulates

45

very important for him to consider that the earth is round? A\Ithough we know that the earth is round, a street map of the city is drawn as if it were flat. Houses are shown as rectangles or trapezoids. In one study, the earth is taken as a sphere, and in another, it

is considered to be a plane.

Three-dimensional geometry is used

in One case, and two-dimensional

geometry in the other.

Geom-

etry is used as a simplified model of the real thing, bringing the problem under study into focus by omitting unimportant detail. Whether some detail is important or not depends on the problem and determines the choice of the model. The advantage of substituting some simplified model for the real thing is the elimination of unimportant and confusing detail. Suppose you wish to make a street plan for some part of the city. First you have to measure the length of the streets between intersections, and the angles between them at these intersections. Then you start to draw a sketch with these measurements at a reduced scale. In trying to represent a triangular city block by a triangle with the three angles as measured, you may discover that this is not possible because these angles do not add up to 180°. Something seems wrong and you go back to measure again. The new figures are slightly different, but the angles still do not add up to 180°. What should you do now? You may reason in the following way: You have been considering the simplified model of an ideal triangle whose angle sum is 180° to represent the city block in question. Since the observational errors in measuring the angles are too small to show up in your small map, they are essentially unimportant detail. Hence there must be some other reason why the model does not fit. The model itself cannot be wrong because you studied and proved the theorems about triangles in your geometry course. Maybe the choice of this particular model was unfortunate. A model is chosen by omitting the unimportant details of the real problem. Maybe you overlooked or misjudged some detail that is actually more important than you thought it to be. This error may have caused you to choose an inadequate model. Let us look at these details again by walking around this city block and watching very carefully for any detail that just might make a difference in the problem. Sure enough, by watching more carefully than before, you discover that along one of the streets there is a slight turn in direction which

46

Unit1

had escaped your earlier attention because it occurs at a point where there is no intersection with another street. Admitting this detail now as an important one, you have to change your model. What model should you choose? Instead of a straight line for that whole street, you will have to take two lines with a measured

angle between them. The resulting model will be a quadrilateral, not a triangle. Perhaps this model will be good enough for your purpose — drawing a small map of the neighborhood. If not, then further research may force you to refine the model even more. This example shows that using a model is a method of analyzing real problems. We reason: If the chosen model represents the real situation (apart from unimportant

detail), then certain state-

ments must be true because they are true for the model. If they are not true, then we must have overlooked some important feature of the real situation, and we had better look again. This reasoning makes us understand the real problem better. Of course, the chosen model is only useful if you know it thoroughly. The geometry of ideal figures and bodies provides us with a variety of such models. Their properties are expressed in a system of interesting and useful theorems, which can be derived by logical reasoning from a relatively small number of basic assumptions (postulates). The application of this system to problems of the physical world is the task of the physicist and the engineer. Our concern will be primarily to study the system itself. We have discussed several very difficult ideas. You will understand them better as you go on with your studies in geometry and in science. There are quite a few non-technical books available on subjects such as: What is science? What is mathematics? The nature of geometry, the foundations of mathematics, and so on. You might enjoy reading some of these. What were the most important ideas discussed in this Section? > A definition is an agreement of how to use anew word. The definition may be given by equivalent, already known words, or by a whole set of basic statements about the usage of the new word (such as the rules of the game). A definition by equivalents must be reversible. (See page 42.)

> A basic statement (axiom, postulate) is accepted proof; it is an assumption which is agreed upon.

without

> A theorem is proved if it can be deduced by logical reasoning from the accepted basic assumptions or from other theorems,

A System of Postulates

47

which were previously deduced from the same basic assumptions. Jf certain basic assumptions are accepted, then the derived theorem is true. > The basic statements and the theorems together form a deductive system.

> In problems about the physical world, a geometric model is useful in two ways: (1) it simplifies the problem by omitting unimportant detail; and (2) it helps in analyzing the problem by comparing theorems about the model with physical facts. b> The choice of the model depends on the problem and the importance or unimportance of certain detail.

©

2

WHICH

BASIC ASSUMPTIONS?

In the Preparatory Chapter we discussed various geometric facts from a practical viewpoint, somewhat like the way in which the Sumerians and the Egyptians engaged in geometry. From now on we shall attempt to organize these more or less isolated facts into a coherent

system

as the Greeks

did with the facts

learned from the Egyptians. Following Euclid’s example, we should start by adopting a set of basic statements (postulates).

What are they?

We could look

at Euclid’s book The Elements and study the axioms and postulates which he used, as has been done century after century. The difficulty is that Euclid’s text is very lengthy and technical, and that the way ideas are expressed there is somewhat different from our modern way. What is the difference? For many centuries mathematicians were mainly interested in expanding the deductive system of geometry by deriving new remarkable

theorems.

The basic statements,

on which the system

rested, were taken as unquestionable truth since they had been established by-the great teachers of the past: Euclid (geometry) and Aristotle (logic). In the last two hundred years, however, several great mathematicians * have focused more of their attention on the foundations and the nature of mathematics and science. Their discussions have made us aware that the postulates and axioms, considered “self-evident truths” for centuries, are neither self-evident nor necessarily true, but mere theoretical assumptions 1 Gauss, Bolyai, Lobachevski, Riemann, Hilbert, Peano, to name just a few.

48

Unitl

on which a deductive system may be based. It was this changed viewpoint which has led to the understanding of geometry as an abstract model rather than as an exact description of the real physical world. As a further consequence of this new thinking, Such Euclid’s axioms were reinterpreted and reformulated. which by such, as geometry Euclidean the affect not do changes we mean the collection of all theorems deducible from Euclid’s axioms. But it became clear that there are more ways than one in which a set of axioms may be chosen and formulated to serve as a foundation for the Euclidean system. We shall follow David Hilbert’s ideas in a general way. We shall try to construct a deductive geometrical system stepby-step by deducing theorems from postulates. Gradually we shall add one postulate after another until we reach the goal of Euclidean concepts. To start the construction strictly from scratch, however, we would have to define every English word we

use, and to list rules of logical thinking. But since this is a course in geometry, we shall concentrate on geometric statements only. In other words, we shall accept as understandable and correct not

only English and logic, but also the vocabulary and rules of arithmetic and algebra. Since we are setting out to construct a system of Euclidean geometry, we know that we shall have to deal with geometric concepts such as points, straight lines, planes, and so forth. Here again we encounter the question: What is a straight line? We know now that an answer to such a question is a definition and may be given in two ways (see Section 1): either in the form of an equivalent, already known, expression (usually, the word definition applies to this form only), or as a set of rules about usage. Since we are just starting out, we have no known equivalent expressions. Therefore we must set up some rules about how to use these first (initial) terms. The rules are our basic assumptions or (since we deal with geometry) postulates. The initial terms are: space, plane, straight line, point. The fact that the initial terms are not defined by equivalents is often expressed by calling them undefined terms. Strictly speaking, this is a misnomer because they will be defined by the system of postulates.

At the moment, however, they are blanks, and we

pretend not to know what they stand for unless told by the postulates.

A System of Postulates

49

The purpose of this pretense, and the purpose of this Unit, is to make you aware of the many complicated notions contained in seemingly simple concepts such as plane, straight line, etc. On the one hand you know from the informal geometry of the Preparatory Chapter what these words should mean, on the other hand it will take several postulates to define them. In the discussions of these postulates we must therefore be careful not to confuse two things: 1. the familiar geometric concept of informal geometry (the goal) and 2. the concept gradually built up from one or two or three —

and so on — postulates,

as our discussion

proceeds.

To avoid this confusion, we shall mark the “unfinished product” with the prefix pseudo. Thus, a pseudo-plane may satisfy several postulates for planes, but not all of them. The more postulates are satisfied, the less difference there is between a pseudo-plane and a plane as we know it from informal geometry. By the end of this Unit there should be no difference between the two, and the “pseudo” may be dropped.

Postulate I: (Post. I) (a) Space is the universal set consisting of infinitely many points. (b) Every plane is a proper subset of space; there are infinitely many planes. (c) Every straight line is a proper subset of a plane; there are infinitely many straight lines on each plane. (d) Every point lies on a straight line; there are infinitely many points on each straight line.

This first set of postulates sounds like a set of definitions, but it is not.

Why not?

(See page 42 on definitions.)

If (b) were a defi-

nition of the word plane, then this word should be interchangeable (equivalent) with the expression “a proper subset of space.” In that case (b) would be reversible, leading to a statement:

proper subset is not true. This set is a of space, but plane. Since

“Every

of space is a plane.” But this statement obviously Think of a cube as a set of infinitely many points. proper subset of space, since it is formed by points not by all points of space. Yet it is certainly not a this set satisfies Post. I, we may call it a pseudo-

50

Unit 1

plane with respect to Post. I. It is clear that several more postulates are needed to make the term plane mean the same thing as it means in the informal geometry of the Preparatory Chapter. Although Post. I does not tell us a great deal about the initial terms, it tells us enough to change these terms from complete blanks to something we do know a little about. And by logical reasoning we can deduce even more: From (a) and (b) together, we know that a plane is a set of points, and that there is at least one point not belonging to a given plane. From (c) and (d) together, we know that each plane contains in-

finitely many points. If there are infinitely many points on each straight line, (d), and if there are infinitely many straight lines on each plane, (c), then there must be infinitely many points on each plane. From

(c) and (d), we also know that every point lies in some

plane. Since any given point lies on some straight line, (d), and every straight line lies in some plane, (c), the given point must lie in that plane.

From Post. I we know that planes and straight lines are sets of points; and these sets may or may not have points incommon. In a Venn diagram (see Supplement I on Sets), let the set II represent a plane, the set L a straight line, and P a point;

all are contained

in the universal set U. Let us use a square pyramid as an example to demonstrate the meaning of the two Venn diagrams in Figure I. Let IT be the face BCV; if the edge BV is L and the vertex V is P,

then we have the case of the first Venn diagram, since V lies on BV, and BV belongs to BVC. If Il = BCV, L = AD, and P = A,

then we have the case of the second Venn diagram.

A System of Postulates

51

EXERCISES 1. What is a proper subset (see Supplement I on Sets)? What do we mean by saying that a straight line is a proper subset of a plane? 2. What is a definition (see p. 42)? Why is Post. I(c) not a definition of a straight line? What do we mean by saying that a definition must be reversible? Give an example. 3. Give an example of a set of infinitely many points that is not a straight line. Find two more examples.

4. Does it follow from Post. I that every straight line must lie in some plane? that it lies in only one plane? 5. Does it follow from Post. I that, if L is a straight line lying in a plane II, then there is at least one point P in II that is not on L? 6. Can you tell from Post. I whether or not a plane is endless? Can the face of a pyramid be a plane according to Post. I? 7. Can you tell from Post. I whether or not a straight line is endless, or could it be the edge of a cube?

8. Draw a pictorial view of a cube and label the vertices. Find an interpretation of the following Venn diagrams, if II is some face of the cube, L an edge, and P a vertex.

*9, Consider a coordinate axis and imagine that all points corresponding to the set of integers are removed;

many points have been taken out.

thus infinitely

Could the remaining set

52

Unit 1

of points still be considered a pseudo-line with respect to Post. I? *10. Consider the set of points that correspond to the set of integers on a coordinate axis. Could 4 this set be considered a pseudo-line e I? with respect to Post. *11. Suppose Post. I did not contain the words “infinitely many,” and suppose we had only four points in all. Then

our universal

set, a pseudo-

space, consists of only these four

ry

e 2

1

points. We shall call it S,. What interpretation could you give to a pseudo-plane and a pseudoline? How many points would each contain? How many pseudo-planes and how many pseudo-lines exist in this S,? *12. Repeat Exercise 11 for a pseudo-space S, consisting of only three points.

o

3

BETWEEN

Postulate I actually says very little about our initial terms. If you think of a straight line in the ordinary sense, then the statement that it contains infinitely many points (Id) and that it lies in a plane (Ic) does not distinguish it from a set of infinitely many points picked at random in a plane, nor from the set of points inside a circle, nor from the set of points on a coordinate axis corresponding to the set J of all integers (see Supplement II on number systems). You can readily see that the ordinary straight line has many properties that need to be stated by further postulates. Name a few such properties. Since we cannot talk about all characteristic properties at once, we have to make a choice as to the sequence of the discussion. The fact that this choice is so arbitrary may make one geometry book appear different from another, but in the end all characteristic properties should be accounted for. The example above of a set of infinitely many random points in the plane makes us aware that we associate a certain order of points with the concept of a line. In tracing a line with a pencil any point is reached at a definite stage before or after certain other points.

LPI Straight

line

Broken

ANI, OR line

Curved

line

A System of Postulates

53

This property is shared by straight lines, broken lines, and curved lines, but not by the plane. Since it seems to be the most characteristic distinction between lines and planes, we shall formulate

the order postulates next. And because these apply also to other than straight lines, we shall drop the word “straight.” For the sake of brevity we shall drop the word “straight” also in the following discussions and use the word line for straight line where no misunderstanding is expected.

Postulate Il: Order Postulate (a) Of three points belonging to the same line, one and only one lies between the other two. (b) Between any two points of a line there is at least one more point on that line. (c) Every point on a line is between two other points of that line.

The order postulates establish a relation (betweenness) concern-

ing any number between line as

three points of a line, which reminds us of the order in our system. Of three numbers we can always tell which is the other two. This analogy makes it possible to use a a graphic representation of numbers. Accordingly, we

say that pointA is before point B (A < B) and B is afterA (B > A),

making use of the arithmetic symbols for smaller and larger.

B

Postulate I(a) does not ap-

ply to a closed line such as the circle. Each of three points

A

A, B,C on acircle can be con-

sidered as lying between the other two. Any broken or curved line that does not pass through the same point twice may be considered as a pseudo-line with respect to Il(a). The closed line such as

the circle is excluded.

C Curved,

closed

line

54°

Unit 1 much more

Postulate II(b) means

There is a

than it sounds:

point B between any two points A and C; hence there also is a point D between A and B, and a point E between A and D, and so A

pf

D

E

B

C

$—t+4—__+—_}—

on. It follows that there are infinitely many points on a line between any two of its points. (This is an equivalent form of II(b).)

In other words, no point has an immediate neighbor. Thus the points on a line are not comparable to a string of pearls, where each pearl has an immediate neighbor. In the set of integers the number 6 has two immediate neighbors, namely 5 and 7. There are no other integers between 6 and 7. The set of integers could have been considered as a pseudo-line with respect to Post. I (it has infinitely many members),

and with respect to Post. I(a) (of three

integers one and only one is between the other two), but not after we accept Post. II(b). Postulate II(c) in essence says that there is no last nor first point on a line, since every point is between two others.

The set of in-

tegers satisfies this postulate. The fact that there is no smallest and no largest number may seem difficult to grasp. Yet, if you name a number as large as you wish, you can always name at least one larger number. How many points are there on a line? How many is infinitely many? Are there more points in the plane than on a line? How many more? If you are interested in questions of this type and want to spend more time on them, then you are in for fun and sur-

prises. (See Suggestions for Special Projects.) The concept of between may be used to define a segment and a ray: Definition I-1: Given two points A and B ona line s, then the set consisting of A, B, and all points of s that are between

A and B is called the segment AB (also called the closed interval AB). A and B are called the endpoints of the segment AB. The set of points betweenA and B is called the interior

of the segment AB (also called the open interval AB). ey

A System of Postulates

55

i

Definition +2:

If AB isa segment on a line s, then the subset of s consisting of AB and all points C beyond B (this means all points ¢C for which B is between A and C) is called the ray AB. The point A is the initial point of the ray. The straight line s is called the carrier of the ray. The subset of s consisting of A and all points not belonging to AB is also aray. The two rays are said to be in opposite directions.

In a drawing it is sometimes useful to distinguish between a segment and a ray by marking definite endpoints and placing an arrow where there is no endpoint. AB means a segment with,the endpoints A and B. AB means a ray with the initial point A, passing through B (but not

ending there). AB means a straight line passing through A and B.

EXERCISES 1. What is the difference between a segment, a ray, a straight line? What are their symbols? What is an open and a closed interval? Does an open interval have a first point? 2. Does Postulate II(a) apply to any three members of the following sets? a. the integers marked along a coordinate axis b. temperature degrees on a thermometer

c. letters in the order of the alphabet d. children in a gym class lined up according to size e . trees in a garden

56

Unit1 f. times within a day; photos of a clock

days within a week;

time shown on

g. if A, B, C are points of a circle, each can be considered as being between the other two; therefore II(a) does not apply. If we mark a point O as starting point and add a direction by arrow, we could imagine winding a thread around the circle, marking off O, A, B, C on the first round, O,, A,; B,, G, on the first repetition, O,, A,, B,, C, on the second repetition and so on. Will II(a) apply now?

3. Does Postulate II(b) apply to the set J of integers, arranged in their natural order? Between 2 and 10 there is at least one other integer, in fact there are... (how many?);

does this

mean that II(b) applies?

4. Does Postulate II(b) apply to the set F’ of fractions between 0 and 1 in their natural order, as marked on a coordinate axis?

Can you give some arithmetic procedure which will always yield a rational number (see Supplement II) between any two given ones? Is it true that between ; and $ there are infinitely many rational numbers? Can you name at least three? (Remember from arithmetic that every fraction can be expressed as a decimal fraction;

what is ; and 4 in the form of

decimal fractions? What decimal fraction is between these two? Could you also find such a fraction without changing to decimals?) 5. Does Postulate I(c) apply to a closed interval AB? Explain. Does it apply toa ray? Does it apply to the setJ of integers? 6. Think of a set M of all numbers between 0 and 1 ona coordinate axis, the numbers 0 and 1 excluded. This set corresponds to an open interval. Does M havea largest number? (Name a number N, of M larger than 2, (N Le ); then name

a number N, > N,; then N, > N, and N, > N,, etc.) 7. Does set M of the previous exercise have a smallest number?

(Find examples of numbers N'Y < N''< NU If any two points of a set M can be joined by a line segment that lies entirely within M, then M is said to be connected. (Here a line segment is not necessarily a straight line segment, since Postulates I, II, and III(a) hold also for broken lines and curved lines.)

> A connected set M consists of pieces joined to a point in (In short: M-—S is

is separated by its subset S if the set M—S such that a point in one piece cannot be another piece by a line entirely within M-S. not connected.)

Example 1: A river flows through a town, dividing it into two sections A and B. It is possible to go from A to B only if there are

A System of Postulates

59

bridges or some other means of crossing the river. If there were no way of crossing, the river would be a barrier, separating the town. If there is a pond in the middle of town instead of the river, then a bridge across the pond is not necessary, because people could always walk around the pond in order to get from any point in town to any other. The pond is no barrier; it does not separate the town. Example 2: The first and second floors of a house are connected by a single staircase. If that staircase is blocked, the connection is severed and the two floors are separated. Since our informal ideas of planes and space certainly include the property that there is some link between any two points, we adopt the following postulate:

Postulate III: (b) Every plane is connected. (c) Space is connected.

EXERCISES Group I: The line 1. Show that the set consisting of all points of a line s except one point B is a pseudo-line with respect to Postulates I and II. 2. Show that the set consisting of all points on a number line except the points corresponding to integers is a pseudo-line with respect to Postulates I and II.

3. Show that the set consisting of all points on a number line except those of the closed interval 0 = x = 1 is a pseudo-line with respect to Postulates I and II. 4. Would you say that a segment is connected? interval?

a ray? an open

5. Does the initial point of a ray separate the ray? other point of a ray separate the ray?

Does any

6. Do the endpoints of a segment separate the segment? any other point of a segment separate the segment?

Does

7. Is there a point on an open interval that does not separate that open interval?

60

Unit 1

8. Is a circle connected? If one point on a circle is removed, is the remainder connected? If two points on a circle are removed, is the remainder connected?

Group II: The plane

Lb If a single point P of a plane II is removed, is the remainder still connected? If one hundred points of II are removed, is the remainder connected? (Think of a sheet of postal stamps, perforated between the stamps.) . If a sheet of paper is slit between two points so that the cut never reaches the edges, does the slit separate the paper? . Make two diagrams representing the two cases of Example 1: the town as universal set U, the river as subset R in one case, and the pond as subset P in the other. What do we mean by saying that R separates U, but P does not?

. Consider the set of points in the xy-plane whose coordinates x and y are integers. Is this set connected? . If the sets A,B,C

within the universal set U are each connected, in which of the diagrams a,b,c,d is the set A U BUC connected? In which is it a union of separated sets?

(c)

(d)

6. On a long rectangular strip of paper draw a straight line s as shown in the diagram on the next page. It is obvious that line s separates the strip: if you cut along it you get two pieces.

Imagine pasting the ends together, placing S, on S,, A on C,

A System of Postulates

61

A

C

B

D

and B on D: if you cut along s, you again get two pieces. Now paste the ends to-

vy

gether so that.S, falls on S,, but at the same time turn the strip so that D falls on A, and C on B, as shown. Now cut along s. What do you get?

:

Moebius

Band

Group III: Space 1. The points filling a certain cube may be considered as a subset C of space. Would you say that the set C is connected? The complement of C is the set of all points of the universal set that do not belong to C. Would you say that this set, U-C, is connected? 2. Suppose the top and bottom faces of the cube C in Exercise 1 are extended in unlimited planes so that instead of C we now have an unlimited layer C’ between these planes. Would you say that the set U-C’ is connected? Does C’ separate space? 3. On each of the six faces of a cube sits a quadratic pyramid pointing outward. Is the union of the six pyramids a connected set? 4. Does a point separate space? Does a set of a thousand points (a segment, a line) separate space?

o

4.2 The Postulates of Separation

Postulate III establishes the property of connectedness for our initially undefined terms line, plane, and space. In the postulates of separation, we want to formulate how the connectedness of one

of these terms is affected by another. Keeping our everyday experience in mind (how a rope, a sheet of paper, and an apple may be cut into pieces), we adopt the following properties:

62

Unit1

Postulate IV: (a) A line is separated by any of its points into two connected pieces. (b) A plane is separated by any of its lines into two connected pieces. (c) Space is separated by any of its planes into two connected pieces.

Postulate [V(a) is closely connected with the order postulates. We already know from these that all points of a line s other than a given point B of s form two subsets s, and s,, such that all points of one subset come before B and before all points of the other subset. Postulate I[V(a) adds that these two subsets are connected. Without IV(a) a line with

branches such as Figure I can be considered as a pseudo-

S,

line, if we agree that all points

of s, come before s,'. If B is removed,

Figure I falls into

two subsets s, and the union Of-sseandss.., ©DUteso coo als not connected. Thus B sepa-

B

rates Figure I into three, not

two, connected pieces. Postulate I[V(a) excludes lines with branches from our concept of a line. Without Postulate [V(b) an open interval is a pseudo-line. (Show that it satisfies all postulates from I up to and including IV(a).)

But since an open interval does not separate a plane (think

of a slit in the paper), it is excluded by Post. IV(b). In other words, a line satisfying IV(b) must extend without limit in both directions throughout the plane in which it lies.

says something about the term plane.

But IV(b) also

The example of the Moe-

bius Band (Exercise 6, page 61) shows that even if a line extends

without limit in both directions throughout the plane, that plane may still not be separated by that line. Postulate IV(b) makes it certain that what we want to call a plane will not be like a Moebius

Band.

A System of Postulates

63

Postulate [V(c) extends the same ideas to three dimensions.

A

plane must be without end in order to separate space; and there are exactly two connected parts for any separating plane. The two connected parts mentioned in Post. IV are sometimes called half-lines, half-planes or the sides of the separating line, and half-spaces or the sides of the separating plane. The point, the line, and the plane causing the separation are called the Doundary between the two parts. Thus all points of a plane IT which are not on the line s of IT are either on one or on the other side of s (in

one or the other half-plane). The boundary s severs the connection between two points A, B on opposite sides,

S

but not between A, C on the same side. In other words,

to go from A to B one must cross the boundary;

but one

B

can go from A to C without crossing the boundary. It is easier to identify the two sides of a boundary if they are given specific names. Thus we speak of a positive and a negative part of a line with respect to a point O; of a right and a left halfplane with respect to a line; and of up and down in space with

Up

——

—

| —

ee

Down

(a)

(b)

(c)

T

respect to a plane. It is we who arbitrarily assign these names, and we could just as well assign the opposite names. If you look at Figure IIb again after turning the book around, then you might prefer to exchange the names.

64

Unit 1

EXERCISES . Consider the set M of all points on the x- and y-axes within a plane. Into how many connected pieces is M separated by the origin? by any other point of M? . Does the endless broken line in the Figure below separate the plane in which it lies? Does the endless line of repeated semicircles separate the plane in which it lies? Are these two examples pseudo-lines with respect to Postulates I to 1V?

ON

RO MESIO

. Do the lines in Exercise 2 separate space?

. Make a Moebius Band. Imagine the line s in the middle of the Band is a river flowing between the land of the good men at its right bank, and the land of the bad men at its left bank. There is no bridge across the river. Mark a direction on s and imagine walking along the right bank; prick the paper with a needle as you walk, to mark the places where you have been, until you return to the place where you started. Then look: have you been in the land of the bad men too? . Make

another Moebius

Band without the line s.

Paint one

side of the paper green and the other brown. Start with the green paint and finish the whole side before starting with brown. What happens? . Step in front of a large mirror and explain to the person in the looking glass which is your right hand and which is the direction “to the right.” He seems to agree with you perfectly. Does he? Which is the direction “‘to the right’? . Is arocket on its way to the moon going up or down? Which direction is “up” at the north pole? Which at the south pole? If somebody said: ‘“‘On the other side of the earth from us people are walking with their feet up,’ what would be your comment?

. Suppose that II is a plane and that A and B are points lying in II. Which of the following sets separate space? at Ele d. I1— AB

b. II— AB c. II—AB

e. II— {A,B} f=

{4}

. Suppose that two adjoining faces of a cube lie in planes I], and II, respectively. If S is the set of all points in space, is S— (II, U II,) connected? Into how many pieces is space separated by H, U I,? Is space separated by H, |iet

A System of Postulates

65

10. Suppose B is a connected set and A is a proper subset of B. If A is not connected, is it possible for A to separate B into two pieces, each of which is connected?

5

INCIDENCE

Incidence is a technical name for the relation between a set of points and a subset or a member of that set. It is usually expressed in a variety of equivalent ways such as: a line / (a point P) belongs to, is a line (point) of, lies on or in a plane Y; > contains, passes through, lies on*1(P). The postulates of this Section deal with two aspects of this relation: (1) the identification of a set by some of its members, and (2) the intersection of two or more sets.

5.1 The Incidence Postulates Let us see what we know about the incidence relation from our informal studies in the Preparatory Chapter. In order to draw a pictorial view of a pyramid, for example, the vertex is connected with each corner of the base to represent the lateral edges. If you connect the two endpoints of a lateral edge by free hand, you may not draw the connection quite straight at first; in that case you might try again or use aruler. You know that there should be only one straight line segment between these two points, although more than one pencil line may connect them. This is one example of an incidence relation which we will incorporate into our postulates. Another example is contained in our informal experience that the center of a square is a common point of its two diagonals. We know that there should be only one common point. We also know that not all lines have a point in common. The opposite sides of a square are parallel and should have no point in common, even if their carriers are prolonged on a very large paper. There also are lines in space that have no point in common and are not parallel; e.g. the altitude of a regular pyramid and one of the base edges. Such lines are called skew lines. Point out pairs of skew lines in this room. 1 In geometry, a plane > that cuts a line / at one point does not pass through / (does not contain the whole line); and a plane through / does not cut /. The expression “‘lies on” is used in both ways: a point lies on a line, a line lies on a point.

66

Unit 1

Definition I-3: Two planes are parallel (symbol |), if they have no point in common. A line and a plane are parallel, if they have no point in common. Two lines within the same plane (coplanar lines) are parallel, if they have no point in common. Two lines are called skew lines, if they have no point in common and do not lie in the same plane.

The common members of two sets are called the intersection of these sets. Accordingly, we speak of a point of intersection between two lines, and a line of intersection between two planes. In geometry, however, to intersect usually means more than having common members. In Figure Ia the line ¢ through a vertex of the triangle has a point in common with the triangle, but we do not say that it intersects the triangle. Neither do we say that in Figure Ib the line ¢ intersects the circle. We prefer to say that t touches

(is tangent to) the circle.

In Figure Ic the lines s and t

intersect; this means not only that they have the point P in common but also that they pierce each other.

t

P

t

P

P

t S

(a)

(b)

(c) |

Definition I-4: Two lines are said to intersect, if they have a point in common which separates each into two parts that lie on opposite sides of the other. (continued on next page)

A System of Postulates

67

A plane and a line are said to intersect, if they have a point in common which separates the line into two parts that lie on opposite sides of the plane. Two planes are said to intersect, if they have a line in common which separates each plane into two parts that lie on opposite sides of the other.

As you read the following incidence postulates show their meaning by letting a sheet of paper, a pencil, and small objects represent an unlimited plane, an unlimited line, and points.

Postulate V: /ncidence

(a) Two points determine one and only one line to which they both belong.

(b) Two planes determine one and only one line of intersection, unless they are parallel.

From these statements we can easily derive the following corollaries: }

Corollary V: From (a) follows: (c) If two lines belong to the same plane, they determine one and only one point of intersection, unless they

From (b) follows: (d) If two lines intersect at one point, they determine one and only'

one plane to which they both belong.

are parallel. From (a) and (d) follows: (e) A line and a point does not lie on line determine one only one plane which they both long.

that that and to be-

From (a) and (d) or (e) follows: (g) Three points that do not belong to the same

From (b), (c), and (e) follows: (f) A line and a plane that does not contain that line determine one and only one point of intersection, unless

they are parallel. From (b) and (f) follows:

(h) Three planes that do not intersect in the

1 Corollaries are theorems which require a very short proof.

68

Unit 1 line (not collinear) determine one and only one plane to which all

same line (not coaxone ial) determine of point one and only

three belong.

intersection,!

unless

they are parallel to the same straight line.

Note: Instead of “one and only one’ we may also say “exactly one” and “not more and not less than one.” The incidence postulates and their corollaries have been worded and arranged in a special way to help you remember them. If you compare the statements of one column with the corresponding statement in the other column, you will find that the words plane

and point are interchanged, and so are the words belong and intersect. Line in one column corresponds to line in the other. This correspondence is called the principle of duality. It is not a perfect duality, because we have to add “unless parallel” in statements

about intersections, but not in those about belonging. Mathematicians have found a way to make the duality perfect. If you are curious to know how this is done, you may find out about it in a special project. (See Suggestions for Special Projects.) The following corollaries are useful. (For proof see Exercises 15 and 16, page 71.) Can you formulate dual statements?

Corollary V: (j) If a line and a plane have two points in common, the line lies on the plane. (k) Two parallel lines determine one and only one plane in which they both lie.

After the acceptance of Postulate V the initially blank terms have acquired most of the properties which we associated with them in the informal geometry of the Preparatory Chapter. For example, a broken line is a pseudo-line with respect to Postulates I through IV; but such a line may have more than one point of intersection

with

some

other

line.

1 Explain this term as derived from Definition 1-4.

Also,

there

are

several

A System of Postulates

69

broken and curved lines between two given points, but there is only one straight line. These pseudo-lines are now excluded. Postulate V gives us an abundance of points, lines, and planes. Within a plane (Figure Ila) we may connect any one of the infinitely many points of a line s with a given point P outside of s, and thus get infinitely many straight lines through P (a flat pencil of lines is the set of all lines through P within a plane) with infinitely many points on each. In space (Figure IIb) we may connect any one of the infinitely many points of a line p with a line g skew to p, and thus form infinitely many planes containing the same line q (an axial pencil of planes with the axis q is the set of all planes containing q). If an axial pencil of planes is cut by a plane that does not contain the axis, then the intersection is a flat pencil of lines. The axis is cut at one point and each plane is cut in one line through that point. In the special case when the cutting plane is parallel to the axis q, the intersection is a special flat pencil, consisting of parallel lines. q

:

tin aie (a)

(b) il

EXERCISES 1. Draw a pictorial view of a cube and label the vertices. Demonstrate the meaning of each of the parts of Postulate V and its corollaries by using the vertices, edges, and faces of the cube. 2. Repeat Exercise 1 using some of the other geometric objects studied in the Preparatory Chapter.

70

Unit1 . Repeat Exercise 1 using the walls and corners of the room. . In an xyz-coordinate system, which line is common to the xy-plane and yz-plane? Where do the xy-plane and the xzplane intersect? Do the xy-plane, yz-plane, and xy-plane have points in common? Demonstrate with the floor and two walls of the room as coordinate planes. . Can three planes have a line in common? (Demonstrate with a pencil as a line and your hand as a plane.) . How many; planes can pass through the same example, through the x-axis?

line?

For

. Three planes that do not have a line in common have either One or no point in common. Why? . Four points that are not coplanar (not all four in one plane) determine how many planes and how many segments? I]lustrate with a pictorial view of a triangular pyramid. . What do we mean by the principle of duality? . What is the difference in meaning between “exactly one” and “‘one and only one’’?

. What are skew lines? Choose a geometric object from the Preparatory Chapter and name several pairs of skew edges. all48 Use two pencils as a model of a pair of skew lines. If you look at the model from various sides, the skew lines will seem to intersect at different points. Is there also a direc-

tion in which the skew lines seem to be parallel? Try to make a plan and front view of a pair of skew lines. In the Figure below one pair of lines intersects and the other is skew;

which is which?

b'

dq"

a!

|

|

b

|

|

a

)

|

eae

|

|

Exae

cl!

z ee,

A System of Postulates

71

13. Does it follow from some postulate that every line lies in some plane?

14. Does it follow from the postulates and definitions that two planes cannot have only one point in common? 15; Cor. V(j): Ifa line and a plane have two points in common, the line lies in the plane. Why? (A line may intersect a plane, be parallel to it, or lie in it.

common points in each case?

What is the number of

Which case fits here?)

16. Cor. V(k): Two parallel lines determine one and only one plane in which they both lie. Why? (Use the definition of parallel to show that they lie in the same plane; then use V(b) to show that they cannot both lie in more than one plane.)

lye S, is a pseudo-space consisting of four points only. (Compare Exercises 11 and 12 in Section 2.) Does S, satisfy all of Post. V? What is the meaning of “‘parallel”? Does S, have pseudo-parallel pseudo-lines? (You may give S, a plane interpretation as the vertices of a quadrilateral, and also a space interpretation as the vertices of a pyramid, a tetrahedron.)

® 4 We We

2 EXeli(

we

Ex.18

*18. Repeat Exercise 17 for S,, the pseudo-space consisting of three points only. Do you have the same choice of interpretation as in Exercise 17?

Some Properties of the Triangle: An Application of the Postulates Although our system of postulates is not yet complete, a multitude of geometric properties can be deduced from the postulates adopted so far. For example, what do Postulates I to V tell about a triangle?

72

Unit1

eee

Definition I-5: Three non-collinear points 4, B, C together with the

three segments AB, BC, CA are called a triangle. The

three

points are

called the vertices,

and the

three segments the sides of the triangle.

1. From V(a) we know that any two points are collinear, but from I(c) follows that there is at least another point that is not on the line through these two points. Thus we know that there are triples of non-collinear points. These three vertices lie in exactly one plane according to V(g). Post. V(a) makes the segments between the vertices unique; and from V(j) we deduce that these three segments lie in the same plane. Thus we know: > A triangle is a plane figure.

2. The carrier of each segment separates the plane into two half-planes (IVb), one of which contains the third vertex and the

other two segments. Let us call this half-plane “the triangle halfplane” of the segment, and the other half-plane “the empty halfplane.” Then the interior of the triangle is the set of points belonging to all three “triangle half-planes.”. The exterior of the triangle is the set of points belonging to any of the three “empty half-planes.” Thus the plane of the triangle is the union of the triangle itself, its interior and its exterior. Since each halfplane

is connected

(IVb)

we

know

that their intersection,

interior of the triangle, is connected. The exterior is also connected, because the

rt SEN

empty half-planes overlap so

that there is some line (not necessarily a ae line) joining a point of one empty half-plane to a point of anOtner (@—>-D) —> fF).

the

Cc

: Nee

;

4

:

cas B Nee

A

> A triangle separates the plane into two connected sets, called the interior and the exterior of the triangle.

3. A connected set M is called convex, if it contains the straight line segment between any pair of its points. From Post. V(a) we

A System of Postulates

73

know that there is a straight line segment between any two points. But not all connected sets contain this straight line segment between any pair of their points. For example, if A and B belong to the interior of a triangle, the segment AB lies entirely in the interior; butif C and E belong to the exterior of the triangle, the segment CE may not lie entirely in the exterior. Thus a connected set may or may not be convex, but a convex set is connected. > The interior of a triangle is convex; is not convex.

the exterior of a triangle

> Lines, planes, half-planes, space, half-spaces are convex sets.

4. A theorem about the triangle:

Given a triangle ABC anda straight line s that does not pass through any of the three vertices; if s intersects one side of the triangle, then it also intersects a second, but not all three.

In other words, if a line enters the interior of a triangle, it must also

leave it: it cannot curl up inside forever. And when it leaves, it must leave through a second side, it cannot turn back through the first; nor can it go out through two sides, nor can it come back through the third. How can we be so sure about so many statements? To prove a theorem means to show that it follows from the accepted system of definitions and postulates. Not all consequences of adopting such a system are apparent right away. To trace these consequences until a particular theorem is reached 1s the purpose of the proof — and its fascination. (More about proofs ime Nite2. Sec. 125;)

* The proof of the above theorem may be divided into two parts: (a) show

that s must

intersect a second

side;

(b) show that s

does not intersect all three sides.

a. Suppose s intersects the side AB. This means (Def. I-4) that A and B lie on opposite sides of s. Since s separates the

74

Unit 1

plane (IVb) and does not pass through any vertex, the third vertex C lies on the same side as either A or B, say A. Then B and C lie on opposite sides and their connection BC must intersect the boundary s. | b. Show that s cannot intersect AC also.

From V(c) and Def. I-3

we know that if s is parallel to AC, then there certainly is no intersection;

and if s is not parallel to AC, then there is exactly

one point P of intersection. Since s does not pass through any vertex, P is either between A and C or outside of AC. If P were between A and C, then A and C would be on opposite sides of s. But it has been stated above that they are on the same side; therefore this cannot be our case and P cannot be between A and C.

Thus, s does not intersect AC.

EXERCISES 1. What is the definition of a triangle? How does it follow from the definition and the postulates that a triangle is a plane figure? 2. What do we plane?

mean

by saying that a triangle separates

its

3. What is a convex set of points?

4. Give examples of several point sets that are convex and of several that are not convex. 5. Which of the shaded areas below is convex? is connected?

Which of them

© 1 @ ha oo € (a)

(c)

(e)

6. What isSie interior and the exterior of a triangle? these terms by using the separation postulates.

(f) Define

7. Define the interior and the exterior of a cube by using the separation postulates. 8. Is the 9. If two in the 10. If two

interior or the exterior of a cube convex? triangles have a side in common, do they necessarily lie same plane? triangles have a side in common and if a line intersects

this common

side and another side of each triangle, do the

triangles necessarily lie in the same plane?

A System of Postulates

o

6

75

PARALLELS

Postulate Vi(a): The Parallel Axiom Through a given point outside a given line there is one and only one line parallel to the given line.

Postulate VI(a) may seem very obvious. We use it very frequently in everyday life. For instance, the edges of your book are made parallel, the paper you write on is cut neatly with parallel edges, your home is built with parallel walls, etc. Yet mathematicians felt that it was not of the same simple nature as the other axioms stated by Euclid.* While they did not doubt that it was true, they thought that it must certainly follow from the other axioms. Many people through the centuries have tried to prove VI(a) as a theorem and were unable to do so.

The great mathema-

tician Gauss was one of them, but he continued his reasoning to a revolutionary conclusion: if it is impossible to prove this theorem, if it is really independent of the other Euclidean postulates, then it should be possible to construct a geometric system that does not contain it!

Gauss

seems

to have been the first one (before the

year 1800) to realize the logical consequences of the existence of a geometry where more than one parallel could be drawn through a point to a given line; but other than in letters to friends he did not dare to write about it. The mere idea that Euclid’s axioms were not necessarily and absolutely true was so revolutionary for his time, that he was

afraid, as he said, of “the clamour of the

Boeotians” (of the boo’s of stupid people). Thus, several decades later, other names were connected with the first published discov-

eries of this so-called non-Euclidean geometry: Bolyai and, independently, Lobachevski. Water Riemann described another nonEuclidean geometry in which no parallel could be drawn through a point to a given line. The geometry of Gauss, Bolyai, and 1 Postulate VI(a) consists of two statements:

parallel.

(1) there is a parallel, (2) there is only one In Euclid’s deductive system, (1) is a theorem and (2) is the postulate (worded

differently) to which the following discussion refers. We have combined both statements into one postulate, because through the acceptance of this Postulate, both hyperbolic and elliptic geometrics are clearly excluded at the same time.

76

Unit 1

Lobachevski is called hyperbolic geometry, that of Riemann elliptic geometry. Today we are used to the thought that geometry is a deductive system based on essentially arbitrary assumptions, although these assumptions are chosen for a purpose: to build a useful model of some aspects of the physical world. We can construct various kinds of non-Euclidean geometries by replacing one of Euclid’s axioms by some other statement. At this point, you may wonder what a geometry without the Parallel Axiom would look like — and yet you know a model of such a geometry quite well, although under another name: the geometry on the surface of a globe. Think of a very big sphere, for example,

the earth.

(Look

at the meridians

and parallels in

Section 6 of the Preparatory Chapter.) If you travel on the surface of the earth from any point to the north pole in a “straight line,” you actually travel along a meridian which is part of a great circle. Any two points on a sphere can be connected through a part of a great circle (not necessarily a meridian). If the sphere is big enough (the earth), the connection between the two points may seem straight (a highway in the Middle West). Let us call it a pseudo-straight-line. We see that Post. V(a) applies: one and only one pseudo-straight-line passes through any two points (with the exception of opposite points such as the north and south pole). The Parallel Axiom VI(a), however, does not apply: through a point external to a given pseudo-line there is no parallel! Suppose the given pseudo-line is a meridian, and the external point is somewhere on the equator. Then a parallel line through that point would seem to be another meridian; yet all meridians intersect at the poles. Or suppose the given pseudo-line is the equator and the given point somewhere north of it. Then a parallel to the equator through that point would seem to be the small circle through that point, and the two lines would not intersect — but the small circle is not a pseudo-line (only great circles were supposed to be pseudo-lines). The geometrical fact that any two great circles on a sphere always intersect can be expressed in Euclidean and non-Euclidean language. In Euclidean language the sphere is a three-dimensional object. Two great circles on the sphere are not in the same plane, but in different planes through the center of the sphere. They constitute therefore a three-dimensional geometrical figure of

A System of Postulates

77

so-called spherical geometry, which will be studied in Unit 6. There are straight lines through the sphere, but not on its surface. The parallel axiom applies to these straight lines, but has no way of application to the surface, since there are no straight lines on the surface. To understand the non-Euclidean language better, think of a person living in a flat part of the country; unless he travels very far from home or is told by others or reads books he would have no reason to think of his land as part of a sphere. A straight road is for him a straight line, and the geometry connected with his roads is two-dimensional geometry. If any two of his straight lines were extended indefinitely, they would intersect twice (meridians

intersect at the north and south poles);

thus the

parallel axiom does not hold. In this course we study Euclidean geometry and use Euclidean language also for spherical geometry. Therefore, we accept the Parallel Axiom VI(a) and its logical consequences which are formulated in the following corollaries. These corollaries may seem obvious to you if you interpret them as statements about vertices,

edges

(unlimited),

and faces (unlimited)

of one of the

geometric objects studied in the Preparatory Chapter. Such an interpretation may help you to understand the meaning of the corollaries. A proof, however, establishes that they follow from the adopted postulates. We shall merely give an outline of these proofs. A note about sketches: To draw parallel or intersecting planes think of the pictorial view of a cube or a rectangular solid whose faces lie in the coordinate planes or are parallel to them. z z

horizontal planes

vertical planes

intersections (hidden lines are dotted)

78

Unit1

Corollary VI(b): If two parallel planes are intersected by a third, the two lines of intersection are parallel.

Interpretation:

If the top and base of a cube are intersected

by any plane &, the lines of intersection are parallel.

Proof: II, ||1, and both are intersected by >. II, and II, have no point in common (Def. of ||); therefore the line p, in II, can have no point in common with the line p, in IT,. Thus p, and p, are either parallel or skew. Since they both lie in the plane >, they are parallel and not skew.

Corollary VI(c): If a line s is parallel to a plane IJ, and if a plane > through s intersects II in a line p, then s is parallel to p.

Interpretation: The edge HG of the cube is parallel to the base. If a plane through HG (or through any other line parallel to the base) intersects the base, then the line of intersection is parallel to HG.

Proof: s and II have no point in common (Def. of ||). > contains both s and p, which’can have no point in com-

A System of Postulates

79

mon (because such a point would be common to s and []);

therefore p ||s by definition.

Corollary VI(d): If a line s which does not lie in a plane II is parallel to a line p of II, then it is also parallel to the plane IT itself.

Interpretation: The edge HG of the cube is parallel to the edge AB which lies in the plane of the base; then HG must be parallel to the base — and to any plane containing AB, but not HG.

Proof:

s and p lie in some plane > but have no point in

common (Def. of ||). p is the only line of intersection between > and II (Vb). We have to find out whether or not there is a common point P for s and I]. If there is, then

it must lie on s and on p; but that is impossible because

s ||p. Hence there is no point P, and s and II are parallel by definition.

Corollary Vi(e): If two lines are each parallel to a third, then they are parallel to each other.

Interpretation in the plane: the lines on a ruled sheet of paper; in space: the edges of a prism.

80

Unit1

Proof for the plane:

If a|| c and b|| c, then (by definition) a and c

have no point in common, and b and c have

no point in common. The lines a and b either do or do not have some point P in common. If they do, then there would be two parallels to c through P. But this is impossible because of Post. VI(a).

Thus

there is no common point P on a and b; and since they are in the same plane, they are parallel by definition.

* Proof for space: Suppose again a ||c and b ||c. There is a plane IT containing a and c (Def. of ||). Since b is not

in IT and since b ||c, we know from VI(d) that b ||TI.

A

plane > through b and some point A of a (V e) intersects I] in a line p (Vb). From VI(c) we know that p ||b; but we do not know yet whether p and a are the same line. To show that they are the same line, we shall first establish that p ||c by using VI(d) and then VI(c) again: since c ||b,

c must also be || { (VI d) and therefore c || p (VIc). Now we know that both p and a are|| c. But through a pointA there can be only one parallel to c (VI a).

Therefore p

and a must be the same line and thus a ||b.

Corollary VI(f): Through a point P outside a plane II there is one and only one plane parallel to IT.

Interpretation: Given the base ABCD of a prism and one lateral edge AE, then E determines exactly one top face parallel to the base.

A System of Postulates

* Proof:

81

We must show that (a) there is such a plane and

(b) there is not more than one such plane.

(a) Let p and q be intersecting lines lying in I].

Through

the external point P there is exactly one line p’ ||p and exactly one line q’ || g (VI a). The lines p’ and q’ are each ||[1 (VI d) and determine a plane II’ (Vd). The planes II and II’ are either parallel or they have a line s in common (V b). We must show that there can be no such line s. If there were a line s, it would intersect at least one of the

lines p’ and q’ (p’ and q’ could not both be ||s because of Vila). If s intersects, say, p’ in a point S, then S would be a common point of p’ and II. But there is no common

point S, since p’ ||II. no line s in common

Thus we know that IT and II’ have and, therefore, are parallel.

(b) If there were two planes II’ and II” through P parallel to II, they could not have both p’ and q’ in common (V d). Suppose they do not have p’ in common, then a plane >

through P and p would intersect II’ inp’ and II inp”. Both p' and p” would be ||p; but that is impossible (VI a). Thus there cannot be two planes through P parallel to II.

Corollary VI(g): If two planes are each parallel to a third, then they are parallel to each other.

Interpretation:

Proof:

book shelves

If T1,|| I, and I, ||Ils, can we tell whether or

82

Unit 1

not II, and II, have points in common? If there were a point P common to I], and then

II,,

we

,

o“=s

‘ ee)

iS

Pp -

would

have here an example

Ay

we

7

eas

of two planes passing through the same point and being parallel to the same plane

II,. But that is impossible VI(f).

according to Therefore II, and I], must be parallel.

EXERCISES 1. Draw a pictorial view cubes one on top of plain the meaning of ments VI(a) through

of a rectangular solid consisting of two the other. Label the vertices and exthe state(g).

2. Repeat Exercise 1 with the room as a model. 3. The adjoining Figure shows the plan and front view of a cube. How does the base ABCD appear in the front view? How

does the edge AD appear in the front view? What do the side faces ADHE and BCGF look like in the front view? List: a. pairs of parallel edges b. pairs of parallel faces

c. pairs consisting of one edge

A

B

and a face parallel to it

4. In the cube of Exercise 3, are the following pairs of segments parallel, skew, or intersecting?

a. AC and EG

b. AC and FH

c.

AG andBH

A System of Postulates

83

5. If two planes II, and II, intersect, is it possible to find a pair of parallel lines p, and p, such that p, is in 0 ,and p, is in II,? 6. If an axial pencil of planes is intersected by a plane parallel to the axis, then all intersections are parallel lines. Why? 7. If two lines p and q are parallel to each other and to a plane II, is the plane > containing p and qg necessarily parallel to I? What are some examples? *8. How can we be sure that VI(b) is true?

*9, How can we be sure that VI(e) for the plane is true? *10. How can we be sure that VI(g) is true? *11. Is the Parallel Axiom VI(a) satisfied in the S, (the pseudospace of four points only)? In the S, (the pseudo-space of

three points only)?

SUMMARY A.

OF PART A

Inthe Preparatory Chapter we discussed various geometric facts. From now on we shall organize these and many other geometric facts into a coherent system. More than 2000 years ago Euclid (about 300 B.c.) organized the geometry known at his time into a system by listing a small number of axioms and deriving from them, by logical reasoning, a collection of theorems. The collection of all theorems following from these axioms is called Euclidean geometry;

it includes

also theorems discovered after Euclid’s time. Inthe 19th century mathematicians began to examine Euclid’s axioms more closely and to reformulate them according to modern viewpoints. While for centuries Euclid’s axioms had been interpreted as the simplest of absolutely true statements, we now consider them as just one set of statements out of several possibilities. Geometries in which one or another of the Euclidean axioms is not valid are called non-Euclidean geometries. In this course we shall study Euclidean geometry. B.

A

logical system consists

of definitions, initial terms ac-

cepted without a definition, axioms, and theorems. A definition is an agreement on the use of a new word or

phrase. Such an agreement may be stated in two ways: either by equivalent, already known words, or by a whole set of statements about the usage of the new word (such as stating the rules of a game). Usually, the word definition is used for the first kind of agreement. Since equivalent words are inter-

84 Unit1 changeable, a definition by equivalents must be reversible. The chain of defining one word by another must begin with some initial words. Initial terms are words used without being defined by equivalents. They are defined by a set of rules about their usage. For our geometric system we accept as initial terms the words space, plane, line, point. Axioms,

postulates,

and basic statements

are equivalent

names for assumptions accepted without proof. The usage of the initial terms is fixed by these assumptions. Usually, postulates refer to geometry, while axioms have a general application. (See examples on page 40.) Theorems are statements derived from the accepted definitions and postulates by logical reasoning. The chain of reasoning is shortened by referring to already proved theorems. A theorem that follows directly from a postulate or from another theorem may be called a corollary of that postulate or theorem. C. Postulates and their corollaries Post. I(a): Space is the universal set consisting of infinitely many points. (b): Every plane is a proper subset of space; there are infinitely many planes. (c): Every straight line is a proper subset of a plane; there are infinitely many straight lines on each plane. (d): Every point lies on a straight line; there are infinitely many points on each straight line. Post. II(a): Of three points belonging to the same line, one and only one lies between the other two. (b): Between any two points of a line there is at least one more point on that line. (c): Every point on a line is between two other points of that line. (There is no first and no last point on a line.) Post. III(a): Every line is connected. (b): Every plane is connected. (c): Space is connected. Post. IV(a): A line is separated by any of its points into two connected half-lines. (b): A plane is separated by any of its lines into two con-

nected half-planes. (c): Space is separated by Say of its planes into two connected half-spaces.

A System of Postulates

85

Post. V(a): Two points determine one and only one line to which they both belong. (b): Two planes determine one and only one line of intersection, unless they are parallel. Cor. V(c): If two lines belong to the same plane, they determine one and only one point of intersection, unless they are parallel. (d): If two lines intersect at one point, they determine one and only one plane to which they both belong. (e): A line and a point that does not lie on that line determine one and only one plane to which they both belong. (f): A line and a plane that does not contain that line determine one and only one point of intersection, unless they are parallel. (g): Three points that do not belong to the same line determine one and only one plane to which all three belong. (h): Three planes that do not intersect in the same line deter-

mine one and only one point of intersection, unless they are parallel to the same straight line. (j): Ifa line and a plane have two points in common, the line lies on the plane. (k): Two parallel lines determine one and only one plane in which they both lie. Post. VI(a): Through a given point outside a given line there is one and only one line parallel to the given line. Cor. VI(b):

(c): (d): (e): (f): (g):

If two parallel planes are intersected by a third,

the two lines of intersection are parallel. Ifa line s is parallel to a plane II, and ifaplane > containing s intersects II in a line p, then s is parallel to p. Ifa line s external to a plane II is parallel to a line p of II, then it is also parallel to the plane II itself. If two lines are parallel to a third, then they are parallel to each other. Through a given point outside a given plane there is one and only one plane parallel to the given plane. If two planes are parallel to a third, then they are parallel to each other.

. Theorems about the triangle 1. A triangle is a plane figure. 2. A triangle separates the plane into two connected sets, called the interior and exterior. 3. The interior of a triangle is convex; the exterior of a triangle is not convex.

86

Unit1 4. Given a triangle and a line that does not pass through any of the three vertices; if the line intersects one side of the triangle, then it also intersects a second, but not all three.

E. Words and phrases Section 1: axiom, postulate, theorem, basic statement, definition, equivalent, interchangeable, reversible, if — then, deduce, deductive system, model Section 2: Euclidean geometry, initial terms, undefined terms, pseudo Section 3: segment, interior of a segment, closed and open interval, ray, initial point of a ray, carrier, opposite direction Section 4: connected, separated, Moebius Band, half-lines, half-planes, half-spaces, sides of a line, sides of a plane, positive and negative, right and left, up and down Section 5: incidence, a plane passes through a line, a line lies on a point, parallel, skew, coplanar, intersection, to intersect, corollary, collinear, coaxial, one and only one, exactly one, not more and not less than one, principle of duality, a flat pencil of lines, an axial pencil of planes, axis, triangle, interior and exterior of a triangle, convex Section 6: non-Euclidean geometries

PART B

7

MEASUREMENT

Up to now we have not used any measuring rules in our system. We discussed several theorems and problems that did not require measuring. There are some branches of geometry that are concerned with such types of problems only. A geometry that also investigates measuring problems is called a metric geometry. > Euclidean geometry is a metric geometry.

In this Section, therefore, we have to clarify what we mean by

measuring, so that we can incorporate appropriate definitions and postulates into our geometric system (next Section). When we say, for example, that the length of a room is 20 feet, then we actually mean that (1) a certain length, called one foot by agreement, has been used as a unit and has been compared with the length of the room;

and (2) that the room is 20 times as long

A System of Postulates

87

as the chosen unit or, in other words, that 20 units connected to-

gether are just as long as the room. The number 20 is called the measure Of the length of the room in terms of feet. Had we chosen an inch as a unit instead of a foot, the measure of the same length of the room in inches would be expressed by the number 240. (Review the example in the Preparatory Chapter, Sec. 1, where John and Margaret measure the same length in different measuring systems.) & A measurement of a length (an angle, an area, a volume) consists of a comparison with another length (angle, area, volume) which has been arbitrarily chosen as a unit. The measure is

expressed as a number in terms of the chosen unit. & The same length is expressed by a large number when compared with a small unit, and by a smali number when compared with a large unit. > A measure in terms of one unit may be transformed into an equivalent measure in terms of another unit by a transformation rule (for example, 12 in. = 1 foot).

7.1 The Measure of Segments The usual units of length are defined by some physical length. A foot originally was the length of the foot of any grown person. But since this length differs among people, there came a time when such a definition was not precise enough any more, and a standardized length in place of the natural length had to be agreed upon. Other units show the same change from a natural to a standardized definition; an inch, originally part of a finger, is now 7% of a foot and equals exactly 2.54 cm;! the meter? was the 10 millionth part of the distance on the earth from the equator to the north pole, and now is a length equal to the distance between two fine lines on a bar of platinum-iridium alloy kept at a constant temperature of 32°F at the International Bureau of Weights and Measures in Sévres near Paris; * for scientific purposes it is now defined in terms of wave lengths of krypton 86.* 1 By agreement of July 1, 1959 among the English-speaking countries. 2 The metric system is preferred in scientific work. Efforts are under way to change to this system also for general use, after a certain period of transition. Only about 10% of the world’s population, among them the United States, are not now using this system. 3 Since 1889, First General Conference for Weights and Measures, at Sévres. 4 Since 1960, Eleventh General Conference for Weights and Measures, at Sévres.

88 Unit1

The increasing requirements of precision in today’s technology make it necessary that basic definitions of units be reviewed, adjusted and agreed upon from time to time at scientific conferences. The next step after agreeing on some unit U is the construction of a scale with equal and numbered units. In the Preparatory Chapter, we marked equal distances along the coordinate axes by means of compass or paper strip. By superimposing these on the first segment,

and

that

we feel certain that they are given the same length,

they

impart

this

length to the next segment through another superposi-

tion.

, pee

Onn

——1

oo

B __1 _i _,

Smee

The third step is the actual

measurement:

If AB is to be

measured (Figure I), the scale

is superimposed on AB so that the zero point coincides with one endpoint; the measure of AB, that is the number of units contained in AB, is then read off at the other endpoint.

If AB is not

an exact multiple of U (suppose U = 1 ft), we may use a smaller unit = (e.g. + of a foot = 1 inch) that will lead to the measure p i or = U.

(For example, a lamp may be higher than 5 ft and

shorter than 6 ft; using an inch as unit we may find the lamp to be 5 ft 4 in. or 64 in.)

Since -is a rational number, we say that the

measure £expresses the ratio between AB and U. One of the great discoveries by Pythagoras (6th century B.C.) was the fact that there are segments which cannot be measured in terms of each other, however small you might want to make the fraction of one of them to serve as a unit. An example is the side s and the diagonal d of a square (Figure II): there is no fracpee tion a of the side s = | such that a multiple p (;)would equal the segment d. Another example is the edge of a cube and the diagonal of the cube. A third example is the radius of a circle and its circumference. Such pairs were called incommensurables, in contrast to pairs that can be measured in terms of each other, the commensurables. Jt turned out that there are not only these three

A System of Postulates

89

but innumerably many such pairs, and the Greek mathematicians were very puzzled by this discovery. The second endpoint of any segment starting at the zero point and commensurable with the unit s (Figure II) can easily be located on the number line by marking off multiples of some fraction of s. But what shall we do with these incommensurables? Which points will correspond to their second endpoints? — Eudoxus (4th century B.c.) found the answer:

Postulate Vil: Postulate of Eudoxus Given any two quantities a and b, then there is a multiple of a that is larger than b.

How does this postulate help to locate the endpoint of the diagonal of the square on the number line? Let us look at Figure Il first: there is a square with side 1 and another square \ with

side d, the diagonal of

the first square. It is easy to see that four halves of the Sialiesquaresiisexactly—into the big square, or two small squares have the same area as the big square — in other VOR Su wc

d= \/2.

20s) == 20

ed

,

< >

\ \ d \ ae

This fact is based

on the famous Theorem,

and

'

Pythagorean

which we will dis-

TT

90

Unitl

cuss and prove in the next Unit. The Pythagorean discovery that the side and the diagonal of a square are incommensurable is equivalent to saying that \/2 is not a rational number. By this phrase we mean that \/2 cannot be expressed as a ratio = where

both p and gq are integers. How can we be so sure? We could not possibly try out all integers! The proof by logical reasoning is really quite simple: Suppose -= \/2 and show that no integers will make this possible. 1. p and qg are not both even. (If they were, the factor 2 could be canceled out, till they are not both even.) 2. p is either (a) even, or (b) odd. (a) if p is even (p = 2p’), then q is odd.

If

-=V2

then

p?=2q?

4p” = 24° 2p'?=q"’,

| which

means

that q? is even;

but the square of an odd number g cannot be even. Thus g must be even. “. p cannot have been even.

(b) if p is odd, then also p? is odd; but p? = 2q? means that p? is even;

then p must

also be even, but should have been odd. 3. Thus p can neither be even nor odd; there is no ra-

tional number 2 equal to V2. q

Returning to our problem of locating d=/2 on the number line, we may use the relation d? = 2 as a definition of the number d by saying that d lies between those positive numbers whose square is larger than 2 and those whose square is smaller than 2. Post. VII states:

ifa = 1 andb = V2, thena sequence la, 2a, 3a, ...can be

formed which has a member that is larger than b. In our case la is smaller, but 2a is already larger than b, since 12 < 2 < 2? and 2 lies between 1 and 2. Now repeat for a smaller unit, e.g.

A System of Postulates

91

a a : ay ; a 7 or even better 10: 14 times ( 1 4 is smaller and 15 times a is larger than V2, since: ee

196

or

VAN

ed

=

5

Repeat with a still smaller unit a and you will find

141 fal The distance between two points A,B (the length of the segment AB) is an essentially positive number.

A System of Postulates

93

This means that the minus sign of a measure will be disregarded unless the direction of the measured segment is important in a particular problem.

EXERCISES 1. The meter (m) is subdivided into 10 decimeters (dm); each decimeter contains 10 centimeters (cm); each centimeter contains 10 millimeters (mm). A kilometer contains 1000 meters. A measuring system in which each smaller unit can be expressed as a decimal part of a larger unit is called a decimal system. The advantage of a decimal system lies in the ease of computation.

a. Add

4 yd.2ft.10in. SvGs ites eine 5 ft. 9in.

b. Change the three measures in a to the meter system. ec. Form the sum of b. d. Check the result of ¢ with a.

2. What is the transformation rule between miles and feet? between miles and kilometers? If a measure is changed from feet to miles, does it become larger or smaller? 3. If a length measures 144 units and the unit U is changed to another unit U’ =e Oe is the measure

unit larger or smaller?

in terms of the new

How much is it?

4. Does our geometry change if scientific conventions agree on a new definition of a unit? 5. Explain what we mean by saying that two segments which are 3 and 4 inches long, are commensurable. 6. Explain why the side of a square and half that side are called commensurable, while the side and the diagonal are called incommensurable.

7. Show how the infinite decimal 2.583945783 ... can be approached by two sequences of finite decimals, so that it is always larger than the numbers of one and smaller than the numbers of the other sequence. larger than: 2.5, 2.58, (continue) smaller than: 2.6, 2.59, (continue)

8. Repeat Exercise 7 for 0.2796350834 ... Show that at each step the difference between the approaching numbers becomes smaller and smaller.

94

Unit1 *9, By the method of approaching a number from the smaller and the larger side, find V3, V5, V'10 to three decimals. (V3 may be defined as that number which separates the positive numbers whose squares are smaller than 3 from those whose squares are larger than 3.)

*10. Try to prove that V3 cannot be a rational number. (Adapt the proof for V2 by replacing ‘even or odd” by “divisible by three or not.”’)

11. The absolute value of —3 is 3 (write |—3 |= 3); the absolute value of +3 is also 3 (write |+3 |=3). What are the absolute values of these numbers:

=; 20,

17, 995-99, Oe—-25; F205 77-04, — O45 08-200.

12. Given three segments a,b,c; on a given line construct a segment AB that measures a at+b+c b. atb—-—c

c. 3b+c d.c—a

e. 3a+2b—c f. c—2a+b

a fen ooo

on

ee

Exale

7.2 The Measure of Angles Two rays with a common initial point V determine an angle in much the same way that two points determine a segment. (Review the definition of a segment, of its endpoints, and of its interior, Unit 1, Section 3.)

Definition |-7:

—>-

>

Two rays VA and VB determine an angle. XV, XAVB, Xa.) —

(Symbols

—

The rays VA and VB are the sides of the angle: their common initial point V is the vertex of the angle. continued on next page

A System of Postulates

95

If V,A,B are non-collinear and if / is a point in the

interior of AB (I is between A and B), then the ray VI is said to be in the interior of xAVB

(VI is between

VA and VB).

The interior of x

AVB is the set of all rays VI be-

tween VA and VB. The x AVB is the union of its interior and its sides.

The exterior of x AVB is the set of all rays VE that do not belong to the interior of the angle and are not sides of the angle. If VA and VB are on the same line in opposite directions, x AVB is called a straight angle.

The discussion of the length of segments applies also to the size of angles. After a unit angle has been established by agreement, any angle may be compared with this unit angle and the measure (a number) expressed as the ratio between the two angles. Obviously,

this ratio refers to the interiors of the two angles and

not to their sides alone, just as the ratio of segments refers to the intervals and not to the endpoints alone. The usual unit angles are not physical angles but related to the straight angle.

It is customary (since the time of the Sumerians,

3000 B.c.) to assign the number 180° to the straight angle. Although it is customary it is nct necessary to use 180°: you could assign any other number to the straight angle. If you do, you must persuade the other people to accept your system, or else it is not of much

use.

At the time of the French Revolution,

when

96

Unit1

the metric system was established, the French tried to persuade everybody to use the decimal system also for angles, assigning 200¢ (grads) to a straight P ; e 4 100 angle instead of 180°, and 90° subdividing each grad into 100° (minutes) and each minute into 100" (seconds). If

you look at a French map, you may find latitudes and longitudes expressed in grads. Unfortunately, the French have

not

yet

persuading

the

succeeded

rest

eant

400°

igo

0°360°

in

270°

of the

300°

world to use their system. Unfortunately, because it is so much simpler to use a decimal system in computations than the Sumerian sexagesimal system

of degrees with its subdivisions into 60 minutes and 60 x 60 seconds. For this reason the decimal system in other measures (meters and kilometers instead of yards and miles, grams and kilograms instead of ounces and pounds) is being used more and more in scientific work. There is a third, intrinsically geometric system, originating with the Greeks and based on their discovery that the ratio between

the circumference

of a circle and its radius is the same,

however big or small the circle may be. Since these two lengths are incommensurable, the Greek discovery is even more amazing. The measuring system based on this ratio is called the radian measure; it will be discussed in Unit 4. Since the measures of angles are numbers, the rules of arithmetic apply as they do for the measures of segments. Equal angles are angles with equal measures. The sum or difference of two angles is an angle whose measure is the sum or difference of the measures of the given angles. A

Example:

Given x.a = 60°,

xB = 20°; and

x6 = XaXB

B' V

A

5

A System of Postulates

(a) Arithmetical method:

Solution:

97

xy = 60° + 20° = 80°

x 6 = 60° — 20° = 40° use a protractor to draw xy and x 6

Y (b) Geometrical

5 method

(independent

of the particular

number of degrees):

. draw a ray with initial point V . transfer xXa@ by compasses: draw part of a circle with the same radius around V and V; then draw part of a circle with radius AA’ around A to find A’; connect V and A’ 3. transfer 4B So that it lies on the other side of the com-

Re NO

mon ray V A’ ia 4. Solution: xy =x AV B'

Vv

A

5. Repeat 1. and 2. and then transfer x 8 so that it lies on es

the same side of the common ray V A’.

6. Solution:

x5= XAV B’

The discussion of angles and their measure refers to the definition (I-7) of angles formed by two rays with a common initial point. Two such rays always lie in a common plane (Post. V) and

98

Unit1

thus the angle is a plane figure. But we also speak of angles determined by two rays that do not have a common initial point. Such rays may be coplanar or skew. Their measure is defined as the measure of a plane angle by the following definition:

Definition I-8: The measure of an angle determined by any pair of rays is defined as the measure of a plane angle with an arbitrary vertex and sides which are parallel to the given rays or lie on their carriers.

EXERCISES 1. What is a decimal system? 2. What is the difference between the measuring system of degrees and of grads? 3. Which is the bigger unit angle, a degree ora grad? Ifthe size of an angle is expressed in both systems, in which system will the measure be the larger number? 4. Which transformation rule will change a measure from degrees to grads?

5.-Add,

“Pi22'55" 23" 67°13) 34"

Multiply

Subtract

“120° 12°25" 34° 20’ 36”

Subtract

1206 12‘ 25“ 346 20‘ 36"

44° 24’ 51” s< 3

62 Adda

t12655.23% 67 13° 34"

Multiply

44¢ 24° 51” containing a point A; what is the set of all points Q in ¥, for which AQ = r? 13. Can you combine Exercises 9 and 12 and formulate a theorem using the word “intersect”? 14. A sphere and a straight line through its center have exactly two points in common. Why?

