Geometry with Coordinates - the CSMP Preservation Project

CEVREL - CSMP LIBRA^ 162 3. ' 4 . ASHINGTON ST, CARBONDALL, ILL. 62901

School Mathematics Study Group

Geometry with Coordinates

Unit 48 CEMREL - CSMP LIBRARY 103 S. WASHINGTON ST. CARBONDALE, ILL. 62901

Geometry with Coordinates J

Student's Text, Part 11

REVISED EDITION

Prepared under the supervision of a

Panel on Sample Textbooks

of the School Mathematics Study Group:

Frank B. Allen Lyons Township High School

Edwin C. Douglas Taft School

Donald E. Richmond Williams College

Charles E. Rickart Yale University

Robert A. Rosenbaum Wesleyan University

Henry Swain New Trier Township High School

Robert J. Walker Cornell University

Stanford, California Distributed for the School Mathematics Study Group by A. C. Vroman, Inc., 367 Pasadena Avenue, Pasadena, California

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Permission to make verbatim use of material in this book must be secured from the Director of SMSG. Such permission will be granted except in unusual circumstances. Publications incorporating SMSG materials must include both an acknowledgment of the SMSG copyright (Yale University or Stanford University, as the case may be) and a disclaimer of SMSG endorsement. Exclusive license will not be granted save in exceptional circumstances, and then only by specific action of the Advisory Board of SMSG.

1965 by The Board of Trustees of the Leland Stanford Junior University. All rights reserved. Printed in the United States of America.

CONTENTS

Chapter

. . . . . . . . . . . . . . . 8 . COORDINATES I N A PLANE 505

. . . . . . . . . . . . . . . I n t r o d u c t i o n

. . . . . . A Coordinate System i n a Plane

. . . . . . . . . . . . . . Basic Theorems

The Se t -Bu i lde r No ta t ion . . . . . . . . . . . . . . . . . . . . Conposite Condi t ions

. . . . . . . . Equat ions and I n e q u a l i t i e s

F inding the Coordina tes of t h e P o i n t s of a Line . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . Slope

. . . . . . . . . Other Equat ions f o r Lines

. . . . . . . . . . . . Perpend icu la r Lines

. . . . . . . . . . . . . . Para l l e log rams

. . . . . . . . Using Coordina tes i n Proofs

. . . . . . . . . Para l l e log rams Continued . . . . . . . . . . . . . . . . Trapezoids

. . . . . . . . . . . . . 8.15 . concur ren t ~ i n e s 595 . . . . . . . . . . . . . . . . . . . 8.16 summary 602

. . . . . . . . . . . . . . Review Problems 603

9 . PERPENDICULARITY. PARALLELISM. AND COORDINATES . . . . . . . . . . . . . . . . . . . . . I N SPACE 607 . . . . . . . . . . . . . . . 9.1 . I n t r o d u c t i o n 607

. . . . . . . . 9.2 . P e r p e n d i c u l a r i t y R e l a t i o n s 609 . . . . . . . . . . . . 9.3 . P a r a l l e l R e l a t i o n s 617

9.4 . R e l a t i o n s Invo lv ing P e r p e n d i c u l a r i t y and . . . . . . . . . . . . . . . P a r a l l e l i s m 523

. . . . . . . . . . . . . . . 9.5 Dihedra l Angles 633

. . . . . . . . . 9.6 Coordinate Systems i n Space 541 . . . . . . . 9.7 . The Dis tance Formula in Space 651

9.8 . Parametr ic Equat ions of a Line i n Space . . 658 . . . . . . . . . . . . 9.9 . Equat ion oF a Plane 662

. . . . . . . . . . . . . . . . . . 9.10 . summary 670 . . . . . . . . . . . . . . . . Vocabulary 672

. . . . . . . . . Review Problems. Chapters 6 t o 9

Chapter

100 DIRECTED SEGMENTS AND VECTORS . . . . . . . . . . . . . 10.1 . Introduction

. . . . . . . . 10.2 . Directed Segments

. . . . . . . . . . . . . 10.3 . Vectors . . . . 10.4 The Two Fundamental Theorems

10.5 . Geometrical Application of Vectors

10.6 . The Scalar Product of Two Vectors . . . . . . . . . . . . . . . 10.7 Summary

. . . . . . . . . . . . . . . 11 POLYGONS AND POLYflEDRONS 727 . . . . . . . . . . . . . . . . 11.1 . Introduction 727

. . . . . . . . . . . . . . 11.2 Polygonal-Regions 728

11.3 . Sum of the Measures of the Angles of a . . . . . . . . . . . . . . Convex Polygon 735 . . . . . . . . . . . . . . . . . . . . . 11.4 Area 743

. . . . 11.5 Areas of Triangles and Quadrilaterals 753

. . . . . . . . . . . . . . . . 11.6 Area Relations 765 . . . . . . . . 11.7 Relations i n Similar Polygons 773

. . . . . . . . . . . . . . . 11.8 Regular Polygons 779 . . . . . . . . . . . . . . . . 11.9 . Polyhedrons 782

. . . . . . . . . . . . . . 11.10 Polyhedral Angles 788 . . . . . . . . . . . . . . . . . . . 11.11 . Prisms 796

. . . . . . . . . . . . . . . . . . . 11.12 Pyramids 803

. . . . . . . . . . . . . . . . . . 11.13 . summary 809 . . . . . . . . . . . . . . Review Problems 811

12 . CIRCLES

12.1 . 12.2 . 12.3 . 12.4 . 12.5 . 12.6 .

. . . . . . . . . . . . . . . . AND SPHERES 819 . . . . . . . . . . . . . Basic Definit ions 819 . . . . . . . . . . . . . . . Tangent Lines 829 . . . . . . . . . . . . . . . Tangent Planes 841 . . . . . . . . . . . . . . Arcs of Circles 847

Lengths of Tangent and Secant Segments . . . 869

The Circumference of a Circle; t h e . . . . . . . . . . . . . . . . . Number . -K 887 . . . . . . . . . . . . . . Area of a Circle 893

. . . . . Lengths of Arcs . Areas of Sectors 899

Inscribed and Circumscribed Circles . . . . 905 . . . . . . . . . . . . . . . . . . Summary 910

Review Problems . . . . . . . . . . . . . . . 910

. . . . . Review Problems. Chapters 10 t o 12 913

Appendix V . How t o Draw P ic tu res of Space Figures . . . . 921

Appendix VII . Surface Area and Volume . . . . . . . . . . . . 929

Appendix VIII . How Eratosthenes Measuped the Ear th . . . . . 941

Appendix IX . Rigid Motion . . . . . . . . . . . . . . . . . 943 Appendix X . Trigonometry . . . . . . . . . . . . . . . . . 963

Appendix X I . Vectors I n Space . . . . . . . . . . . . . . . 973

Appendix XI1 . Applicat ions of Geometric Theory t o t h e Use of St ra ightedge and Compasses i n Drawing . . . . . . . . . . Pic tures of Plane Figures 975

The Meaning and Use of Symbols . . . . . . . . . . . . . . . 983

The Greek Alphabet . . . . . . . . . . . . . . . . . . . . . . 987 . . . . . . . . . . . . . . . L i s t of Chapters and Pos tu la tes 989 . . . . . . . . . . . . . . Lis t of Theorems and Coro l l a r i e s 991

Index

L i s t of Writers

CEMREL . CSMP LIBRAKT 103 S . WASHINGTON ST . ... CARBONDALI, ILL 6?.901

Chapter 8

COORDINATES IN A PLANE

8-1. Introduction.

In connection with our discussion of distance we introduced

the idea of a coordinate system on a line. A coordinate system

on a line is determined by any pair of points on it; with one point of this pair designated as the origin and the other

designated as the unit-point. A coordinate system on a line is a one-to-one correspondence between the set of all real numbers and the set of all points in the line, such that the

coordinates, i.e., the numbers associated with the points, can

be used to determine distances between points.

Problem Set 8-1 -- 4-e

1. On line AB assume a coordinate system which assigns the

coordinate 0 to A and 1 to B . P is a point on - AB with coordinate x . For each listed condition plot

the set of all points P determined by that condition.

(a) x = 5 . (b) x = -3 . (c) x = 3AB . (d) x = 4AB .

1 (e) x = AB . 1 (f) x = t AB and t is an element of (1, 4, 0, ,

(g) x = k * A B ; k < l .

(n) x = k e A B ; k > O .

(1) x = k e A B ; O < k < 1 ,

(j) x = k o A B ; k > 0 .

r 8-1 - 2. A coordinate system on AB assigns coordinates 0, 1, x

to points A, B, P respectively. Plot the set of all P

such that x satisfies the given conditions.

(a) x > 0 and x is an integer. Describe the set.

(b) x < 0 and x is real.

(c) -2 < x < 5 ; x is an integer. How many points

belong to this set?

(d) -3 < x < 1 ; x is real. How many points belong to

this set?

(e) - 5 < x - < - 1 ; x is real.

(f) Using mathematical terms describe the point sets in

(b), (dl, (4. 3. If A, B, C, P are on ray AP and have respective

coordinates 0, 1, 3, x, what is the value or values of

x determined by each of the following conditions?

(a) AP = 2AC . (d) BP = 3BC . (b) AP = 5AC . (e) BP = k BC . (c) AP = k AC . (f ) BP = SAC .

4. Suppose a coordinate system on a line m is given. If

P and Q are points in m with coordinates p and q

respectively, find the distance from P to Q , If: (a) ~ = 5 , q = 8 - (e) p = . r - 3 , q = r + 3 . (b) p = - 7 , q = - 8 . (f) p = r + g , q = r + 1 2 .

(c) p = 3 , q = - 5 . ( g ) p = a , q = - a , a > O .

(d) p = -9 , q = 4. . ( h ) p = a , q = b .

5. Suppose a coordinate system is established on line m , and P and Q are points on m with coordinates p

and q respectively. If P, TI, M, T2, Q are collinear

in that order and represent the midpoint and trisection - points of PQ , find the coordinates of M, T,, T2 in

the following. Record your results for each problem on

a separate number line. (~efer to Theorem 3-6.)

(a) p = 3 , q = 1 2 . (d) p = a , q = b , b > a .

(b) p = - 1 0 , q = - 1 . (e) p = r + a , q = r - a , a < O

(c) p = - 2 , q = 1 3 . (f) p = ( r + b ) - 2 . q = ( r + b ) + 4 .

6 . I n each of t h e fo l lowing problems i n d i c a t e t h e loca t ion o f t h e o b j e c t s l e t t e r e d from A through I-! by -^sing

either a p a i r o r a triple of symbols.

(a) S e a t s i n an auditorium.

~ r a a a a m m a i 2 3 4 5 6 7

SEATS

( L ) Hniisen a t t h e i n t e r s e c t i o n of s t r e e t s and avenues.

( c ) Tables i n rows i n t h e f loor ' s i n a s t o r e .

I TABLES

(d) P o i n t s on t he s u r f a c e of t h e e a r t h .

PRIME MERIDIAN 45' NORTH

20' WEST, 45' EAST

(e ) Using t h e d a t a of P a r t ( d ) , i n d i c a t e t h e p o s i t i o n o f a i r p l a n e s which a r e above each of t h e l i s t e d

p o i n t s . Assume t h e one above A has an e l e v a t i o n of 5000 f t . , and t h a t the e l e v a t i o n of each one from A t o H i s 200 f t . more than t h e preceding one.

8-2. A Coordinate System in a Plane. - --- Suppose that a plane is given and, until further notice,

that all points and lines under consideration lie in this plane.

Suppose further that a unit-pair of points [ A , A ' ) , as discussed in Chapter 3, Is given. All distances are to be

considered as measures of distances relative to this unit-pair.

Let %?and %?be any two perpendicular lines with 0 M

their point of intersection. Let I and J be points in OX and w, respectively, such that 01 = 1 = OJ . There is a - coordinate system on OX with the point 0 as origin and the

point I as unit point. We call this the x-coordinate - system and the coordinate in this system of a point of OX its x-coordinate. Similarly 0 and J are the origin and - unit point of a coordinate system on OY . We call this system

the y-coordinate system and the coordinate in this system of - a point on OY its y-coordinate.

Thus in the diagram I their respective coordinate of both the x- and y-coordinate systems.

The coordinate of point

P is 3 with respect

to the x-coordinate

system and the

coordinate of R with

respect to the y-coordi-

nate system is 2 . Name the coordinates of

points S and T . Is

it necessary to specify

the coordinate system in

each case? Why?

and J are the unit points of

systems. Point 0 Is the origin

- - The !:ke OX i s c a l l e d t h e x -ax i s and OY i s c a l l e d t h e - p-

y-ax i s ; t~ eir poin t oi' i n t e r s e c t i o n , 0 , i s c a l l e d t h e o r i g i n ; -- and t h e p lane determined by them i s c a l l e d t h e xy-plane. -- - Because 0 - 1 ' t h e way t h e s e axes u s u a l l y a r e shown i n p i c t u r e s on a chalkboard, i t i s customary t o c a l l l i n e s p a r a l l e l t o t h e x -ax i s h o r i z o n t a l l i n e s , and l i n e s p a r a l l e l t o t h e y -ax i s v e r t i c a l l i n e s . It i s customary t o t h i n k of I as l y i n g t o t h e r i g h t of t h e o r i g i n and of J a s l y i n g above t h e o r i g i n . This means, then , t h a t t h e p o i n t s on t h e x -ax i s with p o s i t i v e c o o r d i n a t e s l i e t o t h e r i g h t of t h e o r i g i n , w h i l e t h e p o i n t s oc t h e x - a x i s wi th nega t ive coord ina tes l i e t o t h e l e f t of t h e o r i g i n . Where do t h e p o i n t s on t h e y -ax i s w i t h p o s i t i v e c o o r d i n a t e s l i e ? Vhere do t h e p o i n t s on t h e y -ax i s w i t h

nega t ive c o o r d i n a t e s l i e ?

tie a r e now ready t o de f ine a coord ina te system i n the xy-plane which i s determined by t h e x- and t h e y-coordinate systems. We cons ide r a p a r t i c u l a r p o i n t f i rs t . Suppose t h a t Q i s a p o i n t , t h a t t h e v e r t i c a l l i n e through Q c u t s t h e x - a x i s i n t h e p o i n t whose x-coordinate i s 3 , and t h e h o r i - z o n t a l l i n e through Q c u t s t h e y-axis i n t h e p o i n t whose y-coordinate i s 2 . We say i n t h i s case t h a t t h e x-coordinate of Q i s 3 , t h a t t h e y-coordinate of Q i s 2 , and we c a l l t h e o rde red p a i r of numbers ( 3 , 2 ) t h e coord ina tes of Q .

We are now ready for the general case. Let P be any

point in the xy-plane. From our previous work we know that

there is exactly one line through P perpendicular to the

x-axis and exactly one line through P perpendicular to the y-axis. Why? The point P has an x-coordinate and a y-coordinate which we now define. The x-coordinate of P is the - x-coordinate of the projection of P into the x-axis and the - y-coordinate of P is the y-coordinate of the projection of P into the y-axis. We sometimes call the x-coordinate of P and the y-coordinate of P - the coordinates of P . The - - coordinates of the y-coordinate of P are considered an ordered

pair of real numbers in which the x-coordinate Is the first number of the pair and the y-coordinate is the second. If the x-coordinate of P is a and the y-coordinate of P is b , the coordinates of P are written as (a,b) . Note that the

numbers in an ordered pair need not be distinct. Thus (5,5) is an ordered pair of real numbers. Of course (8,3) and

(3,8) are different ordered pairs. In fact, (a,b) = (c,d)

if and only if a = c and b = d .

In the diagrams A is the projection of P into the

x-axis and B is the projection of P into the y-axis. Thus the x-coordinate of P is a and the y-coordinate of P is b . We call the ordered number pair (a,b) the xy-coordinates

of P .

Since the projection of a point into a line is unique, it

follows that there is exactly one ordered pair of real numbers

assigned to each point as its coordinates. Conversely, if

(a,b) is any ordered pair of real numbers there is exactly one point P in the xy-plane which has (a,b) assigned to it as

its coordinates. Indeed, there is a unique vertical line through the point on the x-axis with x-coordinate a , and a unique horizontal line through the point on the y-axis with

y-coordinate b . And P is the unique point in which this

vertical line and this horizontal line intersect. Therefore

there is a one-to-one correspondence between the set of all

points in the xy-plane and the set of all ordered pairs of real

numbers.

Corresponding to any three points 0, I, and J , such -- that 01 1 OJ and 01 = 1 = OJ there is a coordinate system in the xy-plane. This coordinate system is the one-to-one

correspondence which we described above. Although there are

many xy-coordinate systems in a plane, we usually think of

only one of them in a given problem or theorem. Once a

coordinate system has been set up we may use ordered pairs of

real numbers as names for points. The coordinate pair of a

point is a good name for a point in view of the one-to-one

correspondence described above. Tnus we may say that the

point Q has coordinates (-2,4) , or that Q = (-2,4) . Sometimes we simply write Q(-2,4) . Occasionally we use the

symbol xA to denote the x-coordinate of the point A and

the symbol yA to denote the y-coordinate of the point A . Thus (xA,yA) is another name for the point A .

We have used "above," "below," "right," "left," to

describe the position of a point. These words were introduced

for convenience and we can get along without them if we are

cnallenged to do so. Furthermore, there are situations (not

in this book, however) in which it is convenient to take the

positive part of the x-axis as extending to the left, or the

positive part of the y-axis as extending downward, or some

other variation.

In describing the location of a point in the xy-plane it

is sometimes convenient to specify the portion of the plane in - - which it lies. The lines OX and OY form four angles. - * Every point in the plane lies in OX or in OY or in the - interior of one of the four angles whose sides lie on OX and - OY . The interiors of these angles are called quadrants. The

first quadrant is the set of all points whose x- and y-coordi-

nates are both positive. The second quadrant is the set of all

points whose x-coordinate is negative and whose y-coordinate

is positive. The third quadrant is the set of all points

whose x-coordinate and y-coordinate are both negative. The fourth quadrant is the set of all points whose x-coordinate Is

positive and whose y-coordinate is negative. We denote these

quadrants as I, 11, Ill, IV.

Suppose we wish t o describe the locat ion of the point P = (5,-3) without using the words "r ight ," " l e f t , " "above," ''below.'' We might say that P i s i n the four th quadrant, t h a t i t i s i n a v e r t i c a l l i n e which cu t s the x-axis i n a point 5 u n i t s from the or ig in , and t h a t it i s i n a horizontal l i n e which c u t s the y-axis I n a point which i s 3 u n i t s from the or ig in .

I n the following problems we use the words "plot" and II graph." To - p lo t a point means t o draw a picture of the axes and t o mark a dot i n the proper place a s a picture of the point. A name f o r the point i s frequently wr i t ten beside i t s p ic ture . We use the word graph t o mean a s e t of points. To draw ( o r p l o t ) -- a graph i s t o draw a picture which shows the axes and the s e t of points . If there a re i n f i n i t e l y many points i n a s e t , i t s graph i s sometimes drawn by drawing l i n e segments, o r by shading the appropriate region. The picture of a graph always shows the axes, but they a re not a par t of the graph unless it i s so s ta ted . The l abe l X I s placed along the pos i t ive par t of the x-axis; the l abe l Y i s placed along t h e posi t ive par t of the y-axis. It i s usual ly desirable * t o labe l a t l e a s t one point on OX o ther than the or ig in with

i t s x-coordinate, and a t l e a s t one point on T o t h e r than the o r ig in w i t h i t s y-coordinate. I f we wish t o represent a l i n e segment including i t s endpoints, we sometimes emphasize the endpoints as i n the following picture .

8-2

If we wish to represent a set which consists of all points of a line segment except its endpoints, we may de-emphasize the

endpoints as in tne following pictu~e.

If the axes, or a portion of them, are a part of the graph, we may indicate this by making "heavier lines .'I

Example - 1. Plot the points A(-2,0) , ~ ( 3 , - $1 , ~(4,l) . Y

Example - 2 . Draw the graph of the line segment with

endpoints (3,-2) and (3,l) .

Example - 3 . If A = ( 2 , ~ ) , B = ( 0 ~ 2 ) , C = (0,0)

D = (2 ,0) , draw a graph of the s e t of a l l po in t s wnich belong t o t he polygon ABCD o r i t s i n t e r i o r .

Example - 4. D r a w a graph of t he l i n e segment whose endpoints a r e (1 ,0) and (6,0) .

Problem Set 8-2 -- 1. Plot t h e po in t s (1,0) , ( 3 , l ) , (-3,2) , ( - ~ , a ) ,

(49-3-51 , (09-2)

2. Given P = ( 5 ? 6 )

( a ) I f Q i s the point i n which the v e r t i c a l l i n e through P i n t e r s e c t s t h e x-axis, what are the coordinates of Q ?

( b ) I f R i s the point i n which the hor izontal l i n e through P i n t e r s e c t s the y-axis, what a r e t he coordinates of R ?

Without plottingy name the quadrant in which each of the

following points lies: (3y7) (-.Zy-3) (-6y4.3)

(~9-1) Y ( - 4% fl) Explain what I t means to say that our coordinate system

in a plane has established a one-to-one correspondence

between the set of ordered pairs of real numbers and the

set of points in a plane.

If points Py Qy R have coordinates (3y2) ( 3 , 8 ) (3)5) respectivelyy they are collinear in what order?

Describe the set of all points in a plane for which tne

x-coordinate is 3 ; for which the y-coordinate is -5 . Describe tne intersection of these two sets.

Plot the set of all points (xyy) for which x and y

are both integers and x and y satisfy the following

conditions:

(a) x = 2 -1 5 y 5 . How many points belong to this

set?

(b) y = -3 , 2 < x < 6 . How many points belong to this

s e t ?

( c ) -4 < x - < -1 -2 - < y < 3 . This set contains how

many points?

(d) 0 < x 2 -4 < y < 0 . This set contains how many

points?

Plot the set of all points (x,y) in a plane satisfying

the following conditions. Describe each set using

mathematical terms.

(a) If A = (3y0) and B = (7,0) what is the length - of segment AB ? Justify your answer.

(b) If C = (3y4) and D = (7,4) , what is the distance CD ? Justify your answer.

Without plotting, arrange the following poirlts in order

from lowest to hlghest. Ignore the variation in their

dlstance from the y-axis,

Without plotting, arrange the following poincs in order

from left to right. (1gnore the variation in theLr

distance from the x-axis.)

- What is the length of the segment AJ3 , given t.2e coordinates of its endpoints as follows:

Describe the set of all point with coordinates (xpy)

which satisfy the conditions in each of the follo~ing:

(a) x > O , y < O a (b) x is positive, y is non-negative.

(c) x and y are both negative.

( d ) x > - 2 ,

y is any real number.

(e) x is any integer and y is any integer.

(f) x is any real numoer and y is any re& n-mber.

Plot the points listed m d in each part give the

coordinates of the midpoint of the segment determLne6 by

these points.

Give the coordinates of the midpoints of the segnents

whose endpolnts are:

Describe the position of the point (-7,-8) wLthou5 11 using the words "right ," "left ,'I "above, or "beiow. 'I

8-3. &sic Theorems.

Now t h a t a coordinate system f o r a plane has been defined, we may extend our ideas about d i s tance t o include d i s tance between po%nts i n a plane. I n doing t h i s , it i s convenient t o introduce the concept of absolute value of a number.

Suppose A and B a r e d i s t i n c t po in t s on l l n e 1 and t h e i r coordinates a r e a and b r e spec t ive ly , To f i n d AB , tne dis tance between A and B, we take a - b o r b - a , whichever i s pos i t i ve . I f we do not know whether a > b o r b > a , we cannot t e l l which i s pos i t i ve . Therefore t o ind ica te t h a t we wish t o choose the one which i s pos i t i ve , we use the symbol la - bl , read "absolute value of a - b ." O f course we want la - bl = lb - a1 and f o r t h i s reason we make the following d e f i n i t i o n .

DEFINITION. I f a - > b then la - bl = a - and l b - a l = a - b .

Examplel . - If a = 7 and b = 5 , l a - b l If a = 5 and b = 7 , l a - b l

Example 2. Show t h a t i f x > 0 then 1 -x - Solution: 1 - X I = 1 -x - 01 . Since 0 > -x ,

1-x - 01 = 0 - (-x) = x .

Example - 3. Show t h a t 1-6 + 3 1 = 3 . solut ion: 1-6 -t 3 1 = 1-3 1 = 3 . Note how the concept of absolute value s i m p l i f i e s t he

statement of the next theorem.

THEOREM 8-1. I f P and Q a r e po in t s on t h e same v e r t i c a l - l i n e , then PQ = lyp - yQl .

Proof:

1. If P and Q a r e on the y-axis, then t h e theorem Is proved by the use of t he Ruler Pos tu la te .

2. If P and Q a r e , n o t on t h e y-axis, let A and B be the respect ive pro jec t ions of P and Q i n t o the y-axis.

519 C E M R E L - CSMP LIBRARY 103 S. V,:A,S!-I!NSTON ST. - - - CARBONDALE, ILL. 62901

8-2 Then by our definition of y-coordinates we know that

y p = yA and y = yB . Now yA and y are the same as the

B y-coordinates of A and B and therefore AB = lyA - y d . Since ABQP is a parallelogram, it follows that AB = PQ , and hence that PQ = lyp - YQI

THEOREM 8-2. If P and Q are points on the same horizontal - line, then PQ = lxp - xQ1 .

Proof: A proof similar to that for Theorem 8-1 can be given.

THEOREM - 8-3. Every vertical line is perpendicular to every horizontal line.

Proof: his is a special case of Corollary 6-5-2.

We have seen how to use xy-coordinates to measure the distance between two points when those points are on horizontal or vertical lines. We now proceed to develop a method for finding the distance between two points that are on an oblique line. We Introduce the method by means of two examples.

8-3 Example - 1. Find OA i f 0 = (0~0) and A = ( 3 , 4 ) ,

Solution:

Let B be t he OB = 3 and BA

t h e Pythagorean

Example - 2.

pro jec t ion of A i n t o t h e x-axis . Then = 4 . But AOBA is a r i g h t t r i a n g l e . Theorem we ge t

( o A ) ~ = ( o B ) ~ + ( B A ) ~

( 0 ~ ) ~ = 9 + 16 2 (OA) = 25

OA = 5 .

Using

Find PQ if P = (1,-7) and Q = (8,8) . Solution:

8-3

Let R be the point (8, -7) . Then PRQ is a right triangle. P R = I8 - 11 = 7 ; Q R = I8 - (-7)1 = 1 5 ; and

2 (PQ)~ = (PR)~ + (w)~ = 7 + = 49 + 225 = 274 ,

PQ = y ^ T b

We proceed to the theorem, which, once proved, will enable us to find the distance between any two points without reference

to a right triangle. The result of this theorem is often

referred to as the distance formula for points in a plane.

THEOREM - 8-4. ~f P,(x,,Y,) and p2(x2,y2) are two points

in the xy-plane, then

- Proof: Let R = (x,,y,) . If PIP2 is an oblique

segment, then PlRP2 I s a right triangle. Then

8-3 - I f PIP2 i s norizontal , then y, = y2 , R = P, , and

- y2 - y, = 0 . If PIP2 i s v e r t i c a l , then xl = x2 , R =

p2 , and x2 - x., = 0 .

In e i t h e r case the relat ionships

(1) ( P ~ P ~ ) ~ = + ( P ~ R ) ~ and

are s t i l l val id .

Example 1. Find AB if A = (7,15) and B = (7,13) . -

Solution: AB = 1/(7 - 7)2 + (13 - 15) 2=/(-2)2= 2 .

Example 2. Find CD if C = (-1,5) and D = (5,-1) . - i

Example - 3. The ve r t i ces of AABC are A(-1,-2) , 3(4,0) , ~ ( 2 , 5 ) . Prove t h a t AABC Is a r i g h t i sosce les t r iangle .

Proof: We have t o prove (1) A ABC i s Isosceles . (2 ) AABC i s a r i g h t t r i ang le .

We can prove both i f we know AB , BC , CA .

We can see tha t AB = BC and therefore AABC i s isosceles . We can see tha t ( A B ) ~ + ( B C ) ~ = ( c A ) ~ and therefore A ABC

i s a r igh t t r iangle .

THEOREM 0 -5 . If P and Q a r e two p o i n t s i n t n e same

v e r t i c a l l i n e , then t h e midpoint 14 of i s t n e po in t

Proof: Since P, Q, and M l i e I n t n e same v e r t i c a l

l i n e xp = x~ = x~ Let A , B, and C be t h e r e s p e c t i v e p r o j e c t i o n s of P, Q, and M i n t o t h e y -ax i s . Then

yp = yA , yQ = yB and y = y . Since M i s t h e midpoint M - of PQ i t fo l lows from Theorem 7-2 t h a t C i s t h e midpoint - of AB . It then fo l lows from t h e d e f i n i t i o n of a midpoint

YA + YB y~ + YQ t h a t yc = 7 . Therefore yM = Ã‘Ã‘ and

THEOREM - 8-6. If P and Q a r e two p o i n t s on t h e same h o r i z o n t a l l i n e , then t h e midpoint M of 7^ i s t h e

p o i n t

Proof: A proof similar t o t h a t f o r Theorem 8-5 may be

g iven.

THEOREM - 8-7. If P and Q a r e d i s t i n c t po in t s on a l i n e which i s n e i t h e r v e r t i c a l nor horizontal, then t h e mi.dpoint

M of i s the point

Proof: Let A, B, C be the respec t ive pro jec t ions of

P, Q, M i n t o the y-axis. Let D, E, F be the respec t ive project ions of P, Q, M i n t o t he x-axis. It follows from - Theorem 7-2 that C i s t he midpoint of AB and F i s the midpoint of . It follows from the d e f i n i t i o n of a midpoint

t h a t

y~ + YB yc = and xF x~ + Ê -7

Since yA = yp and yB = y then y., = YP + YQ 2

Since xD = xp and xE = xQ then % = x~ + X~ 7.

Theref o re

Since I n Theorem 8-5 xp = xp + Q when xp 2 = x~ and i n

YP + YQ Theorem 8-6 yp = Ã‘Ã‘x when Yp = YQ 3 we may combine t h e

r e s u l t s of t h e t h r e e preceding theorems as follows:

8-3 THEOREM 8-8. If P = (x , ,~ , ) and Q = (x2,y2) a r e any two -

d i s t i n c t p o i n t s i n a p lane , t h e n t h e midpoint M of ?<S i s t h e p o i n t

Example - 1. If P = ( 3 , 8 ) and Q = (7,4) , f i n d t h e - midpoint of PQ .

- Solut ion: Let M be t h e midpoint of PQ .

Example - 2. If R = ( 2 , 5 ) and S = (-5,-3) , f i n d t n e

midpoint M of .

Solut ion:

Example - 3 . If A = (0 ,0 ) , B = (0,6) and C = (8 ,0) , f i n d t h e l e n g t h o f t h e median from A t o .

- Solut ion: Let D be t h e midpoint of BC . Then

8-3

THEOREM 8-9. Let a be any real number. Then the set of all - points in the xy-plane each of which has x-coordinate

a Is a vertical line.

Proof: Let m be the vertical line which cuts the x-axis at the point A whose x-coordinate is a . We must establish two statements:

1. If a point lies in m , then its x-coordinate is a . 2. If a point has x-coordinate a , then it lies in rn .

The first of these statements follows immediately from the definition of x-coordinate. The second statement is proved indirectly. Suppose, contrary to Statement (2), that there is a point P whose x-coordinate is a and which is not in m . Then the vertical line through P contains A . Then we have - two vertical lines containing A : m and PA . But this is impossible. Why? Therefore every point with x-coordinate a

lies in m . THEOREM 8-10. Let b be any real number. The set of all

points in the xy-plane with y-coordinate b is a horizontal line.

Proof: A proof similar to that for Theorem 8-9 can be given.

8-3 Problem Se t 8-3 --

1. Use t h e d i s t a n c e formula t o f i n d t h e d i s t a n c e between

( a ) (o,o) and (6,10) . ( g ) ( 1 0 ~ 1 ) and (49,811 . ( b ) (0 ,0 ) and (-6,10) . ( h ) (-6,3) and (4 , -2 ) . ( c ) (1,2) and (6,14) . (1) (3$,4.) and (-+,o) 1 . ( d ) (8,11) and (15,35) . ( j ) ( 0 , 3 ) and (-;1,0) . ( e ) (3 ,8 ) and ( - 5 , - 7 ) . ( k ) (8.1,6) and (5.9,4.9) . ( f ) ( -2 ,3) and ( - 1 , h ) . (1) (37r.r) and (-27r,-7r).

- 2. Find t h e midpoint of AB If A and B have t h e

r e s p e c t i v e c o o r d i n a t e s

3 . ( a ) Write a formula f o r t h e square of t h e d i s t ance

between (x^ ,Y^) and (x*,y2)

( b ) Write t h e fo l lowing s ta tement a s a n equat ion: The square o f t h e d i s t a n c e between ( 0 , 0 ) and (x,y)

i s 25 . 4. Show t h a t t h e t r i a n g l e s wi th v e r t i c e s a s given a r e r i g h t

t r i a n g l e s . Use t h e d i s t a n c e formula t o f i n d t h e . l e n g t h s of t h e s i d e s of each t r i a n g l e .

5. The v e r t i c e s of a q u a d r i l a t e r a l are ~ ( 0 , 0 ) , ~ ( 5 , 0 ) , ~ ( 5 , 4 ) and D ( o , ~ ) . (a ) Show t h a t AC = BD .

- ( b ) Show t h a t t h e midpoint of AC and t h e midpoint of

- BD i s t h e same p o i n t .

The vertices of a triangle are A(-2,l) , ~ ( 0 , 5 ) and -

~ ( 2 , - 1 ) . Find the midpoint of BC . Find the length of - the median to BC . The vertices of a triangle are R(-4,-2) , ~ ( 5 , 1 0 ) and

- ~ ( 4 , - 2 ) . Find the lengths of the medians to ST and - R S . Find the coordinates of the midpoint C of if

A = (-1,0) and B = (7A) . Then use the distance 1 formula to verify that AC = CB = -p-AB .

(a) Show that ~(6,ii) is equally distant from B(-l,2)

and ~(3,0) . (b) Show that two of the medians in AABC are equal in

length.

Use the distance formula to show that ~(0,2) , ~(\8) and c(6,11) are collinear. (~int: - Show that AB + EC = AC .)

If the distance between ~ ( 6 , - 2 ) and ~(0,y) is 10 , find the possible y-coordinates of F . Find the coordinates of the points on the x-axis whose

distance from (1,6) is 10 . Using the distance formulas, prove that AD = BC if

A = (0,0) , D = (b,c) , B = (a,0) , and C = (a + b, c ) . The vertices of a square RSTP are ~(a,a) , s(-a,a) , T(-a,-a) , ~(a,-a) . Show that its diagonals are

congruent.

8-4 15. There are two coordinate systems in this diagram. One has

axes labeled X and Y . The other has axes labeled XI

and Y1 . All axes have the same scale. Estimate the

coordinates of P and of Q in the xy-system and then

calculate the length of . Then estimate the

coordinates of P and Q in the xlyl-system and again -

calculate the length of PQ . Do you think that the length of is independent of the choice of axes?

8-4. The Set-Builder Notation. -- In our discussion of sets in Chapter 2 we considered the

set of all positive Integers. The underlined phrase clearly -7-

defines a certain set of numbers. In general, a set is

defined by a list of its elements or by a property of its elements. If a set has an infinite number of elements, we

cannot list all of its members so we use a property or

properties of its elements to define it.

Consider the following property of a real number: between 3 and 5 . Some real numbers which have this - - - property are 3.5 , 4 , 4 . 3 , 3.1 , h.9 , 3.001 , and 4.999 .

Some r e a l numbers which do n o t have t h i s p roper ty a r e 3 , 5 , -7 j 46 , 0 , - 4 . 3 , 5.7 , 5.001 , and 2 . The set of a l l r e a l numbers between 3 and 5 i s a c l e a r l y de f ined set. A

symbol which denotes t h i s s e t i s ( x : 3 < x < 51 . We read

It: t h e s e t of a l l x such t h a t x i s between 3 and 5 . There a r e t h r e e p a r t s wi th in t h e braces : t h e s e t of a l l of

something, before t h e colon; t h e colon, which i s r e a d "such

t h a t ; " and a s t a t e d p roper ty a f t e r t i le colon.

Consider next t h e fo l lowing p r o p e r t y of a p o i n t (x,y) :

i t s x-coordinate i s 3 , and i t s y-coordinate i s a r e a l number. - - - - --- - - - Some p o i n t s i n t h i s s e t are (3,5) , (3,-137) , ( 3 , 0 ) , (3,105732.4) , and (3 ,3) . Some p o i n t s not i n t h i s s e t a r e ( 5 , 3 ) , (4,0) , (-7,2) , A symbol which denotes t h i s set I s

((x,y) : x = 3 ) . We read it: t h e s e t of a l l p o i n t s (x,y)

such t h a t x = 3 . Frequently, as i n t h i s example, we under-

s tand t h a t x and y a r e r e a l numbers even i f i t i s n o t

i n d i c a t e d i n t h e symbol.

I n genera l , t h e symbol ( a : p r o p e r t y ) , which we c a l l

t h e - s e t - o u i l d e r symbol o r n o t a t i o n , denotes t h e s e t of a l l

elements a each of which has t h e s t a t e d p roper ty .

Example - 1. Use a s e t - b u i l d e r symbol t o denote t h e s e t of

a l l p o i n t s I n t h e f irst quadrant .

Solut ion: (P : P i s a p o i n t i n Quadrant I ] . A l t e r n a t e s o l u t i o n : [ ( x , ~ ) : x > 0 and Y > 01

Example - 2. Use a s e t - b u i l d e r symbol t o denote t h e s e t of

a l l p o i n t s whose x-coordinate i s '{ and whose y-coordinate

i s a number g r e a t e r than 5 . Solut ion: [ ( x , y ) : x = 7 and y > :)} . Alte rna te s o l u t i o n : ( ( 7 , ~ ) : Y > 5)

Example - 3. Plot the graph of the set {(x,y) : 3 < y < 5) . The set includes all points in the

infinite strip between the two horizontal lines and on these

lines.

^

Example - h . Plot the set {(x,y) : x < 1 or x 3 } . The set contains all points in the two halfplanes which are

suggested by shading, and all points in the edge of one of

these halfplanes.

8-5. Composite Conditions.

In Example 2 above, the set ((x,y) : x = 7 and y > 5) was discussed and in Example 4 , the set ((x,y) : x < 1 or x > 3) . In the set-builder notation used to indicate these sets we note that the property listed In each

case is - a composite condition, that is to say, a combination of conditions. In Example 2 the conditions x = 7 , y > 5 are

connected by the word "and"; in Example b the conditions

x < 1 , x > 3 are connected by the word "or. II

Our purpose is to illustrate briefly how these composite

conditions should be interpreted in so far as they are used in

our work. Let c-, be the statement x = 7 and c2 the

statement y > 5 . Let S, = ( (x,Y) : c.1 9

s2 = ((x,y) : c2) and S3 = (.(x,y) : c1 @ c21

As shown in the diagrams above S-, is a vertical line

and S2 is a halfplane. Do you see the special relationship

which S3 has to S., and Sp ? s3 ' the interior of a ray, is the intersection of Sl and S2 . Now let c3 be the

statement x < 1 and c L the statement x > 3 . Let

1 = ((x,~) : c3) , T2 = {(x,~) : and

T3 = ((x,~) : c3 ~ 1 , )

Describe T-, ; Ty . Now d e s c r i b e T3 . Notice t h a t i n t h i s case T3 i s t h e union of T- and T2 .

1

I n g e n e r a l when working wi th t h e s e t - b u i l d e r n o t a t i o n you should remember t h e fol lowing:

(1) A s e t whose definlnp- p roper ty i s a composite condi t ion

us ing t h e connect ive "and" can be cons idered t h e I n t e r s e c t i o n - o? the s e t s determined by t h e i n d i v i d u a l c o n d i t i o n s of t h e composite.

( 2 ) A set whose d e f i n i n g p roper ty i s a composite condi t ion II us ing t::e connect ive - or" can be considered t h e union of t h e

s e t s determined by t h e i n d i v i d u a l cond i t ions of t h e composite.

Example - 1. What a r e the p o i n t s i n t h e s e t

T h i s i s t h e s e t of a l l p o i n t s (x,y) such t h a t x > 0 , o r y > 0 , o r both x > 0 , y > 0 . The s e t c o n t a i n s a l l p o i n t s i n t h e p lane except those i n t h e t h i r d quadrant , i n t h e nega t ive x -ax i s , i n t h e nega t ive y -ax i s , and tiie o r i g i n . Note t h a t the graph i s t h e union of t h e graph of x > 0 and t h e graph of y > 0 .

This is the set of all points inside and on the

rectangle whose vertices are (3,l) , (3,4) , ( 5 , h ) , and

( 5 9 1 )

When dealing with sets defined by a composite condition

using the connective "and," a comma often is used in place of the word "and." Thus the set ((x,y) : x = 5 , y > 3 } is

understood to be the same as the set ( ( x , y ) : x = 5 and y > 3) .

Problem Set 8-5 -- 1. Ident i fy the quadrant w i t h the proper Roman numeral.

( a ) ( (x ,y) : x > 0 and y < 01 i s Quadrant ( b ) ( (x ,y ) : x < 0 and y < 01 i s Quadrant

( c ) {(x ,y) : x < 0 and y > 01 i s Quadrant (d) ( (x ,y) : x > 0 and y > 0 ) i s Quadrant ( e ) ( (x ,y ) : y > 0 and x < 0) i s Quadrant ( f ) [ (x ,y) : xy > 01 i s the union of Quadrants

and (g) [ (x ,y) : xy < O} i s the union of Quadrants

and

X Y - ^ O ) i s x > 0 and 1 - > 0 and x x < 0 and

the union of

y # 0) i s the union of 1 - < 0) i s Quadrant Y y = 1 x 1 } i s Quadrant

2. Find the coordinates of the endpoints of a l l possible segments which s a t i s f y the given conditions.

- ( a ) AB l i e s on the y-axis with the o r ig in a t i t s

midpoint; AB = 7 . (b ) i s a subset of the x-axis; A, 0, B a re col l inear

i n t h a t order and A0 = +B ; AB = I 2 . ( c ) .m is e i t h e r horizontal or ve r t i ca l ; A Is a t the

or ig in ; AB = r . ( d ) 1 I t o the x-axis and i s 5 u n i t s above the -

x-axis; the y-axis b i sec t s AB ; AB = 8 . ( e ) AB = 5 ; A l i e s on the x-axis; B l i e s on t h e

y-axis; OA = OB . ( f ) i s i n the y-axis; A i s a t the or igin; AB = 6 ; -

CD I I ; CD = AB ; C i s 2 u n i t s above A and 3 u n i t s t o the r igh t of A .

3. Find the coordinates of the ve r t i ces of the indicated polygons :

( a ) A coordinate system places isosceles t r i ang le ABC - - so tha t the or ig in i s the midpoint of base AB , AB

i s a subset o f the x-axis, C l i e s above the x-axis, AB = 6 , OC = 4 .

(b) An isosceles triangle ABC whose altitude is 3

and whose base has length 5 . Tie base Is a subset

of the y-axis, and the opposite vertex, C , is on the positive x-axis.

(c) An isosceles triangle ABC has AB = 6 , AC = BC = 5 . The origin Is at the midpoint of the

base, the x-axis contains the base, and C is above

the x-axis.

(d) A parallelogram ABCD for which AB = 7 , A and

D nave coordinates ( 0 ,0 ) and ( 3 , 5 ) respectively,

and is on the x-axis.

4. In each of the following find the coordinates of the

vertices of the polygon:

(a) A right triangle ABC has /C a right angle,

CA = 21 and CB = 10 . A coordinate system places

C at the origin and B in the negative x-axis. - (b) An isosceles triangle A B C i-ias base -43 of length

4 and altitude to A3 of length 3 . A coordinate system places A at the origin and B in the

positive x-axis.

(c) Same as (b), except that C is at the origin. A - Is in Quadrant I, and AB is perpendicular to the

x-axis.

(d) An equilateral triangle A X has side of length 10 . A coordinate system is established with the x-axis - containing AB and the positive y-axis containing

c . 5. Find the coordinates of the vertices of the polygon

determined in each of the following:

(a) A right triangle ABC has /C a right angle,

CA = a and CB = b . A coordinate system places C

at the origin, B in the negative x-axis, and A in

the positive y-axis.

- An isosceles triangle ABC has base AB of length

b , and altitude of length a . A coordinate system

places A at the origin, B in the positive x-axis

and C above the x-axis.

The triangle in (c) is placed so that C is at the

origin and the altitude lies along the positive

x-axis.

An equilateral triangle has side of length s and

a coordinate system is established so that one side

lies along the x-axis and the opposite vertex is in

the positive y-axis.

8-6. Equations - and Inequalities.

[(x,y) : x = 3} is a line. It contains all those points and only those points in the xy-plane whose x-coordinate is 3

and whose y-coordinate is any real number. We say that x = 3

is an equation of the line or think of x = 3 as a condition

imposed upon (x,y) . The condition x = 3 places a

restriction on the x-coordinate but no restriction on the

y-coordinate. Thu's the line ((x,y) : x = 31 contains

infinitely many points, such as (3,-173.447) , (3,-2) , (3,-1) , (3,0) , (3,25) , (3,127.3) . Of course there are

infinitely many points - not on this line, such as (2,3) , (2-9999-7) (09-3)

( ( x , y ) : x > 3) is a halfplane. We say that x > 3 is an inequality for the halfplane. This halfplane contains all

those points and only those points in the xy-plane whose

x-coordinate is a real number greater than 3 and whose

y-coordinate is a real number. Is (5,-5) in this halfplane? Is (-5,5) in this halfplane? Is (3,3) in this halfplane?

What set-builder notation could you use for the edge of this

halfplane?

In some textbooks the set ((x,y) : x = 3) is called the

locus of the equation x = 3 ; ((x,y) : x > 3) is called the

locus of the inequality x > 3 . In general a locus is a set determined by a condition or a combination of conditions.

8-6

Tnus t h e locus o f a l l p o i n t s i n t h e xy-plane which a r e equl - d i s t a n t from t h e l i n e s [ ( x , y ) : y = - 7 ) and [ ( x , y ) : y = 13) i s t h e l i n e [ ( x , y ) : y = 3 ) . I n our t e x t however we w i l l use t h e express ion " the s e t of a l l p o i n t s such t h a t " r a t h e r than t h e term locus .

Consider t h e s e t s S and T g iven a s fo l lows: S = { ( x , y ) : x = 5) and T = ( ( x , y ) : x + 1 = 6 ) . A p o i n t (x ,y) s a t i s f i e s t h e cond i t ion x = 5 i f and on ly i f i t

s a t i s f i e s t h e cond i t ion x + 1 = 6 . I n o t h e r words, 1C ( a , b ) i s a po in t i n T , it i s a l s o a p o i n t i n S , and converse ly . Hence we may w r i t e S = T . The equa t ion S = T i s an equa t ion involving s e t s (of p o i n t s i n t h e xy-plane) and you should r e c a l l t h a t two s e t s a r e equal i f and on ly i f they nave exactly t h e

same members. This occu.rs i f t h e s e t s a r e def ined by

equ iva len t equat ions . The equa t ions x = 5 and x + 1 = 6 a r e examples of equ iva len t equat ions . I n a l g e b r a you l ea rned how t o d e r i v e equ iva len t equa t ions i n t h e p rocess of s o l v i n g equat ions . Thus

2x + 3 = kx + 13 3 = 2x + 13

-10 = 2x 2x = -10

x = -5

a r e f i v e equ iva len t equa t ions . Each of them becomes a trt-e sentence if x i s rep laced by -5 ; each of them becomes 2 f a l s e sentence i f x i s rep laced by any number d i f f e r e n t f ro i i

Problem Set 8-6 -- 1. Find t h e d i s t a n c e between l i n e s ( (x ,y ) : x = l and

[ ( x , y ) : x = - 4 1 . 2. Write t h e coord ina tes of t h r e e p o i n t s on t h e l i n e

( (x ,Y) : Y = 31

Plo t t he s e t

(a) ( ( x J Y ) : x = 2 ,

(b ) ( ( ~ J Y ) : Y = 31

Describe t h e union o f t h e two s e t s i n Problem 3 . May t h i s be wr i t t en as the s e t ( ( x , y ) : x = 2 - o r y = 31 ?

What i s the i n t e r s e c t i o n of t he two s e t s i n Problem 3? May t h i s be wr i t t en a s the s e t ( ( x , y ) : x = 2 and y = 3 ) ? - A s t h e s e t ( ( x , y ) : x = 2 , y = 3 } ?

( a ) P lo t t he s e t [ ( x , y ) : x = 2 and 0 < y < 3 ) . (b) What geometric ob jec t does t h i s s e t form? ( c ) How many elements does i t contain?

P lo t t he s e t of po in t s whose coordinates a r e giver1 u?iov;

and descr ibe the graph i n each case .

P lo t and descr ibe each of t he graphs given below:

( a ) The union of ( ( x , y ) : x > 3 ) and ( (x ,y ) : y 5 3 ) . Express t h i s union w i t h one se t -bu i lde r symbol.

(b ) The i n t e r s e c t i o n of ( (x ,y ) : x - < 2 ) and ( ( x J y ) : y > -21 . Express t h i s s e t with one s e t - bu i lder symbol.

( c ) The s e t ( ( x , y ) : x > 0 and y < 3 } . ( d ) The s e t ( ( x , y ) : -4 < x - < 2 and -2 < y < 5) . Express i n s e t -bu i lde r no ta t ion the s e t of a l l points i n t h e xy-plane which s a t i s f y t he following condit ions:

A s e t of po in t s a t a d i s tance of 5 from the l i n e whose equation i s y = 2 . A s e t of po in t s 4 u n i t s from the y-axis. A s e t of po in t s 3 u n i t s above the x-axis and.. 5 u n i t s t o t he l e f t of the y-axis. A s e t of po in t s t he same d is tance from the point ~ ( 0 , 3 ) a s from the point ~ ( 0 , - 3 ) . If P i s any point i n t h i s s e t , prove PA = PB . If P i s any po in t such t h a t PA = PB , prove P I s i n t h i s s e t .

10. Which of the following are pairs of equal sets?

(a) [(x,y) : 3x + 6 = x + 81 and ((x,y) : 2x = 21

(b) [(x,y) : x + 3 > 71 and {(x,~) : x > 41 (c) [(x,y) : 5x - 2 < 2x + 4 ) and [(x,~) : 7x < 21) . (d) ((x,y) : -2x + 4 < 8) and ((x,~) : x < 21 .

6 (e) { x : - 2 3 ) and (x: 6 2 3 x 1 . *11. Answer the questions indicated by filling in t h e blanks

in each of the following: -4-D-

(a) If OC, NB, MA are

parallel and AB = 3 ;

E = 6 ; M N = 2 .

Then BC = k AB ; k =

N O = t M N ; t = - AC = k1 AB ; kt = - and MO = - MN .

- - If A A f 1 1 BB then AP = k AB Alp1 = - AtB1 and AtPt = -

(d) If a coordinate system on assigns the

coordinates 0, 2, 8 to A, B, P r e s p e c t i v e l y ,

Alp1 = ? A'B' .

( e ) The (A,B)-coordinate

system on l i n e .l?

a s s i g n s coordinate k

t o p o i n t P . AP = k AB . Then

Alp1 = k AIB' = k 14 - Wny ?

OP' = 9 .

x = ?

(f) AP = k AB . = k 15 - 31 . Why?

o p t = ? .

8-7 ( g ) A coord ina te system on l i n e /C a s s i g n s coord ina tes

0, 1, k t o A , B, P r e s p e c t i v e l y . Where does P

l i e I f k has va lues a s i n d i c a t e d below?

8-7. Finding t h e Coordinates of t h e Po in t s of a Line. - -- --- We have seen t h a t ( ( x , y ) : x = 3 ) i s a v e r t i c a l l i n e .

It i s understood t h a t y m y be any r e a l number.

It I s n a t u r a l t o a s k i f t h e r e i s an express ion something l i k e t i i s f o r an ob l ique l i n e . Actual ly t h e r e Is, and it i s a u s e f u l t o o l i n geometry.

To show how t o f i n d such an express ion we cons ide r a - p a r t i c u l a r l i n e , t h e l i n e AB where A = (1,2) and B = (3,5) .

4-e Since t h e l i n e AB i s determined by t h e p o i n t s A and B , i t seems reasonable t h a t we should be a b l e t o f i n d t h e - coord ina tes of o t h e r p o i n t s i n AB . For example, i f P i s

+ I n AB and i f AP = 2AB we should be a b l e t o f i n d t h e coord ina tes of P .

d 1 If Q is in A 3 and AQ = AB , we shou-ld be able to find

--w the coordinates of Q . If R is in the ray opposite to AB

and if A 9 = AB , we should be able to find the coordinates of 1 R . Actually, P = (5,8) , Q = (2, 3^) , R = (-I.,-!) . Me

can get these coordinates by working an individual problem for

each point. But our objective here is to derive an expression

from which the coordinates of P, Q, R or for that matter any M

other point on AB can be obtained by simple replacements.

In Chapter 3 we studied a coordinate system on a line. At the beginning of the present chapter we defined an xy-coordinate system in terms of two coordinate systems on

lines: the x-coordinate system on the x-axis and the

y-coordinate system on the y-axis. We wish now to consider a - coordinate system on the line AB . We call It the

(A,B)-coordinate system. In this coordinate system, the coordinate of A is 0 and the coordinate of B is 1 . For

the points A, 3, P, Q, R we have coordinates as tabulated.

The expression which we shall derive shows us how to compute

the x- and y-coordinates of a point in terms of Its

(A, B) -coordinates.

x-coordinate y-coordinate (A,B)-coordinate

4 mx - The (A,B)-coordinate system on AB established a one-to-

A

B

P

Q

R

S

one correspondence between the set of all real numbers k and +-e

the set of all points In AB . If k > 0 , the corresponding +

point is in AB (but not A itself); if k = 0 , the corre-

1

3

5

2

- 1

9

sponding point is A ; and If k < 0 , the corresponding point + is in the ray opposite to AB .

Let k be any real number and ~(x,y) the corresponding

2

5

8 1 ^

-1

9

- point in AB . Then AP = k AB if k - > 0 , AP = -k AB if

0

1

2

1 - 2

- 1

k

k < 0 . Let PI, At, B1 be the respective projections of P, A, 3 into the x-axis. Let P", A", B" be the respective

projections of P, A, B into the y-axis. (if P Is in the

x-axis, then P = P1 ; if P is in the y-axis, then P = P" . ) From Theorem 7-3 it follows that the segments formed by A',

Bt , P* on the x-axis and the segments formed by A", B", P", on the y-axis are proportional to the corresponding segments

u in the line AB . Therefore

(1) A t P t = k A'B1

x - 11 = k I 3 - 11

Since x > 1 - X - 1 = 2 k

x = 1 + 2 k

(2 ) ~ I l p ~ ' = k A ~ ~ B ~ ~

I Y - 21 = k I 5 - 21

Since y > 2 4

y - 2 = 3 k

y = 2 + 3 k

- -

AP = -k AB

(I) A'pl = -k AfB'

x - 11 = -kI3 - 11

Since x - < 1

-X 4- 1 = -2k

X = 1 + 2 k

PI A"P" = -k B"

Y - 21 = - k l 5 - 21

Since y - < 2

-y + 2 = -3k

y = 2 + 3 k

It follows that P = (1 + 2k, 2 + 3k) and that M AB = ((x,y) : x = 1 + 2k, y = 2 + 3k, k is real) .

The equations x = 1 + 2k, y = 2 + 3k are called - parametric equations for the line AB ; the symbol k is

called the parameter. Each value of the parameter yields

exactly one point on the line, the point (1 + 2k, 2 + 3k) . The value of k is the (A,B)-coordinate of the point; it tells

--w us that the point is in AB if k > 0 , in the ray opposite to + - AB if k - < 0 , and that P is k times as far from A

B is. The following table shows several values of k and

their corresponding points.

If we think of ~ ( 1 , 2 ) a s (x,,yl) and ~ ( 3 , 5 ) a s 4--w

(x2,y2 ) , then the parametric equations f o r AB can be wr i t ten

Note how these formulas resemble tha t of Theorem 3-6.

Are these formulas t r u e f o r any oblique l i n e determined by two points (xl,y, ) and (x2,y2) ? Although we could prove t h a t they a re by constructing a proof s imilar t o tha t M

f o r AB i n the above i l l u s t r a t i o n , we s h a l l not wri te it out

here. It i s na tura l t o ask whether we can wri te parametric equations f o r horizontal and v e r t i c a l l i n e s . You w i l l f i nd tha t we can i n the next problem s e t . These r e s u l t s a re con- solidated i n the following theorem.

THEOREM 8-11. If P (x, ,yl) and P2(x2,y2) a re any two

points, then

PlP2 = Hx,Y) : x = x1 + k(x2 - xl) , Y = y1 + k(y2 - yl),

k Is r e a l ) .

According t o Theorem 8-11 every l i n e I n the xy-plane can be "represented" by a p a i r of parametric equations. A na tura l question is: Does every p a i r of parametric equations represent some l ine? The answer t o t h i s question i s no. Consider, f o r example, the s e t

S = ((x,y) : x = 1 + kb0, y = 2 + k . 0 , k i s r e a l )

It i s easy t o see tha t x = 1 and y = 2 f o r every value of k and hence tha t S i s a s e t whose only element i s the

point (1,2) . However there i s a method of ident i fying those parametric

equations which do represent a l i n e i n a plane. We s t a t e it

as our next theorem.

THEOREM 8-12. If a, b, c, d are real numbers such that b

and d are not both zero and if

S = ((x,y) : x = a + bk, y = c + dk, k is real) , then S isaline.

Proof: Taking k = 0 and k = 1 we get two points in S : A(a,c) and ~ ( a + b, c + d) . From Theorem 8-11 it follows that:

4-b AB = [(x,y) : x = a + bk, y = c + dk, k is real); -

therefore A . = S and S is a line.

Example.

If T = ((x,y) : x = -2 + 3k, y = 7 + 2k, k is real) , then T is a line.

Proof: If k = 0 , then x = -2 , y = 7 . 1f k = 1 , then x = 1 , Y s 9 . Thus A(-2,7) , and B(l,g) are two points in T . Using Theorem 8-11 we get -

AB = ((x,y) : x = -2 + 3k, y = 7 + 2k, k is real) . -- -

Therefore AB = T , and T is a line.

We can use parametric equations for a line in expressing the coordinates of the points of a line segment or a ray. If

k.,, kp, kg correspond to Pi, Po, Po , respectively, then kg is between k, and k, if and only if P2 is between P, and P., . This follows from the properties of coordinate

systems on a line as discussed in Chapter 3. Thus we get segments or rays simply by restricting the values which k - may have. For example, AB , where A = (1,2) and B = (3,5) is ((x,y) : x = 1 + 2k, y = 2 + 3k, 0 2 k 2 1) . Similarly, - AB is ((x,y) : x = 1 + 2k, y = 2 + 3k, k - > 01 .

Example - 1.

Given A = (3,0) , B = (-1,2) . Using Theorem 8-11 express, using se t -bu i lde r no ta t ion ,

+ (d ) the ray opposite AB ; a l s o f i n d - ( e ) the midpoint of AB , and

( f ) the point P such t h a t A i s between P and B

and PA = AB . Solution: I n t h i s problem we take x, = 3 , y-, = 0 ,

x2 = -1 , y2 = 2 . Thus x2 - xl = -4 , y2 - yl = 2 . Then - ( a ) AB = [ (x ,y) : x = 3 - 4k , y = 2k , k i s r e a l ) . ( b ) n= [ ( x , y ) : x = 3 - 4k , Y = 2 k , 0 2 k < 1)

--b ( c ) A B = { ( x , y ) : x = 3 - 4 k , y = 2 k , k > 0 )

+ (d ) the ray opposite AB Is

[ (x ,y) : x = 3 - 4k , y = 2k , k < 0) 1

(e ) the midpoint of' ( s e t k = Â¥, i s ( 3 - 4 . - 2 ?) 1 = ( 1 , l ) . 2 '

( f ) the point P ( s e t k = -1) is (7 , -2) .

Example - 2.

Given A = ( 0 ~ 4 ) and B = (3,0) . Find the point C on - AB whose x-coordinate i s -2 .

Solution: I n t h i s problem we take xl = 0 , yl = 4 , x2 = 3 , y2 = 0 . Then x2 - x, = 3 , y2 - yl = -4 and -

A B = {(x ,y) : x = 3 k , y = 4 - 4k , k I s real) . We s e t x = -2 . Then -2 = 3k , k = - 7 , and

2 8 20 2 y = 4 - 4 ( - WJ) = 4 + 7 = -,-= 6,r . Therefore C = ( - 2 , g ) .

In Example 1, Part (e), we found the midpoint of a particular segment . The midpoint of is obtained by

1 J. setting k = 5 in the parametric equations. In general

and therefore the midpoint is

Notice that this result checks with the result derived in Theorem 8-8.

Problem Set 8-7 -- 1. Using parametric equations and set-builder notation -- + +

express AB , AB , AB , and the ray opposite to AB if

(a) A = (1,4) , B = (2,6) . (b) A = (-1,3) , B = (2,0) . (c) A = (0,0) , B = (3,2) . (d) A = (1,l) , B = (4,4) . (e) A = (-1,3) , B = (1,-2) . (f) A = (-3,-2) , B = (0,l) . (63) A = (a,b) , B = (c,d) 9 c + a

(h) A = (a,2a) , B = (3a,4a) , a # 0 . -

2. Find the coordinates of the midpoint of AB In Problem l(a) ; l(b) .

3. Using the midpoint formula find the coordinates of the midpoint of the segment with the given endpoints.

4. I n each of t he following the endpoint and midpoint of a segment a r e given i n t h a t order . Find the coordinates of the o ther endpoint.

- 5. Find the coordinates of the t r i s e c t i o n point of AB

nearer A i n Problem l ( c ) ; l ( d ) . -

6. Find the coordinates of t h e t r i s e c t i o n point of AB

nearer B i n Problem l ( e ) . --b 7. Find the coordinates of P i n AB i n Problem l ( b )

such t h a t

( a ) AP = 2AB . (c) AP = < / 3 ' A B . (b) AP = 100AB . (d) AP = TT AB .

* 8. Find the coordinates of P i n the ray opposite t o AB

I n Problem l ( e ) i f

(a ) AP = 2AB . (b) AP = 20AB .

( c ) AP = 3 . 5 AB . 1 ( d ) AP = AB - 9 . Find the coordinates of P i n AB i f A = ( -1 ,5) and

B = (3,-2) , and

( a ) AP = 3PB . (b) BP = 4PA .

( c ) BA = $ BP . (d) PA = 5BA . - 10. ( a ) Let C = (-1,2) , D = (5,2) . Is CD v e r t i c a l ,

hor izonta l , o r oblique? Use Theorem 8-U t o express - CD . Try th ree d i f f e r e n t values of k t o see i f - the th ree po in t s a r e on CD .

(b) Using C = (xl ,a) , D = (x2,a) , xl # x2 , show t h a t Theorem 8-111s t r u e f o r hor izonta l l i n e s .

4 Using E = (a,yl) , F = (a,y2) , Y, # y2 , show t h a t Theorem 8- is t r u e f o r v e r t i c a l l i n e s .

11. Using parametric equations and se t -bu i lde r no ta t ion express the s ides of t h e t r i a n g l e whose v e r t i c e s a re :

12. Draw the graph:

(a) ((x,y) : x = 1 + 2k , y = 2 - k , k is real] . (b) ((x,y) : x = 2k , Y = k , 0 < k < 21 (c) ((x,y) : x = -1 + k , y = - k , k > 0 ) . (d) ((x,y) : x = k , y = -k , k < 0) . (e) ((x,y) : x = 3 , y = k , -2 < k < 3)

* 13. Given A = (-1,3) , B = (2,-2) , and C is on AB ,

(a) find the y-coordinate of C if its x-coordinate

is 5 . (b) find the x-coordinate of C if its y-coordinate

is 8 . (c) find the y-coordinate of C if its x-coordinate

is 29 . (d) find the coordinate of C if it is on the x-axis.

(e) find the coordinate of C if it is on the y-axis.

14. The vertices of a triangle are A(O,O) , ~ ( 9 , 0 ) , ~(3,6) . Find the coordinates of D , the midpoint of AB' ; E , - the midpoint of ; and F , the midpoint of CA . Show

that a trisection point of each median of triangle ABC is ~ ( 4 , 2 ) .

15. Given p = ((x,y) : x = a + ck , y = b + dk , k is real) . (a) Show that p is a vertical line if c = 0 . (b) Show that p is a horizontal line if d = 0 . (c) Show that p- contains the origin if a = 0 = b .

8-8. S l o w .

We are now ready to study one of the important properties of a line which corresponds to the idea of the steepness of inclination of a line in the world of everyday affairs. The

steepness of a stairway depends on the relationship between the

rise and the run of a step.

AP = RISE PB = RUN

If one stairway has s teps w i t h a ce r t a in r i s e and run and

another stairway has s teps w i t h r i s e and run each twice a s large, i s i t c lea r t h a t the steepness of the two stairways i s the same? I n other words, a run of 2 with a r i s e of 1 gives

the same steepness as a run of 4 with a r i s e of 2 .

The steepness o r p i tch of these stairways may be defined a s i n the number obtained by dividing the r i s e by the run, 3

e i t h e r case.

The concept of the slope of a l i n e i s based on the idea of " r i s e divided by run." If we think of one s tep connecting

then the r i s e i s

C E M R E L - CSMP L I B R A R Y 103 S. WASHINGTON ST. CARBONDALL, ILL. 62901

We could define the slope of the segment PiPo a s r i s e divided

by run, i . e . , . But we do not. The slope of a " 2 - xll

-- ---

segment i s defined a s y2 - Y l . The formula without x2 - Xl

the absolute value i s e a s i e r t o handle, and it turns out t o be more useful . The absolute value of the slope conveys only the magnitude of the slope. The sign of the slope conveys the addi t ional Idea of s l o p e s up o r down" a s suggested i n the

f igure .

(4,7)

7-3 4 SLOPE = - z - 4-1 3

(SLOPES UP) / ( 1 $3)

of the slope of a segment we slope of a l i n e . Consider the

S ta r t ing with the concept now develop the concept of the - l i n e AB where A = (1,2) and B = (3,5) . Then -

A 3 = ((x,y) : x = 1 + 2k , y = 2 + 3k , k i s r e a l ) . - Let us compute the slopes of several segments P1P2 on AB . Take P, a s the point corresponding t o k, and Pp a s the

point corresponding t o kp .

Note t h a t t h e s lope of eve ry segment o f %? i s

- slope of PiPo

3 . Note a l s o t h a t 3 and 2 a r e t h e c o e f f i c i e n t s of k i n t h e equa t ions f o r y and x r e s p e c t i v e l y . Let u s check t h e l a s t l i n e of t h e t a b l e . Suppose k1 and k2 a r e any two d i s t i n c t va lues of k . Sub-

s t i t u t i n g i n t h e parametr ic equa t ions we g e t

if lc = kg, x2 = 1 + 2k2, y2 = 2 + 3k2, P2 = ( 1 + 2k2, 2 + 3k2) . Tnen

x2 - x1 = ( 1 + 2k2) - (1 + 2k1) S(k2 - k ) ,

Does every n o n v e r t i c a l l i n e have t h e p roper ty t h a t a l l of i t s segments have t h e same s lope? We show t h a t t h i s i s indeed t h e case.

Let c be any l i n e and l e t cl(x1,yl) and c2(x2,y2) be

any two p o i n t s on c . Tnen

k i s r e a l ) .

A s i n t h e example above we t a k e two d i s t i n c t va lues of k , s a y p and q , corresponding t o two d i s t i n c t p o i n t s P and

Â¥ Ã‘Ã Q i n ClC2 , and f i n d

Then

Before w e d i v i d e yp - y~ by xp - xQ w e should a s s u r e our-

s . 1 v e s t h a t xp - xQ # 0 . If xp - xQ = 0 , then xp = x~

-<Ã‘Ã‘ - and C1C2 i s a v e r t i c a l l i n e . If ClC2 i s a nonver t i ca l

l i n e , xp - xQ / 0 , xl - x2 / 0 , and

Th i s proves t h a t a l l segments of a n o n - v e r t i c a l l i n e have t h e

same s lope . We may t h e n write t h e fo l lowing d e f i n i t i o n and

theorem.

DEFINITION. The s lope of a non-ver t i ca l l i n e i s ---- equa l t o t h e s lope of any of i t s segments; t h e

s lope of a n o n - v e r t i c a l r a y i s t h e s lope of t h e ---- - l i n e which c o n t a i n s t h e ray .

4--b+- Notat ion. The s lope of AB, AB, AB i s denoted

by m-, m A i , m, r e s p e c t i v e l y . AB

THEOREM 8-13. The s lope of a non-ver t i ca l l i n e p i s

' - , where PlP2 is any segment of p and x2 - xl

Using Theorem 8-11 we can write parametric equations for a line which passes through two given points. In the following theorem we see how to write parametric equations for a line passing through a given point and having a given slope.

THEOREM 8-14. If p is the line through (xl,y1) with - , then slope m = -

â‚

1. p = ((x,y) : x = x1 + kg , y = y1 + kf , k 1s real]

and

2. p = ((x,~) : x = X, + k , y = y, + h , k is real) . Proof: Suppose h Is a number such that (xl + g , yl + h)

is a point on p . Then the slope of p is

h f by hypothesis, it follows that - = - Since the slope is - 6 g g

d h = f . Therefore (xl, y,) and (x, + g , y, + f) are rt

two points on 4 and It follows from Theorem 8-11 that

(1) p = ((x,y) : x = x, + kg , y = yl + kf , k is real) . Next let n be a number such that (xl + 1 , yl + n) is

a point on p . Then the slope of p Is

Since the slope is m by hypothesis, it follows that m = n . Therefore (xl,yl) and (xl + 1 , y, + m) are two points

on /Â and It fol lows f rom Theorem 8-21 that

( 2 ) p = [ ( x , y ) : x = x, + k , y = y + ton , k is rea l ] ,

We now consider three possibilities for the slope of a line; it is positive, it is zero, or it is negative. Let p1(xl,y,) and p2(x2,yn) be two points of a line. We

suppose that the points are named that has the larger

x-coordinate. We disregard the possibility x2 = xl , since Â¥4Ã‘Ã‘Ã

this would imply that PlP2 is a vertical line.

Possibility - 1. The slope is positive. Then y2 - yl and x2 - x, are both positive or both negative. Since we named the points so that x2 > xl it follows that x2 - xl and y2 - yl are both positive. This means Intuitively that,

Â¥Ã‡Ã‘ as a particle moves along PIP2 from left to right (from the

point with x-coordinate xl to the point with x-coordinate

x2), It is going uphill.

Possibility - 2. The slope is zero. Then y2 - yl = 0 . This means intuitively that, as a particle moves along the line it is moving on "level ground." h he y-coordina of all the points of the line are the same.)

Possibility - 3. The slope is negative. Then one of the

numbers, y2 - y., and x2 - xl , is positive and the other one is negative. Since we named the points so that x2 > xl it follows that x2 - xl is positive and yp - Yl is

negative, that I s , y2 < yl . This means intuitively that, as

a particle moves along from left to right, it is going

downhill.

This section closes with several examples Involving the

slope idea.

Example 1. - Given A = (5,-8) and B = (-2,8) .

I d - 8 - (-8) - 16 Solution: b= - - .

A 3

Example - 2. A line r passes through (1,3) and has

slope 5 . Find the point on r whose x-coordinate is -3 ,

Solution: r = {(x,y) : x = 1 + k , y = 3 + 5k , k is real)

Set x = - 3 . Then - 3 = 1 + k , k = - 4 ; ~ = 3 - 2 0 = - 1 7 . Answer: (-3,-17) .

Example 3. Find the slope of the line - ((x,y) : x = 3 + 4k , y = 2 , k is real) .

Solution: Set k = 0 and 1 to get two points on the line.

k = 0: (x,y) = (3,2) ;

k = 1: (x,y) = (7,2) . Then the slope is 2 - 2 F 3 = o .

Alternate solution: By inspection of the parametric

equations, x2 - x, = 4 and y2 - y, = 0 . Therefore 0 m = ~ = 0 .

Problem Set 8-8a -- 1. Find the slope of the segment joining the points in each

of the following pairs.

(a) (0,o) and (6,2) 1 (f) ($3) and (3,w) .

(b) (0,0) and (6,-2) . (g) (-2.8,4) and (4.2,-1) . 1 (4 ( 3 4 and (7,121 . (h) ($0) and (o,-^)

(d) (0,0) and (-4,-3) . (i) (1000,-500) and (1001,-499)

( e ) (-5,7) and ( 3 ~ 8 ) . ( J ) (a,b) and ha); (a # b) .

Replace the "? " by a number so t h a t the l i n e through

the two po in t s w i l l have the slope given. (Iiint:

Subs t i t u t e i n t he s lope formula.)

( a ) (5 ,2 ) and (? ,6 ) , m = 4 . 1 ( b ) (-3,l) and ( & Â ¶ ? , m = -

( c ) (6,-3) and (g,?) m = -i

- 3 (d) ( ? , I21 and (5,121 , m = 0 . Plo t t h e po in t s A(-1,0) , ~ ( 6 ~ 2 ) , ~ ( 4 ~ 5 ) , D(-3,3) . Draw AECD . Find t h e slope of each s ide of ABCD . Which two s i d e s have the same slope?

P lo t t he q u a d r i l a t e r a l PQRS with v e r t i c e s P ( o , ~ ) , Q ( 2 Â ¶ 3 , R(-1,-2) S(-3,-1) . Which p a i r s of s ides

have the same slope?

Without p l o t t i n g t e l l whether the slope of the segment

Joining the po in t s i n each of t he following p a i r s has a pos i t i ve , zero, o r negative s lope. Then t e l l how you

would i n t e r p r e t t h e s ign of a s l o p e . '

Which of t he segments jo ining the po in t s i n each of the

following p a i r s i s s teeper?

( 0 ~ 0 ) and (100,101) o r ( 0 ~ 0 ) and (101 ,100)?

Find the s lope of the l i n e segment joining (a,;) a n d

(b,;) i f a + b . - Given: AB = [ (x ,y ) : x = 3 - 2k, y = -1 + 3k, k Is rea l l

M What is t h e s lope of AB ?

Parametric equations of a l i n e a r e usefu l i n p lo t t i ng the

graph of a l i n e when one point and the slope a r e given.

Consider, f o r example, t he l i n e 1 through ~ ( 1 ~ 2 ) with 3 slope p .

A?= ( (x ,y ) : x = 1 + 2k, y = 2 + 3k, k i s r e a l ) . I f k = 0 , then (x ,y ) = (1 ,2) . If k = 1 , then (x ,y) = (3,5) .

If k = 2 , then (x ,y) = ( 5 , 8 ) . If k = 3 , then (x ,y ) = ( 7 , l l ) . Note t h a t as k Is ass igned v a l u e s 0, ly 2, 3 , ... ( success ive i n c r e a s e s of l), t h e corresponding x-values ape 1, 3 y 5, 7, ... ( s u c c e s s i v e s i n c r e a s e s of 2 ) and t h e corresponding y-values a r e 2, 5, 11, . . , , - ( success ive i n c r e a s e s of - 3 ) . Note t h a t 2 and 3 a r e t h e c o e f f i c i e n t s of k i n t h e parametr ic equa t ions , and

i s t h e s lope . The numerator and denominator t h a t p of t h e t ' s lope f rac t ion ' ' t e l l u s how t o g e t from one p o i n t t o ano the r on t h e l i n e a s suggested i n t h e f i g u r e .

Y

Use t h i s method t o p l o t t h e l i n e s determined i n each of

t h e fol lowing: 2

(a) PI = ( - 3 y 2 ) ; s lope = 3 . 3

( b ) p1 = (o,o) ; s lope = 5 .

( c ) Pl = (2 , -4 ) ; s lope = - 4 3

( d ) Pl = (-1, -3) ; s lope = 2 . b

( e ) Pl = (OY0) ; s lope = - . a

10. P l o t t h e graph of l i n e s through t h e o ~ i g i n having t h e fo l lowing s lopes :

3 (4 7 (4 4

3 (b) - 7 (4 7 r - r < o . ,

8-8 11. Write the parametric equations for the lines in

Problem 10.

Our next theorem gives us a relation between the concept of papallel lines and the concept of slope.

THEOREM 8-15. Two non-vertical lines are parallel if and only if their slopes are equal.

Proof: Let two distinct non-vertical lines p and q

be given. We have two things to prove:

(1) If p 1 1 q , then their slopes are 'equal. (2) If the slopes of p and q are equal, then p 1 1 q .

Let ~ ~ ( x ~ , y ~ ) and p2(x2,y2) be two points in p . Let

vertical lines through Pl and P2 intersect q in

+ h) and Q2(x2,y2 + k) , respectively. Then

PlQlQ2P2 is a parallelogram. merefore PIQl = P2Q2 . Since PIQl = lh 1 , P2Q2 = 1 kl , and since h and k are

both positive or both negative, it follows that h = k .

But t he slope of p i s

and the slope of q i s

Therefore t he s lopes are equal.

( 2 ) Next, suppose t h e r e i s a number m which i s the

slope of both p and q . We want t o prove that the l i n e s a r e p a r a l l e l . We do t h i s by showing t h a t i f they have one point i n common, then they a r e t h e same l i n e , Suppose then t h a t they have a point , say ~ ( x ~ , y ~ ) , i n common. Since p i s not v e r t i c a l , i t contains a point p(x2,y2) such t h a t x2 # xl . Since q i s not v e r t i c a l , it i n t e r s e c t s t he l i n e ( (x ,y) : x = x2] ; that is, q conta ins a point ~ ( x ~ , y ~ ) - such t h a t x3 = x2 . Since t h e s lopes of PR and a a r e the same,

Since x2 = x3 , the denominators x2 - x and x3 - x 1 1 ' a r e t he same. Hence y2 - yl = y3 - yl , o r y2 = y3 . T h i s

means that Q = P . I n o the r words, i f p and q i n t e r s e c t , then p and q a r e t h e same l i n e and there fore p a r a l l e l . If

p and q do not i n t e r s e c t then they a r e p a r a l l e l by d e f i n i t i o n . This completes t he proof' t h a t i f t h e s lopes of p and q a r e equal , then p and q a r e p a r a l l e l .

A na tu ra l quest ion t o ask i s t h e following one. I f t he s lopes of two segments a r e equal , and have a point i n common, a r e they c o l l i n e a r ? This suggests t h e t e s t f o r c o l l i n e a r i t y s t a t e d i n the next co ro l l a ry .

Corollary - 8-15. Three points A, B, C are collinear

if and only if m - = m- * or they lle on a vertical line. A B BC

This "lf and only if" statement is a short statement combining the two statements:

(1) If m - = m , then A Â B * C are collinear. - A B EC

(2) If A, By C are collinear, and do not lie on a vertical line, then m - = m-

A B E!C

Proof: Let A, B, C be three points and let ml = m - , = m . Then A B -

m2 x

M E = ((x,y) : x = x + k , y = yB + km2 , k is real] .

B u

If ml = m2 , then %?= BC and A, B, C are collinear. If

A, By C are collinear, then it follows directly from

Theorem 8-15 that ml = m2 .

Problem Set 8-8b -- -* 4-+- Show that AB 1 1 CD and that AD 1 1 BC if

(a) A = (-3,-2) , B = (5,1) , C = ( 6 , 6 ) , D = (-2,3) . (b) A = (5,-3) , B = (15,-2) , C = (26,-2) , D = (16,-3) . (c) A = (-3,0) , B = (1,5) , C = (10,2) , D = (6,-3) .

- Show that is not parallel to CD if A = (6,2) , B = (-I,^) , C = (-1,2) , D = (8,o) . (a) Is the point ~(4,13) on the line Joining ~(1,l)

to ~(5,171 ?

(b) Is the point (2,-1) collinear with (-5,4) and (6,-8) ?

( c ) Given: A = (101,102) , B = ( 5 ,6 ) and C = ( -95, -94). -4--b Determine whether AB = BC .

(d) Given: A = (101,102) , B = (5,6) , C = (202,203) - d D = (203,204) . Are "S f and CD parallel?

Are they equal?

(a) Given: A = (3,8) and the slope of line p

containing A is 7 . Find the coordinates of

three more points on p . (b) Given B = (-1,0) and the slope of line q

3 containing B is - . Find the coordinates of

three more points on q . (a) Write a pair of parametric equations of the line

2 containing (3,4) whose slope is 7 . (b) Write a pair of parametric equations of the line

containing (-1,3) whose slope is -1 . - 1 (~int: -1 = - 1 * ) -

Given: AB = [(x,y) : x = 3 - 2k , y = -1 + 3k , k is real) u -

What is the slope of AB ? Express CD in parametric 4---e- -

equations if CD 1 I AB and CD contains (0,0) . Given a = [(x,y) : x = 1 + 2k , y = 2 - k , k is real] ,

b = ((x,~) : x = 3 + 2h , y = -1 - h , h is real} , show that a 1 1 b . As part of your proof, show that

Given p = [(x,y) : x = 1 + 2k, y = 3 + 4k, k is real) , q = ((x,y) : x = 1 - 4h, y = 3 - 8h, h I s real) .

Show that p = q . Given m = ((x,y) : x = 1 + 2k, y = 2 + 3k, k is real) ,

n = ((x,y) : x = 1 - 2h, y = 2 + 3h, h is real) . (a) Show that m and n intersect in one point.

(b) Find the coordinates of that point.

Four points taken in pairs determine six segments. Which pairs of distinct segments determined by the following

four points are parallel? ~(3.6) , ~ ( 5 , g ) , c(8,2) , ~(6,-i) . Show by considering slopes that a parallelogram is formed by drawing segments joining in order A(-1,s) , ~(5,l) , ~(6,-2) and ~(0,2) . Show that if one of two parallel lines is vertical, then

the other is also.

Given A(-2,-4) , ~ ( 4 , 2 ) , c(6,o) . Let D be the - - midpoint of AB and E the midpoint of E!C . Show - that is parallel to AC . It is asserted that both of the quadrilaterals whose

vertices are given below are parallelograms. Without plotting the points determine whether or not this is true.

Show that the line through (3n,0) and (0,n) is parallel. to the line through (6n,0) and (0,2n) . Assume n # 0 . P = (a,l) , Q = (3,2) , R = (b,l) , S = (4,2) . Prove -- -- that PQ # RS and that PQ 1 1 RS if and only if

8-9. Other Equations for Lines. -- In the preceding sections we have used parametric equations

to express the coordinates of the points of a line. In this

section we find another expression which "represents" a line. We illustrate with a particular line. -

Consider the line AB where A = (2,5) and B = (4,8) . Then ~(x,y) is collinear with A and B if and only if P = A or m = m . Since m - - and - - -

AP AB AP -n 8 - 5 3 m - = F2 = 7 P(x,y) is collinear with A and B if and

AB

- - If (x,y) = (2,5) then x - 2 = 0 T=-* x - 2 y - 5 x - 2 - Y - 5 y - 5 = 0 , and Ã‘5 = Ã‘'3 . Conversely if - - -

- or (x,y) = (2,5) . It follows that ~(x,y) then r2 = 7

x - 2 y - 5 is colllnear with A and B if and only if - = - . Theref ore

w

If we think of A as (xA,yA) and B as (xB,yg) , the expression appears as

This suggests the following theorem. - THEOREM - 8-16. If P = (x1,yl) and Q = (x2,y2 ) and if PQ

is an oblique line, then

Proof: The point ~ ( x , y ) I s col l inear wi th P and Q

i f and only i f R = P o r ^_^= jj_ ., that is, i f and only i f

x , y ) = (xl,yl) o r PR p%

t h a t is , i f and only i f

- Corollary 8-16-1. If PQ i s the l i n e of Theorem 8-16.

then

Proof: To prove t h i s we show that the equation of the

theorem, 1 - J. , i s equivalent t o the equation

x2 - x1 Y^ - Y1

of the corol lary,

To ge t the second form from the first, multiply both s ides of the f i r s t by y2 - y, ; t o get the f i r s t from the second divide both s ides of the second by y2 - y, .

Corollary 8-16-2. I f p i s the l i n e which passes through ~(x- , ,y . , ) w i t h slope m , then

Proof: If rn # 0 , this follows immediately from Corollary 8-14-1 since p also passes through

If m = 0 , the equation reduces to y = yl . x - x i Y - Y l

DEFINITION. The equation - - x g - x , Y g - Y ,

is called the two-point form for the equation --- Â¥4Ã‘Ã

of an oblique line PIP2 , where

DEFINITION. The equation y - y, = m(x - xl) is called the point slope form for the equation --- of a non-vertical line with slope m and

passing through (xl,yl) .

Example 1. If C = (7,-3) and D = (4,-5) , then -

Example - 2. Write an equation for the line through (-5,-2) with slope 4 . Answer: y + 2 = 4(x + 5) .

Example - 3. If A = (2,l) , B = (3,4) , C = (5,-2) , - 4--b D = (0,-1) , find the point of intersection of AB and CD .

8-9 Solution:

u x - 2 - y - AB = [(x,y) : - - '1 = [(x,y) : 3x - y = 51 . 3 - 2 m - x - 5 - y + 2 ) = ((x,y): x + 5 y = - 5 ) CD = ((x~Y) : --rn

4 - 1 Also L= 3- = 3 , -1 + 2 - Since these ^-^-5--y

AB

slopes are unequal the lines intersect In some point (xl,y,) . Then 3x1 - y1 = 5 , x1 + 5y1 = -5 . Multiplying both sides

of the first equation by 5 and adding to the sides of the

5 second equation we get 16x, = 20. Then xl = , yl - - - I T . 5 - 5 Therefore AB and ^Cf Intersect In the point ( r - , - $) .

Problem -- Set 8-9

1. Write an equation in two-point form for the line determined by the given pair of points,

2. Write an equation in point-slope form for the line which contains the given point and has the given slope.

3. Write an equation in point-slope form of the line that contains the given point (5,8) and is parallel to the line found in Problem 2(c).

4. In triangle ABC , A = (0,0) , B = ( 1 ~ 6 ) , and C = (5~2) - (a) Write an equation for AB .

u (b) Write an equation for AC . (c) Write an equation for the line that cont.ains the

median from A .

( d ) Write an equation for the line that contains the - midpoints of AB and AC .

(e) Write an equation for the line "'BC*'. (f) If D = ( 7 , 7 ) , find the coordinates of the point -

of intersection of AD and T. 5. Below are equations of lines. Which of these lines

contains (2,3) ?

(a) 2y = 3x . (b) 3y = 2x . (e) $ + g = l .

(c) y - 3 = 2(x - 2) , y - 2 3 (f) =

6. Write an equation of the line that contains (-2,4) and whose slope is the given number.

7. Given below is a set of four lines. State which pairs of lines are parallel.

8. For each pair p and q determine whether p 1 1 q , p and q intersect in one point, or p = q . (a) p = [(x,y) : x - 2y = 81 and q = ((x,~) : 2x + Y =1} (b) p = [(x,y) : x - 2y = 8) and q = ((x,~) : 2x -%=la ( c ) p = ((x,y) : x - 2y= 81 and q = ((x,y) : 2x-4y=10}

9. Given two non-zero numbers a and b , show that + = 1 is an equation of the line that contains a

(a,0) and (0,b) . This form of a linear equation is called the intercept form.

10. Given two numbers m and b , show that y = mx + b is

an equation of the line whose slope is m and which intercepts the y-axis at a point whose y-coordinate is b . This form of a linear equation is called the slope-

intercept form.

Perpendicular Lines.

We have seen that two non-vertical lines are parallel if

and only if their slopes are equal. In this section we develop

a condition in terms of slopes for the perpendicularity of two

lines. If one of two lines is vertical, then a necessary and sufficient condition that the lines be perpendicular is that

the other one be horizontal. The following theorem is a state-

ment about the perpendicularity of two non-vertical lines in

terms of their slopes,

THEOREM 8-17. Two non-vertical lines are perpendicular if and - only if the product of their slopes is -1 .

Proof: Let the given lines be denoted by p, and p2 and let their slopes be ml and m2 , respectively. We have

two statements to prove.

(1) If pi 1 p2 , then mlma = -1 . (2) If m1m2 = -1 , then p, 1 p2 .

We prove both statements together as follows.

Let q-, be the line containing (0,0) which Is parallel

to p, . Let q2 be the line containing (0,0) which is

parallel to p2 . The slope of ql is ml and the slope of

q2 is m2 . Let q., and q2 intersect the vertical line

((x,y) : x = 1) in ~(1,r) and ~(1,s) respectively.

r - 0 Then m, = m - = vÃ‘Ã‘ = r , OR

S - 0 m 2 = m - = = = s . 0s

Therefore R = (l,ml) and S = (l,m2) .

From the Pythagorean Theorem and i t s converse it then follows t h a t

p, 1 p2 if and only i f ( 0 ~ ) ~ + ( 0 ~ ) ~ = ( R S ) ~ . Using the distance formula w e get

2 2 Then pl 1 pn if and only if 1 + m12 + 1 + mn = (a, - t i g ) ,

2 2 if and only i f 2 + ml + mg = ml 2 - 2 m m + m 2 , 1 2

if and only if 2 = -2m,m2 ,

i f and only i f mlm2 = -1 ,

which completes the proof.

Example - 1. Given A = (2,5) , B(-l,7) , C = (4,l) , -- D = (8,7) , show that A B 1 CD .

Solution: 7 - 5 2 " w - ^ = - T

-- -1 It follows that A B 1 CD , Since - 7 =

Example - 2. Given P = (4,-15) , Q = (-17,3) , R = (0,5) , determine whether or not is perpendicular to .

Solution: 3 + 15 18 - 6 3 - 5 = 2 a =-- = - -7,SE^=-- 17,

-12 4--m

+ += # -1 . Therefore PQ is not perpendicular - to OB. .

Example - 3. If A = (0,0) , B = (4,3) , c = (8,9) , D = (-5,11) , prove that the diagonals of quadrilateral ABCD

are perpendicular.

9 11 - 3 Solution: Since m - = , m- = Tq = - 8 9 , and since AC BD

-8 (Ã‘ = -1 , it follows that 1 TO . â‚

Example - 4. Given A = (5, -7 ) , B = (0,0) , C = (7,5) , determine whether or not triangle ABC is a right triangle.

- 7 5 Solution: Since rn = - , m = 6 , m = 77 , it AB AC BC

follows that m 1 and hence ABC is a right triangle with right angle at B .

Example - 5. Given A = (-1,3) , and B = (5,-1) , find an equation in point-slope form for the perpendicular bisector of - AB in the xy-plane. - - 1 - 3 Solution: Since the slope of AB = it 5+1=--5'

3 follows that the slope of the perpendicular bisector is . -

The midpoint of AB is (-' - + , - - ') = (2,l) . Then the

3 equation of the perpendicular bisector is y - 1 = w ( x - 2) ,

Alternate solution: The midpoint of is (2,l) and -

the slope of the perpendicular bisector of AB is as in

the above solution. Then parametric equations for the perpendicular bisector are

Then 3 x = 6 + 6 k , 2 y = 2 + 6 k ; 3 x - 6 = 2 y - 2 ; 3x - 2y = 4 . it follows that 3x - 2y = 4 is an equation

.- for the perpendicular bisector of AB in the xy-plane.

Problem Set 8-10 -- 2 1. Lines p, q, r, and s have slopes YJ , 1 -4 , -ly , and

respectively. Which pairs of lines are perpendicular? T ' 2. The vertices of a triangle are ~ ( 1 6 , ~ ) , ~ ( 9 ~ 2 ) , and

c(o,o) -

(a) What is the slope of AB ?

(b) What is the slope of a line that is perpendicular

to AB ?

( c ) What is the slope of S5" ?

(d) What is the slope of a line that is perpendicular

to BC ?

3. Show that the line containing (0,0) and (3,2) is perpendicular to the line containing (0,0) and (-2,3) .

-- Show that AB 1 E!C if A = (a,b) , B = (0,0) , and C = (-b,a) where a / 0 and b # 0 . Given the points ~(l,2) , ~(5,-6) , and ~(b,b) , determine the value of b so that /PQR is a right

angle. - Given AB = [(x,y) : x = 1 + 2k and y = 2 + 3k , - k is real] , and CD = [(x,y) : x = 1 - 3k and

y = 2 + 2k , k is real) . - - Prove: (1) AB and CD Intersect In (1,2) .

4---+4-+ (2) AB 1 CD . -

Given AB = [(x,y) : x = -1 + 4 k , y = 2 - 3k , k is real) -- 4-+ - If CD 1 AB and CD contains (-2,2) , express CD with set notation symbols and parametric equations.

The vertices of triangle ABC are A(O,O) , ~(3,2) , and ~(4,-1) . Using parametric equations express:

4--w (a) The line through B that is parallel to AC . - (b) The line through B that is perpendicular to AC . (c) The line through A that is parallel to%?. - (d) The line through A that is perpendicular to BC .

o--b (e) The coordinates of D if D is on BC and -- AD 1 BC .

Using slopes, show that the quadrilateral ~(8,0) , ~(6,4) , c(-2,0) , ~(0,-4) has four right angles. Express in set notation the perpendicular bisector of the

segment that joins the following pairs of points.

Show that if (a,b) and (c,d) are distinct points,

the line p containing them is perpendicular to the

line q Joining (b,c) to (d,a) . Given A = (0,0) , B = (4,2) , C = (3,-3) , D = (x,-2) .

-4-+ (a) Find x so that AB 1 CD . -- (b) Find x so that AB 1 1 CD ,

13. Given: ~ ( 3 . 5 ) and B(-1,-2). Calculate the slope of the line perpendicular to through A and draw the line.

14. Given P(-3.1) , Q(0,-5) , and ~ ( 5 , 0 ) . Calculate m - . Â¥r^r - rw

(a) Through R draw a line parallel to PQ . - (b) Through R draw a line perpendicular to PQ .

15. The slope of a line p through (2,3) is $ . (a) Give the coordinates of two other points on p . (b) Give the coordinates of two other points which are

contained in a line through (2,3) perpendicular

to p . 16. Given a quadrilateral ~(a,b) , ~ ( a + c,b) , ~ ( a + c,b + c),

D(a,b + c) . (a) Prove that AC 2 ,

(b) Prove that AC 1 ,

(c) Prove that AC and have the same midpoint.

This section contains several definitions and theorems

relating to parallelograms. In Chapter 6 we defined a parallelogram as a quadrilateral each of whose sides is parallel to the side opposite it and proved two theorems.

They are

1. In any parallelogram each side is congruent to the side opposite it. h he or em 6-6)

2. If two sides of a quadrilateral are parallel and congruent, then the quadrilateral is a parallelogram.

(Theorem 6-7)

In Problems 2 and 5 of Problem Set 6-7 and Problem 5 of Problem Set 6-8b, we proved statements which we now introduce formally as theorems.

THEOREM 8-18. A quadrilateral is a parallelogram if each of its sides is congruent to the side opposite it.

THEOREM 8-19. A quadrilateral is a parallelogram if and only if each angle is congruent to the angle opposite it.

We now consider cases of special parallelograms which have properties not common to all parallelograms.

DEFINITION. A parallelogram is a rectangle if and only if it has a right angle.

Perhaps you think of a rectangle as a quadrilateral having four right angles. It is possible to start with this as a

definition or the one given above. In either case the other statement becomes a theorem.

DEFINITION. A parallelogram is a rhombus if and only If two consecutive sides are congruent.

DEFINITION. A parallelogram is a square if and only if it has a right angle and two adjacent

sides that are congruent.

You should notice that every square is a rectangle and

also a rhombus. We might say that the set of squares is the intersection of the set of rectangles and rhombuses. We can picture roughly the set relations as follows:

In this diagram the region marked P represents the set of parallelograms; the region marked R, , the set of rectangles; the region Rp , the set ofrhombuses; the region marked S , the set of squares.

The following theorem is a direct consequence of our definition of a rectangle and Theorem 8-19.

THEOREM 8-20. A quadrilateral is a rectangle if and only if it is equiangular.

The proof is left as a problem.

As a direct consequence of the definition of a rhombus and Theorem 8-18, we also prove:

THEOREM 8-21. A quadrilateral is a rhombus if and only if it

is equilateral.

Problem Set 8-11 -- 1. Does a rhombus have all the properties of a parallelogram?

Does a parallelogram have all the properties of a rhombus? Explain.

2. Define a square in terms of: (a) a rhombus, (b) a rectangle.

(a) Write the two parts of Theorem 8-20.

(b) Prove both parts of the theorem.

(a) Write the two parts of Theorem 8-21.

(b) Prove both parts of the theorem.

Identify the following statements as true or false. Be able to justify your answer for each statement.

(a) If a quadrilateral is a rectangle. it is equiangular. (b) If a quadrilateral is equiangular, it is a rectangle. (c) If a quadrilateral is a rhombus, It is equilateral.

(d) If a quadrilateral is equilateral, it is a rhombus. (e) If a quadrilateral is regular, it is a square. (f) The two triangles determined by a diagonal of a

parallelogram are congruent.

(g) If the triangles determined by one diagonal of a quadrilateral are congruent, the quadrilateral is a parallelogram.

8-12. Using Coordinates in Proofs. - We have seen that the xy-coordinate system is a useful

tool in solving problems in geometry. As we pointed out in the beginning of this chapter, there are many coordinate systems in a plane. It is natural to expect that a coordinate system selected to f i t a problem" might be a better tool than one set up without reference to the problem. And this is indeed the case, as we now illustrate.

Example. Prove that if a line segment joins the midpoints of two sides of a triangle, its length is half the length of the third side.

Proof I: -- Suppose a triangle and a line segment joining the mid-

points of two of its sides are given. Label the given triangle ABC so that the given segment Joins the midpoints of sides - AC and . Call these midpoints D and E respectively.