15. A circle and a straight line in its plane passing through its center have exactly two points in common. Why?

8.2 Congruent Angles Angles that can be made to coincide when superimposed (on transparent paper) are called congruent angles. As previously discussed for congruent segments we must avoid the concept of motion contained in this intuitive idea of congruence and find an

104

Unit 1

appropriate reformulation. The usual method of transferring an geometrical method in the Example the (see compasses by angle of Section 7.2) gives rise to the following definition: B

Definition I-10: 14, __If the segments VA and VB on the sides of an angle a are congruent respectively

V

to V'A' and V’B’ on thesides

B'

of an angle a’ and if AB is congruent to A’B’, then xa and xa’ are said to be congruent angles.

y!

nN

This definition makes

use of the previous definition of con-

gruent segments; but it does not explicitly state a connection to the number of units contained in the angles. Therefore we adopt

Postulate VIII(c): Congruent angles are equal and equal angles are

congruent.

Congruent and equal are now equivalent expressions if referred to two segments or to two angles. It follows: » The congruence of angles is reflexive, symmetric, and transitive.

Congruent angles are incorporated into our geometric system by the following postulate:

Postulate VIII(d): Given

x

AVB and a ray V'4’, then in any particular p

=

plane through that ray there is exactly one ray V’ B' on a given side of V'A’ so that x A'V'B’ is congruent to XAVB.

A System of Postulates

105

The wording of Postulate VIII(d) is more complicated than the wording of the analogous Postulate VIII(a), because an angle is a plane figure and a segment is a line figure. Angle AVB always lies in a,plane and so does angle A’V’B’. The carrier of the given ,

ee

ray VA’ separates each of the infinitely many planes through V’4’ into two

half-planes

(Post.

IVb);

Postulate

VII(d)

holds

for

an angle AVB in whatever plane it may lie and angle A’V'B’ in any of these infinitely many half-planes.

(a)

(b)

In Figure Ia the angles a and B lie on opposite sides of a common ray. Such pairs of angles are called adjacent angles. If the other pair of rays VA and VC happen to form a straight angle (Figure Ib), then y and 6 are called adjacent supplementary

angles. Now suppose that in Figures Ib and Ic x y = x y’ (VIIId) and

ate 5" is drawn such that x § = x 6’ (VUlld: there is exactly one ray VC" ..); then x A'V’C’' (Figure Ic) may or may not happen to be a straight angle as is xAVC (Figure Ib). VII(c) that the measure

We know from

y’ + 8’ is equal to the measure

y + 6;

but we do not know whether straight angles always have the same measure. Neither the measure nor the interior of a straight angle has even been defined. Figures Ib and Ic suggest the following agreements:

Definition I-11: If x AVB and x BVC are adjacent supplementary angles, then the union of their interiors and their common side is considered as the interior of the straight angle AVC.

106

Unit 1

Postulate !X(a): All straight angles are equal.

As an example of what it might be like if Post. [X(a) did not apply, think of a highway climbing a mountain straight up to the top and then leading straight down the other side. (The wheels of the car do not have to turn because it is a perfectly straight S

YY

glued

(a)

together

(b)

road.) A cone (Figure Ila) may serve as a model. You can form a cone from a circular piece of paper with a sector cut out (Figure IIb) by gluing the sides of the cut together. The line SB is opposite the line SA (glued); both together represent the straight road ASB. The straight angles along the uphill and downhill road behave like ordinary straight angles, but the straight angle with the vertex S is smaller (Figure IIb).

You will say:

“But that is a

cone, not an ordinary plane!” Precisely! A thing like this does not happen in an ordinary plane, and we do not want it to happen in our geometric system. We want all parts of our plane to be alike. ‘That is why we adopt Post. [X(a). It prevents our concept of “plane” turning out to be something like a conical rather than a plane surface; it excludes special points (singular points) behaving differently from others, such as the vertex of a cone in a conical surface. Several very useful relations between angles follow from our definitions and postulates: In Figure III there are several pairs of adjacent supplementary angles. Which are they? The sum of all four angles a, B, y, 6

A System of Postulates

is called

a round

a perigon.

107

angle or

From Post. [X(a)

follows that all round angles are equal. (In a conical surface the round angle at the vertex is smaller than at other points!) The angles a@ and y in Figure III have the vertex in common,

y

B 5

but the defining

rays of one are in the opposite direction of those of the other. Such angles are called vertical angles. Also x8 and x6 are vertical angles. From Post [X(a) follows directly that: > Vertical angles are equal.

(a +B =y + B because of Post. [X(a). from both sides leads to

pee

(a)

pei

Subtracting the measure 8

a= y.)

TS

(c)

(b)

rage

B*

(d)

IV

From Definition I-8 it follows that an angle between rays that have no common initial point (Figure [Va) is measured by an angle, with any vertex, whose sides are parallel to the given rays or lie on their carriers. Thus x V, and x % (Figure IVb) measure the same angle and therefore must be equal. The angles V, and

V. are called parallel angles, because the sides of one are parallel to the sides of the other. Not all pairs of parallel angles are equal, however; it depends on whether their rays point into the same or opposite directions. Figure [Vc shows the four possibilities of angles parallel to XV, Figure [Vd shows their numerical relations as adjacent supplementary angles or as vertical angles. For simplicity, angles which have one or both pairs of sides on the same carrier are called parallel angles also.

108

Unit 1

> Parallel angles are either equal sides are in the same direction tions) or supplementary (if one in the same direction and the

(if both pairs of corresponding or both are in opposite direcpair of corresponding sides are other in opposite directions).

Two angles whose sum equals a straight angle are called supplementary angles; they need not be adjacent. The relation between parallel angles leads to a very important theorem:

Theorem |-1: The sum of the interior angles of a triangle equals a straight angle.

An interior angle of a triangle is an angle whose sides carry two sides of the triangle and whose vertex is a vertex of the triangle. It is called an interior angle, because its interior contains the interior of the triangle.

(Review Part A,

Section 5.2.) A triangle has three interior angles. Usually we mean an interior angle when we speak of an angle of a triangle. An exterior angle angle 1s the adjacent mentary angle of an angle. In Figure V is an interior angle

of a trisuppleinterior angle a and the

angles a’ and a” are each exterior angles. Since a’ =a” (vertical angles) we usually consider only one of them. A triangle has three pairs of exterior angles.

Vi

A System of Postulates

109

It is not readily obvious that the three angles of a triangle should have anything to do with a straight angle. We can check the plausibility of Theorem I-1 by cutting out a paper triangle, tearing two corners off and placing them next to the third corner (Figure VI). The three corners together seem to fill the interior of a straight angle. For

a

proof,

however,

we

must show that the theorem follows from postulates and definitions. The plan of the proof is based on the experience with the paper model. PROOF OF THEOREM I-1 STATEMENTS REASONS 1. Consider aline p|| AB throughC 1. There is such a line because of 2.a'+y+f' = straight angle

3. a=a’

and B=f'

4,.a+Bt+y=a't+p'+y 5. a+B+y=straight angle

the Parallel Axiom 2. (a' + y) is adjacent supplementary to 8’ 3. parallel Xs with both pairs of sides in opp. dir.

4. substitute from 3. 5. from 4. and 2.

q.e.d. (quod erat demonstrandum)

The Latin phrase means: this is what was to be proved, and signifies the successful completion of the proof. The two-column form of writing the reasons for each statement and numbering the steps in the chain of your reasoning helps to keep your thinking straight. To save time and space, telegram style and symbols are used rather than full sentences. Note that Theorem I-1 depends on the Parallel Axiom, and thus is characteristic of Euclidean geometry. In a geometry where this axiom does not apply, Theorem I-1 is not true. On the globe, for example, we may consider an interval on the equator as one side of a (spherical) triangle, the meridians from the ends of that interval up to the north pole as the other two sides, each forming a

110

Unit1

right angle with the equator, and an additional angle with each other at the pole. This spherical triangle has two, maybe even three right angles. In the numerical measure of degrees a straight angle measures 180° and a right angle 90°. Thus the sum of the interior angles of a triangle on the globe is greater than 180°, while in the plane it is always exactly 180°. This is possible only because the meridians, seemingly parallel at the equator, are not really parallel since they intersect at the poles. In non-Euclidean language the meridians may be interpreted as “pseudo-straight-lines.” Euclid’s Parallel Axiom does not apply since there are no parallels. In this non-Euclidean elliptic geometry Theorem I-1 is not true; here the sum of the interior angles of a triangle is greater than a straight angle! Corollary |-1-1: If two angles of one triangle are equal respectively to two angles of another triangle, then the third angles are equal. Corollary |-1-2: An exterior angle of a triangle is equal to the sum of the two remote (not adjacent) interior angles of the triangle.

In Figure VII two straight lines a and b are intersected by a third

¢ (a transversal)

forming eight angles: As

lle 7, 8sare called er terior angles

xs 3, 4, 5, 6 are called in-

terior angles The

pair

alternate

(1,7)

are

exterior

called

angles;

also (2, 8).

The

pair

(3,5)

alternate also (4, 6).

are

interior

called

angles;

Vil

A System of Postulates

The pair (1, 5) are called corresponding angles;

111

also (4, 8), (2, 6),

(1): In case the two lines are parallel the eight angles are parallel

angles.

Show that if a ||b, then

corresponding angles are equal,

alternate interior angles are alternate exterior angles are interior angles on the same exterior angles on the same

equal, equal, side of ¢ are supplementary, side of t are supplementary.

Theorem I-2: If two straight lines a and b are intersected by a transversal ¢ and if a pair of alternate interior angles are equal, then a and D are parallel.

The first part of the theorem, the “if” part, describes the situation we are dealing with: there are two straight lines cut by a transversal and a pair of equal interior angles. This part is called the hypothesisof the theorem. The second part, after the word “then,” contains the conclusion to be proved. A theorem is not always worded in the “jf — then’ form, but can be so reworded in order

to recognize more clearly what we already know (hypothesis) and what we

should prove (conclusion).

worded:

If a, 8, y are interior angles of a triangle, then their sum

equals a straight angle. A sketch accompanying a proof shortens the wording through the use of symbols and helps guide your thinking. Mark in the sketch what you already know from the hypothesis, and label the parts you are using in the proof. Hyp.: Con.:

a=8 a||b

Theorem

I-1 could be re-

112)

Unit1 PROOF

OF THEOREM

I-2

STATEMENTS

1. there

REASONS

is a line p||a through

P=b ft 2. a= 3. a=B 4. b andp are the same line

1. ||axiom 2. ||Xs, both pairs opp. dir. 3. hyp. 4. VIII(d) (only one ray on one side of ¢ such that the x with ¢ is = a)

5. a||b, q.e.d.

5. from 4. and 1.

Theorem I-2 sounds similar to one of the unnumbered theorems above about parallel angles: Jf (hyp.) a|| b, then (con.) alternate interior angles are equal. But here hypothesis and conclusion are exchanged in comparison with I-2, and therefore these two theorems are different theorems. Such theorems are called converses of each other. The importance of Theorem I-2 lies in the method it provides for determining whether or not two straight lines are parallel by finding out something about angles. It would be impossible to find it out by the definition of “parallel,” since you could not possibly follow infinitely long lines to see whether eventually they would or would not intersect. In case we do know from some other source that two straight lines are parallel, then the converse of I-2 provides a method to find out whether two angles are equal or supplementary without measuring them. These applications of Theorem I-2 and its converse are examples of the very important principle of indirect measurement: if you want to measure or find out about something, and it cannot be done directly, then you might consider measuring or investigat-

ing something connected with it, and using the theoretical connection to solve the first problem. Most of science research problems are treated that way. Think of examples: the distance to the moon is not measured directly by a tape measure extending from the earth to the moon,

angles;

but by observing certain astronomic

mileage driven by a car is measured by the number of

revolutions of the wheel; echo soundings measure a distance by the time the echo takes to return; the air space in a room is com-

puted from measuring the length, width, and height of the room and then applying geometry.

A System of Postulates

113

EXERCISES Group I

1; What is the meaning of reflexive, symmetric, and transitive? Give examples. . The

congruence

of angles

is transitive.

What

does that

mean?

. The equality between angles is a transitive relation.

What

does that mean? What is the difference between this statement and that in Exercise 2? Can one statement be deduced from the other?

. What is the difference between adjacent angles, supplementary angles, and adjacent supplementary angles? . If an angle is 30° (45°, 80°, 100°, 130°, 150°, 165°), how big is its adjacent supplementary angle?

. Is the relation “adjacent supplementary angle of’ symmetric? reflexive?

transitive?

. What are vertical angles? . In the adjoining Figure name pairs of vertical angles. Name pairs of adjacent supplementary angles. . If x1 in the Figure of Exercise 8 measures’ 100° (120°, 155°), can you tell how big the other eleven angles are?

3

4 5/6

9 7 \8

10. If an angle measures a°, how many degrees are in its adjacent supplementary angle? it Show that the vertical angles of a pair of supplementary angles are supplementary. 12. Show that supplementary angles of a pair of vertical angles are equal.

13. If the difference between a pair of adjacent supplementary angles is 20°, how many degrees are there in each angle? (How many unknowns? How many equations are needed? See Supplement III.) 14, If one angle is 32°30’ (64° 20’, 100°) larger than its adjacent supplementary angle, how big are the angles?

13: If an angle is twice (3 times) as big as its adjacent supplementary angle, how big are the angles?

114

Unit 1

16. If an angle is half as big as its adjacent supplementary angle, how big are the angles? 17. Of the angles formed by three lines through one point, the sum of any three non-adjacent angles is always equal to a straight angle.

Why? Eaxiplig

Group II 1. If two interior angles of a triangle are 30° and 60°, how big is the third angle? are the exterior angles of the triangle?

How big

2. Compute the third angle in a triangle, if two angles are given:

a = 30° B= 110° y= ?

Joy DT 22F

? Se) 86 35,

16°45 | 642516 aes)a5 OL ? ? 93°25"

3. Show how Corollary I-1-1 follows from Theorem I-1. 4. If two angles in a triangle are equal and the third is twice as big as one of these, how big are the angles? 5. If all angles of a triangle are of equal size, how big are they? 6. If the angles of a triangle are a, they in degrees?

B= 2a, y = 3a, how big are

7. The sum of the interior angles in any quadrilateral is equal to two straight angles. Why? (An auxiliary line will help.) 8. The sum of two angles in a triangle is 100° and their difference is 40°. How big are the three angles of the triangle? (How many unknowns? How many equations are needed?) 9. Show how Corollary I-1-2 follows from Theorem

I-1.

10. If an exterior angle of a triangle is 130° and one of the remote interior angles is 52°, how big is the other remote interior angle? 11. An interior angle of a triangle is a= 56° 24’, an exterior angle B' = 130° 15’. Compute the other interior and exterior angles.

12. If two angles of a triangle are each 40°, how big is the exterior angle adjacent to the third angle? Group III

1. If two parallels a, b are cut by a transversal t and if x 1 = 50°, how big are the other seven angles formed?

A System of Postulates 2. Complete the following statements with either “‘equal’’ or “supplementary”: If two parallel lines are cut by a transversal, then alternate interior angles are... interior angles on the same side of the transversal are... exterior angles on the same side of the transversal are... alternate exterior angles are...

115

Bolland

3. Given a straight line s and an external point P, construct a parallel to s through P, using Theorem [-2. 4. In a parallelogram two consecutive angles are...

Why?

5. The diagonal of a parallelogram forms equal angles with opposite sides. Why? 6. Use Theorem I-2 to prove its corollaries: b are cut by a transversal ¢ and

If two lines a and

a. if alternate exterior angles are. .., then a ||b; b. if corresponding angles are ..., then a|| b; c. if interior angles on the same side of ¢tare... ., then a ||b; d. if exterior angles on the same side of tare .. ., then a ||b. 7. What is the principle of indirect measurement? Give examples.

9 PERPENDICULAR 9.1 The Right Angle in the Plane

Definition I-12: If an angle is congruent to its adjacent supou lL 2 plementary angle, then it is called a right angle V (symbol 4 or rt. x). Two lines intersecting at right angles are said to be perpendicular to each other (symbol _). A right triangle is a triangle with a right angle. Two angles are complementary, if their sum equals a right angle.

116

Unit1

From Postulate [X(a) follows directly as a corollary:

Corollary IX(b): All right angles are equal.

Using the definition of a right angle as a special case of an angle, some of our statements about angles take on a useful special form: Theorem I-1 has these corollaries:

Corollary |-1-3: Of the interior angles of a triangle at most one can be a right angle.

Corollary |-1-4: In a right triangle two angles are complementary.

Theorem I-2 takes this special form:

Corollary !-2-1: If two straight lines are perpendicular to a third, they are parallel to each other.

Theorem

I-3:

Within a plane there is one and only one perpendicular to a given line at a given point of that line.

A System of Postulates

117

Whenever a theorem contains the phrase “one and only one,” it

really contains two statements, each of which must be proved: (1) there

is a |:

With

the given

point A as vertex and the given line a as one side construct an angle a whose measure is half the measure of a straight angle. By definition, a is a right angle and its second side p is 1 aby definition of 1. (Q)ethere

isj-only

one,

wi:

A

a

There

could not be more than one, because

if there were two 1s through 4, namely p, and p,, then these would have to be parallel to each other according to Cor. I-2-1; but parallels cannot intersect, and hence p, and p, must be the same line.

Theorem I-4: There is one and only one line perpendicular to a given line through a given external point.

(1) there is a 1:

Consider a line

b||a through P (|| axiom) and a transversal t | b at P (I-3).

This

transversal cuts a at right angles

(corresp. ||Xs are equal) and is therefore | to a. (2) there is only one 1: Same proof as for J-3; there cannot be

||s that intersect. ‘he statement. there 1s— , through PF there iS eXactly one p, | RE (I-3). Within >, through PF’ there is exactly one Ps lePE a3): Since PF is | to both p, and P2, itis 1 to the plane IJ determined by p, and p, (I-6). (2) There

is only one

The set of all lines perpendicular to a given line p at a point F of p is the plane perpendicular to p‘at F.

A System of Postulates

123

Note that in this proof the three-dimensional problem was reduced to a two-dimensional one within }, where it was solved by an already known theorem. This method of simplifying a threedimensional problem by finding a plane containing the essential parts of the problem will be very helpful in many cases. We shall refer to it as the & principle of reducing a three-dimensional problem to a twodimensional one.

Theorem I-8: There is one and only one plane perpendicular to a given line through a given point external to that line.

Why?

You must prove two statements!

Theorem I-9: There is exactly one line p perpendicular to a given plane IT through any given point (on or external to II).

Why?

You must prove four statements!

Theorem I-10: If two lines are perpendicular to the same they are parallel to each other.

Q

Hyp.: p, 1 Il and p, 1 I

Con.: p, ||p, Plan of Proof: Two lines either (1) intersect or (2) are ||or (3) are skew. If we can eliminate (1) and (3) by showing that they lead to contradictions, then we have proved (2).

Eliminate (1):

ae

/

plane,

124

Unit 1

STATEMENTS . suppose p, and p, intersect at O 2. through Q there are now two isto Il 3. contradiction in 2. 4. .. there is no Q

REASONS . tentative assumption for (1)

2. from 1. and hyp.

Eliminate (3) by constructing through a point P of p, a line p 1 Il and coplanar with p,:

1. p, and P determine a plane > 2. % intersects II in f through F, 3. As at F, between p, andfand

— Corvie) 2. Post. V(b) 3. hyp. p, 1 II

between p, and some line g in II are rt. Xs

4. within > through P a) . p intersects 6 . through F, i! . & between 8.

3). 0. 1.

exactly

one

p||p,

4. Post. VI(a)

f at F, under rt. Xs line g ||g in IT p and g is rt. x

. corresp. XS with a transversal . Post. VI(a) eeeto: p LI . from S.,7., and I-6 through P now two 1s to II WO eaornan . from 8. and hyp. p, 1 contradiction in 9. 10. I-9 .. p, is the same as p and cannot be skew to p,. P

Definition I-15: The projection of a point P on a plane II is the foot of the perpendicular to IJ through P. The projection of a line s on a plane II is the set of the projections on II of all points on s.

A System of Postulates

125

Theorem I-11: The projection of a straight line s on a plane I] is a straight line s’, unless s 1 II.

Plan of proof:

Show that all 1s lie in one plane

BectSaliemm auline s:

which inter-

If s 1s

not | II, then the | s through

different points of s do not coincide. p (1 II through P of s) and s determine a plane. > (Poste Ved)segieloll

through any other point Q of s) is ||p (I-10) and thus coplanar with p (Definition

of ||). This plane >’ must be the same as & (one point Q and one line p in common). “.gin>. > and J] intersect in one line s’ (Post. V b).

Definition I-16: The angle a between a line s and a plane II is defined as the angle between s and its projection s’ on II.

Note that the angle between a line and a plane is defined as a plane angle. To find it, the plane containing it will be found first; this plane is determined by s and a perpendicular to [J through any one point of s. EXERCISES Group I

1. What is the definition ofa line perpendicular to a plane? How can we find out whether a line is perpendicular to a plane? 2. If you hold your two draftsman’s triangles so that one short side of each rests on the table while the two other short sides coincide, then this common line is perpendicular to the table, unless the triangles lie in a common plane. Why?

126

Uniti

3. To find out whether a given prism is an oblique or a right prism, we can measure two angles; which ones? 4. In the pyramid ABCV there is a right angle at B within each of the two adjacent lateral faces. Can we conclude that VB must be the altitude?

Vv

Cc A

Y = B eyed

: 5. The plane II perpendicular to a given line PF at F contains all perpendiculars to PF at F. How can we be so sure of this fact? 6. What is the principle of reducing a three-dimensional problem to a two-dimensional one? Give an example. 7. What is the projection of a point (a line, a segment) on a plane? What happens if that point (line, segment) lies in the

plane? 8. What is the projection of a line s on a plane II, if s 1 II? 9. What is the projection of an object on a plane II? view (front view) a projection?

Is a plan

10. If a cube (a square pyramid) stands on a plane II, what is its projection on II? 11. Demonstrate with edges, diagonals and faces of a cube, that the projection of a line may be a point, the projection of a plane may be a line, the projection of a segment may be equal to or shorter than the segment itself. 12. In a regular square pyramid what are the projections of the lateral edges on the base? Which is the angle between one lateral edge and the base? (Pictorial view.)

13. Ina pictorial view of a cube show which is the angle between a face diagonal and the base; between the diagonal from one corner to the opposite corner and the base.

Group II

1. If three lines through one point are mutually perpendicular (each one of the lines is perpendicular to the other two), then

a plane containing two of them is perpendicular to the third. Why? Give an example in the room. 2. Given a line p and a point F on it, if three or more lines are perpendicular to p at F, they must be coplanar. Why?

A System of Postulates

127

3. Why can’t there be two perpendiculars to a given plane at a given point? (Use the principle of reducing to a twodimensional problem.) 4. Under what condition can a point be the projection of a line? a line the projection of a plane? 5. Is the projection of a segment always, sometimes, or never shorter than the segment? 6. What are the four statements to be proved in Theorem I-9? I . Try to prove Theorem I-9, using Theorems I-7 and I-8. 8. Draw a pictorial view of a rectangular solid and use edges, diagonals, and faces to demonstrate the meaning of the following four theorems: a. If a line is perpendicular to one of two parallel planes, it is perpendicular also to the other plane. b. If a line is perpendicular to two planes, then these planes must be parallel to each other. c. If a plane is perpendicular to one of two parallel lines, it is perpendicular also to the other line. d. If a plane is perpendicular to two lines, then these lines must be parallel to each other. *Group III 1. Try to prove theorem 8a of Group II. (Use the principle of reduction to two dimensions by passing an arbitrary plane = through the given line and intersecting it with the given planes.) 2. Try to prove theorem 8b of Group II. 3. Try to prove theorem 8c of Group II. for ||4s.)

(Apply I-6 and look

9.3 The Angle Between Two Planes What do we mean by an angle between two planes, a so-called

dihedral angle?

(In Greek di, dyo means two, hedra means side,

plane.) Take a sheet of paper and fold it to represent two half-planes II, and I], with the straight line intersection k. Then look at your paper model from the side so that both half-planes appear as straight lines (rays) II,”, I],” and the intersection k as point k". It is obvious from this view what we mean by the angle between the two half-planes; but how shall we define it? If we cut II],

128

Unit1l

(a)

(b)

(c)

and II, by a third plane II, which is perpendicular to k, then the intersections p, and p. are perpendicular to k (why?) and form the same angle as II, and II, when seen as straight lines I,” and II,”. We may use this fact to formulate a definition:

Definition 1-17:

The angle between two half-planes II,

and II,: If a plane JI, is perpendicular to the intersection k of I], and I],, then the angle formed by the intersections

p, of I, with II, and p, of II, with II, is considered to be the angle between II, and II,,. Or: At any point P on k consider in each plane II, and II, the respective perpendiculars p, and p, to k; then the angle between p, and p, is considered to

be the angle between II, and II,.

Note that the dihedral angle II,I], is defined as a plane angle (angle within a plane). Sometimes a distinction is made between the dihedral angle as the three-dimensional wedge formed by the two half-planes, and the two-dimensional “measure” defined above. In the particular case that p, and p, form a right angle, the planes II, and II, are considered to be perpendicular to each other. In that case the line p, is perpendicular to two lines of II, (p,. and k); and the line p, is perpendicular to two lines of IT, (p, and k). By Theorem I-6 the line p, 1 II, and p, ‘L I],. In other words:

A System of Postulates

129

Theorem I-12: If two planes are perpendicular, then each contains

a line perpendicular to the other.

Theorem I-13: If a plane contains a perpendicular to another plane, then the two planes are perpendicular.

Hyp.: p 1 Il at F; > contains p Con.: > 1 II PROOF STATEMENTS 1. > intersects IT in k DeeAt TAN 2: Derk Sy ater in Ii: gil Kk 4.plq

Seek

Theorems

Ii gerd.

[I-12 and I-13

1. 2. 3. 4. 5.

REASONS Post.? hyp. and def. of p 1 II I[-3 p 1 II by hyp. from 4. by definition

are converses

of each other.

(See

Sec. 8.2.) They may be combined into one statement using the phrase “if and only if’: Two planes are perpendicular, if and only if one contains a perpendicular to the other.

130

Unit 1

Corollary |-12-1: If each of two planes is perpendicular to a third plane, then their line of intersection (if there is one) is also perpendicular to that third plane.

Plan of proof: If there is a point F common to the three planes, then at F there is a perpendicular to I] within }, and within >, (I-12).

Since there can be only one perpendicular to I] at F (I-9),

it must be the intersection between >, and Sp. Applying I-13 twice also leads to a useful corollary:

Corollary |-13-1: If two planes intersect in a line perpendicular to a third plane, then each of them is perpendicular to that third plane.

EXERCISES Group I

1. State two ways (a definition and a theorem) to determine whether two planes are perpendicular or not. . Demonstrate a model.

the two ways in Exercise

1 with the room as

. If a line p is perpendicular to a plane II, is every plane > containing p perpendicular to II? Demonstrate by using a door as a model, opening it to various positions. . If three lines through a point are mutually perpendicular, then also the planes determined by any pair of the lines are mutually perpendicular. Demonstrate in the room. . If three planes are mutually perpendicular,

then the three

A System of Postulates lines of intersection are also mutually perpendicular. Give an example in this room.

131

Why?

6. Make a pictorial view of a regular square pyramid and show what we understand by the angle between any two adjacent faces. In particular, which is the angle between the base and one lateral face? 7. Every plane through the altitude of a pyramid is perpendicular to the base. Why? 8. Use the definition of an angle between two planes to show where in a sphere the angle between two meridional planes appears. (See Preparatory Chapter, Section 6.) 9. Given a plane > and a straight line s in it; how many planes are there through s, which form a 30° dihedral angle with >? a 90° angle? 10. Given a plane > and a straight line s parallel to it; how many

planes are there through s, which form a 45° angle with }? a 90° angle? 11. Three planes that are mutually perpendicular have exactly One point in common. Why? 12. Is Cor. I-13-1 a converse of Cor. I-12-1?

Explain.

*Group II

1. In the pyramid ABCV the foot of the altitude is F. Show that the plane VFM is perpendicular to the

base ABC.

If MB | MF

show that MB is also perpendicular to the plane MFV (Use I-12 and I-13). Now show that x BMV is a right angle.

(e A

2. If a plane is perpendicular M to one of two parallel Exel planes, it is perpendicular also to the other. Why? 3. Given two perpendicular planes II and >. Ifa straight line s within > is perpendicular to the intersection k of II and &, then s | Il. Why? 4. Try to formulate the definition of a dihedral angle using the concept of a projection.

132

Unit 1

10 Larger and Smaller You know from arithmetic what “larger” (>) and “smaller” ( b (6 > 4) or the reverse relation a< b (5< 7). A common name for these two relations is inequalities.

Of course, 6 > 4 may be written also in the re-

verse form 4 < 6. The relation “is equal to” has three characteristic properties: it is reflexive, symmetric, and transitive (Review Section 8).

Also

the relation “is congruent to” has these properties. The relation “is larger than” seems to be different. If p, g, and r are numbers, is p larger than itself? If p is larger than q, is q also larger than p? If p is larger than g and if q is larger than r, is p necessarily larger than r? Show that > the relation transitive.

“is larger than’ is irreflexive, asymmetric

and

In Algebra I you have studied these computation rules: Suppose a > 5, then 1.a+p>b+pand

a—p>b-—p

6>4 6+1>4+1 ed el

. if m> 0, then ma > mb and

m=2 2X6>2%4,

ao}:

pas

mm

. if m < 0, then ma < mb and

CED

m=—2

5 Cae

Z

CD

=

Il

6+2>4+1 [ TO

et

as

sides

are

multiplied

(divided) by the same negative

number,

2

4.

at+c>b+d

. Both sides may be multiplied (divided) by the same positive

3. 1f both x4,

oa

. if c > d, then

may be on both sides of the inequality.

number.

—2X6 y + 2, then 2y — 6 > 2 (rule 1) 2y > 8 (rule 1) y > 4 (tule 2)

Note that the solution is not one number but a set of numbers

Ware ah,

If a second inequality or an equality is given, we may be able to find out more about the solution set. For example: l. if 3y-—6>y+2

andy < 10, then we know that y is

between 4 and 10.

(4< y
y+2 or 3y—6=y+ 2 (written combined as 3y 26=2 y+ 2), then the solution set includes 4 as

amember:

{y|

y= 4}. The sign = (S) is read “larger

or equal” (smaller or equal). An equivalent very useful expression is “not smaller than” (not larger than). Since measures of segments and angles are defined as numbers (namely numbers of units; see Section 7), we can apply to them the same computation rules. For the sake of brevity, we say that one segment (angle) is larger, equal, or smaller than another, when we really mean that the measure of the one is larger, equal, or smaller than the measure of the other. The combination of inequalities with geometric statements leads to various new geometric statements, called geometric a B -/_—+———_+—— inequalities. 7D

are meas-

y

O

ured in terms of the same unit

¢

and

f+

1. If AB

if AB

Anh

and CD

measures

3 and

CD measures 4 of these units, then there is a point P on CD_

9

1

1

2

2

between C and D such that CP also measures 3 units. are equal and congruent (Section 8).

3

p

D

3 Ee.

4

-_

CP and AB

134

Unit 1

> AB is shorter than CD, if there is a point P on CD between C and D such that AB = CP.

Analogously, > x AV,B as

is smaller than x CV, D, if there is a ray V, P between —>

V,C and V,D such that x AV,B = XCVP.

2. Corollary I-1-2 states an equality between an exterior angle in a triangle and the sum of the two remote interior angles. xat+xy=xXp’. If we subtract one remote interior angle on one side but nothing |

on the other, then we arrive

B

(by rule 5) at an inequality: > An exterior angle of a triangle is larger than either remote interior angle.

Theorem I-1 expresses a condition of size for the three angles of a triangle (an equality). The following postulate expresses a condition of length for the three segments forming a triangle (an inequality).

The triangle inequalities:

Postulate X(a): The sum of any two sides of a triangle is greater than the third.

Corollary X(b): The difference between any two sides of a triangle is smaller than the third.

A System of Postulates

135

Show that the corollary AB — AC < CB follows from the postulate AC + CB > AB by transforming the first inequality. (Subtract AC from either side.)

Suppose AC’B (Figure I) were an elastic string fixed at A and B. By holdingit at C’ and | stretching it to C we form a triangle. Stretching AC more thanin BC, C so_much more that AC” — BC"= AB, changes the triangle again to a segment, this

time with C” outside of AB. This illustration shows the meaning of X(a): it tells us under what conditions three points are not collinear.

Thus it provides a means of

i

distinction between the set of points on a straight line and the set of points in a plane; it provides a means of finding out whether three points of known mutual distances are collinear or not.

If X(a) holds, the three points form a

triangle; if the inequality in X(a) is replaced by an equality, the triangle collapses into a segment, and the three points are collinear. & The point set set consisting of ofpoints A, B, and all points C such that AC +CB= AB or AC— BC=AB, is a straight line; in the first case C is between A and B, in the second case C is outside of AB.

Here we have again the fact that in our system a straight line is determined by two of its points (A and B). Statements X(a) and (b) may be combined with the statements

about collinear points into one statement: > For any three points A,B,C the sum of any two segments between them cannot be smaller (2) than the third, nor can the

difference between any two of these segments be larger (S) than the third. (Larger or equal (2) means the same as ‘‘not smaller’; smaller or equal (S) means the same as “not larger.’’)

136

Unit 1

Postulate X(a) which states that going from A to B via C is longer than going through C’ can also be expressed in the following torm:

Corollary X(c): The shortest distance between two points lies on the straight line segment between them.

In some other geometry book you may find X(c) listed as an axiom or postulate, and X(a) as its corollary. Statement X(c) certainly seems obvious enough to be accepted without proof. Yet, when analyzing its meaning, it becomes less obvious: What is the shortest distance from home to school? As the bus goes, or through side streets avoiding traffic, or in a beeline through the neighbors’ back yards and over fences and through houses? What is the shortest distance between the north pole and the south pole? Along a meridian or through the center of the earth? — It all depends! Running to school “as the crow flies” might take you actually much longer than going by car on the conventional roundabout streets. Thus, the “shortest” way is not necessarily the quickest. In geography, in surveying, in navigation etc., we may be interested in the shortest distance between two places on the earth as measured on the surface of the earth, not through the earth. On the surface of the sphere the shortest distance lies on a great circle and not on a straight line. The surface of a sphere does not even

have a straight line, yet it does have shortest dis-

tances. We see that the meaning of “shortest distance” changes with the context, with other conditions to be considered in the particular context. In Euclidean spherical geometry the shortest distance between two points on the surface of the sphere lies on the straight line through the sphere (X(c) applies);

there is a short-

est distance on the surface, but that lies on a great circle and not on a straight line (X(c) does not apply).

We

have noted that

spherical geometry can be expressed in the language of a nonEuclidean geometry, where the great circles take the place of “pseudo-straight-lines.” In this language the shortest distance does lie on a “pseudo-straight-line” (X(c) applies).

We see that the terms “shortest distance” and “straight line” go

A System of Postulates

137

together; depending on the context, one becomes the definition of the other.

Theorem I-14: Two circles in the same plane have two points in common, if the radii and the segment between the centers satisfy the triangle inequalities.

Proof:

A common point P, belongs to either circle, there-

fore has distance

r, from

C, and distance

Zs

r, from C,.

AC,C,P, can only exist if its sides satisfy the triangle inequalities. Then there is one A on either side of C,C,.

To show that there is such a common

point P whenever 7, fo,

and C,C, satisfy the triangle inequalities takes a lengthy proof. In some geometry books the existence of P is therefore simply accepted as a postulate. We shall briefly sketch the ideas involved in a geometric proof:

*First the concept must be established that a circle is a connected line which separates the plane into two parts, the interior and the exterior of the circle. The circle itself is the boundary line between these parts. (Compare Part A, Sections 4 and 5.2 for the line and the triangle.) Then it can be shown from the inequalities between the radii and C,C, that circle ©2 has some but not all of its points in the interior of circle ©, (see Unit 4).

Thus there is a subset

S, of ©z which lies in the interior of ©, and another subset S, of ©, which lies in the exterior of ©,. Suppose a direc-

138

Unit 1

tion is assigned on ©, so that the order Postulate II(a) be-

comes applicable (see Section 3, Exercise 2g). Following this direction along ©, from some point of S, to a point of S, the boundary line ©, must be crossed. Since ©, is supposed to be a connected line, this boundary point must belong to ©,. It thus belongs to both circles. What happens if r, + r, = C,C,, or r, — rT, = C,C,? What Happens ifr 177 C2 EXERCISES Group I

1. What do we mean by saying that a relation R has one of these properties: reflexive, irreflexive, symmetric, asymmetric, transitive? Give examples. 2. Show by numerical examples which of the properties named in Exercise 1 apply if R means “‘is equal to,” if R means “‘is larger than,” if R means “‘is smaller than.” 3. Which of the properties named in Exercise 1 apply if R means “is the father of,” “takes the same geometry course as,” “‘takes some course together with,” “weighs more than’’? 4. Explain the meaning of the five computation rules for inequalities by numerical examples. 5. Show by numerical examples why rule 4 should not include subtraction and division. 6. Show by numerical examples why the phrase ‘‘in the same sense” is important in rule 4.

7. Find the solution set and mark it on a number line:

Example: a.x>0 bx AF onthe same ray, which point is between the other ‘two? . Which of the following number triples can be the numbers of length units in the sides of a triangle?

(3,4,4), (3,4,5), 3,1,1), (1,2,5), (4,2,3) . Check your answers for Exercise 8 by constructing triangles with these sides. Explain the construction starting with the longest segment as base c of the triangle, and considering each of the other two sides as radius of a circle having the third vertex as an element. Study 10. Draw two circles in all possible relative positions. the number of their common points in connection with the triangle inequalities. AC is prolonged to AD. Show why ADB x ak In AABC the side must be smaller than xACB.

5g, il

Exon

12. If P is a point inside of AABC, then XAPB > X ACB. Why? 13. If P is a point in the plane of AABC, show that the sum of its distances from the vertices is larger than half the perimeter of the triangle. (Four separate cases: P is inside the triangle, P is outside the triangle, P is on a side, P is a vertex.) 14. Show that in any pyramid the sum of the lateral edges is larger than half the perimeter of the base. 15. Is there a point for which the sum of its distances from the vertices of a given triangle is smaller than half the perimeter?

A System of Postulates

SUMMARY A.

OF PART

141

B

We have studied definitions, postulates, and theorems concerning the measurement of segments and of angles. The measure of a segment or angle is the number of units it contains. Units are chosen by agreement, but geometric statements are independent of a particular choice. A statement in terms of one unit can be transformed into an equivalent statement in terms of another unit by transformation rules. Congruent segments are defined by a special one-to-one correspondence, which makes it possible to compare segments at different places without measuring them in a specific unit. The definition of congruent angles is based on congruent segments. Since measures are numbers the rules of arithmetic about equalities and inequalities can be applied.

. Postulates and their corollaries

Post. VII: Postulate of Eudoxus. Given any two quantities a and 5, then there is a multiple of a that is larger than b. Post. VIII(a): Given a segment AB and a ray with an initial point A’, there is exactly one point B’ on that ray so

that A’B’ = AB. (b): Equal segments are congruent. (c): Congruent angles are equal, and equal angles are congruent.

(d): Given

—

x AVB and a ray V'A’, then in any particular plane

through that ray there is exactly one ray V’B’ on a given side of V'A’ so that x.A’V'B’ = x AVB. Post. [X(a): All straight angles are equal. Cor. [X(b): All right angles are equal. Post. X(a): Triangle inequalities. The sum of any two sides of a triangle is greater than the third. Cor. X(b): The difference between any two sides of a triangle is smaller than the third. (c): The shortest distance between two points lies on the straight line segment between them. . Theorems and their corollaries

Th. I-1: The sum of the interior angles of a triangle equals a straight angle. Cor. I-1-1: If two angles of one triangle are equal respectively to two angles of another triangle, then the third angles are equal. I-1-2: An exterior angle of a triangle is equal to the sum of the two remote interior angles of the triangle.

142)

Unit1 I-1-3: Of the interior angles of a triangle at most one can be a right angle. 1-1-4: Ina right triangle two angles are complementary. Th. I-2: If two lines @ and} are cut by a transversal and if a pair of alternate interior angles are equal, then a | d. Cor. I-2-1: If two lines are perpendicular to a third, they are parallel to each other. Th. I-3: Within a plane there is exactly one perpendicular to a given line at a given point. Th. 1-4: There is exactly one perpendicular to a given line through a given external point. Th. I-5: Ifthe sides of an angle are perpendicular to the sides of another angle, then the two angles are either equal or supplementary. Th. I-6: A line is perpendicular to a plane, if it is perpendicular to two lines that lie in I and pass through the foot. Th. 1-7: There is exactly one plane perpendicular to a given line at a given point of that line. Th. I-8: There is exactly one plane perpendicular to a given line through a given point external to that line. Th. 1-9: There is exactly one perpendicular to a given plane through any given point. Th. 1-10: If two lines are perpendicular to the same plane, they are parallel to each other. Th. I-11: The projection of a straight line s on a plane [lis a straight line s’, unless s 1 I. Th. 1-12: If two planes are perpendicular, then each contains a line perpendicular to the other. Cor. 1-12-1: If each of two planes is perpendicular to a third plane, then their line of intersection (if there is one) is also perpendicular to that third plane. Th. I-13: If a plane contains a perpendicular to another plane, then the two planes are perpendicular. Cor. I-13-1: If two planes intersect in a line perpendicular to a third plane, then each of them is perpendicular to that third plane. Th. I-14: Two circles in the same plane have two points in common, if the radii and the segment between the centers satisfy the triangle inequalities. Principles 1. of indirect measurement (Section §) Ll. of reducing a three-dimensional problem to a two-dimen: sional one (Section 9)

A System of Postulates

143

E. Words and phrases Section 7: metric geometry, bil elicnirabile; commensurable, measure of a segment, absolute value, decimal system, interior of an angle, straight angle, measure of an angle Section 8: one-to-one correspondence, congruent segments, scflexive, symmetric, transitive, congruent and equal, sphere, circle, radius, center, congruent angles, adjacent, adjacent supplementary, interior of a straight angle, round

angle, perigon, vertical angles, parallel angles, supplementary, interior and exterior angles of a triangle, g.e.d., remote interior angles of a triangle, transversal, alternate interior angles, corresponding angles, hypothesis, conclusion, converse, principle of indirect measurement Section 9: right angle, right triangle, complementary, perpendicular, existence, uniqueness, proof by logical construction, proof by elimination of opposite statements, direct and indirect proof, foot of 4 perpendicular, distance of a point from 4 line, normal, distance of a point from a plane, principle of reducinga three-dimensional problem to a twodimensional one, projection of a point and a line on a plane, angle between a line and a plane, dihedral angle, if and only if Section 10: inequalities, irreflexive, asymmetric, not smaller than, not Jarger than, geometric inequalities, triangle inequalities, shortest distance

REVIEW EXERCISES TO UNIT | 1. Be sure to know the meaning of the words and phrases listed in the Summary. 2. Make 2 correct statement by inserting “always,” “sometimes,” or “never”: a. Two straight lines... have exactly one point in common. b. Two planes... have exactly one point in common. c. A straight line and a plane... have exactly one point in common. d. If two straight lines have no point in common, they are . . parallel. e. If two planes have no point in common, they are . . . paraljel. f. Two points are .. . collinear.

g Three points are . . . collinear. h. Two straight lines are .. . coplanar. i. A straight line andaa point are... coplanar.

144

Unit1 . Does a straight line always lie in some plane? Explain. . What are skew lines? Are there skew planes?

. How do we a. the three b. the three . What is the

know that: vertices of a triangle always lie in one plane? sides of a triangle always lie in one plane? Parallel Axiom? What is its historical impor-

tance?

. Is the sum of the angles in all triangles the same? . How do we know that an exterior angle of a triangle always equals the sum of the two remote interior angles? . Why is a right angle in a triangle always the largest angle of that triangle? 10. Two parallel straight lines are cut by a third; what are the names of various pairs of angles formed? How big are these angles, if x3 = 45°? or 60°, 90°, 40735 207 7/8 Le Prove that vertical angles are equal. 12. If two interior angles of a triangle are respectively Exeal© equal to two interior angles of another triangle, the third angles are also equal. Why? . If one angle of a triangle is 20° (or 40°, 30° 40’, 90°) and the other two are equal, how big are they? . If the three angles of a triangle are equal, how big are they?

. How big are the exterior angles for each triangle in Exercise 13? . How big is an angle whose sides are perpendicular to the sides of an angle of 30° (or 50°, 65°, 40° 25’)?

. If an interior angle of a parallelogram is 55°, how big are the other three? How much is the sum of all four? 18. Construct a triangle with sides a= 3, b=4, c=5; can you construct a triangle from any arbitrary triple of numbers for sides? Can you name a triple that does not form a triangle? 19. Which of the following triples are possible triangle sides and which are not? (7,5,5), (15,24,34), (21,9,30), (10,3,5), (2,3,4) 20. An exterior angle of a triangle is greater than either remote interior angle. Why?

A System of Postulates

145

21. Two exterior angles of a triangle are 115° 38'12” and 78° 50' 20”; how big are the interior angles? 22. In a triangle Xa = 48° 20’ 40"; X8= 2a; how big is y and all exterior angles? 23. Any side of a triangle is smaller than half the perimeter. Why? 24. Two circles in the same plane have radii 3 and 5 inches respectively; do they have points in common, if the distance between their centers is 4 in., 8 in., 10 in., 1 in., 3 in.? 25. Form a true statement by inserting one of these words: acute, obtuse, straight, right. a. The supplement of an acute angle is... . The complement of an acute angle is... The supplement of an obtuse angle is... . The supplement of a right angle is... . The difference between the supplement and complement cane of an acute angle is... The sum of a right and an acute angle is... gQ ™ . The difference between a straight and an obtuse angle

iS see 26. Given a plane II and a straight line s perpendicular to it; how many planes can be passed through s that are perpendicular

to II?

If-s is parallel to II?

Ifs is oblique (neither

nor ||)

to II? *27. Four non-coplanar points form a skew quadrilateral; do the diagonals intersect? (If they do, what follows?) *28. Prove: If each of two planes is perpendicular to the same straight line, the planes are parallel to each other.

1

SIMILARITY AND CONGRUENCE

Have you read Guiliver’s Travels by Jonathan Swift? Gulliver made a trip to an imaginary country, called Brobdingnag, where people are so enormously tall that Gulliver was afraid they might mistake him for a delicacy to put on a sandwich, as we would sardines, and eat him fora snack. You can imagine how enormous their houses, cars, books, dogs, etc. must have looked to Gulliver!

Do you think all these looked just as enormous to the Brobdingnagians themselves? Gulliver also visited a country called Lilliput, where people were so tiny that he had to be very careful not to step on them or _ on their houses. The Lilliputians stared at Gulliver as an unbelievably enormous monster, just as Gulliver had stared at the Brobdingnagians. Do you think the Lilliputians thought of themselves as tiny? Are we ourselves dwarfs or giants? Suppose a Brobdingnagian or a Lilliputian came to visit us! Unless we have a chance to compare through such visits and find out about the relative scale, nobody would know the difference:

the Brobdingnagians would think their houses just as adequate as the Lilliputians theirs and we ours. 1.1 The Ratio of Similitude Suppose the Brobdingnagians were studying Euclidean geometry. Their right angles would be defined in the same way as ours are; and their squares would have the same properties as ours. Only by Gulliver’s description do we know that the side of their square foot compares to the side of ours, as the length 146

Similarity, Congruence, Symmetry

147

of a grown man compares to that of a sardine.’ Suppose they want to subdivide some area in the countryside into property lots measuring 50 feet by 100 feet. The instructions would sound exactly the same as ours, namely to survey rectangles with sides 50 and 100 feet: First measure a length of 50 units (giant feet, our feet), then lay off a right angle at each endpoint (same in both countries), then measure 100 units on each perpendicular (giant feet, our feet), then connect

the new

endpoints and check to be

sure that the two new angles are right angles (same in both countries) and that the new segment measures 50 units (giant feet, our

feet). We know that the instructions for laying off angles produce equal angles in both countries, because the measure of angles is related to the geometric concept of a straight angle and all straight angles are equal (Post. IX). We also know from the instructions that one pair of sides is twice as long as the other within each country (equal ratio of sides), but we do not know exactly how long they are unless Gulliver compares the units (which are based on some physical length rather than a geometric concept). In other words, we know that the shape (angles and ratio of sides) of the

giant figure is the same as the shape of our figure, but we do not know the size.

Definition II-1: Geometric figures are called similar (symbol ~), if they have the same shape.

1 See Supplement IV.

148

Unit 2

Between two figures which consist of straight line segments and angles and which have the same shape (Figure I), we can set up a one-to-one correspondence called similarity such that the sides and angles named in a certain order for one figure correspond to the sides and angles named in the same order for the other figure,

and such that corresponding sides have the same measure (although we may not know the unit) and include corresponding equal angles. Then “ABCDE and A'B'C'D'E' have the same shape” can be stated in mathematical language as follows:

AB? BC = BB Ce eandaneh oe Be also BC:CD = B'C’':C'D' A relation between

and

four numbers

xC=XC'

and so on.

(four measures)

where two

ratios are set equal is called a proportion. The computation rules governing proportions are based on the fact that a ratio is merely a fraction. You will find these rules in SupplementIV. By applying Rule 2 to similar figures we can derive a useful theorem:

Theorem II-1: Geometric figures are similar, if and only if corresponding angles are equal and corresponding sides are proportional (have a constant ratio).

The phrase “if and only if” indicates that we have to prove two parts of the theorem, each part being the converse of the other. (See Unit 1, Sections 8.2 and 9.3.)

We must prove that “if geo-

metric figures are similar, then corresponding angles are equal and corresponding sides are proportional.” We must also prove that “if corresponding angles are equal and corresponding sides are proportional, then the geometric figures are similar.” Theorem II-1 is a combination of these two statements using the phrase “if and only if.” We shall prove the first statement here and let you supply the proof of the second. Hyp.:

(Figure 1) ABCDE ~ A'B'C'D'E’, which means XA =XA',xXB=XB’,...

and

AB? BC =A’ B’:B'C’,

BC: CD —=B'C: GD

=.

Similarity, Congruence, Symmetry Con.:

xA=xXA',

149

XB=XB’, etc.

Be BIC pig CD:C'D'=p:q...,

etc., where p and g are some numbers.

PROOF STATEMENTS 1.

2B.

REASONS

XA =XA', XB=XB', etc.

BG— ALB BC.

trhyp:

2. hyp.

3. AB:A'B’ = BC:B'C'

3. Rule 2, the inner members of a proportion may be interchanged

4, If AB: A’B’ = p:q then

4. from 3.

BGO BICl-—p-¢

Repeat steps 2 to 4 for the next pair of sides SabC CDR C.. CD 6. BC BC!’ =CD:. CD ae Since BC:B'C'=p:q, CD:C'D'=p:q

also

5. hyp. 6. Rule 2 7. from 4. and 6.

Repeat for the next pair of sides Theorem II-1 changes a statement about the ratio between the sides in the same figure to a statement about a constant ratio between corresponding sides in different figures. This constant

ratio is called the ratio of similitude; it is nothing else but what you know under the name of scale. Since the ratio is constant, all segments are affected in the same way; the effect is like looking at a small or large photograph of the same object, or like looking at it with a magnifying glass. The ratio or scale may be given numerically (the photo is enlarged

3:1) or graphically (on a map cee means that the scale of the map is the ratio of the length of this segment to one mile on the surface of the earth). Have you seen a picture postcard of a sequoia tree with a tunnel through its trunk big enough to let a car pass through? If the car were not on the picture, it could be just any tree with a hole. The car is used as a graphical scale. Since we know the size of a car, we get a better idea of the tree’s girth by comparing it with the car. If you take a snapshot of an unusually tall flower, its height will not be noticeable unless you have something or someone of known size standing next to it as a sort of graphical scale. The shape is preserved in a photograph, but not the actual size.

150

Unit 2

In the special case where the ratio of similitude is 1:1, the ratio

between any pair of corresponding sides is 1:1, which means that the sides are equal. a

nnn aU UEEEEEEEEEES!

Definition 1-2: Similar figures with the ratio of similitude called congruent (symbol =).

1:1 are

With this definition Theorem II—1 leads to:

Corollary ll-1-1: Geometric figures are congruent, if and only if corresponding angles are equal and corresponding sides are equal.

The symbol for congruence is appropriately a combination of the symbol for similarity with the symbol for equality. In Unit 1 we defined congruent segments. These were used to define congruent angles. Now we extend the relation of congruence to two-dimensional figures (and three-dimensional objects) and define it as a special case of similarity. The definition is based on the concepts of equal sides and equal angles, which are related to the concepts of congruent sides and congruent angles by Postulate VIII(b) and (c).

The statement that similar figures have the same shape is sometimes expressed in the more technical wording “similarity preserves shape” or “shape is an invariant (unchanged) under the relation of similarity.” Since congruence is also a similarity relation, although a special one, we say “congruence preserves shape and size” or “shape and size are invariants under the relation of congruence.” In Euclidean geometry we study the shapes of figures, and these are invariants under similarity. It follows that the same theorems

will be suggested by a big drawing on the chalkboard or by various small drawings in your notebooks. It also follows that it will be

Similarity, Congruence, Symmetry

151

quite useful to have quick methods of determining whether or not two figures are similar, because then the study of one tells us at the same time everything about the other. Obviously, the same is true for congruence. In the present Unit we shall study methods of recognizing similarity and congruence, and then use these relations to establish interesting geometric facts, and to simplify proofs of theorems. To find out whether or not two figures or two objects are similar (or even congruent) we should compare each pair of corresponding angles and each pair of corresponding sides. This task may amount to quite a chore however; thus it would be of very practical interest to have some laborsaving theory by which we can be sure of similarity (congruence) after only a certain small number of such comparisons. Since many figures can be divided into triangles, we shall concentrate on triangles first. EXERCISES Group I (Study Supplement IV)

1. Write the following ratios in the form of fractions and then find the simplest equivalent fraction:

307 100.2 5250, 8:

4515912528256

2. Which of the following proportions are correct statements?

30: 100 = 15:200, 25:50 =3:6, 8:4= 40: 80, 15:12=5:3,

6:7=7:8

3. What is the ratio of each of the following pairs of measures? right angle and straight angle 6 inches and 1 foot right angle and round angle 1 foot and 1 yard 45° angle and perigon 1 inch and 1 mile 1 mile and 1 kilometer 1 meter and 1 kilometer 1 inch and 1 centimeter

4. Check each of the ten computation rules in Supplement IV by a numerical example for a= 3, b=6,c =5, d=? 5. Solve for x: a. 21°14 =27:x b. 21:27=14:x

ce. 20:32 d. 25:32

=25:x =20:x

e. a:ab=b:x f. ab:b=a:x

6. Solve for x:

a. (x—3):(4+ 1) =3:4 b. (G2): 2x15 1

Cove Oat I= (x + 3) (ee 5) d. (x+ 2):4=(x— 1):3

152

Unit 2 . Solve for x and y:

ane

b.

kV x+y=40

x:y= v=x—

c.

16:24 1

x:y=1:1 3x+4y= 14

. Divide the number 360 into two parts in the ratio 1:2; 4:5; Dee

. How many degrees are in each of two supplementary angles, if their fatio 18° 22320

4°57

1?

. How big is each of two complementary angles, if their ratio Ke BeBe?

Ales

ibe ily

. What is the ratio between two supplementary angles, if one is 45°? 30°? 223°? gofa straight angle? = of a straight angle? < solve fOn-x, Vz: a Xiyi:2=1:233b.

x+y+z=360

«iy 2225554 x+y+z=360

©

412 x+y

3 ee Z2=110

13. How big are the angles of a triangle if their ratio is a Biy= 132737 2324) Le eri ant a, 14. How big are the angles of a triangle if one is half the second and one seventh of the third? if one is equal to the second and twice the third? Group II

1. In what sense is congruence a special case of similarity? 2. What is an invariant? under congruence?

What is invariant under similarity?

. If Gulliver compares himself with a sardine in Brobdingnag, what approximately is the numerical ratio of similitude between us and the giants?

. Use your answer to Exercise 3 to estimate how long (approximately) a pencil would be in Brobdingnag. . If ina floor plan of your room the length of your bed is 1 inch long, what is the scale of the plan? . If a map bears the legend 1: 25,000, what does this mean?

. If a map bears the legend 1 inch to 8 miles, what does this mean?

. Ifa map bears the legend HK}-—————-++__ ; what does this mean? ) SO 100 miles . What is a numerical scale? What is a graphical scale? *10. If the three maps in Exercises 6, 7, and 8 are photographed down to half their size, is the legend on each one still cor; rect?

Similarity, Congruence, Symmetry

153

1.2 Congruent Triangles (= As) We shall label a triangle in a standard manner as in Figure I: the vertices by capital letters in a counterclockwise

order;

the

sides in lower case of the same letter as the opposite vertex; angles by lower case Greek letters corresponding to the letter of the vertex. An angle is sometimes also referred to by the letter of its vertex or by a group of three points, one on each adjacent side and the vertex in the middle. Thus, angle a may be labeled angle CAB or BAC. The horizontal side c is sometimes called the base of the triangle; but since turning the paper around makes another side seem horizontal, each side may qualify to be called a base occasionally. The height, usually denoted by h, which is perpendicular to a specified side of the triangle carries the letter of that side as a subscript. A standard notation is a timesaver, since we will know which side or angle is meant without much explaining. A triangle always lies in a plane (see Unit 1, Section 5.2). Two triangles may or may not lie in the same plane. The question whether or not they are congruent refers to their shape and size, but not to their being coplanar or not. The following discussions are therefore independent of where these triangles are.

Theorem II-2: If three sides of one triangle are respectively equal to the three sides of another triangle, then the two triangles are congruent. 5.5.8. = side-side-side theorem

154

Unit 2

Hyp.: 4B = A’B’, BC=B'C', CA=C'A’ Con.: AABC = AA'B'C'

PROOF STATEMENTS 1. AB =A’B',BO =B'C’,

REASONS 1. hyp. and Post. VIII(b)

CAS CL 2. xBAC = xB'A'C’,

2. from 1. and Def. I-10

ACBA = C-BAl XBCA=ZXB'CA' 3. XBAC=XB'A'C’, XCBA= XC'B'A’, x BCA=XB'C'A'

3. Post. VIII(c)

4. AB= A'B', BC=B'C’,"

4. hyp.

CA=C'A' 5. AABC

= AA'B'C’, q.e.d.

5. from 3., 4., and Cor. II-1-1

Theorem II-2 tells you two things: (1) To find out whether or not two given triangles are congruent, you do not have to compare all six pairs of corresponding parts; you can get away with half the labor by comparing only the three pairs of sides. The three pairs of angles must be congruent, if the sides are equal. 2 If somebody wants you to construct a triangle giving you as specifications six arbitrary numbers as required measurements of the three sides and three angles, you —_Z

should be suspicious:

since the three sides already de-

termine the triangle and its angles, extra information is

not only superfluous, but could be even contradictory. Theorem I-2 tells you the limit of arbitrary specifications.

What is a geometric construction? Just as any other construction, it means putting an object together according to certain specifications. In a practical construction, for instance in a draftsman’s job or a building construction, a variety of tools are used to carry the job to completion as well and as fast as possible. New tools designed to expedite the job are a welcome addition. In a geometric construction, by contrast, there is a restriction in the use of instruments. It is obvious, that the choice of such instruments will

Similarity, Congruence, Symmetry

155

determine how difficult a problem is going to be and whether it can be solved at all. For example, to cut a plain piece of paper along a perfectly straight line is easy, if you are permitted to crease the paper and use a knife. If no knife or equivalent tool is permitted, then it takes some skill to tear the paper neatly along the crease. If folding the paper is not permitted, then it is almost impossible to tear it along a perfectly straight line. The geometric constructions which we are going to study are restricted to the use of a straightedge and a pair of compasses. These restrictions were made by the Greeks, and have been respected throughout the centuries to this day as the definition of a classical construction (Euclidean construction).

You may ask why the Greeks imposed such arbitrary restrictions on themselves and why we should maintain them 2000 years later. The choice of the particular instruments may have happened by chance, but the idea of restricting their number to a minimum when studying theoretical problems is characteristic of the Greek viewpoint. In those days the pursuit of cultural interests was reserved for the leisured class. Work for a practical purpose was looked down upon as the obligation of servants. Since tools have primarily a practical value they have no place in pure science, pursued for self-satisfaction and not for practical returns.* The most

influential exponent

of this viewpoint was Plato, who suc-

cessfully insisted on dispensing with all mechanical instruments, thus laying down the rules for centuries to come. From this viewpoint the straightedge is no tool at all, but a visible demonstration of the postulate “there is exactly one straight line passing through two given points.” And a pair of compasses represents the postulate “on any given ray there is a segment congruent to a given segment.” A classical construction of a triangle thus becomes a demonstration that the existence of such a triangle follows logically from the postulates, applying them a certain finite number of times.

Today, we hold mechanical instruments in high esteem. Modern research depends more and more on electronic computers, for example. Why then should you submit to Plato’s rule when studying geometry? First of all, you may look at it as a “Can-you-do1 A beginner among Euclid’s students asked what gains he could expect from studying

geometry. Euclid called a servant and told him: must make profit out of what he learns.”

“Give a three-pence to this man, since he

156

Unit 2

it?” game, which has been fun for many people throughout the centuries, and it may be fun for you. Second, trying to solve a construction problem will give you a better insight into the sigThe nificance of theorems, their possibilities and limitations. possibility or impossibility of a construction may throw a new light on problems in another context. For example, the impossibility of constructing the point of intersection between two parallel lines explains why two simultaneous linear equations may have an empty solution set. And the other way around, new knowledge in another branch of mathematics may explain why certain construction problems cannot be solved. This happened in the case of three famous problems, which the ancient Greeks thought up and then could not solve. Neither could anyone in the following centuries solve them; and there were many who tried. These problems are: the Squaring of the Circle (to construct a square with the same area as a given circle), the Trisection of the Angle (to divide an angle into three equal parts), and the Duplication of

the Cube’ (to construct a cube having twice the volume of a given cube).

Of course, these problems can be solved, but not with the

classical restrictions. veloped,

It was not until algebra was sufficiently de-

several centuries later, that it was realized that the Tri-

section problem was equivalent to solving a cubic equation (Vieta, 16th century). You can easily verify for yourself that the algebraic equivalent of the Duplication problem also depends on solving a cubic equation, namely x* = 2a*. Since a straight line is equivalent to a linear equation (see Section 2.5, page 226) and a circle to a special kind of quadratic equation (see Unit 4, Summary B, page 422), constructions with straightedge and compasses are equivalent to algebraic problems involving linear and certain 1 In the year 430 B.c. Athens was afflicted by the plague.

The Athenians sent a delega-

tion to the oracle at Delos with an appeal to the god Apollo that he stay the plague. Delos is a small island in the Aegean Sea. The ancient Greeks held it sacred, because Apollo was born there according to legend. The temple of Apollo had an altar of pure gold shaped as a perfect cube. When the Athenian delegation asked the oracle what they should do to end their affliction, the oracle answered: Double the size of the altar without changing its shape. This demand amounted not only to an expensive gift in gold, but also to a difficult problem. Doubling the edge of the cubic altar would make its size (volume) eight times as large. To find the correct length of the edge by the only admissible method of geometric construction defied all attempts of the Greek geometers. Apparently, the devoted efforts, although unsuccessful, appeased the god, as the plague must have left Athens at some time. The geometric problem of doubling a cube in volume WOU changing its shape is known since as the Delian Problem.

Similarity, Congruence, Symmetry

157

quadratic equations. Thus we have an algebraic definition of the set of “possible classical constructions.” Vieta’s statement means that the general Trisection problem is not a member of this set. Neither is the Duplication problem a member of this set, since it depends on a construction of ¥/2. In very special cases, however,

a cubic equation (general form: ax* + bx? + cx +d=0) can be factored into lower degree equations (for instance, if a= 1, Dao) ed 41, then x — |= (x —1)* + x74 1) —0). This explains why angles of a very special size (90° for example) can be trisected by classical construction. You will learn how to construct certain angles in later Sections. It took until the 19th century (Lindemann) before enough was

known about the number 7r to be sure that a construction problem involving zr was not equivalent to an algebraic problem of the first and second degree. This settled the impossibility of the Squaring of the Circle. There are still people who think that if they try hard enough they will “solve” one or the other of these famous problems. These people misunderstand the meaning of mathematical impossibility. If the set of “possible classical constructions” is defined by a certain algebraic feature of its members, then it is mathematically impossible to find a member without this feature, even if you look very hard. In the same way, if “parallel” is defined as “not intersecting,” then it is mathematically impossible to find a point of intersection, even though you extend the parallel lines as far as you wish. A Construction Problem (with straightedge and compass only) Note: Construction problems will be marked as such. For other problems you may use the draftsman’s triangles and the paper strip in order to save time.

Given

three

segments

G— 3.) —»5,-and c = 6; '

construct A ABC with these segments as sides.

— b ee gel

158

Unit 2 C

I. Analysis: s.s.s. II. Construction: 1. Draw line AB=c=6. 2. Use A as acenter and draw a circle with r = b=5. 3. Use B as a center and draw a circle with r = a= 3. 4. The circles intersect at. G andiC.. 5. Connect AC = b, BC = a. 6. Solution:

B

e

AABC in required order. AABC' in opposite order.

Rea es

III. Proof: a = 3 (radius of circle) b=5 (radius of circle) c=6 (Post. VIiI(a): there is a segment = to a given seg-

Monten IV. Discussion: If the two circles in steps 2 and 3 intersect twice, then the tri-

angle inequalities are satisfied and there are two solutions. If the circles have only one point in common (if they are tangent), then

a+ b=c

and there is a segment but no tri-

angle: no solution. If the circles have no point in common:

no solution.

The four items: analysis, construction, proof, discussion represent the general pattern you should consider when doing a construction problem. For the analysis you will line up the given pieces, and make a sketch of what the solution should look like, in order to decide on a plan of how to reach the solution from the given pieces. The connection between them is often found by working backwards from the solution to the given pieces. The sketch need not be true to shape or scale, as it is only meant as a

guide to your thinking; it should show by markings which the given pieces are, but it should not suggest more than is known; e.g. do not sketch an equilateral triangle for a general triangle,

Similarity, Congruence, Symmetry

159

because it might mislead your thinking. After you have made your plan, write it down, using symbols to make it short. — The construction steps will be listed in the order of your doing them; always show enough of the auxiliary construction lines (such as parts of the circles above) to enable someone else to see what you actually did. — The proof should show why the solution satisfies the requirements; it may be short or even omitted if obvious. — The discussion sets the problem into a larger context by stating the number of solutions under certain conditions. It may be omitted if obvious. — In all four items you may use symbols and telegram style instead of full sentences, in order to save space and time.

Theorem II-3: If two angles and the included side of one triangle are respectively equal to two angles and the included side of another triangle, then the two triangles are congruent.

a.s.a.= angle-side-angle theorem

B!

Hyp.: 4’B’=AB, Con.:

xB’A'C'=XBAC,

XA'B'C'=xXABC

statements of the hypothesis clearly in the sketch.) eo “tp: oe AA'B'C’ = AABC

(Mark

= AC and B’C’ = BC, then use s.s.s. Plan of proof: Show that A’C’

the

160

Unit 2 * PROOF STATEMENTS

AB =4B

REASONS

=

.© around A’ with r= AC and © around B’ withr= BC intersect at C* on the same side of

— . hyp. 2. I-14, A inequalities are satis-

fied since there is a AABC

—_

A'B' asC’

3. AA'B'C* = AABC 4. x B'A'C*=X BAC and x.A'B'C* =x ABC 5. also x B'A’C’ = x BAC and x.A'B'C' =x ABC 6. x B'A'C* = x B'A’C’ and XxA'B'C* = XA'B'C’

3 7 SsSsSs irom 1 and) 2. 4 . from 3.

7. A'C*

7 . on same side of A’B’ ray... (VIIId)

—

a

coincides

:

with

5

5 . hyp.

6 . from 4., 5., and Post. VIII(c)

>

A’C’,

B'C* coincides with B’C’ 8. C* coincides with C’

s

'

,

only one

8 . only one point of intersection

(Vc) 9 10. ee 12.

ALG = A.C’, B Ce AiC* AC, Si = CAC. RB’ C = AA'B'C' = AABC,

9. 10 PS 11 . We .

= BC. BC BC g.e.d.

from frOnl from §.8.8.

8. 5: 9. and 10. from hyp. and 11.