We now set up an xy-coordinate system in the plane of triangle ABC which seems to fit the problem. We choose line

4--& * OX as the line AB . We choose point A as the origin. The - line OY is taken as the line in the plane ABC which is - perpendicular to OX at A . Then A = (0,0) , B = (b,~) , C = (c,d) , for some real numbers b, c, d . (we know that

b # 0 since A and 3 are different points. We know that d # 0 , since A, B, C are noncolllnear points.)

Then we use the midpoint formula to get

- - Then DE and A B are horizontal lines and

AB = b - 0 = b l . 1 It follows that DE = A B , and this completes the proof.

Proof 11: -- In the above proof we labeled our figure and set up a

coordinate system to fit the problem. We now give a proof using a coordinate system which is not set up to fit the problem. Scan this proof to see how it compares in difficulty

with the above proof.

Suppose a triangle ABC is given and that D, E, F are the - midpoints of AC, m, m, respectively. Then using the

midpoint formula we find that

Using the distance formula we get

But

1 i Therefore DE = 7 AB . Similarly EF = Â¥3 AC and DF = - PC. 2

3-12 As you can see from these two proofs an xy-coordinate

system which is set up to fit the problem simplifies the

expressions involving coordinates which are used in the proof. d In Proof I we found that y,, = - = 2 YE This proves that DE

is horizontal and hence that 1 1 . It might appear in our first proof that we are proving

only a special case. Actually the proof applies to all cases.

The x-axis and the y-axis can be chosen anywhere in the plane so long as they are perpendicular to each other. For con- - venience we chose AB as the x-axis. We cannot then choose

4--& AC as the y-axis. For then AC 1 and this would mean

that the proof is for the special case of a right triangle.

After selecting a line for the x-axis we may select any point - in it as the origin. We chose A as the origin. Then OY is taken as the unique line in the plane of triangle ABC

which is perpendicular to X at A . The proof Is general

since we can set up such a coordinate system starting with

any triangle and the segment joining the midpoints of two of

its sides.

We state as a theorem what we have proved.

THEOREM 8-22. A line segment which joins the midpoints of two

sides of a triangle is parallel to the third side and its

length is half the length of the third side.

Problem Set 8-12

1. Prove Theorem 8-17 if the coordinates of the vertices of

AABC are: A = (0,0) , B = (2b,0) , and C = (2cY2d) . Is there any advantage in choosing these coordinates

rather than the coordinates in Proof I of the example? If there is an advantage, explain.

2. Given AABC with AB = 6 , EC = 8 , and AC = 10 . Find the perimeter of ADEF , if D, E, and F are midpoints of the sides of the triangle.

It i s desired t o measure the distance between two t r ees on opposite s ides of a building.

-J

I f the two t r e e s a re represented by points X and Y , then / \ loca te a t h i r d point Z from / \ which both X and Y may be seen. Place s takes at M and - N , the midpoints of XZ and - YZ . How can you f ind the distancebetween X and Y T I - - - a f t e r measuring MN ?

/ \ Explain.

In Problem 1, i f c = 0 , then AAEC i s a t r i ang le . Explain.

Prove t h a t the midpoint of the hypotenuse of a r igh t t r i ang le i s equally d i s t an t from the ver t ices of the t r i ang le .

Given i sosce les t r i ang le ABC . Set up a coordinate system with the vertex of the t r i ang le on the y-axis and the corresponding base of the t r i ang le on the x-axis.

Prove the statement: I f a t r i ang le i s isosceles , the medians t o the two congruent s ides of the t r i ang le a re congruent. [Hint: Let ve r t i ces A and B be contained i n the x-axis and vertex C be contained i n the y-axis.]

Prove the statement: I f the medians t o two sides of a t r i ang le a r e congruent, the t r i ang le i s isosceles .

Write a s ingle statement which combines the statements i n Problem 7 and Problem 8.

Given: In triangle ABC , /A is acute and is an altitude.

Prove :

- 11. In AABC , CM is a -

median to side AB . Using coordinates verify:

8-13. Parallelograms Continued.

Many theorems relating to parallelograms can be proved with or without coordinates. The following theorem and corollaries establish the coordinates of the vertices of a parallelogram, a rectangle, and a rhombus. They will be useful

in proving some of the theorems in this section.

THEOREM - 8-23. Given quadrilateral ABCD with A = (0,0) , B = (a,0) , D = (b,c) , then ABCD is a parallelogram if

and only if C = (a + b,c) . Proof: There are two things to prove:

(1) If ABCD is a parallelogram, then C = (a + b,c) . (2) If C = (a + b,c) , then ABCD is a parallelogram.

(1) Suppose ABCD is a parallelogram. Let C = (x,y) . - 4--+ Since AB is horizontal, then DC is also horizontal. There- - - fore y = c . If b # 0 then neither AD nor BC is vertical. Since AD 1 1 it follows that m - = m - and

c Y AD BC hence that 5 = - . But y = c . Therefore x - a = b , x - a x = a + b , and C = ( a + b,c) . If b = 0 ,then D is in - - - the y-axis and AD is vertical. Since BC 1 1 AD , BC is

also vertical and x = a , x = a + b , and again C = (a + b,c) ,

8-13 (2) If C = (a + b , c ) then S5" is horizontal and hence

parallel to m . Also DC = la + b - bl = la1 , AB= la - 01 = la1 . Then DC = AB and T)5" 1 1 IS. It follows that ABCD is a parallelogram.

Corollary 8-23-1. If the coordinates of the vertices of a parallelogram are A = (0,0) , B = (a,0) , C = (a + b , c ) , and D = (b,c) , then the parallelogram is a rectangle if and only if b = 0 .

Proof: There are two things to prove:

(1) If ABCD is a rectangle, then b = 0 . (2) If b = 0 , then ABCD is a rectangle.

PART (2)

(1) If ABCD Is a rectangle, then /A is a right angle and ADlm . Therefore, D is In the y-axis and

b = O .

(2) If ABCD is a parallelogram and b = 0 , then D = (0,c) . Since D is on the y-axis, we know that AD 1 and /A Is a right angle. Therefore, ABCD I s a rectangle.

Corollary 8-23-2. If the coordinates of the vertices of a parallelogram are A = (0,0) , B = (a,0) , C = (a + b , c )

and D = (b,c) where then the parallelogram is a

rhombus if and only if

Proof: There are two things to prove:

(1) If ABCD Is a rhombus, then a =&-. (2) If a = i/a2 + c2 , then ABCD is a rhombus.

t PART ( I

q PART (2)

(1) If ABCD is a rhombus, then by definition AB = AD . By the distance formula AB = a , and AD = Vb . I& the substitution property of equality a = 4 / b T .

(2) It is given that a = Vb- . the distance formula AB = a , and AD = i / b T 2 , I& the substitution property of equality AB = AD . Since two adjacent sides of the parallelogram ABCD are congruent, the parallelogram is a rhombus.

We shall use the results of Theorem 8-23 and its corollaries to prove certain properties of the diagonals of a

parallelogram, a rectangle, and a rhombus. The following experiment will help us to discover these relations.

Experiment

Draw several pictures of a parallelogram, a rectangle, a rhombus, and a square. Use a protractor and a ruler to discover the properties that appear to be true with respect to the diagonals of each of the given quadrilaterals. Record your findings in the chart by checking the quadrilateral which has the listed property.

Diagonals Diagonals Diagonals bisect 1 bisect each other 1 are 1 1 the angles

Rectangle

Rhombus

Square -. - -. -

THEOREM 8-24. A quadrilateral is a parallelogram if and only if the diagonals bisect each other.

8-13 Proof: There a re two things t o prove:

(1) I f ABCD i s a parallelogram, then AC and b isec t each other .

(2) If AC and bisect each other , then ABCD i s a parallelogram.

t PART ( I )

PART ( 2)

8-13

(1) You will be asked to prove Part (1) of Theorem 8-24

in the' next problem set.

(2) Let the coordinates of the quadrilateral be

A = (0,0) , B = (a,0) , C = (x,y) , and D = (b,c) . Since - AC and i% bisect each other they have the same midpoint.

Thus

and

From this we see that x = a + b and y = c . Therefore

C = (a + b,c) and by Theorem 8-23 ABCD is a parallelogram.

You will be asked to prove the following theorems in the

next problem set. You should note that there are two parts

to the proof of each theorem. You should write out the two

parts of the statement which must be proved before beginning

your proof.

Theorem 8-23 and the two corollaries will help you set up the coordinate system for Theorems 8-25 and 8-26.

THEOREM 8-25. A parallelogram is a rectangle if and only if the diagonals are congruent.

THEOREM 8-26. A parallelogram is a rhombus if and only if the diagonals are perpendicular.

THEOREM - 8-27. A parallelogram is a rhombus if and only if a diagonal bisects one of its angles.

Problem Set 8-13 -- 1. Prove Part (1) of Theorem 8-24.

2. Prove Theorem 8-25.



List the properties of a rectangle that are not true of

all parallelograms.

List the properties of a rhombus that are not true of all

parallelograms.

Keeping in mind its definition, may a square be considered

a rectangle? a rhombus? Then a square "inherits" the

properties of 1 , and Make a chart like the following and check which figures

have the listed properties.

opposite sides are 2 I I I I I I , ,

n

opposite / s are =

opposite sides are 1 1

consecutive / s are supp.

diagonals bisect each other 1

diagonals are =

rectangle parallelogram

diagonals are 1 diagonals bisect angles

it is equilateral

it is equiangular

rhombus

It is regular

square

9 . Starting with the set of all quadrilaterals explain how

the set of parallelograms, rectangles, rhombuses and

squares may be considered as subsets.

8-14. Trapezoids.

DEFINITION. A quadrilateral with one pair of sides parallel and the other pair of sides not parallel is called a trauezoid.

BASE

MEDIAN -LEG

1

BASE

DEFINITION. The parallel sides of a trapezoid are called the bases of the trapezoid; the other two sides are called the legs of the trapezoid. -

DEFINITION. If is a base of trapezoid ABCD then A and B are a pair of base angles of the trapezoid.

DEFINITION. A line segment which is perpendicular to the lines containing the bases of the trapezoid and which has its endpoints in these lines is called an altitude of the trapezoid.

DEFINITION. The line segment which connects the midpoints of the legs of a trapezoid is called the median of the trapezoid.

DEFINITION. A trapezoid whose legs are congruent is called an isosceles trapezoid.

Problem -- Set 8-14

(a)

1. Prove t h a t the median of a trapezoid Is p a r a l l e l t o i t s bases and t h a t i t s length i s half the sum of the lengths of I ts bases. Hint: Let ABCD be the trapezoid w i t h

A = (0,0) , B = (2a,0) , C = (2b,2c) , D = (2d,2c) . 2. Using the r e s u l t of Problem 1, f ind the length of the

segment marked x or y In the following diagram. Pa ra l l e l l i n e s a r e indicated by arrows, lengths by numbers, and congruent segments by dashes.

3 . One angle of a trapezoid measures 100 . Can you f ind the measures of i t s remaining angles? If, i n addition, you were t o l d t h a t the opposite angle has a measure of 70 , could you then f ind the measures of the two remaining angles? What a re they?

4. Prove t h a t a p a i r of base angles of a trapezoid a re congruent i f and only i f the trapezoid i s isosceles . ( ~ e c i d e first whether you w i l l o r w i l l not use coordinates.)

8-15 5. Prove that the diagonals

of a trapezoid are congruent if and only If it is Isosceles. (If you use coordinates you might choose coordinates as

shown. Then you have to prove two statements. )

(1) If d = -b , then the diagonals are congruent.

* ( 2 ) If the diagonals are congruent, then d = - b .

6. Prove: The segment joining

the midpoints of the Y 4

diagonals of a trapezoid I is parallel to the bases and equal in length to half the difference of their lengths.

4 Ãˆ- A(0,o) B(2a ,o)

8-15. Concurrent Lines.

In this section we prove several statements which contain the phrase "the set of all points." Membership in the set is

determined by a condition or a combination of conditions. The proof of such a statement consists of two parts. We must

prove that:

1. Any point belonging to the set satisfies the given condition.

2. Any point that satisfies the given condition belongs to the set.

THEOREM 8-28. The set of all points In a plane which are - equidistant from two given points in the plane is the

perpendicular bisector of the segment Joining the given

points.

Proof: Given two points A and B and a plane which 4--+

contains them. Choose AB as the x-axis and the midpoint of - AB as the origin.

(1) If P is in the y-axis, then AP = PB .

4 - 0 A(-o,o)

(2) If AP = PB , and P is in the xy-plane, then P

is in the y-axis.

P x B(a,o)

(1) If P is in the y-axis, then P = (0,b) for some

number b and

v Then there is a real number a , a # 0 , such that A = (-a,0) and B = (a,0) . Then the y-axis is the - perpendicular bisector of AB . There are two parts to the

proof.

and AP = BP . (2) If ~(x,y) is any point such that AP = PB , then

it follows from the distance formula that

4ax = 0 , and since a # 0 ,

Therefore P is in the y-axis.

8-15 For the purposes of the next corollary it is convenient to

have a definition of concurrent lines.

DEFINITION. The lines in a set of lines are called

concurrent if and only if there is exactly one point

which lies in all of them; the segments in a set of

segments are called concurrent if and only if there

is exactly one point which lies in all of them.

According to this definition and our earlier definitions, we note that concurrent rays lie on concurrent lines, or in the

special case of two opposite rays, they lie on the same line.

Corollary 8-28-1. The perpendicular bisectors of the sides of a triangle are concurrent at a point equidistant from the vertices of the triangle.

Given triangle ABC . Let p, q and r be the - perpendicular bisectors of AB, and AC , respectively. Either p and r intersect or p Is parallel to r . If we assume p 1 1 r then AB 1 r . But r 1 AC by hypothesis.

What theorem does this contradict? The assumption that p is

parallel to r is false. Therefore p Intersects r at a

point.

If 0 is the point of intersection of p and r , then OB = OA and OA = OC by Part (1) of Theorem 8-28. Therefore,

OB = OA = OC by the transitive property of equality. here- fore, 0 is in q by Part (2) of Theorem 8-28. This proves

that p, q, and r are concurrent at a point equidistant from A, B, and C, the vertices of the triangle.

THEOREM 6-29. The set of all points in the interior of an angle which are equidistant from the lines which contain the sides of the angle is the interior of the midray of the angle.

Proof: We construct a proof without coordinates. Let + an angle ABC and its midray BD be given.

* (1) If P is in the Interior of BD , then the distance - -

from P to AB equals the distance from P to BC . (2) If P is in the interior of /AX and if the - distance from P to BA equals the distance from

+ P to "EC*, then P is In the interior of BD . 4

(1) Suppose P is an interior point of BD . Since m /ABC < 180 , then m /PBC < 90 and it follows that the foot * of the perpendicular from P to *BC is some point on BC , call it F . Similarly, the foot of the perpendicular from P - * to BA is some point in BA , call it E . ABPE 2 ABPF by S . A . A . and PE = PF . Hence the distances of P from- - and BC are equal.

(2) Since P Is in the interior of the angle and -- 4--b PE 1 BA , 1 BC , and PE = PF , then A PBE 2 APBF and * m /FBP = m /EBP . Therefore BP is the midray of /ABC .

Corollary 8-29-1. The lines which contain the angle bisectors of the angles of a triangle are concurrent at a point equidistant from the sides of the triangle.

++- Let A ABC with angle bisectors AD, BE, CF be given. Now + 4 AD and CF (except for the points A and C) lie in the

+ - + same halfplane with edge AC . Also AD and CF are not

parallel (since the measures of /CAD and /FCA are each less

than 90 ) . Let I be their point of intersection. From

Part (1) of Theorem 8-25 it follows that I is equidistant 4--P - 4-+ from AB and AC and also equidistant from AC and %?. - It follows that I is equidistant from w a n d BC and by

+ Part (2) of Theorem 8-25 that I lies in BE . This means - that and CF are concurrent in the point I , and - I is equidistant from and dB .

Problem Set 8-15 -- 1. Given A(-3,0) , ~(0,4) , C(5,0) . Plot points A, B

and C and show by a drawing how to locate a point D such that DA = DB = DC . Explain your drawing and state

the theorem (or corollary) that suggested it.

2. Given A(-2,0) , ~(0,-6) , and ~(3,0). Using a protractor - find a point D such that the distances from D to AB , - 4--e BC and CA are equal. State the theorem (or corollary)

that suggested your drawing.

(a) Find a point X such that AX = BX and the distances - from X to AC and BC are equal.

(b) Find a point on the x-axis that is equally distant from A and B .

(c) Find a point on the y-axis that is equally distant u -

from AC and BC . -

4. In triangle ABC , D is the midpoint of BC , E the - - midpoint of CA , and F the midpoint of AB . (a) If AB = 12 , CB = 9 , and AC = 10 , find DE, EF,

FD . - (b) Prove that DE 1 1 ; that 1 1 BC" ; that -

FD 1 1 m . - (c) The perpendicular bisector of AB is also

perpendicular to ? . The perpendicular bisector

of BC is also perpendicular to ? . The - perpendicular bisector of CA is also perpendicular

to ? ,

(d) Are the lines that contain the altitudes of triangle

DEF concurrent? Explain.

5. We sketch two proofs of the following statement: The

lines that contain the altitudes of a triangle are concurrent. You are to fill in the missing parts of each proof. Then decide which proof seems to be more

satisfying.

Proof I : -- Let ABC be the triangle. Consider the line through A

parallel to ; the line through B parallel to AC ; - the line through C parallel to AB . Let these lines meet in P, Q, R as shown in the diagram. Show that

APAB 2 ACBA and that PA = CB . Similarly show that CB = AQ . It follows then that PA = AQ . In the same

way PB = BR and RC = CQ . The altitude from A to - EC is contained in the perpendicular bisector of P .

Proof 11: -- Let the triangle be ABC

and choose axes so that

A = (a,~) , 3 = (0,b) , and C = (c,0) . Then the

y-axis contains the altitude from B to AC .

Why? Therefore the slope

of the altitude, ha from c A is 5 and the slope of

the altitude from C , hc , a is 5 . Why?

The line that contains ha

((x,y) : x = a + b k , y = 0 + ck , k is real) ;

the line that contains hc is

{(x,y) : x = c + bp , y = 0 I- ap , p is real} . Setting k = - - a we find that (0, - as.) is contained b

in ha . hhy? Setting p = - - ac we find that (0, - -g-) b

is also contained in hc . Why? Since the x-coordinate

of this point is 0 , the point is on the y-axis, which contains hb . Therefore the lines which contain the

altitudes are concurrent.

6. Prove Corollary 8-28-1 by coordinates.

CEMREL - CSMP LIBRARY' 601 103 S. \.,'ASSlN3TON ST.

CA,-iBOilDALi-, ILL. 62901

8-16. Summary.

In this chapter we defined coordinates in a plane and we

used them as a tool in formal geometry. We have seen some 11 neat" proofs involving coordinates. In other situations we have decided to write proofs without coordinates. In

constructing a proof using coordinates it is usually wise to

set up a coordinate system which makes the expressions involving

coordinates as simple as possible.

We developed several expressions for the coordinates of

the points of a line, with considerable emphasis on the use of

set-builder notation and parametric equations. We defined the

slope of a non-vertical line and used it to get conditions for

perpendicularity and parallelism of oblique lines:

P 1 q if and only if m . m = -1 , P q

p I q if and only if m = m . P Q

We developed several equations for lines:

the two-point form,

the point -slope form.

We developed several formulas:

the distance formula,

the midpoint formula.

The chapter includes several theorems on triangles:

one about a line joining midpoints of two sides,

one about concurrence of angle bisectors,

one about concurrence of perpendicular bisectors of sides.

The following table summarizes several definitions and theorems

which are concerned with quadrilaterals. Each line in the

table yields a statement of the form: An A is a B if and

only if C . Proofs for statements with no reference listed

are easy.

A

quadrilateral

quadrilateral

quadrilateral

?uadril.ateral

quadrilateral

quadrilateral

quadrilateral

quadrilateral

parallelogram

parallelogram

parallelogram

parallelogram

parallelogram

rectangle

rhombus

paralle10";-an

paralleloS '-ram

parallelogram

parallelogram

rhombus

square

trapezoid

rectangle

rhombus

rectangle

rectangle

rhombus

rhombus

square

square

c

opposite sides are parallel

opposite sides are congruent

two sides are parallel and congruent

diagonals bisect each other

all sides are congruent

all sides are congruent and all angles are right angles

exactly one pair of sides is parallel.

all angles are congruent

all sides are congru-ent

diagonals are congruent

all angles are right angles

diagonals are perpendicu-lar

diagonal bisects one angle

all sides are congruent

all angles are congruent

Review Problems

1. Plot t h e graph of each of the following.

(a) p = ((x,y) : x = 2 , 0 < Y < 3 ) . ( 0 ) q = ((x,y) : y = 2 , 0 2 x 5 3) (c) r = ( ( x , ~ ) : x = 1 + 2k , y = 1 - 21c , k is an

integer and -2 < lc < 2 ) . ( d ) s = ((x,y) : x t - y = 3 , 0 < x < 3 and 0 < y < 3 : . (e) t = [(x,Y) : x -1- y = 3 , -3 < x < 0 ) .

2. Is the following information sufficient to prove

quadrilateral A B C D a parallelogram?

A B = D C ; E. A B = D C ; A D = B C . - - A B 1 1 E ; AD 1 1 E .

- A B = D C ; A D 1 1 g . /A 2 /C ; LB 2 /D . - - AC bisects BD . - - - AC bisects ; BD bisects AC . - - - AB _I AD ; DC J_ AD; BC 1 CD .

A A B D 2 ACBD . ( j ) A A B D 2 ACDB . ( k ) /A 2 /C ; AT I I DC . (1) /A ^ L C ; AB = DC .

3. The diagonals of a rhombus are 16 and 30 . Find

the perimeter of the rhombus.

4. The ratio of the lengths of two sides of a rectangle

is 3 : 4 . The length of the diagonal of the rectangle

Is 40 ; find the lengths of the sides of the rectangle.

5. Three vertices of rectangle ABCD are A(-!,-!) , ~(3,-1) , and ~ ( 3 , 5 ) .

What is the f o u r t h vertex? -

What is the midpoint of A B ? -

What is t h e midpoint of AC ?

What is A B ?

What is AC ? Show that AC = BD . - Write AB using parametric equations. - Write AC using parametric equations.

4 Find Q on AC such that AQ = 4 A C .

(1) Write parametric equations for the line through C - that is perpendicular to AC ,

6. An Isosceles triangle has vertices (0,0) , (4a,0) , (2a,2b) . (a) What is the slope of the median from (0,0) , if any? (b) What is the slope of the median from (4a,0) , if any? (c) Find the slope of the median from (2a,2b) , if any.

- 7. In square ABCD , R is the midpoint of BC and S - - -

is the midpoint of CD . AR intersects BS in T . (a) Prove that BS = AR . (b) Prove that 1 .

*(c) Prove that TD = AB . Hint: Let A = (0,0) and B = (2a,0) . -

8. Prove that the median of a trapezoid bisects a diagonal.

9 . (a) What is an equation of the x-axis?

(b) What is an equation of the y-axis?

(c) Show that all points of both axes satisfy the

equation xy = 0 . 10. A rhombus ABCD has A at the origin and "AB in the

positive x-axis, m /A = 60 , AB = 6 , C is in

Quadrant I.

(a) What are the coordinates of C ?

(b) What are the coordinates of D ?

( c ) Find AC . (d) Show that AC = 4/3" BD .

--D (e) Using parametric equations express AC ,

11. Write an equation for the set of points

(a) whose distances to (-3,0) and (5,0) are equal.

(b) whose distances to the x-axis is 3 . (c) whose distances to the x- and y-axes are equal. (d) whose distances to the horizontal lines y = -2

and y = 8 are equal. (e) whose x-coordinates are 12 . (f) whose y-coordinates are -8 .

12. Show that triangle ABC is a right Isosceles triangle if A = (3,4) , B = ( 4 5 ) , C = (-2,l) .

13. Using parametric equations express the set of points equally distant from ~(0,4) and B(-8,0) .

14. The point A(c,~) is equally distant from ~(1,l)

and c (3,5) . Find the value of c .

15. The distance from (h,3) to the x-axis is twice its distance to the y-axis. Find h . (Two answers.)

16. ABCD is a parallelogram. Show that the segment that - - joins D to the midpoint of AB trisects AC .

17. In triangle ABC , D is

in , and E is in AC . AD = 2DB and CE = SEA . - BE and DC intersect in

F . Show that

m - 3 - - DF 1 FE T and - = FC '6

Chapter 9

PERPENDICULARITY, PARALLELISM, AND COORDINATES IN SPACE

9-1. Introduction.

Our first contact with points, lines, and planes in space

was in Chapter 2, but since then our work has been almost

completely restricted to points and lines in a single plane.

Now, having investigated plane geometry in some detail, we are

ready to turn our attention to space geometry. In particular,

in this chapter we extend the ideas of perpendicularity and

parallelism to figures which may not be contained in a plane.

Most of the results we are going to discuss are familiar

to us from our past experience. However, we often miss the

essential features of things we have seen a hundred times, and

certain results which are true in the plane are not true in

space. Moreover, without practice it is hard to visualize

geometric relations in space and harder still to represent them

by drawings on a sheet of paper. To save time, it therefore

seems wise to omit the proofs of most of our theorems and

concentrate instead on getting a thorough understanding of the results themselves. Fortunately, the proofs of the theorems

in this chapter are quite similar to the deductive arguments

we have seen in previous chapters, and a few samples will be

an adequate indication of how the rest can be constructed. Of

course at any time you are free to use any theorem that has

been previously stated, whether it has been proved in the text

or not.

In preparation for the work which follows, it will be help-

ful for you to review the simple space relations introduced in

Sections 2-5, 2-6, and 2-7, and then to go carefully through the exploratory problems which are given below. The ability to

make and interpret drawings of three-dimensional configurations

will be of great value to you through the rest of this course.

9-1

Be sure that you can do these two things. Appendix V offers many suggestions which may be helpful to you.

Exploratory Problems

1. In the following sketch of a rectangular block, certain combinations of edges, considered separately, suggest certain configurations of lines and planes.

In each of the following, copy the drawing of the block and darken the appropriate edges to suggest your idea of

indicated configuration.

Two distinct Intersecting lines.

T w distinct parallel lines.

Two lines which are neither intersecting nor

paralle 1.

Three mutually perpendicular lines.

Three parallel lines which are not coplanar.

A line intersecting one of two parallel lines but

not the other.

Two distinct lines which are perpendicular to the

same line and parallel to each other.

Two lines which are perpendicular to the same line

at different points but are not parallel to each other.

A line parallel to a plane. Two distinct lines which are parallel to the same

plane and parallel to each other.

Two lines which are parallel to the same plane but

not parallel to each other.

Two distinct parallel planes.

9-2 (m) Two perpendicular planes.

(n) Three mutually perpendicular planes.

(0 ) A plane perpendicular t o each of two d i s t i n c t

p a r a l l e l planes. ( p ) Two d i s t i n c t l i n e s perpendicular t o the same plane.

( q ) Two d i s t i n c t planes perpendicular t o t he same l i n e .

2 . Without including any unnecessary l i n e s , make drawings of

your own t o suggest the conf igurat ions i n Pa r t s (d ) , ( h ) ,

(0, ( J ) , (k)? (11, (m), ( n ) , (01, ( P I ? (4 of t h e preceding problem.

9-2. Perpendicular i ty Relat ions .

I n Section 4-8 we defined what we mean by two perpendicular

l i n e s . I n t h i s sec t ion we a r e going t o d i scuss t he perpen-

d i c u l a r i t y r e l a t i o n between a l i n e and a plane. Before we

define t h i s formally, however, you should study t h e following

experiments and from the c lues they give you, you should t r y

t o make up a d e f i n i t i o n of your own.

Experiments

I n Section 4-10 we learned t h a t i n a plane t h e r e Is a unique l i n e which i s perpendicular t o a given l i n e at a

given po in t . I s t h i s t r u e i n space? Hold two penc i l s so

t h a t they appear perpendicular t o each o the r . Can you hold

one of t he penc i l s i n a d i f f e r e n t pos i t i on and s t i l l have

it appear perpendicular t o the o ther a t t he same point?

How many d i f f e r e n t pos i t i ons can one penc i l assume and

remain perpendicular t o the second penc i l a t the same

point? Do you th ink these "perpendiculars" might l i e i n

t he same plane? Would such a plane be perpendicular t o

the o ther penci l?

2. Place a sheet of paper on your desk. Hold your penc i l so

t h a t it appears perpendicular t o t he paper. With a second

penci l , draw a l i n e on the page t h a t appears t o be

perpendicular t o t he f i r s t penc i l .

(a) Can you shift your first pencil so that it remains perpendicular to the line at the same point but is not perpendicular to the paper?

(b) Draw another line, intersecting the first. Now, place your pencil so that it appears perpendicular

to both lines at their point of intersection. Does the pencil appear to be perpendicular to the plane of the paper? Can you hold your pencil so that it is perpendicular to the paper?

( c ) Draw additional lines through the point of intersection. Does the pencil appear to be perpendicular to each of them at that point? Is it still perpen-

dicular to the plane?

(d) What do you think would be a good definition of a line perpendicular to a plane?

The preceding experiments lead us to the following definition:

DEFINITION. A line and a plane are perpendicular - to each other if and only if they intersect and every

line lying in the plane and passing through the point of intersection is perpendicular to the given line.

The following figure suggests the relations described by this definition:

Q

--- PA , PB , PC , . . . all lying in plane & are perpendicular - to PQ .

The r e s u l t s which we o b t a i n i n t h i s c h a p t e r can a l l be der ived without a d d i t i o n a l p o s t u l a t e s . However, our development proceeds much more e a s i l y i f we accep t as p o s t u l a t e s two theorems whose p roofs a r e long and r a t h e r involved.

The f irst , P o s t u l a t e 24, should remind u s of Theorems 4-21 and 5-11 which d e a l wi th t h e e x i s t e n c e of a l i n e con ta in ing a given p o i n t and pe rpend icu la r t o a g iven l i n e . This p o s t u l a t e i s a l l t h a t w e need a t p r e s e n t . The second, P o s t u l a t e 25, we s h a l l in t roduce i n Sec t ion 9-4.

I P o s t u l a t e - 24. There i s a unique p lane which 1 1 con ta ins a given p o i n t and i s pe rpend icu la r t o a 1

given l i n e .

We should understand c l e a r l y t h a t i n t h i s p o s t u l a t e no r e s t r i c t i o n i s placed on t h e g iven p o i n t . It can e q u a l l y we l l be a p o i n t on t h e g iven l i n e o r a p o i n t which i s n o t on t h e g iven l i n e . The p o s t u l a t e says simply t h a t wherever t h e p o i n t may be, t h e r e i s always one and on ly one p lane which c o n t a i n s t h e p o i n t and i s perpendicular t o t h e g iven l i n e .

From t h e d e f i n i t i o n of p e r p e n d i c u l a r i t y , we know t h a t i f a plane i s perpendicular t o a l i n e a t a p o i n t F , on

./Â£ , then every l i n e i n 9 which p a s s e s through F i s

perpendicular t o /^ . However, we do n o t y e t know whether

t h e r e can a l s o be l i n e s pe rpend icu la r t o 1 a t F which do

not l i e i n 9 . The fo l lowing theorem answers t h i s q u e s t i o n f o r us .

THEOREM - 9-1. The p lane which i s pe rpend icu la r t o a g iven l i n e a t a g iven p o i n t c o n t a i n s every l i n e which i s pe rpend icu la r t o t h e g iven l i n e a t t h a t p o i n t .

Proof: Let 1 be any l i n e and l e t be t h e p lane which

i s perpendicular t o a t t h e p o i n t F . What we must show i s

t h a t i f 1' i s any l i n e pe rpend icu la r t o 2 at F , then

Ji' l i e s i n 9 .

Now the intersecting lines and ji\ determine a plane, say , and by Postulate 9 this plane intersects the plane

In a line, say 1" . Moreover, since 1" lies in the perpendicular plane ,

it must be perpendicular to /^ at F . Hence, in the plane both Â £ and 1" are perpendicular to ,& at the point

F .

But by Theorem 4-21, in a given plane there is exactly one perpendicular to a given line at a given point. Hence jl' and Â£ must be the same line. That is, 1' must lie In the perpendicular plane , as asserted.

According to our definition, before we can say that a

line 1 is perpendicular to a plane f i at a point F , we must be sure that /^ is perpendicular to every line in ^> which passes through F . The next theorem tells us that we do not need nearly this much Information to be sure that a line is

perpendicular to a plane.

THEOREM - 9-2. If a line is perpendicular to each of two inter- seating lines at their point of intersection, it is perpendicular to the plane determined by the two lines.

9-2

Proof: Let in te r sec t a t the perpendicular t o plane determined l e t be the

m and m1 be two d i s t i n c t l i n e s which point F and l e t A? be a l i n e which i s both m and m 1 a t F . Let be the by the in tersec t ing l i n e s , m and rn t , and

plane which i s perpendicular f l

According t o the l a s t theorem both m and m' must l i e i n . Hence the planes 7?? and have both m and m t i n

common. Therefore, by Theorem 2-10, ^7 and ^/ must be the same plane; t h a t i s , the plane determined by the two l i n e s , tn and m 1 , i s the plane which i s perpendicular t o d at F , a s asserted.

Postulate 24 assures us t h a t there i s a unique plane which i s perpendicular t o a given l i n e a t a given point, but it does not answer the corresponding question of the existence of a l ine which i s perpendicular t o a given plane a t a given point. However, t h i s i s s e t t l e d by the following theorem.

THEOREM - 9-3. There i s a unique l i n e which i s perpendicular t o a given plane a t a given point I n the plane.

We s h a l l omit the proof of t h i s theorem. The general outl ine i s suggested by the following f igure . # i s the plane which i s perpendicular a t the given point, F , t o any par t icu lar l ine , p , which l i e s i n the given plane, , and passes through F . The required perpendicular 1 i s the l ine i n j?? which i s perpendicular a t F t o the l ine , r , i n which f l and f l i n t e r sec t .

The corresponding theorem dealing with the existence of a line which passes through a given point not in a plane and is perpendicular to the plane is more conveniently handled a little later after we have discussed parallel relations in space.

Problem Set 9-2 -- In each of the following problems, draw your own diagram

as part of the proof.

1. In the figure, if /PQH is a right angle and Q and H are in 5 , should you infer from the definition of a line

perpendicular to a plane that

-1 5 ? Justify your answer.

2. In the figure, points B, R, K and T are in - plane 5 , and AB 1 &. Which of the following angles must be right 4~ angles: /ABR , /ABK , /REP , /TEA , /KBR ? Why?

9-2 3 . I n the f i gu re , plane 7

A 4 , T

contains the noncol l inear po in t s R, S, P, but '771 does not contain T . ( a ) Do po in t s R, S, and

T determine a plane? Why? d

(b) If SP Is perpendicular t o t h e plane of R, S, T , which angles I n the f i g u r e must be r i g h t angles? Why?

4. I n the f i gu re , t he point A

and t h e square FRHB a r e not coplanar; AB 1 FB . (a ) How many planes a r e

determined by p a i r s of segments i n the f igure? Name them.

(b) A t l e a s t one of the segments i n t h i s f i gu re i s perpendicular t o one of t he planes asked f o r i n P a r t ( a ) . Which segment? Which plane? J u s t i f y your answer.

5. In the f igure , point R

and t r i a n g l e ABF a r e not coplanar, A ABF i s i sosce l e s with ver tex 3 , H i s the - midpoint of AF , and

R S [ H B .

( a ) How many d i f f e r e n t planes a r e determined by p a i r s of s e g m e n t s i n t h e f i gu re? A Name them.

(b ) Find a segment t h a t i s perpendicular t o a plane. S t a t e the perpendicular i ty and the theorems which j u s t i f y your statement.

6. I n t h e f i gu re , ? a t P

and ^rl"Â¥PIT ~ u s t %? l i e 4 Q

i n plane & ? Why? -

Plane s ^ and 3 i n t e r s e c t i n A

w, as shown i n t h e f i g u r e . /^\ ""AlTl & , r l i e s i n Â£ - ,

plane ABR I n t e r s e c t s 2 i n / .I-"" d ' y\ 1

n (a) IS "A?'l'*H "-? my? 7 / ( b ) I s - r1=? Why? ( c ) Is -1"BC"?

- 8. I n t h i s f i gu re , FB 1 plane f t

and i n ARAB , which l i e s i n plane ft , BR = BA . prove AABF 2 A RBF and /FAR 2 /FRA

9. Given the cube shown, with

BR = BL . Does KR = KL ?

Prove t h a t your answer i s

co r r ec t .

9-3 (since we have not yet given cube, we state here, for use properties of the edges of a

a precise definition of a in your proof, the essential cube :

The edges of a cube consist of twelve congruent segments, related as shown in the picture, such that any

two intersecting segments are perpendicular.)

9-3. Parallel Relations.

In this section we are going to investigate parallel relations between lines and planes in space, and this requires that we first define what we mean by saying that a line and a plane, or two planes, are parallel. The following definitions

are natural extensions of the definition of parallel lines which we gave in Section 6-2.

DEFINITION. A line and a plane whose intersection does not consist of exactly one point are parallel to each other.

DEFINITIONS. Two planes (whether distinct or not)

whose intersection is not a line are parallel planes, and each is parallel - to the other.

With these definitions In mind, the following experiments should help you to visualize the properties we are going to discuss.

Experiments

1. Draw a line on a sheet of paper on your desk. Now hold two pencils above your desk so that each appears parallel

to the line. Do the pencils appear parallel to each other? Can you hold them so they are parallel to the line and not to each other?

2. (a) If two distinct lines are parallel to a plane, are they necessarily parallel to each other? The

Parallel Postulate tells us that there is a unique line which contains a given point and Is parallel to a given line. Do you think there is a unique line

which contains a given point and is parallel to a given plane? Hold two pencils so that they "intersect." Can you hold them so that both of them are also parallel to the desk top?

(b) Hold the two pencils so that they represent skew (noncoplanar) lines. Can you hold them so that they are both parallel to the desk top?

3. (a) We have learned that, in a plane, if a line intersects one of two parallel lines in a point, it intersects the other in a point also. Is this true in space? Draw two distinct parallel lines on a sheet of paper. Can you hold a pencil so it will intersect one of the parallel lines but not the other?

(b) Suppose a plane intersects one of two parallel lines in a point. Do you think it must intersect the other also?

( c ) Suppose a line intersects one of two parallel planes in a point. Do you think it must intersect the other plane also?

(d) Sketch diagrams to illustrate Parts (a), (b), (c).

4 . Suppose a plane intersects one of two parallel planes. Do you think it must intersect the other plane also? If a

plane intersects each of two parallel planes, what can you say about the lines of intersection? In your classroom consider the parallel walls "intersected" by the floor. Are the lines of intersection parallel? Think of a book- case. The shelves are parallel planes, the end panel an intersecting plane. What about the lines of intersection?

Draw a diagram of two distinct parallel planes intersected by a third plane.

As Experiment 3 (a) in the preceding list suggests, in

space geometry it is not true that if a third line meets one

of two parallel lines It must meet the other one also. However there are analagous theorems which are true In space, and to

these we now turn our attention.

THEOREM 9-4. If a plane intersects one of two distinct parallel - lines in a point, it intersects the other line in a point also.

Proof: Let 4 and j2 be two distinct parallel llnes, contained in a plane , and let f l be a plane which intersects one of the lines, say / , . in a single point, P, .

Clearly, ft cannot contain ln because otherwise, by Theorem 2-9. it would coincide with , and hence contain , , contrary to the hypothesis that it meets ,/c, in Just

A. JL

one point. Therefore can have at most one point in common

with ,& . Now by Postulate 9, since <?< and have a point, Pl ,

in common.

they must have a line, say p , in common. Moreover, since /' meets each of and Jf2 in at most one point, p must be

distinct from both and 4 .

619

9-3

Now i n the plane , the l i n e p meets one of the two p a r a l l e l l i n e s , Jl and 4 , i n a s ingle point, P . Hence

it must a l s o meet the other l i n e , i n a point , say p p . Since the l i n e p Is contained i n the plane 9 , the

point Pp i s a l so contained i n f l . Therefore in te r sec t s

J p i n a s ingle point, as asser ted .

The next theorem follows e a s i l y from the preceding one, and we s h a l l leave I t s proof a s a problem.

THEOREM 9-5. If a plane i s p a r a l l e l t o one of two pa ra l l e l l i n e s , i t i s a l s o p a r a l l e l t o the other .

THEOREM 9-6. If a plane i n t e r s e c t s each of two d i s t i n c t p a r a l l e l planes, the in tersec t ions a re two d i s t i n c t p a r a l l e l l i n e s .

Proof: Let f l and /be two d i s t i n c t pa ra l l e l planes

and l e t f l be a plane which i n t e r s e c t s both @ and d. Since P and Â¥? do not in te r sec t , must be d i s t i n c t from both -<E? and d. By Postulate 9 t h e in tersec t ion of f l and i s a l i n e , say r . Likewise the Intersect ion of f l

Moreover, these lines lie in the same plane, namely , and have no point in common, since the planes f l and Â£/'hav no

point in common. Therefore, the intersections, r and s , are distinct parallel lines as asserted.

We should observe that the preceding theorem contains the hypothesis that the plane f l intersects each of the parallel planes, and d . Actually, it Is possible to prove the stronger result that if a plane intersects - one of two distinct parallel planes (and does not coincide with it) then it intersects the second plane also and the Intersections are parallel

lines.

THEOREM - 9-7. If a line intersects one of two distinct parallel planes in a single point, it intersects the other plane in a single point also.

THEOREM - 9-8. If a line is parallel to one of two parallel planes, it is parallel to the other also.

The assertions of Theorems 9-7 and 9-8 are illustrated by Figures (a) and (b), respectively. We shall omit their proofs, however.

Problem Set 9-3 -- 1. Make a sketch to Illustrate the hypothesis of each of the

following statements. Indicate whether each statement is

True (T) or False (F).

If two distinct lines are parallel, every plane

containing only one of them is parallel to the other line.

If two distinct lines are parallel, every line intersecting one of them intersects the other.

If two planes are parallel, any line in one of the planes is parallel to the other plane. If two planes are parallel, any line in one of the

planes is parallel to any line in the other plane.

If a plane and a line are both perpendicular to the same line, they are parallel to each other. If a plane and a line are both parallel to the same line they are parallel to each pther.

If each of two distinct parallel planes intersects a third plane, the lines of intersection are

perpendicular.

If two planes are parallel to the same line they

are parallel to each other.

Two lines parallel to the same plane are parallel

to each other.

If a plane Intersects two intersecting planes, the lines of intersection may be parallel.

2. Hypothesis: Planes , and -y^ are parallel as Y- shown, with in ^- - and A in 'y , AC intersects %at B and - AE intersects F a t D .

AC = CE . Prove: BD = BA . *

3. Prove Theorem 9-5. (~int: Let 7' be a plane which is parallel to one of two parallel lines, 1, and 4 , say . Then one of the following must be true. (Why?)

(a) r is parallel to l2 . (b) intersects 4 in a single point.

Use Theorem 9-4 to prove that (b) Is impossible.

9-4. Relations Involving Perpendicularity - and Parallelism.

In Sections 9-2 and 9-3 we considered relations in space which involved, respectively, only perpendicularity and only parallelism. In this section we shall investigate configurations which Involve both perpendicularity and parallelism. Since we shall omit the proofs of most of our theorems, you

should perform carefully the following experiments and make sure that you understand and can visualize the relations they suggest.

Experiments

If two planes are parallel to a third plane, do you think they are parallel to each other? Illustrate by holding two books so that each is parallel to the top of your desk. Do the books appear to be parallel? Draw a diagram of

three parallel planes.

Do you think there can be more than one plane which contains a given point and is parallel to a given plane?

Why?

If two distinct planes are perpendicular to the same line, do you think they can intersect? Illustrate your conclusion by piercing two sheets of cardboard (or small

sheets of paper) with a pencil. Draw a diagram of two distinct planes perpendicular to a given line.

Take a piece of cardboard, p ie rce it w i t h your penci l and place it so t h a t it appears perpendicular t o the penci l a t t he midpoint of the penc i l . Mark a point on the cardboard and f i n d t h e d i s tance from t h a t point t o each end of t h e penc i l . Are the dis tances approximately t he same? Choose another point and make a second measurement. Draw a diagram of a plane perpendicular t o a segment a t the midpoint of t he segment.

If a l i n e i s p a r a l l e l t o a plane and i s not contained i n t h e plane, do you th ink a l l the pe rpend icu la r s Jo in ing the l i n e t o t he plane a r e coplanar? Are these perpen-

d i c u l a r segments equal I n length? Are segments which a re perpendicular t o each of two d i s t i n c t p a r a l l e l planes and have t h e i r endpoints i n the planes equal i n length? I l l u s t r a t e with a diagram.

THEOREM 9-9. Two planes which a r e perpendicular t o the same - l i n e a r e p a r a l l e l .

Proof: Let and be two planes each of which i s perpendicular t o a l i n e . There a r e two p o s s i b i l i t i e s t o consider:

( a ) /7 i s p a r a l l e l t o # . (b) 3 f f i n t e r s e c t s /S i n a l i n e .

I f we can prove t h e second case impossible, the theorem w i l l be es tab l i shed .

Suppose then, t h a t f l and ^- i n t e r s e c t i n a l i n e . Thus f t and f l a r e d i s t i n c t . By Postula te 24 the points , say P

and R , i n which /^ i n t e r s e c t s t h e respec t ive planes and /? , must be d i s t i n c t .

Let L be a point i n both of t he planes f l and @ but no t on . From t he d e f i n i t i o n of a plane perpendicular

w t o a l i n e , it follows t h a t LP i s perpendicular t o A a t P - and LR i s perpendicular t o 1 at R . Hence, s ince P and R a r e d i s t i n c t points , we have two l i n e s each containing L

and each perpendicular t o .

This is impossible, according to Theorem 5-11; hence the

possibility that /z7 ana /F7 intersect in a line leads to a

contradiction and must be rejected. Thus and # are parallel, as asserted.

THEOREM 9-10. If a line is perpendicular to one of two

distinct parallel planes it is perpendicular to the other

also.

Proof: Let ft and be two parallel planes, and let 1 be a line which is perpendicular to one of them, sayyZ7,

at the point P . Then by Theorem 9-7, 1 must also intersect in a point, say R . Let R t be any point of j?? distinct from R and let <

be the plane determined by R * and /^ . Then d intersects -

j@ in the line RR" and, by Theorem 9-6, must intersect ? - in a line, , which is parallel to RR . Thus in , is perpendicular to one of two parallel -

lines, namely PP* (Why?) and hence, it must be perpendicular

to the other also. Since R l was any point in distinct ^Ã

- from R , it follows that - is perpendicular to every line

in which contains R . Hence, by definition, J is perpendicular to the plane , as asserted.

We shall omit the proofs of the remaining theorems in this section, but since some of them are asked for in the next

problem set, it is necessary for us to introduce here the second of the postulates we referred to In Section 9-2. You will find that with this postulate, the missing proofs are not

difficult to construct.

Postulate - 25. Two lines which are perpendicular 1 to the same plane are parallel. I

THEOREM 9-11. If a plane is perpendicular to one of two distinct parallel lines, it is perpendicular to the other

line also.

The assertion of this theorem is illustrated in the following figure : 4 4

THEOREM 9-12. If two lines are each parallel to a third line, they are parallel to each other.

This theorem completes the discussion we began in Chapter 6. There we showed that in a plane If each of two lines is parallel to a third line, they are parallel to each other. The present

theorem assures that this result is true without the restriction

that the three lines lie in the same plane.

THEOREM 9-13. Given a plane and a point not in the plane,

there is a unique line which passes through the point and

perpendicular to the plane.

This theorem completes the discussion we began in

Theorem 9-3. These two theorems together tell us that through any pointthere is a unique line which is perpendicular to a

given

The next two theorems describe properties of planes which

are obvious counterparts of familiar properties of lines.

THEOREM 9-14. There Is a unique plane parallel to a given plane through a given point.

THEOREM 9-15. If two planes are each parallel to a third plane, they are parallel to each other.

Theorems 9-12 and 9-15 provide the final steps in

establishing that the relationship of parallelism has

characteristic properties like those of equality, congruence,

and similarity. The relationship of parallelism for lines in

space has the reflexive, symmetric, and transitive properties.

Likewise the relationship of parallelism for planes has the

reflexive, symmetric, and transitive properties.

In Chapter 6 we considered a line and a point not on the line. We proved that the shortest segment joining the point

to the line is the segment perpendicular to the line. As we

might expect, a similar result holds in space.

9-4 Let 6 be a plane and P a point not on 5 . There are

many segments joining P to <S , in fact one for every point on & . By Theorem 9-13, exactly one of these segments is perpendicular to C' .

THEOREM 9-16. The shortest segment Joining a point to a plane

not containing the point is the segment perpendicular to the given plane.

The proof of this theorem we shall leave as a problem. On the basis of this theorem, we formulate the following definition.

DEFINITION. The distance between a point and a ---- plane not containing the point is the length of -- -- the segment joining the given point to the given plane and perpendicular to the given plane.