Theorem II-4: If two sides and the included angle of one triangle are respectively equal to two sides and the included angle of another triangle, then the two triangles are congruent. 5.a.s. = side-angle-side theorem

Hyp.: A’B’ = AB, A'C’ = AC, xX B'A'C' =XBAC Con.: Plan:

AA’B'C' = AABC Show that x.C’B'A' = x CBA, then use a.s.a. * PROOF

1. draw

STATEMENTS xXC*B'A'=XCBA

REASONS in

———

same half-plane of A’B’ as C’

—

a

2. A'C’ and B’C* intersect at C*

iE there is exactly (VIIId and c) De Post.?

one

ray...

Similarity, Congruence, Symmetry

3. AA'B'C* = AABC 4. AC*=AC

3. a.s.a. (from hyp. and 1.) 4. from 3.

5. also A’C’ =AC

5. hyp.

161

6. AC*=A'C'

6. from 4. and 5.

7. C* coincides with C’

7. on same side of A’ (same half-

8. 9. 10. 11. 12.

a

ee

B’C’ coincides with B'C* x C*B'A'=XC'B'A’ x.C*B’A'’=xCBA xC'B'A’=XCBA AA’B'C' = AABC, g.e.d.

8. 9. 10. 11. 12.

plane of A’ B’) there is only one point... (VIIIa) from 7. and Post. V(a) from 8. from 1. from 9. and 10. a.s.a.

Theorems II-2, 3, 4 tell us that certain three out of the six parts of

a triangle are sufficient specifications to determine a triangle in shape and size. Are any three measurements just as sufficient? Are three angles adequate specifications? Could you form a tri-

angle with a = 80°,

B =60°,

y= 70°?

form a triangle with a = 80°, B = 60°?

Why not?

Could you

How big must y be then?

Of the six parts of a triangle s, s, s, a, a, a the following different

groups of three may be formed: sss, ssa, sas, saa, asa, aaa. Discuss them as possible specifications. Which are covered by which theorem? Which are impossible and why? Only ssa is a new case; try to gather your own experience about constructing a triangle from one given angle, one side adjacent and one side opposite to that angle. We shall discuss this case in a later chapter. Congruent triangles have an important application in proving a pair of segments equal, or a pair of angles equal. This method is the basic idea in proving the following theorem (see Unit 1, Theorem I-6):

162

Unit 2

» Ifa straight line p is perpendicular to each of two straight lines a and b through a point F of p, then p is perpendicular to the plane II containing a and b.

Hyp.: pl aatFandpl batF Con.: p 1 II containing a and b Plan: Show that any arbitrary line c of II through F is also | p by showing adjacent supplementary angles equal.

*PROOF STATEMENTS —_

1. make

—_-

FS'=FS,

REASONS

draw

>

AB

through some point A on a and B on b, intersect AB with some line c through F 2. AAFS = AAFS"', because

1. Postulates?

2. s.a.s. (identity, hyp. a 1 p, 1.)

AF =AF, £AFS= XAFS’, ES rs! 3. AS=AS' 4. ABFS = ABFS'

3sfrom 2: 4. S.a.s. as in 2.

52 DSi BS?

5. from 4.

6. AABS'

6. s.s.s. (from 3., 5., identity)

= AABS

7. XBAS' = x BAS 8. ACAS = ACAS'

DGS

7. from 6. 8. s.a.s. (from ?)

1050

9. from 8.

10. AFCS =AFCS' 11. xCFS=xCFS'

122. CFS=tt*

10. s.s.s. (from ?) 11. from 10.

or CP tp:

12. equal to its adj. suppl. x

q.e.d.

EXERCISES 1. Construct a triangle, given:

a. c=3 in., a=2 in, b= 2} in.

b. c=4 in., a=3

in., b= 22 in.

Similarity, Congruence, Symmetry

163

. Are there any restrictions for the numbers chosen for constructions as in Exercise 1? If so, what are they? . Draw three arbitrary segments

Gg

a, b, and

c; then construct a

eee

triangle with sides a, b, c. Is this always possible? Discuss.

ee

. Given an angle a and a ray A’; construct an _ angle a’ = a with that ray as side. (See Example, Unit 1, Sec-

eee Ex.3 A'

a

Ease ety

A

Ex. 4

. Given two segments c and b and an angle a; construct a triangle with these parts.

/—____¢

—____, 5

ee

. Given a segment c and two angles a and 6; construct a triangle with these parts.

a

. If three arbitrary angles are given, is there always a triangle with _ these angles? Discuss.

example

ee

“a

tion 8.2.)

For

Vises

A

(30°,

B

Ex.6

60903) 207,70"; 100%), (50°; 90°, 110°), (70%, 50°, 60°). . If you know the number of degrees in two angles of a triangle, can you compute the third? By what theorem? For example SCO eee/0OA 1G0% 110°), (80°, 90°). . Given two angles a and B; can you con-

struct SUCH

an

angle y,

hat

ay B.e°y

a

B aXe

164

Unit 2 could be angles in the same triangle?

10. Of a triangle ABC,

.

(eo

side c, angles B and y are given; can

eee

you construct the triangle? Which congruence theorem

will be used?

B

11. Is it possible that a triangle has sides 7, 3, and 10? (24, 34, £5?)

uv Bxa©

12. If a triangle has sides 16 and 19 units long, between what limits is the length of the third side?

1.3 On Proofs In the construction of a deductive system of geometry each theorem is derived from previously accepted statements. These are the definitions, postulates, and already proved theorems of the geometric system, and also the rules of arithmetic and the rules of logical thinking. (Compare the three statements in Unit 1, page 40.) A proof is a demonstration of the logical steps which link the new theorem to the already accepted statements. Once a theorem is proved it may be used in turn as a link in the proof of yet another theorem. You have read quite a few proofs by now and you should be able to write some of your own. Let us review and round out what you have learned about the pattern of proofs. 1. Read carefully the theorem to be proved, reformulate it (if necessary) as a ©O7. {ional sentence (if — then) and rewrite in mathematical language its hypothesis (what is given) and conclusion (what is to be proved) (review page 111). Then plan how to reach the conclusion from already accepted statements. The proof itself may have various forms. The two-column form of statements and reasons (see Theorems I-1, 2, 10, 13) helps guide your reasoning from step to step. Some people list the statements of the hypothesis and the meaning of the letter symbols as the first part of the proof, while others refer to an accompanying sketch where this information is clearly marked.

The

outward: form, however,

does not

Similarity, Congruence, Symmetry

165

guarantee a correct proof. The crucial thing is that your reasons for each step are valid and that your reasoning is correct. And even if your proof is correct, it may look different from another correct proof by containing more or less detail. It is also possible to use different steps in a proof and still reach the same conclusion. 2. The proof of a theorem A is not correct if one of the reasons listed is A itself or a statement B which depends on A. For example, to prove Theorem I-1 in the following way would be incorrect: The sum of the angles in a triangle equals a straight angle (Th. I-1), because an exterior angle equals the sum of the two remote interior angles (Cor. I-1-2), and is also supplementary to the third interior angle by definition. This mistake is called circular reasoning. Similarly, a definition is incorrect if it uses the word that should be defined. For example: “A straight line is a line that is straight.” This is called a circular definition. & Avoid the mistake of circular reasoning.

3. A correct pattern of adding new statements to the chain of statements in a proof may be as follows:

C

D

Hyp.: Xa= x8, AC ||BD;

Con: 45= The rules of inference lead us from accepted statements to new ones in the chain of statements constituting a proof. Do not confuse the theorem you have to prove with a theorem quoted in a rule of inference. The latter theorem must have been proved previously, otherwise it could not be quoted as one of the reasons.

4. Symbolic notation. — In algebra you have learned to use certain symbols to express mathematical statements in a short form. For example, a” =a-a means: the square of any number is the product of that number by itself. The advantage of such a symbolism is not only a tremendous saving in time and space, but also a clearer understanding of the connections between various mathematical statements. The success of the mathematical symbolism has been so tremendous that the logicians adopted the idea for the study of correct reasoning. Some of the notations used in symbolic logic are already familiar to you from working with sets. (Review the symbols in Supplement I.) We will add here only a few symbols, so that we can rewrite the discussion under 3 above in symbolic form. The theorem analyzed in the example could have been replaced by some other theorem. To indicate that the discussion does not depend on a particular theorem, we may write its conditional form if , then with blanks left for the specific hypothesis and conclusion of some theorem. Following the algebraic example of using letters we can also write if p, then q with the agreement that p and q are place holders (variables) for sentences, p for the hypothesis and q for the conclusion. (Compare with algebra: if 2x =a, then ee

Pp

x =4a.)

The connection between the two sentences,

—S—

q

:

the words if — then, can be written symbolically as an arrow—. & A conditional sentence (a conditional, an implication) is written symbolically as follows: ifp, then q orasp—.q. You may read: “if p is true, then gq must be true” or “from p always follows q” or “‘p implies q’’ or “‘p always leads to q.”

The pattern of reasoning within a proof can now be written symbolically as a logical formula:

168

Unit 2

meaning in our example about parallel angles theorem | if p, then g | substep (a) if p,, then g, |substep (b) | if @ and 6 are ||angles in the same direction, then

P1

a=6

substep (c) |@ and 6 are || angles in the same direction

ea

substep (d) | ..a=6

A complete proof of a theorem P — — — Q consists in using the rules of inference to establish a logical chain of statements, which starts with the hypothesis and ends with the conclusion. From P>r—-s—-t—@Q

follows

P > Q.

* In a conditional if p, then q the if-part is also called the antecedent (what comes first) and the then-part is the consequent (what follows). These names refer to the logical sequence of the two

sentences,

of'course.

In the conditional

“x is positive,

if

x > 0” the sentence x> 0 is still the antecedent (hypothesis), although it is written after the consequent (conclusion). * The sentence if p, then q is neither true nor false, since p and q are only place holders. If p and g are replaced by a meaningful antecedent and consequent, then we can decide whether the sentence is true or false. It then becomes a statement. A statement is a sentence that is either true or false.

For example, the sentence

“from p follows q” is not a statement; but the sentence “from the truth of ifp, then q and the truth of p always follows the truth of g” is a true statement. Also “four equals five” is a statement, but a false one. 5. The converse of a true theorem (hypothesis and conclusion interchanged; review page 112) is a different theorem which may or may not be true; therefore it needs a separate proof. Each of the two theorems is the converse of the other. If one of them has been proved, it may be used as one of the reasons in the proof of the other.

But be sure that it has been proved;

otherwise

you

would make the mistake of circular reasoning. In symbolic form the converse of the theorem (A) if p, then q

is another theorem (B) if p’, then q', where p’ = q and q' =p; thus (B) takes the form if q, then p. When changing from a theorem to its converse, the direction of the implication is changed. This change is symbolized either by interchanging the places of p and q or by changing the direction of the arrow: q— p and p 2.0nab): i. Draw parallels in an arbitrary direction through A and B

ii. Make AC = a=3, and BD=b=2, and BE= b=2

iii. Connect C with D to find P on AB =cp 3. @ + b* =c(g + p) =C?, g.e.d.

REASONS 1. Theorem [I-12 2. Cor. I-12-1 3. addition, from 2.

This famous Pythagorean Theorem, as simple as it sounds, is one of the most important theorems in geometry. Although it was already known to the Babylonians about 2000 B.c., the first general proof may have been given by Pythagoras. Thus, his name is traditionally connected with this theorem. Legend has it that Pythagoras was so thrilled at having found a simple proof of the theorem that he sacrificed a hundred oxen to the gods; and it has been said that ever since, oxen

have been terrified whenever a great new truth was discovered. The proof given in this section is based on similar triangles and algebraic computation. The amazing content of the theorem is brought out more clearly if you ponder its geometric meaning as shown in Figure I: the areas of the two squares with sides a and b are together exactly as big as

i

Similarity, Congruence, Symmetry

207

the area of the square with the hypotenuse as side. There is a geometric proof showing directly that these areas must be equal. You will understand this direct proof better after you have studied the Unit on areas. There are over three hundred different proofs of the Pythagorean Theorem. If you are interested in collecting and studying several of them, you could report on them as a special project. In many geometric problems it will be a great help to you to look for right triangles first, and to apply the Pythagorean Theorem whenever possible. The converse of the theorem is quite useful in proving a triangle to be a right triangle.

The Converse of the Pythagorean Theorem: If the square of one side of a triangle equals the sum of the squares of the other two sides, then the triangle is a right triangle.

Prove by constructing a right triangle with its legs equal to the two shorter sides given. Then use the Pythagorean theorem to compute the third side, and use s.s.s. to prove the two triangles congruent. How could you formulate the Pythagorean Theorem and its converse as one statement? Corollary !i-13-1: If the hypotenuse and one leg of a right triangle are equal to the hypotenuse and one leg of another right triangle, the two triangles are congruent.

Why?

(Find the other leg, and then use S685)

Corollary Il-13-2: The shortest distance of an external point P from a straight line (or from a plane) is the distance of P from the foot of the perpendicular through P to the line (the plane).

208

Unit 2

Why? (Compare this distance with other distances and look for a right triangle.) Three integers which could be length measurements of a right triangle are called Pythagorean Numbers, because they comply with the Pythagorean Theorem. A right triangle, whose side measures are integers (in any one unit), is called a Pythagorean Triangle. The triple (3,4,5) is a set of Pythagorean numbers, because 32 + 42 = 57: Are there many more?

so is (5,12,13), because Can you name any?

57 + 12? = 137.

Obviously, if you know one Pythagorean triple you can form more through multiplying by a factor. E.g. (6,8,10) = (2 x 3, 20e-4, 25) and (22x 37) + (2* x4") = 27G2 4) (on) ee general, if (a,b,c) is Pythagorean, then (ka, kb, kc) is also Pythago-

rean,-because k?a? +: k*b? = k?(a? + b?) = k’c’?. How can we find Pythagorean number triples that are not just multiples of the two triples above? Verify these two procedures: (1) If a is an odd integer greater than 1, then {a, b =4(a? — i),

c = 3(a? + 1)} is a Pythagorean triple. (2) If m and n are relative prime integers (have no common factor) and m > n, then (a = m? — n?, b = 2mn, c = m? +n’) is a Py-

thagorean triple. Why, in the first procedure, must a be an odd integer and greater than 1? Why, in the second procedure, must m > n? What happens if m and n are not relative prime? Find Pythagorean triples by continuing the following tables: a, b= 3(a@—1),c=3(@4+1) a=m—n?, b=2mn, c=m+rn? a> 1 and odd m > n, relative prime lye = 6 Th a eee We 9 11 13

G48 bee ES»

5

Die 5

OR) 4 5

i aes’ 1 3 1 4 3 2

3. 4" Sei 8 36 i 24 LO eS)

5 els 40 oe Ly

How much is (c — D) in

How much is (c — a) in

this procedure?

this procedure?

Similarity, Congruence, Symmetry

209

A generalization of Pythagorean number triples is considered in a famous theorem by Fermat (1637) about triples of integers (x,y,z) such that x” + y" = z”, where n > 2. Are there such integers? E.g. x*+ y?= 23 or x14 + y!1= 711, Fermat wrote in the margin of a book a note in Latin that it was impossible to find such integers. The note continued: “... and I have found a truly remarkable proof, but this margin is too narrow to contain it.” Ever since, mathematicians have tried to prove this theorem in its general

form, but so far without success, except for special cases of n. EXERCISES (See Supplement V on computations with roots.)

1. If the side of a square is 3 units (5 inches, s inches), how long is the diagonal? 2. If the diagonal of a square is 8 in. (10 in., d in.), how long is the side? 3. If the side of a cube is 3 in. (5 in., s in.), how long is a face diagonal? How long is the diagonal of the cube?

4. If the diagonal of a cube is 8 in. (10 in., D in.), how long is the side? How long is the diagonal of a second cube whose side is twice that of the first cube? 5. If the legs of a right triangle are 3 and 4 in. long, how long is the hypotenuse? How long are the segments of the hypotenuse formed by the altitude? How long is the altitude?

6. Repeat Exercise 5 for: a=2 |5 |4 | Seis B= S847 (12125 7. Given a segment uv as a length unit; how could you construct a segment of length V2, V5, V10, V17 in this unit? (Consider such a segment as hypotenuse of some right triangle; how long are the legs drawn to a hypotenuse of the required length?)

8. The altitude to the hypotenuse of a right triangle is 3 in. and one segment p of the hypotenuse is 4 in., how long are the three sides of the triangle?

9. Repeat Exercise 8 for

h= 4 |5 |2 p=31514

10. If the two segments of the hypotenuse of a right triangle are p and q, how long are the three sides of the right triangle? p=3 ie qg=41115

210

Unit 2

11. Given one leg of a right triangle and the ae how long =Be 5 Ne are the other sides of the right triangle? on 12. Given one leg of a right triangle and meRata segment of the hypotenuse, how long are the other sides of the tri-

angle?

Ks Seve

13: If the aoe of a rectangular solid are a, b, c, how long is the diagonal? How long are the diagonals of each of the faces? 14. How could you use a rope with four knots in it, to lay out a right angle in a garden? How far apart would you place the knots? 15. In the congruence and similarity theorems three given parts were needed to determine a triangle. In the above exercises right triangles were determined by only two given parts. Is this a contradiction? 16. Compute the height of an isosceles trapezoid with bases 10 and 4 in. and legs 5 in. (The perpendiculars to the longer base from the endpoints of the shorter base form a useful rectangle.) Li: In the . figure

to

the

right,

PF 1 AB and AF > BF. Prove AP > BP. 18. a. Prove: If two sides of a triangle are unequal, then the

foot of the altitude to the

)

a

third side forms unequal segments with the vertices on that side; the larger of the segments is adjacent to the larger of the two sides. b. Formulate the converse of a and prove it.

Exot

19. In the figure to the right, AM=MB, MD | AC, ME 1. BC.

a. if MD

18)

rs

©

Prove:

=ME,

D

then AD =BE

m

b. if MD < ME, then AD > BE 20. Prove the converse of the Pythagorean Theorem.

Ex. 19

Similarity, Congruence, Symmetry

211

21. Prove Cor. II-13-1.

22. Prove Cor. II-13-2.

23. Prove: In a right triangle the side opposite the right angle is the longest. 24. Prove (in the figure to the right) that

OJ =30A.

1

2.4 Trigonometric Functions in a Right Triangle: An Application of Similar Triangles LT] ey

;

i oa

im

I

Figure I shows various right triangles having the same angle a. How big are the other two angles in each triangle? From Cor. II-10-1 we know that all these triangles must be similar. If you are told that x.@ of a right triangle is 45°, do you know the shape of that triangle? What do we mean by “shape”? Although the size of the similar triangles in Figure I is different, the ratio of any two sides in one right triangle is the same as that of the corresponding two sides in any of the other right triangles. These fixed ratios have special names:

212

Unit 2

; F sine a=sina=

opposite leg

hypotenuse adjacent leg

: cosine

@=COS

tangent

a=

@ ~

opposite

hypotenuse

tan a=

leg

opposite leg

Adjacent leg

adjacent

leg

The reciprocal ratios also have special names: 1 i ] cosecant a@ = =— secant a = » cotangent a = sin @ COS @ tan a

We shall use the first three ratios only. ing relationships: }

sin @

na= é cos a sin?at+cos?a=1'!

in

sin

Try to derive the follow-

(90°

—a)=cos

a

cos (90°—a)=sina

What are some values of these ratios?

For instance, what is the

value of sin 45°?

In Figure II,

the square of side s is divided by a diagonal into two congruent triangles,

whose

sides

and

a 2

we know. Using the Pythagorean theorem we can show that the length of the diagonal is sV2. Then

S

angles

Ase

S =

ee ee een D, sV/2 WW)

TT]

(Le

tao

Is it true that sin? 45° + cos? 45° = 1? This fact can be demonstrated better by using the \/2 expression rather than the decimal. Why? If a is different from 45°, the numerical values for sin a, cos a,

and tan @ will naturally be different. In Table I (see page 575) these values are listed to the first four decimal places. (Only rarely are these numbers integers or even rational.) Note the construc1 The square of sin a is customarily written as sin? a.

Similarity, Congruence, Symmetry

213

tion of the table and the range of values of each ratio.

How

do you find sin 75°, cos 75°, tan 60°? How do you find a if sin a = 0.6018? (Usually this is written as .6018.) The ratios sin, cos, and tan and their numerical values are very

useful for solving triangles. Solving a triangle means computing all sides and angles that are not given. It is from this practical application that these ratios were given the name trigonometric functions. (tri = three, gonia = angle, trigon = triangle, metrein = to measure, trigonometric = triangle measuring.)

Solution of right triangles: Given:

a, b; then tana = = B=90

— a,

c=b-cosa+a-cos B= Va’+ Db’ (Draw the altitude h.)

a a, a, then b=——-; c and B tan a

as above c,a;

a thenb=?, cos B=—; ¢

O b

a

a=90° —B ea.

then b= Cc Cos a;

C

B=90° —a, a=c cos B= GaSiuOa NG =D" The trigonometric functions can be applied to a general triangle by considering it to be divided into two right triangles by one of its altitudes. In Figure III 4 sin a =>

and

h,=bsina=a

h sin B= oe then

b

a

sin B; or

a:b=sin a:sin B.

Use h, to show that b:c = sin B:sin y; and use h, to show that

a:c = sin a: Sin y.

Wl

214

Unit 2

*In case X a is obtuse the altitude h, falls outside the triangle and one of the right triangles contains the supplementary angle of a instead of a. But sin (180°—q) =sina, as you will learn in Unit 3, Section 3.

Law of Sines: The ratio between two sides of a triangle equals the ratio between the sines of the opposite angles.

a:b:c=sin a: sin B:sin y

* Theorem Il-14:

Theorem of Ceva (\st version) (Ceva, 17th cent.) Three straight lines through the vertices of a triangle are concurrent if and only if they divide the vertex angles such that the ratios between the sines of the angles formed satisfy, in cyclic order, the following condition:

“Cyclic order” means in the order indicated by a circular arrow (see Figure IV).

“If and only if” means there is a theorem and its

converse. We shall give the proof of one and leave the proof of the other to you; the proof of a converse is simpler because it may make use of the already proved theorem. Proof of: Three straight lines from some point P through the vertices divide a, B, y such that the condition holds.

Similarity, Congruence, Symmetry

1. Drop Ls P, DEP. Pie Sin RS‘ 1

my

215

from P to a, b, c fry

rs

=~7p and sin a. = aE

sin a Ch ee ae (internal ratios are marked negative; SIN Ap P was assumed inside the /\) g

"

4. In cyclic order also =F? EE tesed mM Be

5. Multiply STUMPS -—— SIN Q@,

UW

Sin B,

et

wp

ace

SIN y2

P

we Pp

CFL 2 (-En\-ale z| SiN yp

=—],

g.e.d.

Note that in step 3 every point P on the same line AP leads to the same negative ratio, and that another line AP passing through the interior of x _A leads to another negative ratio. If AP does not pass through the interior of the given angle, the ratio of the perpendiculars to the sides b and c is considered positive. Compare Section 2.2, internal and external ratios with respect to a given segment. Combining this theorem following form:

with the Law

of Sines leads to the

*Theorem 11-45: Theorem of Ceva (2nd version) Three straight lines through the vertices of a triangle are concurrent if and only if they divide the opposite sides such that the ratios between the segments formed satisfy, in cyclic order, the following condition:

Proof of: If these lines pass through some point P, then the condition holds. AI _ sin B

ad;

Jesinia,

1. law of sines

216

Unit 2

Al — sin eek eel a, sina, 3. Ala

a,

; 2. law of sines

sin B sin a,

sina,

=a,

sin y sin a,

siny

3. from 1. and 2. al qhxoyan 3),

5. repeat 1-4. cyclically

6.

2 =?

-

6. repeat 1-4. cyclically 7. multiply 4., 5., and 6. ques.

2

8. by first version of Ceva’s theorem

Ceva’s theorem provides a very useful method of finding out from certain ratios whether three straight lines are concurrent. In a previous Unit you derived a method of finding out from certain ratios whether three points are collinear; what theorem provided that method? What does the sign of —1 or +1 in these theorems indicate? Why is it important to watch the cyclic order in these theorems? EXERCISES 1. From the Table of Values of Trigonometric Functions find the numerical values of: sin 20°, cos 30°, tan 15°, sin 60°, cos 65°, tan 50°.

2. Find a if sin a= .6561 (.6820, .7547, .9976) cos a= .9986 (.8192, .7771, .6018) tan a = .7813 (.2867, 1.0000, .3443)

Similarity, Congruence, Symmetry

217

3. Without looking at the Table, compute cos a and tan a, if sin a= .42262 (.69466, .58779). Table to check your answers.

After finishing, use

the

4. Solve a right triangle from

a. c= 30 in., a= 19° b. a= 70.3 in., b= 59 in. c. a= 225 in., B=37° d. a= 50 in., a= 25° i. c= 26 in.,

e. c= 41 in., B= 50° f. b= 168 in.,.a= 85° g. c= 16 in., a= 13.11 in. h. b = 87 in., B=75° b= 10 in.

5. If a diagonal of a rectangle is 10 inches long and forms a 40° angle with the other diagonal, how long are the sides of the rectangle?

6. If a 23 ft pole casts a 42 ft shadow, how high is the sun? (This means: what angle do the sun rays form with the horizontal ground?)

EXxaG6

Ex. 8

7. How high is a tower that casts a shadow of 34.3 ft, if the sun is 35° high? (Its rays form 35° with the horizontal.)

8. John sits in a tree 10 ft above ground and looks down at Mike who is lying in the grass about 17;ft away from the tree. His line of sight forms with the horizontal direction

the so-called angle of depression. If Mike looks up at John, his line of sight forms with the horizontal direction the so-

called angle of elevation.

Show that these two angles must

be equal. How big are they in this case? the nearest degree.)

(Find angles to

9. If a tower is seen under an angle of elevation of 70° from 100 ft away, how high is it? 10. A balloon, 1280 ft away from us, rises straight up. After a while we see it under an angle of elevation of 65°, and a little later 72°. How much did it rise between these two times?

11. A ship is seen under an angle of depression of 2° from a point on the shore 190 ft above sea level. How far out on the water is it?

218

Unit 2

12. If a road climbs 10 ft over a horizontal distance of 600 ft, vertical distance _ rise 10 ft the ratio of

horizontal distance

run

600 ft

is a measure

of the steepness of the road. It is called the slope. The angle between the climbing road and the horizontal is called the angle of inclination. The slope is the tangent of the angle of inclination. How big approximately is the angle of inclination of this particular road?

13: A ladder of 25 ft length leans against the wall of a house with its foot 16 ft away. What is its approximate angle of inclination with the ground (to the nearest degree)? What is the slope of this ladder? 14. From the top of a tower the foot and top of a 14 m high flag pole are seen under depression angles of 28° and 22°; how high is the tower and how far away is the flag pole? (Make a sketch and find right triangles. Then form two equations in two unknowns; see Supplement III-E and B.) 15. To

determine the altitude of a moun-

tain peak, a 113 m long horizontal

base line is laid out

J.

ae

in the direction to143i wards the mountain Ex. 15 peak; at the ends of the base line the angles of elevation are measured as 20° and 24°. How high is the mountain? (Form two equations in two unknowns and eliminate the unknown which is not requested.)

16. How high is a tower if the ends of a 5 m long vertical pole at its top is seen from the ground under 61° and 67° elevation? 17. The center of a tower clock is seen by an observer 27 m away from the tower under an elevation angle of 433°. Under what angle does he see the hour hand of 2.2 m length at 12 o’clock and at 6 o’clock?

18. To determine the width of a river a base line of 48 ft is measured along one shore, and a pole C is set up on the other shore such that x CAB = 90°. From B the pole

Similarity, Congruence, Symmetry C is seen under 32° from the base line. river?

219

How wide is the

19. In an old manuscript (Levi ben Gershom, 14th cent.) there is a description of an instrument called Jacob’s Staff, for measuring distances or distant altitudes. A stick CD ofa given length is so fastened to a measuring rod AB that it is

perpendicular to AB and can glide along AB parallel to itself. To measure the height of a distant tower, CD is pushed to a position where AC falls into the line of sight towards the top and AD towards the foot of the tower. C Explain how the height of the tower can now be deter-

mined

if its

dis-

A

tance is known. How can the instrument be used to determine the distance from the Improvise such an instrument You can also use a pencil of arm’s length; how would this

B

tower if its height is known? for yourself and try it out. known length and hold it at help?

20. What is the law of sines? If two angles in a triangle are 30° and 70°, can you tell the ratio between any two of its sides? PA |e What are the theorems of Ceva and of Menelaus? Why are they useful? Why is there a value —1 in the one and a value +1 in the other theorem? ABC are bern aly,divided by points *22. The sides of a triangle

I, II, II] in the ratio —3, —£, —3 (or —3, —3, —%). If these points are connected with the opposite vertices, do the connecting segments pass through one point? Why are all these ratios given as negative numbers?

e205: In a triangle of sides

a= 3, b= 4, c= 6 units, point J is on a 2 units from B, point I/ is on b 2 units from C. If point P is the intersection of AJ and BIJ, where does CP intersect c? (Check your finished computations by construction.)

*24, For the triangle in Exercise 23 place another triple of points so that J is on the extended side a 1 unit beyond B, // is on the extended side b 6 units beyond A. Where does the straight line through J and // intersect side c (extended, if necessary)? (Check your finished computations by construction.)

220

1

Unit 2

2.5 Computing with Coordinates: An Application of Similar Triangles

Problem 1: If two points A and B are represented by ordered number pairs (x,,y,) and (x,,y,) within a Cartesian coordinate system, can the distance AB be computed from these numbers? Form a right triangle ABB’ (Figure I) by drawing parallels to the coordinate axes through A and B. Suppose A(1,2) and B(5,4), how

long are the legs of this triangle? How long is the

hypotenuse

AB?

Which

theorem makes this computation possible? Does the theorem require that the legs of the triangle be parallel to the coordinate axes? Why would we want them to be parallel to the axes? Triangle ABB’ was quite useful in the solution of Problem 1, and it will be useful in several other problems.

Let us call it the coordinate triangle of AB.

The Distance Formula:

The distance d between two points A(x,,y,) and

B(x,,y,) is d=V

(x, — x,

+, —y,)?

Note: Up to now we used coordinates as a means to describe the position of figures and to make graphs. In this Unit we shall use them to find geometric properties by computation.

Similarity, Congruence, Symmetry

227

EXERCISES 1. How far is the point (3,2) from the origin? 2. How long is the segment AB between the following points?

a. A(0,0), B(4,6) b. A(2,3), B(6,9) c. A(1,1); BG,7)

d. A(1,2), B(5S,8) e. A(0,0), B(3,4) f. A(1,1), B(4,5)

Note: The important thing in the formula is the difference between the coordinates, not the size of each coordinate. 3. Given a triangle ABC,

how long are its sides?

a. A(3,2), B(O,0), C(1,2) b. A(2,4), B(O,2), C(2,3) c. A(0,5), B(O,0), C(4,2) 4. The

endpoints

of the diagonal of a square are A(2,3) and

C(5,7); how long is the side AB of the square?

5. The following points are the vertices of a quadrilateral. i. Compute the perimeter of each quadrilateral. ii. How long are the diagonals of each quadrilateral?

a. A(0,0), BU,5), C(7,0), D1, —9) b. A(7,2), B(O, —9), C3, —1), D(O,4) c. A(—1,7), B(4,11), CC, —10), D6, —2)

Problem 2: If (x,,y,) and (%,y.) are the coordinates of A and B, can the coordinates (x3,y3) of a point C on AB be computed

that divides AB in a given ratio? The

coordinate

triangles

of AC and AB are similar (why?)

fore

(Figure

IL);

there-

AC: AB =AG

COBB

AB’ =

Ifthe

AC:AB =1:3

(or

B

‘ratio 2:5

or

C

y.-y

p:q), then the legs of AACC'

Aa.

must be if or 2 or A]of those

of

5) Sie ¢, AABB’, which

(a

§are

(Yo — V1)

and

Therefore

AC’ =4(x%2 — x)

and CC’=4(y2—y:). and

?)

The

two

oy):

(or ?

distances

T

222

Unit 2

AC’ and CC’ are added to the coordinates of A to give the coorj

dinates of C.

Dividing segment AB such that AC = ;(AB): If A(x,,y,) and B(x,,y,), then the coordinates of C are: x2 =e

me ety)

Vet +602 =r In the special case that -=5 the point C is the midpoint of AB: The midpoint M of AB has the coordinates: Xy

*1 Tey (Xa gia ota

ch)

Vu = V1 +32 — Yi) = 201 + Yo)

EXERCISES 1. Given en ACI, 2) and BC, >), compute the coordinates of point C on AB such that AC = 5 of AB; compute point D such that AD=+ : of AB. 2. Given 42,1) and B(10,13), divide AB into five equal parts and compute the coordinates of the dividing points. 3. Given the points A(0,0) and B(6,3), find the coordinates of a a c point C on AB which divides AB eas ae in the ratio 1:2. _Wote: | If ee AC; CB= 12+then AC = 14B.)

4. Compute the coordinates of the midpoint a.

b.

=(0,0)-3C52)) B=(4,2))

(5,4)

C:

d.

B

ee 7 2

ee

M of AB: €

23) sO Aye (13) (OS)

4C2))

43)

5. For triangle A(1,1), B(O,5), C(3,6) compute the coordinates of

the midpoints of the three sides and the perimeter of the triangle formed by them. Compare with the perimeter of AABC. 6. In triangle A(0,0), B(4,0), C(4,6) compute the length of the three medians. (A median in a triangle is the segment between a vertex and the midpoint of the opposite side.)

Similarity, Congruence, Symmetry

223

Problem 3: If two segments AB and CD are given by the coordinates of their endpoints, can we tell from these coordinates

whether AB ||CD? Suppose x, =x, and x, = x,, then we have the simple case of

vertical segments (L x-axis) and we know that AB ||CD. If non-vertical AB ||CD,

Why?

or if they are on the same carrier (Figure III), then their coordinate triangles are similar (why?); therefore

BB AB’ =DD'-CD' or

D pall Apa DeoC an 4a eG

This ratio is called the slope and usually denoted by the letter m;

it is a measure

of

steepness with respect to the direction

pressing

of the x-axis,

ex-

nT

change in y __ rise

change inx

run

The angle q@ between the direction of AB and that of the x-axis is called the angle of inclination. Note

that the ratio is meaningless

if the denominator is zero,

that is in the case of vertical segments. > The slope is the trigonometric tangent of the angle of inclination.

» If non-vertical segments AB and CD are parallel or on the same carrier, their slopes are equal.

(m, =™m,)

e Conversely, if the slopes of two segments are equal, the segments are either parallel or on the same carrier.

EXERCISES

1. What is the slope of AB and what is the angle of inclination? a.

A—A12)r B= (3,4)

b.

c.

d.

e.

f.

(1,2), (6,7)

(0,0) (4,2)

(0,0) (0,3)

(2,1) (5,6)

30) (6,4)

224

Unit 2

2. If A(x,,y,) and the slope m of AB are given, can you compute the missing coordinate in B(x,, ?)?

a. d= Cl)

i

Oe

No

aD)

b. OO

c. d. Gea)

ee

2

1

6

10

3. Without plotting can you tell whether or not AB ||CD? a.

A =(00) B=(0;3) C=(1,1) D=(1,4)

b.

Cs

@Q,)—@G22)" (3,2) 6,6) 0,4 (€,1) (5,3) (4,5)

d.

(20) G2) (0,4) ©,2)

‘4. Without plotting find out whether these four points are vertices of a parallelogram.

a. A(0,0), B(3,0), C(0,4), D(3,4) b. A(—2,3), B(2,3), C(0,4), D(2,4) c. A(0,0), B(S,1), C(1,3), D(6,4) 5. What point D would make A(1,1), B(6,2), C(2,4) a parallelogram? Is there more than one solution?

Problem 4: If two segments AB and CD are given by the coordinates

of1their endpoints,

can

we

tell from

these

coordinates

whether AB | CD? If one of the segments is vertical, what can you say about the other? What can you say about the coordinates of their endpoints? If none of the segments is

vertical,

then

x BAB’ =

y

x DCC’ (Figure IV); the coordinate triangles are similar, but corresponding sides are not parallel as in the previous problem.

The slope of AB is 7 B

A

The slope of CD is See

ae D

IV G

Similarity, Congruence, Symmetry

225

From the similarity of the triangles it follows that Ya

V4 _ pa

Va At

Xe.

Con OD

Thee eee | —— —

Then

Yo™

Xp

ee),

[UN —|

me

a=

We

ois

> If AB | CD and neither is vertical, the slopes are negative reciprocals of each other.

("= ~)

Formulate the converse and prove it.

EXERCISES 1. Find the slope of CD 1 AB, if the slope of AB is: a. 3 Caae e. —5

b. 4

d. 2

f.

2. If the coordinates of A, B, C are given, find the coordinates of

some point D such that CD 1 AB. a. A(0,4), B(3,5), C(0,0) 3. Intriangle 4(0,0), B(2,1),C

Problem 5: P,(%1,y1)

and

If two

b. A(0,0), B(4,6), C(U,0) (1,3) find the slopes of the altitudes.

points

P(x2,y2)

are

given by their coordinates, can we tell whether another point P(x,y) belongs to the straight line through P, and PS? Obviously, the coordinates

of any point lying on PP, must satisfy some condition with respect to the two given pairs of coordinates. This condition is called the equa-

tion of P,P2-

V

226

Unit 2

Form the coordinate triangles for various segments P,P’, P,P", P,P (Figure V), where P lies on P,P, but P’ and P” do not. The ) pp angle of inclination at P, is equal to that of P,P, if and only if P,P lies on P,P,. In the language of coordinates this means that: VSN

ae

ENE oh

i eee If the coordinates of some point P(x,y) satisfy this condition, then we know that P belongs to P,P,.

A straight line may be given by two of its points or by one point and its direction (expressed as slope m or as tan w). Considering these possibilities explain the following equations: A straight line

1. given by P, and P,

Equation Moy

1. BWSR ets 2

2. given by P, and m 3. given m and intersection y-axis

oy

wah CieBS, il

2. Y= = mes x) with | 3. y=mx+b, where b is called the y-intercept (slope-intercept form) 4. through origin in given direction | 4. y= mx 5. y-axis 5. x=0

6. ||to y-axis in distance a

6. x=a

7. X-axis

7.

y=0

8. ||to x-axis in distance b

8.

y=b

9. general form

9. Ax+ By+C=0

Because these equations represent straight lines they are called linear equations. They are first degree equations in x and y. EXERCISES 1. Write the equation of a straight line passing through two points A and B: a b. c. d e. f. g A=(2,3) (2,3) 23) (2,3) (0,0) (—3,0) (-3,—2) B=(4,5) (—45) (45-5) 4.5) G3) Of ly G3) 2. Write the equation of a straight line of slope m passing through point P,: d. c b. a. $ 1 3 m=3 P, = (2,2) ee) (5) C4) 3. Without plotting find out whether any of the given points lies on any of the given lines: A(0,0), B(O,3), C(—2,—5)

Similarity, Congruence, Symmetry a. 2x+ 3y=9 b.a2x— 3y=5

227

c. x+y=0 d. y=4x+3

4. The vertices of a triangle are A(0,0), B(4,0), C(1,4). the equations of the three sides.

Write

5. Write the equation of the straight line through P(1,1), parallel

to a given straight line:

a. y=2x+3

be y=3xt2

« y=-ix+5

dad y-x=1

6. Write the equation of a straight line through P(1,2), perpendicular to a given straight line:

a y=2e4+8

bey=x

c. y=—ix

d. y+x=2

7. Find slope-intercept forms equivalent to the following general equations: a. 2x+4y=5 b. 2y—x=3 ec. 4y+x—-6=0 d. y—3x=5

Problem 6: If two straight lines are given, what are the coordinates of their common point? The coordinates of a common point must satisfy both equations. (Review, how to solve two simultaneous equations; see Supplement III.) Two straight lines in the same plane have exactly one point in common, unless they are parallel. How can you tell from the two equations whether they represent parallel lines? How can you tell whether two equations represent the same straight line? EXERCISES 1. Compute the intersection of a. y=3x+4 b. 4x + 3y =2 y= 5x 2x + 5y=4

c. y—4x=5 Sk +2y—3=0

2. Compute the intersection of straight line s with a perpendicular to it through point P(3,4). The equation for s is a. 3x +2y—1=0

b. 2x -—y+3=0

3. For triangle A(0,0), B(4,0), C(3,5) compute of the feet of the three altitudes.

c y=—4x—-1 the coordinates

4. Through each vertex of triangle A(0,0), B(3,0),

C(1,4) pass a

straight line parallel to the opposite side. Determine the coordinates of the vertices of the new triangle formed by these three lines. 5. Show that the three perpendicular bisectors of the sides of triangle A(—2,0), B(2,0), C(1,4) are concurrent.

228

Unit 2

The analytic method: to compute

In the six problems we used coordinates

a distance, a ratio, a direction, a point.

We can take

advantage of such computations to prove a geometric statement. This method, called the analytic method, consists of first translating

the geometric concepts of point, midpoint, line, parallel, perpendicular, etc. into the language of ordered number pairs and equations between them, then solving equations, and then translating the answer back into geometric concepts. In contrast, the method of using the geometric concepts throughout the problem (as we do in this course) is called the synthetic method.

Example: In any triangle ABC the segment between the midpoints of two sides is parallel to the third and half as long. You have proved _ this statement before (as an exer-

cise in Section 2.1); you devised a synthetic proof, using similar triangles but not coordinates. Now we shall give an analytic proof: First we translate the problem into the analytic language of coordinates. Let the coordinates of the vertices of AABC be

A(x,,y,),

B,¥2),

C(%3s¥3).

Then the coordinates of M, are [$(x, + x3), (7, + ys)], and the coordinates of M, are res + x3), 302 + ys)]. Now we prove

UE ||AB by proving that the slopes of these two segments are equa

slope of AB = 22-1, X_ — en

+Ys) slope of M,M, = 2a + Ys) —31 a Xo

Xs) met)

ony Gs mee Xetace eet ees Vem) ¢

Similarity, Congruence, Symmetry

Lastly we prove that

229

M,M, =4AB by calculating the distances in

question.

distance from A to B = V(y, — y,)? + (% — x,)? distance MM,

may e@e =

ate tye)

G)* 2 ays

=

yy

[ay tite)

V2). at (a): (CBS

OR ey

2 t tg) ate)

WOK = Snare C= eee

Choice of coordinate system: To solve the geometric problem by the analytic method we had to impose a coordinate system so that the vertices could be translated into the language of ordered number pairs. The choice of this coordinate system is entirely arbitrary; it does not affect the shape of the triangle nor the property to be proved. We could have used system O}.0;>..0,,.0G)

many

others

(Figure VI). If we had placed the origin at vertex A and the x-axis through AB, the computations would have been simpler. Try to choose a coordinate system that simplifies the computations. Let us repeat the problem

Vl

for coordinate system O,: A(0,0), B(x2,0), C(x3,ys) M,(GXxs,

33)

M, [ae

+ Xz), 33] 1

1

— —— 2)3— 2) slope of AB = 0, slope of M,M, = Setar a 0 3X2

“. M,M, ||AB

distance AB = xX», distance M,M, = V/(0)? + (4X3)? = 3X5

.. MM, =4AB

230

Unit 2 EXERCISES

Prove analytically: 1. If two sides of a quadrilateral are parallel and equal, the quadrilateral is a parallelogram. (Put the origin at one vertex A and the x-axis through one side AB; then choose C; then the coordinates of D must be ...; now prove that AD BC.)

2. If the diagonals of a quadrilateral are perpendicular to each other and also bisect each other, then the quadrilateral is a rhombus. (Put the coordinate axes through the two diagonals;

choose A and B, then C and D follow from that. that all sides are equal.)

Now prove

3. If the diagonals in Exercise 2 are also equal, then the quadrilateral is a square. (Leave A and C as in Exercise 2 and change B and D. What is the difference between a square and a rhombus? If you proved Exercise 2, then you need only a proof for the additional property.)

4. In a right triangle the midpoint of the hypotenuse is equidistant from the three vertices. (Choose the coordinate axes through the legs of the right triangle.) 5. The diagonals of a rectangle are equal and bisect each other.

6. The median of a trapezoid is parallel to the bases and equals half their sum.

3

CONGRUENCE

AND BILATERAL

SYMMETRY

3.1 Direct and Inverse Congruence Figure I shows two right triangles ABC and AB'C with one pair of legs coinciding in b and the other pair on the same straight line, of equal length a, but in opposite directions. According to the s.a.s. theorem each of the two triangles is congruent to the same right triangle A”B”C” with legs a and.b, and they are congruent

Similarity, Congruence, Symmetry

to each other. Yet they look different, as they point in opposite directions. The vertices of AABC are named in positive (counterclockwise) order as indicated by the cy-

231

(oll

clic arrow (arrow in a circle);

the order of the corresponding points AB'C is negative (clockwise) as indicated by the cyclic arrow in the opposite direction. We say: the triangles are symmetric with respect to s, the carrier of the common side b; or we say: they are reflections of each other with respect to s. The straight line s is called the symmetry axis.

The word reflection may remind you of a reflection in the mirror. If you look at yourself in the mirror, you do not see yourself exactly as other people see you; right and left is reversed, as you notice by details in your dress or hairdo. If you put your hand flat on the mirror, you superimpose it on its mirror image; but while you put up your right hand, the image puts up its left hand. You are standing as far in front of the mirror as the image seems to stand behind. Coming nearer seems to make the image come nearer too, and by exactly the same amount.

The mirror takes the

place of the symmetry axis in the two-dimensional case of the triangle in Figure I. The right and left shoe of a pair, the gloves of a pair, your hands, etc. cannot be made to coincide, but they can be brought into a relative position, where they are reflections of each other with respect to a symmetry plane. Another name for this relationship between the two sides of an. axis (two-dimensional case) or between the two sides of a plane (three-dimensional case) is bilateral symmetry. The word bilateral distinguishes this kind of symmetry from other kinds, which will be discussed in Unit 3. If A B'BC in Figure I is considered as one triangle, then its reflection with respect to s is the triangle itself. We say that this triangle has a symmetry axis, that it is bilaterally symmetric, although it may not be reflected into itself with respect to some other symmetry axis (e.g. one passing outside the figure).

232

Unit 2 —————

ee

Definition lI-5: If an axis s or a plane > can be found, with respect to which a figure or object is its own reflection, then it is called a bilaterally symmetric figure or object.

Give some examples of bilaterally symmetric figures and objects. The study of such geometric figures obviously can be simplified by investigating only half of them. Bilateral symmetry is a special kind of congruence. If two triangles are symmetric to each other, they are also congruent, but two congruent triangles are not necessarily symmetric to each other. In Figure I AABC is symmetric to and also congruent to /A\AB'C; but AABC is congruent without being symmetric to

AA"B"C".

These two kinds of congruence are sometimes dis-

tinguished as direct and inverse congruence.

congruent to AA”B"C”

/\ABC is directly

(same cyclic order of corresponding points);

ABC is inversely congruent to AAB'C (opposite cyclic order of corresponding points); also AA”B”C” is inversely congruent to AAB'C.

Moreover,

AABC

is a reflection of AAB’C (there is a

symmetry axis, making them reflections of each other); AA”B"C” is not a reflection of AAB'C (there is no symmetry axis between them, unless one triangle is turned around), but is congruent to some triangle that is such a reflection. All this is a technical way of saying that AABC and AA"”B"C" could be made to coincide by merely moving them within the plane (directly congruent); but /\ABC and AAB'C, or AA"B"C" and AAB'C, could not be made

to coincide by merely moving, unless the paper is also folded and

flipped over (inversely congruent).

If no moving is needed or if

the figures are already in a position where folding the paper only will make them coincide, then they are reflections. EXERCISES 1. How many symmetry axes has a square, a rectangle, a circle? 2. How many symmetry planes has a cube, a regular square pyramid, a right circular cone, a sphere? 3. Draw a pictorial view of a regular octahedron and list its symmetry planes.

Similarity, Congruence, Symmetry

233

. Review the names of all quadrilaterals and group them by the number of their bilateral symmetries.

. Given any two points A and B in space; can they be considered reflections of each other with respect to a plane? Where would that plane be? . Given any two points A and Bon a sheet of paper; can they be considered reflections of each other with respect to an axis on that paper? Where would that axis be?

. Given a point A and a plane =; how would you find a point B symmetric to A with respect to 5?

. Ina plane, if a point A and a straight line s are given, where is a point B symmetric to A with respect to s? . How many symmetry axes are in each of these figures?

10. If two capital A’s are printed next to each other, then there is an axis of symmetry between them which reflects one A into the other. Which other capital letters of the printed alphabet have this property?

3.2 The Symmetry Axis; The Symmetry Plane The discussion of AB’BC in Figure I on page 231 could be repeated for AP,P;Q or AP,PQ in Figure II on the next page, if P and P’ have the same distance from the symmetry axis s (or the

symmetry plane &), and if Q is a point of s (or of X). Any such pair of points P, P’ are symmetric points with respect to s (or X). It follows that any pair of symmetric points is connected by a straight line perpendicular to the symmetry axis s (symmetry plane %); the symmetry axis s (symmetry plane %) bisects the distance between the pair of symmetric points. It also follows that the connecting straight lines (let us call them connectors) must always be parallel to each other. Why?

234

Unit 2

eas

ae

gi

P, a 7

p!

es

ie

=

is / if Pp, ————

|

PS

P,

It further follows that any point Q on the symmetry axis s (or symmetry plane >) is symmetric to itself (is a reflection of itself); and conversely, any point that coincides with its symmetric point must lie on the symmetry axis (or symmetry plane). Why? (If it did not, then there would be a connector . . .)

If a straight line p intersects s in a point Q, then the symmetric straight line p’ must intersect s in Q also. Why?

The distance P,Q must be equal to the distance Pi{Q.

Why?

(Use corresp. parts in congruent triangles.) Since the same reasoning holds for any point Q on s (or 3), and only for points on s (or 2), we may say that the symmetry axis (symmetry plane) is the set of all points equidistant from a pair of symmetric points; M as the midpoint between P and P’ obviously belongs to the set. Stressing the fact that s (or ) is also perpendicular to PP’, it is called the perpendicular bisector of PP’. &> The geometric locus of all points equidistant from two given points P and P’ is the perpendicular bisector of the segment PP’. (a straight line in a two-dim. problem, a plane in a three-dim. problem)

The construction of a perpendicular bisector in the plane follows from its definition as a symmetry axis. Given A and B (see Figure III), we need two points, R and S, that are equidistant from A and B. (Two points determine the straight line through them.) The requirement for R, for example, is only that it shall have the same distance from A and B, but it is unimportant how big that

Similarity, Congruence, Symmetry

235

distance is; so we are free to

choose tance

some r.

The

arbitrary

dis-

set (locus) of

all points at a distance r from A is a circle with A as center and r as radius. The set (locus) of all points at a distance r from B is a circle around B with radius r. The set of all points at distance r from both, A and B, must then be the set of common elements of the two circles, which in general is a

set of two points; these two points are sufficient to determine the symmetry axis we are looking for. Is it necessary to draw both circles in full? Is the choice of r really quite unimportant or do you have to observe certain limits? Could it happen that instead of two points R and S$ you might get only one or none? When? (Think of A, B, R as a triangle.)

The idea of this construction can be applied to the following three construction problems:

x A

pea

B

x (a)

(Cc)

(a) How to construct the midpoint of a given segment AB:

Looking at AB you can imagine where approximately the midpoint M is going to be. Open your compasses from A to more than half of AB, and with that radius describe parts of circles about A and about B where you approximately

expect to intersect the symmetry axis of AB on either side of AB; connect these points of intersection and mark point M. (b) How to construct a perpendicular to a given straight line at a given point M of that straight line: With compasses at M mark two points on the given line

236

Unit 2

equidistant from M; enlarge the opening of the compasses and draw with this radius parts of circles about each of the two new points, to intersect where the symmetry axis is expected; connect this point of intersection with M. One intersection

point is enough,

but the second intersection

point on the other side of the given line could be used to insure a more accurate construction. (c) How to construct a perpendicular to a given straight line from an external point P: With compasses at P mark two points on the given line equidistant from P; with the same opening as radius draw parts of circles about the new points, to intersect on the other side of the given line; connect the point of intersection with P. If the symmetry axis s (or the symmetry plane >) is given, it is easy to find the corresponding point, line, plane to any given point, line, plane by remembering these facts: 1. The connectors ElMtOns (ll atOm> 2. Corresponding 3. Corresponding 4. Corresponding 5. Corresponding

between

a pair of corresponding points are

points lie on opposite sides of s (&). points are equidistant from s (>). lines intersect at s (2). Why? planes intersect at >. Why? EXERCISES

(Use drafting aids.) 1. Given a straight line s and a segment AB on one side of s; draw segment A’B’ symmetric to AB with respect tos. (Find the symmetric points A’ and B’, and connect them.) 2. Draw the figure symmetric to the polygon ABCDE spect to a given axis s, outside of the polygon.

with re-

3. Repeat Exercise 2 for an axis ¢ passing through AD. 4. Draw the body symmetric to the wedge ABCDEF with respect to the xz-plane. Repeat for the xy-plane. Repeat for the yz-plane. Repeat for the ADFE-plane. 5. What are the coordinates of points symmetric to A(3,4,5), B(O,5,2), C(0,0,6), D(0,0,0) with respect to the xy-plane; the xz-plane; the yz-plane?

Similarity, Congruence, Symmetry

237

s t

A

B

E

D

pe C A

X B Ext 1

Ex. 2 and 3

Ex.4

. Draw a triangle symmetric to triangle A(—3,0), B(3,0), C(0,5) with respect to the x-axis; to a straight line s through

B ||y-axis; to the y-axis. . In pictorial view draw a pyramid symmetric

to the given

pyramid A(0, —3,0), B(3,0,0), C(0,3,0), D(—3,0,0), E(0,0,5) with respect to the xy-plane;

to the xz-plane;

to the yz-plane.

. Given two points A and B; how can you find a symmetry axis and a symmetry plane, with respect to which A and B are reflections of each other? . Ina bilateral symmetry, is it possible that a point corresponds to itself? If it is, where is the geometric locus of such points? Discuss the two-dimensional and the three-dimensional cases.

10. Where is the symmetry plane that makes A and B reflections of each other? a. b. Cc. d. e. A = (3,0,0) (4,0,0) (4,0,0) (4,0,0) (4,0,0) B= (-3,0,0) (—4,0,0) (—2,0,0) (0,0,0) (2,0,0)

11. Given a pair of straight lines a and a’ in a plane; is there a symmetry axis that would make a and a’ symmetric lines? 12. If two planes are given, where is a symmetry plane? Can there be more than one? 13. Given three points A, B, C; are there any points within the plane containing A, B, C which are equidistant from all three? (The locus of all points equidistant from two given points is... Points satisfying two conditions must lie . .) 14. Given three points A, B, C; are there any points in space equidistant from all three? 15; Given a straight line s and two points A and B on s; how many points are there that have a distance of 2 in. from s and also are equidistant from A and B?

238

Unit 2

Construction problems (compasses and straightedge only) 1. Given a segment AB;

construct:

a. its perpendicular bisector, b. a perpendicular at A, c. a perpendicular through an external point P.

2. Construct a right triangle given: a. two segments a, b as legs, b. one leg b and the hypotenuse c (start with the leg and the

rt. X), c. one leg b and the adjacent angle a. a (a b

b C

oOo,

+

(a)

(b)

IE, 2e

ESxteZib

3. Find by construction

a

b a ie area

(c) Exxeeaic

all points (how many?) which are at a

given distance d from a given straight line s and which are also equidistant from two given points A and B, if d these points are ons; if these points are not ons. 4. Given three points not on a straight line; find by construction all points (how many?) equidistant from the three given points. What happens if the three given points are collinear?

3.3.

SYMMETRIC TRIANGLES: APPLICATIONS OF BILATERAL 3.3.1 The Isosceles Triangle

SYMMETRY

A triangle with two equal sides is called an isosceles triangle. Is it a symmetric triangle? That is, can you find a symmetry axis, in respect to which it is its own reflection? The fact of the two equal sides gives a hint of where to look: if they are to be reflections of each other, their endpoints A, B must be a pair of symmetric points, connected by a segment, whose perpendicular bisector is the symmetry axis. The common point C, being equidistant from the pair (A,B) must correspond to itself, and therefore

lies on that axis. What other properties can we deduce from knowing that two sides are equal?

Similarity, Congruence, Symmetry

239

Theorem Ii-16: In an isosceles triangle the base angles are equal. The angle opposite the base is bisected by the perpendicular bisector of the base. (The angle bisector coincides with the perpendicular bisector of the base.)

ke Hyp.:

AABC with

Con.:

a=

a=b

b

G

bisector of xC coincides with

1 bisector of AB A

M | ls

B

PROOF STATEMENTS . draw L bisector s of AB

REASONS 1. make AM=MB; xM=90°

eAC —BC G=D)

2. hyp.

(construction)

eC. Ons

AAMC = ABMC .a=B q.e.d. RWW An -¥,=%%, @.e.d.

3. s is the set of all points...

4. s.a.s., from 1. and identity MC 5. corresp. parts in = As 6. corresp. parts in = As

Theorem IJ-16 gives a method to construct an angle bisector: Construct an isosceles triangle with given xA as vertex angle (Figure I); make AP = AP’;

find the | bisector of PP’, of which

240

Unit 2

you already have one point A; find a second point B on the other side of PP’ from A by intersecting a circle about P with a circle about P’ of the same radius;

this line is the bisector

connect AB;

of XA. j Choose any point Q on the angle bisector of angle a and drop a perpendicular onto the sides a and b; then QP =QP’. Why? (Figure II, page 239) » The angle bisector is the locus of all points equidistant from the sides of that angle. Why? » The angle bisector is the symmetry axis of that angle. Why? What additional segment would help to demonstrate the correctness of this statement?

> Symmetric straight lines form equal angles with their symmetry axis. Why?

Theorem II-17: If two sides in a triangle are equal, the angles opposite these sides are also equal; if two sides are unequal, the opposite angles are aiso unequal, and specifically, the larger angle is opposite the larger side.

\c Hyp.:

In AABC, a=b ora>b

Con.:

a=Bora>B

b A

f

Ue

Bee A'

a

aaa

\ \s

PROOF STATEMENTS 1. ifa=b, thena=£8 2. if a # b, suppose a > b, y, = y, 3. constr. AMCA' = MCB 4. a’=8

i 2. 3. 4.

REASONS Theorem II-16 construct x bisector reflection with respect to s symmetric angles

‘

Similarity, Congruence, Symmetry a. Qe-Q"

6. a> B, q.e.d.

241

5. ais exterior x of AAA’'M, therefore > one non-adjacent interior X. 6. from 5. and 4.

Converse of Il-17: If two angles in a triangle are equal, the opposite sides are equal; if two angles are unequal, the opposite sides are unequal, and specifically, the larger side is Opposite the larger angle.

(Prove indirectly:

suppose the conclusion were not true,

(hens)

Theorem II-18: The interior and the exterior angle bisectors in a triangle divide the opposite side internally and externally in the ratio of the adjacent sides.

Se (a) Internally Hyp.:

In AABC, y, =y,

Con.: c,:c,=b:a Plan:

find ~As

Ae!

Sec)

Ce

B

242

Unit 2 PROOF

STATEMENTS 1. extend BC, so that CD=b

REASONS 1. constr.

ADAC isosceles 2 Vt Ve hl ae 3. ¥,—¥y, and 4.1 = 42 4. 7,=41

2. exterior X of A 3. hyp. and isosceles A 4. from 2. and 3.

SCS, ||DA

5. y, and x 1 are alternate int. xs

6. make CE ||c,

6. by constr.

ii Ck

qe

PeaCS Be DEC 9G DC, 0,5 Ged.

8. corresp. sides ||or coincident 9. ratio of similitude

PAS CE

(b) Externally

Use the same idea to prove S$,A:S,B=b:a. show AAFC

(Make AF ||S,C;

isosceles, CF = b.)

Converse of II-18: If a straight line through the vertex of a triangle divides the opposite side of the triangle internally or externally in the ratio of the respectively adjacent sides, it bisects the interior or exterior vertex angle.

(Prove indirectly.)

Theorem I-18 gives a method to divide a given segment internally and externally in a given ratio: Consider the given segment AB as base of a triangle; draw the other two triangle sides in the given ratio (intersection of two circles with radii in that ratio);

bisect the

vertex angle; intersect the base line with the angle bisector and its perpendicular to find C and D.

x

Prove this construction. Why are the internal and external angle bisectors perpendicular to each other? What other construction

Similarity, Congruence, Symmetry

243

for that problem did we learn earlier? What is a harmonic division? The construction using I-18 requires that the two sides of the triangle be chosen in the given ratio, but their actual length is arbi-

trary. Thus many triangles could serve the purpose; where are the vertices of all these triangles? Draw some of them and make a good guess. Ina later chapter we shall discuss this locus. What happens if the given ratio is 1:1? EXERCISES 1. Construct the angle bisector of a straight angle; with the construction of a perpendicular bisector.

compare

2. Use the angle bisector to construct a right angle, 45°, 225°. (Start with a straight angle and keep bisecting.) 3. In an isosceles triangle construct the angle bisectors of the base angles; where do they intersect? Why? 4. In an isosceles triangle construct the perpendicular bisectors of the two equal sides; where do they intersect? Why?

5. In an isosceles triangle the three angle bisectors are concurrent (pass through one and the same point). Why? The point of intersection is equidistant from the three sides. Why? 6. In an isosceles triangle the three perpendicular bisectors of the sides are concurrent. Why? The point of intersection is equidistant from the three vertices. Why? 7. How big are the base angles of an isosceles triangle with a vertex angle of 90°? If the base is taken as symmetry axis, what does this triangle together with its reflection form? How long is the altitude of the original triangle, if the equal sides are 6 inches long? C

8. Prove: The angle bisectors of two adjacent supplementary angles are perpendicular to each other.

9. Prove: In an isosceles triangle the segment between the vertex C and any point D of the base AB is shorter than a leg.

A

B

D Eas

10. Prove: The bisector of an exterior angle at the vertex of an isosceles triangle is parallel to the base.

244

Unit 2

11. An isosceles triangle ABC with vertex angle C of 36° is divided into two triangles by AD, the angle bisector of a base angle. Show that the small triangles are also isosceles. Show that D divides the leg BC.in the golden ratio. 12. a. If X41 = 42, prove AP=AQ b. If X1 > X2, prove AP ? AQ

13. If AC=BC, PD 1 AC, PE 1 BC, prove PD:PE = PA:PB for any point P on AB 14. If AP=PB, PD 1 AC, PE | BC, PD=PE, prove AABC is isosceles C C Q D

A P

z

A

P

Exli2

E B

A

P

ESXi

B

Ex. 14

15. AB = 24, AC = 30, BC = 16, X1 = X2; how big is AD? 16. Repeat Exercise 15 for dB= 6, AC=

6, BC=

AB = 16, AC = 20, BC = 12 Ab

=10, 46 —

S bC—12

17. AB= 60, AC = 40, BC =60, AD =24, XDCE=90°; how long is EA? c

Cc

B

E

A

D

B

D

A

Ex. 15,16

Ex.17

3.3.2 The Equilateral Triangle In an equilateral triangle, according to its name, all three sides

are equal; therefore we can consider it as a special case of an isosceles triangle. It follows that the base angles are equal; since

Similarity, Congruence, Symmetry

245

each of the three sides could be the base, this means that all three angles must be equal, the triangle is equiangular, Since the sum of the angles in any triangle is / 2...«, each of the angles

in

an

equilateral

triangle

MIUStathense Kes elses:

A symmetry axis divides the equilateral triangle into two congruent triangles; how

big are the angles of AAMC in Figure

III? mirenG

1a,

how much is MB? How much is h? (Use the Pythagorean theorem.)

It is very useful to remember the special measurements of half an equilateral triangle. Any 60°-30° right triangle (e.g. one of your draftsman’s triangles) may be considered half an equilateral triangle. Whatever the hypotenuse measures, the leg opposite the 30° angle measures half of it, and the other leg (the symmetry axis, the altitude in the full equilateral triangle) must be

ve ahs

IV

A 45° right triangle (your other draftsman’s triangle) may be considered as half a square of side s. The hypotenuse is the diagonal of the square and measures s\/2. (See Figure IV.)

246

Unit 2 EXERCISES

Group I

ii Is it true that all equilateral triangles are: a. equiangular? b. similar? c. congruent? . How many sides and/or angles do you have to compare in two given equilateral triangles to know whether or not they are congruent? similar? . Construct an equilateral triangle with side a = 2 inches. . Construct the following angles by starting with an equilateral triangle: a. 60° er 120° b. 30° d. 90° (60° + 60° — 30°) . How big are the angles in an equilateral triangle?

Prove your

statement.

6. How many symmetry axes are there in an equilateral triangle?

. Find by construction the point of intersection of the angle bisectors in an equilateral triangle. Can you prove whether they should intersect in one point? . Find by construction the point of intersection of the 1 bisectors of the sides of the same triangle used in Exercise 6. Should they intersect in one point? . Prove that the solutions for Exercises 7 and 8 must coincide.

10. The equilateral triangle is used in the following con-

structions of Exercises 10 and 11: Explain and repeat the construction of a_ perpendicular to a given straight line s at a given poini P: 1. © about P with arbitrary

Iv a Im

I

radius r

2. intersects s at J 3. © about J with same radius r, intersects at //

s P

I

4. © about // with same radius r intersects at IJ], make arc

at probable place of /V 5. © about //I with same radius r intersects arc at IV 6. connect IV with P 11. Explain and repeat the construction of a parallel to a given straight line s:

Similarity, Congruence, Symmetry 1. semi ©

247

about

some point P of Ss, with arbitrary radius r 2. intersects s at A and B 3. © about A with

arene

oe

B

same radius r intersects semi © at /

4. repeat 3. for B, gives I/ 5. connect J with IJ

12. Explain and repeat the construction of a parallel to a given straight line s through a given point Q: 1. © about some point P of s with

r=PO

2. intersects and B

Q

“I

s at A

3. © about B with

r=AQ intersects at I] 4. connect

Q with

P fe

e

II This construction uses symmetry but not the equilateral triangle; what is the difference between this construction and the one in Exercise 11.

13. Using the Pythagorean theorem compute the sides of a 60°30° right triangle (half an equilateral triangle), if: a. the hypotenuse s is given (s=4, s=5, s=10, s in general). b. the leg a adjacent to the 60° angle is given (a= 2, a in general). c. the leg h adjacent to the 30° angle is given (h=6, h in general).

a=3, h=S,

14. Using the Pythagorean theorem, compute the sides of a 45°45° right triangle (half a square), if: a. one leg s is given (s= 1, s=5, s in general). b. the hypotenuse d is given (d= 2, d=5S, d in general). 15. Why are there only two cases in Exercise 14 as compared to three cases in Exercise 13? Give a proof for your statement. 16. In Exercises 13 and 14 the sides of the triangles are labeled s, h, d on purpose, to remind you — of what?

248

Unit 2

Group II: Using trigonometry 1. Explain and then memorize the following statements:

(i

sin 30°=5

ie

cos 60° = 5

cos 30°=5V3

C08 4S Sen meoe

tan 60° = V3

tan 30° =—==—__

Lae v3 «3

Ee

AVES

sin 60°=3V3

ieee

tan 45°=

1

2. Use Exercise 1 to compute: a. the height of an equilateral triangle of 65 in. side; b. the side of an equilateral triangle of 5 in. height; c. the side of a square with a 10 in. diagonal. Group III

1. Equilateral triangles of the same size may be arranged to fill the space between two parallels, as in the Figure. Think Exe | of triangular tiles for a floor covering along a border strip. a. Why is this possible? b. If the side of one such triangle is s, how far apart are the parallels? c. If the parallels are 3 inches apart, what size triangles may be used? 2. If equilateral tiles of the same size are arranged around one point O as in the Figure, the sixth tile will have a common side with the first one: why? A Use the definition of a circle (all points which...) to prove that a circle can be circumscribed about this hexagon. (Show that all vertices lie on a circle.) 3. In Exercise 2, what is the dis-

tance of O from AB?

Where is

the distance of a point from a line measured? Where is the other endpoint of the segment representing the distance of O

from AB?

Similarity, Congruence, Symmetry

249

4. Prove that a circle may be inscribed into the hexagon of Exercise 2. (Show that the distances of O from all sides of the hexagon are equal.) 5. Why can hexagonal tiles be used as a uniform floor covering? (Reduce to a problem about equilateral triangles.) 6. If the center O of a hexagon has a distance of 6 in. from the sides, how long is the distance of O from the vertices? (Reduce to a problem about equilateral triangles.) Group IV:

The tetrahedron

1. The regular tetrahedron has four equilateral triangles as faces (see Preparatory Chapter, Section 3). How many edges does it have? Are all edges necessarily equal? How big are the angles within each face? 2. What is the distance of the vertex D from the midpoint M of

the base edge AB, if the edge of the tetrahedron is s? Ss =7, s in general)

(s=3,

3. If the length of an edge is s, and

if N is the midpoint of DC, how long is BN and how long is AN? What kind of triangle is AABN? Does it have a symmetry axis?

*4, If M is the midpoint of AB and N is the midpoint of DC, how long is MN? (Study Exercise 3 and use the Pythagorean theorem.) Prove that MN 1 DC. (Use the converse of the Pythagorean theorem.)

3.4 Circumscribed, Inscribed, Escribed Circles Is it true for any scalene triangle (no two sides equal) that the three perpendicular bisectors of the sides intersect in one point? Each perpendicular bisector is the geometric locus of all points which... ; the point of intersection between two such bisectors is equidistant from... (how many is the locus of... ; therefore...

vertices?); the third bisector

If there is a point that is equidistant from all three vertices, then could it be considered the center of a circle passing through these three vertices? Try it by drawing one. (What is the definition of a circle?)

250

Unit 2 Theorem II-19: In any triangle the three perpendicular bisectors of the sides are concurrent. The point of intersection is the center of the circumscribed circle through the vertices.

The center of the circumscribed circle is called the circumcenter,

Theorem iI-20: In any triangle the three bisectors of the _ interior angles are concurrent. The point of intersection is the center of the inscribed circle touching the sides.

The center of the inscribed circle is called the incenter

It is important that you say interior angles in II-20, because there is another theorem involving exterior angles (supplementary adjacent to the interior angles) that you can easily prove yourself: Theorem II-21: In any triangle the bisectors of two exterior angles on one side of the triangle and the bisector of the opposite interior angle intersect in one point. This point is the center of a circle touching the triangle from the outside by touching two of the three sides in their extensions. There are three such circles for any triangle.

Ee

Similarity, Congruence, Symmetry

251

These three circles are called escribed circles; their centers are the excenters of the triangle. *The angle bisectors in a triangle are an example of three concurrent straight lines through the vertices of a triangle, therefore should be covered by Ceva’s theorem. This theorem says that for such lines certain relations hold (a) between angles, and (b) be-

tween segments.

Let us check for Theorem I-20:

(a) Each vertex is bisected: a, = a2, B, = Bo ¥1 = Yo Equal angles have equal sines: sin a, = sin ag, sin B, = sin Bo, sin y, = SiN yo. The ratio of sines at each vertex is negative for internal division: sin a,

sin By

SIN a>

:

SIL) 6p

The-product of these ratios:

sin y,

Sa

SiN y.

aly

(—1)(—1)(—1) = -1, ¢.e.d.

(b) Each side is divided in the ratio of the adjacent sides, according to Theorem IJ-18 about angle bisectors; ternal ratios are taken as negative,

Weis = — (CD).

therefore, if in-

Dy. D,=—(a:C), . €,:C,=—(D:a).

The product of these ratios: Ope

Nees

Ve

Cc

Se ee) Gre Dein C. b

a

teh) ©

|

b

a

aed.

& The theorem about the concurrency of the angle bisectors ina triangle is a special case of Ceva’s theorem.

EXERCISES 1. Construct AABC with sides 2, 3, 4 inches. Then construct the circumcenter, incenter, excenters. Check your construction by drawing special circles about these points; what should these circles show? . Repeat Exercise 1 for AABC with sides 3, 4, 5 inches.

. Repeat Exercise 1 for an isosceles triangle. . Repeat Exercise 1 for an equilateral triangle. WN. WwW b&b nan

Given three non-collinear points; construct a circle passing through these points. (Think of the circumcenter of a triangle.) Why should the three points be non-collinear?

252

Unit 2

6. Given three non-concurrent, non-parallel straight lines; there a point that is equidistant from these three lines?

is

*7, Explain why the theorem about the concurrency of the angle bisectors in a triangle is a special case of Ceva’s theorem.

*§. Show that Theorem II-20 can be derived from Ceva’s theorem;

from its 1st version, and also from its 2nd version.

9. Prove: The circumcenter of a right triangle is the midpoint of the hypotenuse. (Use ~ rt. As to find out where the L bisectors of the hypotenuse and of one leg meet.) 10. Prove: A triangle inscribed in a circle such that one side is a diameter must be a right triangle. (Use the 1 bisector of a second side and look for ~ As to prove an angle equal to art. X.) This famous theorem is known as the Theorem of Thales of Miletus (6th cent. B.c.): A triangle is a right triangle ifand only if the midpoint of one side is the circumcenter; this side is the hypotenuse.

3.5 Remarkable Points of a Triangle The fact that in any triangle the angle bisectors meet in one point is surprising, because we could not have guessed it from just looking at a triangle. The proof is relatively simple and obvious, once you understand it; but that does not make the fact any less remarkable. The same goes for the fact that the perpendicular bisectors of any triangle sides meet in one point. These points have been called remarkable points of a triangle. There are more such surprising points, e.g. the excenters, and others; but traditionally the name has been reserved for these four: circumcenter incenter

orthocenter (intersection of altitudes, orthos means right) centroid (intersection of medians)

Theorem I-22: The three altitudes in a triangle are concurrent.

Similarity, Congruence, Symmetry

253

The point of intersection is called the orthocenter.

PROOF STATEMENTS REASONS 1. Draw ||s to each side through the 1. parallel axiom opposite vertex

2. nOUABCF,AB=CF,CB=AF _ 2. opp. sides of 0= N 3. ADEF ~ AABC

in ratio 2:1

3. using 1. and 2. repeatedly gives

DE = 2AC, EF = 248,

put

FD =2CB

4. h, 1 bisector of FD 5. h,, h,, h, are ADEF

1 bisectors

4. | transversal of ||s in

6. h,, h,, h, are concurrent

5. repeat 4. 6. 1 bisectors are concurrent

Theorem II-23: The three medians in a triangle are concurrent. The point of intersection, the centroid,' divides each median in the ratio 2:1, the larger part lying toward the vertex.

A median of a triangle is the segment between a vertex and the midpoint of the opposite side. PROOF STATEMENTS 1. M,, M,, M, are midpoints of

REASONS 1. by constr.

a,b,c

2. medians m, and m,intersectinS

2. by constr.

1 The centroid is called also the center of gravity. footnote on p. 467 in Unit 5, Section 3.3.)

(See Special Project XIII, and the

254

Unit 2

TEX eM

3. M,D ||m,, M,E ||m,

eo

3. by constr.

4. AE=EM,=M,D=DB

&

. AAEM, ~ AAMC, 1:2 ADBM, ~ AM,BC, 1:2

5. AS=25M, 6. CS=2SM,

5; AASM pay AM 6. same reasoning

D235

Now prove that the third median also goes through S: 7. suppose BM, = m, not 7. indirect proof through $

8. m, intersects m, in S' # S§

8. intersects somewhere (V,)

9. S' lies = of m, from C

9. repeat above proof of 6. for m,,

m

c

10. contradiction to S' 4 §

* Theorem 11-24:

Euler’s Theorem (18th century)

The orthocenter O, centroid S, and circumcenter U of any triangle are collinear. The distance between the first two points is twice that between the last two points.

The line through O, S, and U is called Euler’s line.

Similarity, Congruence, Symmetry

255

PROOF STATEMENTS PP AACOR AMIN. 2s 2. 5 — 4 2,CO=2UM. 3. ACOS ~ M,US

REASONS VAG | M M2 CHUM; AH,|| UM, 2. corresp. parts 3, 42 =44,CO=2UM | CS =ZSM-s.a:s;,

4. OSU collinear

2° |

4. corresp. Xs at S must be equal, must be vertex angles 5. ratio of similitude

5. OS = 2SU

* Theorem 11-25:

Nine-point Circle (also called Feuerbach circle) In any triangle these nine points lie on a circle: the midpoints of the three sides, the feet of the three altitudes, the midpoints of the three upper segments of the altitudes. The center of the Nine-point Circle lies on Euler’s line, halfway between the orthocenter and the circumcenter.

The radius of the Nine-point Circle is > the radius of the circumscribed circle.

Yoo. The one ance (See

proof is not difficult, but takes some time and patience. Someamong you may wish to demonstrate his scientific perseverin studying this proof and reporting on it as a special project. Suggestions. )

256

Unit 2

* Theorem !I-26: Feuerbach’s Theorem (1822) The Nine-point Circle of a triangle is tangent to the inscribed circle and to each of the escribed circles.

The proof needs a little more geometry than you now have. So we will believe Feuerbach! * The three altitudes and the three medians in a triangle are two more examples of three concurrent straight lines through the vertices of a triangle, therefore should also be covered by Ceva’s theorem.

Knowing that these lines are concurrent, we also know

something about certain angles and segments; what is it? You could state it in four special theorems. The concurrency of the medians in a triangle can easily be derived directly from Ceva’s theorem in its second version, since we know the ratio of the segments formed on each side. What is that ratio? What is the product of the three ratios? It is not difficult to derive the concurrency of the altitudes also directly from Ceva’s theorem; this is left for a special project (see Suggestions for Special Projects), which will also include the derivation for the perpendicular bisectors. Thus it can be shown that ® The existence of the four remarkable points in a triangle may be derived as special cases of Ceva’s theorem.

EXERCISES 1. Construct a triangle with sides 6, 9, 13 cm. four remarkable points.

Construct all

. Repeat Exercise 1 for a right triangle. . Repeat Exercise 1 for an isosceles triangle. Repeat Exercise 1 for an equilateral triangle. . Repeat Exercise 1 for an obtuse triangle. & Nn WN . Construct

the nine-point circle, incircle, and the three excircles for a right triangle, and also for a scalene triangle. See whether Feuerbach’s theorem holds.

Similarity, Congruence, Symmetry . Prove:

257

Of the set of four points, namely the three vertices

A, B, C of a triangle and its orthocenter O, each is the orthocenter of the triangle formed by the other three.

. Given an equilateral triangle of 4 inch sides; how far is the orthocenter from a vertex?

. Given an equilateral triangle of 5 inch sides; how long is the radius of the inscribed circle?

. Prove that the four remarkable angle lie on its symmetry axis.

points of an isosceles tri-

. Prove that the four remarkable points of an equilateral triangle coincide at one point.

12. Show that the concurrency of the medians in a triangle follows from Ceva’s theorem. 13; Prove analytically, that in triangle A(—3,0), B(3,0), C(2,7) a. the medians are concurrent. b. the altitudes are concurrent. c. the perpendicular bisectors are concurrent.

SUMMARY

OF UNIT 2

. We studied three geometric relations between pairs of figures: similarity, congruence, bilateral symmetry. Congruence is a special case of similarity; bilateral symmetry is a special case of congruence. Shape is an invariant under all three relations; size is an invariant under the last two relations. . Theorems

and Corollaries

II-1: Geometric figures are similar, if and only if corresponding angles are equal and corresponding sides are proportional. Cor. II-1-1: Geometric figures are congruent, if and only if corresponding angles are equal and corresponding sides are equal. II-2: s.s.s. — theorem for congruent triangles II-3: a.s.a. — theorem for congruent triangles II-4: s.a.s. — theorem for congruent triangles II-5: A deltoid is divided into two congruent triangles by the diagonal drawn from a vertex between two equal adjacent sides. Cor. II-5-1: The angle between two equal sides of a deltoid is bisected by the diagonal drawn from its vertex.

258

Unit 2 Cor. II-5-2: The two diagonals of a deltoid divide it into two pairs of congruent triangles. Cor. II-5-3: The diagonal drawn from a vertex between equal sides of a deltoid bisects the other diagonal. Cor. II-5-4: The diagonals of a deltoid are perpendicular to each other. II-6: A parallelogram is divided into two congruent triangles by either of its diagonals. Cor. IIl-6-1: Opposite sides of a parallelogram are equal. Cor. II-6-2: Parallel segments between parallels are equal. Cor. IIl-6-3: Opposite angles in a parallelogram are equal. Cor. II-6-4: If both pairs of opposite sides in a quadrilateral are equal, the quadrilateral is a parallelogram. Cor. II-6-5: If both pairs of opposite angles in a quadrilateral are equal, the quadrilateral is a parallelogram. Cor. II-6-6: The diagonals of a parallelogram bisect each other. Cor. Il-6-7: If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram. Cor. II-6-8: If two sides of a quadrilateral are parallel and equal, the quadrilateral is a parallelogram. Cor. II-6-9: A parallel to a given straight line through an external point can be constructed with only straightedge and

compasses. Cor. Il-6-10: Two parallels are everywhere equidistant. II-7: The median of a trapezoid is parallel to the bases and equal to one half the sum of the bases. II-8: If two or more parallels intersect the sides of an angle such that they intercept equal segments on one side of that angle, they intercept equal segments on the other side of that angle also. Cor. II-8-1: If three or more parallels intercept equal segments on one transversal, they intercept equal segments on any other transversal. IIl-9: s.a.s. — theorem for similar triangles Cor. IIl-9-1: If two sides of a triangle are divided (extended) proportionally starting at the common vertex, then the segment between the two new points is parallel to the third side of the triangle. I[-10: a.s.a. — theorem for similar triangles Cor. I1-10-1: Two triangles with mutually equal angles are similar. II-11: s.s.s. — theorem for similar triangles

Similarity, Congruence, Symmetry

259

Theorem of Menelaus: Three points /, JJ, IJ] on the (extended) sides of a triangle ABC are collinear, if and only if

IA UB _ 2IC =] IBC

Tit

Il-12: A right triangle is subdivided by the altitude to the hypotenuse into two triangles which are similar to each other and to the original right triangle. Cor. II-12-1: In a right triangle h?= p - gq and a2=q-c and b?=p-c. If-13: The Pythagorean Theorem Converse of the Pythagorean Theorem Cor. I1-13-1: If the hypotenuse and one leg of a right triangle are equal to the hypotenuse and one leg of another right triangle, the two triangles are congruent. Cor. II-13-2: The shortest distance of an external point P from a straight line (or a plane) is the distance of P from the foot of the perpendicular through P to the line (or the plane). Fermat’s Theorem: x"+y"=2z”" is impossible for integers 1Ga aa Oj Ne eee Law of Sines: a:b:c=sin a:sin B:sin y II-14: Theorem of Ceva (1st version): Three straight lines through the vertices of a triangle are concurrent if and only

sina,

sinf,

sin y,

sina,

sin 8,

sin y,

’

Il-15: Theorem of Ceva (2nd version): Three straight lines through the vertices of a triangle are concurrent if and only

II-16: In an isosceles triangle the base angles are equal. The angle opposite the base is bisected by the perpendicular bisector of the base. I[-17: If two sides in a triangle are equal, the angles opposite these sides are also equal; if two sides are unequal, the opposite angles are also unequal, and specifically, the larger angle is opposite the larger side. Converse of II-17: If two angles in a triangle are equal, the opposite sides are equal; if two angles are unequal, the opposite sides are unequal, and specifically, the larger side is opposite the larger angle. II-18: The interior and the exterior angle bisectors in a triangle divide the opposite side internally and externally in the ratio of the adjacent sides.

260

Unit 2 Converse of II-18: Ifa straight line through the vertex of a triangle divides the opposite side of the triangle internally or externally in the ratio of the respectively adjacent sides, it bisects the interior or exterior vertex angle. II-19: In any triangle the three perpendicular bisectors of the sides are concurrent. II-20: In any triangle the three bisectors of the interior angles are concurrent. II-21: In any triangle the bisectors of two exterior angles on one side of the triangle and the bisector of the opposite interior angle are concurrent. These three points are the centers of the escribed circles. Theorem of Thales: A triangle is a right triangle, if and only if the midpoint of one side is the circumcenter; this side is the hypotenuse. I[-22: The three altitudes in a triangle are concurrent. II-23: The three medians in a triangle are concurrent. The centroid divides each median in the ratio 2:1, the larger part lying toward the vertex. II-24: Euler’s Theorem: The orthocenter, centroid, and circumcenter of any triangle are collinear. The distance between the first two points is twice that between the last two

points. II-25: The Nine-point Circle: In any triangle these nine points lie on a circle: the midpoints of the three sides, the feet of the three altitudes, the midpoints of the three upper segments of the altitudes. The center of the nine-point circle lies on Euler’s line, halfway between the orthocenter and the circumcenter. The radius of the nine-point circle is one half the radius of the circumscribed circle. II-26: Feuerbach’s Theorem: The nine-point circle of a triangle is tangent to the inscribed circle and to each of the escribed circles.

C. A geometric locus is a set of points (or lines) characterized by a specifically geometric requirement. In any locus problem two things must be considered: (1) that all members of the locus meet the given requirement, (2) that only members of that locus meet the given requirement.

So far we have discussed the following loci:

Similarity, Congruence, Symmetry

261

In the plane, the locus of all points which a. have a given distance from a given point is a circle; Db: are vertices of triangles with a given segment as base and a given length as height is a parallel to the base; Cc. have a given distance from a given straight line is a pair of parallels; d. are equidistant from two given points is the perpendicular bisector of the segment between them; e. are equidistant from the sides of a given angle is the angle bisector. In space, the locus of all points which a. have a given distance from a given point is a sphere; b. have a given distance from a given plane is a pair of parallel planes; Cc. have a given distance from a given straight line is a cylinder;

d. are equidistant from two given points is the perpendicular bisector (a plane) of the segment between them. A combination of two loci can be used for construction problems; for instance? The idea of geometric locus is useful in certain proofs; for instance? (circumcenter, incenter, excenter)

D. Inaclassical geometric construction the only two instruments permitted are the straightedge and compasses. In a construction problem these four items should be considered: analysis, construction, proof, discussion. So far we have discussed the following basic constructions: a triangle from s.s.s.; an angle with one given side, congruent to a given angle; a triangle from s.a.s.; a triangle from a.s.a.; eno op a parallel to a given straight line through an external point (idea of 1 or equilateral A); dividing a segment into equal parts; rh dividing a segment in a given ratio, internally or exG2 ternally (the ratio may be given by numbers or by segments);

perpendicular bisector of a given segment; a circle passing through three non-collinear points; the midpoint of a given segment; ee a perpendicular to a given straight line at a given point of the line (idea of | bisector or equilateral A);

262

Unit 2 a perpendicular to a given straight line through an external point; m. an angle bisector; = using angle bisectors; to divide a given segment internally and externally in a given ratio; o. using angle bisectors, to construct angles of certain sizen(907, 45... 30>) eter): |.

E. Useful methods 1. Look for congruent triangles to prove segments equal or angles equal. 2. Look for a parallelogram to prove segments equal, angles equal, lines parallel. . Look for alternate interior angles to prove lines parallel. . Look for right triangles to use the Pythagorean theorem. Look for special right triangles, such as 30°-60°, 45°-45°. DAnpRW . Look for similar triangles to prove angles equal or seg-

ments proportional. . Look larger . Look larger

for unequal angles in a triangle to prove one segment than another. for unequal sides in a triangle to prove one angle than another.

F. Trigonometry in a right triangle ; opposite leg SD SSS hypotenuse __ adjacent leg

tan

ae

sin a COS @ rye

cos

a= hypotenuse

;

__ opposite leg

sin (90° — a)=cos a

an o~ adjacent leg

cos (90° — a) = sin a

Law of Sines (in any triangle):

sin*

VS

a+ COS?

a=

a:b:c=sin a:sin B:sin y

Similarity, Congruence, Symmetry

263

G. The analytic method consists of translating the geometric concepts of point, straight line, parallel, perpendicular, etc. into the language of ordered number pairs and equations between them, solving the problem in this language, and then translating the answer back into geometric concepts. In contrast, in the synthetic method the geometric concepts are used throughout the problem. As in any translation from one language into another, we must know which words correspond to each other in these languages.

Synthetic language point distance between two points A,B midpoint M between A and B

Xy — nee + Xp), Vy =30, tag)

point

Wee

C

on

AB,

such

that

Analytic language

(x,y) V(x,

—x,)?

+ (5 —

+6 (%p—

4)?

4)»

AC =" - AB q

a

,»

rise

slope of P,P, (orratio = parallel perpendicular straight line

A

2

a

m,

linear equation

if given by two points

a

by one point and the slope by slope and y-intercept through the origin

ViVa Xe)

||y-axis in distance a ||x-axis in distance b

x=aand x=—a y=b and y=—b x=0, y=0 Ax+ By+C=0

coordinate axes in general

ee

Os.)

y=mx+b y=mx

The choice of the coordinate system does not affect the geometric problem; therefore it may be so chosen that the computations with the coordinates are simplified.

H. Words and phrases Section 1: shape, similar, similarity, proportional, ratio of similitude, scale, numerical and graphical scale, congruent figures, preserves shape, invariant, S.s.s., classical geo-

264

Unit 2

metric construction, four parts of a construction problem (analysis, construction, proof, discussion), a.S.a., S.a.s., circular reasoning, circular definition, universal and singular statement, rules of inference, rule of substitution, rule of detachment, conditional, implication, antecedent, conse-

quent, sentence and statement, direction of an implication, biconditional, necessary and sufficient, not-g, negation, denial, law of excluded contradiction, mutually

middle, jointly exhaustive, law of exclusive, law of contraposition,

contrapositive, logically equivalent, inverse, conjugate sentences, proof by logical construction, bases, legs, and median of a trapezoid, isosceles trapezoid, geometric locus Section 2: mutually equal, mutually equiangular, divided internally or externally, negative value of a ratio, divided harmonically, golden ratio, golden section, golden rectangle, algebraic method, positive order, counterclockwise, the mean proportional, Pythagorean numbers, Pythagorean triangles Section 3: cyclic arrow, negative order, clockwise, symmetric with respect to, reflection of, symmetry axis, a symmetric figure, direct and inverse congruence, same or opposite cyclic order, symmetry plane, symmetric points, connector, perpendicular

bisector,

bilateral

symmetry,

a bilaterally

symmetric figure or object, angle bisector, interior and exterior angle bisector in a triangle, concurrent, equiangular, circumscribed circle, circumcenter, inscribed circle, incenter, escribed circle, excenter, othocenter, centroid, center of gravity, median of a triangle, remarkable points of a triangle, Euler’s line, nine-point circle, Feuerbach circle

REVIEW

EXERCISES

TO UNIT 2

. Make sure you know the meaning of all the new words and phrases listed for each of the three sections. . How many independent parts determine a triangle? An isosceles, an equilateral, a right triangle? Are two angles of a right triangle sufficient?

. What is the definition of a deltoid? What properties does it have? Can you prove them? (synthetically and/or analytically) . The angles of a quadrilateral are in the ratio 3:4:5:6. How big is each angle? . Prove your answer to these questions:

Similarity, Congruence, Symmetry

265

a. Does the diagonal of a parallelogram bisect the angles through which it is drawn? b. Do the diagonals of a trapezoid bisect each other?

. Prove: If two angles and the bisector of one of them in one triangle are equal to two angles and the bisector of one of them in a second triangle, then these triangles are congruent. . Prove: If two parallels are cut by a transversal the bisectors of the four interior angles form a rectangle. . In an isosceles trapezoid the bases are 36 and 20 units, a leg

is 10 units; compute its height.

. Prove: In two similar triangles the perimeters have the same ratio as a pair of corresponding altitudes. . Achimney pole casts 11. Prove: In same ratio

casts a 36 ft shadow while at the same time a 2 yd a 15 yd shadow. How high is the chimney? similar triangles corresponding medians have the as corresponding sides.

12. Given a plane and a perpendicular to it; prove that segments between the same point on the perpendicular and points in the plane at equal distance from the foot are equal; and conversely, if segments with the same endpoint on the perpendicular are equal, their other endpoints have the same distance from the foot. 13. A 25 ft vertical pole shall be braced by three wires, each 25 ft long, such that they are fastened to the pole 5 ft from the top, and to the ground by stakes placed at the vertices of an equilateral triangle. What is the side of that triangle?

14. Four wires are attached to the top of a 16 ft vertical pole and fastened to stakes on the ground, which form a square of 20 ft side. The wires are equally long; how long are they? Is it true that the medians of a triangle cannot intersect out15. side the triangle? 16. How long is the side of a square if its diagonal is 12 in.? 17. The hypotenuse of a right triangle is 18 in. and one leg is 7 in.; how long is the other leg? 18. Prove: If one angle of a right triangle is 30°, then the opposite leg is half as long as the hypotenuse. 19. How long is the side of an equilateral triangle if its altitude is 9 in.? 20. Prove: If the non-parallel sides of an isosceles trapezoid are extended until they meet, they form an isosceles triangle with either base.

266

Unit 2

21. Prove: Ifa parallel is drawn to one side of an angle through a point on its bisector, an isosceles triangle is formed by the parallel, the angle bisector, and the other side of the angle. 22. Prove: If an isosceles triangle ABC has a vertex angle C of 36°, then the base equals the larger part in the golden section of the leg. (Draw the angle bisector of base angle A and find similar triangles.) 23. In an isosceles triangle ABC draw a perpendicular to the base through a point P on one of the legs, intersecting the other two sides in D and E. Show that APEC is isosceles. (Prove base angles equal, after proving ADBP ~ AADE.)

E

C

D

B

A

EM, BS

P

B

Ex. 24

24. Prove: If perpendiculars are drawn from any point P of the base of an isosceles triangle to its legs, their sum equals the altitude to a leg. (Draw a | through P to the altitude and find = As.) *25. Prove: If perpendiculars are drawn from any point P inside an equilateral triangle to its sides, their sum equals the altitude. (Draw a parallel through P to one side of the triangle and use Exercise 24. Then compare with the altitude of the larger triangle and find a way to prove that the missing length equals the third perpendicular.)

26. Prove: If a perpendicular to the base of an isosceles triangle is drawn through the midpoint of a leg, it cuts off one fourth of the base. 27. Prove: A right triangle is divided into two isosceles triangles by the median to the hypotenuse. (Find symmetry axes.) 28. Prove: If the midpoints of two parallel sides of a parallelogram are connected to opposite vertices, the diagonal is trisected. (Prove the segments parallel, then find similar As to

show DM = MN and MN = NB.)

Similarity, Congruence, Symmetry

267

c D

E

C

B D E

‘

F

A

F

B

exis

A

Ex 29

29. Prove: If one side of any triangle is extended to twice its length and the new point connected with the midpoint of the nearer second side, then this line will cut off one third of the

third side.

(Draw a|| to this new line through the extended

vertex and find two pairs of ~ As.) 30. In a right triangle, the hypotenuse c equals which: asin a,

pare tana

>

eu

bcos a,

y

sina sina 31. Compute the legs of a right triangle with hypotenuse 15 in. and an angle of 40°. 32. Derive the values for sin a, cos a, tan a if

a= 45°, 30°, 60°.

33. If two angles of a triangle are 60° and 50°, what is the ratio of the three sides?

34. What relations between the trigonometric functions do you know?

35. If sin a=, then does 36. In the plane, what is have: a. a one inch distance b. equal distance from c. equal distance from d. equal distance from

cos a= or 2 or = or 2? the geometric locus of all points that from two two two

a given straight line? parallel straight lines? intersecting straight lines? given points?

37. Find all points P that satisfy the following conditions, and discuss the number of solutions: a. given two intersecting straight lines, P has a 2 in. distance from one and a | in. distance from the other. b. given points A and B, 3 in. apart, P is 4 in. fromA and 6 in. from B. c. given a straight line s and a pointA 2 in. from s, P is 3 in. from s and 4 in. from A.

268

Unit 2 d. given two parallels and a transversal, P is equidistant from all three.

38. Given a straight line s and two points A and B, find P on s equidistant from 4 and B. ~How could you place A, B in relation to s to get: a. no solution? b. exactly one solution? c. infinitely many solutions? 39. Given an angle a and a straight line s, find P on s equidistant from the sides of the angle. How could you place s to get infinitely many solutions? 40. What are the coordinates of the centroid in triangle A(—4,0), B(4,0), C(3,3)? 41. What are the coordinates of C in equilateral triangle ABC, if A(—4,0), B(4,0)? 42. Which point has the largest distance from the origin? (8,5), G2,9)5 (0; —9) 7.6) 43. Compute the coordinates of the orthocenter in triangle A(0,0), B(O,5), C(451). 44, Does any system of two simultaneous equations always have a solution? If the system does not have a solution, what is the geometric significance? Write some equations as typical examples.

1

WHAT

IS A MAPPING?

A given symmetry axis s (or symmetry plane >) groups all points of the plane (of space) in pairs such that the members of a pair are reflections of each other with respect to s (3). If one member of a pair is given, do you know how to find the other? Any other choice of s (or }) will pair the points differently by giving the connectors another direction. The chosen symmetry axis (or symmetry plane) defines a procedure of connecting symmetric points. Whenever we set up a procedure (not necessarily for reflection) by which for any one point the corresponding point can be found unambiguously at a definite place, we have defined a

mapping (or a transformation). Reflection is such a mapping or transformation. We use expressions such as: point P is mapped (transformed, carried) into P'; P goes into P’ (symbol P— P’'); P’ is the image of P. If P’ is the image of only one point P (as in a reflection), then we have a one-to-one mapping. In particular, reflection is a one-to-one mapping that carries a geometric figure (a set of points) into an inversely congruent figure. If the vertices of figure A,B,C,D,E, (Figure I on the next page) are connected to some point S, and the distances from S are enlarged

(or reduced) in a fixed ratio m:n, then a similar figure is formed with the ratio of similitude m:n. The relationship between the two figures is called an expansion.

(See Unit 2, Section 2.1.)

The

point S is the center of similitude. The choice of another point as center of similitude would have placed the second figure somewhere else (try it); so would the choice of another ratio of similitude.

A

particular point S together with a particular ratio m:n defines a procedure of finding the image of any given point. Expansion 269

270

Unit 3

| in a given ratio is another one-to-one mapping or transformation. Expansion carries a geometric figure into a similar figure. Both mappings, reflection as well as expansion, relate specifically placed figures to each other. In Figure Ila, AA,B,C, is the image B,

S A,

A,

va

B.

(a)

A,

(b)

of AA,B,C,. If AA,'B,'C,' lies at some other arbitrary place, it is still inversely congruent to AA,B,C,, but it is not related to it by the mapping called “reflection with respect to s.” Likewise,

AA,'B,'C,' in Figure IIb is similar to AA,B,C, with ratio of similitude m:n, but it is not related to it by the mapping called “expansion from S.”

In general, there is no axis s (center S) for

a pair of arbitrarily placed inversely congruent figures (similar figures) which would map them directly into each other, unless they are first “moved” into a special .relative position. (Which?)

Mapping

271

We have discussed earlier, why “moving” cannot be part of an abstract geometric system (why not?), but that it may be considered as a figure of speech for a one-to-one correspondence between directly congruent figures. If we can set up a definite _ procedure of connecting corresponding points of two congruent figures in a particular case, then we shall have defined a mapping. Looking at Figure II you will realize that definite instructions of how to “move” /AA,'B,'C,' into the position of AA,B,C, may involve two simple steps: (1) “move AA,'B,'C,/ parallel to itself’ until A,’ coincides with A,; (2) “rotate” the new position about A, = A,’ until B,’ coincides with B,. It can easily be proved that then C,' will coincide with C,, and any other point of the system will coincide with its corresponding point. The mapping called parallel shift will technically be defined as a one-to-one correspondence where corresponding points have a specific distance in a specific direction from each other. The mapping called rotation will technically be defined as a one-toone correspondence where corresponding points have equal distance from a fixed point, the center of rotation, and the carriers of these equal distances form a fixed angle with each other, the angle

of rotation. The details of these two mappings and various geometrical insights they provide will be discussed in Sections 4 and 5. Either of the two mappings, parallel shift as well as rotation, carries a geometric figure into a directly congruent figure. Two arbitrarily placed congruent figures cannot, in general, be mapped into each other by one of these mappings alone; a combination, however, will provide such a mapping. Demonstrate by examples that in the plane any figure can be “moved” into an arbitrarily placed directly congruent figure by some parallel shift and some rotation about a point, applied successively. Demonstrate by examples that in space any figure or object can be “moved” into an arbitrarily placed, directly congruent figure or object by some parallel shift and some rotations about an axis (the axis of rotation, in 3-dim. problems). In space it may take up to three rotations about different axes. So far we have mentioned four mappings: expansions (EX), reflections (RE), parallel shifts (SH), and rotations (RO).

One

specific mapping alone will relate only specially placed pairs of similar figures. (Explain.) Appropriate combinations of these

272

Unit 3

mappings (composite mappings) will relate any arbitrarily placed pair of similar figures. Composites of SH and RO are equivalent to a mapping called “moving to any other place”; composites of such a “motion” with RE will relate any arbitrarily placed pair of inversely congruent figures; composites of “motion” with EX, or with both EX and RE will relate any pair of similar figures. These mappings have two properties in common: (1) they are one-to-one correspondences between pairs of points; (2) they preserve geometric shape. There are many other interesting and useful mappings that do not have these two properties. Some branches of geometry study one-to-one correspondences between points and lines, or between lines and lines; some others study geometric properties independent of shape, and therefore are interested in transformations preserving these properties even though they may change the shape. For instance, the property of having at most two points of intersection with a straight line is shared by the circle and the ellipse. A transformation changing a circle into an ellipse (something like looking at the circle sideways) will still preserve the property about the two points of intersection. In the Appendix you may find some suggestions for a project of your own, finding out about interesting transformations. In this course we are interested specifically in those transformations that preserve shape, because the study of shape is considered the primary task of Euclidean geometry. Since all geometric properties describing shape (parallel, perpendicular, specific angles, specific ratios, etc.) are invariant under EX, RE, SH, RO, it has

been said that: > Euclidean geometry is the study of the invariants under certain transformations, namely EX, RE, SH, RO and composites of these.

Sometimes it is said also that Euclidean geometry is the geometry of “rigid motion,” meaning that it deals only with those geometric properties that do not change while the object is moved about. This is not quite true, however. If we translate the obvious meaning of the expression “rigid motion” into the technical form of transformations, only parallel shift and rotations together can describe it, because only these relate directly congruent figures to each other. Expansion changes the size, which is not changed by “rigid motion.” Yet, the Pythagorean theorem has the same

Mapping

273

meaning for us as for the Lilliputians, if they should be interested in Euclidean geometry. Also, a geometric object reflected with respect to a plane cannot be related to its image by “rigid motion,” that is, by composites of SH and RO. However you “move” your left hand, it cannot appear as your right hand.

2

EXPANSION

2.1 The Center of Similitude > Expansion is a one-to-one mapping between pairs of points such that the connectors of corresponding points pass through a fixed point S (the center of similitude), and the distances of cerresponding points from S are in a fixed ratio (the ratio of similitude).

In Unit 2, Section 2, we studied similar figures; expansion relates similar figures which are in a special relative position.

Theorem Ill-1: In an expansion, corresponding segments are parallel.

Why?

274

Unit 3

Corollary Ill-1-1: An expansion similar figure.

carries

a geometric

figure into a

Why?

Theorem III-2: If corresponding segments of two similar figures are parallel, then there is a center of similitude. (Then the connectors of corresponding points are concurrent.)

Why? (Start the proof with one pair of parallel sides; the connectors of the endpoints intersect in some point S$; now consider the adjacent pair of parallel sides and prove that the third connector must also pass through S; look for similar triangles.) If the center of similitude S lies between corresponding points (A and A” in Figure I), then the ratio of similitude g is considered negative; if S lies outside the segment connecting corresponding points (A and A’), then q is considered positive. The two possible positions of S with respect to the segment between corresponding points are indicated by the names internal center S, or external center S,, of similitude. S, divides AAS ternally (negative ratio); S,, divides AA' externally (positive eee A

B

Mapping

275

In Figure II, squares in parallel position are mapped into each other by an external and an internal center of similitude. But the point A is mapped into two different points by the two different centers of similitude:

S, maps A > A’, B > B’, AB— A'B' S, maps A > A”, B > B", AB > A"B". EXERCISES (Use drafting aids.)

1. Draw an irregular quadrilateral ABCD. Choose a point S inside and draw an expansion of ABCD from S in the ratio 2:1. Examine how the position of the image changes with a different choice of S, by repeating the exercise for some other S’' and S”, inside or outside of ABCD. Draw the three images in different colors. 2. Repeat Exercise 1 for the ratio —2: 1.

3. Draw two regular hexagons of different size in a parallel position. Find a center of similitude § which makes one figure appear as an expansion of the other.

Could there be

S

two such centers?

4. Prove: Ifa pyramid is cut by a plane parallel to its base, the figure of intersection is similar to the base. (The whole pyramid may be considered an expansion of the small pyramid.) 5. Prove: Ifa pyramid is cut Ex. 4 by a plane parallel to its base, in half its height (or in ; of its height above the base), then the figure of intersection is similar to the base with the ratio of similitude 1:2 (2:3). 6. Prove: If two triangles have corresponding parallel sides, then the connectors of corresponding vertices are concurrent.

7. How can the theorem in Exercise 6 be used as a basis for a construction connecting a given point P to an inaccessible point OQ? Suppose Q is the intersection of two given straight

216

"Units lines s, and s, which cannot be produced beyond a certain line.

8. If a figure F is an expansion of figure F’, is F similar to F’? If a figure G is similar to’ figure G’, is G necessarily an expansion of G’? 9. If the perimeter of the base of a square pyramid is 24 in., Exe what is the size and shape of the figure of intersection with a plane cutting the pyramid parallel to its base in j of its height above the base? 10. Inscribe a square into a given triangle ABC such that one side

lies on AB and the other two vertices

on

one

side

each.

(Make a square under AB, connect its vertices D and E with

C:; then CD and CE will cut out of AB one side of the required square. Why? Is the square ABED the only possible auxiliary square? What other squares could have been used?)

EBxa®

11. Can you solve the problem of ; Exercise 10 by the algebraic method? (Assume that you know the length of base c and of height h,, and set up an equation for the unknown x = side of required square.)

12. In a given triangle ABC another triangle shall be inscribed with sides parallel to a given third triangle EFG. (Draw an arbitrary line parallel to GF,

intersecting AC in G' and CB in F’;

complete the auxiliary AG’E'F’

simi-

Bae

Mapping

lar to AGFE;

277

connect CE’ to find the vertex on AB of the re-

quired A, and complete it by appropriate parallels. the construction.)

Explain

13. In a given triangle, inscribe a rectangle that is similar to a given rectangle.

2.2 The Center of Symmetry In the special case where the ratio of similitude q = —1, the

center of similitude S$ lies exactly halfway between corresponding points. Corresponding figures are congruent (why?). Are they directly or inversely congruent? (See

Unit 2, Section 3.1.)

.

A

jn

ee

®.

Since each point maps into an ae image in exactly the same distance Beal . from S on the other side of S, this ae Ne special mapping is calleda reflecg27 _/ ays tion in the point S, OF central symmetry with respect to S This mapping must not be confused with a reflection or symmetry with respect to an axis or a plane, I which are bilateral symmetries. In central symmetry the plane or space is not separated into two parts such that corresponding points could not be both in the same part. Prove that centrally symmetric pairs of segments are parallel. Give two separate proofs, (1) using expansion, (2) using Figure I directly. Use one of your draftsman’s triangles as a model for a plane triangle, and the point of your pencil as a symmetry center S. If point S is in the same plane as the triangle, where is the centrally symmetric image of the triangle? If S is not in that plane, where is the image then? In Figure II on the next page triangle A(2,1) B(3,1) C(2,4) is centrally symmetric with respect to the origin O of the coordinate

system to AA’’B'’C"” in the third quadrant. nates of its vertices?

What are the coordi-

/AABC is bilaterally symmetric with respect

to the y-axis to AA"B’C” in the second quadrant;

AA”B"C" in turn

278

Unit 3

is bilaterally symmetric to [AT BIC! AWith® Tespect™ to Thus we find that two successive bilateral symmetries about mutually _perpendicular axes are equivalent to (give the same result as) one central symmetry about the point of intersection of these axes. Can you use this fact to find out whether centrally symmetric figures are directly or indirectly congruent?

il

Choose any point on a globe; where is its reflection with respect to the center of the globe? If all points of the globe were mapped into their reflections with respect to the center of the globe, what is the set of all these images? What is the image of the globe? wu Consider a _ regular square pyramid; where is its centrally symmetric image with respect to its vertex? Now consider this pyramid and its image together as

one object; what is its image with respect

to

the

same

sty A

symmetry

center? > If a geometric figure or object can be mapped into itself under reflection about some fixed point, it is called a centrally symmetric figure or object; the fixed point is its center of symmetry. .

Give some examples of centrally symmetric figures and objects.

EXERCISES 1. What is the meaning of a positive or a negative ratio of similitude? 2. Draw the centrally symmetric figure to ABCDEFGH with respect to S.

Mapping

279

Prove that each side of the figure ABCDEFGH is equal and parallel to the corresponding side of its centrally symmetric image.

. Try to visualize SABCDEFGH as a 3-dimensional object, a pyramid of irregular base with Ex. 2-4 S as vertex. Its reflection with respect to S$ is then another pyramid with the same vertex S. Draw the visible edges of both objects with heavy lines, and the invisible edges with dotted, less heavy lines. . Is a regular hexagon or a square a centrally symmetric figure? If so, where is the symmetry center?

. Give several examples of centrally symmetric figures and objects, identifying the symmetry center. Are they also bilaterally symmetric? If so, give the symmetry axis or plane. . Is an isosceles or an equilateral triangle centrally symmetric? oO.

Which of the quadrilaterals are centrally symmetric?

. Are you centrally symmetric? Are you _ bilaterally symmetric?

10. Given two equal and parallel segments a and b; can they be considered centrally symmetric

P

bore. b ei

to each other? If so, where is the symmetry center?

11. Using central symmetry, prove that: a. the diagonals of a parallelogram bisect each other; b. if the diagonals of a quadrilateral bisect each other, it is a parallelogram; c. if two sides of a quadrilateral are equal and parallel, it is a parallelogram.

12. Prove: In central symmetry a pair of parallel straight lines maps into a pair of parallel straight lines. 13. Is an octahedron centrally symmetric? A tetrahedron? 14. On the globe, where are the centrally symmetric points to the

following

points:

A(4°N

C(90° N 0° E); D(O° N 180° W)

10°F);

B(60°N 100° W);

280

Unit 3

15. Prove: A right angle is an invariant under central symmetry. (Use art. A.)

16. In a Cartesian coordinate system in the plane, find the coordinates of those points which are centrally symmetric to: (2,2), (4,1), €2,4), 3; 3), (4,—2), 2, —5). “Im general; which quadrant is centrally symmetric to which? 17. In a Cartesian coordinate system in space, find the coordinates of those points which are centrally symmetric to: (23,4); (4,3,0) (4253 4 2, ee) eee

o

3

REFLECTION

> Reflection with respect to an axis s or a plane & is a one-to-one mapping between pairs of points such that the connectors of corresponding points are perpendicular to and are bisected by s or >.

In Unit 2, Section 3, we studied reflections (bilateral symmetries) and some applications. Here are some more applications:

|

3.1 Trigonometric Functions of Any Angle: An Application of Reflection About an Axis

We have studied trigonometric functions of angles in a right triangle (see Unit 2, Section 2.4), defining them as ratios between two sides. Since an angle in a right triangle is between 0° and 90° (why?), the functions were defined only for these angles. The slope of a straight line is the trigonometric tangent of its angle of inclination with the x-axis. If that angle q@ (counted from the positive direction of the x-axis counterclockwise) is larger than 90° (Figure I), what is the slope of the line? The original purpose of trigonometric functions, the compu| tation of sides and angles of triangles, is today only a small part of their usefulness. Defined

Mapping

281

for any angle, these functions are more appropriately called goviometric functions (angle measuring) rather than trigonometric functions (triangle measuring). The general definition of goniometric functions for any angle

makes

use of the wnit circle (Figure II).

This is a circle whose

radius is taken as the unit of A pair of perpenlength. dicular straight lines through center

the

O,

customarily

called x- and y-axes, divides the circle into four quadrants. The radius to a point P in the first quadrant forms a right triangle with the coordinate segments of P. If a is its angle of inclination, then y x BIO

rah eC OS.0—

2 (alae

= The advantage of the unit

Il

circle lies in the simplification for r = 1; then sin a = y and cos a@ = x, which gives a simple graphic representation of these two functions. The same simplification can be achieved for tan a, if instead of rt. AOPP’ the similar rt. AOQA is used, whose legs are the unit radius OA on the positive x-axis and the tangent to the unit circle at A. Then

ee OAL, Xx

tangent with the extended radius through P. The graphic illustration of tan a@ is the “length” of the tangent to the circle, from A to Q.

(This graph explains the name “tangent” for this function. Tangere in Latin means to touch; a tangent to a circle is a line that touches the circle. The name “sine” comes from sinus in Latin, meaning a bay; it probably refers to the

fact that the sine is a measure of the opening produced by the angle. Co-sine is the sine of the complementary angle.)

282

Unit 3

The simple expressions for the trigonometric functions in terms of the unit circle are extended to the other quadrants:

Definition I1I-1: If P is a point on the unit circle with OA as radius on the positive x-axis, and XAOP = Xa, then: sin a= y-coordinate of P cos a= x-coordinate of P tan a = length of tangent to the circle, from A to the intersection Q with the extended radius OP.

What happens to point Q when a approaches 90° or 270°? Definition III-1 cover tan 90° and tan 270°?

Does

Figure [II illustrates the meaning of this definition for each quadrant. Note that the tangent always passes through A. It is positive in the same direction as the positive y-axis. What is the sign of each function in each quadrant? Complete the following table:

Mapping

283

Reflection with respect to the coordinate axes or the origin makes it possible to relate the functions of any angle to those of an angle in the first quadrant, so that its numerical value can be taken from Table I and combined with the appropriate sign. Examples: Table values sin cos tan

.5000 5000 9397

sina

cosa

tana

.8660 .5774 :3'—=.866 —.5774 .8660 .5774 —.5 +.866 —.5774 .3420 2.747|—.9397 —.342 +2.747

(a) Reflection with respect to the y-axis (Figure IVa): 180°-a> AA’, BB’, CC’, actually by infinitely

Mapping

287

many more between all other pairs of corresponding points such as

DD’. Since all these vectors have the same length and direction, one of them is sufficient to describe the shift. It makes no difference which one of them is chosen as the representative nor whether another vector V outside the triangle is so used, as long as it has the

same length and the same direction. A vector of different length and/or different direction would produce a different parallel shift. > Two vectors A and B are equivalent (symbol = ), if they have

the same length (|A| = |B]) and the same direction.

Note:

Equivalent means having the same effect if substituted for

each other.

In arithmetic, 5 and (2 + 3) are equivalent since they

give the same result in a computation if substituted for each other. Equivalent numbers are called equal. Equivalent vectors are often also called equal — therefore the equal symbol — but the word equivalent is preferable because vectors are more than numbers; they are numbers and directions. In contrast to vectors, a number is also called a scalar. The length (absolute value) of a vector is a scalar.

2)

(a)

(b)

(c)

The parallel shift taking AABC into AA’B’C’ (Figure Ia) is described by vector P. Another vector Q takes AA’B’C’ into AA"B'C". A third vectorR could have taken ABC into A A"B"C" directly. (Prove this by showing that the segments 4A”, BB", CC" are actually equal and parallel. Look for parallelograms.) Thus R is equivalent to P and Q combined. This fact suggests the use of algebraic language, calling such a combination a sum; then R=P+Q. The graphic representation in Figure IIb, tracing the shift of any one point, is called a vector diagram, Ob-

288

Unit 3

viously, a shift along Q first and P afterwards (Figure IIc) leads to the same result. What geometric figure is formed by the two vector P + Q and Q+ P?

diagrams for

Definition III-2: The sum of two vectors P and Q is the diagonal vector R in the parallelogram formed by these vectors P and Q. P and Q are called components of R.

The fact that P+ Q=Q+P called the commutative law

addition. dition

is of

This law applies to adof

numbers

also,

e.g.

3+4=4+3. The associative law of addition for numbers says that, e.g., (3-+4) + Ss=3-+ (4-4-5).

see

whether

vectors

this

also,

law

that

Let us

applies is,

to

whether

(PQ) -|2R =P + (O-_R).: In the vector diagram in Figure III the vectors are interpreted as shifts: Left side P takes

Q takes

P+Q

Right side

A— B

P takes A— B

B—> C

Q takes

takes A > C

R takes

C — D

(P+Q)+R

takes

B>C

R takes C >D (Q+R)

A > D

Thus (P + Q) + R is equivalent to

takes B— D

4 — D +R) P+(Qtakes

P + (Q + R).

& The addition of vectors is commutative and associative.

What happens to the vector parallelogram (Figure IIc) if the components P and Q have the same direction? What is the length of R in that case? What is the direction of R?

Mapping

289

»> If two vectors P and Q have the same direction, their sum R=P+ Q is a vector R in the same direction whose length

|R|/=|P|+|Q|].

Explain why: > The product of a vector P with a positive number n is a vector in the same direction as P, with a length n times the length of P

(|nP|=n|P|);

the quotient of a vector P by a positive

number n is a vector in the same direction as P, with a length 7,of the length of P (2\=5[P))n

n

A

(By interpreting vectors as shifts show in a vector diagram the

meaning of P + P, P +P + P, 3P, nP, also 5P, 4P, a2 How can we define the difference between two vectors?

If Q

takes A’ — A” (Figure Ila), then the opposite shift taking A” > A’ may be called —Q, the additive inverse of Q.

If

P+ Q=R, then

P=R-—Q. This means that going from A to A” along R and then from A” to A’ along the additive inverse of Q gives the same result as going directly from A to A’ along P. & The difference of two vectors is formed by adding to the first vector the additive inverse of the second vector.

Vectors find useful applications in various aspects of science wherever a change is studied that is due to the combined effect of direction and length. In geometry, vectors can be used to describe geometric transformations. In mechanics, vectors are used to describe forces. If you push a marble in a certain direction, then the harder you push the farther it goes; the magnitude of the force applied is represented by the length of the vector. If, at the same time, your little sister gives that marble a side push, it cannot follow both orders simultaneously; it will make a compromise move, obeying your stronger push P more, but not neglecting her little push Q either; it will go as the vector R indicates. Note that the marble goes farther than if you alone had pushed it, but not as far as the actual sum of your strength and hers combined. If she had pushed in the same direction, however, then...

290

Unit 3

The simultaneous » Forces may be symbolized by vectors. application of two forces in different directions results in a force symbolized by the diagonal in the vector parallelogram. The latter is called the parallelogram of forces, and its diagonal is the resultant;

the two forces are the components of

the resultant.

_ EXERCISES 1. Given quadrilateral ABCD and a vector V (a segment with a direction); draw A'B'C'D' = ABCD through a parallel shift along the given vector.

Em

|

Exe2-andas

2. AABC is subjected to a series of successive parallel shifts along P, Q, R. Draw the various positions. 3. In Exercise 2, which single shift could have substituted for P+Q? Which forP+Q+R? Which for Q+ R? 4. Demonstrate a 3-dimensional parallel shift by using two (supposedly congruent) books as corresponding objects, and a pencil held in a given direction as vector.

5. What is the commutative law? Demonstrate it for the addition of numbers and the addition of vectors. Is there a commutative law for multiplication of numbers? Find some non-mathematical examples where the commutative law holds and where it does not hold. 6. Repeat Exercise 5 for the associative law. 7. If P+ Q=R, show that the length of R is smaller or at most equal to the combined length of P and Q. (This reminds you of what geometric axiom?)

8. What is a vector parallelogram? resultant?

A force parallelogram?

A

Mapping

291

9. If A and B have the same direction, and if the length of B is half that of A, what is A+B? What happened to the vector parallelogram? What is A—B? 10. Interpret

R—R

by a. b. c. d.

a parallel shift of a triangle, pushing a marble, pulling a drawer open, climbing stairs.

(In technical language the result of R—R vector.’’) 11. If P and Q are vectors drawn from point O, what is the meaning of the following? a. P+Q c. P—O e. 3Q

b. —P

1

d. 2P

f Pio

is called a ‘‘zero

Q

°

:

4.2 Components of a Vector

Suppose you find these directions in a treasure hunt: “From where you found this note go 40 steps in the northeast direction, then turn through 35° counterclockwise and go another 30 steps, and you will find IT.” How far away in what direction is N NE this treasure? The required walk can be symbolized in a vector diagram (Figure I). Instead of By going from O to P, to P, one could go from O to P, directly.

—

—

The vectors OP,=A

and P,P, = B are components

edt of the vector OP, = C whose length and direction we want to compute.

Obviously, this

|

vector could be composed of many other pairs of components. The instructions could have been, e.g.: “Go... steps east, then turn north and go ... steps.” In that case the components would be the legs of a right triangle, and Pythagoras would help us to compute the length of the hy-

292

Unit 3

potenuse; the direction could be found by looking up a trigonometric tangent (how?). Thus there is a definite advantage in using a pair of perpendicular components where possible. Does every vector have a pair of perpendicular components, one pair or many? If two perpendicular directions within a plane are designated as east and north, can every vector V be expressed as the sum of an east and a north component VES Nee Explain the following relations between a vector and its east and north components, if q is the angle between the first two:

| V,| =| V |cos @ | V,| =| V| sina tan

a=

1 Vie|

eves

Ves

el eave

How does all this help us to compute the place of the treasure? Obviously, P, in Figure I is as much east of point O as A, and B, together, and as much north of O as A, and By together. These two sums are the east and north component of C, whose length and direction can now be computed.

Computation of

C= A+B:

A: length |A |= 40 steps

B: length |B |= 30 steps

a, = 45° from east

|A, | =40 cos 45°

|A. | = 40-sin 45° C:

OQ, = 45° + 35° = 80° from east

| B,, | =30 cos 80° | B,, | =30 sin 80°

| C, | = 40 cos 45° + 30 cos 80° = 40(.7011) + 30(.1736) = 33.252

Mapping

293

| Cv |= 40 sin 45° + 30 sin 80° = 40(.7011) + 30(.9848) = 57;588

length | C | = V33.3? + 57.67 = V4426.65 = 66.5 steps 57.588

tan a = 33.252 =

ba F/S 9

a = 60°

Note: East and west are used here as coordinates in a Cartesian system. In the more general language you would use x instead of E, and y instead of N. The components of the vector V are then denoted by V,, and V,.

EXERCISES 1. Find the east and north component of these vectors:

a. |A|=5, a= 40°

d. |D|=9,a=0°

b. |B|=3,a= 65° ¢. |C|=20, a= 90°

e. | E|= 16, a= 20° f. |F|=12, a=400°

2. Find the length and direction of a vector from the given east and north components:

a. Dae ee d.

ey

vel

3 225 0) 16

3 20 pa) 30

ve| Watany

G25]

3. Given the direction and the east component of a vector, find its length, and explain the computation.

294

Unit 3

4. Given the direction and north component of a vector, find its length, and explain the computation.

Paley a. b. c. d.

F Women oy eeVv

45° 10 20° 15 60° 2a 90° | 100

5. Given the length of a vector and its east component, find its direction, and explain the computation.

6. Given the length and direction of two vectors A and B, find the direction and length of the vector C representing the sum A+B; make an approximate diagram for each case:

|A |

3

ROTATION

IN THE PLANE

5.1 Rotation Around a Point > Rotation around a point is a one-to-one correspondence between pairs of points that have equal distance from a fixed point O (the center of rotation), while the carriers of these distances form a fixed angle a (the angle of rotation).

It follows that: (1) corresponding

points lie on

a circle around

O;

(Why?

What is the definition of a circle?)

(2) all pairs of corresponding triangles with O;

points form

similar

isosceles

Mapping

(3) the

symmetry

axis

between

any pair of corresponding points passes through O; (4) corresponding segments AB and A’B’ are equal; (Why? Use S.a.S.)

¢!

\_] tee

(5) corresponding congruent.

(Why?

295

triangles

are

Uses.s.s.)

Are

B

bes 0

A

they directly or inversely congruent?

Theorem III-3: In a rotation through angle a corresponding straight lines form the angle a.

Hyp.: x4OA' =a, x BOB' =a,

AB—

CSS

A'B'

Con.: xF=a Plan: prove AAGF ~ AA'OG by two pairs of = Xs

PROOF 1. 2. 3. 4. 5.

STATEMENTS AOAB = AOA'B' XOAB=XOA'B' XGAF=xXOA'F xXxG=xXG xF =a, gq.e.d.

REASONS PSs 2. corresp. Xs in = As 3. suppl. adj. to = Xs 4. identity 5. sum of Xs in any A is the same

Corollary Ill-3-1: If the sides of two angles are mutually perpendicular, the angles are equal if they are both acute or both obtuse; otherwise they are supplementary.

296

Unit 3

Compare the derivation of this important corollary with another proof, studied in Unit 9.1.

1, Section

EXERCISES . Draw

a square and rotate it around a vertex A through 45°,

then again through 45°, and so on. How many _ different C positions can you draw? . Draw a figure OA'B’C’ that corresponds to the given figure c B OABC in a rotation through 60° about O. Rotate the result again through 60°; how many different figures can you get fe) by further such rotations? Ex.2 . Rotate an isosceles triangle with vertex angle 120° around this vertex through 120°; keep rotating the result. How many different triangles can you get? . Draw a triangle and rotate it around its circumcenter through 120° and 240°. 2).

Repeat Exercise 4 for an equilateral triangle. . Rotate an equilateral triangle about one vertex A through 60° and multiples of 60°. . What is a rotation through 180°? Is it correct to say that this rotation is equivalent to a central symmetry? What happens to Theorem III-3? What is a rotation through 0°?

. Prove

Theorem

III-3

for the

special case of a= 90°. . If right triangle ABC

flected image

’

in

S

s,=AC

reflected

is re-

and

the a

in s,=B’O,

where O is a point of s,, could the resulting triangle be produced also through a rotation of AABC?

Mapping

297

10. A rectangle is rotated through 60° around a point O lying on the extension of one of its sides. Could the image be produced also through two successive reflections?

11. If in Exercise 9 the axis s, is parallel to s, through B’ instead of intersecting s,, what single mapping could replace the two reflections?

5.2 Regular Polygons A polygon with equal sides and equal angles is called a regular polygon. If it has n sides, it is also called an n-gon. Which of the following are regular polygons: square, rectangle, trapezoid, parallelogram, rhombus, kite, equilateral triangle, isosceles triangle? We shall now prove that the vertices of a regular polygon lie on a circle, so that we can consider any two of them as corresponding points under a special rotation.

Theorem III-4: A regular polygon has a circumscribed and an inscribed circle; their centers are at the same point.

Hyp.:

4BCDEF has equal sides and equal angles. Con. (a): There is a point O equidistant from all vertices. (b): Point O is equidistant from all sides. Proof of (a): There is a point equidistant from three vertices (the circumcenter); show that it has the same distance from any other vertex:

STATEMENTS 1. construct AABC

circumcenter

DOB =O0C—O0A

REASONS O

of

1. intersect 1 bisectors of AB and BC

2. from 1.

3. XB=Ay

3. ABOC is isosceles, from 2.

4. xB=xXC

4. hyp.

298

Unit 3

5. XP'=xy’

5. from 4. and 3.

6. CD=AB

6. hyp.

7. KOCD

Te Sass

= SOAB

8. OD=OA

8. from 7.

9. repeat for next vertex: instead of AABC consider ABCD with circum-

center O and prove that OF =OB =OC= OD. Proof of (b):

1. As ABO, BCO, CDO,...isosceles 2. these As = 3. all base angles are = AXOA, OB; OClOD, 2x bisec=) tors 5. O is equidistant from all sides

1. O is circumcenter 2. from hyp. and s.s.s. 3. from 1. and 2. 4 fronra: 5. intersection of all x bisectors

Definition III-3: The radius and center of the circle circumscribed around a polygon are called the radius and center of the polygon. The radius of the inscribed circle is called the apothem of the polygon. The angle at the center, between radii of consecutive (neighboring) vertices is called the central angle of the polygon. The angle between consecutive sides is called an interior angle of the polygon. The angle between one side and the extension of a consecutive side is called an exterior angle of the polygon. The sum of the sides is called the perimeter of the polygon.

Corollary Ill-4-1: A regular polygon of n sides is divided by the radii to the vertices into n congruent isosceles triangles. The vertex angle in such a triangle, the central angle, 360° iS ni (Prove by: s.s.s.)

Mapping

299

Corollary WI-4-2: An interior angle of an n-sided polygon is 180°—

360°

; an exterior angle is equal to the central angle.

(Prove by computation)

By considering any two vertices of a regular polygon as corresponding points, we find that a rotation around the center through the angle formed by the radii to these vertices will carry the whole polygon into itself. Are there several different angles of rotation to choose from? In the regular hexagon (see Figure I) we obviously may start with any arbitrary vertex, say A, and consider (A,B) as a pair of corresponding points with a = 60°, or (A,C) with a = 120°, or (A,D) with a = 180°, or (4,£) with a = 240°, or (A,F) with a = 300°, or (A,A) with a = 360° = 0°. The last rotation is not a real rotation since it makes every point correspond to itself; such a correspondence is called an identity. We could

go on and rotate through 420°, 480° and so on; but although this is perfectly correct, it does not yield anything new, since 420°

= 60°, 480° = 120°, in general 360° + a= a. A!

D B

B

B! E

a=60°

A ao =180°

a=120°

B

Be

A

A

B'

a~=300°

|

A' =

360°

identity

300

Unit 3

We only have as many different rotations, including identity, as there are vertices. Note that the different angles of rotation can all be expressed as multiples of the smallest (here 60°), which is

the central angle between the radii to two consecutive vertices: 200) gives the numa, 2a, 3a,..-.Na = 360°. The number n= ber of different rotations possible for a saricule figure. If this figure is a regular polygon, then n is also the number of vertices or sides. A figure that maps into itself by rotation is sometimes called a

rotationally symmetric figure.

The number n = =

of possible

different rotations carrying such a figure into es is called the order of the rotational symmetry; here a is the smallest angle of rotation. A rotationally symmetric figure is not necessarily a regular polygon. Figure I is rotationally symmetric, but consecutive vertices are not corresponding points; there are 12 vertices but only 6 different rotations. What are the angles of these 6 rotations? Which vertex corresponds to point | in each of the 6 rotations? What about the rotations carrying point 2 into any of the vertices 4, 6, 8, 10, 12?

What is the

order of the rotational symmetry of Figure I? Which of the following figures are rotationally

which

order:

symmetric

square,

and,

if so, of

equilateral

tri-

angle, isosceles triangle, trapezoid?

A central symmetry is a rotational symmetry of order 2. What does that mean? The fact that a circle can be circumscribed around a regular polygon yields the simplest construction of such a polygon (Figure III) if the radius is given: First draw the circumscribed circle. Then determine the central angle a = oe

using a protractor. The side AB is a Ge of the polygon, and is transferred by compasses (or paper strip) on the circumference to BC to CD and so on, until A is reached again. If the polygon

Mapping

has special properties such as parallel or perpendicular sides or diagonals, it is advisable to use and check these properties in the process of construction. If a regular n-gon has been constructed,

can

you

construct

from

it a

C

301

B

\ oD

A

regular 2n-gon and a regular 5-gon? A

“classical”

construction

(with

compass and straightedge only; no Hl protractor) is only possible for special sets of regular polygons. We know how to construct a central angle that is 90° or 60° or one which is derivable from these angles by bisecting, adding, or subtracting. Thus, we can construct the set of regular 4-, 8-, 16-, 32-gons etc., and the set of regular 3-, 6-, 12-, 24-gons etc. In a regular decagon the central angle is 36° and the side s,, divides the radius in a golden section. (See Fig-

ure IV;

also see Unit 2, Section 2.2, and exercises about the 36°

isosceles triangle in Unit 2, Section 3.3.)

From the proportion it

follows that s,>) = 5(V5 — 1); since \/5 can be constructed as the

hypotenuse of a right triangle whose legs are | and 2, the whole algebraic expression for the side s,, of the decagon can be constructed (see Exercise 9 in Group II). From the regular decagon (central angle 36°) the regular pentagon (central angle 2 x 36°) and /5-gon (central angle 24° = 60° — 36°, see Figure V) can be

302

Unit 3

derived. Thus we can construct two more sets: the regular 5-, 10-, 20-gons etc. and the regular 15-, 30-, 60-gons etc.

Gauss

(18th cent.) showed

that

also a 17-gon can be constructed. Using trigonometric functions the side of any regular polygon can be computed from the right triangle formed by the radius, the apothem, and half the

side (Figure VI). This triangle contains half of the central angle a, then

sin 5= se and s = 2r sin5-

Vi

|

With the compass opened to the

length s, Figure III may be completed. EXERCISES Group I

1. How big is an interior angle of decagon?)

aregular hexagon?

(pentagon,

2. How big is the sum of the interior angles of a regular octagon? (15-gon, 20-gon?)

3. How big is an exterior angle of aregular hexagon? pentagon, decagon?)

(octagon,

4. How big is the sum of the exterior angles of a regular hexagon? (8-gon, 5-gon, 10-gon?) 5. How many sides does a regular polygon have, if each interior angle is 140°? (108°, 156°?) 6. Prove that for any n-gon the sum of the exterior angles equals 360°. J.

Prove that for any n-gon the sum of the interior angles equals

(n — 2)180°. 8. Which regular polygons can be used for tiles as a uniform floor covering?

9. Prove that the opposite sides of a regular hexagon are parallel to each other and to the diagonal connecting the remaining two vertices.

10. How big is the radius of a circle circumscribed about a square of side 6 in.? 11. How long is the side of a square inscribed in a circle whose radius is 5 in.?

Mapping

303

12. How long is the radius of the circle circumscribed about an equilateral triangle whose side is 12 in.? (s in.?) 13. How long is the radius of a circle that can be inscribed in an equilateral triangle whose side is 12 in.? (s in.?) 14. If the radius of a circle is 6 in., how long is the side of a circumscribed equilateral triangle? 15. How long is the side of an equilateral triangle that can be inscribed in a circle of radius 6 in.?

Group II 1. Construct a square, a regular octagon, and a regular 16-gon inscribed in the same circle whose radius is 2 in.

. Construct an equilateral triangle, a regular hexagon, and a regular 12-gon, inscribed in the same circle of radius 2 in.

. Construct a regular octagon inscribed in a given circle by using the diagonals of two _ squares. Show that these two squares are circumscribed about another smaller regular octagon. . Prove that opposite sides regular octagon are parallel. . Prove

that ABCD

in a

in the Figure is an isosceles trapezoid;

also ABDE.

. Compute the angles EFG, EGF, and DFG. 7. Construct a regular octagon that is circumscribed about a given circle. (See Figure on next page.)

a. construct a pair of perpendicular diameters MN and OP. b. construct perpendiculars at the endpoints of these diameters to form a circumscribed square. c. draw the diagonals of the square. d. construct perpendiculars to these diagonals at their intersections with the circle and intersect them with the sides of the square. e. Prove that ABCDEFGH is a regular octagon. . For the regular hexagon

prove that:

(See Figure on next

page.) a. ABCD

is an isosceles trapezoid.

b. AD is a diameter.

c. BF | AD, CE 1 AD.

304

Unit 3 d. BCEF is a rectangle. e. Compute the angles ABC, ACE, ACD. GOs Fk

9. Construct a regular decagon inscribed in a given circle of radius r: a. Draw _ perpendicular radii AB and CD b. M= midpoint of OD

: ma VGF-e d. © hee M with radius MB intersects e. Tye S=5

OC in S

5 Els5)

ot)

Why should s,, be equal to AG Os 1)?

10. In how many ways is a regular hexagon a symmetric figure? a bilaterally symmetric figure? How many symmetry axes does it have? Is it rotationally symmetric? of what order? 11. Repeat Exercise 10 for a regular octagon and a regular pentagon.

Group Il 1. Using trigonometric functions compute the length of the side of a regular pentagon, decagon, 15-gon of radius 2 in. 2. Using trigonometric functions compute the length of the radius of a regular 20-gon (30-gon) whose side is 6 in.

Mapping

305

3. Using trigonometric functions compute the length of the apothem of a regular pentagon whose radius is 3 in. 4. Compute the ratio between the sides of two regular pentagons which are inscribed and circumscribed polygons for the same circle. 5. Prove by computation that the sides of a regular pentagon, regular hexagon, and regular decagon inscribed in the same circle are related as are the sides of a right triangle: Ss

Ree 2 S¢ + S16

6. Show that in the construction in Exercise 9 of Group II the side of the pentagon is BS. Construct a regular pentagon inscribed in a given circle.

5.3 The Circle as a Limit

Theorem III-5: A circle can regular n-gons.

be inscribed

and circumscribed

Proof by possibility of logical construction:

by

The central angle is

fe)

a=

a

; its sides meet the circle in

two vertices A and B of the inscribed n-gon. Rotation of order n gives all vertices. The radius bisecting a is the symmetry axis between consecutive vertices and the apothem of a circumscribed n-gon. If the radius of a circle is given, can you compute the side and the perimeter of an inscribed and a circumscribed regular n-gon? The latter is an expansion

of the first, in what ratio of similitude?