In Chapter 6 we proved that two parallel lines are every- where equidistant, and the same property holds for parallel planes. More precisely, we have the following theorem.

THEOREM 9-17. All segments which are perpendicular to each of

two distinct parallel planes and have their endpoints in

the planes have the same length.

In Chapter 8 we showed that in a plane the set of all points which are equidistant from two given points P and Q - is the line which is perpendicular to the segment PQ at its

midpoint. The corresponding result in space geometry is the

following:

THEOREM 9-18. The set of all points which are equidistant from

the endpoints of a given segment is the plane which

contains the midpoint of the segment and Is perpendicular

to the line which contains the segment.

The proof of this theorem is deferred to Problem Set 9-7 where it will be an exercise in the use of coordinates in proof.

Problem Set 9-4 --

1. Assuming here that

why are W, X, Y, and

Z coplanar?

2. Hypothesis: A, C in ;

B, D in 77 , W ~ T S ,

Prove: ?7 1 VS .

9-4

2. Hypothesis: In the figure

^ l l ^ 9 TS\ , ^1^. Prove: AD = CB .

4. Plane $ l a the perpen-

dicular bisecting plane - of AB , as shown in the figure .

- '\< (a) AW = . - -

AK =

(b) Does FW = FK = FR ?

Explain.

Problems 5-8 are concerned with geometric projection. The

following definitions are needed.

DEFINITION. The projection of a point into a plane ------ is the point of intersection of the given plane and

the line which contains the given point and is

perpendicular to the given plane.

Consider two examples in the diagram: P is the projection

of A into <$ ; the point Q is in $ , and the projection of Q into <S, is Q itself.

DEFINITION. The projection of a set of points Into ---- a plane is the set of all points which are projections -- into the plane of points contained in the given set.

5. Using projection as defined above, answer the following.

Is the projection of a point always a point? Is the projection of a segment always a segment? Can the projection of an angle be a ray? a line?

an angle?

Can the projection of an acute angle be an obtuse

angle?

Is the projection of a right angle always a right angle?

Can the length of the projection of a segment be

greater than the length of the segment?

If two segments are congruent will their projections

be congruent?

If two lines do not intersect can their projections

be two parallel lines?

If two lines do not intersect can their projections

be two intersecting lines?

If two segments are parallel and congruent, will

their projections be congruent?

6. Let the projection of point A into plane be A t - I distinct from A . Let A P be the ray opposite to A A I . - Let B be a point such that the length of A B is 6 - inches. Draw a diagram showing the projection of AB - into ^t , and find the length of the projection of AB into f in each of the following situations.

(a) SB I I Z . (b) TB\W. (c) m /PAB = 30 . (d) m /PAB = 45 . (e) rn /PAB = 60 .

7. Given the figure with - A B not in plane t , - XY the projection of - A B into plane , 0 the midpoint of , and N the projection of 0 into . Prove that N is the mid- - point of XY .

- 8. Hypothesis: BD is the -

projection of BC into - plane i . AB lies in plane Â¥ and /ABC Is a right angle.

Prove: /ABD is a right angle. (~int: Let be perpendicular to

plane .)

9 . Prove Theorem 9-16.

9-5. Dihedral Angles.

In Section 4-13 we introduced the notion of a dihedral

angle via the following definition:

DEFINITION. A dihedral angle is the union of a line

and two halfplanes having this line as edge and not lying in the same plane.

At that time we were unable to assign measures to dihedral

angles, but now that we have discussed perpendicularity and

parallelism In space we can do so easily. First, however, we

must reduce the problem to one involving plane angles, for

which measures have already been defined.

The following figure shows a dihedral angle, namely

/P-QR-s , and a plane, S> , which is perpendicular to the edge of the dihedral angle. We observe, In the diagram, that

the intersection of the plane and the dihedral angle is the

union of two rays, and furthermore that these two concurrent + 4

rays, namely BA and BC , are not collinear. Thus the

intersection is an angle.

DEFINITION. The i n t e r s e c t i o n of a dihedral angle and any plane perpendicular t o t he edge of the given d ihedra l angle i s c a l l e d a plane angle of --- t he d ihedra l angle . -

If a l l plane angles of a d ihedra l angle were congruent, it would be na tu ra l t o take t h e i r common measure as the measure of the d ihedra l angle i t s e l f . The next theorem guarantees t h a t t h i s can be done.

THEOREM 9-19. Any two plane angles of a d ihedral angle a r e - congruent.

Proof: Let S and V be the v e r t i c e s of two d i s t i n c t plane angles of t he d ihedra l angle /A-PQ-B , ( ~ i g u r e ( a ) ) . Let U and W be po in t s d i s t i n c t from V on d i f f e r e n t s ides of /V . I n plane UVS , apply the Poin t -P lo t t ing Theorem and l o c a t e point R on / S such t h a t

(1 ) U V = R S .

I n plane WVS , l oca t e point T on /S such t h a t

To prove the theorem, we must show t h a t /y 2 / S . I n order t o do t h i s , we shall apply the S.S.S. Congruence Postula te t o show t h a t AUVW 2 ARST .

9-5 Since /V and /S a r e plane angles of the d ihedra l angle, - each of the planes UVW and RST i s perpendicular t o PQ .

By Theorem 9-9, the two planes a r e p a r a l l e l t o each o ther . Ly 4--F

Theorem 9-6, W and %!? a r e p a r a l l e l . This f a c t , together w i t h (1 ) , shows t h a t UVSR i s a parallelogram. Hence the - segments UR and a r e both p a r a l l e l and congruent.

A s imi l a r argument shows t h a t WVST i s a parallelogram, - and there fore t h a t the segments VS and a r e both p a r a l l e l and congruent. By t he t r a n s i t i v e proper ty of pa ra l l e l i sm anc. congruence, the segments UR and a r e both p a r a l l e l and congruent. In o the r words, UWTR i s a parallelogram. Hence

( 3 ) UW = P . Combining (1 ) , (2), (3 ) , the S.S.S. Congruence Pos tu la te t e l l s u s t h a t

Final ly , ,& 2 / S , and our proof i s complete.

With the l a s t theorem es tab l i shed , t h e measure of a dihedral angle can now be defined.

DEFINITION. The measure of a d ihedra l angle i s the - - number which i s the measure of any of i t s plane

angles.

DEFINITION. A r i g h t d ihedra l angle i s a dihedral angle whose measure i s 90 .

DEFINITION. The planes determined by t h e f aces of a r i g h t d ihedral angle a r e s a i d t o be perpendicular .

The proofs of t h e following theorems about perpendicular planes a r e not d i f f i c u l t . Some have been l e f t as problems.

THEOREM 9-20. I f a l i n e i s perpendicular t o a plane, then any plane containing t h i s l i n e i s perpendicular t o t he given

plane.

THEOREM 9-21. If two planes are perpendicular, then any line

in one of the planes which is perpendicular to their line

of intersection is perpendicular to the other plane.

THEOREM 9-22. If two planes are perpendicular, then any line perpendicular to one of the planes at a point on their line of intersection lies in the other plane.

THEOREM 9-23. If two intersecting planes are each perpendicular

to a third plane, then their line of intersection is

perpendicular to this plane.

The assertion of the preceding theorem is illustrated in

the following figure. Planes and f l are each perpendicular to the plane 0 , and their line of intersection, 1 , is therefore perpendicular to 0 .

Problem -- Set 9-5

How many dihedral angles are formed by the floor, walls, and ceiling of your classroom?

If two planes are perpendicular, what kind of

dihedral angles are formed?

Give a definition of an

(1) acute dihedral angle,

(2) obtuse dihedral angle.

Draw three figures showing, respectively, an acute, a right, and an obtuse dihedral angle.

( d ) Give a definition of adjacent dihedral angles.

Illustrate with a drawing.

(e) Give a definition of supplementary dihedral angles.

Illustrate with a drawing of a pair of adjacent supplementary angles.

( f ) Give a definition of complementary dihedral angles.

Illustrate with a drawing of a pair of adjacent

complementary angles.

2. Each of m, m, and is perpendicular to the

other two.

m /a = m /b = m / c = 45.

What is the measure of LC-PA-B ? of /CAB ?

Prove Theorem 9-21.

Hypothesis: Referring to

the figure on the right,

Prove AB1 & . w-

(~int: Take BC 1 PQ in

4 . Prove Theorem 9-23.

(~int: Referring to the illustrative figure in the text, Â¥4Ã‘Ã‘ÃˆÂ¥-

in plane d draw XN 1 NC and

5. Planes & and 3 are 4-e perpendicular to AB .

-4Ã‘ Lines 'BK and 3H , in <fÃ

plane 3 , determine - with AB two planes A -

which intersect & in - - AD and AC . Certain

lengths are given, as in

the figure. Are BKDA

and BACH parallelograms?

Can you give a further

description of them? Is "V ABHK = AACD ? Can you

give the length of CD ?

6. Prove: If a plane is perpendicular to the edge of a dihedral angle, then it is perpendicular to each face

of the dihedral angle.

Review Problems

Chapter - 9, Sections 1 to 5 - - - 1. In this problem, the symbol 1 always denotes a line and

the symbol 7^ always denotes a plane. Fill in each blank with the one of the following words

always, sometimes, never

which makes the resulting statement true. In each case

make a sketch to Justify your answer.

(a) If ll is parallel to l2 and l2 is parallel to 1, , then 1 is parallel to 1, .

(b) If 1. is perpendicular to 1 and l2 is perpendicular to 1 , then JI' is

perpendicular to J r, .

(c) If pl is perpendicular to / and , is perpen-

dicular to 39, , then ftl is perpendicular

(d) If J? is perpendicular to ft, and fi. is parallel

to/^ , then 1 Is perpendicular to pp . -

(e) If 1, is parallel to and is parallel to

(f) If 1 is parallel to < and fi is perpendicular to p2 , then is parallel to p2 .

( g ) If ft. is perpendicular to 1 and 4 is perpendicular to fi , then ftl Is perpendicular to p2 .

(h) If is perpendicular to and 1 is parallel to f12 , then < is perpendicular to p2 .

(1) If ft, is perpendicular to / i? and /s? is

perpendicular to p3 , then ft! I s

parallel to Â£7 . (j) If f i is perpendicular to Ll and ,dl is

perpendicular to 4 , then ft is parallel to Jn .

2. Mark each of the following statements true (T) or false (F) ,

(a) If a line Is perpendicular to each of two distinct

lines in a plane, It is perpendicular to the plane.

(b) If three distinct lines are perpendicular to the same line at the same point, the three lines are

coplanar.

(c) Through a point not on a line, more than one plane

can be passed perpendicular to the given line.

Through a point not on a plane, only one line can be

drawn parallel to the given plane.

Through a point not on a plane, only one plane can

be passed perpendicular to the given plane.

If a plane is perpendicular to the edge of a dihedral angle, it is perpendicular to each face of the

dihedral angle.

If two planes are perpendicular, a line in one of

the planes is perpendicular to the other plane.

If two planes are perpendicular, a line perpendicular

to one of the planes will lie in the other plane.

If a plane intersects one of two distinct parallel lines, it intersects the other also.

If two distinct lines are parallel, one and only one

plane can be passed through one of these lines

parallel to the other.

If a line is parallel to one of two intersecting planes, it Is parallel to their Intersection.

Two planes parallel to the same line are parallel.

Through a line not perpendicular to a plane, a plane

perpendicular to the given plane can be passed.

The projection of a segment into a plane is a segment.

3. Which of the following lines or planes must be parallel?

Which of them must coincide?

Lines through the same point parallel to the same line.

Lines perpendicular to the same plane.

Lines perpendicular to the same line.

Lines parallel to the same line.

Lines parallel to the same plane.

Planes perpendicular to the same line through the

same point.

Planes parallel to the same plane.

Planes perpendicular* to the same plane.

Planes through the same point parallel to the same

plane.

Planes through the same point perpendicular to the

same plane.

Planes parallel to the same line. 640

9-6. Coordinate Systems in Space. --

In Chapter 3 we introduced the fundamental idea of a coordinate system on a line, or a one-dimensional coordinate - system, as it is sometimes called. In Chapter 8, we extended this idea to coordinate systems in a plane, or two-dimensional - coordinate systems. Now that we have the necessary information about perpendicularity and parallelism in space, we are in a position to discuss coordinate systems in space, or three-

dimensional coordinate systems. As we should expect, our

development here will be very much like the development of

two-dimensional coordinate systems In Chapter 8 and for this reason we shall omit many of the details and concentrate instead on the results themselves. -

Let OX and %? be any two perpendicular lines and let 4--e OZ be the unique line (Theorem 9-3) that is perpendicular to * u the plane of OX and OY at their intersection, 0 . --- Clearly, then, each of the lines OX , OY , OZ is perpen-

dicular to each of the other two, so that we have in fact three

mutually perpendicular lines. Let I, J, and K be points on 4--e- OX , OY and )Z* respectively such that - On the line OX there is a one-dimensional coordinate system

with the point 0 as origin and the point I as unit point. M

On OY , there is a one-dimensional coordinate system with the - point 0 as origin and the point J as unit point. On 02

there is a one-dimensional coordinate system with the point 0

as origin and the point K as unit point. We shall refer to

these coordinate systems as the x- , y- , and z-coordinate

systems, respectively.

The line %?is called the -- x-axis, the line %?is M

called the y-axis -9 the line OZ is called the -- z-axis. --- Collectively OX , OY , OZ are called the coordinate axes.

The point 0 , which is common to the three coordinate axes, is called the origin. The plane containing the x-axis and the

y-axis is called the xy-plane, the plane containing the x-axis -- and the z-axis Is called the xz-plane, the plane containing -- the y-axis and the z-axis is called the yz-plane. Collectively -- these planes are called the coordinate planes.

From the theorems we proved in Section 9-4, it is clear that all lines parallel to the z-axis are perpendicular to the

xy-plane. Similarly, all lines parallel to the y-axis are

perpendicular to the xz-plane, and all lines parallel to the

x-axis are perpendicular to the yz-plane. Using the convenient

set-builder notation, these important observations can be

summarized as follows:

"~ictorial" descriotion

'~arallel" iescription

[n:n 1 1 z-axis)

(n:n 1 1 y-axis)

[n:n 1 1 x-axis)

' perpendicular" description

[n:n 1 xy-plane)

(n: n 1 xz-plane)

(n:n 1 yz-plane)

It should be clear also that (a) all planes parallel to

the xy-plane are perpendicular to the z-axis, (b) all planes

parallel to the xz-plane are perpendicular to the y-axis,

(c) all planes parallel to the yz-plane are perpendicular to

the x-axis.

9-6 As you will remember from Chapter 3, a coordinate system on

a line is a one-to-one correspondence between the line and the set of real numbers. Similarly, as you learned in Chapter 8, a coordinate system in a plane Is a one-to-one correspondence between the plane and the set of ordered pairs of real numbers.

Now we are going to establish a coordinate system in space as a one-to-one correspondence between space and the set of ordered

triples of real numbers.

To do this, let P be any point in space. Then through P there passes a unique plane which is perpendicular to the x-axis (postulate 24). This plane intersects the x-axis in a point which has a coordinate, say x , in the one-dimensional - coordinate system established on OX by the ordered pair (0.1) . This number, x , we define to be the first coordinate, or - x-coordinate, of P .

x Similarly, through P there passes a unique plane which is perpendicular to the y-axis and this plane Intersects the y-axis In a point which has a coordinate, say y , in the coordinate - system determined on OY by the ordered pair ( 0 , ~ ) . The number y we define to be the second coordinate, or

y-coordinate, of P . Finally, through P there passes a - unique plane perpendicular to the z-axis, and this plane intersects the z-axis in a point which has a coordinate, say - z , in the coordinate system determined on OZ by the ordered pair (0,~) . The number z we define to be the third

coordinate, or - z-coordinate, of P .

Conversely, if any ordered triple, say (x*,yl,z*) is

given, there is a unique point P* having (x*,y*,zl) as its

coordinates. In fact, there is a unique plane perpendicular to

the x-axis at the point whose x-coordinate Is x * , and there is a unique plane perpendicular to the y-axis at the point

whose y-coordinate is y* . These two planes cannot be

parallel (Why?), hence they must intersect in a line, m , which by Theorem 9-23, is perpendicular to the xy-plane.

Finally, there is a unique plane which is perpendicular to the z-axis at the point whose z-coordinate is z * . Since m is

perpendicular to this plane (Why?), It must intersect it in a

point P* , whose coordinates are clearly (xl,y*,zl), as

required.

Figure (a)

As our efforts in this chapter have already illustrated,

It is difficult to represent space configurations by drawings

on a sheet of paper. You ought to practice plotting In a three- dimensional coordinate system so that you can make drawings and visualize the space relations which they suggest. Of course,

you should begin your practice with simple situations, such as

plotting a single point, a pair of points, a segment, or a line.

9-6 Figure (a) shows how a single point may be plotted. Figures (b) and (c) below show a segment and a line. Notice that an essential technique in plotting is the ability to draw a line

parallel to a coordinate axis. Notice also the significance of

perspective along the x-axis. You may wish to refer again to

Appendix V for further help.

Figure (b ) Figure (c)

Postulate 24, it is clear that the procedure which determines the coordinates of a point, P , works just as well when P is in one or more of the coordinate planes as it does when P does not lie in any of the coordinate planes. Hence,

it should be easy for you to verify the results which are

summarized in the following table.

~quation(s) satisfied by the coordinates of any point of the set

Set of points

origin

'~orm of the coordinates of any point of the set

I --

(0,090) x = 0 and y = 0 and z = 0

x-axis

y-axis

z-axis

xz-plane yz-plane

I I

From the definition of the x-coordinate of a point, P ,

(x,O,o)

(O,Y,O)

(o,O,z)

xy-plane

it is clear that all points which lie in a particular plane, /C? , perpendicular to the x-axis have the same x-coordinate,

y = 0 and Z = O X = O and 2 = o X = O and y = 0

2 = 0

say x = x-, . It is also clear that, conversely, any point

whose x-coordinate is xl must lie in the plane /^ . In

other words, the coordinates of every point in a plane which is perpendicular to the x-axis satisfy an equation of the form x = xl , and conversely, any point whose coordinates satisfy this equation lies in this plane.

Similarly, we can say that all points which lie in a plane perpendicular to the y-axis have the same y-coordinate, say

y = y, , or in other words have coordinates which satisfy the equation y = y1 , and conversely.

What do these observations tell us about the coordinates of points which lie on the line of intersection of a plane

perpendicular to the x-axis and a plane perpendicular to the y-axis? Do you see that if a point P lies simultaneously In

a plane whose points have coordinates satisfying theequation x = xi and in a plane whose points have coordinates satisfying

the equation y = y, , then the coordinates of P must satisfy

both of these conditions? If you understand this, it should not be hard for you to verify the assertions in the following table.

9-6 Set of points

plane 1 x-axis plane 1 y-axis plane 1 z-axis

Form of the coordinates of any point of the set

line 1 xy-plane line 1 xz-plane line 1 yz-plane

~quatlon(s) satisfied by the coordinates of any point of the set

x = xl and Y = Y1

x = xl and z = z 1

Y = Yl and z = z 1

Problem Set 9-6

1. Using the same set of axes, plot the points P, Q, R, S,

T, U, V : ~(0,1,0) ; Q(-3,0,0) ; R(-3,l,0) ; s(-3,1,4) ;

~(3,1,4) ; ~(3,-l,4) ; ~(3,-1,-4) . 2. Using the same set of axes, plot the" points A, B, C, D, E :

~(0,-l,3) ; ~(3,4,6) ; c(-4,2,-7) ; ~(1,-3,0) ; ~(5,2,-4) . 3. Describe the location of all the points in space for which

(a) x = 0 . (b) x = 2 . (c) x = -3 . Illustrate with a sketch for each part.

4. Sketch the set of all points in space which satisfy the given condition.

(a) y = 0 . (b) y = 2 .

(c) z = 0 . (d) z = -4 .

5. Describe the set of points represented by each of the

following:

(a) ((x,y,z):y = z = 0) . (b) ((x,y,z):x = y = 01 . (c) {(x,y,z):x= z = 0) . (d) ((x,y,z):x = y = z = 0) . ( e ) ((x,y,z):y = 2 and x = 0) . (f) [(x,y,z):x = 2 , y = 1) .

6 . Suggest a convenient set of coordinates for the eigiit

vertices :

(a) of a cube each of whose edges has length a ;

(b) of a rectangular solid (parallelepiped) having

mutually perpendicular edges of lengths a, b, c , re spec tively .

7. Where are all the points in space for which x + y = 2 '?

Sketch the graph.

9-7. - The Distance Formula in Space. -- In Section 8-2, we proved two theorems heor or ems 8-1 and

8-2) which enable us to determine the distance between any two

points on a line parallel to either of the coordinate axes.

Similar results hold in space, and we have, specifically:

THEOREM 9-24. If P, and Pn are points on a line parallel

to the x-axis, then PIPo = lxl - x2 1 , where x and 1 x2 are the x-coordinates of Pl and P2 , respectively.

THEOREM - 9-25. If Pi and Pg are points on a line parallel

to the y-axis, then PIPp = lyl - y21 , where y, and

y2 are the y-coordinates of Pl and P2 , respectively.

THEOREM - 9-26. If P, and P2 are points on a line parallel

to the z-axis, then PIPo = lzl - z21 , where zl and

z2 are the z-coordinates of Pl and P2, respectively.

The proofs of these theorems a r e very easy and we sha l l omit them. Two appl icat ions of these theorems a re i l l u s t r a t e d i n t h e following f igures .

(a) If A = (6,3,1) and 3 = (3,3,1) , then AB =

(b) If C = (3,5,1) and B = (3,3,1) , then BC = I3 - 51 = 2 .

As one of the important applications of plane coordinate systems, we developed the so-called distance formula

which enables us to find the distance between any two points, ~,(x~,y,) and p2(x2,yp) , In a plane. It is now natural to seek a formula, analogous to the distance formula in a plane, which will express the distance between any two points In space In terms of the coordinates of the points.

To do this, let ~,(x,,~ ,z ) and P~(~~.Y*.z~) be any two points in space, and consider the figure which is formed by the three planes through P1 which are respectively perpendicular to the coordinate axes and the three planes through Pg which are respectively perpendicular to the coordinate axes. Let A and F be the points (xy,y1,zl)

First, suppose that P1 and P2 , do not lie in a plane which is perpendicular to one of the coordinate axes. Clearly, - PiA is perpendicular to the yz-plane (Theorem 9-23) and hence - parallel to the x-axis (postulate 25). Similarly A F is perpendicular to the xz-plane and hence parallel to the y-axis, and 7 is perpendicular to the xy-plane and hence parallel to the z-axis. Therefore / P ~ A F and / P ~ F P ~ are right angles (why?). Hence, applying the Theorem of Pythagoras to the two right triangles APIFPo and APIAF , we obtain respectively

and

Now substituting for ( F P ~ ) ~ from the second of these equations into the first, we get

But according to Theorems 9-24, 9-25, and 9-26, respectively,

Therefore, substituting,

2 O P since lq12 = q for every real number q ,

If Pl and P2 lie in a plane which is perpendicular to one of the coordinate axes, then either x2 = xl , y2 = yl ,

- zl , and one or more of the terms under the radical in or z2 - the last formula Is zero. It is easy to see that the formula is still valid. For instance, if z2 = zl , then P2 = F , and clearly the correct expression for Popl is

Likewise, if P, and P2 determine a line which Is perpendicular to one of the coordinate planes, then two of the coordinates of Pi must equal the corresponding coordinates

of Po , and only one of the terms under the radical in the formula for P2Pl is different from zero. Again the formula

is valid. For instance, if y2 = yl and z2 = zl , then P2 = A and the correct expression for PpPl is

Finally, If P2 = P, , then x2 = x1 , Y2 = Y1 9 Z2 = z1 and every term in the formula for PpPl is zero. Thus P2Pl = 0 , as of course it should if P2 = Pl . Hence, in every case, the three-dimensional distance formula

gives us the distance between the points Pl(xl,yl,zl) and

P ~ ( x ~ , ~ ~ , z ~ ) . We state this result as a theorem.

THEOREM - 9-27. The distance between the points P (xl,yl, zl)

Example - 1

What is the distance between the points p1(2,4,0) and

p2(1,2,-2) ?

By direct substitution into the three-dimensional distance formula, we obtain

Example - 2

Find the points which lie on the line perpendicular to

the xy-plane at the point (2,3) and are at a distance of 7 from the origin.

Clearly, any point which lies on the line perpendicular to the xy-plane at the point (2,3) has x-coordinate 2 and

y-coordinate 3 . Hence, a point on this line is determined

as soon as Its z-coordinate is known. Thus, we must determine

the value, or values, of z such that the distance from the

origin, 0(0,0,0) , to the point ~(2,3,z) is 7 . Using the

distance formula we thus have

Thus there a re two points which meet the requirements of the problem, namely p1(2,3,6) and p2(2,3,-6) , as can be checked immediately.

Problem Set 9-7 --

A l i n e m i s perpendicular t o the yz-plane a t the point ~ ( 0 , 3 , 4 ) . Find the points which l i e on l i n e m and a re a t a distance of 13 from the or ig in .

A l i n e q i s perpendicular t o the xy-plane a t the point ~ ( 6 , 8 , 0 ) . Find the points which l i e on l i n e q and a re at a distance of 10 from the or ig in .

A l i n e 1 i s perpendicular t o the xz-plane and contains the point ~ ( 1 , - 2 , l ) . Find the points of 1 which a re a t a distance of 4 from the or ig in .

Show tha t A ABC with ve r t i ces ~ ( 2 , 4 , l ) , ~ ( l , 2 , - 2 ) , C(5,0,-2) i s a r igh t t r i ang le .

Is the t r i ang le with ve r t i ces ~ ( 2 , 0 , 8 ) , ~ ( 8 , -4,6) , c(-4,-2,4) isosceles? J u s t i f y your answer.

Given the ver t ices of two t r i ang les , AABC and ADEF ; fo r each of the t r iangles , determine i f it i s equ i l a t e ra l .

(a) Show t h a t the opposite s ides of the f igure ABCD

with ver t ices ~ ( 3 , 2 , 5 ) , ~ ( l , l , l ) , ~ ( 4 , 0 , 3 ) , ~ ( 6 , l , 7 ) a re congruent.

( b ) Does t h i s prove t h a t the f igure i s a parallelogram? Explain.

(a) Show that the opposite sides of the figure ABCD with vertices ~(5,l,l) , ~(3,1,0) , ~(4,3,-2) , ~(6,3,-1) are congruent.

(b) Show that the angles of the figure in Part (a) are all right angles.

(c) Do the results in (a) and (b) prove that the figure is a rectangle? Explain.

Using coordinates, prove Theorem 9-18. (~int: This

follows closely the proof of the corresponding theorem in a plane. )

9-8. Parametric Equations of a Line in Space. ----- In Section 8-7, we obtained what we called parametric

equations of the line determined by two distinct points,

P ~ ( ~ ~ , Y ~ ) and P2fx29~2) 9

For every value of k , the corresponding numbers, x and y , are the coordinates of a point on P , P: and,

Â¥4Ã‘Ã J. c

conversely, to every point on PIPp there corresponds a unique value of the parameter k such that these equations give the coordinates of the point. By an argument similar to the one which establishes the result in the plane, it is possible to establish the corresponding result for space.

z ) are any THEOREM 9-28. If P~( X ~ , Y ~ , Z ~ ) and p2(x2y2,

two distinct points, then for every value of k the

point whose coordinates are

x = x, + k(x2 - x,) Y = Y, + k(y2 - Y,) z = zl + k(z2 -

lies on= and, conversely, to every point on=

there corresponds a unique value of k such that these equations give the coordinates of the point.

6 58

Example - 1 What are the coordinates of the point in which the line

determined by ~,(3,7,2) and p2(l,l, -2) intersects the xy-plane?

We know that a point is in the xy-plane if and only If Its z-coordinate Is zero. Our problem is to find the point on ,PiP/, whose z-coordinate is zero. Now, by the last theorem, the coordinates of any point on are given by the equations

Hence, for the point whose z-coordinate is zero, we must have

2 - 4 k = 0 or 1 k = - 2

Substituting this value into the formulas for x and y , we find

The required point is the point with coordinates (2,4,0) .

Example - 2

Show t h a t t he diagonals of a cube ( a ) have equal lengths, ( b ) b i s e c t each o the r and ( c ) a r e not perpendicular t o each o the r .

Proof: Let t he leng th of each edge of the cube be 2a ,

Choose the coordinate axes so t h a t one ver tex, A , i s a t t he

o r i g i n and t h r e e edges l i e i n t he pos i t i ve x- , y- , z-axis. - Then, t h e endpoints of diagonal AB a r e ~ ( " 0 0 ~ 0 ) and ~ ( 2 a , 2 a , 2 a ) . Another diagonal, CD , has endpoints ~ ( 0 , 0 , 2 a )

and ~ ( 2 a , 2 a , 0 ) . 2 2 2 - + 4 s a 2 = 2 a t / 3 ' , ( a ) AB = n/f2a - 0) + (2a - 0 ) + (2a - 0) -

CD = & - o12 + (2a - o ) ~ + ( 0 - 2a)2 = 1/1Z= 2 a f l . Therefore, AB = CD . Simi la r ly , AB = GH = El? ,

The leng th of each diagonal of a cube i s A/T times the length of each edge. - ( b ) AB = ( (x ,y , z ) :x = 0 + 2ak , y = 0 + 2ak , z = 0 + 2ak ,

k i s a r e a l number] . By tak ing k = 6 , we f i n d the midpoint of t o be

(a ,a ,a ) - Similar ly , CD = [(x,y,z):x = 0 + 2ah , y = 0 + 2ah ,

z = 2a - 2ah , h i s a r e a l number) . - , we f i n d tne midpoint of CD t o be By tak ing h =

( a , a , a )

9-8 - Similarly, we find that the midpoint of EF is (a,a,a) - and the midpoint of GH is (a,a,a) . Thus, any two of the diagonals bisect each other.

(c) If M is the common midpoint of the diagonals, then, by

AB = and CM = a1/3' . Thus Part (a), AM = -y 2 2 AM)^ = 3a2 and (CM) = 3a2 , but (AC)~ = (2a)2 = la . - - Therefore, AM)^ + (14~)~ # ( A C ) ~ . Hence, AM and CM

are not perpendicular.

By this same reasoning all other pairs of diagonals can

be proved not perpendicular.

Problem Set 9-8 --

1. In Example 2 above, prove GH = EF = AB . 2. Given points A(-2,0,4) , and ~(8,2,-2) , use set

notations and parametric equations to express - (b) TB . 4

(a) AB . (c) AB . -

3. (a) Find the midpoint of A 3 in Problem 2. - ( 0 ) Find the trisection point of AB nearer A . - ( c ) Find the trisection point of AB nearer B .

d (d) Find P if P is in AB and AP = 3AB . - (e) Find P If P is in the ray opposite to AB

and AP = 3AB . (f) Find the coordinates of the point in which %?

Intersects the xy-plane; the xz-plane; the yz-plane. H

(g) Find the coordinates of the point in which AB inter-

sects the plane whose equation is z = 3 ; whose

equation is y = -2 ; whose equation is x = -3 . 4. Prove that the diagonals of a rectangular solid are equal

in length and that they bisect each other.

5. Show that A(-l,5,3) , ~(l,4,4) and ~(5 ,2 ,6 ) are

collinear.

6. What are the coordinates of the point P in which the line determined by ~,(2,1,3) and ~~(3,-2,1)

intersects the yz-plane?

7. What are the coordinates of the point P in which the

line .determined by p1 (-1,2, -1) and ~ ~ ( 3 , -2,2)

intersects the xz-plane?

8. A rectangular solid has three adjacent faces in the coordinate planes. One vertex is at the origin and another has coordinates (2a,2b,2c) . What are the

possible relationships among a, b, c if two of the

diagonals are perpendicular to each other?

9. (a) Given the points ~(7,1,3) , ~(4,-2,3) , find the coordinates of the midpoint, M , of 7R? .

(b) Consider the points ~(5,0,0) and ~(6,y,z) . Find y and z so that the midpoint of is the same point M as in Part (a).

( c ) Is figure ABCD a parallelogram? Explain.

10. Using ideas of midpoints, as In Problem 9 above, show that the figure in Problem 8 of Problem Set 9-7 - is a parallelogram.

11. Using ideas of midpoints, as in Problem 9 above, show that the figure in Problem 9 of Problem Set 9-7 - is a rectangle.

9-9. Equation of a Plane. --- Our study of the equations of lines in both two-and three-

dimensional coordinate systems raises the question of whether or not planes in space can likewise be characterized by equations. The answer is Yes , and we shall conclude the chapter by finding an equation corresponding to a plane. First, however, it is convenient to introduce the following

definition.

9-9 DEFINITION. If /^ is any plane, then an equation of ^y is any equation with the following properties: - (a) The coordinates (x,y,z) of every point of f l

satisfy the equation;

(b) Any values (x,y,z) which satisfy the equation

are the coordinates of a point of . Consider the following example.

Example - 1. Let F be the point (3,2,4) and let f l be the plane containing F and perpendicular to OF.

A point ~(x,y,z) lies in /^ if and only if R = F or -- PR 1 OF . By the Pythagorean Theorem, /OFR is a right angle

if and only if

(1) OR^ = OF)^ + (FR)~ . 2

Now (OR) = (x - 0 ) + (y - 0 ) + (z - 0 ) = x + y2 + z 2

and ( 0 ~ ) ~ = (3 - o ) ~ + (2 - o ) ~ + (4 - o ) ~ = 9 + 4 + 16 = 29 2

and ( F R ) ~ = (x - 3)2 + (y - 2)2 + ( z - I ) ~ = x + y2 + z 2 - 6~ - 4y - 8~ + 29 .

Hence, (1) becomes 2 + y 2 + ~ 2 S 2 9 + (x 2 + y 2 + z 2 - 6 x - Q y - 8 2 + 2 9 )

Thus

9-9 As a check, we observe that the coordinates (3,2,4) of F satisfy this equation, since 3 3 + 2 2 + 4 4 = 29 .

Note that the numbers 3,2,4 which appearas the respective coefficients of x,y,z in the equation of the plane

are the same as the coordinates of the point F , and that the number 29 is the sum of the squares of the coordinates of F ,

We use the discussion in Example 1 as a guide in treating the general case. Suppose that is any plane not containing the origin 0 . Let ~(a,b,c) be the point where intersects the line

containing 0 and perpendicular to . By Theorem 9-1, a point ~(x,y,z) lies in the plane /c? if and only if R = F or /OFR is a right angle. But by the Pythagorean Theorem, AOFR has a right angle at F if and only if

Since we know the coordinates of 0, R, and F , it Is a simple matter to obtain the distances OR , OF , and FR by means of the three-dimensional distance formula. Hence the last equation

can be written

or, squaring the binomials and collecting terms,

or finally, 2 2 a x + b y + c z = a 2 + b + c .

This equation is satisfied by the coordinates of every point of

including F , and by the coordinates of no other points. In other words, this equation is an equation of the plane.

If contains the origin, the above derivations must be

modified a little. In this case, let m be the line which is

perpendicular to at the origin and let ~(a,b,c) be any point on m except the origin. A point ~(x,y,z) will now

lie in if and only if R is the origin or /LOR is a

right angle.

But A LOR will have a right angle at 0 if and only if

9 -9

Evaluating these distances by means of the three-dimensional distance formula and simplifying as we did in the previous case, we now find that the coordinates (x,y,z) of any point in /E? , including the origin, must satisfy the equation,

a x + b y + c z = O .

The only distinction between this equation and the equation of a plane which does not contain the origin is the value of the

constant term.

Our discussion thus far has not touched on the related question: Is every equation of the form

ax + by + cz = d

an equation of some plane? The answer to this is Yes , but we shall not take time to prove this fact. Instead, we merely summarize our observations in the following theorem.

THEOREM - 9-29. Every plane has an equation of the form ax + by + cz = d , where one or more of the numbers a, b, c is different from zero; and every equation of this form is an equation of a plane.

Example 2. - What is an equation of the plane which is determined by

the points ~,(2,0,0) , p2(0,1,0) , ~,(0,0,3) ?

By Theorem 9-29, we know the required plane has an equation of the form

which is satisfied, in particular, by the coordinates of Pi , P2 , and Pi . If the coordinates of Pi satisfy this equation, then substituting,

9 -9

Similarly, since the coordinates of Po satisfy the equation,

a o O + b . l + c * O = d or b = d

and, since the coordinates of P3 also satisfy the equation,

d a - 0 + b * O + c b 3 = d or c = ~ .

Substituting for a, b, and c we obtain

d ^ x + d y + p = d , 2

or, multiplying both members by 6 and dividing by d , 3 x + 6 y + 2 z = 6 .

This is the required equation.

Example - 3. What is an equation of the plane whose points are equi-

distant from the points A(1,-3,0) and ~(2.0,-5) ?

Let ~ ( x , y , z ) be any point. The condition that P lie

on tne required plane is expressed by the equation PA = PB . That is,

(x2 - 2x + 1) + (y2 + 6y + 9 ) + z2 = (x2 - 4x + 4 ) + y 2

+ (z2 + 10z + 25) or (by rearranging terms)

Finally, by combining terms, we obtain the equation of the plane in simple form:

2x + 6y - 102 = 19 .

Example - 4 .

Sketch the plane f i given by

f l = ( (x ,y ,z) :2x - y + 32 = 61

The coordinates of each point of i n t e r s e c t i o n between and a coordinate a x i s a r e r e a d i l y determined, as follows. If

a point i s on the x-axis, i t s y-coordinate i s zero and i t s

z-coordinate i s zero; thus t he In t e r sec t ion of and the x-axis i s the point (3,0,0) because 2x - 0 + 3 0 = 6 y i e l d s x = 3 . Similar ly , t h e y-axis I s [ (x ,y ,z) :x = 0 = z )

and it i n t e r s e c t s /?? a t t h e point (0 , -6 ,0) . The point of

i n t e r s e c t i o n of f l and the z-axis i s (0,0,2) . A "sketch" of the plane f t i s conveniently made by p l o t t i n g the t r i a n g l e whose v e r t i c e s a r e the t h r e e po in t s

(0,092)

x Example - 5.

Find an equation of the plane which contains the th ree po in t s (1 ,2 ,3) , (2,1,-3) and (-1,-2,1) .

A point I n the plane must have coordinates (x,y,z) such t h a t ax + by + cz = d . That is , i f the point (1,2,3) i s i n t h e plane, then

9-9 Similarly, if the point (2,1,-3) is in the plane then,

a 9 2 + b l+c(-3)=d

and, if the point (-1,-2,l) is in the plane then

a(-1) + b(-2) + c 1 = d . We find values of a, b, c, in terms of d , which

satisfy all three of these equations, to be

lid a = ~ , 7d b = - T - , c = ~ . 3d Substituting these values in

the equation ax + by + cz = d yields

or

llx - 7y + 3z = 6 , an equation of the required plane.

Problem Set 9-9 -- 1. Write an equation of the plane determined by three points

whose coordinates are

2. Determine an equation of the plane whose points are

equidistant from pl(l,2,3) and p2(2,5,4) . 3. Sketch a diagram of' the plane represented by each of

the equations:

(a) 5x + 4y = 20 ;

(b) x + 2 y + z = 5 .

4 , Find an equation of the plane which contains the point

~(1,-2,2) and is perpendicular to the line containing Q and the origin.

Find an equation of the plane which contains the three points whose coordinates are:

(a) (0,-2,1) ; (2,0,-1) ; (-2,-32)

(b) (1,-2,1) ; (2,03-1) ; (-29-332)

(c) (1,-2,1) ; (293,-1) ; (-2,-32)

Find the coordinates of the point of intersection of

the plane ft and the line A? , if p= {(x,yYz):3x + 5y + 142 = 11)

and

= [(x,y,z):x = 2 - 3 k , y = 1 + k , z = 4 - 2k, kreal)

9-10. Summary.

In this chapter we have studied properties of parallelism

and perpendicularity for lines and planes. The relationship

of parallelism for lines in space is reflexive, symmetric, and

transitive. The same three fundamental properties hold for

parallelism of planes. The relationship of perpendicularity for lines in space is symmetric, but neither reflexive nor

transitive. The same three remarks apply to perpendicularity

for planes.

If a point and a line in space are given, there are:

a unique line containing the given point and parallel to

the given line,

many planes containing the given point and parallel to the given line,

a unique line containing the given point and perpendicular

to the given line,

a unique plane containing the given point and perpendicular to the given line.

If a point and a plane in space are given, there are:

many lines containing the given point and parallel to the

given plane,

a unique plane containing the given point and parallel to

the given plane,

(c) a unique line containing the given point and perpendicular

to the given plane,

(d) many planes containing the given point and perpendicular

to the given plane.

Given a unit-pair, a postulate in Chapter 3 described for us the distance between two points. In Chapters 4, 5, 6 our theorems on perpendicularity and parallelism enabled us to

introduce the distance between a line and a point and the

distance between two parallel lines. In Chapter 9 our study of perpendicularity and parallelism permits us to extend the

notion of distance again. We can speak of the distance between

a point and a plane, the distance between a line and a plane that are parallel to each other, and the distance between two

parallel planes.

The ideas of parallelism, perpendicularity, and distance

play a basic role in developing a three-dimensional coordinate

system. In a one-dimensional system a point is identified by a single real number, in a two-dimensional system by an ordered

pair of numbers, and in a three-dimensional system by an ordered triple of numbers. The formula for the distance between two

points in space is a natural extension of the formula in

two-dimensional geometry. The parametric equations of a line

in space are a natural extension of the parametric equations

in two-dimensional geometry. The first-degree equation in

x, y, z, representing a plane in space, is a natural extension of the first-degree equation in x, y, representing a line in two-dimensional geometry. The coordinate method for proving

theorems or analyzing problems is fully as useful and

convenient in three-dimensional situations as in two.

VOCABULARY LIST

parallel planes

perpendicular planes

plane angle of a dihedral angle

measure of a dihedral angle coordinate system (in space)

coordinate plane

distance formula (in space)

equation of a line (in space)

equation of a plane

Review Problems

Chapter 9, Sections 6 to 9

1. Plot the points A, B, C, and D if the coordinates of the points are: A(2,-2,5) ; ~(2,3,4) ; ~(2,3,-4) ; D(-2,0,4) .

2. Find the distances between the following pairs of points:

3. Find the midpoint of each of the segments determined by

the pair of points in Problem 2.

4. Write parametric equations of the line determined by each

pairs of points In Problem 2.

5. The coordinates of the midpoint of a segment are (8,-4,l) . If the coordinates of one of the endpoints of the segment are (4,-1,3) , find the coordinates of the other endpoint of the segment.

6 . The coordinates of the vertices of a triangle are given in each of the following problems. Classify the triangle in each of the problems.

7. The coordinates of three points are listed in each of the following problems. Tell whether the points are collinear or noncollinear.

8. Given four distinct points A, B, C, and D . If

AB = CD and AD = BC , is ABCD a parallelogram? Explain.

Given four noncollinear points A, B, C, and D . 1f - the midpoint of ^5" is the midpoint of BD , is ABCD

a parallelogram? Explain.

Does the plane whose equation is x + y + z = 4 contain the point whose coordinates are (3,-1,2) ?

What is the intersection of the xy-plane and the plane whose equation is 2x - 3y + z = 6 ?

Describe the following: ((x,y,z):y = 5 ) . What is the equation of a plane whose points are equidistant from the endpoints of a line segment with coordinates (-2, -4,7) and (4,5,1) ?

M is the midpoint of an edge of the rectangular solid

shown in the figure below. Prove by means of coordinates

that MB = MC .

REVIEW PROBLEMS

C h a p t e r s 6-9

Write + i f t h e s t a t e m e n t i s t r u e ; 0 i f t h e s t a t e - ment i s f a l s e :

1. The measure o f a n e x t e r i o r a n g l e o f a t r i a n g l e i s g r e a t e r t h a n t h e measure o f a n y i n t e r i o r a n g l e o f t h e t r i a n g l e .

2 . Two a n t i p a r a l l e l r a y s a r e d i s t i n c t .

3. The a n g l e o p p o s i t e t h e l o n g e s t s i d e o f a t r i a n g l e i s t h e a n g l e t h a t has t h e g r e a t e s t measure .

4. A set o f p a r a l l e l l i n e s i n t e r c e p t s c o n g r u e n t segments on a n y t r a n s v e r s a l .

5 . If E l = , t h e n A B < A C .

6 . T h e r e i s a t r i a n g l e whose s i d e s h a v e l e n g t h s 351, 513, and 135.

7. Two l i n e s a r e p a r a l l e l i f e a c h o f them i s p e r p e n d i - c u l a r t o t h e same l i n e .

8. Given two l i n e s and a t r a n s v e r s a l of t h e l i n e s , i f one p a i r o f a l t e r n a t e i n t e r i o r a n g l e s a r e c o n g r u e n t , t h e

o t h e r p a i r a r e a l s o c o n g r u e n t .

9. Given two i n t e r s e c t i n g l i n e s and a t r a n s v e r s a l o f t h o s e l i n e s , no p a i r o f c o r r e s p o n d i n g a n g l e s d e t e r m i n e d by

t h e g i v e n t r a n s v e r s a l a r e c o n g r u e n t .

10. The b i s e c t o r s o f a p a i r o f c o n s e c u t i v e i n t e r i o r a n g l e s a r e p a r a l l e l .

11. A t a p o i n t on a l i n e , t h e r e a r e i n f i n i t e l y many l i n e s p e r p e n d i c u l a r t o t h e l i n e .

12. The d i s t a n c e between a l i n e and a p o i n t n o t on t h e line i s t h e l e n g t h o f a n y segment c o n n e c t i n g t h e

p o i n t and t h e l i n e .

13. The p l a n e s which c o n t a i n t h e r e s p e c t i v e f a c e s o f a r i g h t d i h e d r a l a n g l e a r e p e r p e n d i c u l a r .

If two a n g l e s o f one t r i a n g l e a r e c o n g r u e n t r e s p e c t i v - e l y t o two a n g l e s o f a n o t h e r t r i a n g l e , t h e n t h e t h i r d a n g l e s a r e c o n g r u e n t .

The a c u t e a n g l e s o f a r i g h t t r i a n g l e a r e complementary.

An e x t e r i o r a n g l e o f a t r i a n g l e i s t h e supplement o f o n e o f t h e i n t e r i o r a n g l e s o f t h e t r i a n g l e .

One o f t h e a n g l e s o f a r i g h t t r i a n g l e may be a n ob- t u s e a n g l e .

Two r i g h t t r i a n g l e s a r e c o n g r u e n t i f t h e hypo tenuse and

a" l e g o f o n e a r e c o n g r u e n t r e s p e c t i v e l y t o t h e hypoten- u s e and a l e g o f t h e o t h e r .

I f a l i n e i n t e r s e c t s one o f two p a r a l l e l l i n e s , i t

i n t e r s e c t s t h e o t h e r .

Two l i n e s t h a t a r e e q u a l a r e n o t p a r a l l e l .

Any two c o n s e c u t i v e a n g l e s of a p a r a l l e l o g r a m a r e s u p p l e m e n t a r y .

I n A ABC, i f m LA = 50 and m L B = 40, t h e n i s t h e l o n g e s t s i d e o f t h e t r i a n g l e .

I f x, y, and z a r e three l i n e s such t h a t x 1 1 y

and y I z, t h e n x 1 1 z .

If x, y , and z a r e three l i n e s such t h a t x 1 y ,

and y l z , t h e n x l z .

The c o n t r a p o s i t i v e o f a s t a t e m e n t i s l o g i c a l l y e q u i - v a l e n t t o t h e c o n v e r s e o f t h e s t a t e m e n t .

A t r i a n g l e h a s a r i g h t a n g l e i f t h e l e n g t h s o f t h e s i d e s o f t h e t r i a n g l e a r e p r o p o r t i o n a l t o 7, 29, 25.

If one p a i r o f o p p o s i t e s i d e s o f a q u a d r i l a t e r a l a r e p a r a l l e l and c o n g r u e n t , t h e n t h e q u a d r i l a t e r a l i s a p a r a l l e l o g r a m .

The l e n g t h o f t h e d i a g o n a l o f a s q u a r e c a n b e found by m u l t i p l y i n g t h e l e n g t h o f a s i d e by fi.

A d i h e d r a l a n g l e i s t h e u n i o n o f two h a l f p l a n e s .

G iven three d i s t i n c t c o ~ l a n a r p a r a l l e l l i n e s and two

d i s t i n c t t r a n s v e r s a l s , t h e s e g m e n t s formed on o n e o f

t h e t r a n s v e r s a l s a r e p r o p o r t i o n a l t o t h e c o r r e s p o n d -

i n g s e g m e n t s formed on t h e o t h e r t r a n s v e r s a l .

I n A A B C , i f m L A < ~ L B , t h e n A C < BC.

If t h e l e n g t h s o f t h e s i d e s o f a t r i a n g l e a r e 2 0 , 2 1 , and 31, t h e t r i a n g l e i s a r i g h t t r i a n g l e .

If t h e m e a s u r e of o n e o f t h e a n g l e s o f a r i g h t t r i a n g l e

i s 30, t h e n t h e l e n g t h o f t h e l e g o p p o s i t e t h a t

a n g l e is e q u a l t o o n e - h a l f t h e l e n g t h o f t h e hypo-

t e n u s e .