Note that the

apothem of the circumscribed n-gon is the radius of the inscribed n-gon.

306

Unit 3

Theorem III-6: For the same circle, the perimeter of an inscribed regular polygon is larger fora 2n-gon than for an n-gon; the perimeter of a circumscribed regular polygon is smaller for a 2n-gon than for an n-gon.

Proof (a) inscribed: AM + MB > AB, then

show compute

that and

compare the perimeters.

(b) circumscribed: TC is 4 of the side of the n-gon; RS is 4 of the side of the 2n-gon; show that RS < CS < CT, then compute and compare the perimeters. It follows from Theorem III-6 that for a given circle the perimeter of the inscribed regular polygons increases if n, the number of sides, in-

creases. These polygons approach the circle more and more from the inside, while the circumscribed polygons, whose perimeter decreases with increasing n, approach the circle from the outside. The difference in perimeter between a circumscribed and inscribed n-gon becomes smaller and smaller for increasing n, yet the circle is enclosed between them. Thus we can approximately compute the circumference of a circle by enclosing it in smaller and smaller brackets between the perimeters of inscribed and circumscribed n-gons of increasing n. For a circle of diameter d, the perimeter p, of the inscribed n-gon and P, of the circumscribed n-gon is given in the table on the next page. Check the first few of these numbers by your own computation. Since the diameter d may be any number, we see that: > The ratio c:d between circumference and diameter of any circle is the same fixed number, which is called 7.

This fact was Archimedes

already known

to Hippocrates

(5th cent. B.C.).

(3rd cent. B.c.) found that the numerical value of 7

is approximately #. The value of 7 is not a rational number, which means it cannot be expressed as a ratio of two integers.

Mapping

307

n

Pp

Cc

P

6 12 24 48 96 2 384 768 1536 3072

d-3.00000- - @3:10583 >=. d-3.13263--de 313939 coe d 3 14103 = * d-3.14145--d-3.14156- > d-3.14158--d- 3.14159 d-3.14159---

Oper ee G3, Painai CsEMSA Vai Sule d-3.14:----Oe Sse d-3.141d-3.141---d 3141 < o* d*3.14159:-:

d-3.46410:-CIS aed WokelePoet dS 5966)>\.< d-3.14609--d*3:14271-~* Ce Stal 8 fat d-3.14166:-:d- 3.14161-d-3.14160:-ase Waleok! Dic

The decimal fraction derived in the table for the first five decimals has no end, however far you may compute it. 7 is one of those

irrational

numbers

that puzzled

the Greek

geometers,

when they discovered that two lengths, such as the circumference of a circle and its diameter,

could be incommensurable

(review

Unit 1, Section 7). With electronic computers today there is no difficulty to compute as many decimals as one could wish, but there also is no need for that. Mostly a7 = 3.14159 --- or evena = 3.14 +--+ is quite sufficient; usually 7 may be left as such in the answer. Some people impress us by knowing many decimals of 7 by heart; they probably have not memorized the digits as such, but have a little trick of their own, such as knowing a sentence whose

words each represents a digit through the number of its letters. You can do that too, if you wish to impress your younger brother or sister. Maybe you can even think of a poem! Up to 30 decimals

a = 3. 141592 653589 793238 462643 383279... Now I need a verse recalling pi: Digits found for ratio diameter, perimeter. Ancient geometers got pi. Now remember this series of digits With use for geometry! But pi extends endlessly! The idea that the circle can be approached from the inside and the outside by polygons with increasing number of sides is sometimes expressed by saying that the circle is the limit of an n-gon as n becomes infinite (n — ©); or briefly, the circle may be con-

308

Unit 3

sidered as a polygon with infinitely many sides and infinitely many vertices. Infinite (symbol ©) is a short word for a complicated idea: n> means that however big 7 is taken, there are bigger ones, without ever reaching a biggest. Thus « does not mean a biggest number (there is no such) but it means a process of approximation. In advanced mathematics it can be proved that with increasing n the difference in perimeter between any sequence of outside and inside polygons approaches zero, which means that the perimeters tend to be more and more the same length. This same length that can only be approached but never reached by polygons is defined as the circumference of the circle. In technical language the word limit (lim) is used to express these ideas. We write:

lim(P,—p,)=0 and limP, = lim p, = c From c= ar follows: d em» The circumference of the circle is

c= md= 27r.

EXERCISES 1. a. If the radius ofa circle is 5 in. (26 in., 15.3 in.), compute its circumference. b. If the diameter of a circle is 6 in. (17 in., 23 in.), compute its circumference. c. If the circumference of a circle is 127 (94.25 in.), compute its radius. 2. If the radius of a circle is 4 in., how long is the side of a circumscribed square and of an inscribed square?

3. What is the ratio of similitude between the two squares of Exercise 2? 4. If the radius of a circle is 6 in., how long is the side of an inscribed regular hexagon and of a circumscribed regular hexagon?

5. What is the ratio of similitude between the two regular hexagons in Exercise 4?

6. How long is the side of an equilateral triangle, formed by every other vertex of a regular hexagon of radius 5 in.? 7. Prove that in Exercise 6 the side of the triangle is a perpendicular bisector of the radius it intersects.

Mapping

309

8. Show that the sides of an inscribed and a circumscribed regular hexagon for the same circle are corresponding parts of similar triangles with the same vertex O. If the radius of the circle is 7 in., how long are the apothems of the two n-gons? 9. Show that for a given circle and any pair of regular inscribed and circumscribed n-gons this proportion’ holds: the side of the smaller n-gon is to the side of the larger n-gon as is the apothem of the smaller n-gon to the radius of the circle.

10. Show how the proportion in Exercise 9 can be used to compute the side of one of these n-gons if the side of the other is known. 11. Given a circle of radius R and

the side AB of an inscribed

C

regular polygon;

AVM

compute the

side AM of another inscribed

ow,

regular polygon, with twice as many sides. (Usert. AOFA to

VES 2

compute OF, then express MF in terms of R and OF; compute AM from rt. AMFA.) 12. In Exercise 11, how long is the

By lil

side CD of the circumscribed regular polygon?

13. Compute the first few digits of 7 by comparing the perimeters of inscribed and circumscribed squares, octagons, 16-gons. 14. Is the circle rotationally symmetric? centrally symmetric? bilaterally symmetric? Can any two points of a circle be considered corresponding points in a bilateral or rotational symmetry, carrying the circle into itself? 15. Can any two points A, B on a given circle be considered vertices of an inscribed regular polygon? *16. Review the axiom of Eudoxus (Unit 1, Section 7.1) and its usefulness in locating an irrational number on the number line. Can you see a relationship between his idea and the idea of enclosing a circle between n-gons of increasing n?

6

ROTATION

IN SPACE

6.1 Rotation Around an Axis Suppose a regular polygon is the plan view of a prism or, together with its diagonals, the plan view of a pyramid (Figure I).

310

Unit 3 0

0!

o!!

Then its center O is the plan view of a vertical axis o about which the prism or pyramid is rotated whenever the polygon is rotated about O through angle a. In front view this axis appears as o" perpendicular to the polygonal base, whose front view lies on

the straight line through A”B”.

Another plane section perpendic-

ular to the axis (a cross section) also contains a rotation about

a point, namely around its intersection with the axis. A plane through the axis (an axial section) and through A is carried into an axial section through B, relating the edge a to the edge Db. © Rotation about an axis (the axis of rotation) is a one-to-one correspondence between pairs of points in space such that any pair of corresponding points lies in a plane perpendicular to the axis (a cross section) and is related through a rotation about the point of intersection through the same angle of rotation. °

What we have learned about the order of rotation n =

, the

inscribed and circumscribed circles in the plane, and the circle as

a limit can easily be adapted to space: interpreting the circles as plan views of cylinders or as base circles of cones we see, for instance, that: & A cylinder (cone) can be circumscribed around or inscribed in a regular prism (regular pyramid);

> A regular prism (regular pyramid) is rotationally symmetric about its axis, with order n = number of lateral faces or lateral edges; > A cylinder (cone) is the limit of sequences of prisms (pyramids), if n becomes infinite.

Mapping

311

Figure I shows rotationally symmetric objects of order 6. In problems about such objects, it is often sufficient to study the 6th (nth) part, namely the small triangular pyramid or prism (Figure I) with the axis as height and edge at the same time. It may be called the characteristic part or sector, because it describes the whole object. In the particular case of Figure I the characteristic sector is a bilaterally symM

metric object. Thus one half of it (4 of the original object) already describes the whole

(0)

object, if a combination of bilateral reflection

and rotation is considered; a characteristic part also. Of particular practical

it may be called interest

are

faces of these characteristic sectors: rt. AAOV connects height and edges;

the

M

A

rt.

AMOV connects height and lateral height (slant height); rt. AAMV connects lateral height and edges; rt. AAMO connects base edge, angle of rotation, foot of height.

WA

Vv

Leal

iS

A

Hl

These characteristic triangles make it possible to reduce threedimensional problems to two-dimensional ones. This helpful principle was discussed already in Unit 1, Section 9.2. The axial triangles AOV and MOV may appear in the same axial section or not (Figure II); in a triangular pyramid they do, in a hexagonal pyramid they do not. In the latter case the axial sections are bilaterally symmetric figures. Figure IV shows the connection between rotation about an axis and reflection with respect to an axis. In the two-dimensional

312

Unit 3

(a)

(c)

(b)

case (a) a pair of inversely congruent triangles, reflections of each other, may be interpreted as related also through a three-dimen-

sional rotation through 180°. (See intermediate rotation angles in the plan view; here the triangle appears as a segment, because vertices B and C have the same plan view.) In the three-dimensional case, (b) and (c), this is not possible. Adding points V, and VW, to make a pyramid out of the triangle gives us two possibilities: in Figure IVb there is a rotation about an axis, as seen from the plan view, but then the pyramids are not reflections of each other. In Figure IVc they are reflections of each other with respect to the symmetry plane > (seen here from the side as a straight line), but they cannot be mapped into each other by a rotation. EXERCISES 1. For the regular square pyramid S-ABCD explain what we mean by axial sections and cross sections. Are all cross sections similar figures? Are all axial sections similar figures? Are some axial sections congruent figures?

S

N c L A

K 8 Ex.1-8

Mapping

313

. In the Figure, K, L, M, N are the midpoints of the base edges. Prove that: a. ASAC is isosceles d. ASNL is isosceles b. ASDB is isosceles e. ASAC = ASDB c. ASKM is isosceles f. ASKM = ASNL 3. Prove that ASFA and ASFK cannot be congruent.

4. If the base edge AB is 6 in. and the height SF is 4 in., compute: a. the slant height SK, b. the lateral edge SA, c. the radius of the inscribed cone, d. the radius of the circumiscribed cone. . Given AB =6 in. and AS= 10 in., compute SK and SF.

. Given AB=6 in. and SK =4 in., compute SA and SF. . Given SF= 24 in. and SK = 26 in., compute AB and AS and the radius of the inscribed cone and the radius of the circumscribed cone.

. Which are the characteristic triangles in the Figure?

Which

is a characteristic sector? . A

cone

whose

radius

is 5 in. and

whose altitude is 12 in. is inscribed in a regular square pyramid. How long are the base edge, slant height, lateral edge, and height of the pyra-

mid?

)

ae

a 10. Repeat Exercise 9 for the case that the cone is circumscribed about a Ex. 9 and 10 regular square pyramid. What is the order of rotational symmetry of your pencil? 11. 12. Describe objects:

(a)

the rotational

and bilateral

Ex. 12

symmetries

of these

314

Unit 3

13. A regular hexagonal pyramid of 12 in. height and 5 in. base edge is to be represented by a wire model. How much wire is needed? (Which characteristic triangle will help you find the lateral edge?) 14. What is the height of a regular square pyramid with equilateral triangles as faces and a base edge of 3 in.? 15. Compute the height of an octahedron with 3 in. edges. 16. What is the order of rotational symmetry of a tetrahedron? of an octahedron? 17. Compute the height of a tetrahedron of edge 5 in. (and in general s units). Where in the base triangle is the foot of the altitude? (What kind of triangle is the base?) How far is this point from each base vertex? Which characteristic triangle will help? 18. Compute the edge s of a tetrahedron if its height h is given. (Use experience and answers from Exercise 17.) 19. Compute the height of a lateral face in a tetrahedron of 5 in. edge. 20. A regular square pyramid is inscribed in a cube such that the bases coincide. Compute the lateral edge and the lateral height of the pyramid if the cube is given.

6.2 Regular Polyhedra The three-dimensional

counterpart to regular polygons in the

plane are the regular polyhedra.

A polyhedron

is a solid with

many plane faces; and a regular polyhedron is one whose faces are congruent regular polygons, and whose vertices are surrounded by the same number of such faces. A cube is a regular polyhedron (how many and what faces at each vertex? how many faces in all?) and so is the tetrahedron (four equilateral triangles; how many at each vertex?). Although there are infinitely many regular polygons in the plane (the only condition being na = 360°), there are only five regular polyhedra. It is not difficult to show this, and at the same time to find out what they must be like. At each vertex there are m congruent angles a belonging to the m congruent faces forming this vertex. Obviously, there must be

at least three faces to form a corner at all; thus [m = 3]. On the other hand, the sum of these angles around one vertex must be less than a full angle of 360°, otherwise the faces would spread out in

Mapping

315

a plane and not form a corner; thus Ima < 360°). Now, what values can a take? Since each face is a regular n-gon, let us examine all possibilities for n. The smallest n is obviously 3; then the faces are equilateral triangles and a = 60°. Together with the inequalities m 2 3 and ma < 360°, we find that only 3 x 60°, 4 x 60°, 5 x 60° can form a corner of a solid, because 6 X 60° = 360° violates the second in-

equality. Thus, there are these three types of regular solids with equilateral triangles as faces: the tetrahedron (with 3 As at each vertex, 4 Ass in all), the octahedron (with 4 As at each vertex, 8 As in all), the icosahedron (with 5 /\s at each vertex, 20 As in all).

For n = 4 the faces must be squares and

a = 90°.

Then only

3 x 90° can form a corner, because 4 X 90° = 360°; thus the only

regular solid with square faces is the cube or hexahedron (with 4 squares at each vertex, 6 squares in all). For n = 5 the faces are regular pentagons and a = 108° (is it?). Only 3 X 108° is possible; the pentagon-dodecahedron has 3 pentagons at each vertex, 12 pentagons in all. For n = 6 the faces are regular hexagons and a = 120°; but since 3 x 120° = 360° there is no such solid. The five regular polyhedra are sometimes called the Platonic solids (Figure I, on page 316). There is a fixed relationship between the numbers of vertices V, faces F, and edges E of a regular polyhedron:

Theorem IIl-7: Euler’s Theorem The sum of the number of vertices and faces of a polyhedron is two more than the number of edges.

(V+F=E+2)

Euler’s theorem holds not only for regular polyhedra but even for any convex polyhedron (a polyhedron, where a straight line segment between any two points of different faces lies entirely in the interior of the solid).

Verify Euler’s theorem

by counting the vertices, faces, and

edges of the five Platonic

solids in Figure I, and of some other

solids.

Complete the two tables on the next page:

316

Unit 3

\ PENTAGON-DODECAHEDRON

aN

ee

4 5)

3. 4 5. 3 3.

HEXAHEDRON

OCTAHEDRON

TETRAHEDRON

tetrahedron octahedron icosahedron hexahedron pentagondodecahedron

ICOSAHEDRON

PSS

we

Prism, 3 4 n Pyramid, 3 4

sided sided sided sided sided

n sided

* To prove Euler’s theorem in general, let us build up a convex polygon from one face by adding one adjacent face at a time while counting vertices, faces, and edges. The first face is a polygon of v, vertices and e, edges and f, = 1 face; since in any polygon v = e, we may say v, + f, = e, +1. When adding one adjacent face, we must add one more edge than vertices, because the two faces have two vertices but only one edge in common; if v vertices are

Mapping

added, then v, =v, +v, Via Ve Cpa Coa — stituted into the equation —l+lorv.+f,=e,+1. equation as before. The

317

eg=e,+v+1, f=f,t1, or |. fi — f,1, which.can be.subabove: v,-—v+f,—1l=e,—v Thus, we get the same form of next face to be added may have

one or two edges in common

with our structure so far, but

we still must add one more edge than vertices. The same computation leads to v, +f, =e,+ 1, and so on till we reach the last face. Adding this last face seems like putting the iid on a many-sided box; whatever n-gon we have to put on, all its vertices and edges are already there and counted, only the number of faces is increased by 1; Vee eo Ce) |) ee dean alter substitution into the previous equation v, ,+f, ,—€,_,+ 1, we get EE) lee Comin ROL AY ot ae 1) 0G. C.C-

Theorem Ili-8: A regular polyhedron can be inscribed in and circumscribed about a sphere.

Proof: Each face is a regular polygon with acenterC. All points equidistant from the vertices of one face lie on a perpendicular to that face through C. Two adjoining faces J and IJ have an edge

AB in common, and all points equidistant from A and B must lie in the symmetry plane

of AB, that is the plane > per-

0

pendicular to AB through its _ midpoint M. Plane > must therefore contain the two perpendiculars at C, and C,, which intersect at O. This point O is equidistant from all vertices of J and JJ and, by repeating the reasoning for the next face, also from all other vertices.

of the circumscribed sphere.

O is the center

From symmetry reasons the right

318

Unit 3

triangles MC,O

and MC,O are congruent and OC, = OC, = OC;

and so on; therefore point O is equidistant from all faces. O is the center of the inscribed sphere. Point O is called the center of the polyhedron. A relationship between the radii R, and R; of the circumscribed and inscribed spheres and the radius r, of the circle circumscribed about a face can be read from the right triangle C,DO: RAS Re ar U2, Uu 1

A regular polyhedron of n faces may be considered as consisting of n congruent regular pyramids with common vertex in the center of the polyhedron. Sometimes such a pyramid is quite helpful in analyzing a problem about a polyhedron. The axis of rotation of such a pyramid is also an axis of rotation for the polyhedron; what is the order of the rotation? An axis through any vertex and the center of the polyhedron is also an axis of rotation; what is the order of rotation?

cuboctahedron

CONGRUENT

octahedron with cube

VERTICES,

DIFFERENT

REGULAR

°

cube with octahedron

FACES

i rhombic dodecahedron

CONGRUENT

tetrakishexahedron

IRREGULAR

?

FACES,

hexakisoctahedron

DIFFERENT

triakis— octahedron

VERTICES

Although there are only five regular polyhedra, there are many partially regular polyhedra (see Figure I). The so-called semi-

Mapping

319

regular polyhedra are of two types: they have either congruent vertices but more than one kind of regular faces (Archimedean solids) or they have congruent faces but not always the same number of them around each vertex. Also there are many combinations of faces which form polyhedra of various degrees of regularity. Many beautiful polyhedra are formed by nature when certain chemicals are solidified from a liquid state to crystals. Diamonds and other precious stones are cut in various polyhedral forms. Crystallography describes the geometric shapes of crystals formed by specific minerals. If you are interested, you could make a special report on crystal forms in nature, or on various forms used in cutting and facetting diamonds. (See Suggestions for Special Projects.) EXERCISES 1. Make a net and a model of an icosahedron (see Figure I on page 316). 2. Make a net and a model of a pentagon-dodecahedron (see Figure I on page 316).

3. Describe the five Platonic solids. Are all of them bilaterally symmetric? centrally symmetric? rotationally symmetric about some axis? Is the order of rotational symmetry different for different axes? Taking two adjacent vertices as corresponding points, is there an axis of rotation mapping them into each other? The cube as a regular polyhedron

4. Compute the radius of the inscribed and the circumscribed sphere for a cube whose edge is 6 in. Which right triangles will help? (Make a sketch.) 5. How long is the edge of a cube that is circumscribed about (inscribed in) a sphere with a 3 in. radius?

6. Demonstrate the relationship R,?=R,? +r,” in a pictorial view of the cube. The tetrahedron with edge s 7. Describe the tetrahedron as a regular polyhedron: a. What are the shape and number of its faces?

b. Where is the center of a face?

How big is its radius r,?

c. Is the face rotationally symmetric? What order? d. Is the altitude of the tetrahedron a rotational Where is the foot?

axis?

320

Unit 3 e. Does the center O of the tetrahedron lie on this altitude? Does the center of the circumscribed and inscribed sphere lie on this altitude also? f. What is the shape of an axial section through a lateral edge? . Compute the lateral height of aface. (Which rt. A will help?) . Compute the height of the tetrahedron for the edge s in general and for s=5 in. (Reduce to two dimensions by axial section; which rt. A will help?)

*10. Compute the radius R,, of the circumscribed sphere. duce to two dimensions by axial

section through AD.) = es

(Re-

OA =OD

the altitude DF was com-

puted in Exercise 9; how big is

AF?

Usert. AAFO to set up an

equation

for the unknown

R,.)

boWr Compute the radius R, of the inscribed sphere. glPA Show that the center O of a tetraExa© hedron lies at ; of the altitude above the centroid of the base. 13. Using Exercise 9 and the statement in Exercise 12 check the relationship R,?= R,?+1,? for the tetrahedron. The octahedron with edge s

14. Describe the octahedron: a. What are the shape and number of faces? b. Where is the center? c. What is the shape of an axial section through a lateral edge? d. Show that each of the three axes (segments between opposite vertices) is a diagonal of a square and equals s V2. e. Show that the octahedron consists of two congruent

square pyramids, each with height =1sV2. 15. How high is an octahedron, if its edge is s (or s=3 in.)? 16. How long is the edge s of an octahedron, if its height H is known? (H = 10 in.) 17. How long is the radius R,, of the sphere circumscribed about an octahedron of edge s? 18. Compute the slant height in a face of an octahedron. (Which rt. A will help?)

Mapping

321

19. Use the relationship R,?= R,?+r,? to compute the radius of a sphere that is inscribed in an octahedron. Polyhedra

20. Explain the meaning of Euler’s theorem for the cube; for one of the polyhedra in Figure II.

and

21. Which of the polyhedra in Figure II are Archimedean solids? 22. Give an example of a semi-regular polyhedron that is not Archimedean. 23. Make a pictorial sketch of an Archimedean solid.

| |

7

GROUPS

OF TRANSFORMATIONS

7.1 Combining Transformations

Given a two-dimensional Cartesian coordinate system and a geometric figure F in its first quadrant; if F is reflected in the yaxis, and the image reflected in the x-axis, then the image of this image could have been produced also directly by a central symmetry. It could also have been produced by three successive rotations through 60°. Show how. It could be produced in many other ways. Name some. In mathematical language these facts can be expressed in the

following form:

T,7,F =T,F and T,°F =T,F;

it follows that

Tol =TY. The code for this message is easy to break: F = geometric figure T, = reflection in the y-axis (RE,y) T, = reflection in the x-axis (RE,x)

T; = central symmetry with respect to the origin (CS)

T, = rotation about origin through 60° (RO,60°) fAF ed be ABWES T,T, = first apply T, and then T,. (Customarily the later operation is written at left of the previous one.) T,T, =T,° means:

applying (RE,y) and then (RE,x) gives the

same result as applying three (RO,60°). The mathematical form is short and precise; it suggests that computation rules may be set up for transformations, similar to those in arithmetic. It is customary to write transformations next to each

322

Unit 3

other, or, if that is not clear, with a dot between them.

If we com-

pare the combining of transformations to adding or multiplying numbers, then we have immediately a few questions to settle: (1) Is there an identity transformation

J? Are there inverse trans-

formations T-* such that T-!T = 1? Is the operation closed within a certain set S of transformations? (If T, € S and T, € S, is ToT, € S also?) What do these questions mean for multiplying numbers and how are they answered in arithmetic? Instead of a set of numbers, we have here a set of transforma-

tions. Let us take a specific example, e.g. the set SH of all possible parallel shifts within a given plane (shifts in any direction and by any distance). The identity shift is obviously a shift by a zerodistance. To any specific shift there is an inverse shift in the opposite direction by the same distance, so that applying both shifts one after the other amounts to not shifting at all — the identity shift. Any two shifts combined are equivalent to some single shift. Demonstrate by examples. This means the set SH is closed under the operation “combining shifts.” Investigate the answers to the same three questions for the following sets:

(a) the set RO of all possible rotations in a given plane about a given point; (b) the set (RE, axis) of reflections, in the plane, with respect to

any axis; (if two such reflections are applied successively, is the resulting figure a reflection of the original figure? three reflections? four reflections?) (c) the set EX of expansions, in the plane, from any of its points. (Can you prove whether or not two successive expansions from different points can always be replaced by some single expansion?)

Suppose figure F, (e.g. a triangle) is mapped into F, by an expansion EX; after that F, is carried into F, by a SH; then F, is rotated about some given point (e.g. one vertex) into F,. We could write this procedure as: F,—>

F,—>

F,— >

F,

or as

RO-SH:-EX-F,=F,

or as

TF, =F ,, where T=RO-SHEX

Mapping

323

We know something about this composite mapping T, if we know from which simple mappings it was formed. We know, e.g., that T preserves right angles (why?), and even that F, must be similar

to F’,, without even looking at figures F, and F,, between.

(How

do we know all that?) Is it true that “shape is an invariant under 7”? If we know that F,, = [SH(||y by 5 in.)] - [RO(60° about P)] -

[SH(||x by 4 in.)]F,, what can we say about F, and F,, without even carrying these mappings out? If we know that F,, and F,, are inversely congruent, what type of transformation (simple or composite) will certainly map F, into F,? Is more than one type possible? Is it true or not that any pair of similar figures or objects can be mapped into each other by a transformation T of the type T = EX RE -RO-SH? Ina specific case the mappings contained in the composite T could be given with more detail. A bilateral figure (e.g. an isosceles triangle) can be mapped into itself by RE alone; RE can be written as a special case of T, namely T = EX(/) - RE (symmetry axis of A) - ROW) - SHU), where J stands for “iden-

tity.” ROU)

Accordingly, EX(W/) means “expansion in the ratio 1:1,” means “rotation through angle zero,’ and SH(/) means

“parallel shift by distance zero.” means

Then the procedure TA, = A,

“don’t shift the A, don’t rotate it, reflect it with respect to

its symmetry axis, don’t change its size” and the result will be Ag.

Definition IIl-4: A set S= {T} is called a group, if an operation is defined combining any two members T such that:

. the set is closed under this operation (7,7, = T,); . the operation is associative: (7,7,)T,=T7,(T,T,); . there is an identity member (7T/ = T and IT= 7); . for each member T there is an inverse member 7-! = RWYN such that T7-!=J/; and T?3T=1I.

If the operation is also commutative (7,7, =T7,T,), then the set is a special kind of group, Abelian group. (Niels Henrik Abel, 19th cent.) i

called an

324

Unit 3

Is the set of all integers a group under addition? (Which four questions must be investigated?) Is it an Abelian group? Is the set of counting numbers a group under addition? Is the set of all parallel shifts in space a group? An Abelian group? . Do all reflections with respect to planes form a group? (Are two successive reflections replaceable by a single reflection?) Do the transformations T= EX - RE - RO - SH form a group? What do the following statements mean? (1) Shape and size are invariants under the group of transformations T= RE: SH: RO. (2) Euclidean geometry is the study of the invariants under the group of transformations T= EX - RE - SH- RO. EXERCISES

1. Given AABC and two vectors: V, in the direction of AB but twice as long as AB, and V, in the direction 1 V, and as long

as V,. These two vectors define two transformations, the parallel shifts T, and T,. Construct 7,A and 7,7,A. Which single transformation T,=T,7T, could have led to the same result?

hee

=

Exec

2. Given are the axes s,, S,, 5,, 8, through point O, forming 45° angles. T,=RE,s,, T,=RE,s,, etc. Given AOAB, draw be Cs a I EM RT Wg I 0S Be ITE LIEN, Si sine Dg te ake

SAREWE I WY DY YN EY TY 1

ns Or

2d

SV

4.)

BE IESE Soon

tee

VETS T: dh, Si

oad

3. To get the same result as in Exercise 2, what other transformations could have been used? Are there more possibilities? Demonstrate.

Mapping 4. In Exercise 2, what would fiesBAOadteal Bed Road Od IES RAR, eee sa NS Os

325

be a shorter way to describe

5. In Exercise 2, can you name some single transformation that is inverse to T,T.T,T,A? feNair Saar eT | 6. Given equilateral triangle ABC, which transformation will produce Figure (a), and which Figure (b)?

Rack. C

C

B

B

a

(a)

(b)

Ex.6

7. What is the inverse transformation T-! to:

a. b. c. d. e. f.

T T T T T T

= turning about a point through +180°? = turning about a point through (first +90°, then —180°)? =enlarging to double size? = reducing to % in scale? =reflecting in the y-axis? =reflecting in the xz-plane?

8. If 7,=rotate about O through 90°;

T,=rotate about O

through 45°; 7T,=reduce in the ratio 1:2, from point O; what happens to any triangle ABC under the composite trans-

formation TA=T,T,T,A? meaning of T=

(Make a sketch.)

7,~T,“T,4?

What is the

Of

dB) Bd Meld (at 9 lead BS Up Wes (Remember that a later transformation is written at the left of the previous one.) 9. What is a group?

An Abelian group?

10. Do all “rigid motions” (parallel shifts and rotations and all combinations of them) form a group? Which four questions do you have to investigate?

1

7.2 Repetitions

Of special interest are sets consisting of one transformation and its repetitions. Take, for example, T =rotation through 90° about the fixed point O. The repetition TT (also written 7) is

a rotation through 180°, T* = rotation through 270°, T* = rotation through 360° or no rotation at all, 7° = T is not a new element.

326

Unit 3

\ The set RO,90° = {7,T?,7?,T*} has four members. Combining any two members gives another member. (7°T* = ?,T°T* = ?) Does it make any difference whether (T7*)T* is applied or T(T’T?)? Obviously, T* =J gives identity. Is there a member that complements a given member to a full rotation or identity? Does the set RO,90° form a group? An Abelian group? 360° If the angle a of rotation is the nth part of 360°, then n = is the order of rotational symmetry, and also the number of different members in the set of repeated rotations. The fact that T" =I and T"+1=T, starting all over again, is expressed by the name cyclic group of ordern. RO,90° is acyclic group of order 4. RO,60°, the set of rotations through multiples of 60° around a

fixed point, is a cyclic group of order 6. cyclic? Why order 6?)

(Why a group?

If T = (RE, y-axis), what is T?, T*, T*, T°?

Why

How many dif-

ferent members are there? Do they form a cyclic group? Of what order? Is there an identity member? An inverse member? Now, let us discuss a parallel shift SH and its repetitions. Suppose a wallpaper design is needed to form a strip around the walls of a nursery. It might show animals repeated over and over again. In the figure, the distance between

any point on one duck and the corresponding point on the next duck gives the vector (length and direction) defining the parallel

eee

shift. The set of shifts is {7, T?, T*, T*,...7"}, where T means “shift by one duck,” JT? means

“shift by two ducks,” etc. and n depends on the length of the strip needed to go around the walls and to connect with the first duck. Then 7” yields the first duck again (T” = 1), T"*+1 the second, and

Mapping

so on.

327

Show that this set forms a group, specifically a cyclic group

of nth order, just as the rotations above; if the room were round, the J’s would be rotations. If n = 100, what is the inverse of T°?

But if we did not paste the strip around the walls to connect with

the first duck,

if we

considered

it as a strip theoretically

without end, would it then demonstrate a group? Any two shifts will still be replaceable by one shift through a longer distance than either of the two; but 50 or 200 repetitions will not lead to the first duck again. There are no inverse members in this {T”}. Examples for repetitions of shifts in non-mathematical context are found in the repetition in time of a regularly recurrent grouping of accents

on words, of steps, of tones, etc. (poetry, movement,

music). Rhythm is the periodic repetition in time of some feature. In poetry there is a rhythmical pattern of accents called the

verse meter: Twinkle, twinkle, little star, How I wonder what you. are! Have you been there every night? Could you be a satellite? 1 wv

Up above the world se) high Like a station in the sky Tuning iin to near and far, Could you be a TEL star?

The metric foot “ ~ is the unit that is repeated.

Note how an un-

expected break of the pattern, the pause in the last word, is used

for special emphasis. In movements, the simplest example is the marching rhythm of {234,1234,.... Other examples are play-school singing games (Here we go round the mulberry bush), where children form a circle, move along the circle and perform certain movements at certain periodic intervals, indicated by the musical rhythm. “Pat-a-cake” is a pattern of hand movements, symmetric within the repeated unit by using the left and right hand alternately. Also dancing steps, e.g. the waltz steps, are examples for repetition of motion with bilaterally symmetric units alternating left and right. In music the rhythm is the accent pattern within the 1 Courtesy WWDC

Radio, Inc.

328

Unit 3

repeated measures. In canon singing two or more identical voice parts are shifted by a fixed number of measures. In arithmetic any fraction may be expressed as a decimal in which, from a certain place on, the digits or groups of digits repeat indefinitely. Divide the numerator of a fraction by its denominator, and you will get such a repeating decimal number; 4= 0.50000: - -, $= 0.33333 - + -, 2= 0.285714 285714 285714 - - -. An arithmetic progression is a sequence of numbers, where the operation

T = (add the number

N) produces

the next number;

for example, 2, 5, 8, 11, -- + is an arithmetic progression. A geometric progression iS a sequence of numbers, where the operation T = (multiply by the factor NV) produces the next number; for example 2, 6, 18, 54, -- + is a geometric progression. In higher mathematics periodic functions play a great role. As an example the function “y = 1 if x is an integer, otherwise y = 3” is plotted, so that you can see the pattern of this periodic function as alternating dots and segments. The length of the repeated unit, a dot and a segment, is the period. The goniometric functions are very important periodic functions. You will see their periodicities best from their graphical representations.

Each function is referred to a unit circle, where

the value for a particular angle is represented by a particular segment.

(Review

Section 3.1.)

On the abscissa of the graph for

902 180°

180°

270°

360

20s

90°

SINE

180°

180°

720°

ZiOe

COSINE

Mapping

329

each of the three functions equal segments are marked off to represent equal numbers of degrees (here for every 30°). The length of the ordinate at each mark is taken from the corresponding segment in the unit circle. Then the plotted points are connected to a smooth curve. Note that the tangent function approaches an

asymptote at 90° and at 270°.

TANGENT

| | | | |

After constructing these graphs for angles from 0° to 360° it is easy to imagine what their continuation would look like for another 360°, and yet another. You do not even have to construct these curves point by point, since you can tell right away that, for instance,

sin.(a + 360°, 720°,:--)=sin-cos (a + 360°, 720°,- - -)=cos::tan (a + 360°, 720°,---)=tan--also tan (a + 180°, 540°,-- -)=tan---

If a function repeats its values at periodic intervals, it is called a periodic function. The smallest interval of repetition is called the period of the function. ®& The goniometric functions are periodic functions. The period of the sine and the cosine function is 360°. The period of the tangent function is 180°.

330

Unit 3 EXERCISES

L Describe the patterns in the following designs by their symmetries and units of repetitions.

. Find examples of periodic design on your family’s china, on vases, in flowers, in architecture, especially on domes, columns, and portals. Describe their symmetries and periodicities. . Design some patterns of your own and describe their symmetries.

4. What is

acyclic group?

What is the order of a cyclic group?

. How many different members does the set {7 and repetitions of T} have, if T = reflection with respect to the x-axis? . Write five members of an arithmetic progression starting with 3, if T = (add 4). . Write six members of a geometric progression starting with

1, if T = (multiply by 5). . What is a periodic function? What is the period of such a function? Make up some examples of your own. . Construct graphs of the sine, cosine, and tangent functions. Discuss their symmetries. What are their periods? Can you draw these functions from memory?

10. Here are some Gothic ornaments, frequently seen in Gothic architecture. Construct them by circle arcs with centers as marked. Discuss their symmetries. Are there cyclic groups of transformations?

aS

(a)

Mapping

331

ob 2,

SUMMARY

TO

OF UNIT III

A. Mappings A mapping is a correspondence between two sets of points, which is defined by a definite procedure of how to find corresponding elements. We have discussed four simple mappings (expansion, reflection, parallel shift, rotation) and composites of these mappings. From the procedure of finding corresponding points it can be seen that these mappings are one-to-one correspondences. EX:

All connectors pass through a fixed point, the center of similitude, which divides the segment between corresponding points in a fixed ratio. The division may be external (external center of similitude, positive ratio) or internal (internal center of similitude, negative ratio). The special case when the ratio is —1 is called central symmetry. RE: Connectors between corresponding points are perpendicular to and bisected by a fixed axis (or plane) of symmetry. SH: Connectors are parallel and of equal length (represented by a vector). RO: In the two-dimensional case, corresponding points lie on some circle around one fixed center of rotation, and the corresponding radii form a fixed angle of rotation. In the three-dimensional case, corresponding points lie in some plane perpendicular to a fixed axis of rotation and are related to each other by a two-dimensional rotation about the intersection with the axis. Composite mappings: Corresponding points are connected by successive application of the simple mappings which form the composite mapping.

If two geometric figures or objects are connected through one simple mapping, namely:

332

Unit 3 EX, RE, SH, RO,

then then then then

they they they they

are are are are

similar; inversely congruent; directly congruent; directly congruent.

A group is a set whose members can be combined by an operation satisfying four conditions: (1) a combination of two members can be replaced by one member of the set (the set is closed), (2) the operation is associative, (3) there is an identity member, and (4) for every member there is an inverse member. A set of transformations may form a group. A set of repetitions of one transformation forms a cyclic group, if a certain number n of repetitions is equivalent to the identity transformation. The number zn is the order of the cyclic group and also the number of different members in the set. Euclidean geometry is the study of geometric shape. Shape is an invariant under the group of transformations T= (EX - RE RO - SH).

B. Applications The study of the various types of symmetries helps us to see patterns. These, in turn, help us to analyze problems and to isolate characteristic sections containing the essential parts of the problem. Rotational symmetries simplify the study of regular polygons, pyramids, prisms, and polyhedra. Bilateral and central symmetries of the unit circle relate the trigonometric functions of angles in the second, third, and fourth quadrant to those of the first. Only the latter are usually tabulated, but all are needed (slope of straight lines, polar coordinates, vectors). Parallel shifts are symbolized by vectors. Any magnitude connected with a direction may be symbolized as a vector (e.g. a force). Computation rules for vectors: The sum of two vectors is the From this follows diagonal vector in the vector parallelogram. how to form the difference between two vectors, the product and the quotient of a vector by a number (a scalar). In a coordinate system the sum of two vectors can be computed from the sums of components in the direction of the coordinate axes. (How?) Polar coordinates locate points in the plane by ordered number pairs (r,a), the radius vector and the angle of inclination.

Mapping Conversion

formulae

between

these

and Cartesian

333

coordi-

nates (x,y) permit a change from one system to the other.

. An important idea: approaching a limit ‘““A sequence of polygons with increasing number of sides approaches the circle as a limit’? means all the following statements together: 1. If polygons inscribed in the same circle are arranged in a sequence of increasing number z of sides, then their perimeters p, form a sequence of increasing lengths. 2. If polygons circumscribed around the same circle are arranged in a sequence of increasing number n of sides, then

their perimeters P, form a sequence of decreasing lengths. 3. If the same circle is circumscribed by a polygonA and inscribed by a polygon B, then P,, the perimeter of A, is longer than p,, the perimeter of B. 4. Although the p, in (1) are becoming larger and larger, they cannot become larger than any P, of the same circle. 5. Although the P, in (2) are becoming smaller and smaller, they cannot become smaller than any p,, of the same circle.

6. The differences P,, — p,, for the sequences in (1) and (2) for the same circle become smaller and smaller; there is no

smallest (P,, — p,,). 7. Expressed in decimal numbers the sequences of P, and p,, for the same circle have more and more digits alike (beginning at the left end of the number) without ever being the same number throughout. 8. The circumference of a circle is defined as a length expressed in terms of its diameter by the decimal number of equal digits in the sequence in(7). Itis impossible to write this number out. . Theorems and Corollaries IIJ-1: In an expansion corresponding segments are parallel. Cor. I1J-1-1: An expansion carries a geometric figure or object into a similar figure or object. IlI-2: If corresponding segments of two similar figures are parallel, then there is a center of similitude. III-3: In a rotation through angle a corresponding straight lines form the angle a. Cor. I1I-3-1: If the sides of two angles are mutually perpen-

334

Unit 3 dicular, the angles are equal, if they are both acute or both obtuse; otherwise they are supplementary. Ill-4: A regular polygon has a circumscribed and an inscribed circle; their centers are at the same point. Cor. I1I-4-1: A regular polygon of n sides is divided by the radii to the vertices into n congruent isosceles triangles. The vertex angle in such a triangle (the central angle) is 360 . Cor. IlI-4-2: 180° — 360°

An

interior angle of an n-sided polygon is

; an exterior angle is equal to the central angle.

III-5: A circle can be inscribed and circumscribed by regular n-gons. III-6: For the same circle, the perimeter of an inscribed regular polygon is larger for a 2n-gon than for an n-gon; the perimeter of a circumscribed regular polygon is smaller for a 2n-gon than for an n-gon. III-7, Euler’s Theorem: The sum of the number of vertices and faces of a polyhedron is two more than the number of edges. (V+ F=E+2) IlI-8: A regular polyhedron can be inscribed in and circumscribed around a sphere. E. Words and phrases

Section 1: mapping, transformation, image, expansion, center of similitude,

parallel

shift, rotation,

center

of rotation,

angle of rotation, rigid motion Section

2: internal and external center of similitude, center

of symmetry, reflection in a point, central symmetry, a centrally symmetric figure or object Section 3: goniometric functions, unit circle, polar coordinates, initial straight line, pole, radius vector Section 4: vector, length of a vector, absolute value, scalar, commutative law, associative law, additive inverse, parallelogram of forces, resultant, components of the resultant, components of a vector

Section 5: n-gon, apothem, a rotationally symmetric figure, order of rotational symmetry, infinite, limit Section 6: rotation around an axis, axial section, cross sec-

tion, a rotationally symmetric body, polyhedron, Platonic solids, convex polyhedron, Archimedean solids, semi-

regular polyhedron

Mapping

335

Section 7: identity transformation, inverse transformation, composite mapping, group, Abelian group, cyclic group of order n, rhythm, repeating decimal number, arithmetic progression, geometric progression, periodic function, period

REVIEW

EXERCISES

TO UNIT III

1. Explain the meaning of the following expressions: a. Reflection (or bilateral symmetry) is a transformation that “preserves shape and size.” b. Shape and size are “invariants under rotation.” c. Reflection is a transformation that ‘carries a geometric figure into an inversely congruent figure.” d. An isosceles triangle is a bilaterally symmetric figure. e. A bilaterally symmetric figure may be ‘‘mapped into itself.” f. An equilateral triangle may be ‘“‘carried into itself by three different bilateral symmetries.” g. A circle may be carried into itself by “infinitely many symmetries.”’ h. A given straight line “‘determines a bilateral symmetry in the plane.” i. A given plane ‘‘defines a bilateral symmetry in space.” j. A symmetrical figure is its own image. k. Euclidean geometry is the study of the invariants under certain transformations. 2. What is an external or internal center of similitude? What is the meaning of a positive or a negative ratio? What is a center of symmetry?

3. What is the unit circle?

Why is it a useful concept? What is the difference be-

4. What is a vector parallelogram? tween a vector and a scalar? 7).

What is the difference between rotation and reflection?

6. What is a group of transformations? ~l.

In polar coordinates, what is the equation of a circle around the pole, with a 3 inch radius? Change this equation to Cartesian coordinates. 8. In Cartesian coordinates, what is the equation of a circle around the origin with a 5 inch radius? a 6 inch radius? (Use your answer to Exercise 7.)

336

Unit 3

9. In polar coordinates, what is the equation of a straight line through the pole forming a 45° angle with the initial line? What is the equation of a second straight line perpendicular to the first? 10. In Cartesian coordinates, what are the equations of the two straight lines in Exercise 9?

11. If DEFG is a square inscribed in a triangle ABC as shown, and if from some point P on AF parallels are drawn toJ EF and |GF intersecting AB and AC in Q and

R, then PO= PR. 12. Construct a square inscribed in a given tfiangle. 13. Two triangles with the same

Ex. 1

base AB and of equal height are cut by the same straight line parallel to the base; show that the segments on it within the two triangles are equal. (Two expansions.)

C,

C,

at 1S

Ex. 14

14. A square of side 2 in. is inscribed in a right triangle with hypotenuse of 3V/5 in., such that the right angles coincide. Compute the legs of a right triangle. 15: How many sides does a regular polygon have if an irterior angle is 140°? 156°?

16. What is the sum of all interior angles in a regular polygon? 17. Prove: If two regular polygons have mutually parallel sides, then the ratio between their perimeters is the same as the ratio between a pair of corresponding sides.

Mapping

337

18. Prove: In a regular polygon the radius drawn to a vertex bisects the angle at that vertex. 19. Prove: The central angle of a regular polygon is supplementary to an interior angle. 20. How long is the apothem of a regular hexagon of 5 in. side? 21. What is the perimeter of an equilateral triangle inscribed in (circumscribed around) a circle of radius 2 in.? 22. A square of 3 in. side is inscribed in a circle. What is the perimeter of a regular octagon inscribed in the same circle? 23. What kinds of regular polygons are centrally symmetric? 24. The perimeter of a polygon is 150 in.; its smallest side is 12 in. The shortest side of a similar polygon is 18 in.; how long is its perimeter?

25. If the number of sides in a regular polygon is doubled, each interior angle is increased by 15°; what was the number of sides? (Form an equation between the degrees in the interior angles of an n-gon and a 2n-gon.)

26. For which regular n-gon does an interior angle increase by 16° if the number of vertices is tripled? 27. For which regular n-gon does an interior angle increase by 6° if the number of sides increases by 2? 28. In which regular polygon is the sum of the interior angles equal to 900°? 2700°? 29. Given a triangle, find a point on one side that is equidistant from the other two sides.

30. Show that a rotation of a figure is equivalent to two successive reflections where the two axes intersect in the center of rotation. 31. Show that a parallel shift of a figure is equivalent to two successive reflections with respect to two parallel axes. a2. A regular square pyramid of 5 in. height has base edges 3 in. long; how long are the lateral edges and the slant heights? 33. The base of a regular pyramid is an equilateral triangle of 4 in. side; the lateral edges are 6 in. long; how high is the pyramid? A regular square pyramid of 15 in. height and 12 in. base edge 34. is cut into three layers by two planes parallel to the base. How long are the edges of the two figures of intersection if the altitude has been divided into three equal parts? 55: The edge of a regular octahedron is 3 in.; how long is its diagonal?

338

Unit 3

36. What is the sum of all face angles in each of the Platonic solids? a7. Is it true or not that any rotational axis of a regular polyhedron must pass through its center? 38. In a cube, what is the order of rotational symmetry for an axis through a vertex? an axis through the center of a face? an axis through the midpoint of an edge? 39. In an octahedron, what is the order of rotational symmetry for an axis through a vertex? Are there axes with a different order? 40. Draw the graphs of the: sine, cosine, tangent function from memory, then check with the book. What are the periods of these functions? 41. Are there any values that the sine, cosine, tangent cannot take? 42. What are the signs of the sine, cosine, tangent functions in the four quadrants?

CUMULATIVE

TEST

Group I: Form a true statement by inserting always, sometimes, or never.

. If the diagonals of a quadrilateral are perpendicular, the quadrilateral is .. . a rhombus. . The centroid of a triangle . . . lies in its interior. . The orthocenter of a triangle . . . lies in its interior.

. Three segments in the ratio 5:9:3 will... form a triangle. nan NY WwW &.

If two planes are perpendicular to a third plane, they are... parallel to each other.

. If two lines in space are perpendicular to a third line, they are ... Skew to each other.

. If two lines in space are perpendicular to the same plane, they are... skew to each other. . If two coplanar figures can be mapped into each other by a mapping composed of one rotation, one reflection in a line, and one parallel shift, they are . . . directly congruent. . Two directly congruent coplanar figures can .. . be mapped into each other by a composite of a parallel shift and a rotation. 10. If a pyramid is cut by a plane, the figure of intersectionis . similar to the base.

Mapping

339

Group II: Write the letter of the correct answer on a separate sheet of paper. 1. An interior angle of a regular decagon is: 2050.

b. 72°

c. 144°

d. 108°

2. An exterior angle of a regular n-gon and the central angle are: a. equal

b. supplementary

c. complementary

3. Two regular polygons with the same number of sides: a. have the same perimeter c. have the same radius

b. are similar

4. The total number of diagonals in an n-gon is:

a.n—3

b. n(n— 3)

c. 5n(n—3)

5. The diagonal of a regular octahedron with edge 3 in. is:

Bea

2

eve)

ce. 3V3

6. The altitude of a regular tetrahedron with edge 2 in. is:

a. 3V3

b. 2V2

c. 2V6

7. The legs of a 30°-60° right triangle are in the ratio:

a. 1:V3

b. $:V3

d. he

e. 1:5V3

8. The altitude to the hypotenuse of a right triangle divides the hypotenuse into the segments p and-q. If the adjacent legs are a and b respectively, the ratio p:q equals: a. a:b

b: a*:b?

c. (a?+ b2)2 (a? — b?)

9. The circumcenter of a right triangle is: a. inside 10. The

b. outside

diagonal of a rectangular

a. Vabc

b. Vatbt+c

c. on the triangle solid with edges a, b, c is:

¢« Va?+b2+c?

d. Vab+act+be

Group Ili: Explain the meaning of each of the following statements. 1. Rotation and reflection preserve the size, but expansion does not. 2. Euclidean geometry is the study of the invariants under mappings composed of expansions, rotations, reflections, and parallel shifts.

340

Unit 3

3. A two-dimensional reflection in a line is equivalent to a three-dimensional rotation about that line. But a threedimensional reflection in a plane is not equivalent to a rotation. . If a geographic map is photographically reduced or enlarged, a graphic scale in its legend remains correct, but a numerical scale becomes incorrect.

. The side of a regular decagon divides its radius in the golden ratio.

. The mapping “‘parallel shift’ is defined by a vector. . The reflections with respect to a given plane form a group of order 2.

. The bilateral symmetries form a proper subset of the congruences. The congruences form a group, but the bilateral symmetries do not. . The phrase ‘‘exactly one” in a theorem indicates that two proofs are required. . The sides of AABC are 3, 4, and 5 in. This is a sufficient but not necessary condition for AABC to be aright triangle. . The converse of a true theorem is not necessarily true.

. A conditional and its contrapositive are logically equivalent. . The converse and the inverse of a theorem are contrapositives of each other.

*14, An indirect proof of a theorem consists in proving its contrapositive. Group IV:

Miscellaneous

1. If x1 > x4 in the adjoining figure, what relation exists between x2 and x3? between x5 and x 6? . Prove: The lines joining the midpoints of the sides of a triangle divide the triangle into four congruent triangles. . The

bases

of

an _ isosceles

trapezoid are 12 in. and 6 in.; its height is 4 in. are the other sides? 4. Prove:

How long

The diagonals of an isosceles trapezoid are equal. . How high is a regular triangular pyramid if its base edges are 3 in. and its lateral edges are 6 in.?

Mapping

341

. Prove: The bisector of an exterior angle at the vertex

between the equal sides of an isosceles triangle is parallel to the base.

. If xAOB=xCOD=90° in the adjoining figure, what relation exists between and x BOD?

. The

radius

scribed

circle

of the

x AOC

circum-

EX. 7

of a 30°-60°

right triangle is 5 in.

How long are the sides of the triangle?

. If the angles of a triangle are in the ratio 1:2:3 and if the shortest side is 2 in., how long are the other two sides?

10. The diagonals AD and EC ina regular pentagon ABCDE tersect at F. Prove that ABCF is a rhombus.

in-

. In a regular square pyramid s is the base edge, A is the height, h is the slant height, and e is the lateral edge. If the length of two of these four segments is known, how can the others be computed?

12. Prove: If the circumcenter and the incenter of a triangle coincide, the triangle is equilateral.

13. If the hypotenuse of a 30°-60° right triangle is 12 in. long, how long is the perimeter of a second triangle formed by joining the midpoints of the sides in the first triangle? 14. Prove: Two right triangles are congruent, if the hypotenuse and an adjacent angle in one are respectively equal to the hypotenuse and an adjacent angle in the other. ————

15; Given:

Prove:

SS

a

OB | OD, OB=OD,

es

AB=CD

AD= BC

16. The sides of a square are divided in the ratio a:b as shown. Prove that the four dividing points form another square. b

a

342 17.

Unit 3 . To prove segments parallel, look for... (2 methods) . To prove segments equal, look for... (2 methods) To prove angles equal, look for... (4 methods) . To prove one angle larger than another, look for.. sa oe (2 methods) e. To prove one segment larger than another, look for. . (2 methods)

18. Given: AB =CD and AD = BC Prove:

AAEB

= ACDE

(An auxiliary line will help.)

D

B

C

E D C A

A Ex.

18

19. Four points 4, B,

ExauS)

C, D may or may not be coplanar.

In

either case, if AB = BC and AD = CD, prove that XA = XC. What can be said about the diagonals in the coplanar and the non-coplanar case? 20. Prove:

The sum of the interior angles of a regular n-gon

is (n— 2)180°. 21. Prove: The sum of the exterior angles of a regular n-gon is 360°. 22. The edge of a cube is 3 in. long. Compute the length of the diagonal of the cube. Find the shortest path on the surface of the cube between the endpoints of that diagonal and compute its length. Is there more than one shortest path? There are various ways of arranging the surface squares of the cube for a net; investigate whether these different arrangements influence the length of the path between the two vertices in question. 23. Given two coplanar non-parallel lines a and b, whose point of intersection is not accessible, and a point P in the same plane. How can similarity be used to draw the line through P which passes through the inaccessible point? 24. If the top of a pyramid is cut off by a plane parallel to the base in 2 of the height above the base, and if that base is a

Mapping

343

triangle with the sides 3, 5, and 6 in., what do you know about the upper base of the frustum?

25. Why can there be only five Platonic solids?

*26. Given a line s and a point Q on it, what is the geometric locus of all points P, for which the distance from s equals one half the distance from Q? *27. In AABC the midpoint E of

D

the median AD is connected with C.

Prove

vides AB

that CE

in the ratio

di-

A

1:2.

(Draw an auxiliary parallel to ss CE through D.)

B Esxmeh

*28. Given two intersecting lines p and q, what is the geometric locus of all points P, for which the distances from p and q

are equal to } in. and 1 in. respectively? *29. Construct AABC, given its perimeter p = 23 in., Xa = 60°, xB= 45°. (Construct first an auxiliary triangle that has p as a side whose adjacent angles can be derived from the given angles.)

et, AE

Ze,

OO

*30. Given two segments p and h and an Xa; construct AABC with p as perimeter, h as height, and xa opposite h.

Group V: Using coordinates 1. The vertices of AABC are A(0,0), B(5,1), C(3,4). a. How long is the perimeter?

b. What are the coordinates of the midpoint D of BC? c. Find the equations of the three medians. d. Find the coordinates of the centroid.

344

Unit 3 . What is the slope of the line through A(2,1) and B(4,6)? What is the equation of the symmetry axis of AB?

. Prove analytically that the perpendicular bisectors of any triangle are concurrent. (Choose a convenient coordinate system.) . Prove analytically that the altitudes in a triangle are concurrent.

. The vertices of AABC

are A(0,0), B(4,0), C(3,2).

Find the

coordinates of a point D that forms a parallelogram with the given three points. How many possibilities are there? . Prove analytically that the median of any trapezoid equals the arithmetic mean of the bases.

. Prove analytically that in any right triangle the median to the hypotenuse equals half the hypotenuse. Which is the most convenient coordinate system for this problem?

. Prove analytically that in any triangle the segment joining the midpoints of two sides is parallel to the third side and half as long. . What is the geometric meaning of a linear equation in x and y? Under what circumstances will two simultaneous linear equations have no unique solution? Give some examples.

10. What is the geometric locus of all points in the first quadrant, for which the sum of their coordinates equals 4? What is the equation of this locus? 11. Repeat Exercise 10 for the condition that the sum of the squares of the coordinates equals 4. 12. If P(3,2) and Q(1,4) are endpoints of two vectors P and Q, both starting at the origin, what are their absolute values? What are theirx- and y-components? What is the length and the direction of the vector V=P+Q? Group VI:

Using trigonometric functions

1. What are the definitions of sin a, cos a, and tan a in a right triangle? 2. The lateral edge of a regular square pyramid is 6 in. long and forms a 50° angle with the base. How high is the pyramid? How long is the base edge? (Use the Table of Values for Trigonometric Functions.) . A point P in the interior of xXAOB = 74° is at a distance of 3 in. from either side of the angle. What is its distance from the vertex?

Mapping

345

4. The projection of a 5 in. long segment onto a plane is 3 in. long. What is the angle of inclination to the nearest degree? - How long is the shadow of a 10 ft pole if the sun is at an elevation of 50°?

. How

big is the angle between the diagonal of a cube and

its base?

. From a 200 ft high lookout tower at the shore of the ocean a boat is seen under an angle of depression of 15°. How far away is the boat? . Show whether or not it is true that 1 Lh ————=|= Gs wv V1 +tan2a

tana SSS SI Vi-+tan2a

. In AABC XA =S3° and x B= 85°; what is the ratio of the three sides? If the perimeter of AABC is 30 inches, how long are the sides?

10. What are the definitions of sin a, cos a, and tan a in the unit circle? How do these definitions relate to the definitions as ratios in a right triangle? 11. If a is in the third quadrant, how can sin a, cos a, and tan a be found in a table of trigonometric functions in the first quadrant? 12. Why are trigonometric functions considered to be periodic functions? What is their period?

13. What are polar coordinates? What is the simplest equation of a circle in polar coordinates and in rectangular Cartesian coordinates? 14. What are the transformation formulae between the systems of polar and Cartesian coordinates?

1

THE CIRCLE

The curved surfaces which we are going to study in this Unit are closely related to the plane figure of a circle. Therefore, we shall study the circle first. A circle is defined as the set of all points within a plane which are at a given distance from a given point of that plane (see Unit 1, Section 8.1). The circle is a highly symmetric figure: it is bilaterally, centrally, and rotationally symmetric. From the definition and the symmetries of the circle many fascinating theorems about its relation to straight lines and other circles may be derived. Some of these we have noted in connection with the remarkable points of a triangle (Unit 2, Section 3) and in connection with regular polygons (Unit 3, Section 5). This Section deals systematically with the most important theorems about the circle.

1.1 The Chord The straight line segment between any two points of a circle (sphere) is called a chord. The diameter is achord passing through the center of the circle (sphere).

/

P

a

In Figure I, if the chord PP’

Q

is considered a connector between corresponding points in a bilateral symmetry, where is the symmetry axis? What direction does it have

compared to PP’?

Does it pass 346

1

Curved Surfaces

347

through the center of the circle? Does it have to pass through the center? Does it contain a diameter? How do you find the point bilaterally symmetric to some given point Q of the circle? The symmetry axis bisects the chords PP’ and QQ’. > The geometric locus of the midpoints of all chords of the same circle in a given direction is the diameter perpendicular to that direction.

Theorem IV-1: Exactly one circle passes through three non-collinear points. The center of the circle is the intersection of the symmetry axes between any two of these points.

Why?

(Think of a triangle and

its circumscribed circle.)

A problem:

In a given circle

of known radius r, is it possible to tell how far away center the chord PP’ length

is

known?

Sep,

from the is if its

(Figure

iy.

If)

Look for a right triangle and a chance to use the Pythagorean theorem. Since the perpendicular bisector (the symmetry axis) of the chord must contain a diameter,

i

there is a right triangle OMP, of which the hypotenuse is the radius r, one leg is PM,

half the chord, and the other legis the dis-

tance OM we are looking for. Pythagoras says: r? = PM? + OM?;

therefore OM = Vr? — PM?.

Note that only half of the chord is

part of that helpful right triangle. The other half of the chord forms a symmetric triangle. Thus, there is an isosceles triangle with the chord as base, whose sides are as long as the radius. The Pythagorean relationship between OM, MP, and r can, of course, also be used to find the length of the chord if its distance

348

Unit 4

and the radius are given, or to find the radius if the length of the chord and its distance from the center are given. For example: In a circle of 5 in. radius, an 8 in. chord must have a 3 in. distance from the center. Correct? In the same circle a chord that has a 4 in. distance from the center

must be 6 in. long.

Correct?

If the center of a circle has a 12 in. distance from a 10 in. chord, its radius must be 13 in. Correct?

A three-dimensional problem:

What is the relationship in a

sphere between the length of a chord, its distance from the center, and. the radius? (Figure III.)

Using the principle of reducing a three-dimensional problem to a two-dimensional one (see Unit 1, Section 9.2) whenever possible, we pass a

plane through the two given

YAS,

Se,

endpoints of the chord and the center of the sphere (three points determine a plane, remember?) which must inter-

Hl

sect the sphere in a circle of the same radius. (Why? What distance do all these points have from the center?) Then dealing with this circle only, we have the same three problems and the same solutions as before.

Theorem IV-2: Of two chords in the same circle the longer chord is nearer to the center of the circle.

Why? (Use the rt. AOMP leading to OM =... and see what happens to OM, if PM becomes larger or smaller.) Can Theorem IV—2 be used to prove that the diameter is the longest chord?

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349

Corollary IV-2-1: In a given circle, of all chords through a given internal point P the chord perpendicular to the diameter through P is the shortest.

Why?

(Show that PO is longer

than the distance of O from any of the other chords through P. These distances appear as legs of rt. As with PO as hypotenuse.)

Vio | v

EXERCISES Group I 1. Can you prove from the definitions of a circle and of a sphere, that a plane passing through the center of the sphere intersects it in a circle of the same radius?

. In a given circle, what is the maximum length of a chord? Does the length of a chord have anything to do with its distance from the center?

. Construct a circle from three given non-collinear points. Draw the circle and see how accurate your construction is: does the circle actually pass through all three points? . Given two points P and P’; what is the geometric locus of the centers of all circles passing through P and P’? . Given two pointsA, B; construct a circle of radius 2 inches passing through 4 and B. Is the construction always possible? Compare with the construction of an isosceles triangle.

. Given a straight line s and an external point A; construct a circle of radius r= 2 passing through A, which has its center on Ss.

. Given a straight line s and two external points4, B; construct a circle through A and B which has its center on s.

350

Unit 4

a A point P is 1 unit inside of the circumference of a circle with radius r= 13. How long is the shortest and the longest chord through P? A chord of 24 cm length has a 5 cm distance from the center of the circle. How long is the radius?

10. Repeat Exercises 8 and 9 after replacing the word “circle” with the word “sphere.” . Can you find, by construction, the center of the circle of which only part is given? 12. Given a circle and a chord in it; what is the locus of the midpoints of all chords parallel to the given chord?

Bet

13. Given a circle and a chord of it; what is the locus of the midpoints of all chords of the same length as the given one? 14. What is the locus of the centers of all circles with radius 3 inches, that pass through a given point? iS: What is the locus of the centers of all circles passing through two given points 4 inches apart? 16. Repeat Exercises 14 and 15 after replacing the word “‘circle”’ with the word ‘“‘sphere.”’ Group II 1. What is the set of points (x,y) represented by the following equations and inequalities? Ei oa whVarma Dixy? 16

Cy HSN OAS

ya 9 ya

. Explain why x? +y? =r? is the equation of a circle whose center is the origin and whose radius is r. (Use the concept of geometric locus and the Pythagorean theorem.) . What is the radius of the circle whose equation is x? + y? = 10? x?+ y?= 18? . Draw graphs of the following equations:

ar(x—3)+ GV — 47 =1 b. (x— 3? +y2=1 c x2 +(y —4)? =1

d. (x — 1)?+(Q—-2)?=9 e. 4 +17 +0 +27? =9 fi. ~—1)* +O +27 =4

. Given C(3,4), what condition must be satisfied by the coordinates of any point on a circle with C as center and r=5

units?

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351

6. Show that (x — a)? + (y — b)? =?" is the general equation of a circle whose center is the point (a,b) and whose radius is r.

7. Show that the condition b=r in the general equation of a circle characterizes a circle tangent to the x-axis. 8. Write the equation of a circle which is tangent to both the x- and y-axes. *9, Show how a quadratic equation of the form x?+ y?+Ax+ By + C =0, where x? and y? have equal coefficients, can be transformed into the form of a general equation of a circle. Use numerical examples for A, B, C and determine the coordinates of the center and the length of the radius. What happens if r? turns out to be zero or negative? *10. Find center and radius of the circle:

a. x?+y?— 6x+4y+9=0

c. x7

box4ey? = 2% = 24

d. 2x2 + 2y? +12x — 8y = 24

y2— 10x =0

1.2 The Arc and the Central Angle The circumference of a circle is divided into two parts by the endpoints A and B of a given chord AB (Figure IV). Each of the two parts is called an arc (symbol AB); the larger one is the major 2 arc,

and

minor

the smaller

arc.

one

is the

Either

arc together

with the chord AB

forms a seg-

A

ment of the circle. (Do not confuse this circle segment with the segment of a straight line!) Either arc together with the radii at A and B forms a sector. (An apple pie is usually cut into sectors.) The

sector

angle

at the

is its central

center

angle.

of the

IV

We

re}

say: the central angle AOB subtends (or intercepts) the arc AB (and the chord AB). What relation exists between the central

angles subtending the major and minor arcs of the same chord? The length of the circumference of a circle is defined as the limit of a sequence of perimeters of inscribed polygons with increasing

number

of sides.

(See Unit

3, Section

5.3.)

Obvi-

ously, the same reasoning can be applied to only a part of these

352

Unit 4

perimeters, for example the part between two given radii. Then the length of an arc is the limit of ... By reducing statements about the circle to statements about polygons it can be shown that: m In the same circle, or in equal circles, equal central angles subtend equal arcs; of two unequal central angles the larger one subtends the larger arc.

The correspondence between central angle and subtended arc may be used to compute one from the other. The round angle of 360° corresponds to the circumference 27rr; 180° corresponds to the semi-circle mr, etc.

round

If a central angle a is the nth part of the

angle, then the subtended

arc is the nth part of 27r.

(a: 360° = arc: 2777)

In the unit circle (r= 1), 360° corresponds to the number 27 = 6.28 ..., 180° to 7 = 3.14..., etc. The angle corresponding to the number | can be computed from the proportionx:360° = 1:27 fe}

leading to x = 6.28 ~ 57° 18' approximately. a radian.

This angle is called

In scientific work it is customary to measure angles in T

terms of radians.

For example, 180° = z radians, 90° = 7 radians,

60° =4 radians.

(You may leave 77 in the result, unless asked to

multiply out by its approximate value 3.14.) > A radian is the central angle subtending an arc equal to the radius of the circle.

(It is easy

to remember

that a

radian is just a little less than 60°, when comparing the arc of a radian with the chord of a 60° angle. Since the arc is a curved line, it does not reach as far on the

circumference as the straight line segment of the same length.) The length of an arc in a circle of radius r can thus be computed in three ways:

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353

1. from the number of radians contained in the subtending central arc, by multiplying this number with the length of the radius. (Example: if r=4 inches and xAOB measures 3 radians, then the length of the arc AB is 12 inches.)

. from the number of degrees in the subtending central arc, by setting up a proportion. (Example: if r=4 inches _and

XAOB measures 60°, then 60°:360° = 4B:2mr and AB =

¢° 2774 = 4,19 inches.) 3. from its ratio _to the circumference. (Example: if r= 4 inches and AB is 2 of the circumference, then AB = 2 - 2774 = 10 inches.) In the third century B.c. Eratosthenes of Alexandria, Egypt, used the relationship between an arc and its subtending central angle to compute the size of the earth. Eratosthenes knew of a deep well at Syene (Assuan), south of Alexandria and on the same

meridian, where the rays of the sun reached to the bottom during the summer solstice (June 20-22).

He concluded that at this time

the sun was directly overhead, and the vertical pointer of a sundial would

cast no shadow

at noon.

In Alexandria, at noon on

the same day, the pointer of a sundial does cast a shadow as the sun is not directly overhead. Eratosthenes found that the angle a between the rays of the sun and the pointer of the sundial was z of around angle. EraMe,; Q, tosthenes now reasoned in the following way: The sun Alexandria aS is so far away that its rays in Alexandria and Syene may be considered parallel (see Fig‘gd;

/

ure).

The vertical directions

at the two places form an angle at the center of the earth,

to

the

which

must

angle

Alexandria

be equal

measured

(alternate

at inte-

rior angles). The length of the arc between Alexandria and Syene was approximately known from the time it took to travel from one place to the other. It was

estimated to be 5000

stadia.

(A stadium is an old unit of

354

Unit 4

length, equal to about 185 meters or one tenth of a nautical mile.*) If the subtending central angle a is go of the round angle, then the arc is 4 of the circumference.

circumference

Thus, Eratosthenes concluded, the

of the earth must be 250,000 stadia.