G iven a c o r r e s p o n d e n c e be tween two t r i a n g l e s , i f two

a n g l e s o f o n e t r i a n g l e a r e c o n g r u e n t t o t h e c o r r e s -

p o n d i n g a n g l e s o f t h e o t h e r , t h e c o r r e s p o n d e n c e i s

a s i m i l a r i t y .

I f a , b , c , a r e t h e l e n g t h s o f t h e s i d e s o f o n e

t r i a n g l e , i f k i s a p o s i t i v e number, and i f

a k , bk , c k , a r e t h e l e n g t h s o f t h e sides of a n o t h e r

t r i a n g l e , t h e n t h e t r i a n g l e s a r e s i m i l a r .

G iven two t r i a n g l e s , i f a n a n g l e o f o n e t r i a n g l e i s c o n g r u e n t t o a n a n g l e o f t h e o t h e r , and two s i d e s of

o n e t r i a n g l e a r e p r o p o r t i o n a l t o two s i d e s of t h e

o t h e r , t h e t r i a n g l e s a r e s i m i l a r .

I f t n e l e g s o f a r i g h t t r i a n g l e h a v e l e n g t h s a and b,

and i f t h e h y p o t e n u s e h a s l e n g t h c , t h e n 2 o = ( c - a ) ( c + a ) .

G iven a c o r r e s p o n d e n c e be tween two t r i a n g l e s , i f two

a n g l e s and a s i d e o f o n e t r i a n g l e a r e c o n g r u e n t t o

t h e c o r r e s p o n d i n g p a r t s o f t h e o t h e r , t h e c o r r e s -

o o n a e n c e is a c o n g r u e n c e .

Given a c o r r e s p o n d e n c e be tween two t r i a n g l e s , i f two

s i d e s and a n a n g l e o f o n e t r i a n g l e a r e c o n g r u e n t t o t h e c o r r e s p o n d i n g p a r t s of t h e o t h e r , t h e c o r r e s -

pondence i s a c o n g r u e n c e .

Two i s o s c e l e s t r i a n g l e s a r e congruent i f t h e v e r t e x

a n g l e and t h e base o f one t r i a n g l e a r e congruent r e s p e c t i v e l y t o t h e v e r t e x a n g l e and t h e b a s e of t h e

o t h e r .

Any two r e c t a n g l e s a r e s i m i l a r .

2 If ( a , x) = ( x , b ) , t hen ab = x . Congruent convex polygons a r e s i m i l a r w i t h a propor- t i o n a l i t y c o n s t a n t o f 1.

Any two e q u i l a t e r a l t r i a n g l e s a r e s i m i l a r .

I f x and y a r e two d i s t i n c t p o s i t i v e numbers, i f t h e l e n g t h s of two s i d e s o f a r e c t a n g l e a r e x and y, and i f t h e l e n g t h s o f t h e s ides of a second r e c t a n g l e a r e x + 6 and y + 6, then t h e r e c t a n g l e s

a r e s i m i l a r .

If a , b a r e p r o p o r t i o n a l t o c , d w i t h propor- t i o n a l i t y c o n s t a n t k, and c , d a r e p r o p o r t i o n a l t o e, f w i t h p r o p o r t i o n a l i t y c o n s t a n t g , then a , b a r e p r o p o r t i o n a l t o e, f w i t h p ropor t ion -

k a l i t y c o n s t a n t - . g

If a l i n e i n t e r s e c t s t h e i n t e r i o r s o f two s i d e s o f a t r i a n g l e s o t h a t co r r e spond ing segments a r e propor- t i o n a l , t h e l i n e i s p a r a l l e l t o t h e t h i r d s i d e .

The r a t i o o f t h e p e r i m e t e r s of two s i m i l a r t r i a n g l e s i s e q u i v a l e n t t o t h e r a t i o of any p a i r of c o r r e s -

ponding s i d e s .

I f p, q a r e p r o p o r t i o n a l t o a , b w i t h propor- b a t l o n a l i t y c o n s t a n t k, then - = - P q -

50. Any r e a l number is p e r m i t t e d t o be a c o n s t a n t o f p r o p o r t i o n a l i t y .

51. The l i n e s which c o n t a i n t h e r e s p e c t i v e b i s e c t o r s o f

t h e a n g l e s of a t r i a n g l e a r e c o n c u r r e n t a t a p o i n t e q u i d i s t a n t f rom t h e v e r t i c e s o f t h e t r i a n g l e .

52. Any p a i r o f o p p o s i t e a n g l e s o f a n i s o s c e l e s t r a p e z o i d a re supplernenta r y .

If t h e d iagona 1s o f a q u a d r i l a t e r a l a r e p e r p e n d i c u l a r and b i s e c t e a c h o t h e r , t h e q u a d r i l a t e r a l i s a rhombus.

If t h e c o o r d i n a t e s o f a q u a d r i l a t e r a l ABCD a r e

A(-5, -2), B(-4, 2 ) , ~ ( 4 , 6 ) , ~ ( 3 , I ) , t h e n t h e q u a d r i l a t e r a l i s a p a r a l l e l o g r a m .

The p o i n t s (0, 6, -2), ( 3 , 4, -2), and (1, 4, -5) a r e t h e v e r t i c e s o f a n e q u i l a t e r a l t r i a n g l e .

If two segments a r e c o n g r u e n t , t h e i r p r o j e c t i o n s on a g i v e n l i n e a r e c o n g r u e n t .

If a p l a n e i s p a r a l l e l t o o n e o f two d i s t i n c t p a r a l l e l

l i n e s , i t i s p a r a l l e l t o t h e o t h e r .

Given A(-1, 01, ~ ( 6 , 2), c ( 4 , S), D(-3, 3 ) , <Ã‘

t h e n AB I I %. The d i s t a n c e between a n y two d i s t i n c t p o i n t s i n a p l a n e i s a p o s i t i v e number.

The i n t e r s e c t i o n o f a l i n e and a p l a n e i s a p o i n t .

If a d i a g o n a l o f a convex q u a d r i l a t e r a l s e p a r a t e s i t i n t o two c o n g r u e n t t r i a n g l e s , t h e q u a d r i l a t e r a l i s a p a r a l l e l o g r a m .

If e a c h p a i r o f o p p o s i t e s i d e s o f a q u a d r i l a t e r a l a r e c o n g r u e n t , t h e q u a d r i l a t e r a l i s a p a r a l l e l o g r a m .

The o p p o s i t e a n g l e s of a p a r a l l e l o g r a m a r e c o n g r u e n t .

A d i a g o n a l o f a p a r a l l e l o g r a m b i s e c t s two o f i t s a n g l e s .

The p l a n e whose e q u a t i o n i s x + y + z = 4 c o n t a i n s

t h e t h r e e p o i n t s ( 1 , 2 , l ) , ( 3 , -1, 2), and ( 5 , -3, 2 ) .

The perimeter o f t h e t r i a n g l e formed by j o i n i n g t h e m i d p o i n t s o f t h e s i d e s o f a g i v e n t r i a n g l e i s h a l f t h e p e r i m e t e r o f t h e g i v e n t r i a n g l e .

If t h e d i a g o n a l s of a q u a d r i l a t e r a l a r e p e r p e n d i c u l a r and c o n g r u e n t , t h e q u a d r i l a t e r a l i s a rhombus.

68. ~f a l i n e i n t e r s e c t s o n e o f two p a r a l l e l p l a n e s i n a

s i n g l e p o i n t , i t i n t e r s e c t s t h e o t h e r p l a n e i n a s i n g l e p o i n t .

69. If t h e c o o r d i n a t e s o f p o i n t s A , B, C a r e (5, 9 , l l ) ,

(0, -1, -4), (5, -11, l ) , r e s p e c t i v e l y , t h e n A ABC

i s a r i g h t t r i a n g l e .

70. ( ( x , y ) : x = 4-4k, y = 1-2k, k i s r e a l ) c o n t a i n s

t h e p o i n t ( 3 , -1) .

71. ( ( x , Y ) : x = 1 - 3k, y = 7k , k i s r e a l ) and f ( x , Y): x = 9k, y = 1 + 21k, k i s r e a l ) a r e p a r a l l e l l i n e s .

72. ( ( x , y ) : x = 3k, y = 3-k, k i s r e a l ] i s a l i n e p a s s - i n g t h r o u g h t h e o r i g i n .

73. I f e a c h o f two p l a n e s i s p e r p e n d i c u l a r t o a t h i r d p l a n e , t h e y a r e p a r a l l e l t o e a c h o t h e r .

74 . The p r o j e c t i o n o f a l i n e i n t o a p l a n e i s a l i n e .

75. If a r a y i n one f a c e o f a d i h e d r a l a n g l e i s perpen- d i c u l a r t o t h e e d g e o f t h e d i h e d r a l a n g l e , t h e l i n e c o n t a i n i n g t h e r a y i s p e r p e n d i c u l a r t o t h e p l a n e c o n t a i n i n g t h e o t h e r f a c e o f t h e a n g l e .

76. Through a p o i n t n o t i n a p l a n e , there i s e x a c t l y one l i n e p e r p e n d i c u l a r t o t h e p l a n e .

77. If a p l a n e i n t e r s e c t s two o t h e r p l a n e s i n p a r a l l e l l i n e s , r e s p e c t i v e l y , t h e n t h e two p l a n e s a r e p a r a l l e l .

78. I f a l i n e i s p e r p e n d i c u l a r t o a p l a n e , t h e n any p l a n e c o n t a i n i n g t h i s l i n e i s p e r p e n d i c u l a r t o t h e g i v e n p l a n e .

79. A q u a d r i l a t e r a l w i t h t h r e e r i g h t a n g l e s i s a r e c t a n g l e .

80. If A = (-2, -4, 7 ) , B = (2, 2 , 3 ) , and C = ( 4 , 5, l ) , t h e n A , B, and C a r e c o l l i n e a r .

81. ((x, y, z ) : 2 x - 2y + z = 4) c o n t a i n s t h e p o i n t

( 3 , 2, - 1 ) .

82. T h e r e a r e i n f i n i t e l y many p l a n e s p e r p e n d i c u l a r t o a g i v e n l i n e .

83. If a p l a n e i s p e r p e n d i c u l a r t o e a c h o f two l i n e s , t h e two l i n e s a r e c o p l a n a r .

84. If e a c h o f t h r e e n o n c o l l i n e a r p o i n t s o f a p l a n e i s e q u i d i s t a n t f r o m two d i s t i n c t p o i n t s P and Q,

t h e n PQ i s p e r p e n d i c u l a r t o t h e p l a n e .

85. If e a c h o f two p l a n e s i s p a r a l l e l t o a l i n e , t h e

p l a n e s a r e p a r a l l e l t o e a c h o t h e r .

86. If e a c h o f two i n t e r s e c t i n g p l a n e s i s p e r p e n d i c u l a r t o a t h i r d p l a n e , t h e n t h e i r l i n e o f i n t e r s e c t i o n i s p e r p e n d i c u l a r t o t h i s p l a n e .

67. The i n t e r s e c t i o n o f t h e x z - p l a n e and t h e p l a n e whose e q u a t i o n i s 2 x - 3y + z = 6 i s

( (x , y, 2): 2 x + z = 6 and y = 0) .

88. I f A , B, C, and D a r e f o u r d i s t i n c t p o i n t s i n - - s p a c e s u c h t h a t AB ff CD and 3 AD, t h e n ABCD i s a p a r a l l e l o g r a m .

89. ( ( x , y ) : x = 3k, y = 2k , 0 < k < 1 ) i s a segment . - - 90. ( ( x , y ) : x = 2 + 3k, y = 2 k , k i s r e a l ) i s a

3 l i n e h a v i n g t h e s l o p e 2 .

91. ( ( x , y ) : x = 3 + k, y = 1 + 2k , k > 0 ) i s c o n t a i n e d - i n Q u a d r a n t I.

92 [ ( x , y ) : x = 3k, y = 2k, k i s r e a l ) and ( ( x , y ) : x = 3 ) I n t e r s e c t a t t h e p o i n t ( 3 , 2 ) .

93. The i n t e r s e c t i o n o f { ( x , y ) : x = 3k, y = 2 k , k i s r e a l ) and ( ( x , y ) : x = 2h, y = 3h, h i s r e a l ) i s t h e

o r i g i n .

94. [ ( x , y ) : x = 3k, y = 2k, k is r e a l ) i s p e r p e n d i -

c u l a r t o ( ( x , y ) : x = 2k , y = 3k, k i s r e a l ) .

95. If two a n g l e s h a v e t h e p r o p e r t y t h a t t h e s i d e s o f o n e a r e a n t i p a r a l l e l t o t h e c o r r e s p o n d i n g s i d e s o f t h e o t h e r , t h e a n g l e s a r e s u p p l e m e n t a r y .

7

96. If ABC i s a r i g h t t r i a n g l e and CD i s t h e a l t i t u d e t o t h e hypo tenuse o f t h e t r i a n g l e , t h e n A ABC i s s i m i l a r t o A ACD.

97. The u n i o n o f t h e set of a l l rhombuses and t h e set o f a l l r e c t a n g l e s i s t h e set o f a l l s q u a r e s .

- 98. If ABCD i s a q u a d r i l a t e r a l and AB TÎ) and AD 1 1 E,

t h e n ABCD i s a p a r a l l e l o g r a m . -

99. If ABC is a r i g h t t r i a n g l e and CD i s t h e a l t i t u d e to t h e h y p o t e n u s e o f t h e t r i a n g l e , t h e n t h e s q u a r e o f t h e l e n g t h o f CD i s e q u a l t o t h e p r o d u c t of t h e - l e n g t h o f AD and t h e l e n g t h of m.

100. The a l t i t u d e of a n e q u i l a t e r a l t r i a n g l e each of whose i l d e s h a s l e n g t h s i s 2 s .

Chapter 10

DIRECTED SEGMENTS AND VECTORS

10-1. Introduction.

Many of the quantities which we encounter in life are adequately described by a unit of measure and a number. Expressions of this sort, such as 5 inches, 20 degrees, 75 cents, 2 hours, 10 cubic feet, or 15 miles per gallon, can be represented as the distance between two points on the appropriate scale, and for this reason are called scalars. On the other hand, there are numerous quantities

such as displacement, velocity, acceleration, and force, for which more information than this must be given before they are adequately specified.

Consider the simple idea of a displacement, for instance. If we are told that a boy walked two miles, we really don't know very much about what he did unless we are also told the direction in which he walked. Even if we know that he walked two miles northeast, say, we don't have a complete description of what he did. He might have started at school and walked two miles northeast, or he might have started at his home, or any number of other points:

1 School

/ Home

We have a n adequate d e s c r i p t i o n of t h e boy's walk only i f we know

( a ) t h e p o i n t from which he s t a r t e d ,

( b ) t h e d i r e c t i o n i n which he walked, ( c ) t h e d i s t a n c e he walked;

o r , e q u i v a l e n t l y , i f we know

( a ) t h e p o i n t from which he s t a r t e d ,

( b ) t h e p o i n t a t which he ended.

Speaking i n somewhat more a b s t r a c t terms, It appears t h a t a displacement can be represen ted i n e i t h e r of two ways:

( a ) By a segment extending a given d i s t a n c e i n a given d i r e c t i o n from a g iven p o i n t .

( b ) By a p a i r of p o i n t s , one i d e n t i f i e d as a s t a r t i n g p o i n t , t h e o t h e r a s a t e rmina l p o i n t .

A segment, a s we de f ined i t i n Sect ion 3-6, i s J u s t a set of p o i n t s and has no d i r e c t i o n a s s o c i a t e d with i t . S i m i l a r l y , t h e s e t c o n s i s t i n g of t h e endpoints of a segment has no d i r e c t i o n a s s o c i a t e d w i t h i t . Hence n e i t h e r t h e - segment AB nor t h e s e t (A,B] i s adequate t o desc r ibe t h e displacement from A t o B because n e i t h e r d i s t i n g u i s h e s between t h i s displacement and t h e displacement from B t o A , which i s q u i t e a d i f f e r e n t t h i n g . C lea r ly , we can spec i fy t h e displacement which starts a t t h e p o i n t A and ends a t t h e p o i n t B by u s i n g t h e ordered -- p a i r of p o i n t s (A ,B) . However, i f we w i s h t o desc r ibe a displacement by means of a

segment we must extend our o r i g i n a l d e f i n i t i o n :

DEFINITION. A segment i s a d i r e c t e d segment i f

and on ly i f one of i t s endpoints i s des ignated a s i t s i n i t i a l p o i n t , o r o r i g i n , and t h e o t h e r endpoint i s des ignated as i t s t e rmina l po in t o r -- - terminus.

2 Notation. The symbol (A,B) is used to denote the

directed segment whose origin is A and whose

terminus is B . In a drawing a directed segment is shown by placing a half arrow-head at its terminal

point, thus:

DEFINITION. Directed segments which have the same initial points and the same terminal points are said to be equal.

DEFINITION. By the length of the directed segment 2

(A,B) we mean AB , that is, the length of the - associated segment AB . In the next section we develop some of the properties of

directed segments, using the concept of displacement to motivate

our work. Then in later sections we introduce the important generalization of a directed segment known as a vector.

10-2. Directed Segments.

In the last section we introduced the idea of a directed segment by considering the displacement of a single object from one point to another. Let us now consider a number of objects which move equal distances in the same direction along parallel lines, as for example a group of planes flying in formation or the linemen of a football team charging down the field together in their pregame warm-up:

The displacement of the planes, in the,first case, and of the players in the second, are all different because no two begin and end at the same points. None the less, in each of the two

cases there are characteristics common to all the displacements; specifically, the displacements take place in the same direction along parallel lines and involve movement through equal distances In many applications it Is convenient to be able to refer concisely to directed segments with these characteristics, and to provide forthis we introduce the following definition:

DEFINITION. Two directed segments, (0) and (C,D) are equivalent if and only if AB = CD - + and AB 1 1 CD .

(we suggest that you review Section 6-7 regarding parallel rays at this point. ) ' This definition is illustrated in the following figures:

A A In each case the directed segments (A,B) and (A',B') are

+ equivalent because AB = AIBf and AB 1 1 A*. In these

2

figures, are the directed segments (A,B) and (m) equivalent? Why?

a Notation. We indicate that (A,B) is equivalent

by writing

If you understand the definition of equivalent directed segments you should have no trouble verifying the following

statements.

Properties - of Directed Segment Equivalence.

1. Directed segment equivalence is reflexive: 3

(A~B) - (A,B) . 2. Directed segment equivalence is symmetric:

2 a 2 2 If (A,B) A (c,D) , then (c,D) (A,B) ,

3. Directed segment equivalence is transitive: 2 2 3 A

~f (A,B) (c,D) and (c,D) = (E,F) , 2 2

then (A,B) g (E,F) . A fundamental property of directed segments is given by

the following theorem:

THEOREM 10-1. There Is one and only one directed segment which

is equivalent to a given directed segment and has its origin at a given point.

Proof: By definition, the directed segment which has its 2

origin at P and is equivalent to (A,B) must lie on the 4--e unique line /^ , which contains P and is parallel to AB .

Moreover, the required directed segment must lie on the unique ray of 1 which has P for its endpoint and is parallel to + AB . Finally, on this ray, the terminal point, Q , of the required directed segment must have the property that AB = PQ , and by the Point Plotting Theorem, there is one and only one such point. Hence the theorem is proved.

If we return to the idea of displacement for a moment, and consider the problem of getting from one point to another in a city, it is clear that only rarely can one go directly from one point to another. Usually, because of the buildings which are in the way, one must walk down one street a certain distance, then turn a comer, and continue on another street

to reach his destination.

Instructions for getting around in a city reflect this fact,

and we are all accustomed to being told to go so many blocks in one direction and then go so many blocks in another to get

to some desired address.

The observations suggest, first of all, that any displacement can be achieved in various ways by a succession of

other displacements. In the second place, they suggest that

it may be convenient when speaking of displacements to use a

coordinate system in describing simple displacements which

together produce a given displacement. For instance, it is

clear from the following figure that the displacement from the

point ~(2,l) to the point ~ ( 7 , 4 ) can be achieved by first performing the displacement from the initial point ~(2,l) to

the point ~(7,l) and then continuing with the displacement

from C to the terminal point B ( 7 , 4 ) . It can also be

accomplished by first performing the displacement from the

initial point A to the point ~ ( 2 , 4 ) and then continuing

with the displacement from D to the terminal point B . A

displacement to the right of 7 - 2 = 5 and an upward dis-

placement of 4 - 1 = 3 , regardless of the order in which they are performed, thus combine to give precisely the dis-

placement from A to B . In terms of the given coordinate system, the numbers 5 and 3 associated with the displace-

d + ments parallel to ray OX and the ray OY , respectively, are thus fundamentally related to the displacement from A(2,l)

to ~ ( 7 , 4 )

The extension of the foregoing observations to directed

segments is contained in the following definitions.

points in the xy-plane, the number x2 - x, is - called the x-component of (Pl,p2) and the

number y2 - yl is called the y-component - of

The ordered pair of numbers [x2 - x, , yp - yl] A

are called the components of (P~,P~) . (Note

the use of brackets to indicate components.)

Although a given directed segment determines a unique pair of components, it is not true that a given pair of

components defcermjnes a unique directed segment. For instance, as we nave just seen, the points ~(2,l) and ~ ( 7 , 4 ) determine a directed segment with components [5,3] , but so do the points ~(4,2) and ~ ( 9 , 5 ) . What is another directed segment with components [5,3] ? There are many more, but it is possible to derive the following theorem.

A A THEOREM 10-2. Two directed segments (P,,P~) and (p3,p4)

are equivalent if and only if they have the same components.

To establish this theorem, we must show two things: &

1. If (Pi,P2) and have the same components,

they are equivalent. A 4

2. If (p1,Pp) and (p3,P4) are equivalent, they

have the same components.

two

are the

are

To prove these assertions requires the consideration of

cases, according as the four points P, , P2 , P3 , P4 or are not collinear. To save time, we shall give only outline of the proof in the general case where the points

noncollinear .

1. Let the coordinates of Pl, P2, P,, Pi be,

respectively, ( x l , ~ l ) , ( x 2 9 ~ 2 ) , (x3 ,~ '3) 9 ( x 4 9 ~ 4 > Then the

hypothesis of the f i r s t pa r t of the theorem i s t h a t

x* - Xl = x4 - X3 and Y2 - Y1 = Y 4 - Y3

If x2 - x = x - x = 0 , then the l i n e s and=

are ve r t i ca l and hence p a r a l l e l . If x2 - x1 = x 4 - x 3 # 0 , then by divis ion

Y^ - Yl Y4 - Y3 - - x2 - X1 '4 - ^3 '

Therefore, the slopes of PiPp. and a re equal and hence, by Theorem 8-12, the l i n e s a r e p a r a l l e l . In every case, then, when the'given points a re not co l l inear , i s p a r a l l e l t o

. Now from the hypothesis t h a t

we obtain a t once

Hence, by an argument analogous t o the one we have jus t given, it follows t h a t *P,P' i s p a r a l l e l t o .

Therefore, if Pi, Pg, P3, Pb a r e n o t c o l l i n e a r , they form a para l le logram, PlP2P3P4 . Hence

p1p2 = ^P4 ,

and P2 and Pi, l i e on t h e same s i d e of t h e l i n e x. J 4

Therefore ( P ~ , P ~ ) and (p3,P;.) are equ iva len t , as a s s e r t e d .

A A 2. Now suppose t h a t (Pl,P2) (p3,p4) . Let

Pr = (x2 - x1 + x3 , y2 - y1 + y3) . Then t h e components of 2

( p 3 . p ) are

A But t h e s e a r e a l s o t h e components of (p.,,P2) . It fol lows, then , from t h e f i r s t p a r t o f t h i s theorem (which we have proved) t h a t

But t h e hypo thes i s of t h i s p a r t t e l l s u s t h a t

Because of t h e t r a n s i t i v i t y p roper ty of d i r e c t e d segment equivalence w e conclude t h a t

By Theorem 10-1, it fo l lows t h a t P = PL , o r 5

x2 - x + x3 = x4 and y2 - yl + y3 = y,, , 1

o r

This completes the proof of t h e theorem.

A s an immediate consequence of t h i s l a s t theorem, t o g e t h e r wi th the Pythagorean Theorem, we have t h e fo l lowing r e s u l t :

THEOREM 10-3. If P, and Pg have coordinates ( x l , y ) and

(x2,yo) , respec t ive ly , the length of any d i r ec t ed A

segment equivalent t o (p,,P2) i s

We have, i n e f f e c t , def ined equivalent d i r ec t ed segments

a s d i rec ted segments having the same length and d i r ec t ion ,

regardless of t h e i r i n i t i a l po in t s . Another i n t e r e s t i n g c l a s s of d i r ec t ed segments cons i s t s of those having the same i n i t i a l

point and d i r ec t ion , regard less of t h e i r lengths . This l eads

t o the idea o f the product - - of a d i r ec t ed segment and a number, -- which i s made prec i se i n t he .following de f in i t i on :

DEFINITION. Let (A.B) be any d i r ec t ed segment and A

l e t k be any r e a l number. The product, ~ ( A , B ) , 2

i s the d i r ec t ed segment (A,x) where X i s t he

point whose coordinate i s k i n the coordinate - system on AB , with o r i g i n A and u n i t point B .

3 2 DEFINITION. The d i r ec t ed segment -1 (A,B) = - ( A , B)

A

i s ca l l ed the opposite of the d i r ec t ed segment (A,B) .

The following f igu re i l l u s t r a t e s t h e mu l t i p l i ca t ion of a d i rec ted segment by the numbers 2 and -2 :

4 We note t h a t , i f k > 0 then X i s i n AB ; i f k < 0 ,

+ then X i s i n t he - r ay opposite t o AB ; i f k = 0 , then A

X = A . This l a s t r e s u l t in t roduces tile p o s s i b i l i t y of (A ,A) , a zero d i rec ted segment.

As we might expect, when a directed segment I s multiplied by a number, k , the components of the directed segment are both multiplied by the same number. More precisely we have the following theorem.

THEOREM - 10-4. If the coordinates of P, and P2 are (xl,y,)

and (x2,y2) , respectively, then the components of the A

directed segment (P1,P3) wnicn is k times the directed A

segment (Pl,P2) are k(x, - x,) and k(y2 - y.) . 2

proof: Let ( ~ ~ 1 ~ 3 ) tx3 - x1 , y3 - y1J . By the Two Point Theorem

x3 - x1 = k(xg - x ) and y3 - Y = k(y2 - Y,) Thus

Problem Set 10-2a -- 1. A and B are two points. List all the directed line

segments they determine.

2. A, B, C are three points. List all the directed line segments they determine.

3. In the figures below list the equivalent directed line segments.

A ' FIGURE o

_________^Ã

E F

FIGURE b

FIGURE c

694

2 4 4. I f A, B, C , D a r e d i s t i n c t points and (A,B) A (c,D) ,

A show f o r each of the following cases t h a t (ATc) (3,~) :

( a ) A, B, C , D a re co l l inear i n that order. ( b ) A, C, B, D a re co l l inear In t h a t order. (c ) No three of A, B, C , D a r e co l l inear .

(d) Why did we not consider three points co l l inea r i n P a r t ( c ) ?

( e ) Why do we not consider A, C , D, B co l l inea r i n tha t order?

3 2

5. If B, F, G, H a re d i s t i n c t points, (B,F) A ( G , H ) , and a l i n e /*' i s not perpendicular t o %? and does not - A i n t e r sec t BF , show t h a t the project ions of (B,F) and J

(G,H) on /c a re equivalent d i rec ted segments. Let the projection of B on 2 be B* , and consider three cases:

(a) G i s contained i n w. ( b ) G i s i n the same halfplane a s F with respect

t o BB'. ( c ) G i s i n a d i f f e ren t halfplane from F with

respect t o BB'. 6. Determine k so t h a t each of the following statements

Is t rue .

A 2 3 2

(a) (A ,c ) A ~ ( A , B ) . (f) (A,D) G ~ ( A , F ) . 2 2 2

( b ) ( A , E ) = ~ ( A , B ) . & A 2 >

(g) ( A , F ) ^ ~ ( A , B ) ( c ) (A,F) = ~ ( A , E ) . ( h ) (B,c) = ~ ( A , D ) .

A 2 & (d ) (D,A) = ~ ( A , F ) . = ~ ( A , F ) .

2 - (e) (A,E) A ~ ( A , c ) .

7. A, B, X are collinear points. Find r such that 2 2 2 A

(A,x) ~(A,B) and s such that (B,x) A S(B,A) , if -

(a) X is the midpoint of segment AB ; - (b) B is the midpoint of segment AX ; ' - ( c ) A is the midpoint of segment BX ; (d) X is two-thirds of the way from A to B ;

(e) B is two-thirds of the way from A to X ; (f) A is two-thirds of the way from B to X .

8. In triangle ABC , X Is the midpoint of AC and Y is the midpoint of . Determine k so that- each of

the following statements is true. A 2

(a) (B,x) A ~(B,Y) . B

2 2 (b) (B,Y) = ~(B,x) . 2 2

(c) (A,c) ~(c,A) . 2 J

(d) (A,c) A ~(c,x) . & 2

(e) (c,x) A ~(x,A) . 2 2

(f) (c,A) A ~(A,x) . A 2

(g) (x,B) = k(B,Y) A X C

- 9. In the parallelogram ABCD , E is the midpoint of CD

and trisects at F as indicated below.

Determine k so each of the following statements is true. 3 & 2 A

(a) (D,F) ~(D,B) . (e) (F ,B) A ~(B,D) . 3 J A J

(b) (D,E) ~(c,D) . (f) (A,B) ~(D,c) . 2 A 4

(c) (B,D) = ~(B,F) . A A

(g) (A,@ = ~ ( c , B ) (d) (F,B) = ~(D,B) .

Having now discussed briefly the product of a directed

segment and a number, it is natural to ask If the sum of two directed segments can be defined. Since the displacement from

B to C following, or in a sense "added to," the displacement from A to B accomplishes exactly the same thing as the

displacement from A to C , c

A

B

it seems natural to write

However, as a possible definition of the sum of two directed segments, this expression has serious limitations, for it

pertains only to two directed segments for which the terminus

of the first is the origin of the second. With the idea of

equivalent directed segments in mind, we might go further and 2 A

say that to add (c,D) to (A,B) where B and C are

different points, first determine the unique directed segment, 2 A

(B,D') , which is equivalent to ( C , D ) h he or em 10-1) and &

then add it to (A,B) , according to the above geometric definitions.

This procedure permits us t o form the sum of any two directed

segments. However, the r e su l t ing process i s a very curious

type of addition, f o r , a s the following f igure shows, 2 2

(A,B) + (C ,D) i s n o t the sameas

From the preceding f igure , i t appears t h a t although A 2 4

(A,B) + (c,D) i s not the same as (cTD) + ( A , B ) , these two directed segments a re equivalent. That t h i s i s ac tua l ly the

case follows from the next theorem.

A - THEOREM 10-5. The components of ( P . , , P ~ ) + (p3,pii) a re the

J sums of the corresponding components of (p1,Pp) and

To prove t h i s , l e t the coordinates of Pl, Pp, Pn, Ph be

J components of (p1,Pp) a re [xn - x, , y2 - yl] and the

2 components of (p3,p4) a re [x4 - x3 , y4 - yg] . Now t o add 2 2

(p3,p4) t o (p1,P2) we must f i r s t determine t h e directed

segment, ( P ~ , P ~ ' ) which i s equivalent t o (p3,Pii) . Since

equivalent d i rec ted segments have the same components A

(Theorem 10-2) It follows t h a t the components of (P2,Pn1) a re

x , ~ ' - xg = X^ - Xg and Y ' - Yg = Y;, - Y3

Hence the coordinates of P^,f are

Therefore the components of the sum

and

A L Clearly, the components of the sum, (p1,P2) + (P3,P4) are

the sums of the corresponding components of (P,,P> and A

(p3,pti) , as asserted.

2 A From this theorem, It is apparent that (P1,pg) + (p3,P4) A A

and (p3,pi.) + (pi,P2) nave the same components. Hence, by

Theorem 10-2 they a r e equivalent. Thus we have the following

theorem:

The following theorem is also an immediate consequence of

Theorem 10-5.

On the basis of the preceding discussion, it should occur

to us tnat instead of focusing our attention on directed

segments, it might be better to consider as fundamental

entities the various sets of equivalent directed segments.

This is really not difficult to do, even though each set

contains infinitely many members, for according to Theorem 10-2,

each such set is characterized by a unique pair of components,

and conversely. In other words, there is a one-to-one corre-

spondence which matches each ordered pair of real numbers with

each set of equivalent directed segments. Moreover, if we

define the sum of two sets of equivalent directed segments,

Sl and S2 , to be the unique set which contains the sum of any directed segment from Sl and any directed segment from

Sy , we have a process of addition in which, by Theorem 10-5 and Theorem 10-6, It is true tnat

In the rest of this chapter we shall adopt the point of view we have just described. Sets of equivalent directed

segments, or the ordered pairs of components which are in

one-to-one correspondence with these sets, we shall call

vectors. A directed segment is thus - not a vector, although it

clearly determines the vector consisting of all the directed

segments equivalent to the given one. Each directed segment is

thus

each

is a

a representation of a vector, in somewhat the same way that

member of a set of equivalent fractions such as

representation of a unique real number.

I n the next sect ion we s h a l l Introduce formal de f in i t ions of vectors and t h e i r propert ies . However, these a re a l l motivated by the propert ies of directed segments which we have discussed i n t h i s section. I f you keep the l a t t e r i n mind, the work ahead of you should seem a na tura l extension of what we have already done.

Problem Set 10-2b -- 1. You may r e c a l l from the algebra of r e a l numbers the

following defini t ion:

DEFINITION. If a, b a r e two r e a l numbers, then a - b i s the r e a l number c such t h a t b + c = a . The operation of finding c where a, b a re given i s subtraction.

Using the above de f in i t ion a s a guide wri te a de f in i t ion f o r the subtraction of two d i rec ted l i n e segments.

d

( a ) (A,B) + (B,c) A ? 2 A

(b) ( B , A ) + ? (B,c) . 2 2

( c ) ? + (B,A) = (B,c) . 2 A

(d ) ? + (A,B) = (A,A) . 2 A

( e ) (A,B) + (B,c) + (c,A) A ? 2 A 2

( f ) (B,A) + (A,c) + (c ,B) = ? 2 2

(g) ( c , A ) + ? A ( c , B ) .

2 2 Given two d i r ec t ed ll'ne segments (A,B) , (c,D) and hor izonta l l i n e s 4 and 1, , as indicated below.

B

What i s t r u e about the two sums? & & A

Given (A,B) , (c ,D) , ( E ~ F ) as shown below; 1 , 1, , A

j2 a r e hor izonta l l i n e s .

Find graphica l ly

2 2

(d) (A,B) + ((c,D) + (eJ) . Do this by adding the sum 2

2 2 in (b) to (A,B) .

2

(e) ((A^) + (C,D)) + (E,F) . Do this by adding (E,F)

to the sum in (a).

What appears to be true of the two sums in (d) and (e)?

6. Letting 1 inch represent 2 miles, find graphically the resultant displacement if an automobile travels 4

miles due north then 5 miles northeast.

7. ABCD is a parallelogram. E is the midpoint of ; - AC and trisect each other at F .

2 2

(a) (D,A) + (A,E) A ? 3 2 2

(b) (A,E) + ?(D,E) = (A,F) . 2 - 3 J

(c) $D,E) + ?(A,c) * (D,c) . 3 2

(d) ?(CF) + (A,c) - ( A ~ A ) . 3 A 1-

e ) ?(A,B) + ?(E,D) $A,c) .

10-3. Vectors.

Motivated by our discussion at the end of the preceding section, we begin our work in vectors with the following definitions.

DEFINITION. Any real number is called a scalar.

DEFINITION. A vector is an ordered pair of real numbers, called the components of the vector.

Notation. A vector w i l l o f ten be denoted by a s ingle lower case l e t t e r w i t h a half arrow above

2 it, thus: u . I f a and b a re the components

of a vector we may a l s o denote the vector by the symbol [ a ,b ] . If the components of a vector a re the same a s the components of a directed

2 segment (P,Q) we may denote the vector by the

2

symbol PQ . We should note t h a t square brackets, r a the r than parentheses a r e used i n the symbol [ a ,b ] , f o r the vector whose components a re a and b . This i s done t o avoid confusion w i t h ordered p a i r s of r e a l numbers such a s (x,y) which a re the coordinates of a point. We should a l so be careful not t o

3

confuse the symbol, PQ , f o r the vector determined by the d

point P and Q w i t h the symbol, PQ , f o r the ray determined by P and Q . The former has only a half arrow above the l e t t e r s ; the l a t t e r has a f u l l arrow.

DEFINITION. If Tf = [a, b ] , the number I/a2 + b 2

2

i s ca l led the magnitude, o r length, of u .

Notation. The magnitude, o r length, of 6 I s denoted by the symbol 161 . DEFINITION. The ordered p a i r [0,0] i s ca l led the zero vector.

2

DEFINITION. I f u = [a ,b] and i f k i s any r e a l number, the vector ,

2

i s ca l l ed the product of the vector u and the -- - -- sca la r k . -

DEFINITION. Two vectors a re equal i f and only i f t h e i r respective components a re equal.

DEFINITION. Non-zero vectors whose components are

proportional are said to be parallel. -

The next theorem follows immediately from the definition of parallel vectors and the concept of slope.

A 4

THEOREM 10-8. If P P 2 and P3Pb are parallel vectors, then

and are parallel.

The following simple but very important theorem Is an

immediate consequence of the definition of parallel vectors, the definition of the product of a vector and a scalar, and the definition of the magnitude of a vector.

THEOREM 10-9. If "i? and $ are parallel vectors, then

where

A DEFINITION. If u = [a,b] and ? = [c,d] , t h e vector

[a + c, b + d.] = [a,b] + [c,d] = i?+ $

Is called the sum of u' and Y? . - *-*"

DEFINITION. The vector i? + (-l)? is called the difference between and ? . - - -

Notation. The difference between '3' and $ - - - is written "S i? .

The following important theorem is an immediate conse-

quence of Theorems 10-2 and 10-5.

10-3

A THEOREM 10-10. The sum of the vector P1P2 and the vector A 2 P3Ph i s the vector PIX where X i s the unique point

3 2 such t h a t PgX = P3P4 .

The geometrical significance of t h i s theorem i s I l l u s t r a t e d i n the following f igure.

Since vectors a re not numbers, there i s no reason t o believe t h a t they obey the same laws tha t govern the operations of ar i thmetic . Actually the addi t ion of vectors and the mult ipl icat ion of vectors by sca la r s do obey the fami l ia r laws of ar i thmetic . For t h i s reason w e sha l l merely l i s t these propert ies f o r reference, and ver i fy i n only two cases tha t they do, indeed, hold f o r vectors.

Prone r t i e s of Vectors

If i? , $ a r e vec to rs , then '5' + v' i s a vec to r . A 2 I f u, v, T, a r e any t h r e e vec to r s then

A _^ There e x i s t s a vec to r , 0 , such t h a t f o r any vec to r u

2 ^& For ' every vec to r u t h e r e e x i s t s a vec to r -u such t h a t

a If 5 , v a r e any two vec to rs , then T? + $ = ? + ^T . If , $ are any two vec to r s and k i s any s c a l a r , then

If 6 i s any vec to r , then ku= i? when k = 1 . 8. If i s any vec to r and h , k a r e any two s c a l a r s , then

(h + k)i? = h + ki? . 9. If 6 i s any vec to r and h , k a r e any s c a l a r s , then

h(ku) = hkiT= k(hi?) . 10. I f i? i s any vec to r and k i s any s c a l a r , then

k u l = lkl . Property 1 i s a n immediate consequence of t h e d e f i n i t i o n

of t he sum of two vec to rs . a

To prove Property 2, l e t '3' = [ a ,b l , v = [ c , d l , and A w = [e,f] . Then

(z?+ $) + $ = ( [ a , b ] + [ c , d ] ) + [ e , f ] = [ a + c , b + d ] + [ e , f ] = [ ( a + c ) + e , ( b + d ) + f ]

= [ a + ( c + e ) , b + ( d + f ) ] = [ a , b ] + [ c + e , d + f ]

= [ a , b l + ( [ % d l + [ e , f l ) = i? + (i? + i?) , as a s s e r t e d .

Properties 3 and 4 follow immediately from the definition of the sum of two vectors and the definitions of the zero vector and the difference between two vectors.

Ã‘I

To prove Property 5, let u = [a,b] and i? = [c,d] . Then

A A u + v = [a,b] + [c,d]

= [ a + c , b + d ] = [ c + a , d + b ]

= [c,dl + [a,bl A A

= v + u , as asserted. The proofs of Properties 6 to 10 are very much like the

two proofs we have given, and to save time we omit them. In each case, it is the corresponding property of the real numbers

which appear as components that plays the decisive role in the proof.

Problem Set 10-3 -- 1. ~f A, B, C are respectively (1,2) , (4,3) , (6,1)

express each of the following vectors in component form.

(a) AÂ . (b) BA . (c) A . (a) AC .

2. Same as Problem 1 if A, B, C are respectively (-1,2) , (4,-3) , and (-6,-1) .

3. ~f A, B, c are respectively (1,2) , (4.3) , (6,1) find X so that

2 3 2 (a) AB = CX . (c) XA = CB . (b) A ? = & . (d) 5(? = @ .

4. Same as Problem 3, if A, B, C are respectively (-1.2) , (49-3) , (-6,-l)

2 5. Given 2 = [3,2] , '6'= [-4,3] , c = (5,-61 ,

Determine the following.

(a) 2 + S . A A

(b) a - c . - -A (c) o - a . _ & A _ &

(d) a + b - c .

2 - 2

(e) a - b - c . -A. A

( f ) b + c - a . - - A

(g) -a - b - c .

6. Using the vectors In Problem 5, determine A

(e) b - 2? . A A 1

(f) -a+3b+-if. . (g) $-$+F.

7. Using the vectors In Problem 5, find the real number

which expresses each of the following.

8. Determine a and b so that

9. Physicists have found that forces and velocities obey the

law of vector addition. Physicists call this sum the

resultant. Using this knowledge and a scale of 1 inch

to represent 2 miles per hour, solve the following

problem graphically.

A r i v e r has a 3 mile per hour cur ren t . A motor boat moves d i r e c t l y across t h e r i v e r a t 5 miles per hour. How fast and i n what d i r e c t i o n would the boat be t r a v e l i n g i f t h e r e were no cur ren t and the same power and heading were used i n cross ing the r i v e r ?

10-4. The Two Fundamental Theorems. -- Many of t h e app l i ca t ions of vec tors depend upon one o r

t he o the r of two theorems, which we shall now prove.

You w i l l note i n t h e proofs of these theorems t h a t we r e f e r t o diagrams of geometric f i g u r e s when we speak of vectors . While a vec tor i s an ordered p a i r of numbers and not a s e t of po in t s , t he f a c t t h a t a d i r ec t ed segment determines a vector and t h a t a vector toge ther with an i n i t i a l point determines a d i r ec t ed segment, enable u s t o t h ink of a d i r ec t ed segment as a vec tor .

2 THEOREM 10-11. If and OB a r e two non-zero vec tors

which a r e not p a r a l l e l and i f 0 i s any vector i n the plane OAB , then t h e r e e x i s t s c a l a r s h and k such that

2 If OP i s the zero vector , it i s obvious t h a t h = k = 0 .

L A 2 If 6 i s p a r a l l e l t o e i t h e r OA o r OB , t he a s s e r t i o n of

2 t he theorem follows immediately from Theorem 10-9. If OP

2 i s n e i t h e r t he zero vector nor a vec tor p a r a l l e l t o OA o r

2 OB , let m be the line which contains P and is parallel to - OB and let n be the line which contains P and is parallel u

to OA .

H Let M be the Intersection of m and OA and let N be the - intersection of n and OB . Then by Theorem 10-9,

A 2 2 2

OM = hOA and ON = kOB . 2 2

Finally, since ON = MP , It follows by Theorem 10-10 that

as asserted,

Theorem 10-11 has an interesting algebraic interpretation. 3 2 2

~f OA = [ai,a21 , OB = [bl,b21 and OF = [pl,p21 , then the 3 2 2

assertion OP = hOA + kOB is true if and only if there exist

numbers h and k such that

= [ha, + k b , ha, + kb2] . This in turn requires that

ha, + k b = pl

and

Now we know that these equations have a unique solution for h and k unless their coefficients in one of the equations

are zero or unless their coefficients in the two equations are

proportional. If the coefficients are proportional, then

- = b1 a2 '5 or equivalently - = - .

1 b2

But this is precisely the condition that % and 6 should be parallel, a situation ruled out by the hypothesis of the

2 2 theorem. Thus, if OA and OB are non-zero, non-parallel

vectors, whose components are known, it is possible to express

a third vector, O? in terms of and OB in a purely algebraic way.

Example a 2

Express w = [5,2] in terms of u = [2,3] and $ = [-I,^]

To do this, we must determine h and k so that

This requires that

2 h - k = 5 and 3 h + 4 k = 2 .

Solving these two equations simultaneously, we find

h = 2 , k = - 1 .

Hence A A 2 w = 2 u - v .

The second of our fundamental theorems is the following.

THEOREM 10-12. If T? and i? are non-zero, non-parallel vectors,-and if x, y, z, w are scalars such that

a 2 a x u + yv = zu + w ,

then

x = z and y = w .

To prove t h i s , we observe t h a t by adding the vector

- (& + snT) t o both s ides of the given equation,

we obtain

or , using Property 9 A

(x - z)u = ( w - y)v . If x - z # 0 we can wri te

From t h i s we conclude e i t h e r t h a t "if i s the zero vector 2

( i f w - y = 0) o r e l s e t h a t T? and v a re p a r a l l e l (s ince one i s a sca lar multiple of the o ther . ) However each of these a l te rna t ives contradicts the hypothesis of the theorem. Hence x - z cannot be d i f fe ren t from zero and so x = z . But if

x = z , then it follows that "A Ã‘Ã

0 = (w - y)v

and since ? Is not the zero vector, by hypothesis, it follows tha t w = y . Hence

x = z and w = y , as asser ted.

Problem Set 10-4 -- Determine x and y so t h a t each of the following s t a t e -

ments i s t rue.

10-5. Geometrical Application - of Vectors.

Many theorems i n geometry can be proved by means of vectors . I n t h i s s ec t ion we s h a l l present several t yp i ca l examples of vec tor proofs of geometrical theorems.

THEOREM 10-13. The midpoints of the s ides of any quadr i l a t e r a l a r e t he v e r t i c e s of a parallelogram.

Proof: Let A, B y C, D be the v e r t i c e s of t h e quadri- l a t e r a l and l e t P, Q, R y S be the midpoints of the; s ides

SE , B5" , â‚¬ , T5ff , respec t ive ly .

- 1- 2 l> 2 l a 2 12. By hypothesis , SD = -yKD , DR = yDC 9 PB = ÂB , BQ = -pE!C . - a 2 1 - Hence SR = SD + DR = y ( ~ ~ + 6S)

2 - 2 1 2 2 and PQ = PB + EQ = y ( ~ ~ + BC) .

2

Moreover 6 + 2 = 6 + , s ince each i s equal t o AC . 2 2

Therefore S R = PQ , - 4 - w

which Implies t h a t SR = PQ and SR 1 1 PQ . Hence PQRS

i s a parallelogram, as asser ted .

THEOREM 10-14. The segment Joining the midpoints of two sides of a triangle is parallel to the third side and the length of the segment is one half the length of the third side.

Proof: Let A, B, C be the vertices of the triangle and - let D and E be the midpoints of AB and T^ respectively.

A A - 1^ By hypothesis DA = 'BA and AE = -rfC . ?

Hence - 1- DE = wBC --

which Implies that DE = $E?C and DE 1 1 E3C as asserted.

THEOREM 10-15. A quadrilateral is a parallelogram if and only if its diagonals bisect each other.

Proof: Let ABCD be the parallelogram and let M be

the intersection of its diagonals.

10-5

2 A 2 A - A -

Let A D = u and DC = v . Then AC = u + v and - & A 2

D B = v - u . Since A, M, C are collinear, AM is some A

scalar multiple of AC , say x ( u + v ) . Similarly 2 A a 2 2 2

DM = y(v - u) . Since AD + DM = AM , we have

or, collecting like terms, A A

(1 - y - x)U + (y - x ) v = 0 . Theref ore 1 - x - y = 0 and y - x = 0 .

Solving these simultaneously we find

Hence

These imply 1 AM = -pAC

- la and DM = $lB . 1 and DM = Ã‘D , as asserted.

Now let A B C D be any

bisecting each other at M A & - 2

Let t = AM = MC and w = 2 - 2

DC = w + t ; therefore 6 -- and A B 1 1 DC . Hence ABCD is a parallelogram.

quadrilateral with its diagonals

so that DM = ME! and AM = MC . 4 2 A A A

DM = MB . Then A B = t + w and 2

= DC , which implies that AB = DC

Problem Set 10-5 -- 1. The segment joining the midpoints of the non-parallel

sides of a trapezoid is called the median of the

trapezoid. Prove that the median of a trapezoid is

parallel to the bases and has a length equal to one-half the sum of the lengths of the bases.

4--+- 2. Let ABCD be a trapezoid, with A B 1 1 CD , and E, F - -

the midpoints of AC , B D , respectively. Prove that

3. Prove that the medians of a triangle are concurrent at

the point which trisects each median.

4. Let ABCD be a parallelogram, with E the midpoint of - - - A B , and DE Intersecting AC at F . Prove that F

7

is a point of trisection of AC . 5. Let ABCD be a parallelogram, with E the point on "S5

1 such that AE = Ã‘A , with intersecting A? at F . 1 Prove that AF = =AC .

10-6. - The Scalar Product of Two Vectors. -- In Section 10-3 we defined what we meant by two parallel

vectors. It is now convenient to introduce the ideas of perpendicular vectors.

A 2

DEFINITION. Two vectors, PIQl and PoQo are

said to be perpendicular if PiQ is perpendicular

to " Q-

In many applications it is important to be able to tell whether or not two vectors are perpendicular. To develop a procedure for deciding this question, consider two non-zero vectors 4 2 OP = [pl,pg] and OQ = [q1,q2] . These will be perpendicular if and only if APOQ has a right angle at 0 . By the Pythagorean Theorem, this will be the case

if and only If 2 2 - 2 P Q I = lop1 + la1 .

Hence

and t h e r e f o r e

Now, r e c a l l i n g t h e d e f i n i t i o n of t h e magnitude o r l eng th of a - 2 vec to r , we can write 1 % 1 2 = 1 0 ~ 1 + 1Gl2 i n t he form

o r , expanding and c o l l e c t i n g terms,

Hence,

Thus, s i nce t h e preceding s t e p s a r e a l l r e v e r s i b l e , we have

e s t a b l i s h e d t h e fo l lowing important theorem.

THEOREM 10-16. Two non-zero vec to r s a r e perpendicular i f and only If t h e sum of t h e products of t h e i r r e spec t i ve components i s zero.

The number p1ql + p2q2 obta ined from t h e components of

t h e vec to r s [pl,p2 1 and [ql,q2] i s a very important quan t i ty ,

and i t i s convenient t o have a name f o r it.

.A A DEFINITION. ~f u = [ p p 2 1 and v = [q,,qnl , t h e number plql + p2qo i s c a l l e d t h e s c a l a r

2 A product of u and v .

2 Notation. The s c a l a r product of u and $ i s _& _&

denoted by t h e symbol u * v ( r ead "u dot v") ,

We should understand t h a t t he s c a l a r product of two vec tors i s a s c a l a r and - not a vec tor . The name s c a l a r product i s used t o emphasize t h i s f a c t .

There a r e s eve ra l important a lgeb ra i c p r o p e r t i e s of t h e s c a l a r product of two vec to r s with which we should be f a m i l i a r . These a r e not hard t o prove, and we leave t h e p roofs of t he f i r s t two a s exe rc i s e s .

A A A A Property 1. u v = v u .

A A A A Property 2. u ("\?"+ i?) = u v + G w .

A a Property 3 . If k i s a s c a l a r , u (&) = (&) w

= k(\? ^) . A A 2

Property 4. u u = lul . We have a l ready seen ('Theorem 10-16) t h a t two non-zero

vectors a r e perpendicular i f and only i f t h e i r s c a l a r product i s zero. However, whether two vec to r s a r e perpendicular o r n o t , t h e i r s c a l a r product has an i n t e r e s t i n g geometrical i n t e r - - pre ta t ion . To discover t h i s l e t OP and 00 be two non-zero vectors and l e t OF be a s c a l a r mul t ip le of OP , say 2 A OF = kOP . Then

A A 2 A

We note t h a t FQ = 0 if and only i f 0% = kOP , which means t h a t 0$ and O? a r e p a r a l l e l . Now I f 8 # 5 , F$ and 6 w i l l be perpendicular i f and only i f

o r , using P rope r t i e s 2 and 3 f o r s c a l a r products,

Let k t be t h e va lue of k determined by t h i s equat ion. Then

To i n t e r p r e t t h i s r e s u l t it i s convenient t o in t roduce t h e fo l lowing d e f i n i t i o n .

2 DEFINITION. By t h e p r o j e c t i o n of a v e c t o r AB on

2

a v e c t o r 6 we mean t h e v e c t o r MN , where M

and N are, r e s p e c t i v e l y , t h e f e e t of t h e perpen- d i c u l a r s from A and B t o t h e l i n e %?.

The fo l lowing f i g u r e i l l u s t r a t e s t h i s d e f i n i t i o n .

MN i s t h e p r o j e c t i o n 2 a

of AB on CD .

2 2

Now k 1 i s t h e va lue assumed by k when FQ 1 OP . Hence klO? i s t h e p r o j e c t i o n of 0$ on O? and 1 k t 6 l i s t h e

A A

l e n g t h of t h i s p r o j e c t i o n . Moreover, i f FQ = 0 , then 6 2 2 a

i s p a r a l l e l t o OP , and OQ i s i t s own p r o j e c t i o n on OP , 2 A

Hence i n a l l cases , kfOP i s t h e p r o j e c t i o n of 6(? on OF . 2

Returning now t o t h e express ion f o r OF % which we

de r ived above, and n o t i n g t h e symmetry of t h e s c a l a r product guaranteed by Proper ty 1, It fo l lows t h a t , except f o r s ign , t h e s c a l a r product of two v e c t o r s , and 00, , i s equal t o

e i t h e r :

(a) The length of 6 multiplied by the length of the 2 A

projection of OQ on OP , or (b) The length of 08 multiplied by the length of the

A projection of OP on .

The sign of the scalar product is positive if kt > 0 , + that is, if F lies on the ray OF , and negative If kt < 0 ,

+ that is, if F lies on the ray opposite to OP . As an example of the use of the scalar product in coordinate

geometry, let 0(0,0) and ~(a,b) be two distinct points in the xy-plane and let J^ be the line which is perpendicular to * OF at F . If F(x,y) is any point of ,d distinct from F , - will be perpendicular to OF if and only If

2 A A 2

OF FP = 0 . Now OF = [a,b] and FP = [x - a , y - b] . (Why?) Hence OF @ = 0 can be written

a(x - a) + b(y - b) = 0 or 2 a x + b y = a 2 + b .

By direct substitution, it is easy to verify that this equation

is also satisfied by the coordinates of F . Hence this

equation is an equation of the line /^ . By an almost identical argument it can be shown that if /^ contains 0 , an equation of ,d is

It Is interesting to compare this discussion with the derivation

of an equation of a plane in Section 9-9.

Problem Se t 10-6 -- I n each of t he following problems determine the s ca l a r

product and from i t t e l l whether t he two vectors a r e perpendicular . L-

12. Using t h e s c a l a r product, show t h a t ~ ( 5 , 7 ) , Q(8,-5) , and ~ ( 0 , - 7 ) a r e t he v e r t i c e s of a r i g h t t r i a n g l e .

13. By using Proper t ies 1 and 2, show t h a t A A A A ( 6 - 3 ( v ? - " ^ = u . w - u - 2 - v * w + v ~ z .

14. Show t h a t an equation of a l i n e through the o r i g i n i s a x + b y = O .

15. Prove Proper t ies 1 and 2 of t h e a lgebra ic p roper t ies of s c a l a r products.

10-7. Summary.

A d i r ec t ed segment I s t he mathematical e n t i t y which corresponds t o a displacement I n the physical world. It d i f f e r s from a segment i n t h a t one of i t s endpoints i s i d e n t i f i e d as an

o r i g i n and t h e o the r as a terminus. A d i r ec t ed segment there- f o r e t e l l s both a length and a d i r e c t i o n . After def ining

equivalent d i r ec t ed segments we Introduced a vec tor a s a set of equivalent d i r ec t ed segments. Since equivalent d i r ec t ed segments have the same components we can consider a vector t o be an ordered p a i r of numbers, and t h i s i s how we defined a vec tor . We used vec tors t o prove some geometric theorems. These proofs were sometimes simple due t o the f a c t t h a t the a lgebra of mult iplying vec tors by s c a l a r s i s similar t o the

algebra we studied in previous grades. The chapter ended with scalar multiplioation which enables us to prove two lines perpendicular and to find the projection of one vector on another.

Review Problems

1. Given ABCD is a parallelogram and E and F trisection -

points of AC , such that E is between A and F . Prove that DEBF is a parallelogram.

2. Given parallelogram ABCD , and E and F so chosen

that AB + BE = AE and CD + DP = CF and BE = FD . Show that AECF is a parallelogram.

3. Show that the points ~ ( 6 , 8 ) , ~(0,-2) , R(-3,-7) are

collinear.

4. Show that ~ ( 4 ~ 0 ) , ~ ( 7 ~ 8 ) , ~(0,10) and s(-3,2)

are the vertices of a parallelogram. 2

5. ~f = [4,01 , '6 = [-3,21 , c = [7,81 find

(a) a + $ . A A

(b) a - c . A A A

(c) a + b + c . A

(d) t - c . 6. In the figure D and E are -

midpoints of AB and AC A

respectively. A A

(a) = ? ( A , B ) A A 2

(b) ?(A,D) + (B,c) = ?(A,E) . 2 2 2

(c) (A,D) + (D,E) = ?(A,c) . 2

(d) (0) + (c,A) = ? B c A 2 2

(e) (D,B) + (B,c) = (DYE) + ?(A,c) . 7. A, B, C, D are vertices of a parallelogram. List all

the directed line segments they determine, and indicate

which pairs are equivalent.

9. ABCD is a parallelogram and P, Q, R, S are the

midpoints of the s i d e s .

10. Determine x and y so that

~f A , B, C a r e r e s p e c t i v e l y ( 4 , 2 ) , ( 6 , 3 ) , ( 2 , l )

express t h e fo l lowing v e c t o r s i n component form.

( a ) Ag . ( d ) CB . ( b ) BA . ( e ) BC . ( c ) A5 . Determine t h e s c a l a r product of

I n a cube what i s t h e maximum number of equ iva len t

d i r e c t e d l i n e segments?

I n a t r apezo id what Is t h e maximum number of e q u i v a l e n t

d i r e c t e d l i n e segments?

A man i s s t and ing on t o p of a h i l l . He weighs 200

pounds. Represent as a v e c t o r each of t h e fol lowing:

(use a s c a l e of 1 Inch = 200 pounds.)

( a ) The downward p u l l of t h e e a r t h ' s g r a v i t y on him.

(b) The upward push of t h e h i l l on him.

An o b j e c t i s suspended by ropes a s shown i n t h e f i g u r e .

If t h e o b j e c t weighs 10 pounds, what i s t h e f o r c e

exer ted on t h e f u n c t i o n C by t h e rope ?

18. A weight of 1000 pounds is suspended from wires as shown i n the figure.

( a ) What force does the wire AC exert on the Junction

C ? - (D) What force does the wire BC exert on C ?

19. A 5000 pound weight i s suspended a s shown i n the figure. - - Find the tension i n each of the ropes CA , CB , and .

20. A ship s a i l s eas t a t 20 miles per hour. A man walks across i t s deck toward the south a t 4 miles per hour. What i s the man's ve loc i ty r e l a t i v e t o the water?

Chapter 11

POLYGONS AND POLYHEDRONS

11-1. In t roduc t ion .

I n t h e phys ica l world n a t u r e abounds i n geometric shapes.

Many of t h e s e shapes a r e r e p r e s e n t a t i o n s of polygons and poly-

hedrons. For example, t h e s e c t i o n s of a honeycomb a r e hexagonal;

each snow c r y s t a l i s i n t h e shape of a t i n y hexagon; diamonds

a r e i n t h e form of r e g u l a r octahedrons; salt c r y s t a l s appear t o

be t i n y cubes; and q u a r t z c r y s t a l s have t h e shape of hexagonal

pyramids.

Man uses t h e shapes of r e g u l a r polygons i n des ign ing

formal landscapes, i n making b o l t heads, chickenwire, s t o p

s igns , and linoleum t i l e s . Box c a r t o n s , b u i l d i n g s , and sky-

sc rapers t ake tile form of prisms and o t h e r polyhedrons.

I n t h i s c h a p t e r , we cont inue our s tudy of polygons wi th

s p e c i a l emphasis on t h e a r e a of polygonal - regions . It i s

i n t e r e s t i n g t o no te t h a t one of t h e first p r a c t i c a l u s e s of

geometry was t h a t of f i n d i n g a r e a . Many people t h i n k t h a t

geometry had i t s o r i g i n i n t h e f o u r t e e n t h cen tu ry B. C . a long

t h e banks o f t h e Ni le River . A t t h a t t i m e t h e king of Egypt

d iv ided t h e l and i n t o p l o t s and ob ta ined h i s revenue from t h e

annual r e n t which t h e l andho lde rs were r e q u i r e d t o pay. Each

year t h e Nile River overflowed and c a r r i e d away p o r t i o n s of

s o i l . Tnis n e c e s s i t a t e d a remeasurement of t h e l a n d s o t h a t

t h e r e n t demanded of a n i n d i v i d u a l t h a t y e a r would be

p ropor t iona l t o t h e land which he he ld .

It Is also i n t e r e s t i n g t o n o t e t h a t t h e word geometry comes from two Greek words meaning "ea r tn" and met re in

meaning " t o measure." Hence t h e f i r s t meaning of t h e word

geometry was " e a r t h measurement. II

Today t h e s tudy of a r e a i s a l s o important . Land i s bought

and so ld by t h e a c r e ; t h e f l o o r space of a bu i ld ing i s

considered in determining the rent of an office, factory, or storeroom; the area of the wing of an airplane is important In designing the airplane; painters, bricklayers, surveyors, map makers, and interior decorators must know how to calculate the

area of simple geometric figures.

In the latter part of this chapter, we Introduce figures

in three dimensions which are closely analogous to the polygons we have studied in two dimensions. Each of these figures is

called a polyhedron. We shall investigate some of the interesting properties of this set of surfaces. However, the study of the measure of a polyhedral-region will be deferred.

11-2. Polygonal-Regions.

A triangular region consists of a triangle and Its

interior. Each of the following diagrams represents a triangular-region.

A polygonal-region is a figure In a plane, like one of these

four :

Notice in particular that a polygonal-region may have one or

more "holes" in it. A polygonal-region can be "cut up" into

triangular regions. For example, each of the first two

polygonal-regions shown above is "cut up" in the diagrams

below.

DEFINITIONS. A triangular region is the union of a triangle and its Interior.

A polygonal-region is the union of a finite number

of coplanar triangular regions.

Each of the following figures pictures a polygonal-region as a

union of triangular regions.

The preceding two pairs of pictures suggest that a polygonal-region can be considered as a union of triangular-regions

in more than one way. Note that we often do not shade a

polygonal-region in a picture, in case the context makes clear

that we are considering the polygonal-region rather than the

polygon which is the "boundary" of the polygonal-region.

One of the above diagrams shows five diagonals of a convex

polygon with eight sides. These five diagonals have a common

endpoint and they cut u p the polygonal-region so that we see

the polygonal-region as a union of six triangular-regions.

Noting that 5 = 8 - 3 and 6 = 8 - 2 , we are ready to consider the general situation.

Consider any convex polygon, say PIP o...P n We wish to

observe that the union of the convex polygon and its interior

is the union of n - 2 triangular-regions and is therefore a polygonal-region.

-

(in the figure, the dots indicate other possible vertices and sides, because we do not know what the number n is.) The

number of sides of the polygon is n . Since the polygon is a

convex polygon, the n - 1 rays x, n, .. . , PX, are concurrent in that order. Two of the corresponding segments,

namely PIPp and PIPn , are sides of the polygon, while the - - remaining n - 3 segments, PIP3 , P A , . . . , PIPn-l , are all of the diagonals with one endpoint at P, . (some of

these diagonals are shown in the figure, and other possible /

diagonals are suggested by - - -'""- )

These diagonals and the s i d e s of the polygon give u s t r i ang le s ,

APlPpP3 , APlPnP4 , ... , APIPnlPn , the number of which

i s n - 2 . The union of these t r i a n g l e s and t h e i r i n t e r i o r s

i s the union of PIP ... Pn and i t s i n t e r i o r . Thus the union

of a convex polygon and i t s i n t e r i o r i s a polygonal-region.

Furthermore, we observe t h a t the n - 2 t r i a n g l e s

mentioned above have the property t h a t the i n t e r i o r s of no two

of them i n t e r s e c t . Hence, i f a convex polygon has n s ides ,

the union of t h e polygon and i t s i n t e r i o r i s the union of

n - 2 t r i a n g l e s and t h e i r i n t e r i o r s such t h a t the i n t e r i o r s of

any two of t he t r i a n g l e s do not i n t e r s e c t .

DEFINITIONS. If a polygonal-region i s t h e union of a convex polygon and i t s i n t e r i o r , then the polygon i s

c a l l e d the boundary -- of the polygonal-region and the

i n t e r i o r of t he polygon i s c a l l e d the i n t e r i o r of the -- polygonal-region.

I n t h i s chapter , we make use of t r iangular-regions i n two

ways: (1 ) t o determine the sum of t h e measures of the angles

of a convex polygon, and ( 2 ) t o study the a r eas of various

polygonal-regions.

Problem Set 11-2 -- 1. Show t h a t each of t he following i s a polygonal-region.

More s p e c i f i c a l l y , show t h a t each i s a union of t r iangular-

regions such t h a t t he i n t e r i o r s of any two of the tr iangular-

regions do not i n t e r s e c t . Try t o f i n d the smal les t number

of t r i angula r - reg ions i n each case . ( ~ o t e : - The boundary

i n Par t (d ) i s a star-shaped polygon, and each s ide of the

polygon i s c o l l i n e a r with another s ide of the polygon.

The boundary i n Par t (g) i s a polygon having two noncon-

secu t ive s ides which a r e c o l l i n e a r . )

2. In the following figure, A, B, C, D, E, F, G are called - vertices, the segments AB, E, m, m, m, m, m, m, are called edges, and the polygonal-regions ABE, FED, BCDF are called faces. The exterior of the figure will also be considered as a face.

Let the number of faces be f , the number of vertices be v , and the number of edges be e . In a theorem which was originated by a famous mathematician, Euler, and which refers to figures of which the above figure is an example, there occurs the number f - e + v . Using the figure, let's compute f - e + v . You should see that f = 4 , v = 7 , e = g , andthisgivesus f - e + v = 2 . Using the two figures below, compute f - e + v . Notice

that the edges are not necessarily segments.

(b) Suppose this figure to be

a section of a map showing counties:

What do you observe in the results of the three computations? In Part (a) take a point in the interior of the quadrilateral and draw segments from each of the four vertices to the point. How does this affect the number f - e + v ? Can you explain why? Take a point in the exterior of the figure of Part (a) and connect it to the two nearest vertices. How does this affect the computation?

If you are interested in this problem and would like to pursue it further, you will find it discussed in h he Enjoyment of ~athematics" by Rademacher and Toeplitz and in "Fundamental

Concepts of ~eometry" by Meserve.

11-3. Sum of the Measures of the Angles of a Convex Polygon. --- -- -- In Chapter 6 we proved that the sum of the measures of the

angles of a triangle is 180 . AS an application of this important theorem we studied the sum of the measures of the angles of a convex quadrilateral. Let us review the method which we used (see Theorem 6-13 and Its proof), but let us express the ideas with the aid of the new terminology Introduced in the preceding section.

If the quadrilateral ABCD is a convex quadrilateral,

then polygonal-region ABCD is the union of the two triangular-regions ABC and ACD .

We showed in the proof of Theorem 6-13 that the sum of the

measures of the four angles of the quadrilateral is the same as the measures of six angles, three from each of the two

11-3

triangles. Thus we obtained the number 2 180 , or 360 , as the sum of the measures of the angles of the convex quadri-

lateral.

We wish to extend this discussion to the case of a convex polygon of any number of sides. The following exploratory problem utilizes our observations in the preceding section concerning the representation of a polygonal-region as the union of triangular-regions, no two of whose interiors intersect.

Exploratory Problem

Consider the diagonals from A in each of the convex polygons pictured below. 'By

used with the quadrilateral, the angles of each polygon. table as shown.

a procedure similar to the one we

find the sum of the measures of Summarize the results in a data

This exploratory problem leads us to the following

important result.

Number of sides of convex polygon

4

THEOREM 11-1. The sum of the measures of the angles of a

convex polygon of n sides is (n - 2) 180 . Proof: Let V be any vertex of the given convex polygon

Number of diagonals from A

1

with n sides, and let the polygon be VABC...GH . There are

n - 3 diagonals from the vertex V . The union of the tri-

Number of triangular regions

2

angular regions AVB , BVC , ... , GVH is the polygonal-region

Sum of measures of angles of the polygon 2 x 180 = 360

VABC...GH . There are n - 2 of these triangular-regions, and

the interiors of no two of them

intersect. The total measure of

all the angles of these triangles

is (n - 2) 180 . On the B other hand, the total measure of

all the angles of these triangles \ \ is the same as the sum of the \

measures of all the angles of +%

polygon VABC . . . GH . (Why? ) v

Corollary

polygon of n

11-1-1. The

sides is

measure of each angle of a regular

Proof: A regular polygon of n s ides has n angles, and a l l of these angles have the same measure. Hence each has

1 measure -(n - 2)(180) . NOW

The notion of an ex te r io r angle of a tr iangle, a s described i n Chapter 5, may be extended i n a natural manner t o polygons of more than three s ides .

DEFINITIONS. Let V be any vertex of a convex polygon.

The angle of the polygon with vertex V i s sometimes ca l l ed the i n t e r i o r angle of the polygon

(Â ? at V . Either angle which forms a l i n e a r p a i r with the

i n t e r i o r angle of the polygon a t V i s ca l led an ex te r io r angle of the polygon a t V .

THEOREM 11-2. For any convex polygon of n sides, the sum of - the measures of ex te r io r angles, one at each vertex of the polygon, i s 360 . Proof: A t each vertex of the

polygon, choose an ex te r io r angle. The chosen ex te r io r angle and the i n t e r i o r angle a t t h a t vertex a re supplementary; the sum of t h e i r measures i s 180 . The sum of the measures of a l l the i n t e r i o r angles and a l l the chosen ex te r io r angles Is n 180 . The sum of the measures of the i n t e r i o r angles i s (n - 2)18o . By subtraction, the sum of the measures of the selected ex te r io r angles i s

Corollary 11-2-1. The measure of each exter ior angle of 3 60 a regular polygon of n s ides i s - . n

Proof: This statement i s an immediate consequence e i t h e r of Theorem 11-2 o r of Corollary 11-1-1. Why?

Problem Set 11-3 -- 1. Find the sum of the measures of the i n t e r i o r angles and

the sum of the measures of the ex te r io r angles of a

polygon, one ex te r io r angle a t each vertex, i f the number of s ides of the polygon is:

2 , The sum of the measures of the i n t e r i o r angles of a ce r t a in regular polygon i s 1080 . By Theorem 11-1,

Hence

and

Thus the number of s ides of the polygon i s 8 ,

Find the number of s ides of a regular polygon i f the sum of the measures of the i n t e r i o r angles is:

3. Consider a regular nonagon (nine s ides) . The measure of 360 o r 40 . The i n t e r i o r angle each exter ior angle i s - 9

and an exter ior angle a t each vertex a re a l i n e a r p a i r of angles.

( a ) What i s the measure of each i n t e r i o r angle of t h i s

polygon?

(b) What i s the sum of the measures of a l l the i n t e r i o r angles?

Use two methods t o f i n d the measure of each angle of a regular polygon of 12 s ides .

We know t h a t an i n t e r i o r angle and an exter

i n t e r i o r

l o r angle a ver tex of a polygon a r e a l i n e a r pa i r of angles. Thus i f

the measure of an I n t e r i o r angle of a regular polygon i s 120 , t he measure of each ex te r io r angle i s 60 . From Corollary 11-2-1, it follows t h a t 360 - - 60 . Therefore n = 6 .

Find the number of s ides o f a regular polygon if the measure of each i n t e r i o r angle of the polygon

( c ) 144 4

(d) 128 , Complete the following chart:

Name of regular polygon

measures of the i n t e r i o r angles

Sum of the measures of ex te r io r angles, one a t each vertex

Equi- l a t e r a l Triangle

Square

Regular Hexagon

Regular Pentagon

Regular Octagon

b

Regular Decagon

Measure of each i n t e r i o r angle

is:

Measure of each exter ior angle

7. Consider a r e g u l a r polygon of twenty s i d e s . Find t h e

measure of :

( a ) Each i n t e r i o r ang le of t h e polygon;

( b ) Each e x t e r i o r ang le of t h e polygon;

(c) The sum of t h e measures of t h e i n t e r i o r ang les of

t h e polygon;

( d ) The sum of t h e measures of - a l l t h e e x t e r i o r a n g l e s

of t h e polygon.

8. I n a c e r t a i n r e g u l a r polygon, t h e measure of a n e x t e r i o r

angle i s o n e - f i f t h t h e measure of a n i n t e r i o r ang le .

Find t h e number of s i d e s of t h e polygon.

9. The sum of t h e measures of e l even a n g l e s of a polygon of

twelve s i d e s i s 1650 . ( a ) What i s t h e measure of t h e remaining angle?

(b ) Do you have enough informat ion t o decide whether

t h e polygon i s r e g u l a r ? Explain. 10. Is i t p o s s i b l e t o have a r e g u l a r polygon wi th t h e measure

of each i n t e r i o r ang le equal t o 153 ? Why?

11. The s tar -shaped f i g u r e i s formed by extending t h e sides of

a r e g u l a r pentagon. Find t h e measure of t h e ang le a t each

po in t of t h e s t a r .

12. Given a pentagon ABODE such t h a t m /a = 150 , m /b = 60 , and t h e measures of /c , /d , /e a r e p r o p o r t i o n a l t o - 4 , 3, '1 . Prove t h a t A 3 1

'[ . 4 CEMREL - C S M P LIBRAR'? 103 S. WASHINGTON ST. CARBONDALE, ILL. 62901

13. I n a regular polygon ABCDE...J of a t l e a s t 5 s ides , - prove that diagonal AD i s p a r a l l e l t o s ide .

1 4 . I n the f igure , we have given a regular pentagon ABODE

D

and a rectangle ABXY , where C, X, Y, E a re co l l inear i n E C that order. Find m /CBX , m /DCX , and m /XBD .

15. Consider the problem w i t h non-overlapping t i l e s have a s ide I n

of how t o cover a polygonal f loor t i l e s such t h a t any two adjacent c ornmon .

Suppose t h a t each t i l e has the shape of a square and a l l t i l e s a r e congruent t o one another. How many t i l e s a r e needed t o cover the surface around a point which i s a t a corner of t i l e s ?

I f the t i l e s a re i n the shape of congruent equi lateral t r i ang les , how many a re needed t o cover the surface around a point which i s a t a corner of t i l e s ?

Could the t i l e s be shaped l i k e other regular polygons of the same number of s ides and cover the surface around a point without any overlapping? How many t i l e s of any one polygonal shape would be needed?

If two t i l e s have the shape of a regular octagon and another has the shape of a square, the three t i l e s would cover the surface around a point without overlapping. What other combinations of three regular polygons (two of which a re a l i k e ) w i l l do t h i s ?

Hint: Find solut ions of the equation 2x + y = 360 where x and y a re the measures of the i n t e r i o r angles of regular polygons having a d i f f e ren t number of sides. In the i l l u s t r a t i o n x = 135 and

y = 9 0 .

( e ) Investigate the p o s s i b i l i t y of other combinations of t i l e s shaped l i k e regular polygons which would be

su i tab le f o r use i n covering a f loor .

16. Consider a sequence of regular polygons w i t h the number of s ides as follows: 3 , 4, 5, ..., n, ... Choose the expression which correc t ly completes each of the following sentences . (a) The sum of the measures of t he i n t e r i o r angles of the

polygons (increases, decreases, remains the same.)

(b) The sum of the measures of the ex te r io r angles, one a t each vertex of the polygon, ( increases, decreases, remains the same.)

( c ) The measure of an i n t e r i o r angle of the polygon (increases, decreases, remains the same. )

(d) The measure of an ex te r io r angle of the polygon (increases, decreases, remains the same. )

11-4. Area.

I n Chapter 3 we introduced i n t o our formal geometry the notion of the distance between two points. Guided by our experiences from the physical world, we selected postulates and def in i t ions which describe precisely the basic propert ies of distance i n our geometry. We then deduced by logica l reasoning other propert ies of dis tance and the connections between distance and re la t ed topics . I n pa r t i cu la r , we discussed segments. A segment i s a c e r t a i n s e t of points; i t s "size," commonly ca l led i t s length, we defined t o be the same as the distance between i t s endpoints. The notion of congruence f o r two segments we described i n terms of t h e i r

lengths.

In a similar manner in Chapter 4, after describing a different type of set of points, namely an angle, we stated, by means of postulates and definitions, exactly what is meant

by the measure of an angle. Additional properties of angle measure we deduced as theorems.

We now wish to discuss how to measure a polygonal-region, 11 that is, how to determine its "area. A polygonal-region is a

set of points of a quite different type from the segment or the angle. We follow an approach like that used before; namely, we select postulates and definitions which formalize In our geometry the corresponding notion from everyday life. Notice the resemblances between the postulates in this section and those describing distance or angle measure. The first one

says that every polygonal-region has a unique measure relative to any standard "unit .'I

Postulate - 26. If R Is any given polygonal- 1 region, there is a correspondence which associates

to each polygonal-region In space a unique positive number, such that the number assigned to the given polygonal-region R is one.

DEFINITIONS. The given polygonal-region R mentioned in Postulate 26 Is called the unit-area. -- Relative to a given unit-area, the number which corresponds to a polygonal-region, in accordance with Postulate 26, is called the area of the polygonal-region.

Postulate 26 does not tell us what number the area of any particular polygonal-region Is (except the unit-area), nor

does it tell us how the areas of various polygonal-regions

compare. We need more postulates to give us this information.

If a segment is the union of two segments whose interiors do not intersect, then the measure of the given segment is the

sum of the measures of the other two segments. In the figure,

Recall the corresponding situation for angles, as shown in the + + diagram, where VB is between VA

4 and VC . The interiors of the two adJacent angles, LAVB and v< /BVC , do not Intersect, and the measure of /AVC is the sum of

the measures of the two angles

/AVB and /BVC . We wish to have a similar property for the areas of polygonal-regions. Thus, for example, If R is the polygonal-region consisting of the parallelogram ABCD and its Interior, as shown in Figure a ,. then we want the area of R to be the sum of the areas of the two triangular-regions R, and R2 . The following

postulate guarantees this. c

A Figure a

Postulate - 27. Suppose that the polygonal- region R is the union of two polygonal-regions R, and Rp such that the intersection of R1

and Rp is contained in a union of a finite number of segments. Then, relative to a given unit-area, the area of R is the sum of the

areas of Rl and Rp .

In Figure a, the two triangular-regions R, and R2 intersect in a segment. Other illustrations are given in

Figures b and c.

Figure b Figure c

In Figure b, the intersection of the polygonal-regions Rl and

Rp is the union of three segments. In Figure c, the intersection (marked heavily) consists of one segment and two other points; it is contained in the union of a finite number of segments. In each case the sum of the areas of R, and Rp

is the area of the entire polygonal-region.

On the other hand, consider the polygonal-region shown in Figure d. It is the union of triangular regions T, and Tp .

Figure d

Their intersection is not contained in a union of a finite number of segments, but instead is the cross-hatched polygonal-

region whose boundary I s a quadrilateral. Thus Postulate 27 is not applicable to this case. If we tried to calculate the area of the entire polygonal-region by adding the areas of Ti and Tp , then the area of the polygonal-region which is the intersection would be counted twice. Of course if we cut the entire polygonal-region in a different way, we may be able to apply Postulate 27.

We recall that two segments are congruent if and only if they have the same measure. Two angles are congruent if and

only If they have the same measure. We wish to compare the notions of congruence and area for polygonal-regions. Since a polygonal-region is the union of triangular-regions and since we have extensively studied congruence for triangles, we consider triangular-regions in particular. On the basis of our

experience in the physical world, two triangular-regions whose respective boundaries are congruent triangles have the same size and

shape ." Being of the same "size, their areas seem to be the same.

The next postulate guarantees this.

Postulate - 28. ~f two triangles are congruent, then the respective triangular-regions consisting

1 of the triangles and their Interiors have the same 1 area relative to any given unit-area.

Thus two triangular-regions with congruent boundaries have the same area. Notice that the converse is not valid. If two triangular-regions have the same area, we do not know whether the triangles which are their respective boundaries are congruent or not. The picture shows two triangles which have

different "shapes," although the ''sizes," that Is, areas, of the corresponding triangular-regions

appear to be the same.

For any convex polygon, the union of the polygon and its

interior is a polygonal-region. This polygonal-region has an area relative to a given unit-area. It is customary and very convenient to speak of ''the area of the polygon' when we really mean "the area of the associated polygonal-region." Thus, as examples, we speak of the "area of a triangle" when we mean the area of the union of the triangle and its interior; the area

of a parallelogram" is a conveniently short phrase for the 'area of the polygonal-region consisting of the parallelogram and its Interior."

In the physical world the notion of area is closely

related to the notion of distance. If an inch is chosen as a unit of distance, then the customary choice !It

for a unit of area Is the "square inch."

This is the area of a polygonal-region 1

consisting of a square and Its interior such that each side of the square is one

Inch long. Although some other type of 1 'I

polygonal-region can be chosen as the unit-area in our geometry, we prefer, in this book, to adopt as our unit-area

the so-called "unit-square," which is defined as follows:

DEFINITION. Given a unit-pair for measuring distance, a unit-area is called a unit-square If and only if the unit-area consists of a square and its interior such that the measure of a side of the square is one.

The diagram pictures a unit-square

Our fourth postulate concerning area tells us how to determine the area of certain polygonal-regions. It connects

the concept of area with the concept of distance developed in Chapter 3.

I Postulate - 29. Given a unit-pair for measuring 1 distance, the area of a rectangle relative to a 1 unit-square is the product of the measures (relative to the given unit-pair) of any two consecutive sides of the rectangle.

DEFINITIONS. Any side of a parallelogram is a base of the parallelogram.

An altitude of the parallelogram relative to the base is any segment which is perpendicular to the

base and whose respective endpoints lie on the parallel lines containing the base and the side opposite to the base.

In particular, any side of a rectangle is a base of the rectangle, and any side which is consecutive to the base of the rectangle is an altitude of the rectangle (relative to the

base).

Our work in the following sections is largely concerned with the areas of certain polygonal-regions and the lengths of

ce r t a in segments r e l a t ed t o the polygonal-regions. It i s

customary t o shorten the phrase "the length of a side" of a polygon and say simply " the side," whenever the context makes c lear t h a t we mean a number r a the r than a segment. In a l i k e manner, a base of a parallelogram o r a diagonal of a polygon i s a segment, t h a t i s , a s e t of points; sometimes, however, we use the word "base" o r "diagonal" t o mean the number which i s

the length of the segment; we do t h i s only i n case there i s no danger of confusion between the two d i f fe ren t uses of the same word.

If the lengths of two consecutive sides of a rectangle a re 6 and 3 , then we may consider the base t o be 6 and the a l t i t u d e 3 ; o r we may choose 3 a s the base, In which case 6 i s the a l t i t u d e . For e i the r choice, the area of the rectangle i s 18 .

Using the terminology given by the last def in i t ions , Postulate 29 t e l l s us tha t :

The area of a rectangle i s the product of i t s base and i t s a l t i t u d e .

If the area, the base, and the a l t i t u d e of a rectangle a re denoted by A , b, h, respectively, then

As a special case, the area A of a square each of whose s ides has length s i s given by

11- 4

Problem Set 11-4 -- 1. Complete each of the following tables and answer the

questions pertaining to each.

1 II 1 Base 1 Altitude 1 Area 1

I Base I Altitude I ~reaI

n Ease Altitude Area

a 5 100

en Base 1 Altitude 1 Area I I I

Consider a set of rectangles

with equal altitudes. If

these rectangles are arranged so that the bases of any two consecutive rectangles have the ratio of 1 to 2 , then the ratio of the areas

of any two consecutive rectangles is to

Consider a set of rectangles

with equal bases. If these rectangles are arranged so that the altitudes of any two

consecutive rectangles have

the ratio of 1 to 3 , then the ratio of the areas of any

two consecutive rectangles is to Â

Consider a set of.rectangles with equal areas. If these rectangles are arranged so that the bases of any two consecutive rectangles have

the ratio of 1 to 2 , then the ratio of the altitudes of any two consecutive rectangles is to

What is the ratio of the bases of any two consecutive

rectangles in the table? What is the ratio of the corresponding altitudes? What is the ratio of the corresponding areas? The four

rectangles are members of a set of rectangles.

11-4 (e ) Complete the following sentences:

If the ratio of the lengths of a pair of corresponding

sides of two similar rectangles is 1 to 3 , the ratio of the areasis to

If the ratio of the lengths of a pair of corresponding

sides of two similar rectangles is 2 to 3 , the

What is the ratio of the areas of each of the following pairs of rectangles?

ratio of the areas is to . 2. L K J I H

(a) Rectangle AN to rectangle AK . e ere we name a rectangle by naming a pair of opposite vertices.)

(b) Rectangle AJ to rectangle AH . (c) Rectangle A 0 to rectangle AF . (d) Rectangle BI to rectangle CI . (e) Rectangle BF to rectangle CF . ( f ) Rectangle BO to rectangle ND .

M -

(a) We are given two rectangles with equal bases. If the ratio of the altitudes is 1 to 3 , the ratio of the areas is to

(b) If the bases of two rectangles are In the ratio of 1

to 4 and the corresponding altitudes are in the ratio of 1 to 2 , the ratio of the areas of the rectangles is to

(c) If the areas of two rectangles are equal and the ratio of the bases is 1 to 3 , then the ratio of the altitudes is to

(d) If the bases of two rectangles are equal, and the altitude of the second is 25 per cent more than the

altitude of the first, then the ratio of the areas of

the first to the second is to

A B C D E

This figure is

separated into " twelve rectangular

regions. Let each

F small region be k

units long and one unit high.

N

0

4. The r a t i o of the lengths of two consecutive s ides of a rectangle i s 4 t o 5 . If the area of the rectangle i s 5780 , f ind the length of each s ide.

5. Let a and b be pos i t ive numbers. Show by a drawing

t h a t the area of a square whose s ide measures a + b i s the same a s the sum of the areas of:

( a ) a square whose s ide measures a , ( b ) a square whose s ide measures b , and

( c ) two rectangles each of which has s ides measuring a and b .

- D C 6. I n the f igure , AC i s a

diagonal of rectangle ABCD . The polygonal- Rl region ABCD i s cut i n t o 6 polygonal- regions: the boundary of Rl i s a square; the boundary of Rp Is a rectangle; R3, R4, Rr, Rg b -dB a re t r i angu la r regions.

( a ) The a rea of AABC i s the sum of the areas of the polygonal-regions -y - 9 -*

(b ) The area of AADC i s the sum of the areas of the polygonal-regions -9 -3 --

( c ) The area of AABC i s equal t o the area of AAIX . Why?

(d) The areas of R5 and R6 a re equal. Why?

(e ) The areas of R3 and R4 a r e equal. Why?

( f ) Therefore, the areas of Rl and R2 a re equal. Why?

11-5. Areas of Triangles and Quadrilaterals. -- - On the basis of the four postulates concerning area in the

preceding section, we can calculate the areas of triangles,

parallelograms, and various other quadrilaterals.

THEOREM 11-3. The area of a right triangle Is one half the product of the lengths of Its two legs.

Proof: Let triangle PQR have a right angle at R . Let the lengths of its legs be a and b , and let A be the

area of the triangle. he diagram above shows two pictures of the same triangle PQR . ) Let T be the Intersection of the line parallel to PR through Q and the line parallel to

^QR through P . Then QTPR is a rectangle, and APQR 2 AQPT . By Postulate 28, the area of AQPT is A . By Postulate 27, the area of the rectangle QTPR is A + A , because the two triangular regions intersect only in the - segment PQ . 'By Postulate 29, the area of the rectangle is ab . Therefore

From this we can derive the formula for the area of any

triangle. Once we obtain this formula. it will include Theorem 11-3 as a special case.

THEOREM 11-4. The area of a t r i ang le i s one-half the product - of any base and the a l t i t u d e t o that base.

Proof: Let A be the area of the given t r i ang le XYZ . Consider the a l t i t u d e t o the s ide of the t r iangle . Let b = YZ and h = XD . Let the distances, bl and b" , between D and the endpoints of the side opposite X be chosen so that b1 - < b" . There a re three cases t o consider.

(1) I f D i s between Y and Z , then 335 cuts the given t r i ang le i n t o two r igh t t r iangles , wi th bases bt and b" , as indicated. Furthermore, b = b1 + b" . By the preceding theorem, these two

1 r igh t t r i a n g l e s have respective areas wb'h and $b"h . Hence, by Postulate 27,

(2 ) If D i s one of the endpoints of , then AXYZ 7

i s a r i g h t t r i ang le . Therefore, A = $bh , by

Theorem 11-3.

( 3 ) If D i s not on the segment , there a r e again two r igh t t r i ang les , namely AXDZ and AXDY . I n t h i s case, bl + b = b" . Hence

Why? Solving the above equation f o r A , we obtain

Explain how.

Since the length of any side of a t r i ang le can be chosen as the base, Theorem 11-4 can be applied t o any t r i ang le i n three d i f fe ren t ways. The f igure below shows the three choices

1 f o r a s ingle t r i ang le . Any of the three formulas, A = Â¥^Â¥bi , 1 A = - b h 2 2 2 ' A = Â¥'Â h , gives the area of t h e t r i ang le . 7 3 3

Corollary 11-4-1 . The area A of

whose s ide has length s i s given by:

The proof i s l e f t as a problem.

THEOREM 11-5. The area of a rhombus i s

an equ i l a t e ra l t r i ang le

one half t h e product of the lengths of the diagonals.

Proof: Let M be the point of intersection of the diagonals of the rhombus EFGH , namely and .

Let d = EG and d1 = FH . The diagonals are perpendicular to each other. Since is the altitude to side EG of

triangle E N , the area of AEFG is

+(m)

In a like manner, we note that the area of AEGH is

Hence, by Postulate 27, the area of the rhombus is

+(m) + ~ ( H M ) = +(FYI) = &I dl .

Corollary 11-5-1. The area A of the square whose

diagonal has length d Is given by


THEOREM 11-6. The area of a parallelogram is the product of - - any base and the altitude to that base.

Proof: Let A be the area of the parallelogram PQRS . Let b and h be a base and the corresponding altitude.

- The triangles PQR and RSP , which have the diagonal PR of the parallelogram as a common side, are congruent. Hence the

triangular-regions PQR and RSP have the same area, by

Postulate 28. Hence the area of PQRS is twice the area of APQR . Since b and h are a base and a corresponding

1

altitude of APQR , the area of APQR is $bh . Therefore 1 the area of PQRS is 2(?bh) , or

Since the length of any side of a parallelogram can be taken as the base, Theorem 11-6 can be applied to any parallelogram in two ways. The figures following illustrates

the two choices for a single parallelogram. In one case, we obtain A = bh , and in the other, A = blhl . Either of these two expressions gives the area of the parallelogram.

THEOREM 11-7. The area of a trapezoid is one-half the product - of its altitude and the sum of its bases.

Proof: Let A be the area of the trapezoid, h its altitude, and b, and b2 its bases.

A diagonal of the trapezoid cuts the polygonal-region into two triangular-regions whose respective areas are $blh and

b h h he dotted lines on the right In the diagram indicate 7 2 why the two triangles have the same altitude.) By Postulate 27, the area of the trapezoid is

Algebraically, this is equivalent to the formula

DEFINITION. The median of a trapezoid is the segment which joins the midpoints of the two non-parallel sides.

Corollary 11-7-1. The area of a trapezoid is equal to the

product of Its altitude and the length of Its median.


Summary of Formulas: - Area of a rectangle:

Area of a parallelogram:

Area of a triangle:

Area of an equilateral triangle:

11-5

Area of a rhombus:

Area of a square:

Area of a trapezoid:

Problem Set 11-5 -- 1. Find the area of a right triangle if the lengths of the

legs of the triangle are 6 and 10 ,

2. Find the area of an isosceles right triangle if the

length of each of the congruent sides of the triangle

is 12 . 3. Find the area of a 45-45-90 triangle if the hypotenuse

of the triangle is 12 . 11 . Find the area of a 30-60-90 triangle if the hypotenuse

of the triangle is 12 . 5. If h is the hypotenuse of a 45-45-90 triangle, find:

(a) The length in terms of h of the side opposite an angle whose measure is 45 .

(b) The area of the triangle In terms of h . 6. If n is the hypotenuse of a 30-60-90 triangle, find:

(a) The length in terms of h of the side opposite the

angle with measure of 30 . (b) The length in terms of h of the side opposite the

angle with measure 60 . (c) The area of the triangle in terms of h .

7. Find the unknown In each of the following triangles if A is the area, b the base, and h the altitude.

-- - -

8. In the diagram, the line containing A, C , D, E, B is

parallel to the line containing X, Y, 2, and is perpen- - dicular to AX . AC = 3 ; C D = 1 ; EB = 6 ; AX = 4 .

In each of the following, find the ratio of the area of the first-named triangle to the area of the second-named triangle.

(a) A A Y C ; AAYD (c) A A Y C ; AEZB

(b) A A Y C ; AEYB ( d ) A E Y B ; AEZB

9. Refer to the diagrams Y

at the right and find the area of each of the following:

A (a) AABC 8 8

(b) AXYZ

( c ) ARST

(d) ADEF

10. Prove Corollary 11-4-1.

11. In the diagram to the right, ABCD is a quadrilateral with - diagonals AC and perpendicular to each other. - BD bisects T(5 . AC = 20 and BD = 24. Find the area of the quadrilateral.

If ABCD is a rhombus with diagonals 20 and 24 , find the area of the rhombus.

Find the area of a rhombus if the length of one side of the rhombus is 15 and the longer diagonal of the rhombus is 24 . The area of a rhombus is 1600 . Find the length of each diagonal of the rhombus if one is twice as long as the

other.

Prove Corollary 11-5-1.

Find the area of a square if the diagonal of the square is 8 . Find the area of a parallelogram if the base of the parallelogram is 12 and the altitude of the parallelogram is 7 . The area of a parallelogram Is 8430 and the altitude of the parallelogram is 150 . Find the base.

Find the area of a parallelogram ABCD if AB = 10 and AD = 1 4 , and: (a) m /A = 30 . ( c ) m /A = 60 . (b) m /B = b5 . (d) m /D = go .

20. The sides of a parallelogram are 8 and 10 respectively.

If the shorter altitude is 4 , what Is the longer altitude?

21. Find the unknown in each of the following trapezoids if A is the area, h the altitude, and bl and b2 the

bases of the trapezoid.

- 22. ABCD is an isosceles trapezoid with AB 1 1 DC and

m / ~ = 30 . AB = 10 and DC = 6 . Find the area of the trapezoid.

23. Prove Corollary 11-7-1.

24. Find the side of a square if the area of the square is equal to the area of a rectangle 16 feet by 9 feet.

25. In quadrilateral ABCD , D? 1 1 and ml'Ec .

E If AB = 10 , DC = 14 , c

EB = 7 , find the areas of AADC and AABC . 0 0

0

0

^ /' # ^

A B

The points P, S, Q, R, collinear in that order, and the parallelogram ABCD lie in the same plane; and , , - CR and T)S are perpendicular to PR . AP = 12 , E Q = 6 , D S = l 6 , Q R = 5 , C R = 1 0 , S Q = 2 , P S = 5 .

Find the area of ABCD .

The vertices of a. triangle have coordinates

(-4,l) , and (4 ,5 ) . Prove that the triangle is a right triangle. Find the area of the triangle.

Three of the vertices of a rhombus ABCD are: A(O,O) , B(-69-2) c(-8,-8) . (a) What are the coordinates of vertex D ?

(b) Find the area of the rhombus.

The vertices of a trapezold have the following coordinates: ~(0~0) , ~(12~0) , ~(17,6) and ~(2,6) . Find the altitude and the area of the trapezoid.

The vertices of a quadrilateral ABCD have the following coordinates: A(-3,0) , ~ ( 2 ~ 4 ) , ~ ( 6 ~ 0 ) , and ~(3,-5) . Find the area of the quadrilateral. Hint: Consider the - altitudes of AABC and AADC . The coordinates of the respective vertices of rectangle

ABCD are (3,2) , (10,2) , (10~7) and (3,7) . In the same coordinate system, the vertices of AEFC are

(5,2) , (3,5) , and (10,7) . Find the area of AEFC .

32. In the quadrilateral ABCE , point D is between C and -4--b E ; EC 1 1 A B ; A B = B C = C D = D E = E A .

Prove: AC BD = EB AD .

33. The hypotenuse of right triangle A X is , and is the altitude to the hypotenuse.

Prove: AB AC = BC AD .

34. Prove: If the diagonals of a quadrilateral are perpendicular, the area of the quadrilateral is equal to one-

half the product of the lengths of the diagonals.