(Correct?)

EXERCISES 1. What is the definition of a radian? How much is one radian expressed in degrees? How many radians are in a straight angle? in aright angle? 2. How big is an angle of 20°, 45°, 90°, 60°, 135° in radians? (Use a proportion.)

3. How big is an angle of 7,

7 7

34 NIA

7 : Ts a expressed in de-

grees? 4. Through how many radians does the hour hand of a clock rotate in 4 hours? in 6 hours? (It is not necessary to substitute a number for z in the result.)

5. In a circle of 3 inch radius, how long are the arcs subtended by the angles in Exercise 2? (Use 7=3.14.) 6. In a circle of 5 inch radius, how long are the arcs subtended

by the angles in Exercise 3?

(Use 7=3.14.)

7. Can a circle be mapped into itself by a rotation through 30°? through 60°? through any angle? through 27? through 47? through —7?

8. If the arc of a sector is twice as long as the radius, how big is the central angle? 9. How long is the perimeter of the segment of a circle, if the subtending central angle is 60° and the radius is 3 inches? 10. If the radius of the earth is 6378 km, how long is the circumference of the equator? How long is an equatorial arc of 1° (to the nearest km)? I nautical mile =length of 1 minute of arc at the equator = 1.15 statute miles, approximately. 11. How long is an equatorial arc of 1° in nautical miles? in statute miles? How long is one nautical mile in kilometers? (R = 6378 km)

12. Compute the distance on the surface of the earth between a point A(2° 12’ N, 70° W) and a point B(38° 35’ N, 70° W) in ‘Many different units of length were called a stadium. which one was used by Eratosthenes.

The experts do not agree on

Curved Surfaces nautical miles and in statute miles. minutes of arc are in that distance.)

355

(First find how many

13: Compute the radius of a sector if its arc is 2 in. and its central angle is 45°. 14. What was the size of the angle (in degrees and minutes) which Eratosthenes measured?

1S; If one stadium equals 185 m, how long to the nearest kilometer is the circumference of the earth, according to Eratosthenes? How long is the distance between Alexandria and Syene to the nearest kilometer, according to Eratosthenes’ estimate? 16. With more precise modern measurements the distance between Alexandria and Syene is about 800 km. How large (in km and %) was the error in Eratosthenes’ estimate of the size of the earth?

Using trigonometry:

7, ait ee rete

1. How much is sin 7, sin=

2 What is the period of the sine, cosine, and tangent functions expressed in radian measure? (Leave answer in terms of 77.)

1.3 The Inscribed Angle

If the endpoints of a chord AB in a given circle (Figure V) are connected with a point P on the circumference (periphery), the angle APB is called an inscribed angle. We say: x APB_subtends (or intercepts) the arc AB (or the chord AB). There is a numerical

relationship between an inscribed angle and the central angle subtending the same chord: Theorem IV-3: An inscribed angle is equal to half the central angle subtending the same chord, if both angles are on the same side of the chord.

356

Unit 4

PROOF STATEMENTS

1. Connect OP;

APOA

REASONS

is isosce-

les 2. Xa= Xa’ Bee — OO 4. y=2a 5. 6=28 6. y+ 6=2(a+ B) = 2XP, g.e.d.

1. OP and OA are radii 2. 3. 4. 5. 6.

base xs of isosc. A exterior x of APOA from 2. reasons 1.-4. addition

Corollary IV-3-1: An inscribed angle is equal to one half the central angle subtending the same arc.

Corollary IV-3-2:

Theorem

of Thales of Miletus (6th

cent. B.C.)

The chords drawn from a point on the circumference of a circle to the endpoints of a diameter form a right angle.

This famous corollary is often referred to briefly as Thales’ theorem of the angle in a semi-circle. It is quoted also in these forms: “An angle inscribed in a semi-circle is a right angle,” or “The geometric locus of the vertices of all right triangles on the same side of a given hypotenuse is the semi-circle with that hypot-

Curved Surfaces

enuse as the diameter.”

357

In Unit 2, Section 3, p. 252, we used the

wording: “A triangle is a right triangle if and only if the midpoint of one side is the circumcenter; this side is the hypotenuse.” Thales’ theorem provides useful methods of construction:

nee onsinuesarientirianele ai) if the hypotenuse c and one leg b are given: 1 AN = VIB +c

b

2. semi-circle with

C

r= AM

3. © aroundA withr = b intersects semi © at C

4. Connect CB Why gle?

is XC a right an-

A

£

M

5

(2) Construct a right triangle if the hypotenuse c and the altitude h are given: Cc

eh A

M

B

1. AM = MB =3c 2. semi © with r= AM

3. draw a ||to AB in distance h 4. intersection C, and C, of ||with semi © 5. solutions: AABC, and AABC, Discuss the number of solutions. Can it happen that there is no solution? (3) Construct

the mean proportional between

two given

segments: From Cor. II-12-1 we know that in a right triangle (a) the altitude is the mean proportional between the segments on the hypotenuse, and (b) each leg is the mean

B

358

Unit 4

proportional between the hypotenuse and the adjacentseement,, is. S1V6S) us two methods:_ (a) 1.

Draw AB,

= cen! aren Giese eee

such

that AP =m, =n

PB

Zola (OFA Bala 3. M=midpoint

C

of

AB

h

4. semi © around M with r =3(AB) 5. intersection C of 1 with semi © 6. solution

4

Uv

3;

=F

h = CP = \/mn

(b) 1. Draw AB = m (the longer segment) 2. On AB

mark

m

AP

=n

peel

nes

3. 1 to AB at P

4. M=midpoint

of

AB 5. semi © around M

with r =4(AB) 6. intersectionC of 1 with semi ©

;

Ta

7. solution b=AC mn

=

AC

a

/P

n

yy

eS

Thales’ theorem is needed to prove the following:

Theorem IV-4: All inscribed angles intercepting the same chord are equal to each other if on the same side of the chord, or they are supplementary if on opposite sides of the chord.

Curved Surfaces

Proof:

(a) on the same side: x P, (b) on opposite sides: show that xP, + X.P, = 180°; choose P,

on

359

=4xO = xP,, g.e.d.

diameter

through P;; A P,P3A must be: a5 tt.» 7Z\ (Thales)

fore 20,

and there-

XxXy=90° — also /\P,P.B

is art. A and x6 =

90° — x B; together XP, =y +6 = 180° Sie Bee 180" = XP,; from xP, = XP, (proved under (a)) follows that xP;

Set See

MPa

Pe

OL

XP, = 180°,

q.e.d. Theorem IV-—4 is very useful when worded as a geometric locus: m Given one side AB of a triangle and the opposite angle y, the geometric locus of the third vertex C is the arc of a circle with AB as its chord, subtended by a central angle 2y.

Corollary IV-4-1: Inscribed angles subtending the same arc are equal.

Compare this corollary with Theorem IV-4. Both refer to the same fact. Why can the corollary be worded so much shorter?

Corollary IV-4-2: If a quadrilateral is inscribed in a circle, its opposite angles are supplementary.

360

Proof:

Unit 4

Each diagonal is subtended by a pair of opposite inscribed angles.

Thales’ theorem

5 ae

NS B

is also needed

A

to prove the following:

Theorem IV-5: The Circle of Apollonius (3rd cent. B.C.) If one side AB of a triangle and the ratio of the other two sides are given, the geometric locus of the third vertex P is a circle; it passes through the points C and D which divide AB internally and externally in the given ratio.

Do you remember the theorem that the angle bisector of a triangle divides the opposite side in the ratio of the adjacent sides? (Review Theorem I-18 and its converse.) Do you remember that the interior and exterior angle bisectors are perpendicular to each other? Why are they?

The proof of Theorem IV-5 ; consists of two parts: (a) prove that P lies on the circle, if AP: BP =b:a

1. constr.

STATEMENTS some AABP_

A

b A

Ci“

Ba.0

=]

REASONS with

1. s.s.s.

AP: BP in the given ratio 2. x bisectors at Pintersect AB at C and D; C, D belong to the locus

. XCPD=tt.

Ww

x

4. P lies on semi © with CD as diameter, q.e.d.

2. Theorem II-18 (xX bisectors)

3. bisectors of adj. suppl. xs

4. Cor. IV-3-2 (Thales)

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361

(b) prove that for any point P on that circle AP: BP =b:a

1. given AB, constr.

C, D with

the given ratio b:a; constr. semi © with CD as diameter;

assume

P on

it and

PC and PD

connect

ee

2. through B draw||/sto PC and PD,

CONS: 2. constr.

intersect AP at E, F

3. PC 1 PD Aer — tek i 5. PF=PB=PE

3. Thales 4. rectangle with PB as diag. 5. Thales

6-620 =AC*CB=AP:PE

6. similar As

=AP:BP

b:a =AD:BD=AP:FP = AP: BP, q.e-d.

> A given triangle has three circles of Apollonius.

A special case of the Apollonian circle:

The geometric locus

of the vertices P of all isosceles triangles with the same base AB is the perpendicular bisector of AB. (Why?) In this case the ratio of the sides AP: BP is 1:1 and the interior bisector of x P intersects AB at its midpoint C. The exterior bisector of xP is parallel to AB and thus does not intersect AB at all. We might say that Theorem IV-5 does not apply to isosceles triangles, since the locus of P is a straight line and not a circle. However, we also might present the following argument: If the sides of a triangle are unequal, the ratio of the smaller side to the larger one is smaller than One. Increasing the smaller side will increase this ratio until it approaches the value One as a limit. At the same time the external angle bisector approaches more and gore a parallel to AB; the diameter of the Apollonian circle on AB becomes larger and larger; and the Apollonian circle itself — where we need it for the triangle — seems to straighten out, approaching the perpendicular bisector as a limit. We have used the concept of a limit before: it helped us to define the circumference and the area of a circle by a sequence of polygons; it made us see a tangent as a special case of a secant, and a cylinder as a special case of acone. The concept of a limit makes it unnecessary to state exceptions of a theorem.

362

Unit 4

Since the perpendicular bisector of an isosceles triangle may be considered as the limit of the Apollonian circle, Theorem IV—-5 applies to all triangles without exception. EXERCISES 1. If acentral angle in a circle is 30°, 40°, 70°, 90°, how big is an inscribed angle on the same side of the common chord, and how big on the opposite side?

2. What is the geometric locus of all 30° angles subtending the same chord? Construct it for a chord of 2 in. length. 3. What is the geometric locus of all 90° angies subtending a

straight line segment of 15 in. length? 4. Construct a right triangle with hypotenuse 4 units and equal legs. How long is its altitude? 5. If the distance from the midpoint of the hypotenuse of a right triangle to the vertex of the right angle is 5 units, how long is the hypotenuse?

6. Construct a triangle in which the base is 3 units long and the opposite angle is 45°. How many solutions are there? What is the locus of the third vertex? 7. Repeat Exercise 6 with the additional condition that the height of the triangle is 2 units. How many solutions are there? 8. Repeat Exercise 6 with the additional condition that the triangle is isosceles. 9. Construct

a triangle with base c=8

units, and a:b=

units, altitude h,=2

1:3.

10. Construct a triangle with c= 3 units,

a= 30°,

a:b =3:4.

11. What happens to the circle of Apollonius if the ratio of the triangle sides is 1:1? 12. Construct the three circles of Apollonius for an equilateral triangle, a right triangle, and an acute scalene triangle. 13. If a quadrilateral is inscribed in a circle and if one interior angle is expressed as 90°—x°, how much is the opposite angle?

14. Prove that a circle can be circumscribed around any rectangle.

15. Construct the mean proportional of two given segments by two methods. 16. Compute

the mean

proportional between m=9

and n= 4.

Curved Surfaces

363

Measure the given segments in Exercise 15 and compute the mean proportional; then check how accurate your construction was. 17: How long is the diagonal of a rectangle inscribed in a circle of radius 6 units?

18. If two chords AB and CD cle, show that AAPC and 19. Repeat Exercise 18 for a 20. Prove: If a rectangle is are diameters.

intersect at a point P inside the cirABPD are similar. point P outside the circle.

inscribed in a circle, the diagonals

21. Prove: The geometric locus of the midpoints of all chords drawn from a point A ona circle with center O is another circle passing through A with OA as diameter. (Connect a midpoint of such a chord with the center O and see whether Thales might help.) *22. The following theorem gives a relation between the sides and the diagonals of an inscribed quadrilateral. Prove:

Ptolemy’s Theorem (2nd cent. A.D.): In any quadrilateral inscribed in a circle the sum of the products of the opposite sides equals the product of the diagonals.

(Draw an auxiliary line DK so that x 1= x2; then prove x3 =xX4 as inscribed angles. Show that AAKD ~ ABCD and set up a proportion involving one pair of opposite sides and one diagonal. Then show that x5=x6and AABD

~AKCD. This leads to a second proportion involving the other pair of opposite sides and the same diagonal. The two proportions should lead to the required theorem.)

act+bd=ef

364

Unit 4

1.4 Secants and Tangents A secant is the straight line carrier of a chord. Parallel secants intersect the circle in symmetric points with respect to the symmetry axis s, of the chords (Figure VI).

The farther away from the center the secant lies the nearer to each other are the corresponding symmetric points. A limit is reached when the secant has a distance from the center that is equal to the radius. Then the two symmetric points coincide at T on the symmetry axis, and the secant becomes a fangent Thus we see from symmetry reasons that: & The tangent forms a right angle with the radius at the point T of tangency (or point of contact).

Sometimes it is quite helpful to think of this point of contact as actually being a pair of infinitely near points; it helps you to see the tangent as just a special (limiting) case of a secant. A straight line in greater distance from the center than the length of the radius does not intersect the circle at all; it is neither a se-

cant nor a tangent. > A straight line intersects a circle in two, one, or no points, depending whether its distance from the center is smaller than, equal to, or greater than the radius.

From symmetry with respect to the secant s, passing through the center (Figure VI) follows: > There are two tangents to a circle in a given direction.

Through an external point P (Figure VII) there are infinitely many straight lines which may intersect a given circle in two, one, or no points, depending on their distance from the center. One of the secants through P contains a diameter (there is exactly one straight line through P and O). Considering this particular

Curved Surfaces

365

secant aS a Symmetry axis, we can find the reflection of any

secant,

since

we

know

that a pair of symmetric straight lines must be equally distant from the center and must form equal angles at P with the symmetry axis. Where is the reflection of the special secant through P, called the tangent? It follows that:

Vil

> There are two tangents to a circle through an external point.

Definition IV-1: If a tangent through an external circle at T, then the distance PT of the tangent from P. If a secant through an external the circle in the points A, B, and

point P touches the is called the length point P intersects if AP contains the

chord AB, then the distance AP is called the length of the secant from P, and BP is the external segment of the secant.

Theorem IV-6: The two tangents to a circle from an external point P have equal length and form equal angles with the secant through the center.

Prove by showing that AOPT = AOPT"’.

ibe

366

Unit 4 Corollary IV-6-1: If a quadrilateral is circumscribed about a circle, the sum of one pair of opposite sides is equal to the sum of the other pair of opposite sides. (See Fig. p. 365)

Why? (Use Theorem IV-6.) Compare this corollary with the corollary about inscribed quadrilaterals.

Theorem IV-7: The angle between a tangent to a circle at point T and a secant through T is equal to half the central angle subtending the chord on the secant.

Why? (Draw the symmetry axis of the chord and look for equal angles with mutually perpendicular sides.)

Theorem Two of one other, larger

Hyp: Con.:

a—a AABC

IV-8: The fourth congruence (similarity) theorem. triangles are congruent (similar) if two sides are equal (proportional) to two sides of the and the corresponding angles opposite the of the two sides are equal. S.s.a. — theorem

cc .a@—a.a— © |for similarity, = A4’B'C' [As ~]

c:e-—a-a |

PROOF STATEMENTS 1. draw h= BH 1 AC; h'=B'H' 1 A'C' 2. tt. AABH = [~] AA'B'H’

REASONS

1. constr. 2. equal

4 [= c) in Theorem IV-8 becomes even clearer, if you try to construct a triangle from QC: 1. 2. 3. 4.

AB=c XA =a © around B with radius a C = intersection of © with —_>

AH Discussion:

Since

a circle

and a straight line may have two, one, or no intersections,

this construction

may

have

two, one, or no solutions, if the condition a > c is not required.

368

Unit 4

If a > c, there are two intersections C and C, of which C is not a solution because A ABC does not contain xa. C is the solution based on IV-8.

If a =c, then AABC is isosceles (Xa =Xy, no

XB = 180° — 2a)

need for an S.s.a.-

and can be constructed from a.s.a.; theorem. a /h, then there are two triangles ABC, and ABC, If An S.s.a.with a, c, a, and these triangles are not congruent.

theorem for this case would be ambiguous and useless. If a < c anda =A, then there is one solution (AH is tangent to the

circle);

the triangle is a right triangle and can be constructed

from a.s.a.; If

no need for an S.s.a.-theorem.

a< cand a~ c, to be use-

ful. This condition is expressed symbolically as S.s.a. by capitalizing the one S$ that is not next to a. EXERCISES Group I 1. Given a point T on acircle, construct the tangent to the circle at T. 2. Given a circle and an external point P; construct the tangents through P. (Use Thales’ theorem. Where is the center of the semi-circle?)

3. The distance of a point P from the center of a circle with

radius 25 units is 65 units.

How long is the tangent to the

circle from P? 4. The tangent from a point P to a circle of 3 units radius is 4 units long. How far is P from the center? How far is P from the circle? (This means: from the nearest point of the circle.) 5. Prove: If two circles are concentric (have a common center), the chords of the larger circle which are tangent to the smaller circle are equal.

6. What is the geometric locus of all points from which the tangents to a given circle are equally long? 7. Given a circle and a straight line; construct the tangents to the circle that are parallel to the given line. (You must find the points of contact first.)

Curved Surfaces

369

8. How far is a point P from the center of a circle if the tangents from P are as long as the diameter?

9. Given a circle of radius 2 units and a straight line s; find a point P on s, such that the tangents from P to the circle are 15 units long. Discuss the number of solutions. (How far must P be from the center of the circle? What is the locus of all points with that distance from the center?) 10. Prove: A chord in a circle forms equal angles with the tangents at its endpoints; the sum of these angles equals the central angle subtending the chord. 11. Construct a triangle, given a = 45°, c= 2 units, a= 2} units. (Start with c.) Repeat the construction for various values of =

live 2, We, l units.

Discuss your solutions and compare

with Theorem IV-8. 12. What are the four congruence (similarity) theorems? Why is the phrase “‘opposite the larger side’’ important in the fourth congruence (similarity) theorem? Is it ambiguous or impossible to construct a triangle from any two sides and an angle that is not the included angle? 13. Construct a triangle if given: a.

b.

(%

d.

e.

a = 60° Coal i

60° 15 12

30° 2 1

30° 2 13

30° 2 23

14. What is the geometric locus of all points P such that the tangents drawn from P to a given circle are perpendicular to each other? (What is the angle between the radii of contact?

How long is OP?) 15. Prove: If two chords intersect inside the circle, the angle formed equals one half the sum of the two central angles cor-

+E $(4a+48) Exaal5

4 EF (4a - 48) Ex. 16

370

Unit 4

responding to the two intercepted arcs. (Draw AD and determine the size of the angles in AADE as inscribed angles; then think of x E as an exterior angle.) 16. Prove: If two secants intersect outside the circle, the angle formed equals one half the.difference between the two central angles corresponding to the two intercepted arcs. (Draw AD and determine the size of one interior and one exterior angle of AADE as inscribed angles; then find an equation for xE.) 17. Prove: Ifa secant and a tangent intersect, the angle formed equals one half of the difference between the two central angles corresponding to the two intercepted arcs. Give two proofs: one as an independent theorem and one as a corollary of which theorem?

4-4 (4a-48) Em, Wi

4E+3 (40-48) Ex.18

18. Prove: The angle formed by two tangents equals one half the difference between the two central angles corresponding to the two intercepted arcs. Give two proofs: one as an independent theorem and one as a corollary of which theorem? Group II: Using coordinates 1. Review solving two simultaneous equations, one of which is quadratic. (See Supplement III.)

2. What geometric interpretation can be given to a set of two simultaneous equations, if one is linear and one quadratic? What can be said about the solution set? 3. Write the equation of a circle whose center is the origin and whose radius is 3. List the coordinates of the points where

Curved Surfaces

371

this circle intersects the coordinate axes. Write the equations of the tangents to the circle at these points.

4. Determine the points of intersection x? + y?= 16 and the following lines: a. y=x

b.

y+x=0

between

c.y =3x+2

the circle

d. 2x+5y=0

5. How can you express analytically that the tangent of a circle at a point P is perpendicular to the radius through P?

6. How can you find out without plotting whether the lines represented by the following equations intersect, touch, or bypass the circle x? ++y?= 9?

a y=2x+1

b. y=x+6

ce. 3x+2y=0

d.y+3=0

7. If P(4,y) is a point on the circle x? + y? = 25, what is the slope of the tangent to the circle at P? Write the equation of this tangent and of the second tangent in the same direction. Then draw a graph to check your answer.

8. Write the equation of a straight line parallel to the y-axis, which

a. is tangent to the circle x? + y? =10, b. intersects this circle, c. bypasses this circle.

9. Write the equation of a circle tangent to both the y-axis and the line y= 8. 10. How long is the chord of the circle (x — 1)? + (y + 3)?= 13, which lies on the x-axis? 11. Write the equation of a circle whose center is C(3,6), if the y-axis is a tangent.

12. Prove that the line x +y=5 is a tangent of the circle (x— 1)? +(y — 2)? =2. Show that there is only one point of intersection.

©

1.5 Power with Respect to a Circle

Theorem IV-9: If two secants

are drawn

from the same external

point P to the same circle, the product of one and its external segment equals the product of the other and its external segment.

372

Unit 4 A

Proof:

Consider two arbitrary secants ABP and A'B'P, and show that

AP - BP=A'P -B’P.

REASONS

STATEMENTS

1. draw AB’ and A’B 2. AAB'P

~ AA'BP, because

XP=XP A

. constr.

2A”

XB=XB'

identity inscribed Xs intercepting same arc BB’ sum of XsinaA... Ww ANS . proportion

Corollary IV-9-1: If a tangent and a secant are drawn from the same external point P to the same circle, the tangent is the mean proportional _between the secant and its external segment.

Why?

(A tangent is a special case of a secant, . . .)

Curved Surfaces

373

Theorem IV-10: If two chords are drawn through the same internal point P of a circle, the product of the segments formed on one equals the product of the segments formed on the other.

A!

Proof:

Consider two arbitrary chords APB

A Vas

ots B

and A’PB’, and show that AP - BP = A'P:B'P. B!

__ STATEMENTS 1. draw AB’ and A’B 2. AAB'P ~ AA'BP, because xP=xXP XA=XA',XB=XB'

REASONS 1. constr. Me vertical angles inscribed Xs subtending arc

3. AP: A'P=B'P: BP 4. AP- BP=A'P - B'P, q.e.d.

3. ~ As 4. proportion

Corollary IV-10-1: If a diameter is drawn through an internal point P of a circle, then half of the chord perpendicular to the diameter at P is the mean proportional between the segments of the diameter formed by P.

Prove as a special case of Theorem IV-10.

same

374

Unit 4

Theorems IV-9 and IV-10 and their corollaries are statements about a fixed length connected with a given point P and a circle. If P is an external point, this length is the length of the tangent drawn from P; if P is an internal point, the length is half of the chord that is perpendicular to the diameter through P. In each case the square of this fixed length is equal to the product between the segments on any other line through P formed by P and the intersections with the given circle. The square of this fixed length k is called the power k? of point P with respect to the circle. The power k? can be used to construct the mean proportional between two given straight line segments m and n: (a) using the tangent, we assume m and n to be a secant and its external segment.

(Cor

IV-9-1)

1. AP = mM, BP = n,

where B is between A and P

2: AO

08

3. © around O with radius AO

4. semi © with

OP

as

diameter (Thales)

5. T = intersection of © and semi ©)

6. TP = tangent = \/mn

(b using the chord, we assume m and n to be segments of a di—

ameter AB. (Cor. IV-10-1) 1. AP =m, BP =n, where P is between A and B

2. AO = OB: 3. © around O with radius AO 4. | toAB atP 5. C = intersection of | with ©

ee es

ee C

6. PC =$ chord = Vmn Note:

Construction (b) can be

interpreted also as finding the altitude of a right triangle as the mean proportional between the segments on the hypotenuse.

B

Curved Surfaces

375

The power k? is used also to construct a golden section. Unit 2, Section

2.2, golden

section;

Unit

agon.)

(See

3, Section 5.2, dec-

Divide a given segment AB in a golden section.

(Find

C on AB such that AB: BC = BC:(AB — BC).) 1. OA AB | and OA

= x(AB) 2. ©

around

iE On

O with

A

er

B

.

3. secant OB intersects © atM and N 4. BC=BM, _ solution C

Proof:

N

rta

1. BN:AB BM -BN =AB:BM = AB?or

‘

ot jacee

2. substitute BM =BC

A

C

B

and BN = 2r + BM

Pabst

AB

ABC

3. (AB + BC):AB = AB: BC 4. (AB + BC)—AB

AB 5,

_AB—BC

(See

AO

Proportions.)

6

Supplement

IV,

AB: BC= BC:(AB ABO) G20,

EXERCISES 1. Cut a cake for the family along lines of equal power with respect to a center piece. How to cut a chocolate cake with power: Suppose it is your 16th birthday. There should be 16 13 3 candles equally spaced aN around the edge of the ee ON cake, and one candle in S : the center to grow on. 11 Cs Mark the center piece as a small circle first. (A S i small juice glass. will

376

Unit 4 help.) Place a candle holder in the center, then place the other candle holders in symmetrical order: first no. 1 and 9, and 5 and 13, then bisecting with 3, 7, 11, 15, then bisecting again with the BeSt Now cut along the marked center circle first. Then cut along the tangents to the small circle from between two candles. These tangents are equally long. The serving pieces are rotationally symmetric to each other.

2. Divide a given segment AB in the golden ratio. Write another proof for this construction, showing OB=rV5 and OM=r(V5—1). Review the construction given in Unit 3, Section 5.2, Exercise 9.

3. Prove: In an isosceles triangle with vertex angle of 36° the bisector of a base angle divides the opposite leg in a golden section. (Look for similar As.) 4. Construct an isosceles triangle with vertex angle 36° if a leg AC is given. 5. Construct an inscribed regular decagon for a given circle. (How big is the central angle?)

6. Construct an inscribed regular pentagon for a given circle. (How big is the central angle? Show that two sides of the decagon form an isosceles triangle with the side of the pentagon.)

7. Construct an inscribed regular 15-gon for a given circle. (How big is the central angle? Can it be constructed from a 36° and a 60° angle? Show that the characteristic triangle of a 15-gon can be constructed from the characteristic triangles of a decagon and a hexagon in the same circle; the side of the 15-gon can be obtained if the sides of a decagon and a hexagon are drawn from the same point on the circle.) 8. Construct the mean proportional between two given segments a and b, using Cor. IV—9-1. 9. How many constructions of the mean proportional do you know?

Curved Surfaces

377

10. If two chords AB and CD of a circle intersect at P and if the segments formed on AB by P are 3 and 6 units long, how long is the chord CD, if CP is 9 units? 11. The diameter AB of a circle is 9 units. Froma point P on the extension of AB a tangent is drawn to the circle with T as point of contact. If TP is longer by 3 units than the external

segment BP, how long is BP?

(Set up an equation with the

unknown x = BP.)

12. If a chord AB is bisected by a chord CD at point P, and if CP=4 and PD=9, how long is AB? 13. A tangent and a secant are drawn to a circle from an external point P. If the tangent is 10 units long and the external segment of the secant is 5 units, how long is the chord lying on the secant? 14. What is the locus of alJ points 100 that have equal power with respect to a given circle? 15. If you look out at the ocean from a tower of 100 m height, how far can you see? (Radius R of the earth R = 6,400,000 m)

1.6 Two Circles

Vill

Two circles may intersect each other or not; if they do not intersect, one may be inside or outside of the other. If they have the same center, they are called concentric circles. They may be tangent to the same straight line ¢ at the same point of contact 7; then they are said to be tangent to each other, or to touch each other. They may lie on different sides of the common tangent (externally

378

Unit 4

tangent circles) or on the same side (internally tangent circles). Describe the relative position of various pairs of circles in Figure VIII. Make a separate sketch for each possible type of relative position. Is there a symmetry axis for each pair? One of the famous problems of antiquity is that of Apollonius of Perga (3rd cent. B.c.): Given three circles, find a circle that

is tangent to all three. There are eight solutions in general, considering that the required circle may touch one, two, or three of the given circles from the outside or the inside.

Count, why there

should be eight possibilities. In special cases of relative position and relative size of the given circles there may be less than eight solutions. (Here is a special project for some of you: make a report on this famous problem.) Can we tell from the distance between the centers of two circles and their radii, whether these circles are tangent, entirely outside

each other, one entirely inside the other, or intersecting?

Theorem IV-11: If two circles (O,,r,) and (O,,r,) are tangent, their line of centers

O,O, the and the this

passes

through

point of contact is perpendicular to common tangent at point.

Curved Surfaces

379

Prove Theorem IV-11.

Corollary IV-11-1: If two circles are externally tangent (internally tangent), the distance between their centers equals the sum (the difference) of their radii.

> If two circles are entirely outside of each other (one entirely inside the other), the distance between their centers is larger (smaller) than the sum (difference) of their radii.

e(

©

A;

[\ Az

If two circles intersect, then the two radii and the distance be-

tween the centers form a triangle on either side of the line of centers. In that case the triangle inequalities (review Unit 1, Section 10) must be satisfied:

Theorem IV-12: Two circles of radii r, and r, intersect in exactly two points, if the distance between their centers is smaller than the sum and larger than the difference of their radii.

The straight line through the two points of intersection A, and A, is a common secant of the two circles, carrying the common chord A,A,; it is perpendicular

to the line of centers. Why? How many circles may have the same segment A,A2 as a common chord? What is the geometric locus of their centers?

A;

Az

380

Unit 4

Definition IV-2: The angle between two intersecting circles is defined as the angle between their tangents at a point of intersection.

Theorem IV-13: Two circles iatersect at right angles (orthogonal circles) if and only if the tangent of each circle at a point of intersection passes through the center of the other.

Why? EXERCISES Group I 1. If two circles are tangent, what is the relation between their radii and the distance between their centers, if they touch (a) from the outside, (b) from the inside? . Given a circle and a point C outside; if of two circles around C one is tangent to the given circle from the outside and the other from the inside, which has the larger radius? . What is the geometric locus of the centers of all circles that are tangent to a given circle at a given point? . What is the geometric locus of the centers of all circles with radius 5 units, that are tangent to a circle of radius 12 units (a) from the outside, (b) from the inside? . Repeat Exercise 4, if the given circle has a radius of 4 units.

Curved Surfaces

381

- Construct a circle of radius 2 units that is tangent to a given straight line and to a given circle with a 3 unit radius. Discuss the number of possible solutions. (Intersections of two loci.)

. Given two circles of radii r, = 1 unit, r, = 2 units with a distance of 4 units between their centers; construct a circle of 3 unit radius that touches the two given circles from the outside. (What distance must its center have from the given centers?)

. Repeat Exercise 7 for various pairs of given circles and various distances between them. Discuss the number of possible solutions.

. Repeat Exercise 7 for the case where the required circle touches one of the given circles from the outside and the other from the inside. 10. Does Exercise 7 also have solutions if the required circle touches both given circles from the inside? 11. How can you tell from the radii of two circles and the distance between their centers, whether these circles have any points in common? 12. Given a circle of radius 3 units and a point C ina 5 unit distance from its center O; if a circle around C intersects the given circle, between what limits of length may its radius be?

iS: What is the geometric locus of the centers of all circles intersecting in the same pair of points? 14. What is the geometric locus of the centers of all circles with radius 2 units, intersecting at a given point A? 15. Prove: The angle between two intersecting circles is supplementary to the angle between their radii through the point of intersection. (In Greek orthos means 16. What are orthogonal circles? right, and gonia means...) Prove: The triangle formed by the centers of a pair of orthogonal circles and a point of intersection is a right triangle. 17. Given a circle and two points A and B on it; construct a second circle passing through these two points and intersecting the first circle at an angle of 60°. 18. Under what circumstances is the angle between two circles equal to zero? 19. Given a circle and a point T on it; construct a second circle equal and orthogonal to the first, passing through T.

382

Unit 4

20. Given a circle and a point T on it; what is the geometric locus of the centers of all circles passing through T orthogonal to the given circle? 21. If two circles have a chord AB of 2 unit length in common

and if from a point P on the common secant 3 units outside of AB tangents are drawn to the circles, how long are they? 22. Prove: The tangents drawn to two intersecting circles from a point P of the common secant are equal in length.

P

Ex. 22

Exec

23. Prove:

If two circles have a common tangent ft, then the other tangents drawn from any point P on ¢ are equal.

24. Given three circles, of which the first intersects the second, and the second intersects the third; is there a point from which the tangents to all three circles are equally long? Group II: Using coordinates

i: Find

the points of intersection

between

the two

circles

x?+ y?=16 and (x—3)?+y?=13. How many points at most can you expect? (Review quadratic equations,, Supplement III.)

. How long is the common chord of the two circles in Exercise 1?

. Write the equation of a circle which x? + y?=9 from outside.

touches

the circle

. Write the equation of a circle which x? + y?= 16 from inside.

touches

the circle

. Write the equation of the geometric locus of all points which are outside the circle x? + y?= 4 at a distance of 1 unit. . How

can you tell without solving two simultaneous quad-

)

Curved Surfaces

383

ratic equations whether or not the following two circles intersect, or whether they touch?

a. ?+y2=9 + y2= 12

ce. (x— 1? +(—-3)=4 A= 2) (y= 27 =9

b. x?7+y?=9

d. (x¥— 32+

(x—6)/?+y?=9

GY—4)7=9

(x2)? + yr 1) = 4

. Write the equation of the geometric locus of the centers of

all circles which have the segment AB as a common chord: A(0,0), B(2,1).

. What is the distance between the centers of the following circles?

Do these circles intersect?

ay) (vy — 35) = 16 (1) (y —- 6) 7 — | DAG 10) y= 1

Cox

+ yy? = 121 (x — 15)?+ y?= 100

(x—7" +y?=9

9, What is the distance between the centers of these two circles?

a. x? +y?— 14x—4y+44=0 ete OX Ay rc =)

b. x?+y?—6x+ 10y = 66 aay — 18x —6y + 50=0

*10. Write the equation of the common secant for the following two circles.

0

What is the length of the common chord?

ye = ae a1 yt LO

OFand x4 734 = yi 8 =)

1.7 Common Tangents of Two Circles

Theorem IV-14: Given

two

unequal,

non-concentric

circles and a

pair of parallel radii; the straight lines through the outer endpoints of any such pair pass through the same fixed point S, on the line of centers if the radii are in

the same direction, and S, if in opposite direction. These two fixed points divide the distance between the centers externally and internally in the ratio of the radii.

384

Proof:

Unit 4

Find similar triangles

(a) S,, AP,OS,~ AP,O,S, (Why?) Tail, = OS,:0,5_ (Why?) (DSS

F. Cys

sae

ry! 21, = O,S;: 0,8; Where are the points S, and S_, if the circles are of equal size? tangent? concentric? Considering tangents as special secants we derive:

Corollary IV-14-1: (a) If two circles are outside each other, they have a pair of external and a pair of internal tangents in common; these pairs pass respectively through S,,

and S...

(a)

(b)

(b) If two circles touch from the outside, the pair of common internal tangents coincide in the common tangent.

Curved Surfaces

385

(c) If two circles touch from the inside, the pair of common external tangents coincide in the common tangent; there are no common internal tan-

CD © gents.

(e)

(d) If two circles intersect, they have only a pair of

external tangents in common. (e) If one circle is entirely within the other, then there are neither external nor internal common tangents.

Definition IV-3: The length of a tangent common to two circles is the distance between the two points of contact.

Since any inscribed n-gon may be reduced to n isosceles triangles with radii as legs, Theorem IV-14 leads to:

Corollary IV-14-2: Given two circles, for any polygon inscribed in one there is a corresponding polygon inscribed in the other that is an expansion of the first in the ratio of the radii as ratio of similitude.

The fixed points $, and S, are the external and internal centers of similitude. a

we

SS as J eo

ar, LP =

os Se

ae

(/

x

Ss (>

=

ie Ss

386

Unit 4

Since any point of a circle may be a vertex of an inscribed polygon, there follows: Corollary IV-14-3: Two circles are carried into each other by expan-

sions from S,, and S,, their external and internal center of similitude, with the ratio of their radii as ratio of similitude. The intersection of the common external tangents is the external center of similitude; the intersection of the common internal tangents is the internal center of similitude.

Where are the centers of similitude of the special pairs of circles in Cor. IV-14-1? Theorem IV-14 provides a construction to divide a given segment AB internally and externally in a given ratio m:n. (See also Unit 2, Section 2.2.)

1. At A and B draw |ls in an arbitrary direction 2. AC=m, BD=BD' =n 3. CD intersects AB in S,; CD’ intersects AB in S,. EXERCISES 1. Given two circles in various relative positions; draw the common tangents.

for each case

2. What is the external and the internal center of similitude of two circles? How can these centers be found? Do two

Curved Surfaces circles always have such a center? have a center of similitude?

387

Do concentric circles

3. How can Theorem IV-14 be used to divide a given segment AB externally and internally in a given ratio? 4. What other constructions do you know, by which a given segment can be divided externally and internally in the same given ratio? (See Unit 2, Sections 2.2 and 3.3.1.) . What is a harmonic division? . Prove: The common internal tangent of two externally tangent circles bisects each of their common external tangents. (How long are the tangents to each circle drawn from P, the intersection of an external and the common internal tangent?)

Exo

Ex.7

. Prove: The common secant of two intersecting circles bisects each of their common external tangents. (What is the power with respect to the circles of the point P of intersection be-

tween the common secant and the common tangent?) . Given two externally touching circles with radii R and r; compute the length of the common external tangent. (Draw a || to the external tangent through one of the centers, to form a rectangle; then use Pythagoras.)

7

“ses Ex.8

Exe

388

Unit 4

9. Given two externally touching circles with radii R and r; compute the length of the common internal tangent between its intersections with the pair of common external tangents. (Use Exercises 6 and 8.)

10. Two circles of radii 1 and 2 units have a distance of 4 units between their centers. Compute the length of their common external tangents. (Find a helpful rectangle and a right triangle.) iWI Compute the length of the common internal tangents of the circles in Exercise 10. (There is a helpful rectangle and a right triangle.)

Ex.10

12. Two circles of radii 10 and 2 units have a common external tangent that is 15 units long. Compute the distance between their centers, and the length of their common internal tangents.

13. Prove: If two circles have equal radii, their external common tangents are parallel to the line of centers.

14. Prove: If two circles have equal radii, their internal common tangents bisect the line of centers. 15. Explain the following construction of the common external tangents to two given circles:

a. Draw a circle around O, with radius (R — r);

b. Construct the tangents to this small circle from O,, using Thales’ semi © to find the points of contact;

Curved Surfaces

389

ce. Connect these points of contact with O, and extend to intersect the first circle at T, and T,’;

d. ||s to O,T, and O,T,’ through O, intersect the second circle at. and 147; e. T,T, and T,'T,’ are the required common tangents. 16. Explain the following construction of the commonexternaland internal tangents to two given circles, based on Cor. 1V-14-3:

a. Use an arbitrary pair of ||radii to find S$, and S,. b. Find the point of contact on one of the Os by intersecting

it with Thales’ semi ©, which has OS, and OS, respectively as diameter. c. Draw the radii of contact. d. 1 to the radii of contact are the required tangents through S, and $,; they are common to both circles. e. By symmetry find the other points of contact and the common tangents through them. What happens if the circles are equal in size? construct their common external tangents?

2

SURFACES

How will you

OF REVOLUTION

2.1 Axial and Cross Sections

Describe the geometric objects in Figure I. They have one thing in common: there is a straight line s (an axis s) such that any plane = which passes through s intersects the object in congruent sections (axial sections). Figure II shows these sections; they are bilaterally symmetric figures. To each connector between

390

Unit 4

S

S

P

p!

P

P fot P'

i P'

p

p!

a pair of symmetric points P and P’ intersecting s in a point $ there are congruent connectors through the same point S in any of the other axial sections. The set of all points P related through the same point S$ is therefore a circle with radius SP; in other words, the cross sections of these objects (sections in a plane perperpendicular to s) are circles. (Compare with rotation about an axis, Unit 3, Section 6.) One axial section characterizes the object; it is a characteristic section. A cross section is not a characteristic section, because

all cross sections of these objects are circles. The last object in Figure I is a sphere. Interpreting it as a globe and using geographic language, the axial sections are the meridians and the circular cross sections are the parallels. (Review Preparatory Chapter, Section 6.) This geographic language may be applied also to the other objects; then Figure II shows the meridians of the objects in Figure I. One half of the bilaterally symmetric meridian is called a meridian also. It looks as if the objects in Figure I were generated by rotating (or revolving) the meridional sections around the axis s. Em-

Curved Surfaces

391

ploying “rotation” or “revolution” as a figure of speech, such objects are called solids of revolution, and their surfaces are called surfaces of revolution.

Definition IV-4: A surface (an object) is called a surface (a solid) of revolution, if there is an axis s (axis of revolution) such that all axial sections are congruent bilaterally symmetric figures and all cross sections are circles.

Definition IV-5: The axis of a circle is a straight line through the center of the circle and perpendicular to its plane.

Can you think of a reason for this definition? Which surface of revolution is characterized by the meridian m below? m = a Straight line parallel to the axis; m = a Straight line intersecting the axis at an angle different from 90°;

= a straight line intersecting the axis at a right angle; I a circle with a diameter on the axis;

= a square with a diagonal on the axis; 2S m = a square with one side on the axis. > The geometrically simplest surfaces of revolution are the cylinder, the cone, the plane, and the sphere.

Explain how the plane can be considered as a surface of revolution. The connection between a bilaterally symmetric figure and a solid or surface of revolution provides a method of connecting a two-dimensional and a three-dimensional problem. & The principle of reducing a three-dimensional problem to a two-dimensional one is used when a problem about solids or surfaces of revolution can be solved by considering an axial section. For example, to compute the height of a cone if the diameter of its base and the length of its side are given, we con-

392

Unit 4 sider an axial section, an isosceles triangle, and solve a twodimensional problem about a right triangle. (Which?)

& Conversely, the principle of three-dimensional interpretation of a two-dimensional theorem is used when a theorem about a bilaterally symmetric figure (or half of such a figure) leads to a theorem about a solid or surface of revolution. For example, the two-dimensional theorem: “If an isosceles triangle is cut by a straight line through the midpoint of its altitude parallel to its base, then a similar triangle is formed with a base half as long as the original base”’ leads to a three-dimensional theorem through revolution: “If a cone is cut by a plane through the midpoint of its altitude parallel to its base, then a similar cone is formed with the diameter of its base circle half as long as that of the original base circle.”

In the next two Sections we shali use these two principles for problems about spheres, planes, cylindrical and conical surfaces (solids) of revolution. EXERCISES 1. Describe the solids of revolution generated by the meridians as shown:

s

.

3

S

S 3

5

(a)

3

4

(b)

(c)

s

S

s

FEI (_| 2

Sin

eal

3

(d)

(e)

S

(g)

2. Draw the meridian of these solids of revolution: a

.

4 2]

(a)

(b)

a> =

(c)

s

NEO 2

2 (f)

S

(7.0

ee)

Curved Surfaces

393

3. What is the axis of a circle?

4. Is a solid of revolution rotationally symmetric? Is a rotationally symmetric object necessarily a solid of revolution? 5. Prove by reduction to a two-dimensional problem using an axial section: The locus of all points equidistant from two given points 4 and B is the symmetry plane. (Take the straight line through A, B as axis of revolution; each axial section contains which 2-dim. problem? The solution to the 2-dim. problem is now rotated to give the 3-dim. solution.) i

6. Can a doughnut (the kind with a hole) be considered a solid of revolution? If so, what is its characteristic section? What are the cross sections? Are all cross sections exactly alike? (The geometric model for a doughnut with a hole is called a torus. Give a formal definition of a torus.)

2.2 Spheres If we consider the sphere as a surface of revolution generated by rotating a circle around one of its diameters, then many important theorems about the sphere may be derived from theorems about the eIrcle. Which two-dimensional theorems lead to the following threedimensional theorems? > A plane intersects a sphere in a circle, a point, or not at all, depending on whether its distance from the center is smaller, equal, or greater than _ the radius.

> A plane tangent to a sphere (a plane with only one point in common with the sphere) is perpendicular to the radius through the point of contact (the common point).

S;

4

Sp

394

Unit 4

»> A sphere has two tangent planes perpendicular to a given direction. m> Ifa plane intersects a sphere in a circle, the radius of that circle is larger the nearer the plane is to the center of the sphere. The largest circle lies in a plane through the center; its radius equals the radius of the sphere.

Definition IV-6: On a given sphere, a circle with the same radius as the sphere is a great circle; its center is the center of the sphere. A circle with a smaller radius is a small circle.

On the globe, the meridians and the equator are great circles, the parallels are small circles.

> Infinitely many spheres pass through a given circle; their

y

.

centers lie on the axis of that circle.

Ss

Theorem IV-15: Exactly one

sphere

passes

through

four

non-

coplanar points.

Proof:

Show that the construction of the center has exactly one solution.

Curved Surfaces

395

. No 3 of the 4 non-coplanar points are collinear—

(Why?) . Exactly one © through 3 of the 4 points (IV-1).

. Infinitely many spheres through this ©; locus of the centers is the axis s of that ©. . Center of sphere equidistant from 4th point and, say,

the

3rd,

therefore

Cpa

ie

SD

must lie on symmetry plane between 4th and 3rd point. There is exactly one such plane. . Solution C = intersection of symmetry plane with axis s. Exactly one intersection between a plane and s, unless they are || or s lies in the plane.

. Indirect:if s were in the symmetry plane or ||to it, then the connector 3, 4 would be 1 to s and, therefore, the 4th point would

be in the plane of the circle; but then the 4 points would be coplanar, which contradicts the hyp. By considering a plane section of a sphere, a three-dimensional problem may be reduced to a two-dimensional one: > A straight line intersects a sphere in two, one, or no points depending on whether the distance of the line from the center is smaller, equal, or greater than the radius. (Pass a plane section through the straight line and the center of the sphere. Which 2-dim. theorem gives the solution?)

> A tangent to a sphere (a straight line which has exactly one point in common with the sphere) is perpendicular to the radius of contact.

> The tangent plane at point 7 of the sphere is the geometric locus of all tangents at T.

How can you adapt the statements about two circles to statements about two spheres?

396

Unit 4

®& Two spheres are tangent, if...; then the distance between their centers equals... if they are externally tangent, and... if they are internally tangent.

> Two spheres have no pom in common, if the distance between theif centerstis|c or... » Two spheres intersect in..., if the distance between their centers iS...and....

EXERCISES 1. What is a great circle and a small circle? 2. How many spheres pass through a given circle? locus of their centers?

What is the

3. What is the geometric locus of the centers of all spheres with radius 4 units, passing through a given point A? 4. If M is the center of a small circle of radius r, belonging to a sphere of radius R and center O, what is the relation be-

tween R, r and the distance OM? Which three types of problems about the sphere can be treated through this relation?

equator

Ex.4

5. On a 12 inch globe, how big is the radius of the small circle at 30° N latitude? (Reduce to a 2-dim. problem by meridional section. The diameter of the small circle appears as a chord in this section, and the plane of the equator as a straight line through O. Latitude 30° N means that the angle between the equator and radius R leading to any point P of the small circle is 30°. Then AOMP must have what angles?) 6. Repeat Exercise 5 for 60° WN latitude. 7. What is the radius of the small circle at 60° N latitude for a 20 inch globe?

Curved Surfaces 8. How many points are needed to determine a sphere? restrictions are placed on these points, and why?

397 What

9. A sphere of 13 inch radius is intersected by a straight line such that a 10 inch segment of that line is inside the sphere. Can you tell by how much the line misses the center of the sphere? 10. A straight line s intersects a sphere of 3 inch radius in the points A and B. How long is the chord AB, if a point P on the extension of AB is 2 inches from B and 4 inches from the center of the sphere? 11. Formulate

Theorems

[V-9,

10, and

their

corollaries

for

spheres and straight lines. (Consider plane sections through the spheres and the given lines.) 12. What relative position may two spheres have? 13. If two spheres intersect, what do they have in common?

14. If two spheres touch, what do they have in common? 15. If you know the radii of two spheres and the distance between their centers, can you tell whether they have points in common?

2.3 Cones and Cylinders If a right circular cone, its frustum, or a right circular cylinder is regarded as a surface of revolution, what is the meridian or characteristic section? Using the figurative expression “rotating,” we may say: & A conical surface of revolution is generated by rotating a straight line e (element, generator), which intersects a fixed axis s

wp A cylindrical surface of revolution is generated by rotating a straight line e (element, generator), which is parallel to a fixed

axis s, around this axis. at a fixed point V (vertex), around this axis. (See Figure on next page.)

The element e is here considered as an unlimited straight line, generating an unlimited surface. The two parts of the conical surface on either side of the vertex V are called the nappes. These unlimited surfaces are to be distinguished from the familiar solids: the cone with one nappe only and a base; the cylinder with two parallel bases (see Preparatory Chapter). Very often the unlimited surfaces are also called “cone” and “cylinder” for short, and you must watch the context to know whether they

398

Unit 4

s

s

or the solids are meant. The characteristic section of a cone is a triangle, of a conical surface a pair of intersecting straight lines. The characteristic section of a right cylinder is a rectangle, of a cylindrical surface a pair of parallel straight lines. Using the expression “geometric locus,” we may say: em» A conical surface is the geometric locus of all straight lines e passing through a fixed point V and through exactly one point of a fixed curve d (the directrix).

®& A cylindrical surface is the geometric locus of all straight lines e parallel to a fixed direction and passing through exactly one point of a fixed curve d (the directrix).

ZH

|

> The directrix of a conical surface of revolution is a circle whose axis passes through V.

® The directrix of a cylindrical surface of revolution is a circle whose axis is parallel to e.

In the general definition of a conical (cylindrical) surface the directrix d may be any curve, it need not be a circle. If dis a circle, its axis does not necessarily pass through V (is not neces-

Curved Surfaces

sarily parallel to e).

399

Thus the surface of revolution is a special

case of a conical (cylindrical) surface.

What happens if the angle at V between the axis s and the generators e of a cone becomes smaller and smaller, while the directrix

d remains the same? What happens if angle V becomes larger and larger? What do the following statements mean: > If xV — 0° for a fixed directrix, the cone — clyinder. > If xV — 90°, the cone —

plane.

> The cylinder and the plane are special (limiting, degenerate) cases of the cone.

What two-dimensional theorems lead to the following statements about conical and cylindrical surfaces of revolution by applying the principle of three-dimensional interpretation? > All tangents drawn to a sphere from an external point are equally long. Their geometric locus is a conical surface of revolution.

> From a point outside a sphere a tangent cone may be circumscribed around a sphere. The two objects have a circle of contact in common. & The geometric locus of all tangents to a sphere parallel to a given direction is a tangent cylinder. The two bodies have a circle of contact in common. > The tangent cylinder and the tangent plane may be considered as special (limiting, degenerate) cases of the tangent cone. em The tangent plane at a point T of a sphere is the geometric locus of all tangents at T. > Infinitely many spheres may be inscribed in a conical (cylindrical) surface of revolution; their centers lie...

By axial section we may reduce the following types of problems about cones and cylinders to two-dimensional problems: m Ifa plane parallel to the base of a circular cone passes through the midpoint of the altitude, the figure of intersection is . . . and the ratio between... and...is.

400

Unit 4

» The radius of a sphere (O,R) inscribed in a cone of revolution V) is related to the sides of the cone VT and the dis(vertex tance OV by the equation...

®& The radius of the circle of contact between circumscribed cylinder equals*...;

> The radius p of the

a sphere and a

A

circle of contact between a sphere and a circumscribed cone can be computed from a proportion between the sides of two similar right triangles; [Oe ee ear

=|

V

EXERCISES 1. What is the geometric locus of all points in space that have a distance of 3 units from a given straight line? 2. How many tangents can be drawn to a sphere from a given point? Does your answer depend on the position of the given point in relation to the sphere?

3. How many tangents parallel to a given direction can be drawn to a sphere? What is the locus of their points of contact? 4. Give a three-dimensional interpretation to the theorem: The bisector of an angle is the geometric locus of all points that are equidistant from the sides of the angle. 5. What is the geometric locus of all straight lines in space passing through a given point S$ on a given straight line s, that form a given angle a with s?

6. A sphere of radius R= 3 units is circumscribed by a cone, whose vertex is 9 units from the center of the sphere; how far is the vertex from a point of contact? How big is the radius p of the circle of contact? 7. A sphere of radius R = 3 units is circumscribed by a cylinder. How big is the radius of the circle of contact? 8. A sphere of radius R =5 units is inscribed in a cone such that the circle of contact has a radius p= 4 units. How far is the vertex of the cone from the center of the sphere?

Curved Surfaces

401

9. A circular cone of 6 units height has a base radius r= 2 units. If it is cut by a plane parallel to the base at a height of 5 units above the base, how big is the radius of the circle of intersection? 10. A frustum is formed by cutting a right circular cone of 10 inch height at 4 inches above the base. If the radius of the base circle is 3 inches, how big is the radius of the cut?

11. What is the radius and height of a right circular cylinder that can be inscribed in the frustum of Exercise 10? *12. If you have an unobstructed view from a lookout tower on a 1000 m (1 km) high mountain, how far can you see? (The lines of sight form a tangent cone to the surface of the earth. Radius of the earth

R = 6,400 km)

*13. In Exercise 12, how long is the line of the horizon? (The horizon is the circle of contact between the cone of sight and the surface of the earth. Compute the radius of the circle of contact and then its circumference. m= 3.14)

1

3

PLANE

SECTIONS

OF CYLINDERS

AND CONES

Any plane section of a sphere may be considered as a cross section or an axial section of a surface of revolution, but there are plane sections of cylinders and cones that are neither the one nor the other. A plane intersecting a cylinder or a cone may pass through a generating element and not through the axis; or it may have a general position containing no generating element. These two cases are discussed in this Section.

|

3.1 Planes Containing a Generator

If a plane > intersects a cylindrical surface of revolution and contains an element e but not the axis s, it must be parallel to s and to all other elements.

(Why?)

A cross section (Figure Ia)

relates a circle to the cylinder and a secant of that circle to the plane >. Each point of the circle is also a point of an element; through each point of the secant there is a parallel in } tos. Since the secant has two points in common with the circle, we see that > must have a second element in common with the cylinder. The distance between these two elements depends on the distance of > from

the axis s.

comes to your mind?

(How?)

Which

two-dimensional

problem

What is the maximum distance that } may

402

Unit 4

have from s and still have points in common with the cylinder? What happens to the two elements in that case? As in the twodimensional problem, we distinguish three cases, depending on the distance of > from s:

Theorem IV-16: A plane parallel to an element of a cylindrical surface of revolution either intersects that surface in two elements, or is tangent to it along one element of contact, or has no point at all in common with it.

(a)

(b)

Modify the discussion above to fit a conical surface of revolution intersected by a plane that contains an element of the surface. What will you say instead of “parallel to an element”? (Figure Ib.)

Theorem IV-17: A plane through the vertex of a conical surface of revolution either intersects both nappes in two elements, or is tangent to both nappes along one element of contact, or has no point other than the vertex in common with it.

Curved Surfaces

403

EXERCISES 1. What is a tangent plane of acone? ofacylinder? points does it have in common with the surface?

How many

2. If a plane intersects a cylindrical surface of revolution of radius r =5 units in two parallel straight lines that are 6 units apart, how far away from the axis is the plane? 3. Could an axial section be considered as a special case in the discussions of this Unit? What is an axial section of the cylinder in Exercise 2? What is the difference between an axial section of a cylinder and one of a cone? 4. In the discussion concerning the cone, how do we know what happens in the second nappe?

1 3.2 The Ellipse If a plane & intersects a cylindrical or conical surface of revolution and does not contain an element of the surface, what is the

intersection? Obviously, the intersection is a plane curve, because it lies in the intersecting plane. What else can be said about a point P belonging to this curve? In Figure Ila two spheres are

(a)

(b)

404

Unit 4

inscribed in a cylinder and are tangent to the intersecting plane >. The respective circles of contact with the cylinder are c, and cy; the points of contact with } are F, and F,. The segment PF, isa tangent from P to the first sphere; the segment PC, is also a tangent from P to the first sphere. .Thus PF, = PC,. (Why?) For the same reason PF,=PC,. By addition, we find PF, + PF, = PC,+ PC, =C,C,. The segment C,C, lies on the cylinder element through P and equals the distance between the two circles of contact c, and c,. For any other point Q (instead of P) the cylinder element through Q must contain a segment of equal length between these two circles. Why is that length constant? (G. P. Dandelin, 1822.) Thus we find:

Theorem IV-18: If a plane intersects all elements of a cylindrical surface of revolution, the intersection is a plane curve such that for each point of the curve the sum of its

distances from two fixed points of the plane is constant.

This curve is called an ellipse;

the two fixed points are its foci

(plural of focus).

Compare Figure I[b with Ila and prove that the intersection between a conical surface of revolution and a plane intersecting all elements in one nappe is an ellipse also. According to the definition, PF, + PF, =a constant, an ellipse can be drawn by using two thumbtacks

(foci) to tack

a

string of a certain length to the paper; if the pencil

keeps the string taut while moving, it traces an ellipse.

2

go: Sa

A gardener uses the same idea when laying out an elliptical flower bed with the help of a rope and two poles. With compasses and straightedge alone an ellipse cannot be con-

Curved Surfaces

405

structed; but as many points as necessary to see its shape may be constructed and then smoothly connected. By analyzing the symmetries of the curve special important points may be found. _ Construction of an ellipse, given the foci and the fixed length PP, + PR, = 2a:

(a) a general point P i. divide 2a arbitrarily, li. OF, with r=p,

p + q = 2a

OF, with r = q

iii. ©)s intersect at P;, P.; sym. w.r. to FF, iv. OF, with r=q, ©F, with r=p =e V.-@s imtetsect at P,, P,; sym. w.r. to FF.

Show that P,, P, sym. to P3,, P, w.r. to | bisector of FF, through O.) vi. Another arbitrary choice of p and g ini. leads to another point Q and symmetrically related points. (b) special points on the two symmetry axes i. F,;C=F,C =a and (ESD) NEADS 3h (choose p = gq=a) i. A sym. to B, therefore AO = OB =a (Show that AF, + AF, = 2a, BF, + BF, = 2a.)

Properties of the ellipse: iM Bilaterally symmetric with respect to the major axis AB= 2a and the minor 2. AO = OB = a= semi-major axis, CO . axis, F,O = OF, = e =linear

two perpendicular axes: axis CD = 2b. =OD = b = semi-minor f5 e

eccentricity,

the ratio - is the

numerical eccentricity. Cure C=O. — att. AF .OC yields at—=yb* +e".

406

Unit 4

4. a> e (from 3.), a necessary condition to permit the construction. 5. If e— O, the ellipse becomes a circle. Centrally symmetric with respect to O. [on. 7. Any straight line through the center O (a diameter) intersects the ellipse in exactly two points, symmetric with respect to O. The ellipse has many surprising properties. You may enjoy finding out about some of them in a special project of your own. (See “Suggestions for Special Projects.”’) Suppose the vertex of the cone in Figure II (on page 403) is a light source casting the shadow of the sphere onto the plane 3; then the contour of this shadow is geometrically the intersection of > with the tangent light cone — an ellipse! The shape of the particular ellipse depends on the position of the light source with respect to the sphere and the plane. Only in the special case where the axis of this cone (the straight line through the vertex and the center of the sphere) is perpendicular to >, is the shadow rotationally symmetric and therefore a circle.

Mie

If the light rays are parallel, as are the rays of the sun, then the tangent light cone becomes a tangent light cylinder and the shadow of the sphere on & is the intersection of this cylinder with 5 — again an ellipse! Only in the special case where the rays are perpendicular to > is the shadow a circle. In both cases the contour of the shadow actually is produced by the light rays through the circle of contact between the sphere and the cone

(a small circle) or the cylinder (a great circle).

This

Curved Surfaces

407

circle of contact separates the illuminated portion of the spherical surface from the dark portion. If the light rays are interpreted as lines of sight, then the above discussion gives some insight into the two methods of drawing three-dimensional objects on paper, which we have used in this course. In the pictorial view the object is seen sideways; the lines of sight are parallel to each other and under some arbitrary (oblique or right) angle to the projection plane > (the paper). In the front view the lines of sight are parallel to each other and perpendicular to the paper. In the corresponding plan view the lines of sight are parallel to each other and perpendicular to another plane, namely the “first projection plane” which is perpendicular to that of the front view, the “second projection plane.” In order to see the plan view we turned the “first projection plane” down into the paper. (See Preparatory Chapter, Section 4, Figure I.) In both cases the lines of sight (the projectors) are parallel to each other;

only their direction with respect to > is different.

A com-

mon name for both techniques is: Parallel Projection. The special case where the projectors are perpendicular to & is called an Orthogonal Projection (orthos = right, orthogonal = right angled). Ifthe angle of the projectors is oblique, we have an Oblique Projection. If the light rays emanating from a point source are interpreted as lines of sight (projectors), then the geometric projection is called a Central Projection or Perspective. The vertex of the cone of projectors is the center. This projection is informally quite familiar to you from photographs and from just seeing. That is why perspective drawings seem much more “true” to us, although parallel lines intersect. (If you drive on a straight highway, its parallel shoulders seem to meet in front of your eyes.) We have not used central projection in this course because its geometry is less simple than that of a parallel projection. > The contour of a sphere in parallel or central projection is an ellipse. For a special direction of the projectors this ellipse becomes a circle. Only in orthogonal projection is this circle the view of a great circle of the sphere.

EXERCISES 1. A definition:

forswhich

An ellipse is the geometric locus of all points

408

Unit 4

a= 2 inches and 2. Construct an ellipse with semi-major axis linear eccentricity e =1 inch. How big is the semi-minor axis b? 3. Construct an ellipse, if the two axes are given.

4. Construct an ellipse, if the foci and one point of the curve are given. 5. Construct an ellipse, if one focus and two points of the curve are given and also the length of the major axis. 6. Construct an ellipse, if the minor axis CD and the length of the linear eccentricity are given. 7. Construct an ellipse, if one focus F,, one endpoint D of the minor axis, and the length but not the direction of the minor axis are given.

8. In a given ellipse, what is the geometric locus of all midpoints of chords parallel to the major axis? 9. On an ellipse with major axis 10 units and minor axis 6 units, are there any points that have a distance from the center equal to the linear eccentricity? 10. How many points can there be on an ellipse that have a distance equal to the linear eccentricity from one of the foci? Discuss the various possibilities. 11. Explain why a circle can be considered as a special case of an ellipse. What can be said about the axes and the linear eccentricity in the special case? 12. Let a ball cast a projection) and Watch the shape tion of the paper cular shadow?

shadow on white paper in sun light (parallel in the light of a lamp (central projection). of the shadow change as you vary the posior of the ball. How can you produce a cir-

13. Look at the views of a sphere in the Preparatory Chapter, Section 6, Figures I, II, III. All of these onal projections. (What does that mean?) duce the pictorial effect in Figures Ia, Ila, (not the projectors!) was tipped forward, pole and a region around it could be seen. pole could not be seen.

|

views are orthogIn order to proand III the sphere so that the north

But then the south

3.3 The Hyperbola

If a plane > intersects both nappes of a conical surface of revolution, the curve of intersection is different from an ellipse. It is called a hyperbola. We see from Figure III, that it has

Curved Surfaces

409

two separate __branches. What other properties does it have? In Figure III two “Dandelin spheres” tangent to > are inscribed in the cone,

one in each nappe. F, and F, are the points of contact with >. From a general point P of the curve of intersection two tangents are drawn to the first sphere (PF, = PC,, why?) and two

tangents to the second sphere (PF, =PC,, why?). But now the constant distance between the two circles of contact c, and cz, is the difference PC,—PC,, which equals PF,— PF, by sub-

stitution.

uy

& The distances of any point P of a hyperbola from two fixed points F, and F, (the foci) have a constant difference.

Since the plane } cuts both nappes of the cone, its angle of inclination (X.1) towards the axis s of the cone is smaller than that of each element towards s (x.2). If you look at the cone and >

from the side such that © is seen as a straight line, then x1 and Xx.2 are seen as in Figure [Va. A plane II parallel to > through the vertex V obviously is seen as a straight line also, forming the same X1 with s. By contrast, the plane > in Figure IVb cuts all elements in one nappe (in an ellipse) and x_1 is larger than x2. According to Theorem IV-17 the plane II contains two elements in the hyperbolical case a and none in the elliptical case b. Every point of the hyperbola is, by definition, a point of intersection between > and an element of the cone. Not every element, however, intersects >. The two elements lying in II (the red lines in Figure V) are parallel to } and therefore do not intersect >. Considering the cone and © as unlimited, a sequence of elements approaching one of the red lines (1,2,3 — 4) corresponds to a

410

Unit 4 | | | | | |

|

| | | | | a. hyperbola

in 2

sequence of points of the hyperbola, which leads farther and farther out without ever reaching a last point. The elements 5, 6, 7 beyond line 4 intersect > in the other branch of the hyperbola. Approaching and _ passing the second red line (8), we

switch back to the first branch until the circuit is complete when reaching element 1. The directions in which the two switches from one branch to the other occur are given by the directions of the red lines. We shall

see later on the hyperbola is a centrally symmetric figure. The two straight lines in >,

which pass through its center

b. ellipse

in =

Curved Surfaces

411

and have the same directions as the red elements (are parallel to the red elements), are called the asymptotes Construction of a hyperbola, given the foci and the fixed length Pio — Ph

A= OG

Kc

Sh =a q

tt 2a =

(a) a general point P

i. choose arbitrary segments such that p — g = 2a

li. OF, with r =p, OF, with r= q ili. @s intersect at P,, P., sym. to FF,

iv. OF, with r=q, OF, with r =p Ves antersect at Ps, P,, sym. to FF, (Show that P,, P, are sym. to P;, P, w.r. to the | bisector of FF, through

O.) (b) special points on the two symmetry axes i. no points on the perpendicular bisector (why not?)

ii. make OA = OB =a (Show that AF, — AF, = 2a, BF, — BF, = 24a.) (c) construction

of

the

asymptotes (It is easier to sketch a hyperbola, if the asymptotes are drawn first.)

i. ©O withr = OF, =e ii. Ls to AB at A and at B intersect © at C and C’

Pn"

Pp

eee

r2- 2.59807 --r2 +3 .00000 - : roe 1 OS830." r2-3.13262--T2793; 1S939I* Path 102 r2-3.14146--:r2-3.14154--ie SONS WIL O° c To VASO. eae

Ap

eee

Aer,

r2-3,°¢+3° iP uaBIal OO. OND P22) Slee re P2312 r2-3.141- > Ta eee oA ler r- 3.141 -% r2- 3.14159

r2+-3.46410--:r2-3.21539--r-3.15966-:r2- 3.14609 --r2-3.14271--:Toa AnOime ck r2-3.14166::PS 141613 r2+-3.14160-:r2 3.14159 °°:

Ag =ra

r-

J foe

7

J 7

The computation of this table is quite easy if we make use of the already computed perimeters, listed in the table on page 307, and of the ratio of similitude between the n-gons. This ratio is s,/S,; it can also be expressed as p,/r and can be computed as Pr/Pn (column 2). It follows that the apothem p, = r(p,/P,). Considering that the diameter d = 2r,

the column for a, is computed from the column for p, on page 307 by multiplying it by 3p; and the column for A, by multi-

plying P, by 3r. We see from the table that a, and A, approach the same limit as n increases.

Definition V-1: The area A of a circle is defined as the common limit

of the areas a, of the inscribed n-gons and A, of the circumscribed n-gons as n approaches infinity.

The table shows that with increasing n more and more digits in a, and A, are the same, giving more and more precision to the area of the circle between a, and A.

Area and Volume

455

In technical language the meaning of the table may be expressed in various forms: “As n approaches infinity” is an equivalent phrase to “as n increases without bound.” [written n+ o| As n approaches infinity, (A, — a,) approaches zero. [for n > 0, =) 0] As n “becomes infinite,” the ratio of similitude p,/P,, approaches 1.

ap

1

As n “goes to infinity,” p, approaches r. [for n > %, p —>r] With increasing n, A, and a, approach the same limit. [lim 4, = lim Cua Al 2. A sector In a circle of radius r, the area

of a sector with central angle a compares to the area of the circle

as

a

to 360°

(or 27).

a . 7r (Here a 360° is in degrees.) a A (sector) = “re Tr? = 407" T

A (sector) =

(Here qa is in radians.)