Area Relations.

I n the study of a rea it i s in te res t ing and important t o compare the areas of two o r more f igures when they d i f f e r i n one o r more dimensions. Proport ional i ty i s one of the most e f fec t ive methods of studying t h i s change. Before continuing, you may wish t o review the de f in i t ion and fundamental properties of proport ional i ty and of proportions, a s presented i n Chapter 7.

Consider two t r i ang les . Suppose t h a t one of the t r iangles has base b1 , a l t i t u d e hl , and area A, ; suppose tha t the other t r i ang le has base b2 , a l t i t u d e h2 , and area A2 . Then

and

Hence, by division,

If the two t r i ang les have the property t h a t bl = b2 , then

h2 - = - . I n other words, the areas of two t r i ang les w i t h hl

equal bases a r e proportional t o the corresponding a l t i t u d e s . We note t h a t A2 = k hp and Al = k h, , where the constant

of proportionali ty k i s one-half the base of each t r i ang le , 1 1 namely k = Ã‘b = yb2 .

If the two t r i ang les under consideration have the property hl = h2 , then

I n other words, the areas of the two t r i ang les a re proportional t o the bases.

On the other hand, if the two triangles have the property that A, = A* , then blhl = b2h2 . In other words, the product

of the base and altitude is the same for one triangle as for the other. This situation suggests a notion which is related to the concept of proportionality and which we wish to discuss

now, namely "inverse proportionality."

An Important property of proportionality is the following, expressed for the case of three numbers: If the positive numbers q, r, s are proportional to the positive numbers

a, b, c then the largest of the numbers a, b, c corresponds

to the largest of the numbers q, r, s . By contrast, as the definition below shows, if the positive numbers q, r, s are inversely proportional to the positive numbers a, b, c , then the largest of a, R , c corresponds to the smallest of the numbers q, r, s . With this introduction, we are ready for the definition.

DEFINITION. Suppose that to the positive numbers q, r, s, ... there correspond the positive numbers a, b, c, ... (that is, q-a , r - < ~ ~ - b , S-<Ã‘Ã , ...). The numbers q, r, s, ... are inversely proportional to the numbers a, b, c, ... if and only if all the products of corresponding numbers are the same (that is, qa = rb = sc = ...).

As an example, the numbers 2 , 6 , 15 , 12 are because inversely proportional to the numbers 9 , 3 , 5 , 2 ,

each product of corresponding numbers is 18 . As another example, find the numbers x and y such

that 2 , x , 5 are inversely proportional to 6 , 4 , y . By the definition, the products 2 6 , x 4 , and 5 y are all the same. Thus 12 = 4x = 5y . The desired numbers

12 are x = 3 and y = - 5

We now extend our preliminary remarks about two triangles

to the case of any number of triangles.

THEOREM 11-8. Consider a set of two or more triangles.

(a) If the bases of all the triangles are equal, then the areas of the triangles are proportional to the corresponding altitudes.

(b) If the altitudes of all the triangles are equal, then the areas of the triangles are proportional to the corresponding bases.

(c) If the areas of all the triangles are equal, then the bases of the triangles are inversely proportional

to the corresponding altitudes.

Proof: For definiteness, we prove the theorem for a set of three triangles; the method applies to any number; by

choosing three, we avoid complications of notation in discussing

many triangles.

(a) By hypothesis, all the bases are the same number, say b . Let the areas of the triangles be A , At , A" , and let the corresponding altitudes be h , h1 , h" .

Now A = b b $I, A 1 = r f i t , A" = khll 2 . Hence the numbers A , A 1 ,

A" are proportional to the numbers h , h1 , h" , with the b non-zero number - as the proportionality constant. 2

(b) 'By hypothesis, all the altitudes are the same number, say h . Let the areas of the triangles be A , A * , A and let the corresponding bases be b , b* , b'' .

h h n NOW A = b , A t = ^bI , A" = -b" 2 . Thus A , A * , A" are

h proportional to b , b* , b" , with the non-zero number -y as the proportionality constant.

( c ) By hypothesis, all the areas are the same number, say A . Let the bases of the triangles be b , bl , b and

let the corresponding altitudes be h , h1 , h" .

products bh , b l h l , b"htl are equal, since each of them is equal to 2 A . Thus, b , b1 , b" are inversely proportional to h , i l l , h" .

Analogous to Theorem 11-8 is the following theorem for

parallelograms.

11-6

THEOREM - 11-9. Consider a set of two or more parallelograms.

(a) If the bases of all the parallelograms are equal, then the areas of the parallelograms are proportional to the corresponding altitudes.

(b) If the altitudes of all the parallelograms are equal, then the areas of the parallelograms are proportional

to the corresponding bases.

(c) If the areas of all the parallelograms are equal,

then the bases of the parallelograms are inversely

proportional to the corresponding altitudes.


A special case of Theorems 11-8 and 11-9 occurs when the number of triangles or parallelograms is two. In fact the case of two triangles has already been mentioned. Nevertheless It is worthy of repetition. If b , h , A are the base, altitude, area, respectively, of one triangle or parallelogram and if b l , hi , At pertain to the other, then the respective parts of the two theorems tell us the following:

(a) If b = bt , then A , A1 are proportional to A h h , hl , and hence -,-,- = v-\ . A b (b) If h = hl , then TCT = Â¥CT .

(c) If A = At , then bh = blh' .

Problem Set 11-6 -- 1. Prove Theorem 11-9.

2. In the quadrilateral ABCG , ABDF is a rectangle, and ABCE and ABEG are parallelograms. Compare the areas

of the three parallelograms. Explain your answer.

- 3. As shown in the figure, AD is divided into three

segments whose measures are proportional to 1 , 2 , 3 ,

Compare the areas of the three triangular-regions

RP RO J R3

- 4. It is given that d \ \ l , 114 ; AC Il-; CAD= 30 ;

AB = 4x ; BC = 6x . What is the ratio of the areas of

ADEB and ADFC ?

5. Prove: The diagonals of a parallelogram divide the

parallelogram and its Interior into four triangular-regions of equal area.

6 . Prove that each median of a triangle cuts the triangular- region into two triangular-regions of equal area. - -

7. AE , CD , and are medians of AABC Intersecting at point 0 . Prove that the areas of AAOB , ABOC and ACOA are equal. Hint: Use Problem 6 to compare the - areas of:

(a) AACD and ADCB ; AABF and AFBC . (b) AAOD and ADOB ; ABOE and AEOC .

Then prove that the areas of AAOB , ABOC , and ACOA

are equal.

8. The following experiment illustrates the fact that the point of Intersection of the medians of a physical triangle is the center of gravity of the triangle.

Cut a model of a triangle from cardboard and draw the three medians of the triangle. Try to balance the triangle on the head of a pin at the point of intersection

of the medians. Use the results in Problem 7 to explain why the intersection of the medians is the balance point

or center of gravity.

9. If the area of AABC in Problem 7 is 216 , find the area of each of the following triangles: AABO , ABOC , AAOC , AODB , ABOE , AAOF .

*lo. Given: AABC -v.AA'BtC1 .

(a) If A B = 1 2 , AÃ§B = 7 , and h = 10 , find hÃ . (b) If h = 9 , hÃ = 3 , and AÃ§B = 4 , find the

length of . 1 (c) If AB = 7 , BC = 8 , AC = 6 , and AÃ§B = 3 ? ,

find the perimeter of AA'BIC1 . -

In AABC , E l m ; DF 1 1 m ; CF= 2 0 ; CE= 10 ; CB = 30 , DF = 18 . Find:

c (a) The length of ;

(b) The area of ADFC ;

(c) The area of AABC ;

(d) The area of quadrilateral ABFD . B

The areas of two triangles are equal. What is the ratio of the base of the first to a base of the second if the

corresponding altitude of the second is:

(a) Three times the corresponding altitude of the first.

(b) One-fourth the corresponding altitude of the first.

(c) Three-fourths the corresponding altitude of the first.

(d) One hundred fifty per cent of the corresponding altitude of the first.

(e) Ten per cent more than the corresponding altitude

of the first.

Are the areas of two triangles equal if a base of the second is 5 units more than a base of the first, and the corresponding altitude of the second is 5 units less than the corresponding altitude of the first?

Explain your answer.

What is the ratio of the areas of two rectangles if the base of the second is 25 per cent more than the base of the first, and the altitude of the second Is 25 per cent less than the altitude of the first?

11-7. Relations - in Similar Polygons.

THEOREM 11-10. Every similarity between triangles has the property that the measures of the three sides and any altitude of the one triangle are proportional to the measures of the corresponding sides and the corresponding altitude of the other triangle.

Proof: &et PQR-<Ã‘^P*Q*R be a similarity between triangles. Let k be the proportionality constant. Then

p = kpÃ , q = kql , r = krÃ . Let RI3 and R'D' be the respective altitudes from R in APQR and from Rl in

APIQÃ§R . Let h = RD and hÃ = RID1 . If D / Q , then consider the correspondence

Rl%-RID1Q* between right triangles. Since /RDQ 2 /R~D*Q* (why?) and /DQR 2 /DIQIRI , the correspondence is a similarity. The proportionality constant for the similarity

RDQ-<Ã‘Ãˆ-R*DÃ is also k , since RQ = p = k p Ã = k RIQI . Hence h = kh* .

On the other hand, if D = Q , then h = RD = RQ = p and h* = p* ; in this case also, h = khl , since p = kpl .

Thus, in every case, p , q , r , h are proportional to pl , qÃ , r* , h1 , with proportionality constant k .

THEOREM 11-11. Every similarity between triangles has the property that the areas of the triangles are proportional

to the squares of the lengths of any pair of corresponding sides.

Proof: Let PQR-PIQIR* be a similarity between triangles. Consider any pair of corresponding sides, say ? and P ' and let r and r1 be the respective lengths of these sides. Let h and hl be the lengths of the altitudes to these sides in the respective triangles.

Let A and At be the respective areas of APQR and APtQIR1 . By Theorem 11-10, (r,h) 5 (r1,hI). Thus

Now

By substitution,

Thus, (A,A') - (r2,!-12) , as asserted. As an example, suppose that DEF-LMN Is a

similarity between triangles such that an altitude of ADEF

is three times as long as the corresponding altitude of ALMN . Then, by Theorem 11-10, every side of ADEF is three times as

long a s the corresponding side of ALMN , and every altitude of ADEF is three times as long as the corresponding altitude of ALMN . By addition, the perimeter of ADEF is three times

the perimeter of ALMN . Furthermore, by Theorem 11-11, the area of ADEF is nine times the area of ALMN .

We now turn our attention from triangles to polygons with any number of sides.

Theorem 11-12 is concerned with perimeters and of course

applies to triangles as well as to other polygons. Theorem

11-13 generalizes Theorem 11-11 to the case of polygons with

n sides.

THEOREM 11-12. Every similarity between convex polygons with

n sides has the property that the lengths of the n

sides and the perimeter of one polygon are proportional

to the lengths of the corresponding sides and the perimeter

of the other polygon.

Proof: Let the lengths of the sides of one convex polygon

be a, b, c, .. ., and let the perimeter a + b + c . .. be p . Let the lengths of the corresponding sides of the other convex

polygon be a1 , bl , c ' , ..., and let the perimeter a1 + b * + ct + ... be pl . Let k be the proportionality

constant for the similarity.

Then a = kal , b = kbl , c = kct , ... Hence

Thus a , b , ..., p are proportional to a' , b' , ..., pl with proportionality constant k . THEOREM 11-13. Every similarity between convex polygons with

n sides has the property that the areas of the polygonal-

regions (consisting of the polygons and their interiors,

respectively) are proportional to the squares of the

lengths of any pair of corresponding sides.

Proof: We outline the proof; you are asked to supply the details as a problem in the next problem set. Let

PQR. .. - Ã‘Ã PIQIR1. .. be a similarity between polygons with n sides. Let k be the proportionality constant. The diagonals from P cut the polygonal-region PQR.. . into

triangular-regions; let the areas of these triangular-regions be A , B , . .. In a like manner, let A' , B1 , ... be the areas of the corresponding triangular-regions into which the

diagonals from P' cut the polygonal-region PIQIR1. ..

Prove that APW "- APIQ'R1 , APRS APIRtS1 , etc.

By Theorem 11-11,

Hence

2 = k (A' + B' + C' + ...) .

Problem -- Set 11-7

1. The lengths of a pair of corresponding sides of two similar triangles are 4 and 5 . What is the ratio of the areas of the triangles?

2. The areas of two similar triangles are 64 and 100 . What is the ratio of the lengths of corresponding sides? the ratio of corresponding altitudes? the ratio of perimeters?

3. Two similar triangles are such that the area of the first triangle is 16 times the area of the other triangle. What is the ratio of the length of a side of the first

triangle to the length of a corresponding side of the

second?

4. The areas of two similar triangles are 64 and 100 . If a side of the first measures 24 , find the measure of the corresponding side of the second.

5. The altitude of an equilateral triangle is equal to the length of a side of a second equilateral triangle. What

is the ratio of the lengths of corresponding sides? the

ratio of the areas?

6 . Cut a triangle into three polygonal-regions of equal area by drawing lines parallel to a base.

By hypothesis, we have two

similar pentagons, ABODE and AIBICIDIE1 . We are to prove that their areas

squares of the lengths of are proportional to the

any two corresponding A s B

sides. 0

area ABODE - sc Restatement: area A,BIc,D,El - 7 . (Draw diagonals from A and A1 of the polygons.)

8. Problem 7 asks for the proof of Theorem 11-13 for the case of pentagons. Use the same ideas and give a proof of Theorem 11-13 for polygons with any number of sides.

9. The areas of two similar polygons are 144 and 256 . If a side of the first measures 9 , what Is the measure of the corresponding side of the second?

The lengths of the corresponding diagonals of two similar

polygons are 7 and 10 . What is the ratio of the

areas? the perimeters?

Find the ratio of the perimeters of two regular octagons

if the areas are 25 and 50 . Prove that the area of a square having the diagonal of a given square as a side has twice the area of the given square.

Two similar polygons RSTUV and R t S 1 T ' U 1 V are such that /R coincides with /R' . The coordinates of

R = R I , of s , of ~1 are (2,2) , (2,11) , (2,8) , respectively. Find the ratio of the lengths of corresponding sides of the polygons; the ratio of perimeters;

the ratio of areas.

The areas of two similar triangles are 1 4 4 and 81 . If a side of the former measures 6 , what is the length of the corresponding side of the latter?

In AABC , the point D Is on side AC , and AD is - twice CD . Let the line DE parallel to %? intersect %? at E . Compare the areas of triangles ABC and DEC . How long must a side of an equilateral triangle be in order that its area shall be twice that of an equilateral triangle whose side measures 10 ?

If similar triangles are drawn having, respectively, the side and the altitude of an equilateral triangle as corresponding sides, prove that the ratio of their areas

is 4 to 3 . Two pieces of wire of equal length are bent to form a square and an equilateral triangle respectively. What Is the ratio of the areas of the two polygonal-regions

bounded by the respective polygons?

11-8. Regular Polygons.

THEOREM 11-14. The bisectors of the Interior angles of a regular convex polygon of n sides intersect at a point.

Proof: Given a regular convex polygon ABCDEF... with

etc.

Let the bisectors of /A and /B intersect at point V . 1 1 Then AAVB I s isosceles, because m /at = pn LA = ifs LB = m /b .

Thus AV = BV . Now m Lav = m /b = m /bl . Hence the correspondence AVB-<Ã‘Ãˆ-B between triangles Is a congruence, by

1 1 S.A.S. (Why?) Therefore m/c = m/b = T ~ / B = wm/C . That is, C? Is the bisector of /C . In a like manner, we can prove that A B V C is isosceles, that the correspondence

+ BVC-<Ã‘^CV is a congruence between triangles, and that DV + bisects /D . The same procedure shows that EV bisects /E ,

etc. In summary, all the bisectors meet at the point V . DEFINITIONS. The center of a regular polygon is the point of intersection of the midrays of any two angles of the polygon.

Any triangle whose vertices are the center and two consecutive vertices of the polygon is called a central triangle of the regular polygon.

A radius of a regular polygon is any segment joining the center and a vertex of the polygon.

An apothem of a regular polygon is any segment which joins the center and a side of the polygon and is perpendicular to that side.

1 As an example, the center i

I

of the regular hexagon in the

diagram is C . There are six central triangles, one of which A

is AABC . The segment is - a radius and the segment CR is an apothem of the regular hexagon.

Theorem 11-14 tells us that the center of a regular polygon is the point of intersection of all the bisectors of - angles of the polygon.

THEOREM 11-15. Every central triangle of a regular polygon is Isosceles and is congruent to every other central triangle.

Proof: These statements, expressed now in the new language of "central triangle," were actually established in the proof of Theorem 11-14. Indeed, using the notation of that proof, we showed that each of the central triangles A V 3 , BVC , CVD , etc., is isosceles and that AAVB 2 ABVC 2 ACVD 2 . , .

THEOREM 11-16. The area of a regular polygon is one-half the product of the apothem and the perimeter of the polygon.

Proof: Let ABC... be a regular polygon with n sides.

Let V be the center of the polygon, let a be the apothem,

and let e be the length of one side of the polygon. The segments joining the center V and the vertices of the polygon

determine n central triangles.

Each of these cen t ra l t r i ang les has base e and a l t i t u d e a 1 and hence area p a . The area of the regular polygon i s

1 1 therefore n ( p a ) = ~ ( n e ) . Since ne i s the perimeter of the polygon, the theorem i s proved.

Problem Set 11-8 -- 1. Does a perpendicular segment from the center of a regular

polygon t o a s ide b isec t the s ide? Why?

2. The apothem of a regular hexagon i s 10 '/5' . What i s the length of each s ide of the hexagon?

3 . The diagonal of a square has length 6-/? . What I s the radius? the perimeter? the apothem? the area?

4. Given an equ i l a t e ra l t r i ang le whose s ide measures s , f ind the radius and the apothem of the t r i a n g l e i n terms of s .

5. The perimeter of a regular hexagon i s 12 . Find the apothem, the radius, the area .

6. The radius of a square Is r . Find t h e apothem, the length of a s ide, the perimeter, and the area of the square a l l i n terms of r .

7. The apothems of two equ i l a t e ra l t r i ang les a r e 8 and 1 2 ,

(a) What i s the r a t i o of the r a d i i ? of the lengths of t h e i r s ides? of the perimeters? of the areas?

(b) Find the area of the smaller t r i ang le by two d i f fe ren t methods.

8. ( a ) Each s ide of a regular hexagon I s 8 A/^ . Find the area of the hexagon.

( b ) The apothem of a regular hexagon i s 12 . What i s the perimeter of the hexagon? the area?

(c) U s e another method t o f ind the area of the hexagon i n ( b ) .

11-9. Polyhedrons.

Pictures of various polyhedrons look like the following:

DEFINITIONS. A polyhedron is the union of a finite number of polygonal-regions, each of which consists of a convex polygon and its interior, such that (1) the interiors of any two of the polygonal-regions do not intersect and (2) every side of any of the

polygons is also a side of exactly one other of the polygons.

Each vertex of any of these polygons is called a vertex of the polyhedron.

Each side of any of these polygons is called an edge of the polyhedron. - Each of the polygonal-regions is called a - face of the polyhedron.

E

As an example, consider the polyhedron in the above diagram. It has five vertices. It has eight edges, two of

which are and . It has five faces, one of which is the shaded triangular-region CDE .

A polyhedron is named according to the number of faces which it contains. Since the number of sides of a polygon is the basis for naming a polygon, we expect some resemblance

between the names of polygons and the names of polyhedrons. The following table shows this analogy.

11-9

Name of Polygon

Triangle

Quadrilateral

Pentagon

Hexagon

Heptagon

Octagon

Nonagon

Decagon

Dodecagon

20-gon

Number of Sides

Tetrahedron

Pentahedron

Hexahedron

Heptahedron

Octahedron

Nonane dron

Decahedron

Dodecahedron

Icosahedron

Name of Polyhedron

Prisms, pyramids, and frustums of pyramids are examples of

special kinds of polyhedrons. Other examples are the so-called

regular polyhedrons.

Number of Faces

DEFINITIONS. Any non-empty intersection of a polyhedron and a plane is called a section of the polyhedron.

(No polyhedron has three faces.)

A polyhedron is a convex polyhedron if and only if every section of it which contains at least three non-collinear points is either a convex polygon or a face of the polyhedron.

A regular polyhedron is a convex polyhedron such

that:

(1) each face is the union of a regular polygon

and its interior;

(2) all these regular polygons have the same

number of sides; and

( 3 ) all vertices of the polyhedron belong to the

same number of faces.

It is interesting to note that there are only five types of regular polyhedrons: the regular tetrahedron, the regular hexahedron (also called the cube), the regular octahedron, the

regular dodecahedron, the regular icosahedron. This fact will

be discussed again later In the chapter. Pictures of these five types of polyhedrons are shown below.

Tetrahedron

Octahedron

Dodecahedron L/^

Hexahedron or cube

@ ICO sahedron

Models of regular polyhedrons are not difficult to make,

and they are very helpful in studying the properties of the

regular polyhedrons. The plans for making these models are given below. They should be constructed from stiff paper, using dimensions that are at least five times as large as the

dimensions of the pattern.

Tetrahedron

Octahedron

Hexahedron

Icosahedron

11 -9

Problem Set 11-9 -- 1. Make a t a b l e similar t o t h e fol lowing and f i l l i n t he blanks

f o r t h e ind ica ted r egu la r polyhedrons.

Tetrahedron

Octahedron

Icosahedron

Hexahedron

Dodecahedron

Regular Polyhedron

Number of

Faces

Boundary of

Face

Number of Faces ( o r ~ d g e s ) a t a Vertex

Number of

Edges

2 . From the preceding t a b l e , v e r i f y t h e formula f - e

Number of

Ver t i ces

where f i s t he number of f aces of t h e r egu la r polyhedron,

e i s t h e number of edges, and v i s t he number of v e r t i c e s .

Do you th ink t he formula i s a l s o t r u e f o r polyhedrons which

a r e not regu la r polyhedrons?

3. Explain why the re i s no polyhedron with t h r ee f aces .

I f you would l i k e t o know more about t h e r e l a t i o n s t h a t

" e x i s t among regu la r polyhedrons, o r i f you a r e i n t e r e s t e d i n

const ruct ing models t h a t use r egu la r polyhedrons a s a ba s i s

f o r t h e i r cons t ruc t ion , the fol lowing books w i l l be of

i n t e r e s t t o you.

Steinhaus, Mathematical Snapshots

Cundy and R o l l e t t , Mathematical Models

11-10. Polyhedral Angles.

In Chapters 4 and 9 we studied plane angles and dihedral angles. In this section we introduce another type of angle

known as the polyhedral angle. We also study some important properties of polyhedral angles.

A picture of a polyhedral angle is the following:

This polyhedral angle is determined by the convex quadrilateral PlPpP3P4 and the point V not In the plane of the quadri-

+ + + 4 lateral. The rays VP, , VPn , VP, , VP4 , are edges of the polyhedral angle. Each of four angles at V , namely /p1vPp , LP~VP, , /p3vp4 , ZP;,VP., is a -- face angle of the polyhedral

angle. A - face of the polyhedral angle is the union of a face angle and its interior; for example, in the plane VPlP4 , the union of /PbVPl and its interior is a face. The polyhedral

angle itself is the set of all points belonging to any of the faces. This illustration leads us to the following definitions.

DEFINITIONS. Let a convex polygon and a point V not in the plane containing the polygon be given; the union of all the concurrent rays which have

endpoint V and which contain a point of the

polygon is called a polyhedral angle.

The point V is the vertex of the polyhedral angle.

Each ray with endpoint V and containing a vertex of the polygon is an - edge of the polyhedral angle.

An angle with vertex V and containing two

consecutive vertices of the polygon is a face angle of the polyhedral angle.

A face of a polyhedral angle is the union of a face angle and its interior.

A polyhedral angle of three faces is called a trihedral angle.

Notation. If a polyhedral angle is determined by

the convex polygon PIP 2..,Pn and the vertex V , +

if Ql is an Interior point of VPl , if Qg is 4 an interior po1nt.of VPg, ..., and if Qn is an

+ interior point of T O , then the polyhedral angle is denoted by the symbol /y - QIQo.. .Q^ .

In particular, the polyhedral angle may be denoted by p - Ill2. .P,

Other pictures of polyhedral angles are the following.

/V - RSTXZ

Exploratory Problems

1. How many trihedral angles are formed by the walls, floor, ceiling of your classroom? What do you think is the measure of each of their face angles?

2. Can you make a model of a trihedral angle with exactly one of the face angles a right angle? With exactly two of the face angles as right angles? With every face angle a right angle? Is it possible to make a model of a polyhedral angle with four faces such that each of the face angles is a right angle? Explain.

3. Make a model of a polyhedral angle with five faces so that each face angle measures 60 . Can you make a model of a polyhedral angle with six faces if each face angle measures

60 ? Explain.

4. Do you think it is possible for a polyhedral angle to have four face angles whose respective measures are 50 , 120 , 90 , 100 ? Explain.

5. Complete the following statement: The sum of the measures

of the face angles of a polyhedral angle is

Construct a model of a trihedral angle, say /V - ABC , such that the measures of the face angles /AVB , /BvC , /CVA are 80 , 4 0 , 100 , respectively. h he pattern of such a model is given in the diagram below. The suggested distances are measured in inches. As you complete the model by bringing A and A' together, keep the rays

"pointing downward from" the vertex V and keep face AVB toward your right. )

Compare your model with those of your classmates. Do you think that all the trihedral angles represented by these models are congruent to each other?

7. Construct, a s i n Problem 6, a model of another t r i h e d r a l

angle, say /W - DEF , where the measures of /DWE , /EWF , /FWD a re 40 , 80 , 100 , respectively. Compare your model with those of your classmates. Do a l l these t r ihedra l angles appear t o be congruent?

8. Does the t r ihedra l angle whose model you constructed i n Problem 7 appear t o be congruent t o the t r i h e d r a l angle whose model you constructed i n Problem 6 ?

9. The models which you constructed i n Problems 6 and 7 give an example o f a p a i r of s y m m e t r i c t r i h e d r a l angles. I I

What do you think i s meant by saying t h a t two t r i h e d r a l I angles a r e symmetric - t o each other?

Draw pic tures of a p a i r of v e r t i c a l t r i h e d r a l angles. Do they appear t o be congruent? Do they appear t o be symmetric? (You should be able t o guess the meaning of ' v e r t i c a l " t r i h e d r a l angles by analogy with v e r t i c a l angles. )

Try and make models of t r i h e d r a l angles wi th face angles measuring:

( a ) 40 , 50 , 100 , respectively;

( b ) 40 , 50 , 90 , respectively;

( c ) 40 , 50 , 80 , respectively.

Explain the r e s u l t of Problem 11.

Complete the following sentence: The sum of the measures of two face angles of a t r ihedra l angle i s

The preceding exploratory problems lead us t o the following two theorems, whose proofs we omit.

THEOREM 11-17. The sum of the measures of any two face angles of a t r i h e d r a l angle, i s grea ter than the measure of the t h i r d face angle.

THEOREM 11-18. The sum of the measures of a l l the face angles of any polyhedral angle i s l e s s than 360 . A s an appl icat ion of the preceding two theorems, consider

the following s i tua t ion . Suppose t h a t the measures of two of the face angles of a t r i h e d r a l angle a re known t o be 75 and 115 . We ask what Information can be deduced about the measure of the t h i r d face angle of t h i s polyhedral angle. Let the measure of the t h i r d face angle be denoted by x .

(1) By Theorem 11-17, we f ind tha t :

x + 75 > 115 , x + 115 > 75 , and

75 + 115 > x

The first of these inequalities tells us that

since x is positive, the second of the inequalities gives us no new information; and the third of the inequalities, namely x < 190 , also does not provide any new information about the number x (why?) .

(2) By Theorem 11-18, we find that

x + 75 + 115 < 360 . Hence

x < 170 . (3) Since Part (1) tells us that x > 40 and Part (2)

tells us that x < 170 , we finally conclude that

Problem Set 11-10 -- 1. In each of the following, the measures of two of the face

angles of a trihedral angle are given. Find two numbers such that the measure of the third face angle is between

them, in accordance with the information provided by Theorems 11-17 and 11-18.

(a) 80 , 105 ( d ) 3.45 , 175 (b) 100 , 125 (e) 50,135

(4 60 , 135 (f) 80 , 95 2. True - False statements. Write + if the statement is

true; write 0 if the statement is false.

(a) Each of the three face angles of a trihedral angle

can be obtuse.

(b) A polyhedral angle can have four face angles that

are right angles.

(c) The measure of the face angles of a polyhedral angle with four faces can be 50 , 65 , 100 , and 110 .

The measures of the face angles of a t r i h e d r a l angle can be 140 , 130 , and 120 . If the measures of two face angles of a t r ihedra l angle a r e 100 and 120 , the measure of the t h i r d face angle i s l e s s than 20 . If each face angle of a polyhedral angle measures 60 , t he polyhedral angle must be a t r ihedra l angle. If the measure of each face angle of a polyhedral angle i s 90 , the polyhedral angle must have four faces. If a plane Is perpendicular t o one edge of a polyhedral angle, it i s perpendicular t o two faces of the polyhedral angle.

Corresponding t o each vertex V of a convex polyhedron, there i s a polyhedral angle, whose vertex i s V and whose edges a re the rays conta in ing those edges of the polyhedron t h a t have an endpoint a t V. In the A - --- - D i l l u s t r a t i v e diagram a t the r igh t , the polyhedral angle associated with vertex V of the polyhedron VABCDE 0 x /

i s /V - A N D . The faces of the polyhedral angle w i t h vertex V contain the respective faces of the polyhedron that contain the point V .

I n the preceding sect ion we described the so-called regular polyhedrons. By p ic tures and models we found f i v e types of regular polyhedrons. The number of respective faces i s 4 , 6 , 8 , 1 2 , 2 0 .

The length of an edge of a cube (regular hexahedron) may be any pos i t ive number. So, although cubes can occur i n any II s i ze , " they a l l have the same "shape," i n other words, they a re similar t o one another. I n a l i k e manner, regular tetrahedrons of d i f f e ren t ' s i z e s " a re nevertheless s imilar t o each other. I n general, regular polyhedrons of any of the f i v e types we have studied a re s imilar t o one another. The

remarkable f a c t i s t h a t these f i v e types a re the only types of regular polyhedrons tha t ex i s t . We formulate t h i s a s the next theorem, whose proof we merely sketch.

THEOREM 11-19. There a re no more than f i v e types of regular polyhedrons.

Outline of proof: -- (1) A polyhedral angle has a t l e a s t three face angles.

(2) The sum of the measures of the face angles of a polyhedral angle i s l e s s than 360 ,

(3) The face angles of the polyhedral angle corresponding t o each vertex of a regular polyhedron have the same measure.

( 4 ) Therefore the measure of each face angle must be l e s s than 120 .

(5) The measure of each angle of a regular polygon with 6 or more s ides i s a t l e a s t 120 .

( 6 ) Hence every face of a regular polyhedron has l e s s than 6 edges; i n other words, a face of a regular polyhedron i s a polygonal-region whose boundary has e i t h e r 3 o r 4 o r 5 s ides .

(7) Suppose t h a t each face has 3 edges. Then,

( a ) each face angle has measure 60 ; (b ) each polyhedral angle can have 3 o r 4 o r 5

faces, by p a r t s (1) and (2) ;

( c ) no more than three types of regular polyhedrons have faces which a re tr iangular-regions.

(8) Suppose t h a t each face has 4 edges. Then,

(a) each face angle has measure 90 ;

( b ) each polyhedral angle has exact ly 3 faces, by pa r t s (1) and (2 ) ;

( c ) no more than one type of regular polyhedron has faces which a re squares and t h e i r i n t e r i o r s .

( 9 ) Suppose that each face has 5 edges. Then,

(a) each face angle has measure 108 ;

(b) each polyhedral angle has exactly 3 faces;

(c) no more than one type of regular polyhedron has

faces which are pentagons and their interiors.

(10) In summary, there are no more than five types of regular polyhedrons.

11-11. Prisms.

We now study another type of polyhedron, namely the prism.

DEFINITION. A prism is a polyhedron such that two of its faces (called bases) have boundaries which are congruent polygons in parallel planes and each of the remaining faces has a boundary which is a parallelogram with two sides in the parallel planes.

Prisms are classified according to their bases: A prism

each of whose bases is a triangular-region is called a triangular prism; a prism each of whose bases is a rectangular-

region Is called a rectangular prism; and so on. Of particular importance among the prisms each of whose bases has a quadrilateral as a boundary are the following:

Parallelepiped

Rectangular Parallelepiped

Cube

DEFINITIONS. A parallelepiped is a prism such that the boundary of each of its bases is a parallelogram.

A rectangular parallelepiped is a parallelepiped such that the boundary of each of its faces is a

rectangle.

A cube is a parallelepiped such that the boundary of each of its faces is a square.

Notice that each parallelepiped is a polyhedron with six faces, that is, is a hexahedron. In particular, the cube is the regular hexahedron.

An ordinary box I s a model of a rectangular parallelepiped. A prism such as a rectangular parallelepiped has three pairs of faces, each of which may be considered as a pair of bases. Is

this also true of a parallelepiped which Is not rectangular? Why? By contrast, only one pair of faces of a triangular prism

may be considered as the two bases. Why?

DEFINITIONS. With reference to a selected pair of bases of a prism, we define the following:

any one of the remaining faces is called a lateral face of the prism;

the union of the lateral faces is called the lateral surface of the prism (sometimes known prismatic surface) ;

any edge which is the intersection of two lateral faces is called a lateral edge of the prism;

the prism Is said to be a right prism if and only if -- a lateral edge is perpendicular to a base.

The left-hand figure below is a picture of a prism that is not a right prism. Face ABYX is a lateral face and is a lateral edge of the triangular prism. Each of the other

two diagrams below Is a picture of a right prism.

Triangular prism Right tr iangul.ar pr i sm y4 . A,-;- d-lt pentagon21 p r i s a

Problem Set 11-lla - 1. Explain why all the lateral edges of a prism are parallel

to one another.

2. Prove that in a right prism every lateral edge is perpendicular to each base.

DEFINITIONS. With reference to a selected pair of bases of a prism, we define the following:

a cross-section of the prism is any non-empty intersection of the prism and a plane which is parallel to, and distinct from, the planes containing the

bases;

a right-section is any intersection of the prism and a plane which is perpendicular to, and intersects the Interior of, every lateral edge.

In the diagram at the right, plane ABCD , plane EFGH , and plane QRST are parallel planes; H

and plane WXYZ is perpendicular - G to AE . Quadrilateral QRST is a cross-section of the

prism and quadrilateral WXYZ Is a right-section

of the prism.

A

DEFINITIONS. With reference to a selected pair of bases

of a prism, we define the following:

any segment whose endpoints lie in the two parallel planes containing the bases and which is perpendicular to these planes is an altitude of the prism;

the sum of the areas of all the lateral faces of the prism is the lateral area of the prism;

the sum of the areas of all the faces of the prism is the total area of the prism. --

A method of computing the lateral area of a prism is to

find the area of each of the lateral faces and then to add

their areas. The following experiments help you to recognize a simpler method for finding the lateral area of a prism.

Experiments

1. Cut a long one of t h e edges of a r i g h t prismatic sur face . Note t h a t t h i s su r face can be f l a t t e n e d i n t o a r e c t a n g l e a s

shown i n t h e next f l g u r e . The base of t h e r e c t a n g l e i s t h e of the of t h e prism, and t h e a l t i t u d e of

t h e r e c t a n g l e i s t h e of t h e prism. Therefore, t h e l a t e r a l a r e a of t h e prism i s t h e product of and

Cut a long t h e l a t e r a l edge of a p r i smat ic su r face t h a t i s not a r i g h t p r i s m a t i c s u r f a c e . F l a t t e n t h i s p r i smat ic su r face i n t o a plane s u r f a c e

Dl a s shown i n t h e fo l lowing f i g u r e .

Draw a l i n e i n t h e plane perpendicular t o one of t h e edges of t h e f l a t t e n e d s u r f a c e as shown. Does t h e l eng th of - R S equal t h e sum o f t h e a l t i t u d e s of t h e para l le lograms which are t h e lateral f a c e s of t h e prism? Why does t h e l a t e r a l a r e a - of t h e prism equa l t h e product of t h e l e n g t h s of RS and a lateral edge o f t h e prism?

Change the f l a t t ened f igure back t o the or ig ina l prismatic surface. The length of i s the same as the perimeter of a

of the prism.

THEOREM 11-20. The l a t e r a l area of a prism i s equal t o the product of the length of a l a t e r a l edge and the perimeter of a r ight-sect ion.

Proof: We are given a prism w i t h bases PIP2.. .Pn and R R 2 . .Rn and r ight sect ion QiiQ ... Q . Let L be the l a t e r a l area of the prism, e the length of a l a t e r a l edge, and p the perimeter of the given r i g h t section. We a r e required t o prove t h a t L = ep .

S t a t e m e n t s R e a s o n s

- - - - 2 . Q n Q l l P I R l ; sif2~2, e t c .

3 . Area of PIR,RnPn Is ~(Q,Q, ) ,

Area of P2RgR,Pn Is ~ ( Q ~ Q . , ) ,

Area of P 3 R A P i , I s ~(Q.,Q!^) , e t c .

4. Sum of the areas of n p a r a l l e l o g r a m s i s

e ( Q f i n + Q2Q3 + Q3Qi, + . . . + \,On + QnQ,) .

5. T h e r e f o r e , L = e p .

1. Why?

2. Why?

3. Why?

'I. Why?

5. Why?

Corollary 11-20-1. The lateral area of a right prism is the product of the length of a lateral edge and the perimeter

of a base.

Problem Set 11-llb - Supply the reasons for the statements in the proof of Theorem 11-20.

Prove Corollary 11-20-1.

Find the area of the lateral surface of a right prism

whose altitude is 10 if the sides of the pentagonal base

measure 3 , 4 , 5 , 7 , 2 , respectively. Find the total area of a right triangular prism if the base is an equilateral triangle 8 inches on a side and the altitude of the prism is 10 inches.

If the sides of a cross-section of a right triangular prism measure 3 , 6 , and 3 f l , then any other cross section will be a triangle whose sides measure 3

Y and , and whose angles measure Y

- f - , and whose area is .

The length of a lateral edge of a right prism is 10 and

its lateral area is 52 . What is the perimeter of the base of the prism?

The apothem of the base of a right hexagonal prism is

8 A/S . The altitude of the prism is 20 . Find the lateral area of the prism.

At one of the vertices of a certain square prism, the associated polyhedral angle has face angles which

measure 90 , 90 , 30 , respectively. Each lateral edge of the prism is 20 inches long, and the perimeter of

the base is 48 inches. Find the total area of the

prism.

Prove by the use of coordinates that the diagonals of a

rectangular parallelepiped have equal length.

Prove by the use of coordinates that the diagonals of a rectangular parallelepiped bisect each other.

11-12. Pyramids.

Pyramids resemble prisms in several respects. Many of the terms such as lateral face, lateral edge, cross-section, and lateral area are the same, and we shall use them without formal

definition.

DEFINITIONS. A pyramid is a convex polyhedron which, except for the interior of one of its faces, is con-

tained in a polyhedral angle.

The vertex of the polyhedral angle is called the vertex of the pyramid.

The face of the pyramid which is not contained in the polyhedral angle is called the base of the

pyramid.

The segment which Joins the vertex and the plane containing the base and is perpendicular to that plane is called the altitude of the pyramid.

In the diagram below, V is the vertex of the pyramid; the polygonal-region CDEFGH Is the base of the pyramid; with

the exception of the Interior of face CDEFGH , the pyramid is contained in /y - CDEFGH .

DEFINITIONS. A pyramid is a regular pyramid if and only if the boundary of its base is a regular polygon and the center of the base is an endpoint of the altitude of the pyramid.

The slant height of a regular pyramld Is the distance between the vertex of the pyramid and an edge in the base of the pyramid.

In the following diagram, the altitude VQ of the

isosceles triangle EVA is the slant height of the regular

pentagonal pyramid.

Associated with the set of all pyramids is another important class of polyhedrons. If we Imagine "cutting off

the top" of a pyramid, the remaining figure suggests a frustum

of the pyramid. In the diagram below, the polyhedron with vertices A , B , C , D , P , Q , R , S is a frustum of the pyramid whose vertex is V and whose base is the polygonal-

region ABCD .

DEFINITION. Given a pyramid, a frustum of the -- pyramid is a polyhedron such that:

(1) one of its faces is the base of the pyramid;

(2) another of its faces is in a plane parallel

to the plane containing the base of the

pyramid, and

( 3 ) each of its other faces is contained in the

pyramid.

The proof of the following theorem is left as a problem.

THEOREM 11-21. Let a triangular pyramid be given.

(a) Every cross-section of the pyramid is a triangle similar to the boundary of the base.

(b) If the distance from the vertex of the pyramid to the plane containing the cross-section is k and if the altitude of the pyramid is h , then the area of the cross-section and the area of the base

2 are proportional to the numbers k2 and h .

Problem Set 11-12 -- Prove t h a t the boundaries of the l a t e r a l faces of a regular pyramid a r e i sosce les t r i ang les which a re congruent t o one another.

Prove t h a t t h e l a t e r a l area of a regular pyramid i s given 1 by the formula A = p p , In which p i s the perimeter of

the base and a i s the s l an t height.

I f p i s the perimeter of the base of a regular pyramid and a i s the s l a n t height, f i n d the l a t e r a l area of the pyramid, i n each of the following cases.

(a) p - 1 8 , a = % .

1 1 (b) p = 27 (yards) , a = 2a ( f e e t ) .

Find the a rea of the l a t e r a l surface of a regular square pyramid i f each s ide of the base i s 8 inches long and

the s l a n t height of the pyramid Is 5 inches long.

What i s the s l a n t height of a regular pyramid i f the area of i t s l a t e r a l surface i s 80 and the perimeter of i t s base i s 20 ?

Find the a l t i t u d e of a regular square pyramid with a l a t e r a l edge measuring 25 and a diagonal of the base measuring 14 . F i l l t h e blank: The boundary of each l a t e r a l face of a frustum of a pyramid i s a - .

8. Derive a formula f o r the l a t e r a l E area of a regular frustum (such a s t h a t shown i n the diagram) i f p i s the perimeter of the lower base, p' i s the perimeter of the upper base, and a i s the a l t i t u d e of a l a t e r a l face of the frustum. B' C'

9. The bases of a frustum of a regular pyramid a re squares 8 inches and 6 inches on a s ide. The a l t i t u d e of a l a t e r a l face of the frustum i s 4 inches. Find the l a t e r a l area and the t o t a l area of the frustrum.

10. I n a frustum of a regular square pyramid, an edge of the lower base measures 1 4 and an edge of the upper base measures 8 . If the l a t e r a l area of the frustum i s 220 , f ind the a l t i t u d e of a l a t e r a l face of the frustum.

11. In a pyramid with vertex V , rectangular base ABCD and a l t i t u d e (as i n the diagram), l e t EFGH be a cross- section s imilar t o the rectangle ABCD such t h a t the

2 proportionali ty constant i s 3 . ( a ) AFVG - ABVC . Why?

(b) _[ . Why?

( c ) AKVG - AOVC . Why?

( d ) What i s the r a t i o of VK t o V O ?

( e ) Suppose the perimeter of the rectangle ABCD

i s 36 . What i s the perimeter of the rectangle EFGH ?

(f) Suppose the altitude of a lateral face of the pyramid

is 18 . What is the altitude of a corresponding

lateral face of the frustum?

(g) What is the area of the lateral surface of the

pyramid?

(h) What is the area of the lateral surface of the f rus turn?

(1) What is the ratio of the lateral area of the pyramid

to the lateral area of the frustum? Explain.


Hint: Let AABC be in plane & and point V a distance - h from 5 . Let plane $. , parallel to & and - - at distance k from V , intersect m , VB , VC In A ' , Bl , Cl , respectively. Then show that AAlBlC' - AABC and that

area of AA1B1C8 k 2 area of AABC =

13. A point of light is 6 feet from a wall. A piece of

cardboard containing 24 square inches of surface area is held between the light and the wall 3 feet from the wall and parallel to it. Find the area of the shadow of

the cardboard on the wall.

14. A point of light is 10 feet from a wall. How far from

the wall, but parallel to it, should a piece of paper be held so as to cast a shadow four times the area of the

paper?

Summary.

There are four major types of measures in our geometry. In Chapter 3, we discussed the measure of the distance between two points, or equivalently, the length of the segment Joining

the points. This is the basis for measuring a figure in one- dimensional geometry. In the present chapter, we treated a measure of convex polygons, or more generally, the area of any polygonal-region. This is the basis for measuring a figure in two-dimensional geometry. The next stage in this development

would be (if we only had time!) to examine a measure of a

convex polyhedron, or more generally, the volume of any "polyhedral-region." This would be the basis for measuring a

figure in three-dimensional geometry. In Chapter 4, we considered the measure of an angle, or if you like, a measure of --- the angular distance between two concurrent rays.

There are extensions of measurement. The measure of a dihedral angle is defined in terms of the measure of an angle. In the next chapter the measure of an angle will permit us to measure an arc of a circle. Also in the next chapter the area of a polygonal-region will permit us to describe the areas of

circular-regions and of>other two-dimensional regions associated

with circles. The volume of a polyhedral-region would permit

us to discuss the volumes of spherical regions and other regions in space.

These types of measure have certain properties in common, (1) A measure is a real number that is not negative. (2) A

measure depends upon a chosen unit, and the measures of the

same geometric figure relative to different units are related in a simple manner. ( 3 ) The measures of two congruent figures are the same. ( 4 ) The measure of a "whole" is the sum of the measures of its nonoverlapping "parts"; that is, the length of a segment Is the sum of the measures of any segments such that

the interiors of any two of them do not intersect and their union is the given segment; the area of a polygonal-region is the sum of the measures of any polygonal-regions such that the interiors of any two of them do not intersect and their union

is the given polygonal-region; a similar remark would apply to the volume of a polyhedral-region; the measure of an angle is the sum of the measures of any angles such that the Interiors

of any two of them do not intersect and the union of the angles and their interiors is the same as the union of the given angle and its interior.

There are connections among the various types of measure.

The area of a two-dimensional region may be related to a product of two lengths. The volume of a three-dimensional region may be related to a product of three lengths, or to a product of a

length and an area. These relationships are the familiar formulas for calculating areas and volumes. Their practical importance depends heavily on the fact that in the physical

world it is often less convenient to measure an area or a volume directly than to compute it from data obtained by measuring appropriate distances and perhaps angles.

A similarity between two geometric figures either directly or Indirectly prescribes many corresponding parts: sides, angles, diagonals, altitudes, medians, faces, bases, and so forth. In a similarity, the lengths of all segments, the

square roots of the areas of all polygonal-regions, and the cube roots of the volumes of all polyhedral-regions in one geometric figure are proportional to the corresponding numbers for the other geometric figure. For the special case in which the constant of proportionality is one, the similarity is a

congruence.

Review Problems

What i s the measure of each i n t e r i o r angle of a regular polygon of f i f t e e n s ides? What I s the measure of each exter ior angle?

I f an ex te r io r angle of a regular polygon has measure 10 , how many s ides has the polygon?

The sum of the measures of the angles of a ten-sided polygon i s 1440 . Is the sum of the measures of the angles of a twenty-sided polygon twice as much? Verify your answer.

The hypotenuse of a 30 , 60 , 90 t r i a n g l e is 16 and the shorter l e g of a second 30 , 60 , 90 t r i ang le is 13 . What i s the r a t i o of t h e i r areas?

Two face angles of a t r ihedra l angle measure 56 and 100 . Between what numbers must the measure of the t h i r d face angle of the t r i h e d r a l angle be?

Which of the following t r i p l e s of numbers can be the measures of the three face angles of a t r i h e d r a l angle?

( a ) 45 , 45 , 90 . ( c ) 140 , 171 , 70 . (b) 60, 60, 60. (d) 150 , 118 , 130 . I n choosing t i l e f o r a f l o o r covering, would congruent regular hexagonal t i l e s give a complete coverage? What

other regular polygons would fit together? How do you know?

R By hypothesis we have an equi la tera l t r i ang le with s ide measuring s , a l t i t u d e h , and area A . (a) Show that h = -1 0 . (b) show t h a t A =+2 .

P

Find the area of an equilateral triangle having the length of a side equal to:

(a) 2 (4 0 (b) 8 (d l 7

The area of an equilateral triangle is 9 0 . Find the length of its side and Its altitude.

The altitude of an equilateral triangle is 8 i/ . Determine the length of its side and its area.

A square whose area is 81 has its perimeter equal to

the perimeter of an equilateral triangle. Find the area the equilateral triangle.

ABCD is a square, find, D-

s .c terms of s and b ,

the area of the star pictured here. The

segments forming the boundary of the star

are congruent.

A

Prove that the area of an isosceles right triangle is equal to one-fourth the area of a square having the hypotenuse of the triangle as a side.

Given: Square ABCD with

points E and F as shown, SO that EC 1 . Area of ABCD is 256 . Area of

ACEF is 200 . Find BE .