A (sector) = arc . $r (irom Gess0U —vare: 2757, See Unit 4, Section 1.2)

Note: This last formula looks very much like a formula for a triangle with the arc as its base and the radius on the symmetry axis as its height—an informal way to remember it.

3. A segment In a circle of radius r, the area of a segment equals the area of the corresponding sector minus the area of the triangle formed by the corresponding chord and the radii at its endpoints.

a

°. “o

456

Unit 5 EXERCISES . Check the dimensions of wr? and 2ar; perimeter and which for the area?

which stands for the

. If the area of a circle is.20 sq in., how long is its circumference? . What is the ratio between the areas of the inscribed and the

circumscribed circles of a square? their circumfereiices? . Repeat Exercise square.

What is the ratio between

3 for an equilateral triangle instead of the

. Two circles touch each other from the outside.

their areas is 747 sq in. and the distance centers is 12 in. Compute their radii.

The sum of

between

their

. The difference of the areas of two circles is 6007 sq in.; the ratio between the radii is 7:5. Compute the areas of the circles. . What is the area of a circular ring (space between two concentric circles) of the radii R and r? If the area of a 2 in. wide circular ring is 207 sq in., how big are its radii?

. A circular and a square swimming pool have an equal perimeter of 100 ft; they are surrounded by an even 4 ft grass strip. Which of the grass areas is larger and by how much? . The radii of three concentric

circles have the ratio 2:3:4.

How big are the areas of the two circular rings formed if the area of the smaller circle is 12 sq in.? 10. A circle is divided into four equal parts by three concentric circles. What is the ratio of the four radii? 11. Prove: Ifa straight circular cone of radius r and height h is cut by a plane & parallel to its base, and if some straight line through the vertex V intersects the plane of the Vase in A and the plane = in A’, then the area of the base circle is to the

area of the intersection as VA?: VA”. 12. Given the radius r of a circle and the areaA of a sector, find the central angle: a Sone A= 47 sq in.

b 6 in. 127 sq in.

c 16 in. 6477 sq in.

13: For a circle of radius r compute the area of a segment with a 45° arc. (60°, 90°, 120°)

Area and Volume

457

14. The moons of Hippocrates (400 B.c.): The semi-circles drawn with the sides of a right triangle as diameters form two crescents; prove that the sum of these crescents equals the area of the triangle.

Ex. 14

Exe

15. The arbelos (sickle) of Archimedes (220 B.c.): If the diameter AB of a semi-circle is divided into two arbitrary parts AC and CB, and if a semi-circle is drawn with each part as a diameter (as shown), then the area between the three semi-

circles equals the area of the circle with CD | AB as a diameter. 16. The salinon (salt cellar) of Archimedes: The diameter AB of a semi-circle is subdivided symmetrically by x = AC = BD and semi-circles are drawn above and below as shown. Prove that the shaded area formed equals the area of the

circle with EF as diameter. What choice of x gives the salinon a maximum or minimum area? F

E Ex. 16

EX

17. The pelekoid (ax): Compute the ratio of the area and circumference of the pelekoid (shaded area) to the area and circumference of the whole circle. (AB =BC = CD = DE)

458

Unit 5

18. Compute the shaded area in the given figures if the radius r of the whole circle is given:

(a)

(b)

(c)

19. Compute the shaded area in the given figures if the side s of the whole square is given:

3. SURFACES 3.1 Developable Surfaces The surface of a square pyramid consists of a square and four lateral triangles. The area of this surface is the sum of these five plane figures and can be computed by the area formulas discussed in the previous Section. The five plane figures can be mapped into congruent figures in a common plane. This fact finds a practical expression in the familiar method of making paper models of some geometric objects from their nets. (Figure I.)

Ce

Area and Volume

459

Surfaces that can be flattened out into a plane without distortion (in technical language: mapped into a plane such that the size of angles and the length of segments are preserved) are called developable surfaces. The pyramid and the prism are developable surfaces. So are the circular cylinder and cone, as you know from informal experience in rolling a rectangular piece of paper into a cylinder, and turning a sector of a circle into acone. The technical explanation for this informal experience must make use of a process of infinite approximation. How? (See Unit 3, Section 6.) The surface of the sphere and most other curved surfaces are not developable. Although the surface of the earth is mapped onto a sheet of paper in a geographic map, all these maps show distortions. As the area of a developable surface is computed as the sum of the areas of plane figures the following formulae are easy to understand.

Prove them

and memorize

them

(to save time at work).

They are symbolically represented by an L?; why?

Notation:

S = total surface (everything that could be painted),

LS = lateral surface (total surface without the base(s)), a base, p = perimeter or circumference

B = area of

of a base, r = radius of a

base circle, a, b, c,... = edges, H = height of the object, height, s = side of a cone or frustum.

= slant

To explain the following formulae, draw a pictorial view of each case (see Preparatory

that each face can base, which faces there are in the Then derive the faces. The cube:

Chapter,

Section 3);

label each vertex so

be identified. Show which face may be called a belong to the lateral surface, and how many faces total surface. Compute the area of each face. following formulae by adding the areas of the

S = 6a? (see Figure I)

The rectangular solid:

LS = 2ac + 2bc=p-H S = LS + 2B = 2(ab + ac + bc)

The right prism:

LS = p - H (see Figure I) S=LS+ 2B

The n-sided pyramid: LS = sum of n triangle areas (see Figure I) S=LS+B

460

Unit 5

The frustum of a pyramid:

LS = sum of n trapezoids S = LSB Bs

The right circular cylinder: LS =p: H =2ar -H (see Figure 1) S=LS + 2B =2nrd + 2ar =2ar(r + A)

The right circular cone:

LS =tp - s = mrs (see Figure I) S=LS+B=ars+ar=ar(rt+s) (The LS of a cone flattened out into a

plane equals the area of a sector of a circle whose radius is the side s. The central angle a of this sector is to 360° as the length of the arc is to the circumference 27rs of the circle. The area of this sector (see page 455) equals 3 arc: s. The arc of the sector equals

informally, Cees eek

the circumference 27rr of the base circle of the cone. Thus LS = 7r -: s.)

The frustum of a right circular cone: LS = 75(7,+-7,);,.(detiveuss as the difference between the LS of two cones.)

S=LS+B,+B,

2

o/ 2

>

= 1 1g) ee ge = 7S Pa) eee aes |

ees 3 rs

" informally, a trapezoid

EXERCISES . Prove: In two similar pyramids (prisms, cylinders, cones) the ratio between their surfaces equals the square of the ratio between their heights.

. Prove whether or not the two bases of a frustum of a pyramid (or cone) are similar figures. . If the surface of one cube is 4 times as large as the surface of another cube, what is the ratio of their edges? . The lateral surface of a regular square pyramid is 4 times as

Area and Volume

large as the Base, whose edge is 3 in. ing: (which rt. A’s will help?) a. The area of a lateral face. b. The length of a slant height. c. The length of a lateral edge. d. The height of the pyramid.

461

Compute the follow-

. A regular square pyramid has a base edge of 12 in. and a lateral edge of 10 in.; it is cut at half its height by a plane parallel to its base. Compute the surfaces of both pieces.

. Repeat Exercise 5 for the general case of base edge a and slant height s. . What is the height of an equilateral cylinder (the axial section iS a square), if its total surface is 5 sq ft? (What is the relation between r and H of an equilateral cylinder?) . A sphere of radius R is inscribed and circumscribed by an equilateral cylinder (see Exercise 7). What is the ratio between the surfaces of these cylinders? . The lateral surface of a right circular cone is 45 sq in. and the radius of the base is 3 in. Compute the slant height and total surface. (Use 7 = 3.14)

10. The total surface of a right circular cone is 8 sq ft and the radius is foot. Compute the slant height and lateral surface. (Use 7 = 3.14)

11. An equilateral cone (the axial section is an equilateral triangle) has a total surface of 150 sq in. Compute its radius. (What is the relation between r and s of an equilateral cone?) 12. A sector of a circle with radius 5 in. and central angle of 120° is formed into a circular cone. What is the radius and height of the cone? 13. If an equilateral cylinder and an equilateral cone have equal lateral surfaces, what is the ratio of their radii and the ratio

of their total surfaces?

14. What is the total surface of a frustum of a right circular cone if its axial sections are isosceles trapezoids of base line 10 in. whose other three sides are each 6 in. long? 15. Compute the total surface of a frustum of a right circular cone, if the radii of the bases are 5 and 2 in. and the height is 4 in. 16. Compute the total surface of each of the five Platonic solids if an edge is 2 in. long. Are these surfaces developable surfaces?

462

Unit 5

17. An equilateral cylinder and an equilateral cone have equal radii; what is the ratio of their total surfaces?

18. A right circular cone of radius 6 units and height 8 units has the same lateral surface as a right circular cylinder whose radius is 4 units. Compute the total surfaces of both objects. 19. An equilateral triangle of side s = 2 units is inscribed and circumscribed by acircle. The larger circle is the base of a right cone, and the smaller circle is the base of a right cylinder. How high are they if the lateral surface of the cone is twice that of the cylinder while their heights are equal? 20. A right circular cylinder of radius 3 units is 8 units high. Two cones are cut out of the cylinder such that their bases coincide respectively with the two bases of the cylinder, while their vertices coincide with the midpoint of the cylinder’s altitude. Compute the area of the remaining object.

o

3.2 The Surface of the Sphere

Theorem V-1: If a 2n-sided regular polygon is rotated around an angle bisector as axis, the area of the resulting surface of revolution equals the product of that axis

AH and the circumference 2p of scribed circle.

the

in-

Proof:

Draw

1s to the axis AH from each vertex and from the mid-

point of each side.

Each side of the polygon generates either a

cone (1) or the frustum of a cone (2) or a cylinder (3). of the inscribed circle. (1) AB generates a cone (also GH). AABB' ~ AMOM' (Why?)

p = radius

Area and Volume

463

AB: AB’ = OM:MM' or AB - MM’ = OM - AB’or

AB -BB’=p - AB’

LS of cone =BB’ - AB - 7 = 2p - AB’ by substitution (2) CD generates the frustum of a cone (so do BC, EF, and FG). Draw CD" | DD’.

ACDD" ~ ANON’ (Why?) CD:CD" =ON: NN’ or CD -NN’ = p - CD" =p -C'D’

LS of frustum= 7s(CC' + DD’)

= 27 -CD -NN’

= 27rp - C'D' by substitution. al (3) DE generates a cylinder.

LS of cylinder = 27p - E’D” In all three cases the LS generated equals the product of 27p and the projection of the generating side on the axis. The sum of all these projections is the length AH of the diagonal used as the axis of revolution. The total surface generated thus equals 27rp = AH;

q.e.d.

Corollary V-1-1: If only part (some adjacent sides) of the polygon in Theorem V-1 is rotated, the area of the resulting surface of revolution equals the product of the corresponding part of the axis and the circumference of the inscribed circle.

Example: If CDE rotates around AH, the area of the surface of revolution equals C’E’ - 27rp. Theorem V-1 in connection with the already familiar concept of approaching a limit in infinitely many steps helps us to find an expression for the area of the surface of the sphere. A circle and a pair of inscribed and circumscribed 2n-gons with a common angle bisector are rotated around this angle bisector and generate . (what?). What happens if n— 0? What becomes of p if 1 1001) » The surface of a sphere is the common limit of sequences of circumscribed and inscribed surfaces generated by rotating

2n-gons around a common symmetry axis as n > ©

464

Unit 5

The sphere:

S = lim 27p - axis = 27R -2R = 477R?

By Cor. V-1-1 parts of the sphere can be computed: The spherical segment (spherical cap):

LS = 27 RH (Why?) S = LS + 7p?

(Here p is the radius of the small circle and H is the height of the segment.)

O axial

section

The spherical zone: LS = 27RH_, (Why?) S = LS + wp, + mp."

(Here H, is the height of the zone.)

axial

The spherical sector:

S = LS of the segment + LS of the cone =27RH

+a7pR

section

Area and Volume

QD)

465

(N/ axial section

Can problems about R, p, H of parts of the sphere be reduced to two-dimensional problems by axial section? Prove and remember these very useful relationships:

p? = H(2R — H)

R? =p? +(R — Hy?

EXERCISES . A regular hexagon of 2 in. side rotates about an angle bisector. Compute the surface of revolution.

="

. A square of side s =3 (in general s) rotates around its diagonal. Compute the surface of revolution: a. using Theorem V-1. b. as two cones. . If the surface of a sphere should be doubled, the radius must

be multiplied by...? If the radius is multiplied by 3, the surface becomes how many times as big? . How will you place parallel plane sections through a sphere in order to divide its surface into five equal parts?

. What is the total surface of a spherical segment with base radius p= 6 units and height H = 2 units? . What is the height of a spherical segment in a sphere of radius R=5 units, if the lateral surface is twice (n-times) the area of the base circle? . For a spherical segment: a. H=2,R=5; computep. . R=52, p=20; . p=

compute H.

15, H= 3; compute R.

466

Unit 5

8. A cylinder is circumscribed around The axis of the cylinder is subdivided into four equal parts H, and through each point of division a plane perpendicular to the axis is drawn. Compare the lateral areas of all segments and zones formed on the sphere and on the cylinder. 9. What is the ratio of the surfaces of three concentric spheres with centers at the

a sphere of radius R.

Front

view

Ex. 8

center of a cube, of which one

is tangent to the faces of the cube, one passes through the midpoints of the edges, and one passes through the vertices? 10. What is the height of a spherical segment, whose lateral surface equals the area of a great circle of its sphere?

|

*3.3

Surfaces of Revoiution

If a segment a is rotated around an axis parallel to a at a distance P, it generates a cylinder whose lateral surface A equals a‘ 27p. In other words, A is the product of the length of the generator times the circumference of the circle described by one point of a during rotation (the path traveled when rotated). Pappus (4th century A.D.) generalized this idea: If a were bent into a curved or broken line, a triangle or some other plane figure, the generated surface of revolution should equal the length of a times the path of some mean point. If a is a segment inclined towards the axis of rotation, then the path of one endpoint is much shorter than the path of the other. One would expect that the midpoint of a is the correct choice for the mean point whose path, if multiplied by the length of a, will give the correct surface of revo-

lution.

If a is an equilateral triangle, the correct mean point turns

out to be the centroid of the triangle (see Unit 2, Section 3.5). It can be shown that in a bilaterally symmetric figure the mean point lies on the symmetry axis; in a centrally symmetric figure it lies at the center of symmetry. These examples show that this point does not necessarily lie on the generator of the surface (the perimeter of the rotating figure). The following theorem by Pappus

Area and Volume

applies to any plane figure in general; centroid.* Theorem V-2:

467

its mean point is called its

Pappus’ Rule for Surfaces (Pappus, 4th century A.D.)

If a plane figure is rotated around an axis in its plane, such that the axis does not pass through the interior of the figure, the surface of revolution generated equals the product of the perimeter of the figure and the path of its centroid.

We shall leave the general proof of this very practical theorem to a calculus course, and use it here after checking it for some simple cases:

Example 1: If an equilateral triangle is rotated around one of its sides as axis, then it generates a surface consisting of the lateral

2

surfaces of two cones with a common base radius 1qv/3 (correct?)

and height $a. S=2-7-$a Is

the

$ c

Thus

3

3- a= TaV3.

dimension

correct

for

2 a

s

surface? Using Pappus’ rule we need the perimeter p = 3a and the path of the center of gravity (the centroid). In an equilateral triangle the centroid C divides each median in the ratio 2:1, therefore the distance p of C from the axis is 4 of 4a\/3.

C during rotation is 277p (is it?). S=p-2mp

The path of

Then

=3a- 2m -4-4aV3 = 7aeV3

as above. 1 The centroid of a geometric figure is also called its center of gravity, in analogy to the center of gravity of a physical object (see a textbook on physics). The location of the physical center of gravity depends on the distribution of mass in the physical object. If this distribution is uniform throughout the object, as for example in a thin cardboard triangle, the center of gravity lies at the geometric centroid. (See Unit 5, Section 2.2, Exercises 16 and 17; also Special Project XIII.)

468

Unit 5

Example 2: A straight line segment rotates around an axis in three different positions as shown. The center of gravity is the midpoint, the “perimeter” is the length of the segment. What surfaces are gen: erated in each case? Compute them (Ne by Pappus’ rule and check the result with the usual surface formulas. Example 3: If the area of a surface of revolution is known, then Pappus’ rule

Se fs

may be used to find the center of gravity

cc

of the axial section.

If you had for-

p

gotten where the centroid of the triangle in Example | is, you could reason as follows: the two cones generate a surface $ = qa?\/3; this must equal S =p - ae where p is the a unknown. With p = 3a

= TaV3 _ 4\/3 = Fh.

we find p = ea = 67a

(Correct?)

EXERCISES 1. An equilateral triangle is rotated around an axis in its plane, passing through a vertex perpendicular to one side. Compute the surface of revolution: S a. by specific surface formulas, b. by Pappus’ rule, using the centroid of the triangle, c. by Pappus’ rule, using each side separately. 2. A square rotates around

one of

its sides. Compute the surface a of revolution it generates, by the three methods of Exercise 1. Ex. | Where is the centroid in a Square? 3. A square rotates around an axis in its plane, perpendicular to a diagonal at distance d from the nearest vertex. Compute the surface of revolution generated. 4. Acircle rotates about a tangent. Compute the surface of revolution generated.

Area and Volume S

469

S

BxXeS

Exe

5. A circle of radius r rotates around an axis in its plane at distance d. Compute the surface of the torus. 6. A regular hexagon of side a rotates about one of its sides. Compute the surface of revolution.

4

VOLUME

A volume is represented by an L?. What does this mean? As you learn the volume formulas in this Section check their dimensions.

4.1 Prisms and Cylinders The cube 1. The volume V of a cube of edge a in terms of any length unit equals a: a: a=a’ in terms of the corresponding cubic unit. 2. If the edge of a cube is increased n times, the volume V is increased n® times. The rectangular solid 1. V=a-~-b--c,

if the edges are a, b, c.

2. If the edges a, b, c of a rectangular solid are increased /-, m-,

n-times respectively, the volume V is increased / - m - n-times. 3. If each edge of a rectangular solid is increased n-times, the volume V is increased n*-times.

Theorem V-3: Principle of Cavalieri (17th cent.) Two objects which have intersections of equal area with any plane parallel to a given plane have equal volume.

470

Unit 5

Such objects are called Cavalieri solids. It is obvious that not all objects of equal volume are Cavalieri solids. Give examples. The proof of Theorem V-3 will be left to a calculus course. By finding a suitable rectangular solid to serve

as a

Cavalieri solid to a prism we derive the following corollary: Corollary V-3-1: The volume B-H.

of any prism equals Base

X height;

(Here the capitalized Base shall mean “base area” as distinguished from base = “base line.” Capital H means height of the object.) By using the process of infinite approximation of a cylinder by a sequence of prisms we further derive: Corollary V-3-2: The volume of a cylinder equals Base x Height (V = nr’H). Corollary V-3-3: If two prisms or cylinders have equal height, then the ratio between their volumes is the same as that between their Bases (Why?) Corollary V-3-4: If two prisms or cylinders have equal base areas, then the ratio between their volumes is the same as that between their heights. (Why?)

Corollary V-3-5: Pyramids and cones with equal Base and equal height have equal volume. (Why?) (Show that they are Cavalieri solids.)

EXERCISES Group I

1. Compute the volume of a right square prism that is 24 in. high and has a base edge of 16 in. (32 in. high, 25 in. base edge)

Area and Volume . The edges of a rectangular solid are a=6 c=9 in.; compute its volume. (13, 25, 46)

in., b=7

471 in.,

. How high is a rectangular solid with a base of 25.5 sq ft, if its volume is 204 cu ft?

. A regular hexagonal prism with base edge a = 3 in. is 10 in. high. Compute its volume. If each base edge is made 4 times as long, what is the volume then? . A cylinder is 10 in. high and has a radius of 3 in.; compute its volume.

. If a cylinder has a volume 3.5 in., how high is it?

of 384.8 cu in. and a radius of

. Compute the volume of a cylinder which is 5.6 in. high and has a base of 2257 sq in.; what is the radius of the base? . A cylinder which has the same base as the cylinder in Exercise 7 is 4 times as high; what is the ratio between the two volumes? . A cylinder which is as high as the cylinder in Exercise 7 has a base of 4507 sq in.; what is the ratio between the two volumes?

10. How long is the diagonal of a cube whose edge s=5 in.? This diagonal equals the edge of a second cube; what is the ratio in volume between the two cubes? 11. What are Cavalieri solids?

Give examples.

Group II (To compute a volume from a surface or vice versa, you must inspect both formulas involved to find out which length (a radius You may have to compute this or an edge) connects them. length first, even if it is not specifically asked in the question of the problem. See example in Supplement III-E on equations and word problems.) 1 The surface of a cube is 150 sq in. What is its volume? (For which unknown will you solve first?) . What is the surface of a cube whose volume is 27,000 cu ft?

. The volume of an equilateral Compute its total surface.

cylinder is 686,0007 cu ft.

. The lateral surface of an equilateral cylinder is 1447 sq ft. Compute its volume. . If the base of an equilateral cylinder is 425 sq ft, how big is its lateral surface?

472

Unit 5

6. A regular hexagonal prism and a regular square prism have equal height and equal lateral surface. What is the ratio between their volumes? . A rectangle with the sides a= 8 in. and b=5 in. is formed into a cylinder by bending once a and once b; compute the volume of the two cylinders. . If the ratio between the surfaces of two similar objects is 1:4 (or 4:9 or 100: 225), what is the ratio between their volumes?

. If the volume of one cube is ; of that of a second cube, how

do their faces compare?

(zag)

10. Are two equilateral cylinders similar objects? * Group III

i If x stands for the edge of a cube and y for its volume, then y is a function of x. Make a graph of that function. 2. In a cube:

a. given S, find V. b. given V, find S.

c. given the longest diagonal D, find S$ and V. d. given the face diagonal d, find S and V. Note: The answers are to be expressed in terms of the given measurement. Thus, in 2d, the expressions for S and V may contain only d and numbers.

ab Compute the volume of a cube whose surface equals that of a rectangular solid with edges 2, 5, 8 in. . The volume of a regular 6-sided prism is 512 cu ft. Its height is twice the length of a base edge. Compute its total surface. . The lateral surface of a regular 3-sided prism is 90 sq ft and its Base is 9V3 sq ft. Compute its volume. . The lateral surface of a right circular cylinder is 20 sq ft and its Base is 9 sq ft. Compute its volume. . A regular 3-sided right prism of base edge 1 foot and height 3 ft is inscribed and circumscribed by cylinders. Compute the volume of the pipe formed by the two cylinders. . A rectangle of sides a and b is formed into a cylinder by bending once a and once b. What is the ratio in volume between the two cylinders?

. A rectangular solid with edges a, b, c may be circumscribed by three different cylinders. What is the ratio of their volumes?

10. Compute the volume of a rectangular solid with longest di-

Area and Volume

473

agonal D = 28 in., if the ratio between the edges is a:b:c =2:3:6. 11. A parallelepipedon has lateral faces of 36 sq in., 60 sq in., and 100 sqin. How big are the lateral faces of a similar parallelepipedon, if: a. The volume of the first is 8 times that of the second, b. The surface of the first is 4 times that of the second?

12. The Base of a prism is 75 sq in. and its height is 15 in. Compute the Base and height of a similar prism whose volume is

; of that of the first. 13. If the ratio of similitude of two similar objects is n: 1, what is the ratio of their volumes? What follows for n:1= 1:1? 14. The volumes of two similar cylinders are 720 and 90 cu ft. One cylinder is higher than the other by 10 ft. Compute their heights and their radii.

15. What relation exists between the radius and the height of a right circular cylinder, if the measures of its surface and its volume are expressed by the same number of sq ft and cu ft respectively?

4.2 Pyramids and Cones The Pyramid

V =3 Base X height

E

A

Ca

:

D

UaRA et A

E

C

B

D

c C

A B

Proof for a triangular pyramid: A prism of the same height and with the same Base as the

474

Unit 5

pyramid ABCS (Figure I) may be split into three pyramids of equal volume as follows: 1. The plane ACS cuts off the given pyramid ABCS. 2. The remaining 4-sided pyramid with the parallelogram ACDE as Base may be split into two equal pyramids by the plane ECS. (Why equal?) 3. One of these two equal pyramids, ESDC, has a base triangle and height equal to those of the given pyramid ABCS. (Why?) Therefore the three pyramids are of equal volume and each pyramid is 3 of the prism. With Cor. V-3-5 this proof may be extended to cover any pyramid and cone.

The Cone V = 3 Base X height =$ar?-H

The frustum of a pyramid or of a cone

= 1H(B, + VB,B, + Bo)

IM\

=~

Prove as a difference between two similar pyramids (cones):

V =4(6.4, — 3

B,

al,

B.H,),-where H,

H, H,

oh

h

ES

—H,=H

Beh VBC

eS

see

and Bi: B,— A": A,”

H,\3 1 J masa

es

I

_ (Hh_,\[ (Hy, & (Gz) +a tt 4

Area and Volume

B =1 ; net a1 V=4i3B,

475

IB a1 pol

=3H [B, + NB

By tes |i q.e.d.

The frustum of a circular cone Seal

WAR

epee)

EXERCISES (V = volume, B= base area,

r=radius of base circle, a = edge

of base, H = height, h = slant height) Group I

1. Compute the volume V of: a. a regular square pyramid, if a= 5 in. and H = 12 in. b. a regular triangular pyramid, if a= 5 in. and H = 12 in. c. a 7.5 in. high pyramid, if B =36.80 sq in. d. a circular cone, if

r=3 in. and H = 8 in.

. How high is: a. a pyramid, if V= 120 cu ft and B =45 sq ft. b. a cone, if V= 2257 cu ft and B= 757 sq ft. c. acone, if V= 1967 cu ft and r=7

ft.

. Compute the volume V of the frustum of a pyramid:

a. if B, = 25 sq ft, B, = 12.25 sq ft, H = 12 ft. b. if H=9 ft and the bases are squares with a,=5 ft and a, = 3 ft. c. if

H=15 ft and the bases are equilateral triangles with

a, = 6 ft and a, =5 ft. . Compute the volume V of the frustum of a cone: a. if r, = 6.5 in., r,=4 in., H = 12 in. b. if B, = 257 sq in., B, = 167 sq in.,

H = 18 in. .

. How high is the frustum of a regular square pyramid: a. if V= 185 cu ft, a, =4 ft, a, =3 ft. b. if V= 51.24 cu ft, B, = 2.56 sq ft, B, = 6.25 sq ft. . What is the geometric locus of the vertices of all pyramids with the same volume and a given square as Base?

. An isosceles trapezoid with base lines 6 and 2 in. long and a base angle of 45° rotates around its symmetry axis. Compute the volume of the object generated. What is the radius of the base circle? . Compute the volume of a regular six-sided pyramid, if the Base B =54V3 sq ft, and if the height is twice a base edge.

476

Unit 5

Group II (To find ratios concerning objects that are not similar, compute the areas or volumes in question and then simplify the ratio, using the additional information given. Reduce to two dimensions and draw a sketch of.the two-dimensional problem involved.) 1. A circular cone is inscribed in a right circular cylinder; what is the ratio in volume between the two objects? (Draw an axial section.)

. A regular three-sided pyramid is inscribed and circumscribed by a cone.. What is the ratio in volume of the two cones? (Note that all three objects have the same height; therefore an axial section would not tell much. Draw a plan view to show the relation of the three bases.)

. A circular cone is circumscribed about a circular cylinder whose height is half the height of the cone and whose Base lies in the Base of the cone. What is the ratio in volume between the two objects? (Draw an axial section to find the relation between the radii.) . If an equilateral cone and an equilateral cylinder have the same radius, what is the ratio between their volumes? their heights first.)

(Find

. If an equilateral cone and an equilateral cylinder have equal lateral surfaces, what is the ratio between their volumes? (Find the connection between their radii.)

Group III ies If two similar pyramids have the ratio of similitude 1:3 (or 1:4, or 2:3), what is the ratio of their heights, of their Bases, of their surfaces, of their volumes?

. If a 12 in. high pyramid with a 36 sq in. Base is cut by a plane parallel to the Base in half its height, what is the ratio between the area of intersection B’ and the Base B? How big is B’? Do you have to compute the volumes of the two pyramids in order to know their ratio? What is this ratio? Compute the volumes of the two pyramids and check. - Compute the volume of the frustum left in Exercise 2, using two methods: as the difference between the volumes computed in Exercise 2, and by the formula for a frustum. What is the ratio in volume between the two pieces into which the pyramid was cut?

Area and Volume

477

. If 4 of the height is cut off from the top of a pyramid by a plane parallel to its base, what is the ratio between:

a. the altitudes of the two similar pyramids? b. the altitudes of the two pieces into which the pyramid was cut? c. the base areas of the two similar pyramids? d. the volumes of the two similar pyramids? e. the volumes of the two pieces? . If the top of a cone is cut off by a plane parallel to its base, and if the volume of the small cone is } (or 4) of the whole cone, what is the ratio between the altitudes of the two cones? between their radii? between their Bases? between their surfaces? between their volumes? What is the ratio between the volumes of the part cut off and the frustum left? . Where was a pyramid cut off, if the frustum has a volume of 520 cu ft, a larger Base B, = 36 sq ft and a ratio between the two Base areas of 9:1? (First compute the smaller Base, then the height of the frustum from the formula, then the ratio of similitude between two similar pyramids, then the ratio between the part of the height that was cut off and the part that was left for the frustum, finally the length that was cut off.) . A cone is cut off parallel to its Base such that one fourth of the lateral surface is lost. What is the ratio between the radii of the frustum? How much of the volume was lost by the cut?

o 4.3 Polyhedra and Spheres The formula for the volume of a pyramid can be used to compute the volume of any polyhedron if there is a point O that is equidistant from its faces (i.e. a polyhedron that can be inscribed by a sphere). In that case the polyhedron may be considered as the sum of as many pyramids of equal height as there are faces, with the point O as a common vertex. (See Unit 3, Section 6.2.) The polyhedron, inscribable by a sphere V =3p

-S, where

p = radius of the inscribed sphere

S = surface of the polyhedron

S

vs

IX

478

Unit 5

The sphere V=ARe

S sphere =47R*®

(Archimedes)

Derivation: As the number n of faces of polyhedra with the same inscribed sphere increases, their surfaces approach the surface of that sphere. Thus:

V...=lim 4R-S,=1R-S_, sphere

sphere

=4R - 4qR? =4nR®, ged.

(Can all of the approaching polyhedra be regular polyhedra?)

Parts of the sphere:

(see Figures on pages 464 and 465.)

The spherical sector VA

+p, ma]

Area and Volume

479

EXERCISES Group I: (compare with Unit 3, Section 6.2) 1. What are Platonic solids? How many are there? we know that there are no more?

How do

2. What is the radius of a sphere that is inscribed in (circumscribed about) a cube of side s? What is the volume of that sphere? 3. How can a cube be considered as the sum (in volume) of pyramids with a common vertex and equal height? Make a pictorial view of a cube, find the common vertex and show one of the pyramids. How many such pyramids are there in the cube?

4. Considering the cube with the edge s as a polyhedron, compute its volume by the formula for polyhedra inscribable by a sphere, and then check with the usual formula. 5. If a sphere can be inscribed in a polyhedron, and if the volume V and the surface S$ of the polyhedron is known, how can the radius of that sphere be computed? Check your answer for the cube. 6. Compute the surface and volume of a regular octahedron whose edge is 3 in. (What is the height of one of the two square pyramids that make up the octahedron?)

7. What is the radius of a sphere that can be circumscribed around a regular octahedron? 8. If you know the surface and volume of an octahedron from Exercise 6, can you compute from them the radius of the inscribed sphere? 9. The center of a tetrahedron lies on the altitude at ; of its height above the centroid of the base (see Unit 3, Section 6.2, Exercise 12). Compute the length of this altitude, the radius of the circumscribed and inscribed sphere for a tetrahedron, whose side is 3 in.

10. Using Exercise 9, compute the volume and the surface of a regular tetrahedron of edge s. Use two methods: the formula for a pyramid, and the formula for a regular polyhedron. 11. Compute the surface S of a sphere if its volume V = 23047 cu ft. 12. Compute the volume V of a sphere if its surface S = 12967 sq ft. 13. Compute the volume of a spherical sector if the height of the segment is 2 units and the radius of its base circle is 6 units.

480

Unit 5

14. Prove the theorem of Archimedes: A cone, a hemisphere, and a cylinder with equal base circle and equal height have a ratio in volume of 1:2:3.

Ex.14

Group II 1. Use Cavalieri’s principle to show that a hemisphere has the same volume as the circumscribed cylinder minus a cone which has the same base circle and the same height.

=, Exe!

2. The diameter of a given sphere is divided into two parts in the ratio 1:2 by a plane perpendicular to it. Compute the volume of the two pieces of the sphere separated by the plane. 3. Four planes perpendicular to the diameter of a sphere divide this diameter into five equal parts. Compute the volumes of the parts of the sphere formed.

4. Two equal spheres of radius R intersect; the distance of their centers is d. Compute the volume of the object common to both (two spherical segments), and the volume of each piece not common to both. (R= 6 and d=3) 5. Show that the volume of a spherical zone equals the arithmetic mean between the two cylinders that can be inscribed in and circumscribed

about

the

zone,

creased by the volume inscribed sphere.

in-

of the

6. The space between two concentric spheres is called a spherical

Area and Volume

481

shell; the difference between the two radii is the width of the shell. Compute the surface and the volume of the spheri-

cal shell of radii R, and R,. (R,=9 and R, = 4) 7. If the radius of a sphere is divided into four equal parts, compute the ratio in volume between the three spherical shells formed. 8. What is the width of a spherical shell whose volume equals the hollow space inside?

|

*4.4 Solids of Revolution

The volume of a cylinder equals the product of its base B times its height H. Suppose, a straight circular cylinder can be bent into a circular tube, a solid of revolution. Then its height H is stretched more at the outside than at the inside. It should be possible to find a mean value H such that the volume of the tube could still be computed by the simple cylinder formula B - H. The mean value H would represent the circumference of a circle passing through some mean point in the plane of B; it could be considered as the path of this mean point, while the solid of revolution is generated by a rotation of B around some appropriate axis. The mean point is called the centroid or the center of gravity of the area B (see FN in Section 3.3, page 467). The reasoning here is similar to that of Section

3.3, and it leads to a similar

theorem:

Theorem V-4: Pappus’ Rule for Volumes If a plane figure is rotated around an axis in its plane such that the axis does not pass through the interior of the figure, then the volume of the solid of revolution generated equals the product of the area of the figure and the path of its centroid.

Let us be consistent and leave the general proof of this second Rule of Pappus also to the calculus course. We shall check it for simple cases, however, before we use it.

482

Unit 5

Example: An equilateral triangle is rotated around one of its sides. The centroid lies on a

median in 3 of its length from the base. = ANG

Then

S

q

Ute

a | ¢

p= 3° ave

the area of an equilatrmja

eral triangle is 4a?\/3 (is it?); then y=70V3 - Qa 7

a=

47a’.

Check by using the specific volume formulas for two cones with

a common

base circle:

V=2-0(5 a3) - fees 3 sre The centroid for this example and the corresponding example for the first Rule of Pappus is the same because of the symmetry of the figure. The perimeter and the area of a figure do not necessarily have the same centroid (see exercises). Each center does lie on a symmetry axis —if there is one —of the figure.

EXERCISES 1. What is the volume of a doughnut (torus) formed by rotating a circle of radius r around an axis s at a distance d? What is the amount of raspberry jelly that fills the doughnut to half the radius of the cross section? 2. A square of side a rotates around an axis (a) parallel to a side ata

distance d, (b) passing through a

Ex. |

vertex and perpendicular to the diagonal through that vertex. Compute the volume of the solid of revolution formed, using two methods: Pappus’ Rule and specific volume formulas.

Area and Volume

483

3. A semicircle rotates around a diameter. At what distance from the center O on the symmetry axis of the semicircle does the centroid of each of the following lie: a. of the area of the semicircle? b. of the circumference of the semicircle?

(Note:

These are not the same.) Ss

O

a

a

a XS

Ex.4

4. An equilateral triangle rotates around an axis in its plane, passing through a vertex and perpendicular to a side. Compute the volume of the solid of revolution. 5. Find the centroid of the perimeter of half a regular hexagon: a. on one side of an angle bisector. b. on one side of a perpendicular bisector of a side.

s Ex.5

s Ex.6

6. Repeat Exercise 5 for the centroid of the area of half a regular hexagon.

SUMMARY

OF UNIT V

A. Formulae

Notation: p, P=perimeter or circumference; A= area; =area of base; LS =lateral surface; S$ =total surface;

B V

484

Unit 5 =volume; d=face diagonal; D=diagonal of the object; r, p =radius of circle, inscribed circle, small circle on sphere; R=radius of sphere, circumscribed circle, great circle on sphere; h=height, slant height; H—=height of the object; a,b,c,..., s=edges; s=half the perimeter of a triangle, side of the cone, p, g = segments on the hypotenuse. The square: P=4s; A=S?; d=svV2 The rectangle: P=2(a+b); A=a:b; d=Va°+b? The parallelogram: P=2(a+b); A=a-h

The rhombus, the deltoid: The trapezoid: The triangle:

A= 3+ d, - d,

P=a+b+c+d;

m=3(at+b);

A= ;(a+b)-h=m-h P=a+b+c=2s;

A=so NW A=. s: A = Oe, A=Vs(s —a\(s — bs —c); A= PAS);

p =1V's(6 =

4S

Se

abc

age

4V's(s — ays — bys —c)

r =F _Vis =a = DS): h. ==ViG =o

The equilateral triangle:

=/aN(s =e)

h=4aV3; A=+a?V3

The right triangle: .c*=a?4_b*’ =p p

Group III

Miscellaneous two-dimensional problems. 1. Prove: If a median of a triangle equals one half the side it bisects, then the triangle is a right triangle. . Prove:

A trapezoid inscribed in a circle is isosceles.

. Prove: A circle can be circumscribed around a given isosceles trapezoid. How can the center of that circle be found by construction? . What is the geometric locus of the vertices C of all triangles

with base AB and a given angleatC? be used to construct AABC,

How can this locus

if AB, x.C, and h, are given?

. If the diameter d of a circle is extended by the length x and if from the new point a tangent ¢ is drawn to the circle, what relation holds between d, x, and t? How can one of the three lengths be computed, if the other two are known? . The

ratio between

the areas of two circles is 4:9.

If the

circumference of the smaller one is 24 in., how long is the circumference of the other?

492

Unit 5

Ue From a point on a given circle two chords are drawn in different directions, cutting off 7 and 2 respectively of the What is the angle between these whole circumference. chords? . Prove: The angle between two intersecting tangents of a circle is twice the angle between the chord connecting the two points of tangency and a diameter through one of the points of tangency. . Prove: If one of the equal sides of an isosceles triangle is extended through the vertex by its own length, then the line joining the new endpoint to the nearer end of the base is perpendicular to the base. (Find two different proofs: one comparing segments, and one comparing angles.) 10. What conditions must be satisfied if a quadrilateral can be circumscribed about (inscribed in) a circle? 11. Define each symmetries. 12. Compute compute: a. b. c. d.

the the the the

conic as a geometric

locus

and describe

the area of a triangle with sides 5, 6, 9.

its

Then

three altitudes radius of the inscribed circle radius of the circumscribed circle ratio between the areas of these two circles

Group IV Miscellaneous three-dimensional problems. t. Compute the volume of a sphere whose surface is 1967. 2. A sphere of diameter 30 in. is intersected by a plane at a distance of 12 in. from the center. What is the diameter of the small circle of intersection? . What is the relation between the radius r, the height H, and

the side s of a right circular cone? How can each of these three measurements be computed from the other two?

. Compute the volume and the total surface of an equilateral cone.

. Compute the volume and the total surface of an equilateral cylinder. . A sphere with a 26 in. diameter is intersected by a plane at a distance of 10 in. from the center.

tween the areas of the two caps?

What is the ratio be-

Area and Volume

7. Find the surface and the volume with a 4 in. edge.

8. The volume surface.

493

of a regular octahedron

of a hemisphere is 1447.

Compute

the total

9. From a circular ring with radii 4 and 12 in. a quadrant is cut out and the frustum of a cone is formed from the remainder. Compute the radii of the bases, the altitude, the total surface, and the volume of the frustum.

10. The radii of two circles are in the ratio 3:5. If each circle is the base of a 6 in. high cone, what is the ratio between the ~ volumes of these cones?

11. Given a plane © and a line s which is neither parallel to > nor contained in it. What is the complete geometric locus of all - points which are at a distance of 5 in. from both > and s? 12. Repeat and discuss Exercise 11 for a line s which: a. lies in >

b. is parallel to >

Group V Using trigonometric functions. 1. What angle does the diagonal of a cube form with the base?

2. A regular square pyramid is 8 in. high. The angle of inclination of each lateral edge with the plane of the base is 53°. How long is a lateral edge and how long is a base edge? 3. The base edge of a regular hexagonal pyramid is 5 in. How high is the pyramid, if the angle of inclination of a lateral edge to the base is 65°? 4. What angle does a lateral edge of a regular tetrahedron form with the base? (Which axial section? Which function makes use of the simplest measurements?)

5. What is the angle between two faces of a regular tetrahedron? (Which axial section?) 6. If a point P is 4 in. outside a sphere of radius 5 in., what is the angle between a tangent from P to the sphere and the secant through P and the center of the sphere? 7. How high is a tower, if its top is seen under an angle of elevation of 57° from the ground 80 ft away from the base of the tower? 8. In AABC x A=30° and xB=120°. its three sides?

What

is the ratio of

494

Unit 5

9. On a globe with a 12 in. radius, how long is the circumference of the parallel at 50° latitude? 10. Use Table I to find the values of cos 200°, sin 310°, tan 100°. (Use the unit circle.) 11. Prove that sin (90° + a)=cosa. 12. What is the period of the tangent function?

Group VI Using coordinates.

1: What is the equation of a circle whose center is at the origin and whose radius is 4? . Prove analytically that every straight line through the origin intersects the circle in Exercise 1 in two points. What are the coordinates of these two points for the particular straight line y = 3x?

. What are the equations of the tangents to the circle in Exercise 1 which are parallel to the x-axis? parallel to the yaxis? . What is the set of points described by x7+ y?>9? x2 +

ys 12?

. Does a quadratic equation always represent a circle? . Is the line 4x + 3y =25 a tangent to the circle x? + y? = 25?

. What is the slope of a line tangent to the circle x2 + y?= 13 at the point 7(2,3)? What is the equation of this tangent? . What is the set of points (x,y) satisfying both conditions xSP y~ =< 4randay al? . What is the set of points (x,y) represented by 9 S x2 +y2 < 16? 10. What is the fastest way to decide whether or not the following two circles are tangent:

a

3) y= x?+ (y— 4)? = 16

b.

@— 1)? + 4+3)-=4 (x+ 27? ++ 3)=4

11. 9x + 25y? =225 is the equation of an ellipse. Investigate whether the coordinate axes intersect this ellipse; and if so, then determine the coordinates of these points. 12. Repeat Exercise 11 for the hyperbola also for the hyperbola 25y? — 9x? =225.

9x? — 25y?= 225;

13. Repeat Exercise 11 for the parabola y?=6x; parabola y? = 6x + 4.

also for the

Area and Volume

14. What radius 1s: What y-axis

495

is the equation of a circle with center C(3,4) and 6? is the equation of a circle with center C(—2,5) if the is a tangent?

16. Which of the following quadratic equations represent circles? What feature characterizes such an equation? » y=4x—-—5 ~ xX +3x-4y =

f. 9x? + 4y? = 36 |

g. 9x? — 4y? = 36

ot? y+ 3x=4 » x2 +4x + y? —3x =9 cRoeD x? +4x —y? —3x =9

h. x?—y?=4 be = 2) (y= 3)? =F jati=—2)?

P33)

=7

PAT Determine the center and the radius of these circles:

ee

ee

vee

OV

=

OD. yoy

x3

*18. Determine

the ordinates of four points A(—S,y), B(-3,y), C(0,y), D(4,y) lying on the circle x?+ y2=25. Formulate

and verify Ptolemy’s theorem for these four points if: (a) none of the ordinates is negative. (b) y, and y,, are negative. How long are the diagonals?

*19, Given A(3,4) and B(7,8); compute the distance AB. With the origin as the center of similitude and 1)3 as the ratio of similitude map A, B into A'(x’,y’) and B’(x',y’).

coordinates of A’ and B’?

What are the

Verify that A’B’ is parallel to AB

and three times as long.

*20. Given an isosceles right triangle A(0,0), B(0,3), C(3,0). Verify that a parallel shift along the vector V (V,, = 5, V,, = 4) maps AABC into a congruent AA’B'C". e271. What transformation formulas represent a reflection with respect to the x-axis (would carry a point P(x,y) into P'(x’,y’) so that P and P’ are reflections of each other with respect to the x-axis)? *22. What transformation formulas represent a rotation around the origin through an angle of 180° (would carry a point P(x,y) into P’(x',y’) by this rotation)?

aZs: Repeat Exercise 22 for 90°.

| | GEOMETR’ ON

Fate bapeigeh oS OF THE SPHERE

6

|

1

THE SPHERICAL

DISTANCE

If you were laying out a big triangle on a ball field, you would expect it to have the same properties as a triangle in Euclidean geometry. You would expect it to have straight lines as sides and an angle sum of 180°. Yet, you know that the earth is not a plane, but is approximately a sphere. The size of the earth (the sphere) in comparison to the size of the ball field is so great that its curvature is not noticeable within the ball field. If you were standing on the shore

of an

ocean,

looking

out at two ships approaching the horizon,

you would

yyy

be

— oe

able to see their upper structure long before you would see their hulls. This observation made the ancient

Y YY U1

Greeks, a seafaring people, aware of the fact that the earth is curved

and not a plane. The lines of sight to the ships and the segment between the ships form a large triangle, whose sides are not exactly on the earth’s surface. We may also speak of a triangle on the earth’s surface between the observer and the water lines of the two ships, even though the sides of this triangle are curved lines. For such a triangle it makes just as much sense to talk about distances and angles as it does for a triangle whose sides are not curved. Looking at a globe, it makes sense to say that Philadelphia, San Francisco, and the north pole form a triangle. For a traveler or a surveyor, the sides of this triangle lie on the surface of the earth and are actually parts of circles. 496

For a scientist, the straight

Geometry on the Surface of the Sphere

497

line segments through the earth might be of interest. These segments are obviously shorter than the corresponding distances on the surface of the earth. How much shorter is the chord distance from Philadelphia at 40° N to the north pole at 90° N? (Compute the difference in length between the arc and the chord subtended by a central angle of 90° — 40° = 50° for a circle of radius 6371 km.) How much shorter is it from the north pole to the south pole through the earth than on the earth? Three points on a sphere determine a triangle on the surface

of that sphere.

This triangle is called a spherical triangle to

distinguish it from the plane triangle formed by the chords through the inside of the sphere. In this Unit we shall study a few properties of spherical triangles. Geometric properties of figures on the surface of the sphere form the subject of spherical geometry. These properties are the same for small and for large spheres; the only difference is the scale. Since spheres of different size are similar and since properties of shape are invariant under similarity, we may study spherical geometry on any sphere, taking its radius R simply as the unit of length.

Theorem VI-1: The shortest

distance between sphere lies on a great circle.

two points on the

There is a branch of mathematics, called Calculus of Variations,

which deals with problems such as finding a shortest line between two given points among various neighboring lines meeting a certain condition, for instance the condition of lying on a sphere. We will leave the general proof of Theorem VI-1 for this college course. Here we will be content with verifying it on the globe. Mark two arbitrary points on the globe and hold a string taut connecting them; the string will align itself along a great circle arc. Every other path will need more string. Theorem VI-1 is used by airlines to find the shortest flight distances between points on the earth. Although San Francisco (38° N) and Moscow (56°N) are both below 60° N, the shortest flight between these points is made across the north pole region.

498

Unit 6

Similarly, the airline route between New York (41° N) and Rome

(42°N) follows a great circle path over Gander, Newfoundland (48° N), even though both places are almost on the same parallel. *Example: You may wonder that a flight curving north should Let us compare the really be shorter than along a parallel. two distances for a simple case. How far is it from a point A (60°N 0° W) west of Norway

to a point B (60° N 90° W) in the

Hudson Bay, Canada, computed along the 60th parallel (a small circle) and computed along a great circle?

The chord distance AB in Figure Ia is on a chord which is common to both circles (the small circle in 60° latitude with center M and radius r, and the great circle with center O and radius R

of the earth).

We will compute the length of the arc AB, with

center at M, and the arc AB, with center at O.

The length of an arc is computed by the formula....

Because

of the special choice of longitudes for A (0°) and B (90° W) the

central angle in the small circle of AB,, is 90° (the difference in longitudes). The subtended arc, therefore, is } of the perimeter 2mr, or 37r. How big is r in terms of the radius R of the earth? This three-dimensional problem can be reduced to a two-dimensional one by an axial section through the earth (Figure Ic). Here r appears as half a chord in a circle of radius R. Because of the special choice of 60° latitude, r is half the side of an equilateral

triangle. Therefore r= $R. From the fact that AB e711 follows that AB, =47R. If you wish to know this length in kilometers, you may substitute R = 6371 km and 7 = 3.14159 and multiply out, which yields 5004 km.

Geometry on the Surface of the Sphere

499

To compare AB, with AB, the two circles with the common chord AB are drawn in Figure Id as if they were in the same plane. Since R is larger than r (in this example R = 2r) the center O is farther away from AB than the center M. After drawing both circles, we can see that the arc Mier to the great circle is nearer to the chord.

Therefore, AB, < AB,,._

Graphically, the

comparison is complete; mide niatically) we il need a little more perseverance. To compute ABo, we need to find the subtending central angle 2a at O. Note that 2a is not the longitude difference of 90° between A and B, because longitude angles are measured in the plane of the equator or of a parallel. The angle a (Figure Id) is determined by a right triangle with hypotenuse R and opposite leg 4AB. The length of AB can be computed from the right triangle AMB

as r\V/2.

Thus,

In the right triangle containing @ we compute:

4$AB_

sin @ =

4RV2

ei

005755

In the Table for the sine function this value is between sin 20° and sin 21°. A more precise value for a can be obtained from a larger Table:

a@ is between 20° 42' 17” and 20°42’ 18”, which is

certainly smaller than 20.705°.

From the proportion 2a@:360°

= AB): 277, we find:

~~ 20.705° = 7R(0.23006) AB, < 7R 90°

Expressed in kilometers, AB, is a little less than 4605 km. AB, is shorter than ABy by about 400 km.

Definition VI-1: The spherical distance between two points A and B on a sphere is the minor arc of the great circle through A and B. a

Thus,

500

Unit 6

Remember that the plane of every great circle passes through the center O of the sphere (why?), and that three non-collinear points A, B, O determine exactly one plane.

Thus, the two points A, B

on the sphere determine exactly one great circle. On this circle there are two arcs between A and B, one smaller and one larger than 180° (or zr), unless A and B happen to be the endpoints of the same diameter. To avoid an ambiguity in the meaning of the distance between A and B, we agree on Definition VI-1. In the special case of antipodes (endpoints of the same diameter), for example the north and the south pole, the three points N, O, S are collinear and there are infinitely many planes through NOS (an axial pencil of planes). Each of these planes contains a great circle, and the spherical distance NS is a semicircle. Although there are many arcs between WN and S, the measure of the spherical distance is unique, namely

180°

In other words, all meridians

are equally long.

According to the definition of the spherical distance there is a maximum length that it can have: half a great circle. In contrast, there is no maximum length for distances in plane and solid geom-

etry. Straight lines are considered “infinitely” long. In spherical geometry the great circle arcs play the role of “pseudo-straight lines” because they contain the shortest distances between points; but their length is limited. > A spherical distance is at most 180°.

Since the spherical distance is defined as the arc in a circle, it is measured as such an arc is measured: > The spherical distance between two points A and B is measured within the great circle through AB in angular measure as the central angle AOB, in nneas measure as a fraction of the circumference of that circle 3=—; ra 27R,or a: Rif ais expressed in radians.

The antipodes A’, B’ of a pair of points A, B lie on the circle as A, B (why?) and are centrally symmetric to respect to the center O. The arc A’B' is theiimage of under central symmetry (or rotation through 180°). know that:

same great A, B with the arc AB Thus, we

> The spherical distance between two points is equal to that between their antipodes.

Geometry on the Surface of the Sphere

501

EXERCISES 1. Compute the spherical distance AB on the unit sphere (R = 1) in angular and in radian measure:

a. A(Q0° N, 0°), B(90° 'S, 0°) b. A(50° N, 15° E), B(40°S, 15° E) c. A(south pole), B(0°, 40° E) d. A(north pole), B(0°, 0°)

e. A(0’, 10°W), B(O°, 20° EB) 2. Repeat Exercise 1 for a sphere of radius 12 in, 3. Repeat Exercise 1 for the earth (use R = 6371 km). 4. What are the latitude and longitude of the antipodes of these points: A(0°, 0°), B(YQO°N, 0°), CU10°N, 0°), D(O°, 90° E), EGON, 20°E), FG0"N, 20° WW). 5. Explain and derive the formula:

a is measured in radians.

length of arc=a-R,

where

What is a radian?

6. On a globe of 20 in. radius how long in inches is the distance between the poles? from New York (40° N) to the north pole? from New

Orleans (30° N) to the south pole?

7. On a globe of 5 in. radius a distance of 3 in. along a meridian is how many degrees in latitude? A distance of 2 in. along the equator is how many degrees in longitude? 8. How big must a globe be to show 5° in latitude as one inch? 9. Prove that the spherical distance between two points is equal to that between their antipodes.

|

2

POLE AND

POLAR

It is easier to talk about specific points on a sphere, if we use the geographic language of latitude and longitude. (Review Preparatory Chapter, Section 6.) This gives us a coordinate system on the surface of the sphere, a little different from the x,y-system in the plane and the x,y,z-system in space, but just as useful. Any great circle may be interpreted as an equator, determining two poles as the endpoints of the diameter perpendicular to the plane of the equator. Any point on the sphere may be interpreted as a pole, determining its antipode as the second pole and the symmetry plane between them as the equator plane. In the same problem it might be confusing to talk about several different equators; in such problems it is customary to talk of a pole and its polar (equator). For each pole (any point) there is

502

Unit 6

exactly one polar (how do you find it?); for each polar (any great circle) there are two poles (where?). We can distinguish between these two poles by calling one positive and the other negative according to the following agreement:

Definition VI-2:

fan

The spherical distance AB on a polar in the direction from A to B is seen in the positive (counter-clockwise) direction from its positive pole, and in the negative (clockwise) direction from its negative pole.

In less technical, but more

graphical lan-

gauge this means: if you “walk” from A to B, the positive pole is to your left and the negative pole is to your right.

Considering this agreement about positive and negative directions, every great circle may be counted as two circles with different directions attached to them. The circle with the positive direction is associated with the positive pole, and the same circle in the opposite direction is associated with the other pole. Thus a one-to-one correspondence is established between poles and polars. This is an example of a geometric correspondence not between pairs of points, but between points and great circles with an attached direction. The spherical distance between a pole and any point of its polar is called a quadrant, Can you think of a reason for this name? The polar of a point P is the geometric locus of all points which have a spherical distance of 90° from P.

Theorem VI-2: If the polar a of a point A passes through a point B, then the polar b of B passes through point A.

Hyp.: a is the polar of A, b is the polar of B, B is a point of a Con.: A is a point of b

v4

Geometry on the Surface of the Sphere

1.

2. .

es)

4.

PROOF STATEMENTS a is the geometric locus of all 1. points which have a spherical distance of 90° from_A spherical distance AB = 90° 2. bis geometric locus of all points 3. which have a spherical distance of 90° from B A is a point of b, q.e.d. 4.

503

REASONS hyp. (a is polar of A)

hyp. (B lies on a) and 1. hyp. (b is polar of B)

from 3. and 2.

Corollary VI-2-1: If the polars a and b of two points A and B intersect in a point C, then the polar c of C must pass through A and B.

Hyp.: a is polar of A, b is polar of B, c is polar of C, C lies on a and b Con.: A and B lie onc PROOF STATEMENTS REASONS 1. C lies on the polar ofA 1. hyp. 2. A lies on c, the polar of C

2. from VI-2 and 1.

3. C lies on the polar of B 4. B lies on c, the polar of C 5. A and B lie onc, q.e.d.

3. hyp. 4. from VI-2 and 3. 5. from 2. and 4.

Definition VI-3: Every small circle parallel to a given great circle is said to have the same poles as the great circle. The spherical distance of a point on a small circle from its nearer pole is called the polar distance.

Theorem VI-3: All points of a small circle on a sphere have the same polar distance.

504

Unit 6

Why? What is the polar distance of points on the 60° N parallel? On the 50° S parallel?

Definition VI-4: The polar distance of the points of a small circle on a sphere is called the spherical radius of that small circle.

What is the definition of a plane circle? Is it correct to say: A circle on a sphere is the geometric locus of all points that have the same spherical distance from a given point on the sphere? Here we see again that great circle arcs (the carriers of spherical radii) play the same role in spherical geometry as straight lines in plane geometry.

The larger the spherical radius becomes, the larger is the circle. The maximum obviously is a great circle with spherical radius 90°. In the technical language of limits we might say: lim small © = great spherical r > 90°

©, or “the small circle changes to a great circle when the spherical radius becomes 90°.” Do we have an analogous case in plane geometry? When the plane radius becomes longer and longer, the circle becomes larger and larger, its curvature becomes less and less, its difference from

a straight line within a limited region becomes less and less noticeable. We say: “A circle approaches a straight line when r becomes infinite,” or lim © = straight line. r>

0

Does this remind you of the triangle on the ball field discussed at the beginning of this Unit? Our daily experience is that of plane geometry, because we operate only within a limited region compared to the size of the earth. Within this limited region the difference between a plane and the curved surface of the earth is too small to be noticeable.

Cg

Geometry on the Surface of the Sphere

505

EXERCISES 1. Is the north pole the positive or negative pole of the equator, if longitudes are counted positive in the east direction? 2. Which is the positive pole of the following meridians considered as polars, if the direction towards north is the positive direction? 0°, 20°E, 90° E, 50°W, 10° W, 180°. 3. What is the spherical distance between a pole and a point of its polar? 4. Explain the meaning of the following statement: There is a one-to-one correspondence between poles and polars. 5. What is a spherical radius? What is the spherical radius of the 65th parallel north? the 10th parallel south? the equator? 6. How do you find the polar to a given point? Where is the po-

lar of A(0°,0°)?

1

B(0°,10° E)?

3

SPHERICAL

ANGLES

Definitions VI-5: A spherical angle is formed by two great semicircles. Either of the two points of intersection may be considered as the vertex; the quadrants starting at the vertex are the sides of the spherical angle. The size of a spherical angle is defined as the size of the angle between the planes of the two great semicircles.

Since any point may be considered a pole we may use geographic language and call the vertex the north pole N. Then the sides of the spherical angle @ are the meridians (semicircles) through A and

B on the equator. The angle between the two meridional planes (see Unit 1, Section 9) appears in the equatorial plane as xXAOB. It also appears in the tangent plane at N as the angle between the tangents t, and f, of the great circles,

506

Unit 6

since these tangents are parallel to OA and OB respectively (sides of the circumscribed square). > A spherical angle ANB is measured as the plane §angle formed by the tangents ¢, andr, to the ue NA and NB at the vertex N;

or by the central angle AOB subtending the arc AB

on the polar of the vertex;

or, in radians, by the arc AB.

What is the angle between the equator and any one meridian?

Definition VI-6: A lune is formed by a pair of meridians. It contains two equal spherical angles, one at each point of intersection.

The equator forms alternate interior right angles with a pair of meridians. Jn plane geometry this would make the meridional lines parallel; but in spherical geometry these lines are not parallel, they intersect twice, forming a lune.

Zoe

2

Theorem VI-4:

A spherical angle is equal or supplementary to the spherical distance between the poles of its sides.

Proof: In the adjoining figure the spherical angle at N is measured by AB or by the side NA

XP,OA=90°

«=x AOB.

The poles of

are P, and Q ao where

and

x0,0A =90°.

The poles of the side NB are P, and Q,, where X PgOB = 90° and XQ,0B = 90°. The spherical distance between ,

the poles of NA

Taw

N fe

as

and NB

ames

=

is

either P,P, = 2,0, or P.O, = EON:

ee Oar = pP___ Ps

Oe iene A

Geometry on the Surface of the Sphere

507

XAOB = X P,OP, (by rotation through 90° around O) X AOB = 180° — x P,OQ,, (the supplementary angle of MaiOR-). You may find it easier to visualize Theorem VI-4 by considering two planes II, and II, with angle a between them, and perpendiculars p, and p, at the same point rigidly attached to them. If II, is rotated into I], through x a, then the rigidly attached p, must turn into p, rotating through the same angle a. *It has been established that the north pole of the earth varies slightly with respect to the crust of the earth. The movements of the pole are determined at observatories around the earth from latitude observations (principle of indirect measurement). How can latitude variations tell us about north pole variations? To remove the ambiguity in Theorem VI-4, we must distinguish between the positive and the negative poles. If we “walk” around the angle a from B to N to A (counterclockwise), then the

positive pole of BN is P, (to your left!) and the positive pole of NA is Q, (again to your left!). Corollary VI-4-1: A spherical angle is supplementary to the spherical distance between the positive poles of its sides.

EXERCISES 1. What is the spherical angle between the following pairs of meridians, and where is it measured? (18°F and 25°F), (18° E and 25°W), (160°E and 160° W) 2. What is the spherical angle between the zero meridian and the equator? Where is it measured? Between any other meridian and the equator? 3. How big are the three spherical angles in the triangle formed by the south pole andA (0°,0°) and B(0°,90° E)?

508

Unit 6

4. What is a lune? What are the angles of the lune formed by the 20th and 30th meridians E? 5. How long in inches is the perimeter of a lune on a sphere of 5 in. radius? 10 in. radius? 6. Show that two great circles form four lunes, and explain their relative sizes. 7. What is the spherical distance between the poles of the sides of a 50° lune? *8. In 1891 the observatories in Berlin and Honolulu, which are almost exactly 180° apart in longitude, determined their latitudes in a series of simultaneous observations. Plot these latitudes on separate graphs for Berlin and Honolulu and compare. (Note that the latitudes vary only in the decimals of

the seconds.)

What do the graphs tell about the north pole?

1891

_| May 31 |June 10 | June 30 | July 20 | Aug. 9 | Berlingsms 0 Honolulu 21° 16’

ea Se lies ON 2AODeed 5A OOM

A) (7".49 2.4229OMe ees

l7 56a) 24”.75 |

|Aug. 29 | Sep. 18 | Oct. 8 | Oct. 28 | Nov. lak Dece/al Dec ce Olen | WA OAR S264" | 6 Sie ile ON alee ee |24".70 | 24".67 | 24”.67 | 24”.68 | 24”.71 24".78 | 24.86

Feb. 5 | Feb. 25 |Mar. 16

:

9. What is wrong with the adjoining pictorial view of an equator and two meridians? How should the meridian

Apr. 25 | May 15

Ene, Se ey

M, be corrected?

1.

4

THE SPHERICAL

TRIANGLE

A spherical triangle ABC is formed by on a sphere, not all on the same great circle, distances between any two of them (Figure Three great circles actually divide the

three points (vertices) and the three spherical Ia). surface of the sphere

Geometry on the Surface of the Sphere

(a)

(b)

509

(c)

into eight spherical triangles, since they intersect also in a second triple of points A’, B’, C’, the antipodes of the first triple. Can you see the eight triangles in Figure Ib? In the spherical case where the planes of the three great circles are mutually perpendicular, they form the familiar Cartesian coordinate planes; each of the eight triangles is in one octant (Figure Ic).

The spherical triangle ABC is identified unambiguously by these two properties: & Every side of a spherical triangle is smaller than 180°.

Why”? (Definition of spherical distance. What would happen to the triangle if one side were increased to 180°?) b> Every

angle of a spherical

triangle

is smaller

than

180°.

Why?

Theorem VI-5: Two diametrically opposite (centrally symmetric) spherical triangles are inversely congruent.

Consider the spherical triangles ABC and A’B'C’ in Figure Ib. Corresponding angles are equal because they are vertical angles of the same pair of planes. In the first triangle the vertices A, B, C follow in counterclockwise (positive) order; A’B’'C’ (when viewed

also from the outside of the sphere) follow clockwise (negatively). Each pair of corresponding sides by itself is directly congruent,

510

Unit 6

mapping into each other through a rotation, yet each of the three pairs around another axis. There is no single rotation carrying the whole triangle ABC into A’B'C’.

Theorem VI-6: Any side of a spherical triangle is smaller than the sum and larger than the difference of the other two sides.

This theorem reminds plane triangle.

us of a corresponding

(See Unit

1, Section

10.)

theorem for a

To prove Theorem

VI-6, we note that the spherical sides a, b, c are measured by the

central angles a, B, y in their respective planes (Figure Ila).

(b)

Thus, Theorem

VI-6 actually says something

OABC (Figure Ifa), a so-called trihedral angle.

about the object (A dihedral angle

is formed by two planes, a trihedral angle by three planes. The intersections of any two of these three planes are the edges of the trihedral angle, considered as rays from the vertex O. The angles

a, B, y are the face angles.

The sectors with the face angles as

central angles are the faces.) The net of this trihedral angle consists of the three face angles laid out in one plane as, for example, in the plane of y (Figure IIb). To form th the trihedral angle from this net the plane of B is rotated around OA, and the plane of a around OB until the outer edges OC coincide.

If 8 +a

=y, this ray oc would fall into the plane of y;

-

Geometry on the Surface of the Sphere

511

if B+a y can atrihedral angle be formed. Thus, the first part of Theorem VI-6 is correct, since any other of the three face angles

could be called y and the reasoning could be repeated. The second part of the theorem follows from the inequalities, as in Unit 1. > A useful principle: Translate a statement about a spherical triangle into a statement about a trihedral angle.

Theorem VI-7: (Menelaus, Ist cent. A.D.) The sum of the three sides of a spherical triangle is smaller than 360°.

Prove by interpreting this theorem as one about a trihedral angle. Use the net discussed above: can a trihedral angle be formed

from a net with

a+ B+ y = 360°?

Definition VI-7: A pair of trihedral angles are called supplementary or polar, if the edges of one are perpendicular to the faces of the other.

Theorem VI-8: In a pair of supplementary (polar) trihedral angles the face angles of one are supplementary to the angles between the faces of the other.

Hyp:

(0,a,b,c) is polar to (O'F,F2F 3); which means OLFE all

plane (a,b), O’F, 1 plane (a,c), O'F; 1 plane (b,c)

512 Con.:

Unit 6 ata’ = 180°, Bt+p' = 180°,

y + y’ = 180°. * Proof:

1. plane

> = O'F ,F,_ inter-

sects a at A

2. > 1 (a,b)

because

O'F;

1 (a,b) (which theorem?)

5. > 1 (aje_because

OF, Ase) . > | intersection OA of (a,b) and (a,c) (which theorem?) x a in > is the angle between the faces in quadrilateral AF,O’F, the sum of the angles is 360°

AF, = XF, = 90° (hyp.)

. ata’ =180°, g.e.d. . repeat for the other planes NAH ON Note: If a parallel shift is applied to (O’F,F,F;) so that O’ is mapped into O, then x a’ is mapped into an equal angle. Thus Theorem VI-8 applies also to polar trihedral angles with a common vertex.

Theorem VI-9: (Menelaus, Ist cent. A.D.) The sum of the angles in a spherical triangle is larger than 180° (> 7) and smaller than 540° (< 37)

This theorem spells out the amazing difference between plane and spherical triangles: the angle sum is not 180° and it is not constant. A triangle formed by the equator and two meridians 10° apart has two right angles and the sum of the angles is 190°. If the meridians are 90° apart, we have three right angles in one triangle, and the sum of the angles is 270°. Theorem VJ-9 gives the upper and lower limits for possible values of the sum of angles in a triangle. The upper limit of 540° follows simply from the fact that each spherical angle in a triangle is smaller than 180°; thus all three together must be smaller than 3 x 180°. But how do we know the lower limit?

ea

Geometry on the Surface of the Sphere

513

The proof of this part of Theorem VI-9 follows from Theorem VI-7, if we translate both into statements about trihedral angles and use Theorem VI-8. The spherical angles a, B, y of a

spherical triangle ABC are measured in the planes of the tangents at A, B, C; each

of these tangent planes is perpendicular to a radius of the sphere which is an edge of the trihedral angle OABC. The three tangent planes meet at O', forming the faces of a

polar trihedral angle O’ABC. From VI-8: a+B+y=(7-a’)+(a7-

B')+(7-y’)

= 37 —-(a’+ Bp t+y’')

from VE7e it follows:

a +B’ +y < 27 a+ B+ y > 3m — 27 (since we took off a larger amount)

Oe Baty71,) q.€.d.

Corollary VI-9-1: A spherical triangle may have one, two, or three right angles; it may have one, two, or three obtuse angles.

Point out such triangles on the globe.

Definition VI-8: The amount by which the sum of the angles in a spherical triangle surpasses 180° is called the spherical excess, €.

514

Unit 6

EXERCISES 1. Find the vertices of the triangle which is centrally symmetric to the spherical triangle ABC: a. b. cy d. A | north pole | 10° N; 60°E | 80°S, 30° W | 10°S, 30°E BUOsO 20°N, 0° 10°S 10°ER 405N eee GC \0°, 30° Fx} 30° N, 20° Bl 1028, 10" Wid DOS 0 ae

. Two sides of a spherical triangle are 100° and 60°; can the third side have any arbitrary length? If not, what are its

limits? . Could

the

sides

of a spherical

triangle

be (45°,90°,180°),

(150°,200°,20°), (70°, 175°, 130°), (60°, 140°,90°), (90°,90°,90°)? . Give three triples of possible lengths of sides in a spherical triangle. . Could the angles of a spherical triangle be (140°,190°,80°),

(50°,90°,40°)? . What is the spherical excess?

What are its limits?

Compute the spherical excess for the spherical triangle ABC: a. A(0°,0°), B(0°,90° EZ), C = north pole; b. A(0°,0°), B(0°,30° W), C =south pole.

Show that in a spherical triangle with two right angles the opposite sides must be 90°; and vice versa,...

. Show that all sides of a spherical triangle with three right angles must be 90°; and vice versa, all angles in a spherical triangle with three 90° sides must be 90°. 10. Change the wording of Theorems VI-7 and 9 to statements about trihedral angles. 11. If two face angles of a trihedral angle are 100° and 130°, can the third angle be 90°, 130°, 30°, 140°? Express your answer also as a statement about a spherical triangle. 12. In a spherical triangle with angles of 90°, 45°, 90° lying ona sphere of 12 in. radius, how long (in inches) is the side opposite the 45° angle?

13. Is it possible that the perimeter of a spherical triangle on a sphere of 12 in. radius is 247 in.? 167 in.? 267 in.? 207 in.? 14. What is the upper limit for the perimeter of a spherical triangle on a sphere with a radius of 5 in.? 14 in.? 24 in.? 15: If one side of a spherical triangle is 100°, between what limits could you choose the sum of the other two sides?

Geometry on the Surface of the Sphere

515

16. If two sides of a spherical triangle are 80° and 130°, between what limits could you choose the third side? 17. Repeat Exercise 16 for (100°,150°); (90°,60°), (30°,110°).

18. On a sphere of radius 12 inches, what is the perimeter measured

in inches

of a triangle

with

the sides

(90°,20°,90°),

(80°,30°, 100°)? 19. On a sphere with radius of 5 inches, what is the perimeter in inches of a triangle with three right angles? 20. Show that a lune is a bilaterally symmetric figure with respect to two spherical symmetry axes.

1

*5

POLAR TRIANGLES

Each vertex of a spherical triangle ABC has a polar. Each

(Where?)

side a, b, c has two poles (where?), one of which is con-

sidered the positive pole (which?) for a given direction on that side.

Definition VI-9: The positive direction around

a

spherical

triangle

ABC

.

is

counterclockwise. (If you “walk” around the triangle in the positive direction, the triangle is always to your left.)

a

x

A

B ——>

Definition VI-10: Given a spherical triangle ABC, the three positive poles of its sides in positive direction are the vertices of the polar triangle A'B'C’.

Theorem Vi-10: If A’B’'C’ is the polar triangle of ABC, then also ABC is the polar triangle of A’B'C’.

Why?

(Use Theorem VI-2 and its corollary.)

516

Unit 6

Theorem VI-11: In a pair of polar triangles each side in one triangle is supplementary to the angle at the corresponding pole in the other triangle.

Proof:

If a spherical triangle with sides a, b, c is the polar triangle

of A’B'C’,

and if A’ is the pole of a, then OA’ | to the plane

through O and a. Translate into statements about trihedral angles: in Oabc and OA'B'C’ each edge of one is | to a face of the other; therefore these are polar trihedral angles and Theorem VI-8 applies. Translating VJ-8 back into a statement about spherical triangles gives Theorem VI-11. For an alternative proof see Exercise 9. Theorem VI-11 puts some more restrictions on the possible values for angles in a spherical triangle: these angles must each not only be less than 180°, and their sum must be between 7 and 377, but their supplementary values must satisfy the conditions for sides, namely each must be smaller than the sum and larger than the difference of the other two. For example, the spherical angles 140°, 75°, 120° are each < 180°, the sum 335° is between 7 and 37r, but the sides of the polar triangle would be 40°, 105°, 60°, which is impossible: 105° is not smaller than (40° + 60°), and

40° is not larger than (105° — 60°). Show that this restriction means that the sum of any two angles minus the third is less than 7.

Theorem VI-12: The spherical distance between corresponding vertices A and A’ in a pair of polar triangles is less than 90°.

(Here A' is the angle opposite the polar a’ of A; andA is the angle opposite the polar a of A'.)

Geometry on the Surface of the Sphere

Proof: imeter

517

“Walking” along the perof ABC in the positive

direction, the positive pole A’ of a lies on the left side, and so does

the opposite vertex A. The spherical distance from any point of a to A’ is 90° by definition. The distance from a point on a to A depends on the triangle; it may be less or more than 90°. Since A and A’ are on the same side of a and since any side of a triangle is less than 180°, A cannot be a distance of 90° beyond A’.

Since the pole A’ of a side a is always 90° away from a, while the vertex A opposite a may be chosen at 90° or nearer or farther than 90°, we note that a vertex A’ of the polar triangle may lie on, outside, or inside of the original triangle. > Two polar triangles may intersect, coincide, or one may lie inside or outside of the other.

What happens in the case of a triangle with three right angles; where is its polar triangle?

EXERCISES 1. What are the vertices of the triangle polar to ABC? a.

b.

c.

A | north pole | south pole | north pole B02. Oe OPO a | O28 iy Corey’. 20° W 10", 40°F O°; 80° E 2. Demonstrate Exercise 1.

Theorems

VI-10

and

11 for the triangles in

3. Prove that the sum of the angles in a spherical triangle is larger than 180°.

4. Ona sphere of radius 12 in., the angles of a triangle are 70°, 90°, 120°; compute the perimeter of the polar triangle in inches. 5. To check whether three angles can possibly be the angles of a spherical triangle, what relations must be satisfied?

518

Unit 6

6. Can the following triples be angles in a spherical triangle?

a. (75°,120°,140°) b. (70°,100°, 150°)

c. (50°,80°, 130°) d. (50°,60°, 100°)

sphere of 5 in. radius is a triangle with angles 90°, 90°, Compute in inches the sides of the polar triangle. 8. On a sphere with a 12 in. radius lies a triangle with the sides 4 in., 27 in., 537 in. What are the angles of the polar triangle? \=). Prove Theorem VI-11 by using Corollary VI-4-1.

7. Ona 20°.

|

6

THE AREA OF THE SPHERICAL

TRIANGLE

To compute the area of a spherical triangle or of any other spherical region, we have to determine two things: (1) What part of the total surface § of the sphere is this region? (2) How big is it on a sphere with a given radius R? For example, the area of a hemisphere is one half of S, where S =477R2 Thus, the hemi-

sphere measures $ - 477R?, which amounts to 27 square feet for R =1

foot, and to 27

-9 = 187 square feet for

R =3 feet.

We

are reminded of what we learned about similar figures (see Unit 5, Section 2.3): The ratio between corresponding areas in similar figures equals the square of the ratio between corresponding lengths. This means that the area of a region on any sphere can be found easily if we know the area of the corresponding region on some given sphere (say, on the unit sphere with R = 1) by multiplying by the square of the ratio between the radii. This answers

question

(2) above if we know

the answer

to question

(1). We shall now study question (1) for the lune, a special spherical triangle, and the general spherical triangle.

The area of a lune with angles of 1° obviously is ae OfcS

oki

the angles are a®, then the area is a times as big. To measure any spherical region as part of the total surface § we should agree on a small part of S as a measuring unit (see Unit 5, Section 1, on measurement).

We could choose a lune of 1° as

a unit, but one half of such a lune is a more useful choice.

Geometry on the Surface of the Sphere

519

Theorem VI-13: The area of a lune with angles a Qa lune

3 60°

sphere

Bs ce

47 R?

360° (a in degrees)

= 2aR? (a in radians)

Definition VI-11: The angle angles 1° is gree.

area of a spherical triwhich has two right and a third angle of called a spherical de(spher. deg.')

ih

4° ig

In geographical language a spherical degree is the area of a long and narrow spherical triangle with an angle of 1° at the north pole and the opposite side of 1° on the equator. This unit of area is simple to apply. The area of a lune with angles of 1° obviously contains two spherical degrees. A hemisphere contains 360 spherical degrees, and the entire surface S of the sphere contains

720 spherical degrees. In other words, a spherical degree is 735 of S. The sum of the angles in a spherical degree is 90° + 90° + 1° = 181°. According to Definition VI-8 we find that a spherical degree has a spherical excess € of 1°. A triangle ABC on the globe with A(0°, 10° £), B(O°, 15° £), C(north pole) contains 5 spherical degrees. The sum of its angles is 180° + 5° and € = 5°. 1 Do not use the ° symbol, to avoid confusing the unit of a plane angle with the unit of an area on the sphere.

520

Unit 6

In such triangles the number of spherical degrees equals the number of the spherical excess, expressed in degrees. > The spherical excess e€° of a triangle with two right angles is the measure of its area expressed in spherical degrees. S =e spherical degrees. (€ in degrees)

This relation between the spherical excess and the area is as amazing as it is practical: amazing, because a spherical area is related to a plane angle by the same number; practical, because one can be found by computing the other. For the special rightangled triangles considered here, it is easy to find the spherical excess, since it is equal to the angle at the pole. This angle may be given in degrees or in radians as part of the two round angles

at the two poles (2 times 360° or 2 times 27 radians).

Question

(1) above can now be answered by a simple proportion: > For ¢ in degrees, use $:47R? =€°:720° to find

ae

= (eocne For ¢ in radians, use S :47R?

=e

2 ,:47_,, to find

S=e_,-R? rad

We shall now prove that these simple area formulas hold also for the general triangle.

Theorem VI-14:

The area of ABC with angles a, B, y

S =e spherical degrees (€ in degrees) € ce}

~ 180° —

rad

qtR? mR

where «°= a° + B° + y° — 180° = spherical excess

Plan of proof: We know how to compute the area of a lune. If we augment the given triangle to a lune beyond each of its three sides, then it can be shown that exactly half of the total surface is covered, while the triangle itself is covered twice extra.

Geometry on the Surface of the Sphere

521

Proof: Augment ABC to a lune in three ways: iE ABC + BCA' = 2a spher. deg. 2. ABC + ACB’ = 28 spher. deg. 3. ABC + ABC’ = 2y spher. deg. 4. Sum = 2ABC + ABC + BCA’ + ACB’ + ABC’ = 2(a+B+y) spher. deg. To each of these three lunes there is a centrally symmetric, congruent lune, which is not included in the sum. Therefore SABC + BCA’ 4 ACB’ + ABC’ =45 sphere = 360 spher. deg. . 2ZABC + 360 spher. deg. = 2(a + B + y) spher. deg. . 2ABC = 2(a + B + y — 180) spher. deg. . ABC = Nn COnaIDA

« spher. deg., q.e.d.

The other two forms given in Theorem VI-14 are found from the same proportions as for the right-angled triangle above.

Corollary Vi-14-1: On the unit sphere (R= 1) the area of a spherical triangle is measured by its spherical excess, expressed in radians.

Corollary Vi-14-2: On the same sphere and on equal spheres spherical triangles with an equal sum of angles have equal areas.

522

Unit 6

Corollary Vi-14-3: On the same sphere and on equal spheres the ratio between the areas of two spherical triangles equals the ratio between their spherical excesses. (The area of a spherical triangle is proportional to its spherical eXcess.)

EXERCISES 1. What is meant by spherical excess? How much is the spherical excess in a spherical triangle with three right angles? How many spherical degrees are in the area of such a triangle? 2. What is the area of a spherical triangle with three right angles, on a sphere with radius R? 3. How much is the spherical area of an octant of a sphere? Compare with Exercise 2. 4. Find the area to the nearest 1000 sq km between the 10th and 20th meridians east on the surface of the earth? (R = 6371 km.) 5. What is the area to the nearest 1000 sq km of triangle A(north pole), B(0°,90° W), C(0°,0°) on the surface of the earth? 6. What is the area to the nearest 1000 sq km of a triangle on the surface of the earth, if the sum of its angles is 210°? Can you use the result of Exercise 5 for a quick computation? 7. Explain the following statement: angle equals its spherical excess.

The area of a spherical tri-

8. The angles of a spherical triangle on a sphere of radius inches are 70°, 90°, 120°.

9. What is the angle of a lune ona is 647 sq in.?

SUMMARY

18

Compute the area in square inches.

sphere of radius 8 in., if its area

OF UNIT VI

A. Theorems and corollaries VI-1: The shortest distance on the sphere between two points lies on a great circle. VI-2: If the polar of a point A passes through a point B, then the polar of B passes through A. Cor. VI-2-1: If the polars of two points A and B intersect in a point C, then the polar of C must pass through A and B.

VI-3:

All points of a small circle on a sphere have the same

Geometry on the Surface of the Sphere

523

polar distance. VI-4: A spherical angle is equal or supplementary to the spherical distance between the poles of its sides. Cor. VI-4-1: A spherical angle is supplementary to the spherical distance between the positive poles of its sides. VI-5: Two diametrically opposite spherical triangles are inversely congruent. VI-6: Any side of a spherical triangle is smaller than the sum and larger than the difference of the other two sides. VI-7: The sum of the three sides of a spherical triangle is smaller than 360°. VI-8: In a pair of polar trihedral angles the face angles of one are supplementary to the angles between the faces of the other. VI-9: The sum of the angles in a spherical triangle is larger than 7 and smaller than 37. Cor. VI-9-1: A spherical angle may have one, two, or three right angles; it may have one, two, or three obtuse angles. VI-10: If A’B’C’ is the polar triangle of ABC, then also ABC is the polar triangle of A’B’C’. VI-11: Ina pair of polar triangles each side in one triangle is supplementary to the angle at the corresponding pole in the other triangle. VI-12: The spherical distance between corresponding vertices A and A’ in a pair of polar Paes is less than 90°.

VI-13:

The area of the lune:

S,,,.= E08 - 47rR?

VI-14:

The area of a spherical triangle ABC is S =

ex

=

R?

750° BSS

* € radians

Cor. VI-14-1: On the unit sphere the area of a spherical triangle is measured by its spherical excess, expressed in radians. Cor. VI-14-2: On the same sphere and on equal spheres spherical triangles with an equal sum of angles have equal areas. Cor. VI-14-3: On the same sphere and on equal spheres the ratio between the areas of two spherical triangles equals the ratio between their spherical excesses. . Words and phrases Section 1: spherical distance, spherical triangle, spherical geometry, shortest distance on a sphere, antipodes, maximum length of a spherical distance

524

Unit 6

Section 2: pole, polar, positive and negative pole, polar distance, quadrant, spherical radius Section 3: spherical angle, lune Section 4: diametrically opposite, trihedral angle, face angle, face, polar or supplementary trihedral angles, spherical excess Section 5: polar triangles, positive direction around a spherical triangle

REVIEW

EXERCISES

TO UNIT VI

. In what way is a distance in spherical geometry different from a distance in plane or solid geometry? 2. What is a spherical angle and how is it measured?

jar

3. What is a trihedral angle and what does it have to do with spherical geometry? 4. What are polar triangles and how do you find them?

. What are the limitations on three spherical distances chosen for the sides of a spherical triangle? . What are the limitations on three spherical angles chosen as angles of a spherical triangle?

. What is the spherical excess and how big can it be? . Why is spherical geometry sometimes mentioned in connection with non-Euclidean geometry?

ii

If a circle M, and an inscribed ellipse E represent the front view of a meridian and the equator of a sphere, what do you have to watch when drawing correctly another meridian M, which intersects the equator at A? Can the meridian M, be the contour of this view of a sphere? (If all meridians were correctly drawn, could the north pole appear on the contour?)

*10. If a circle M, and an inscribed ellipse M, represent the front view of two meridians and their points of contact N and S$ represent the north and the south pole of a sphere, how would you draw correctly the equator?

M,

Ex.9

Supplement I

SETS “Mathematicians are like Frenchmen: what you tell them they translate into their own language, and from then on it is something quite different.” Goethe

A. A sez is a collection of objects, each of which is called a member of the set. In algebra you studied sets whose members are numbers, or pairs of numbers. In geometry we study sets of points and sets of lines and sets ofplanes. A set may be defined in two ways: (1) by listing the members of the set;

(2) by naming a property that is shared by all its members and only by its members. Example for (1): S = (3, 6, 9.} Example for (2): T = {x |x 1s divisible by 3}. ‘““The set T consists of all x, such that x is divisible by 3.” Note that the vertical bar, |, stands

for such that (meaning, which have the property that). A universal set is the set of all objects we agree to deal with. Examples:

In arithmetic the universal set is the set of all numbers.

In plane geometry the universal set is the set of all points within one plane. In solid geometry the universal set is the set of all points in space. B. Relations between two sets. Two sets may or may not have members in common. A Venn diagram is a graphical representation of sets as circular regions (or regions of some other form), which indicates whether or not cer-

tain sets have members in common. If two sets A and B have no members in common, they are said to be disjoint. If two

sets have

the same

mem-

bers, they are said to be the same set 525

(4)

526

Supplement |

or equal sets.

For example, the set

S = {3, 6,9} is the same set as the

set S’ = {all counting numbers less than 10 and divisible by 3}. We write S = S’.

é¥)

or

A set Sis a subset of the set U, if

every member of S is a member of U. (S may or may not be equal to U.) A set S is a proper subset of T, if every member of S is a member of T, but at least one member

of T is

not a member of S. Example:

S = {3, 6,9} is a proper

subset of T = {x |x is divisible by 3}.

C. Set operations. The unionof aset A andaset B(A U B)

is the set of all members that belong to either A or B.

Example:

A = {all girls in a certain

school}, B = {all seniors in that school}; A UB = {all girls and all seniors in the school}

Suppose this school has 1000 girl students and 300 of these are seniors.

There

are 1500 boy students and 400 of these are seniors.

Then A has 1000

members, B has 700 members, and A U B has 1400 members.

Note

that the members of 4 and B are not simply added in this example because there are 300 girls who are also seniors; these 300 must not be counted twice in A U B. The intersection of a set A and a set B (A - B) is the set of all members that be-

long to both A and B. Example:

A = {all girls in a certain

ag

school}, B = {all seniors in that school}; A © B = {all girls who are also seniors}.

With the same enrollment as in the previous example, 4 300 members.

B has

Sets

527

The complement of a set A is the set of all members of the universal set that do

not belong to A. (A = U — A). The difference between two sets is the set of all members that belong to one of the two sets but not to the other.

(4 — B

or A 1 B). Example: school},

A = {all girls in a certain

B = {all seniors in that school};

A — B = {all girls who are not seniors} B — A = {all seniors who are not girls} D. A relation is any set of pairs. Example:

The

set

Q = ({(3,6),

co }

(5,0),

(2,6), (3,8), (0,0)} is a relation consisting of five pairs.

An argument of a relation is the first member of at least one of the relation’s pairs. For the relation of Q above, the arguments are 3, De

aid:

The domain of a relation is the set of all its arguments.

The domain

Oleic 55.5, 2,-0}.

A value of a relation is the second member of at least one of the relations pairs.

The values of Q are 6, 0, and 8.

The range of a relation is the set of all its values. The range of Q is {6, 0, 8}. In a relation between two variables x and y, the set whose members may replace the first variable x is called the domain

of the relation;

the set whose members may replace the second variable y is called the range of the relation. Only if the variables x and y are replaced by members of the domain and the range, does the relation become a true statement. Example:

For the relation “‘is a son of” the domain is the set of all

males, {x |x is a male}.

The range is the set of all people who have

sons, {y | y has a son}.

A true statement: “Bob is the son of his father.” A false statement: ‘“‘Ann is the son of her father.”

A one-to-one correspondence is a relation such that for every member of the domain there is only one member in the range, and for every member of the range there is only one member in the domain.

528

Supplement |

Example:

P = {all points on a given

straight line} R

ie

{all real numbers }

Se 1000 eee mene

On a number line (a coordinate axis) every point is identified by a number, and for every number there is a point (see Figure).

a one-to-one correspondence between the sets symbolically P R

E. Symbols:

6¢€S 2¢S Oe kaumas

U SoT Soy

P and R.

There is

We write

6 is a member of S 2 is not a member of S the set of all x such that... universal set Sis a proper subset of T S is a subset of V (either a proper subset or equal to V) union of A and B intersection of A and B complement of A (not A) difference of A and B (A but not B)

one-to-one correspondence between P and R

Supplement IT

NUMBER

SYSTEMS

A. Counting Numbers 1

2

3

~

Engine, Engine, number nine

5 6 i? 8 Running down Chicago Line.

y) 10 11 12 If the train should jump the track, 13 14 en 6 Do you want your money back?

When starting a game with this little verse, the person who is “IT” is the sixteenth person in a certain order, usually in a circle. To count from | to 16 would have served exactly the same purpose, but the little verse is more fun and easier for children. The essential thing is that a certain counting word (a numeral or a word group of the verse) is made to correspond to a certain person, usually by pointing a finger, so that a definite order is imposed on the persons. The Incas tied knots in strings to be used as a counting system, instead of counting words. A shepherd may count his herd by touching one knot after the other as one sheep after the other passes through the gate. Notches in a counting stick may have served to count the days. p The basic idea of counting is a correspondence between a counting system

(words,

numerals,

knots, notches,

fingers, etc.) and the things

to be counted. p The elements of a counting system are arranged in a definite order, such that it is always quite clear which of two elements follows the other.

In our number system the words “larger” and “smaller” are used to describe the order of the numbers. If there are two names for the same number, we use the word “equal.” Our counting system 1, 2, 3, 4, ... is a sequence of numerals without end. To any number, no matter how large it may be, we can always add 1 and thus get an even larger number. There is no largest number. 529

530

Supplement

Professor Edward

II

Kasner * once

asked his little nephew what he

would call a really BIG number; the child said: a googol. 1 googol = 10! and 1 googolplex = 10#°°8! = 100% For short we will designate the letter W for the set of counting numbers. W

=

{1,2,3,..., googol, ..., googolplex, .. .}

B. Integers

The set of integers consists of the set of counting numbers and zero and the set of negative counting numbers. The counting numbers are a subset of the set of integers; they are called positive integers, to be distinguished from zero and the negative integers. p There is no largest and no smallest integer.

For short we will designate the letter J for the set of integers: J =

te rae ==, = 2;

Oar

lea

Zeno.

E a)

C. Fractional or Rational Numbers What does it mean if a statistical table tells us that in the United States there is a car for every 35 persons, while in Sweden there is a car only for every 5§ persons? What is $ person? If a table shows that the average family increased in size from 4 people to 45 people, does that mean that the average family contains half a person? Is yours an average family? These fractional numbers were derived as the result of a division: the number of people divided by the number of cars;

or the number

of people divided by the number of families. The purpose is a fair comparison. To count the number of cars alone would not give the right picture; two cars owned by a family of seven are not as much a sign of wealth as two cars owned by a single person. The ratio between the two numbers of people and cars makes a comparison meaningful. The results of such divisions between integers are the fractions or rational numbers. The name “rational number’ comes from ratio and has nothing to do with being sane or insane. From the definition we know that a rational number

is formed by a pair of integers, the

dividend and the divisor of a division; it may be written in the form

a:b or -

From arithmetic you will remember that multiplying or

* E. Kasner and J. R. Newman:

Mathematics and the Imagination.

Number Systems

531

dividing both parts of a fraction by the same number gives different forms with the same value. If the divisor b equals 1, we have an integer. Thus the set of integers is a subset of the set of rational numbers. If the divisor is 10, 100, 1000 and so on, we usually prefer the form of decimals: 7§9 = 0.03, z¢5%o = 0.0029. Any other fraction may also be given the form of a decimal by carrying out the division. You will find that the quotient between two integers is a decimal that either ends after a certain number of places (a finite decimal such as 4 = 0.25) or keeps repeating the same group of digits (a repeating decimal such as 3 = 0.33333333...). A finite decimal can be changed to the form of a repeating decimal: 0.25 = 0.24999999999999 . . . For short we will designate the letter F for the set of fractions or rational numbers: — 7 1 ee Oe

1

7 ee

D. Irrational Numbers and Real Numbers

Decimals that are neither finite nor repeating are not quotients between two integers. Since they are not ratios.of two integers, they are not rational numbers, but irrational numbers.

nothing to do with being insane.

Again, this name

has

The Greeks called these numbers

aoyor (alogee), which means “not pronounceable,”

because one can

never live long enough to pronounce all the infinitely many digits of which there is neither a last one nor a periodic pattern. Examples of such irrational numbers are: V2 = 1.414..., V3 = 1.732..., mw = 3.1415926..., and infinitely many more. The rational and the irrational numbers together form the set of real numbers. For short we will designate the letter R for the set of real numbers: R =

{rational and irrational numbers}

Supplement [I

5x+

EQUATIONS IN A NUTSHELL

Sees x

+2=0

This is only a short reminder of how to handle equations. If this reminder is too short for you, then you had better borrow an algebra book for a weekend or two and brush up in a hurry. A. One equation in one unknown (a) A linear equation Example:

ox —- 3 +62

parentheses and fractions

5x —3+12—6x

collect terms solve for x

—x)=2 iC+ 3)

=x4+5

SO eel S29)

check by substituting

3(2)

— 3 + 62 — 2) = 22 4+ 3) 7=7

Vv

(b) A quadratic equation Example:

S = parentheses and

_ 4](2x +5) =4105.-26x) — Ox (9 + 3x — 8\(2x +5) = 15+ 6x —2x

fractions collect terms on each side standard form of

6x7 + 17x +5

5] —5

=10+4+ 4x

quadratic ax*+bx+c¢=0

6x" + 13% = 5 = 0

solution x,,. = mais ee ead Fi

=

—13 + V169 + 120 1192

x, = Zand x, = —22 check by substituting

Gis): [5 - 4]o@ +51 = 405 + 6a — 2@ +5) 53 = 5% V

Equations in a Nutshell

Bee N34) 2 5)

5) Sas Pe =3 25)

25)

533

4) |

Or=10

V (c) A shortcut for the form x? + bx +c=0

Example:

(a =1)

Xe 5x

0 =.0)

factoring

(x — 3)(x — 2) =0

solutions

x1 = 3 and %2 = 2

Rule: If the coefficient of x? is +1, then the product of the solutions equals the constant term and the sum of the solutions equals the negative of the coefficient of the x-term. Ifa = +], then x;x, = c and x,-+ x, = —b

When to use: Although this rule is always true, it is a shortcut only if the solutions are integers so that the factors can be seen easily by inspection. Try the constant term first: 6 = 1 times 6 = 2 times 3 = (—1)(—6) = (—2)(—3).

But since the sum of the

solutions must equal +-5, only the pair 2 and 3 fits. DVS

AN 3) 2X 4.x? + 10x +21 =0 2. 3x(x —2) +4x =5 5. x? — 10x +21 =0 See 5) Seok oLS We One 6xt 11 6910

B. Two linear equations in two unknowns

First arrange the equations in standard form:

{Galicia. ° mx + ny = p

Then choose one of three methods of solution: (a) method of substitution Example:

one unknown from one equation substitute into the other equation solve for the other unknown

ee i Ss 0

yaHox —9 7x — 4(5x — 9) = 10 1 — 20%, 4 36: =a —13x = —26 x =2

substitute to find the first unknown

Vy = 52) —9 y=1

534

Supplement III

9V

-l1 =

5(2)

check by substituting into both orig-

7(2) —-4 = 10/

inal equations (b) method of elimination

eh ica

= — 4y = 10

Example:

express the same unknown

from both

y

—

5)

=

5 eas

equations . eliminate this unknown by setting equal

5x —9 =

9

ay — 10

| 7x — 10 Z ii

20x — 36 = 7x — 10 13x =226

solve for the other unknown

x=

substitute to find the first unknown

y = (2)

9

eset check by substituting into both equations (as before) (c) method of equal coefficients Example:

,

Sk = seed Tene dy om {

multiply the first equation by 4 to give y the same coefficient as it has in the second : equation

subtract one equation from the other to eliminate one unknown solve for x substitute to find y (as before)

20x — 4y = 36 7

A

x —4y

eet

=

10

13x = 26 x= 5(2) =n y=

2 9 1

check by substituting into both original equations (as before)

Note: To get equal coefficients for x, you would multiply the first equation by 7 and the second equation by 5S.

Try all three methods:

l. Sx +3y=7 9x — y=3

2 y=2x4+3 So 2y 33

3. 5x +3)4+9y41) =5 Ty — 3) — 32x +1) +19

=0

Equations in a Nutshell

535

C. Two equations in two unknowns, where one equation is quadratic Example: :

Neher) 0

one unknown from the linear equation substitute into the quadratic equation

a

solve for the other unknown

x = $(3y — 2) OV)

Oy? — l2y +4

— 4y? = 36

Se ely

20)

+12 + V144 + 640 Lip

=

10

y1 = +4 andy, = —1.6

substitute into the linear equation to find the corresponding first unknown: for y, = +4

for y, = —1.6

2x, — 34) +2 =0 x, = +5

2x2 — 3(—1.6) +2 =0 xX, = —3.4

check by substituting each solution pair into the quadratic equation: for (X1,y1) = (+5, +4) for (x2,y2) = (—3.4, —1.6) 5? — 44 =9 (—3.4)? —(—1.6)7 = 9 J 11.56 = 2.56 =9 Try: ee x

Vv yee 2 -y = 1

2. ey 4 Ly?

32 iy = 4 eet oe eet 2)

D. Two quadratic equations in two unknowns (a) one variable is squared Example:

’

eliminate squares by method of equal coefficients; multiply the first equation by 2 and the second by 5; then subtract:

Oe

ce Ver

ee a,

et) 6

10x? +4x + 6y — 26=0 10x? — 5x + 10y — 25 =0 9x — 4y— 1=0

combine with one of the original

y = 40x — 1)

quadratic equations and use C

2x?—x+5(9x —1)—-5 =0

above:

AXA xe =O —7 + V49 + 176 1,2

8

x, = +l and x, = —}

536

Supplement Ill

substitute into the linear equation for x, = +1 forx. =

Solutions:

al

Vial Vi=ence

Sia (1)) VagatZ

(+1,+2) and (—232, —3%)

(b) both variables are squared

Try to eliminate one or both of the squares Example: x? and y? have the same coefficient

subtract to eliminate both squares substitute into one of the quadratic equations simplify and solve for x

x eve == i = 12. yy Oy

12

+ 6y = 60

y = 10 = 2% Ke (LO hs 2X) ees x? + 100 — 40x + 4x? = 25 5x? — 40x x? —8x¥ (x — 3)x XxX; =15 and

+ 75 =0 +15 =0 — 5) =0 919

find the corresponding y from the /inear equation

yi = 10 — 23)

wes

Two solutions:

y2 = 10 — 25)

we

(3,4) and (5,0)

check by substituting into the other quadratic equation Example: x* and y? have different coy2=9 ee a efficients Bn ik ts) en 8x? — 2x —1=0

subtract to eliminate one square

solve this quadratic equationinoneunknown

Ae 2

x,,, = ZEN 16

x1 =% and %2 = —4

substitute into one of the original equations to find the corresponding other unknown:

for x, =%

90) + y7 =9

bee View

3

eV

fore

=

(ey aye ae

Vee aa

+3V15

Equations in a Nutshell

537

Four solutions: (4,33), (4, —3V3), (—4,3V15),(—4, —2v/15) check by substituting into the other quadratic equation When you study Unit 4, try to find a geometric explanation why there are two solutions in one example and four solutions in the other.

Lela EX

tye

le

A SY = 1 E

eX

Ayea

age

als

BE x Ne

By? = 2 =

E. How to set up an equation In this geometry course the word problems will concern segments, areas, and volumes.

1. Make a sketch of the problem showing the geometric relationships involved. 2. Recall (or look up) all the geometric formulas that are concerned with these relationships. 3. Choose and write down the best suited of these forraulas. 4. Mark which quantities within the formula are known and which are unknown. What is the question? What is unknown? — Just be sly: Call it x or call it y!

5. Translate the English wording of the problem into algebraic wording. 6. Substitute the algebraic wording into the formula. Example 1: If the side of a square is shortened by 6 inches, its area becomes smaller by 396 sq. in. How large was the side and the area of the original square? iN

2., 3. area of square, A = s? 4. unknowns, x = original side s y = original area A 5. new side = x — 6 new area = y — 396

x

538

Supplement

III

6. substitute into the formula:

originally new

y = x?

y — 396 = (x — 6)?

Solve two simultaneous equations, x? — 396 = x? — 12x + 36 12x = 432 = 36 y = 1296

Answer:

The original side was 36 in. The original area was 1296 sq. in.

Example 2: How big is the volume of a cube, if its surface area is 24 sq. in.? S = 6a? 6a? = 24 V=a3 a=2 V=@

=8

Answer: 8 cu. in.

The volume

of the cube is

Supplement

ALL ABOUT

IV

PROPORTIONS

I met a little elf-man once,

Down where the lilies blow. I asked him why he was so small And why he did not grow. He slightly frowned, and with his eye He looked me through and through. “T am as big for me,” said he,

“As you are big for you!” John Kendrick Bangs

It is all relative; I am to my yardstick what you are to yours. The ratio between the length of the side s of a small square and its perimeter pis 1:4 or 4; and so is the ratio between the length of the side S of a large square and its perimeter P. sip

=S:P

or

ou P

p> A proportion consists of two ratios set equal. A ratio can be written and treated as a fraction. A proportion is concerned with four numbers.

A proportion contains four terms, numbered in the order of writing. The first and fourth terms (s,P) are called the outer members or the

extremes;

the second and third terms (p,S) are the inner members or

the means.

The first terms of each ratio, that is the first and third terms

of the proportion (s,S) are the antecedents, the second and fourth terms (p,P) are the consequents. A. Computation rules

: ’ a Since any ratio may be written as a fraction («(b= hbthe computa-

tion rules for proportions follow from those for fractions, and you can

easily prove them for yourself: Ia:b

= c:d, then

1. ad = bc, the product of the extremes equals the product of the means; 2. a:c = b:d, the means may be interchanged; 539

540

Supplement

IV

3. d:b = c:a, the extremes may be interchanged; 4. b:a = d:c, the two ratios may be inverted;

5. (a + b):a = (c + d):c, the sum of the first two terms is to the first as the sum of the last two terms is to the third;

6. (a + b):b = (c + d):d, the sum of the first two terms is to the second as the sum of the last two terms is to the fourth;

7. (a — b):a = (c — d):c, the first minus the second term is to the first as the third minus the fourth term is to the third;

8. (a — b):b = (c — d):d, the first minus the second term is to the second as the third minus the fourth term is to the fourth; 9. (a+ b):(a — b) =(c + d):(c — d), the sum of the first two terms is to their difference as the sum of the last two terms is to their difference;

10. (a + c):(6 + d) = a:b, the sum of the antecedents is to the sum

of the consequents as one antecedent is to its consequent. B. Solving for one unknown

If one term of a proportion is unknown, it can be computed from the three other terms through an equation with this one unknown: If x:b = e:d, thenx = °

If a:b = x:d, then

= =

If a:x = c:d, then x = =

If a:b ='¢2x; then'x = e

A special proportion: If the two mean terms of a proportion are equal,

their common value is called the mean proportional between the other two terms; it equals the square root of their product: If a:x = x:d, then x = Vad Another special proportion: If the two mean terms of a proportion are equal, and if the fourth term is the difference between the first and

the mean term, then we have a “‘golden ratio”’: If a:x = x:(a — x), then x? = a(a — x)

Me

Nh

aa A)

X12 = 43(-—a + Va? + 4a?)

ee 5«V5 = [y= a0. 61S) oe

5(V5 + Des en 6leeen)

For a geometric application of the golden ratio see Unit 2, Section 2.2. rye le 33 215°=1325% 2 ok = Laer Ska XeSEES 5

All about Proportions

541

C. Solving for two unknowns

x:y = 2:3 may be written as = :- As you know, the fraction 5may have many forms, such as _ 5 = etc.; in general :- a where k may be any non-zero number. It follows that the above proportion has many solutions of the form x = 2k and y = 3k where k may be any non-zero number. It further follows that you must be given more information, if that proportion should have a particular solution for x and y. The additional information may have the form of a second equation (the proportion being the first equation). p» To solve for two unknowns, we need two equations.

Example 1: The ratio of girls to boys in a certain mathematics class of 21 students is 3:4; how many girls and boys are in that class? Call the unknown answers to the questions x and y: x = number of girls x:y = 3:4 y = number of boys xy = 2] Here are two methods of solving these two simultaneous equations:

(a) usual method for two equations from Ist equation 4x = 3y

2ndequation

(b) method using the factor k Res EN ole

x+y = 21 y=21—-x 4x = 63 — 3X ta.

hed on y 4 4k x = 3k and y = 4k substitute into the 2nd equation:

Sea

3k + 4k = 21

Vea WS

Tk = 21

Je

ig ey 8

x =9andy = 12 check:

9 +12

= 21/7

Which method do you prefer? Example 2: In a certain English class of 27 students two thirds are girls; how many girls and boys are there? Note the difference between examples | and 2: in example | the ratio between the two groups of the class was given; in example 2 the ratio of one group to the whole class is given. The questions are the same.

x = number of girls y = number ofboys

x+y eC

= 27 a= 263

542

Supplement IV

(a) usual method for two equations x +p =27 = (x + ») 3x

x=2°x+typ=3kyp=k

= 227) =

ax

substitute into the Ist equa-

x = 18 0

ears 27

(b) method using the factor & & _2 x 3k 3 x+y

-—

18

tion:

iol

3k = 27

Example 3: If a segment AB is divided into two parts by a point C, you must be careful to see whether the ratio given is that between the two parts (as in Example 1) or between one part and the whole segment (as in Example 2 above). If 4C:CB = 2:3, then AB is actually divided into 5 parts, of

which AC takes up twoand CB

A

C

B

takes up three. Then. AC = 2k, CB = 3k, AB = Sk.

If 4C:AB = 2:3, then AB is actually divided into 3 parts, of

which AC takes up two and CB takes up one. Then AC = 2k, CB

= k, AB = 3k.

Try: 1. x > y =5:7 x+y = 36

2. AC 4+ CB =36in. AC :_CB =4:5

3. AC+ CB = 6m. AC : AB |= Se ss

D. Solving for three unknowns > To solve for three unknowns, we need three equations.

seetepeste How big are the angles a, 8, y in a triangle, if their ratio

is 2:3:42 The proportion as given is a so-called continued proportion, w hich isj a short way of writing two proportions, namely a:8= 2:3 and 8:y

= 3:4.

You could also write ea:y = 2:4, but this gives noth-

ing new, since it follows from the other two. Any two of these three proportions, but not all three, give us independent equations.

Ist equation 2nd equation

a:8 8:y =

3rd equation

a+

Ps

yd

“4

bet sd

8 + y = 180° (from studying geometry)

All about Proportions

Solution:

543

You may use any of the methods of solving three linear

equations in x, y, z (or in a, B, y), but shortest is the method using the factor k, since this method reduces the problem in the three unknowns a, B, y to a problem in the one unknown k.

If a:B:y = 2:3:4, then a = 2k, B = 3k, y = 4k substitute into the 3rd equation: 2k + 3k + 4k = 180° 9k = 180° k = 20° a = 40°, B = 60°, y = 80° check:

40° + 60° + 80° = 180° VY

Try: Ata garden party of 1125 people there were men, women, and children in the ratio of 120:105:150. How many were there of each group?

Supplement V

NUMERICAL

COMPUTATIONS

“The purpose of computing is insight, not numbers.” R. W. Hamming *

In numerical problems we encounter unending decimal numbers more often than not. In the problems of this course such numbers may be the result of a division, of taking a square root or a cube root, and of using 7. A. Rounding of numerical values

If the area of a rectangular garden is 220 sq ft and if one of its sides is 21 ft long, then the length of the other side is 220 + 21 = 10.476 190 4--- ft. Where shall we stop the division? The number lies between 10 and 11 ft., nearer to 10 ft. It lies between 10.4 and 10.5, nearer to 10.5. It lies between 10.47 and 10.48,

nearer to 10.48. Ifyou are told to give the answer “‘to the nearest foot”, you will write 10 ft. If you are told to give the answer to the nearest tenth (hundredth) of a foot, you will write 10.5 ft (10.48 ft).

If you

are not told how many places of the number are required, then you will make your own decision after inspecting the precision of the given numbers. In general, the answer should be rounded to one place more than the given numbers. In the above example the preferred answer is 10.5 ft. Rounding rules: The last digit to be written is left unchanged, if the next digit is less than 5. The last digit to be written is raised by 1, if the discarded remainder is more than 5 followed only by zeros. if the first digit to be discarded is 5 and the digits that follow are either zero or unknown, then the last digit to be written is raised by 1 in case this makes it an even number.

Thus 11.5 is rounded to 12, but

10.5 is rounded to 10. *R. W. Hamming, Numerical Methods for Scientists and Engineers, McGraw-Hill Book Company, Inc., New York 1962.

544

Numerical

B. Roots (a) The symbol V2

Computations

545

represents a positive unending decimal number.

This number is rounded to 1.414 in the Table of Powers and Roots (at the end of this book). (W2)(/2) = 2, but (1.414)(1.414) = 1.999.

Which representation should you use? In algebraic expressions V2 is preferred.

The diagonal of a square

is remembered as s\/2 and not as 1.414s.

;

In numerical problems keep V2 as long as possible.

Multiply out

for the final result, if that is required.

(b) Review computations with roots in any algebra book. The characteristic cases which you will encounter in this course are the following:

(1) V108 = V(4)Q9)3) = 6V3

(find factors that are

(2) (V3\(V2) = V6

(multiplication)

jee sel ye50 _ a5 = 5

Ginko

squares)

ie 12 ee @ WOOL e215

5D 4/15 15

5 Vis

lize the deeas a

Numerical results containing roots should be simplified as shown here. Cer. mq = 3.14159 --- is an unending decimal number. The same general rule applies to 7 as to roots; namely, keep 7 in this form and also in the result, unless needed or requested otherwise. If needed, you may use:

! _ 0.3183 T

AV

Me.

0649 WV 1

SUGGESTIONS

FOR SPECIAL

PROJECTS

The guide lines for these special projects are sketchy to give you a chance to think for yourself. You may modify or expand the topics if you wish. List reference books used and give the pages of the reference. After the Preparatory Chapter has been studied, you may consider

one of these: I. GEOMETRY

IN ANCIENT

TIMES

(historically)

Geometric knowledge of the Sumerians, Egyptians, Indians. do we know about these beginnings? Il. READING

PLAN

AND

FRONT

How

VIEWS

(a) Choose one or more of the examples given (Figures Ia and Ib) and try to visualize what the object looks like. is

(1)

(2)

(3)

Suggestions for Special Projects

(6)

(7)

eck

547

|

QY/

BAH PLAN

VIEWS

ONLY:

|(b)

(b) Draw a pictorial view of the object by first choosing a coordinate system and units of length for each axis. For simplicity choose the coordinate system such that the axes are either parallel to edges of the object or pass through them if possible. You may use a coordinate system as described in the Text or one with the x- and y-axes equally inclined as in Project III. (c) Show that one view alone is not enough to describe an object. Draw several different objects that could have the same plan view.

Ill. PLAN AND FRONT SPECIFICATIONS

VIEWS

OF OBJECTS

WITH

GIVEN

Choose one or more of the examples given in pictorial view (Figures IIa and IIb) and draw a pair of plan and front views. Choose a unit of length and make a scale on the edge of a strip of paper, up to the largest length needed; or choose some part of an inch or of a centimeter as unit.

548

Suggestions for Special Projects

ss (2) 6x8x6

(1) 6x6x6

l(a)| Examples (1)-(4) are cut from rec-

tangular solids with given dimensions. (5) is a cube with a pyramid on each face; the height of the pyramid is 4 of the edge of the cube.

(6)8) are

drawn in a coordinate system, where the x- and y-axes are equally inclined against each other and the z-axis;

the

units on all three axes are equal.

IV. PLAN

AND

FRONT

VIEWS

OF

GEOMETRIC

OBJECTS

(a) Draw a pictorial view of some combinations of geometric objects,

such as a pyramid on top of a cube, a monument consisting of a low rectangular base plate and a high rectangular solid with a short pyramid on top, or whatever you like. (b) Draw a pair of plan and front views of (a).

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549

(c) Imagine that you look at the object from a different side and

draw a second front view to the same plan view. Turn the paper slightly and use another direction for the projectors. (d) By turning the paper again, draw some more front views for the same object and the same plan view. Watch how different edges may become visible or invisible, and indicate the invisible ones by dotted

lines. With several front views you can show the details of your object from all sides. Mark one particular face and show where it appears in the various views. Example: In Figure III an octahedron stands on a vertex E and has a side AB parallel to II,, The plan view is a square with its diagonals. The front view is a parallelogram with only one diagonal. (Why?) If you turn the paper a little and use another plane as II, (another axis and direction of the projectors), the side AB is not parallel to the new axis, and the front view of the octahedron will look different.

various screens show which faces are seen.

The

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Suggestions for Special Projects

After Unit 1 has been studied, you may consider one of these:

V. FAMOUS

PROBLEMS

OF

ANTIQUITY

(historically)

The Greeks formulated some apparently simple geometric problems that puzzled mathematicians for centuries to come (Rectification and

Squaring of the Circle, Trisection of the Angle, Duplication of the Cube). They also were interested in logical paradoxes that are still quoted today. (Are all Cretans Liars? Can a Flying Arrow be in Motion?

Achilles and the Turtle.)

VI. INFINITELY

MANY

POINTS

(Georg Cantor, 19th cent.)

How much is “infinitely many’? Archimedes in ‘““The Sand Reckoner” computed the number of grains of sand it may take to fill a sphere which has the sun as its center and the distance from the sun to a star as radius. He found that this number was less than 10°. He wanted

Suggestions for Special Projects

551

to show that it may take a huge number of grains but not infinitely many to fill up such a vast space. By the first postulate of the Text straight lines contain infinitely many points. What do we mean by that? In the “order postulates” more points were added. The postulate of connectedness to-

gether with the postulate of Eudoxus adds even more. Does that mean that there are different kinds of ‘“‘nfinitely many’? How can these points be counted? Explain how sets can be compared without actually counting their members. A large segment and a small segment can be shown to have the same number of points by establishing a one-toone correspondence between their points. A pencil of lines through S$ cuts out pairs of corresponding points from these segments. Yet, AB is shorter than A’B’ if measured in inches instead of counted

by points. Suppose AB is 3 in. and A’B’ is 5 in. comparing the finite set of 3 inches with the finite set of 5inches. A oneto-one correspondence between finite sets can only be established if these finite numbers

Measuring means

are the same;

the set of 3 can correspond only to a subset of the set of 5. This is quite different, if both sets are infinite. For example, take S =

T = {all even counting numbers}. of S, yet there is a one-to-one correspondence between S and T as shown.

{all counting

numbers}

and

Obviously T is a proper subset

That means, there

are just as many even numbers as positive integers! Explore this crucial difference between finite and infinite sets in more examples. On the number line, are there more points corresponding to

rational numbers (fractions) than points corresponding to integers? Would you believe that there are just as many fractions as there are in-

552

Suggestions for Special Projects

tegers? This question can be settled with a surprising result! It is easy to establish a one-toone correspondence between set J and set F in the following way: A fraction :can be represented by an ordered number pair and by a point in the plane with these numbers as coordinates. Let us consider just positive coordinates, that is just points of the first quadrant. Starting with + and tracing two sides of larger and larger squares

as shown,

a1 (P,d)=q

all positive

rational numbers will be reached eventually. Write down the order in which they are reached! Which is the 7th and which the 15th number? Is the routing shown the only possible one? Any such routing orders the members of set F into a sequence, so that they can be counted. Set F is countable. But it can be shown that no such sequence is possible for the real numbers, not even for the irrational num-

bers between uncountable. forming such suggest, it can not

work;

O and 1! They are Whatever scheme of a sequence you may be shown that it does

there will be some

num-

bers left out. If the irrational numbers are written as unending decimals in any order (the a, b, c,... stand for

any digits), then these decimals cannot appear in it: O.a bc... where a#.0,,

Did,

does this prove?

6

.C,..

ee.

What

Suggestions for Special Projects

Vil. HOW MANY MORE THAN ON ITS SIDE?

POINTS

ARE

WITHIN

553

A SQUARE

Do you think there are more points inside a square than on its side? — Here is proof that there are not! Here is a one-to-one correspondence between these two sets of points: The points on the side can be considered as represented by the set S of all numbers between 0 and 1. Any point inside the square can be represented by an ordered number pair (x,y),

where x and y are between 0 and 1. Let us write x and y in the form of unending decimals. Nowa pair of such vice versa.

rule can be formulated by which a

decimals (x,y) corresponds What rule, for instance,

to one decimal in S, and is used if (0.3490065...,

0.2 034078...) and 0.324 03 94006 07 58... correspond to each other? By this rule every point inside the square corresponds to a point on the side of the square, and vice versa. What does this prove? Could you adapt this idea to three dimensions? Could you prove that there are just as many points on the edge of a cube as in the whole space enclosed by the cube? VIN.

IDEAL ELEMENTS (Desargues, 17th cent., Poncelet, 19th cent.)

Within a plane, let us consider a pencil P of straight lines and a straight line p that does not belong to this pencil (Figure I). Any point

554

Suggestions for Special Projects

of p (A,B,C, .. .) lies on exactly one line (a,b,c, . . .) of the pencil P (which

postulate?), and any line through P intersects p in exactly one point — with the exception of v ||p. There is exactly one such exception (which postulate?). This extra line v is determined by P and the direction of p (namely, ||p). Thus there is a one-to-one correspondence between the set P = {all straight lines through P} and the set p = {all points of p and the direction of p}.

The correspondence between the members of the sets P and p is a correspondence between lines and points, if we agree to call the direction of p a special kind of point, as a figure of speech. It is called the ideal point of p, in contrast to the actual points of p. Discuss why this name is not a very fortunate choice! (Are not all concepts ideas? Are points in geometry dots on the paper, or corners of an object, or theoretical concepts and therefore ideas?) Another name is the point of p at infinity. This name is also misleading by suggesting that there is a definite place “infinity” on p. Both names, however, have the great advantage of keeping us reminded that in the above correspondence of P and p the direction of p behaves as if it were a point. This makes it much easier to formulate statements in connection with such a correspondence. Example 1: In Figure Il a one-to-one correspondence between the points of p (A,B,C, .. .) and the points ofg (A’,B’,C’, . . .) is established

,

by the pencil P.

5

A’ Ponce AS Bite

BOL

The ideal line (line at infinity) of a plane is the set of all ideal

points of that plane.

556

Suggestions for Special Projects

What happens in space, if every plane has an ideal line? What happens if two planes intersect in a straight line, whose ideal point must lie — where? What could be meant by the ideal plane of space? If an axial pencil of planes is the set of all planes through one line, what is an axial pencil of parallel planes? Find several examples for the simplifying use of ideal elements in space! They not only simplify statements; they also help us see that prisms, for example, are special cases of pyramids, cylinders are special cases of cones, etc. IX. THE PRINCIPLE (Poncelet, 19th cent.)

OF

DUALITY

IN THE

PLANE

A report on ideal elements (maybe by someone else in your class) is a prerequisite for this report. What are ideal elements in the plane? Show that the concept of ideal elements simplifies the wording of Cor. V(c). The simplified form permits the following dua/ statements:

Two points

|Two straight lines

lie on (belong to)

| lie on (intersect in, pass through)

exactly one straight line. |exactly one point.

The fact that by interchanging the words “point” and “straight line” along with “belong to” and “‘intersect in’’ we derive one true statement from another, is called the principle of duality in the plane. p Within the plane, “point” and “straight line” are dual concepts; and so are “‘belong to”’ and “‘intersect in (pass through).”’

Of course, this principle applies only to statements about incidence, that is about “lying on” and “‘intersecting.”” Nothing has been said about distances or angles. (a) Find examples! For instance, find the dual to a figure of four points no three of which lie on the same straight line. (The names quadrilateral and quadrangle are sometimes used as dual names.) How is a triangle defined and what would be a dual definition? (b) Make an equilateral triangle of cardboard and find its center C in 3 of its height from the foot. With a pin, fix C on a sheet of paper so that the triangle can still easily be turned around C. Trace the three sides in many positions of the triangle while C is kept fixed, and see what figure is enveloped by these sides as tangents. p> Acurve may be defined as a set of points, but also (in a dual def-

inition) by a set of tangents.

Suggestions for Special Projects

557

(c) Desargues’ Theorem (17th cent.): If two triangles are so placed that the straight lines through the three pairs of corresponding vertices meet in one point, then the intersections of the three pairs of correspond-

ing sides lie on one straight line. Make a drawing of this theorem to show exactly what it says. Formulate its converse. Formulate the dual of the theorem and of its converse. X. THE PRINCIPLE (Poncelet, 19th cent.)

OF

DUALITY

IN SPACE

A report on ideal elements (maybe by someone else in your class) is a prerequisite for this report. (Duality within the plane is not a prerequisite.) Reformulate the incidence postulates and corollaries in Unit 1, using ideal points and ideal lines as figures of speech. Which concepts are dual concepts in three-dimensional geometry? Is there a concept dual to itself? Find examples of dual statements! Duality applies only to incidence, that is to statements about “lying on” and “‘intersecting,” not to distances and angles. What is dual to a tetrahedron? to an octahedron? toacube? to an axial pencil of planes? *XI. EXAMPLES OF SETS SATISFYING SOME ALL OF THE EUCLIDEAN POSTULATES

BUT

NOT

(a) Think up some non-mathematical examples for possible interpretations of point sets: 1. Pseudo-points = people belonging to a chess club; universal set = the membership of this club; pseudo-line = two people playing a game of chess; pseudo-plane = more than two people of the club who meet on a particular evening to play chess. What interpretation could

be given to the words

“parallel,”

“intersect,”

“determine”?

What rules (postulates) similar to or different from Euclidean postulates

would you like to set up for the club ? 2. Pseudo-points = students of a school; universal set = all students of this school. Think up possible interpretations of pseudolines and pseudo-planes (sections of mathematics classes with the same or different teachers, sets of seniors, juniors, girls, boys).

Think up

scheduling rules as ‘“‘postulates.”” How could the parallel axiom or the incidence postulates be interpreted?

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Suggestions for Special Projects

(b) Find some geometrical examples: 1. A finite number of points in the plane; the universal set S, consists of not more and not

6

less than 7 points. Place them as vertices of a quadrilateral

“a

1, 2, 3, 4, the intersection 5 of the

diagonals, and the intersections 6, 7 of opposite sides. A pseudoline is a set of points (how many on each?); tween

them,

3

the connection beas shown

in the

2

figure, is only a help to see which points belong to some line. How many pseudo-lines are there? Do every two points determine a line?

fi

Now add a (broken) line from 6 through 5 to 7 and

call this also a pseudo-line; do every two points determine a line now? Which postulates do or do not apply to this $,? 2. A finite model in space consisting of 10 points (clay balls) and 10 straight lines (sticks or wire) such that 3 points are always on a line and 1

3 lines pass through a point.

This is called a /0; configuration.

You

may think of a triangular pyramid with vertex 1, base points 2, 3,4, and a plane section through 5, 6, 7, such that corresponding sides of the sec-

Suggestions for Special Projects

tion and the base meet in 8, 9, 10.

559

Discuss the order postulates, inci-

dence postulates, and parallel axiom. Remember that each line consists of 3 points only; the sticks between them are only there to hold the model together.

3. An infinite model in the plane; universal set of points = points of the upper half plane only, excluding the boundary line; pseudolines = semi-circles with their centers on the boundary line. If two points are given, how can one find the pseudo-line on which they both lie? (You may have to look ahead into Unit 2, Section 3.) What would be a pseudo-triangle? a pseudo-quadrilateral? a pair of parallels? (Remember that intersections on the boundary line do not belong to our universal set!) Which postulates are valid and which are not? After Unit 2 has been studied, you may consider one of these:

XII. THE

PYTHAGOREAN

THEOREM

Try to find several more proofs of the Pythagorean Theorem, study them and present them. XII. THE

CENTER

OF

GRAVITY

AND

LINES

OF

GRAVITY

(a) Cut a triangle of cardboard, tie a string through a hole very near each vertex, and hang the triangle up at each string in turn. When the triangle hangs motionless, mark the direction of the string on the cardboard. (The edge of a sheet of paper lined up with the string will help to put a pencil mark in the continuation of the string.) This direction of the string shows the direction of the pull of gravity. You will find that this is the direction of the median in the triangle. The medians are lines of gravity.

Their intersection, the centroid, is called the center

of gravity. (b) Suspend the same triangle at various other points near its edges. The direction of the string, continued and marked on the cardboard,

560

Suggestions for Special Projects

will show different lines of gravity; but they will pass through the same center of gravity. The mass of the cardboard triangle should be evenly distributed around this point. If suspended at this point (fasten the string through a hole at this point with a knot on the other side) the triangle should hang horizontally. Instead of suspending it, you may try to balance it at the tip of your pencil. (c) Make cardboard models of various quadrilaterals and find several

lines of gravity for each. They should be concurrent at the center of gravity. Test this center by suspending the model or by balancing it on a pencil. XIV. THE NINE-POINT CIRCLE (see Unit Thales’ Theorem, see Exercises, Section 3.4)

2, Section

3; needs

1. Prove that these 9 points lie on a circle: 3 midpoints of the sides, S; 3 feet of the altitudes, F;

3 upper midpoints on altitudes, M; Plan:

Start with a circle through 3 of the points, and show that it

passes also through one after another of the other points. You will have to use Thales’ theorem about the right angle in a semicircle. (a) Start with the circle through S,, F,, M,. This is art. A; from Thales’ theorem follows that M,S, is a diameter and its midpoint N

is the center of that circle.

(b) If you can find some more rt. As with the same hypotenuse, then

Suggestions for Special Projects

the vertex of the rt. £ must also lie on that circle.

561

Show that S,M, ||CO

(~ As) and S,S;, ||AB (~ As); therefore S,M, 1 S,S;, (since CO He AB) and S, must lieon ©N. Repeat for S;. (c) Likewise prove by ~ As that M,M, 1 M,S; and M,M, 1 M,S;. Thus M, and M; lie on ON. (d) You know now that S,, S,, S;, M,, M,, M;, F, are on ON. Show that AS,M,M, is a rt. A and its hypotenuse must be a diameter of ©N. But then the vertex F, of the rt. AS,F,M, must also lie on ON! Repeat for F; by showing that also M,S, is a diameter (from rt. AM,;M;S,) and that the vertex F,; of rt. AM,F;S; must lie on ON.

2. Show that the radius of © N equals $ the radius of the circumcircle: NM, =4UA. First show AS,S;U2%AM,M,O and then US, = OM, = M,A; also US, || M,A. Thus US,M,A is a parallelogram and 2NM, = UA. 3. From the parallelogram it also follows that O, N, U are collinear

and that N is the midpoint of OU. XV. THE Exercises)

THEOREM

OF

Thus N lies on Euler’s line.

MENELAUS

(a) Prove this theorem for both cases:

(see Unit 2, Section 2.2, a transversal that intersects

the triangle, and a transversal that does not intersect it. Watch the sign of the ratios. “If and only if” reminds you that you must prove a statement and its converse. (b) Formulate theorems for special cases by using special numerical values for the ratios. Make a drawing to check your results. (c) Show that the following theorems about three collinear points follow from the Theorem of Menelaus: 1. The bisectors of two interior angles of a triangle and of the third exterior angle intersect the opposite sides in collinear points.

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Suggestions for Special Projects

2. The three external angle bisectors opposite sides in collinear points.

of a triangle intersect the

*(d) The Theorem of Menelaus also covers those cases where one

of the collinear points is an ideal point (see explanation to Special Project VIII, Example 2); this means

that the trans-

versal is parallel to a side of the triangle. Try to formulate special theorems for such cases.

For example, “A transversal

through the midpoints of two sides of a triangle is parallel to the third.” Show that the ratio of the ideal point with respect to AB must be considered as uct |e XVI.

THE

THEOREM

OF

CEVA

(without trigonometry)

(a) The proof given in the Text made use of trigonometry. Try to prove the second version directly from similar triangles, without

Suggestions for Special Projects

trigonometry.

Through

C draw

563

a parallel to AB (see Figure).

AA III P~ AECP and AB III P~ AFCP.

A007, beeen) 01 agus 16

eer

eet

ae Gs

a SCR ON mB

Likewise, find a and ae and multiply. LC ITA the converse:

equals

LCF

Then formulate and prove

If there are points J, IJ, IJ] such that a certain product

—1, then there are three concurrent lines.

You must show

that, if P is the intersection of two of these lines, the third must pass

through P also; because if it did not pass through P, then the previous statement which we already proved, would be violated (indirect proof). (b) The proof given in the Text employed a point in the interior of the triangle. Now prove the theorem for a point outside the triangle. Two out of the three angles will be divided externally, which makes the ratio positive. XVU.

THEOREMS

DERIVED

FROM

CEVA’S

THEOREM

(a) Show that the concurrency of the three altitudes of a triangle can be derived from Ceva’s Theorem by considering the two small right triangles AOF, and BOF, formed by two of the altitudes (see Fig.); prove a, = 6.

Draw separate sketches for the other pairs of altitudes to prove Bae = 72 andy;

=10,.

Lhe sine

of equal angles is equal, but watch the sign for internal ratios! (b) Show

that

the

concur-

rency of the perpendicular bisectors follows from (a), because

perpendicular bisectors (c) The existence of

the altitudes in one triangle are in a certain other triangle. the excenters of a triangle follows from Special example for a special numerical ratio. numerical ratios you may be able to formulate

Project XVI (b) as an (d) By using special interesting theorems about the concurrency of other triples of straight lines in a triangle. After Unit 3 has been studied, you may consider one of these:

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XVIII. SYMMETRIES OF DESIGN IN NATURE AND IN ART Bring interesting examples in Flora and Fauna, in architecture, in geometric decorations of china, rugs, wall paper, dress material, etc. XIX.

CRYSTALS

(see Unit 3, end of Section 6)

Find a book on crystallography and copy some of the more regularly formed crystals. Discuss their symmetries. Diamonds are cut in certain ways. XX.

HEXAFLEXAGONS

Have fun folding a strip of paper into layers of hexagons. Turn into a magician, showing different hexagons by special foldings (flexing). Read: Martin Gardner:

“‘Flexagons,” Scientific American,

Dec.

1956, March

1957 C. O. Oakley and R. J. Wisner: ‘‘Flexagons,” The American Mathematical Monthly, vol. LXIV, no. 3, March 1957, pp. 143-154 Margaret Joseph: ““Hexaflexagrams,” The Mathematics Teacher, vol. XLIV, no. 4, April 1951, pp. 247, 248 XXI.

FANCY

CURVES

Plot from equations; find special points and symmetries first. Watch beautiful figures emerge! Some of them have a significance in mechanics. 1. lemniscate, r = 2a? cos 2a 2. four-leafed rose, r = a sin 2a

r, a = polar coordinates a = constant, a fixed length,

orr = acos 2a

for example, 5 inch

3. three-leafed rose, r = a sin 3a or7 =acos 3a

4. cardioid, r = a(l1 — cos a) orr = a(1 + cos a) 5. witch, x*y = 4a?(2a — y)

6. cycloid, described by a point on the tire when the car moves x =aa—asina

here a is in radians!

V=4a a4 COs. & a is the radius of the rolling circle 7. epicycloid, described by a point on a circle of radius a, rolling on the outside of another circle of radius b.

If 6 = 3a, then

x = 4a cos a — acos4a y = 4a sina — asin4a

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XXII.

PLATONIC

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SOLIDS

Make a set of cardboard models of the five Platonic Solids.

Use

sticks or wire as axes of rotation. An axis through a vertex, through the center of a face, or through the midpoint of an edge may be of different order of rotation.

XXII. PLANE SECTIONS A Study of a Transformation

Show one axis of each kind.

OF

A PYRAMID.

If a pyramid is cut by a plane parallel to the base, the intersection is a figure similar to the base. The points of one figure may be mapped into the points of the other figure by a one-to-one correspondence called an expansion. If the intersecting plane is not parallel to the base, then there also is a one-to-one correspondence between the points of the intersection and the base, but not an expansion. What can you find out about this correspondence (called a perspective) by reasoning (Figure I)?

Start with the example of a square pyramid V-ABCD, standing on the plane II. The intersection with a plane ~ is the quadrilateral

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A'B'C'D'. The figures ABCD and A’B’C’D’ are said to be in perspective. The connectors of corresponding points meet at V, the center of perspective. Given P in II, where is P’ in 2? Given Q’ in 2, where is Q in II?

Since D and I are not parallel, they intersect in a straight line p (the axis ofperspective). Explain why every point of p corresponds to itself. All points lying on a straight line s in one of the two planes 2, II correspond to points on a straight line s’ of the other plane. Can you prove this? (A straight line s and the point V determine a plane of connectors;

this plane intersects 2 in ? and IJ in ?.)

Do corresponding straight lines (for example, AB and A’B’) intersect?

If so, where?

Application 1: The properties derived above can be used to draw

a plane section of any pyramid. For example, repeat Figure I in the following order: Draw the pyramid V-ABCD. The intersecting plane > may be given in various ways; let it be given by its intersection p with the plane II of the base and the point C’ on VC. (It is not necessary to draw the cut-off lines of 2 and II as in Figure I.) To find B’ on VB use the fact that corresponding lines intersect on p. Thus CB and

C’B’ intersect on p. Therefore, extend BC till it intersects p, connect this point of intersection with C’, and where this new line crosses the connector VB is the point B’. The other neighbor D’ is found by extending CD till p, connecting the intersection on p with C’ and

cutting VD at D’.

Likewise A’ must lie on VA and also on a line

—

corresponding to DA (or BA). You may wish to demonstrate this simple procedure for several different pyramids with any irregular polygon as base. Application 2: Although you have studied a problem in space, namely the plane section of a pyramid, the drawing consisted of lines on paper. What you have learned can therefore be interpreted also as a study of perspective within a plane. Given a point V as center of perspective, a straight line p as axis of perspective, one pair of corresponding points C and C’; find for any figure ABCD... the corresponding figure A’B’C’D’... Draw several examples. (In a space problem hidden edges are drawn by dotted lines. In problems within a plane there are no hidden lines. The two corresponding figures and the axis are of main interest and drawn in full lines. The connectors and extensions are drawn in fine lines or dotted, as in Figure II).

Application 3: Figure II represents the famous Theorem of Desargues (17th cent.): If the corresponding vertices of two triangles lie

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on concurrent

lines, then the intersections

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of the (extended) corre-

sponding sides are collinear. And conversely, if the intersections of the (extended) corresponding sides of two triangles lie on a straight line, YA

then the connectors of the corresponding vertices pass through one point. You may interpret Figure II also as the plane section of a triangular pyramid. How can this theorem be used to connect a given point B with an inaccessible point V (Figure IIT)? XXIV. THE ELLIPSE AS A Study of a Transformation

AN

IMAGE

OF

A CIRCLE.

If we look at the opening of a circular bowl from the side, we see it as an ellipse, but we | know from experience that it is a circle. We can verify this fact by looking straight down at the bowl. We have used this experience in the Preparatory Chapter, when drawing an ellipse as if it were a circle inscribed into a square. If a square is seen sideways it appears as a parallelogram, which can be drawn easily. The relationship between a circle and its elliptical image can technically be described as a one-to-one correspondence between points, a transformation. Here is an example of a transformation that does not preserve the shape. This transformation is quite useful in solving

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problems about an ellipse by transforming them into problems about acircle. Each of the two Figures IIa, b in Section 5 of the Preparatory Chapter represents a specific case of such a transformation, called an affinity. You may wish to start with the simpler case of Figure Ila. The square 1 )2,304, (side 2a) corresponds to the rectangle 1234 (sides 2a, 2b).

Procedure

of finding corresponding points:

Given

C,, we find the