16. Find the area of each of the following polygonal-regions,

If possible:

(a)

Parallelogram

(b)

8 Triangle

Square

(d l

18

Trapezoid

(f)

Rhombus

Triangle

Parallelogram

17. If W , X , Y and Z are midpoints of s ides

. of square ABCD , as shown i n the f igure, compare the area of ABCD witn the area of square RSPQ .

18. The area of a convex quadr i la te ra l i s 126 and the length of one diagonal i s 21 . If the diagonals a re perpen- d icular , f ind the length of the other diagonal.

19. The diagonals of a rhombus have lengths of 15 and 20 . Find i t s area. If an a l t i t u d e of the rhombus i s 12 , f i n d t h e length of one side.

20. Find the area of the polygonal-region which Is the in te r - sect ion of ( (x,y) : -7 < x < 5) and

((x,y): -5 < y 2 -1) . ( D r a w and shade the polygonal- region. )

- Diagonal AD of the pentagon ABODE shown has length 44 and the perpendiculars from B , C , and E have lengths 2 4 , 1 6 , and 1 5 , respectively. AB = 25 and CD = 20 . What i s the area of the pentagon?

D

E

22. Given: Parallelogram ABCD B C with X and E midpoints of and AD respectively

To prove: Area of AECX is

the area of ABCD . '2 A E D

23. If IS is a segment in plane S> , what other positions of P in plane "$Ã will let the area of AABP remain constant? Describe the location P

of all possible positions of P in plane & which satisfy the condition. Describe the location of all possible positions of P in space which satisfy the condition.

A B

24. This figure represents a H G

cube. The plane determined by points A , C and F is shown. If is 9 inches long, how long is - AC ?

c

What is the measure of / FAC ?

What is the area of AFAC ? A B

25. Find the length of the diagonal of a cube whose edge is

6 units long.

26. Explain how to cut a polygonal-region bounded by a trapezoid into two polygonal-regions having equal areas by a line through a vertex of the trapezoid.

27. I n t rapezoid ABCD , base angles of measure 60 include a base of l eng th 1 2 . The - non-paral le l s i d e AD has

length 8 . Find t h e a r ea of t he t rapezoid.

28. Find the area of t rapezoid ABCD .

29. (a) Prove t h e following theorem: The median of a t rapezoid i s p a r a l l e l t o t he bases and equal i n l eng th t o ha l f the sum of t he lengths of the bases.

(b) i f AB = 9 and DC = 7 , then PQ =

and PQ = 7 , then AB =

30. I n surveying the f i e l d ABCD

shown here, a surveyor l a i d - off north-and-south l i n e N S

through B and then located 4--b the east-and-west l i n e s CE ,

M %?, AG . He found that CE = 5 (rods) , DF = 1 2 (rods) , AG = 10 (rods) , BG = 6 (rods), EW = 9 (rods) , FE = 4 (rods) .

Find the area of the f i e l d .

We are t o prove t h a t the area of square AKHB i s equal t o the sum of the areas of square ACDE and square BFGC . Hint: Consider , BE , - 7% such t h a t CM I I . Study AKAC and ABAE . (proof from Euclid. )

D 31. By hypothesis we have a

r ight t r i ang le ABC with the r igh t angle at C . E

F

/ /

I / I

/ /

I / I

/ /

I /

I / I

K M H

32. A proof of the Pythagorean Theorem by the following method i s a t t r i b u t e d t o President Garfield. Let AABC have r i g h t angle a t C . On the ray opposite * t o AC , l e t T be a point such t h a t U b T AT = a . Let U be a point on the - same s ide of AC as B such t h a t - 0 T U l m and TU = b . Find AU , express the area of the trapezoid BCTU i n two ways, A

2 2 2 and deduce t h a t a + b = c . b

0 c

33. I n a cube with A as one vertex, a t r iangular pyramid i s

formed by joining t o A and t o each other the midpoints of the three edges which meet a t A . Find the t o t a l area of the pyramid i f each s ide of the cube i s 12 .

34. A regular r ight hexagonal prism 10 u n i t s high has a l a t e r a l a rea of 480 . Find the apothem of the base and the t o t a l a rea of the prism.

Chapter 12

CIRCLES AND SPHERES

12-1. Basic Definit ions.

I n t h i s chapter we begin the study of s e t s of poin ts not made up of planes, halfplanes, l ines , rays and segments. The

simplest such curved f igures a r e the c i r c l e and the sphere and portions of these. We begin with some def in i t ions .

DEFINITIONS. The s e t of a l l points i n a plane whose distances from a given point i n the plane a r e a given number i s ca l l ed a c i r c l e .

The given point i s ca l led the center of the c i r c l e .

The given number i s ca l l ed the radius of the c i r c l e .

Y

Circle

I f we choose a two-dimensional coordinate system i n t h e plane whose o r ig in i s the center P and i f Q i s any point of the c i r c l e , then PQ = r . Using the dis tance formula of Theorem 8-4, w e can wri te

2 2 2 1 / ( x - o ) * + ( y - 0 l 2 = r o r x + y = r . 2 2 Therefore the c i r c l e i s ((x,y): x2 + y = r ) .

DEFINITIONS. The s e t of a l l points i n space whose dis tances from a given point a r e a given number i s ca l l ed a sphere.

The given

The given

point i s ca l led the center of the sphere.

number Is ca l l ed the radius of the sphere.

Sphere

If w e choose a three-dimensional coordinate system whose o r ig in i s a t the center , P , and Q i s any point of the sphere, then PQ = r . Using the three-dimensional distance formula, we can wr i te

2 2 2 Therefore the sphere i s [(x,y,z): x + y2 + z = r ) . DEFINITION. Two o r more spheres o r c i r c l e s with the same center a r e ca l l ed concentric.

THEOREM 12-1. The In tersec t ion of a sphere with a plane through i t s center i s a c i r c l e whose center and radius a re the same a s those of the sphere.

Proof: We choose a three-dimensional coordinate system with the center of the sphere P a s o r ig in (0,0,0) and the

given plane a s the xy-plane ( i n which every point has zero a s i t s z-coordinate). If r i s the radius of the sphere, the in tersec t ion of the sphere and the plane i s

2 2 2 ((x,y,z) : x + y 2 + z = r and z = 01 . We recognize t h i s t o be the s e t of points i n the xy-plane given by

2 2 2 f (x ,y ) : x + y = r 1 . This i s a c i r c l e whose center and radius a re the same as those of the sphere.

DEFINITION. The c i r c l e of in te r sec t ion of a sphere with a plane through the center i s ca l led a grea t c i r c l e of the sphere.

There a re two types of segments t h a t a re associated with spheres and c i r c l e s .

DEFINITIONS. A chord of a c i r c l e o r a sphere i s a segment whose endpoints a r e points of the c i r c l e o r the sphere. The l i n e containing a chord i s a secant.

A diameter i s a chord containing the center .

A radius i s a segment one of whose endpoints i s the center and the other one a point of the c i r c l e o r the sphere.

The l a t t e r endpoint i s ca l led the outer end of the -- radius.

The p lu ra l of radius i s r a d i i .

Notice that the single word "radius" is used to mean two different objects~a certain segment and also the length of that segment. This should not be confusing because once we know that the word has two meanings we can easily decide which one is intended wherever the word occurs.

The word "diameter" also has two meanings. In addition to meaning a certain kind of chord it also is used to mean the length of such a chord.

DEFINITIONS. Circles with congruent radii are

called congruent circles.

Spheres with congruent radii are called congruent sphere s .

A direct outcome of these definitions are the following two theorems.

THEOREM 12-2. The radii of a circle or congruent circles, or - of a sphere or congruent spheres, are congruent.

THEOREM 12-3. The diameters of a circle or congruent circles, or of a sphere or congruent spheres, are congruent.

It should be clear that the radii and diameters referred

to in these theorems are segments, not numbers.

Problem Set 12-1 --

1. F i l l i n the blanks wi th a word o r words which w i l l best name or describe the indicated pa r t s . Assume t h a t points a re where they appear t o be.

A

Diagram (a) Diagram ( b )

( a ) Refer t o Diagram (a) . Point 0 i s the center of the c i r c l e .

(1) Any point X on any subset of the c i r c l e i s a distance from 0 .

( 2 ) OS is a t h e

( 6 ) Points i n the diagram which are In (on) the given c i r c l e a re .

( 7 ) Points i n the diagram which a re not i n the c i r c l e a re .

(8) S i s the of the radius (9 ) Each point i n the c i r c l e i s the of one

and only one radius. (10) Any point X on the subset of the c i r c l e

between R and M would i n every possible posit ion be a distance from 0 .

(b) Refer t o Diagram (b). Point 0 i s the center of the sphere.

75K i s a Points a r e outer endpoints of given r a d i i . If 0, A, and B a r e co l l inear , then i s a If 0 l i e s i n plane f l , then the c i r c l e wi th

center 0 and radius i n & , i s a of the sphere. -

3M i s a . Every point on the Is the outer end

radius. A l l polnts In which l i e a t a distance from 0 l i e i n wi th center and radius . How many planes may contain any given point such as 0 ? How many great c i r c l e s a re there on any given sphere? A l l great c i r c l e s on a given sphere a re t o each other. I n order t o specify a unique sphere, we must be given . With a given point a s center, it Is possible t o consider (how many) spheres? A l l these spheres a r e ca l l ed spheres.

2. Te l l whether the following statements a re t r u e o r f a l se .

(a) There i s exact ly one great c i r c l e of a sphere.

(b) Every chord of a c i r c l e contains two points of the c i r c l e .

( c ) A radius of a c i r c l e i s a chord of the c i r c l e .

(d) The center of a c i r c l e b isec ts only one of the chords of the c i r c l e .

(e ) A secant of a c i r c l e may i n t e r s e c t the c i r c l e i n only one point .

(f) A l l r a d i i of a sphere are congruent.

(g) A chord of a sphere may be longer than a radius of the sphere.

(h) I f a sphere and a c i r c l e have the same center and i f they in te r sec t , then the in te r sec t ion i s a c i r c l e .

3 . Tel l whether the following statements a re t r u e o r f a l s e .

(a) If a l i n e in te r sec t s a c i r c l e i n one point, it i n t e r s e c t s the c i r c l e i n two points.

(b) The in tersec t ion of a l i n e and a c i r c l e may be empty.

( c ) I n the plane of a c i r c l e , a l i n e which passes through the center of the c i r c l e has two points i n common w i t h the c i r c l e .

(d) A c i r c l e and a l i n e may have three points i n common.

( e ) If a plane i n t e r s e c t s a sphere i n a t l e a s t two points, the in tersec t ion i s a l i n e .

( f ) A plane cannot in te r sec t a sphere i n one point .

(g) If two c i r c l e s in te r sec t , t h e i r in t e r sec t ion i s two points.

(h) The radius of a c i r c l e i s a subset of the c i r c l e .

4 . Consider an xy-coordinate system and yz-coordinate system a s indicated i n the diagrams below.

t Figure ( a ) Figure (b)

(a) Refer to Figure (a).

(1) Express the distance between C and P in

terms of x and y . (2) Write an equation of the circle with center at

P and radius 5 . (3) Write the coordinates of 4 points which you

know lie on the circle in Part (2). (4) Find the distance between B and P . (5) Write an equation of a circle which has P as

a center and contains B . (6) Write an equation of the circle which has P

as its center and contains A . (b) Use Figure (b).

(1) What is the radius of the circle which has P as center and contains B ?

(2) Write an equation of the given circle.

(3) Find the distance RP . How can you tell without a diagram that R is not on the circle?

(4) Write an equation of a circle with center P and which contains A .

5. B is a point on the circle with center A and radius 3 . the xz-coordinate system as indicated.

Express the distance

between B and A . Write an equation of the circle which has

A as center and radius 3 . Write an equation of the circle which has

point (-2,0) as center and which contains the origin.

Write an equation of

a circle in the xz-plane with center at point (h,k)

and with r as its radius.

2 6. Given c i r c l e C = ((x,y) : x + y2 = 25) . Check whether o r not the following points a r e points of C . ( a ) (0.-5) (4 (rn,-A/?) (b) ( -3 .4 ) (a) (12,131

7. Write an equation of the sphere w i t h center a t point (0,0,0) and radius = 3 .

8. Given sphere S = ((x,y,z): x2 + y2 + z2 = 169) . Check whether o r not the following points are points of S . ( a ) (0,13,0) (4 (1,1,0) - ( b ) (-3,4,-12) . ( f ) (dim- 4/^,3)

(4 (4 ,4293) (g) (-4/1^,2,- i /59)

( 4 (o,o,o) 9. Find 5 more points of S i n Problem 8.

10. Using the set notation, wri te an expression f o r the points of the c i r c l e whose center i s (0,0) and whose radius i s :

(a ) What r e s t r i c t i o n on x and y would give only the portion of the c i r c l e i n Quadrant I ?

(b) What portion of the c i r c l e would you be considering under the r e s t r i c t i o n , x > 0 ?

(c) What r e s t r i c t i o n on x and y would give the in tersec t ion of C and Quadrant III?

2 12. Given C = ((x,y): x + y 2 = 9 ) . ( a ) Find x if (x,2) i s a point of C . (b) Find y if (3,y) i s a point of C . ( c ) Can you f ind y so t h a t (4,y) Is a point of C ?

Explain.

2 13. Given S = [(x,y,z): x + y2 + z2 = 25) . (a) Find z if (3,0,z) is a point of S , (b) Find y if (-4,y,3) is a point of S . (c) Find x if (x,0,0) is apoint of S . (d) Can you find z such that (3,5,z) is a point of

S ? Explain.

14. Prove: A diameter of a circle is its longest chord. -

15. Given: AB and CD are diameters.

Prove: AC 2 BD . B

A

16. Prove: If and CD are distinct diameters of a circle, then ACBD Is a rectangle.

17. Prove that the midray of the angle formed by two radii - of a circle, PA and , lies in the perpendicular bisector of ,

18. Consider a sphere whose center is at 0(2,3,-1) . Let

Q(x,y,z) be a point of the sphere. What is the distance OQ , by the distance formula? Is a radius of the sphere? Write an equation of the sphere which has (2,3-1) as center and which contains Q , if OQ = 5 . (~liminate radicals by squaring both members of the

equation. )

12-2. Tangent Lines.

Anyone who looks a t a drawing of a c i r c l e sees t h a t it divides the plane i n t o two regions, one consis t ing of t h e points Inside the c i r c l e and the o ther of the outside points . We now define these terms formally.

DEFINITIONS. The i n t e r i o r of a c i r c l e i s the s e t of a l l points i n the plane of the c i r c l e whose dis tances from the center a r e l e s s than the radius.

The ex te r io r of the c i r c l e i s the s e t of a l l points I n the plane of the c i r c l e whose dis tances from the center a r e g rea te r than the radius.

From these de f in i t ions It follows t h a t a point i n the plane of a c i r c l e i s e i t h e r i n the i n t e r i o r of the c i r c l e , on the c i r c l e , o r i n the ex te r io r of the c i r c l e . (we frequent ly use the more common word "inside" f o r " i n t h e i n t e r i o r of,' ' e t c . ) In terms of an xy-coordinate system whose o r ig in i s the center of a given c i r c l e of radius r , the i n t e r i o r of the c i r c l e i s

2 ( ( x , ~ ) : x2 + y < r21

and i t s ex te r io r i s

Q i s an i n t e r i o r point of C . Q i s an exter ior point

of C .

Problem Set 12-2a -- (Exploratory)

2 1. Given C = ((x,y): x + y 2 = 1 6 ) and M = ((x,y): x - a )

Find the s e t of points in the in tersec t ion of C and M , If a = 3 ; i f a = 4 ; i f a = 5 .

2. Using the r e s u l t s you found i n Problem 1, complete the following.

(a) The in tersec t ion of C and M contains ?

point (s ) i f a < 4 . ( b ) The in tersec t ion of C and M contains ?

po in t ( s ) i f a = 4 . ( c ) The in tersec t ion of C and M contains ?

p o i n t ( s ) i f a > 4 . 3 . What three r e l a t ions between a c i r c l e and a l i n e i n the

plane of the c i r c l e are suggested by Problems 1 and 2?

If a stone i s twir led on the end of a s t r i n g i n a c i r cu la r path and then l e t go, it w i l l " f l y off on a tangent." Try t o see how t h i s use of the word tangent i s re la ted t o the one w e now give.

DEFINITIONS. A tangent t o a c i r c l e i s a l i n e i n the plane of the c i r c l e which i n t e r s e c t s the c i r c l e i n only one point.

This point i s ca l led the -- point of tangency, o r point

of contact, and we say t h a t the l i n e and the c i r c l e - a r e tangent a t t h i s point. -

I n the f igure , 1 i s tangent t o the c i r c l e a t Q ,

Lines and c i r c l e s a r e important subsets of planes. Let u s consider a s ingle plane and study the r e l a t ions of l i n e s and c i r c l e s t o one another. It looks a s i f the following three f igures indicate a complete catalog of the p o s s i b i l i t i e s :

I n each case, P i s the center of the c i r c l e , and F i s the -- foot of the perpendicular from P t o the l i n e . We s h a l l soon --- - - --- see t h a t t h i s point F , t he foot of the perpendicular, i s the key t o the whole s i tua t ion . If F i s outside the c i r c l e , a s i n the first f igure , then a l l other points of the l i n e a re a l s o outside, and the l i n e and the c i r c l e do not in te r sec t a t a l l . If F i s on the c i r c l e , then the l i n e Is a tangent l i n e , a s i n the second f igure, and the point of tangency i s F . If F i s inside the c i r c l e , as i n the t h i r d f igure, then the l i n e i s a secant, and the points of in tersec t ion a r e equidis tant from the point F . To ver i fy these statements, we need t o prove the following theorem:

THEOREM 12-4. Given a l i n e J^ and a c i r c l e C i n the same plane. L e t P be the center of the c i r c l e , and l e t F be the foot of the perpendicular from P t o t h e l i n e .

(1) Every point of i s outs ide C i f and only i f F

i s outside C . (2) 1 i s a tangent t o 0 If and only i f F i s on C . ( 3 ) & i s a secant of C i f and only i f F Is ins ide C .

Proof: Let r be the radius of C and l e t PF = a . We introduce the xy-coordinate system w i t h o r ig in a t P

whose y-axis i s p a r a l l e l t o A? and whose pos i t ive x-axis 2 2 2 contains F . Then F = (a,0) , C = ((x,y): x + y = r ) and

A = ((x,y): x = a1 o r ( ( a ,y ) 1 .

(1) I? Is outside C . (2) F i s on C . ( 3 ) F i s insideC.

(1) Suppose F i s outside C , then a > r . Since a 2 and r a r e pos i t ive numbers, it follows t h a t a > r2 and

2 a + y > r2 . Therefore a l l points (a,y) a re outside C . Since =((a, y)) , then a l l points of /^" a re outside C .

O f course, if every point of -i i s outside C , then F i s a l s o outside C . T h i s proves both pa r t s of (1) .

(2) Suppose F i s on C . Then r = a , and the in te r - 2 2 s e c t i o n o f JC' and C i s ((x,y): x2 + y = a and x = a ) ,

o r [(a ,y) : y2 = 01 . But there i s exact ly one number whose square i s zero,

namely zero. Therefore the only point of In tersec t ion of ,d and C i s I?(a,0) . Therefore A i s a tangent t o C .

If Â£ i s a tangent, it can have only one point i n common w i t h C . That point i s shown t o be F(a,0) . Thus, both p a r t s of (2) a r e proved.

12-2

( 3 ) Suppose F i s inside C , then r > a . The i n t e r - 2 2 2 sect ion of 1 and C i s f (x ,y) : x + y = r , x = a and

2 2 2 2 2 a < r ] o r [(a,y): y = r - a , P >/a; . Zince r d-2a is a posi t ive number, y can be e i t h e r r - a o r - r - a . Therefore the in te r sec t ion cons is t s of (a, dr-) and

(a, - r ) . These a r e d i s t i n c t points. Why? Therefore J!. is a secant.

If ,d i s a secant, i t i n t e r s e c t s C i n two d i s t i n c t points, which w e have shown t o be (a, 4-1 and

2 (a , - dr-1 . This implies t h a t r - a2 > 0

2 2 (Why

can't r - a = 0 ? ) . And because r and a a re posi t ive, r > a . But PF = a ; therefore PF < r , o r F Is ins ide C . This completes the proof of the theorem.

The following t ab le displays some of the f a c t s about F

t ha t we met i n our proof.

Case 1

a > r 2 a t y 2 > r 2

no y f o r which

2 2 2 a + y = r .

No point of JL l i e s on

c .

Case 2

a = r

i f and only

i f y = o .

Only F i s on C .

Case 3

a < r

2 2 i f and only i f y = r - a

,d and C have exact ly two points i n common. They a re

Now we can proceed t o our f i r s t basic theorems on tangents and chords which a re a l l co ro l l a r i e s of Theorem 12-4. To prove them, you merely need t o r e f e r t o Theorem 12-4 and see which of the three cases of the theorem appl ies .

Corollary 12-4-1. Given a c i r c l e and a coplanar l i n e , the l i n e i s a tangent t o the c i r c l e i f and only i f it i s perpen- d icular t o a radius of the c i r c l e at Its outer end.

Corollary 12-4-2. A diameter of a c i r c l e b isec ts a non- diameter chord of the c i r c l e i f and only i f it i s perpendicular t o the chord.

Corollary 12-4-3. I n the plane of a c i r c l e , the perpen- d icu la r b isec tor of a chord contains the center of the c i r c l e .

Hint f o r proof: Use Corollary 12-4-2. ---

Corollary 12-4-4. If a l i n e i n the plane of a c i r c l e i n t e r s e c t s the i n t e r i o r of the c i r c l e , then it in te r sec t s the c i r c l e I n exact ly two points.

Case (3) appl ies . [ In Case (1) and (2), the l i n e does not in te r sec t the i n t e r i o r of the c i r c l e . ]

THEOREM - 12-5. Chords of congruent c i r c l e s a re congruent i f and

only i f they a r e equidis tant from the centers .

Proof: Let P and P1 be the c e n t e r s o f t h e congruent c i r c l e s , l e t and A'B' be the chords, let F be the foot of the perpen-

4--b dicular from P on AB and l e t Ft be the foot of the Ã‘Ã‘

perpendicular from P 1 on A I B q . Then by Corollary 12-4-2, 1 1 we have FB = AB and FIB' = -AIB1 .

By the Pythagorean Theorem

By hypothesis PB = P*Bl so

It follows t h a t PF = P 'F f i f and only i f AB = AIBt . DEFINITIONS. Two c i r c l e s a r e tangent i f and only i f they a r e coplanar and tangent t o the same l i n e a t the same point. Tangent c i r c l e s a r e in te rna l ly o r external ly tangent accordingly a s t h e i r centers l i e on the same side o r on opposite s ides of the common tangent l ine .

In terna l ly tangent Externally tangent

Problem Set 12-2b -- 1. Given: C = [ (x,y): x2 + y2 = 36) . Tel l whether each of

the following points i s i n the i n t e r i o r , on, o r i n the exter ior of C .

2. Given: (3.5) i s on the c i r c l e whose center i s ( 0 ~ 0 ) . ( a ) Find the radius of t h i s c i r c l e .

(b) Find four points on the c i r c l e .

( c ) Find two points i n the i n t e r i o r of the c i r c l e .

( d ) Find two points i n the ex te r io r of the c i r c l e .

3 . Sta te the number of the theorem o r corol la ry which J u s t i f i e s each conclusion below. P i s the center of a c i r c l e . I?, H, B, K, A a re points on the c i r c l e and S, T, and R a re coplanar with the c i r c l e .

If TA = TB , then 1 . 4--P -

If RK 1 , then RK i s tangent t o the c i r c l e . If T i s i n the i n t e r i o r of the c i r c l e , then- w i l l i n t e r s e c t the c i r c l e i n exact ly one point other than point K . The perpendicular bisector of contains P . If and a re equi- d i s t an t from P , then m z m . - If *RS* i s tangent t o the c i r c l e , then _[ RS . If , then AT = TB . - - - If AB = FH , then and a re equidistant from P .

4. I n a c i r c l e with radius of 5 un i t s , how long i s a chord 3 u n i t s from the center of the c i r c l e ?

5. If a chord 4 Inches long i s 1 . 5 inches from the center of a c i r c l e , what i s the radius of the c i r c l e ?

6. How far from the center of a c i r c l e w i t h radius equal t o 12 i s a chord whose length i s 8 ?

7. Chord i s p a r a l l e l t o 1 which i s tangent t o the c i r c l e at Q . P i s the center of - the c i r c l e . AB b i sec t s a t R . AB = 12 . Find PQ .

8. Given: The figure below, with C the center of the circle

and 1 . In the ten problems respond as follows:

Write "A" if more numerical information is given than is needed to solve the problem.

Write "B" if there is insufficient information to

solve the problem.

Write "c" if the information is sufficient and

there Is no unnecessary information.

Write "D" if the information given is contra- dic tory.

(YOU do not need to do the computations.)

(a) K P = 4 , P C = 1 , C T = 6 , K T = ? (b) RP = 5 , RS = ?

(c) CT = 13 , CP = 5 , RS = ?

(d) KP = 18 , RS = 48 , KC = 25 ,

(i) R S = 6 , K C = 5 , P T = ?

( J ) P T = 5 , C S = 6 , R S = ?

9. In a circle whose diameter is 30 inches a chord is perpendicular to a radius at a point on the radius

3 inches from its outer end. Find the length of the chord.

10. Prove that the tangents to a circle at the ends of a diameter are parallel.

11. Write Corollary 12-4-2 as two statements, each the converse of the other. Prove each.

12. For the concentric c i r c l e s of t h e f igure, prove t h a t a l l chords of the l a rge r c i r c l e which a re tangent t o the smaller c i r c l e a r e bisected at the point of contact.

13. One arrangement of three c i r c l e s so t h a t any one i s tangent t o each of the o ther two Is shown here. Make sketches t o show three o ther arrangements of th ree c i r c l e s with each c i r c l e tangent t o each of t h e other two.

14 . Prove: The l i n e of centers of two tangent c i r c l e s contains t h e point of tangency. ( ~ i n t : - Draw the common tangent. )

Case I Case I1

I n the f igure, A, B

and C a re the centers of the c i r c l e s . A B = 1 4 , = = l o ,

AC = 18 . Find the radius of each c i r c l e .

16. Prove: The midpoints of a l l chords congruent t o a given chord i n a given c i r c l e l i e on a c i r c l e concentric with the or ig ina l c i r c l e and with a radius equal t o the d is -

tance of a chord from the center; and the chords are a l l tangent to t h i s inner c i r c l e .

17. ( a ) The distance from P , t he center of a c i r c l e , t o T , an ex te r io r point, i s 20 . A tangent from T

t o the c i r c l e has a point of contact A . If the radius of the c i r c l e i s 12 , f ind AT .

( b ) A second tangent from T has B a s a point of contact. Find AB .

18. In the c i r c l e wi th center D -

at 0 , AB i s a diameter and AC i s any other chord - from A . If CD i s the -- tangent at C ,and DO 1 1 A C ,

prove t h a t %? i s tangent a t B .

2 19. Consider the c i r c l e C = ((x,y): x + y2 = 100) . ( a ) I f l i n e /& , [ (x, y): x = a ) , i s tangent t o c i r c l e

C , f ind the values f o r a . ( b ) Find an equation f o r a l i n e t tangent t o c i r c l e

(Flints: What i s the slope of the radius t o t ?

What must be the slope of t ?

Must t contain (5 'i/2, 5 &) ? )

2 20. Consider the s e t P = ((x,y): (x - 1) + (y + 2)2 = 25) . (a) Can you i n t e r p r e t the equation as specifying t h a t

the dis tance between (x,y) and (1,-2) i s 5 ?

(b) Is the s e t a c i r c l e ? If so, what a re the coordinates of i t s center and the length of i t s radius?

2 (c) Given P - [(x,y): ( x 2 - 2 x + l ) + ( y + 4 y + 4 ) = 25)

2 2 Show t h a t P = ( ( x , y ) : x + y - 2x + 4y = 20). If you were confronted with t h i s l a s t equation, could you complete the squares t o reproduce the or ig ina l? Demonstrate t h i s process.

(d) Find an equation of a l i n e t which i s tangent t o the c i r c l e P a t the point (5, l ) . (Hint: - Find the slope of the radius t o ( 5 , l ) . Use i t s negative reciprocal a s the slope of t . Tangent t must contain ( 5 , l ) .)

2 21. Consider the c i r c l e C = ((x,y): x + y2 = 1) .

( a ) Write an equation of the l i n e tangent to C which contains the point (-1,0) .

(b) Write an equation of the tangent l i n e to C which contains the point T(- 1 T, - 2) . ( ~ i n t : Is

I/? - the tangent l i n e perpendicular to the radius of C

which contains T ?)

( c ) Find t h e coordinates of the point P on the x-axis which contains the tangent t o T determined i n (b) above.

(d) Find the distance PT .

*22. Consider the sphere S

and the plane ^? such t h a t

2 S = ((x,y,z): x + y2 + z2 = 25) % = ((x,y,z): z = a } .

Y How i s re la t ed t o the xy-plane?

How i s re la ted t o the z-axis? I

f l i n t e r sec t s the z-axis a t a point, say F , with coordinates (0,0,a) . Consider the in te r sec t ion of S

and ^2 , when a has the values indicated below.

(a) Assume a = 4 . What geometric f igure I s t h i s in tersec t ion?

(b) Assume a = 5 . How many points a r e i n t h i s in tersec t ion?

(c ) Assume a = 7 . How many points a re i n t h i s in tersec t ion?

(d) What appears t o be the r e l a t i o n between the i n t e r - sect ion of S and 7 and the dis tance PF ?

12-3. Tangent Planes.

We have Jus t studied c i r c l e s and l i n e s i n a plane. We a re now going t o study spheres and planes i n space. We s h a l l see tha t many of the def in i t ions and theorems of the l a s t sect ion resemble the def in i t ions and theorems about spheres and planes.

DEFINITIONS. The i n t e r i o r of a sphere i s the s e t of of a l l points whose dis tances from the center a re l e s s than the radius.

The exter ior of the sphere i s the s e t of a l l points whose distances from the center a re g rea te r than the radius.

In terms of a coordinate system whose o r ig in is the center of a given sphere of radius r , the i n t e r i o r of the sphere i s

and i t s ex te r io r i s

2 ~x,y,z): x + y2 + z2 > r 2 ) ,

Az /

SPHERE o Q ( x l

Â ¥

Q i s an i n t e r i o r point Q Is an exter ior point 2

x + y 2 + z 2 < r 2 X 2 + Y 2 + z 2 > r 2

DEFINITIONS. A plane t h a t i n t e r s e c t s a sphere i n exact ly one point i s ca l led a tangent plane t o the sphere.

I f the tangent plane in te r sec t s the sphere i n the point Q then we say t h a t the plane i s tangent t o - t he sphere at Q . - - - Q is cal led the -- point of tangency, o r the point of contact.

When we Investigated circle-line relations in Theorem 12-4 we found that the key in each phase of the study was the foot of the perpendicular from the center to the line. Our sphere- plane study also has a key. It is the foot of the perpendicular from the center of the sphere to the plane.

The basic theorem relating spheres and planes is the following:

THEOREM 12-6. Given a plane ^Z and a sphere S with center - P . Let F be the foot of the perpendicular from P to^? . 1. Every point of is outside S if and only If F

is outside S . 2. Â¥ Is tangent to S If and only if F is on S . 3. 7? Intersects S in a circle with center F if

and only if F is inside S . Proof: Let r be the radius of S and let PF = a . We introduce an xyz-coordinate system with origin at P ,

whose xy-plane is parallel to 7 1 and whose positive z-axis 2 2 contains F. Then F = (0,0,a), C - [ (x ,y ,z ) i x + y +z2-r2)

and rn, = ( (x ,y ,z) : z = a ) or {x,y,a)j.

* - - - - - -Ã

*Y

SPHERE

(1) F is outside S . (2) F Is on S . (3) F Is inside S .

843

(1) Suppose F i s outside S , then a > r . It follows, 2 since a and r a r e pos i t ive numbers, t h a t a > r2 and

2 2 x + y2 + a > r2 . This t e l l s us tha t (x,y,a) I s outside S . But = [(x,y,a)}. T h e r e f o r e m i s outside S .

O f course, i f every point of 777 i s outside S , then F

i s a l so outside S . This proves both p a r t s of (1).

(2 ) Suppose F i s on S . Then r = a and the Inter- sect ion of /^ and S i s

((x,y,z): x2 + y2 + z2 = a2 and z = a ) or

But there Is only one p a i r of numbers, (x,y) , namely (0,0) , 2 2 sucl-i t h a t x + y = 0 . Therefore ^- and S have only

~ ( 0 , 0 , a ) I n common and i t follows t h a t I s tangent t o S . If /^ i s tangent t o S , they have only one point i n

common and it has been shown that (0,0,a) i s t h a t point. Thus both p a r t s of (2) a r e proved.

( 3 ) Suppose F i s inside C , then r > a . The in te r sec t ion of -'%Â and S i s

Because r2 - a > 0 , we can see from the form of the equation 2 2 2 x + y = r - a , t h a t we have a c i r c l e i n the plane z = a ,

with center (0,0,a) and radius &- . On the o ther hand, i f ^% i n t e r s e c t s S It follows tha t

2 2 2 x + y = r - a has a solution. This implies t h a t r2 > a o r r > a . Therefore ~ ( 0 , 0 , a ) i s i n S .

This completes the proof of the theorem.

Corollary 12-6-1. A plane i s tangent t o a sphere i f and only i f it i s perpendicular t o a radius a t i t s outer endpoint.

Corollary 12-6-2. A perpendicular from the center of a sphere t o a chord of the sphere b i sec t s the chord.

Corollary 12-6-3. The segment Joining the center of a sphere t o the midpoint of a chord i s perpendicular t o the

chord. 844

Problem Set 12-3 -- 1. The sphere with center 0

is tangent to plane < at A . FB and RT are lines of & through A . What I s the relationship * - - of OA to FB and RT ?

2. In a sphere having radius 10 , a segment from the center to the midpoint of a chord has length 6 . How long is

the chord?

3. A sphere has radius 5 . A plane 3 units from the center Intersects the sphere in a circle. What is the

radius of this circle?

4. Prove that circles on a sphere in planes equidistant from the center of the sphere are congruent.

5 . State Corollary 12-6-1 as two statements which are

converses of each other. Prove each.

6. Show that two great circles of a sphere intersect at the

endpoints of a diameter of the sphere.

7. Consider the sphere S = [ (x,y,z): x2 + y2 + z2 = 9) . (a) What is the center of S ? What is the radius of S ?

(b) Write an equation of a plane tangent to S and parallel to the xz-plane. How many such planes are there?

(c) Write equations of all planes tangent to S and parallel to the yz-plane.

8. Consider the se ts :

( a ) Describe the in tersec t ion of S and M ; of S and

N ; of S and R . ( b ) Describe the s e t T.

( c ) What Is the in tersec t ion of S and T ?

9. Two great c i r c l e s a r e sa id t o be perpendicular i f they l i e i n perpendicular planes. Show t h a t , given any two great c i r c l e s , there i s one other grea t c i r c l e perpendicular t o both. I f two grea t c i r c l e s on the ear th a re meridians (through the north and south poles) what great c i r c l e i s t h e i r common perpendicular?

10. Plane & i n t e r s e c t s a sphere whose center i s 0 . A and B a re two points of the i n t e r - sect ion. F l i e s i n

4--b plane & . OF _[ <* . mlm. If A B = 5 and OF = AF , f ind the radius of the sphere and m /AOB . If G i s the midpoint of , f ind OG .

11. Given a sphere and three points on it. Describe the steps you would take t o f ind the center and the radius of the sphere.

12 . Plane $Â i s tangent t o a sphere X a t point T , and plane 3 i s any plane other than & which contains T . Prove :

( a ) t h a t plane - i n t e r s e c t s sphere S and plane & i n a c i r c l e and a l i n e respectively;

(b) that the line of intersection is tangent to the circle

of intersection.

13. Consider the sphere S = ((x,y,z): x2 + y2 + z2 = 100) . (a) Find the intersection of sphere S with the plane

((x,y,z): z = 10) . (b) Consider the plane P = ((x,y,z): z = 8) . In order

for a point (x,y,z) to be in the intersection of

S and P , certainly z = 8 ; what conditions, then, must x and y meet?

2 14. Consider the circle C = ( (x,y): x + y2 - 4x + 6y = 23) . (a) Complete squares and transform the equation of C to

2 the form (x - h) + (y - k)2 = r2 . What are the coordinates of the center and the length of the radius of C ?

(b) Write an equation of the sphere S whose radius has the same length as the radius of C and whose center

is at (2,-3,0) . (c) Write an equation describing a plane tangent to sphere

S and perpendicular to the z-axis. (Two answers are

possible. )

12-4. Arcs of Circles. -- So far in this chapter we have been able to treat circles

and spheres In a similar manner. For the rest of this chapter we confine ourselves exclusively to circles. The topics we discuss have their corresponding analogies in the theory of spheres but these are too complicated to consider in a beginning course.

DEFINITION. A central angle of a given circle is an angle whose vertex is the center of the circle.

DEFINITIONS. If A and B a re two points of a c i r c l e w i t h center P and i f A and 3 a r e not the endpoints

of a diameter of t h a t c i r c l e , then the union of A, B, and a l l the points of the c i r c l e i n the i n t e r i o r of /APB i s a minor a rc of the c i r c l e . -- The union of A, B, and a l l points of the c i r c l e i n the ex te r io r of /APB i s a major a r c of the c i r c l e . --

major

I f i s a diameter, the union of A, B, and a l l points of the c i r c l e i n one of the two halfplanes,

4--w with edge AB , lying i n the plane of the c i r c l e i s a semicircle.

An a rc i s e i t h e r a minor a rc , a major a rc , o r a semi-circle. A and B a re the endpoints of the

a rc .

I n some ways an arc of a c i r c l e i s l i k e a segment of a l ine ; f o r instance, It has two endpoints. However, unlike the segment, an a rc i s not determined by i t s endpoints. I n f a c t there a re i n f i n i t e l y many a rcs which have any given pa i r of points as t h e i r endpoints, as the f igure suggests. This makes

it hard t o f ind a symbol t o denote an arc , b u i l t up from t h e

symbols f o r i t s endpoints. I n s p i t e of t h i s we of ten denote

- an a r c whose endpoints a r e A and B by AB . We must be sure we know what c i r c l e we have i n mind f o r t h i s t o make sense, and a l s o we must know which of the two a r c s on t h a t c i r c l e we have i n mind. Sometimes i t w i l l be p l a i n from t h e context which

arc i s meant. If not , we w i l l p ick another point X somewhere - - i n t he a r c AB , and denote t he a r c by AXB . For example, i n - t h e f i g u r e , AXB i s a minor a r c ; AYB i s t h e assoc ia ted major - a rc ; and t h e a r c s CAB and CYB a r e semic i rc les .

The reason f o r t h e names "minor" and "major" i s apparent when one draws severa l a r c s of each kind. I n such drawings t h e major a r c looks "bigger" than a minor a r c . This r e l a t i o n w i l l

be made more prec i se i n our next d e f i n i t i o n .

DEFINITION. If AXB i s any a r c then i t s degree - measure, ~ X B , i s given a s fol lows:

1. I f AXB i s a minor a r c , then ~AXB i s t h e measure of t h e assoc ia ted c e n t r a l angle .

2. If A= i s a semic i rc le , then ~AXB = 180 . 3 . I f AXB i s a major a r c , and A% Is the cor re -

sponding minor a r c , then

I n t h e f i gu re , m /APB i s 60 . Therefore & i s 60 , - and mAXB i s 300 .

Hereaf ter , rn~z w i l l be c a l l e d simply the measure of the a r c A% . Note t h a t an a r c i s minor o r major according a s i t s measure i s l e s s than o r g r e a t e r than 180 . -

If X i s a point of an a r c AB d i f f e r e n t from A and 3 ,

it determines, w i t h A and B , two o the r a r c s , AX and 5 . It i s na tu ra l t o inqui re how t h e measures of such a r c s , AX and - XB , a r e r e l a t e d t o the measure of ; t he answer i s simple and reasonable, namely, rnAz = rnz + & . T h i s can a c t u a l l y be proved as a theorem but t he proof i s su rp r i s ing ly long and tedious . We p r e f e r t o s t a t e the r e s u l t as a pos tu la te .

Pos tu la te 30. I f and a r e a r c s of 1 - t he same c i r c l e having only t h e point B i n common, and If t h e i r union i s an a rc AC , then

Notice t h a t f o r t he cases i n which AC i s a minor a rc o r

a semicirc le t he theorem follows from the Pro t rac tor Postulate. It i s the o the r case whose proof i s d i f f i c u l t .

In each of the f i g u r e s below, t he angle x i s said t o be Inscr ibed i n t h e a r c AX .

B

Figure a

DEFINITION. An angle I s inscr ibed i n an a r c i f and only i f (1 ) t he angle contains t h e two endpoints of t he a r c and ( 2 ) t he ver tex of the angle i s a point , but not an endpoint, of t he a r c .

More concisely, /AX Is Inscr ibed i n A= ,

I n Figure a , t he angle i s inscr ibed i n a major a r c , and i n Figure b, the angle i s inscr ibed i n a semicirc le .

I n each of t h e f i g u r e s below, t he angle shown i s s a i d t o i n t e r cep t P% . 4

I n Figure c, the angle i s inscribed; i n Figure d, the vertex i s outside the c i r c l e ; i n Figure e , the angle i s a cent ra l angle; and i n Figure f, one s ide of the angle i s tangent t o the c i r c l e .

I n Figure d, the angle shown in tercepts not only PS but a l so

These f igures give the general idea of an intercepted arc . We w i l l now define what it means t o say t h a t an angle in te r - cepts an arc . You should check carefu l ly t o make sure that the de f in i t ion r e a l l y takes care of a l l four of the above cases.

DEFINITION. An angle in tercepts -- an a rc i f (1) the

endpoints of the a rc l i e i n the angle, ( 2 ) each s ide of the angle contains a t l e a s t one endpoint of the a r c and ( 3 ) except f o r i t s endpoints, the a rc l i e s i n the i n t e r i o r of the angle.

The reason why we t a l k about the a r c s intercepted by

angles i s t h a t under c e r t a i n conditions there i s a simple r e l a t i o n between the measure of the angle and the measure of the arc .

I n the f igure above we see three inscribed angles, /x , /y , / z , a l l of which in tercept FC and a re inscribed i n

EC . It looks as i f these three angles a re congruent. That t h i s i s so i s a corol la ry of the following theorem:

THEOREM 12-7. The measure of an inscribed angle i s half the

measure of i t s Intercepted arc .

Proof:

Given: Circle with center P - A i s an inscribed angle intercept ing BC

1 - TO prove: rn /A = ~ B C .

There a re three possible cases: (1) P i s on a s ide of /A , - say AC , (2 ) P i s an i n t e r i o r point of /A , ( 3 ) P i s an exter ior point of /A .

Case (1). P i s on iE . L e t /x and /y be a s shown. By Theorem 6-10,

- * - m /A + m /x = m /y . By Theorem 12-2, AP = BP and therefore

m /A = m /x . By the subs t i tu t ion property of equal i ty ,

2m /A = m /y . By the def in i t ion of the measure of an arc,

rn /y = . By the mult ipl icat ion property of equal i ty we conclude t h a t

Case (2). P l,s an Interior point of A . 'By the Betweenness-Angles Theorem m /A = m /v + m /w ,

and by the Postulate 30, m ~ z = ms + t&? .

1 - By Case (l), m /v = p E D and m /w = 1- Ã‘D . Therefore

Case (3). P is an exterior point of /A .

From this theorem we get two very important corollaries:

Corollary 12-7-1. An angle Inscribed in a semicircle is

a right angle.

This is so because such an angle intercepts a semicircle,

which has measure 1.80 .

Corollary 12-7-2. Angles inscribed i n the same arc a re

congruent.

This i s so because a l l such angles in tercept the same arc .

We now say what we mean by congruent a rcs . Jus t as we already did f o r segments and angles, we s t a t e our de f in i t ion i n terms of tne appropriate measure.

DEFINITION. In the same c i r c l e , o r In congruent c i r c l e s , two a rcs a re ca l l ed congruent i f they have the same measure.

Corollary 12-7-3. ~ o n g r ~ e n t angles inscribed i n congruent c i r c l e s Intercept congruent arcs .

Problem Set 12-4a -- 1. In the c i r c l e s i n each diagram, P i s the center and

points A, B, C , D, M a re contained i n the c i r c l e s as indicated.

( a ) Refer t o Figure a .

(1) Name the cent ra l angles.

Figure a - AB contains P

DP1m

(2) The measures of a l l cent ra l angles a r e posi t ive numbers l e s s than what number?

(3) Name a l l the minor a rcs i n the c i r c l e .

Figure b

AC contains P

Figure c

rnA% = 95

mDC = 40

( 4 ) Name a l l the major a r c s i n the c i r c l e .

(5) Name the p a i r s of congruent a rcs i n the c i r c l e .

( 6 ) Name two a r c s which do not have cen t ra l angles associated with them.

(b) Refer t o Figure b.

(1) How many inscribed angles a re shown? (2) Name any 6 of these and f o r each name the arc

i n which it i s inscribed and the arc it in tercepts .

(3) For which of the inscribed singles can you give the degree measure?

( c ) Refer t o Figure c.

(1) Name the a r c s intercepted by /A , /B , /C , /D . (2) Give the degree measure of each angle i n Part 1.

(3) Name two p a i r s of congruent angles. Jus t i fy your statements i n two ways. Write the def in i t ion , theorem o r corol lary used.

2. Consider angles formed by secant-rays, tangent-rays, and/or chord-rays. Call the vertex of such angles V . Make diagrams which indica te a l l possible pairing of chords, secants, tangents t o form these angles i f :

( a ) V i s I n the ex te r io r of the c i r c l e . (b) V i s on the c i r c l e . ( c ) V i s i n the i n t e r i o r of the c i r c l e .

3 . The center of an arc i s the center of the c i r c l e of which the arc i s a par t . How would you f i n d the

A center of ?

4. Given: P i s the center of A C , m / ~ - 4 5 .

Prove: BP 1 AP .

12-4

5. I f m ~ % = rnG , ( a ) Prove AAHK - ABHF . ( b ) What other t r i ang le

i n the f igure i s s imilar t o ABHF ?

6 . I n the c i r c l e w i t h center P , l e t m fi = 85 , ms = 40 , mf? = 90 . Find the degree measures of the other a rcs and angles indicated i n the f igure .

v

7. An inscribed quadr i la te ra l i s a quadr i la te ra l having a l l of

i t s ver t ices on a c i r c l e . Prove the theorem: The B

opposite angles of an inscribed quadr i la te ra l a re supplementary.

8. The two c i r c l e s i n t h i s f igure are tangent a t A and the smaller c i r c l e passes through

0 , the center of the l a rge r c i r c l e . Prove t h a t any chord of the la rger c i r c l e wi th endpoint A i s bisected by the smaller

c i r c l e .

9. I n t h e f i gu re , ACB i s a semicirc le and CD 1 . Prove t h a t ( c D ) ~ = AD DB . Refer t o Corol lary 7-7-1.

10. Prove t h e following converse of Corol lary 12-7-1: If an angle inscr ibed i n a c i r c u l a r a r c i s a r igh t angle, then B t h e a r c i s a semicirc le .

D

4 11. A, B, C, D a r e po in t s on a c i r c l e such t h a t AC b i s ec t s

/BAD . Prove t h a t m E = m 6 . 12. Prove: A diameter perpendicular

t o a chord of a c i r c l e b i s e c t s both a r c s determined by t h e B chord.

13. Prove: If a l i n e b i s e c t s both an a r c and i t s chord, it contains t h e cen te r of the c i r c l e .

- * - I n t h e diagram AB = AC . Prove that ~AB = mG .

15. Prove Corollary 12-7-1 by using coordinates. [~int: Let

an xy-coordinate system assign (0,0) to the center of

the circle. Find the slopes of the rays forming the

angle. What must the product of these slopes be?] -

16. XY is the common chord of

two intersecting circles - AB and DC are two

segments cutting the

circles as shown in the figure and con-

taining X and Y respectively .

- Prove: AD 1 1 . (~int: See Problem 7.)

- 17. In this diagram A3 is a

diameter and 1 . (a) Indicate a corre-

spondence which is a similarity between the

triangle with vertices A B

A, C , B and the triangle

with vertices B, D, E . (b) Express the relation

between the corresponding

sides of the two triangles in (a) using the symbols

for a proportionality.

( c ) Prove: BD BC = BA BE .

18. In this figure, SB is a diameter of the smaller of

two concentric circles,

both with center 0 , and - - AC and BD are tangent to - the smaller circle. CO

and DO are radii of the larger circle.

Prove that CD is a diameter of the larger circle.

(Hint: Consider AD and . )

We return to congruent arcs and related chords.

THEOREM 12-8. In the same circle or in congruent circles, if two chords, not diameters, are congruent, then so are the

associated minor arcs.

Proof: Referring to the above figure, we need to show - Q - that if AB = AIB1 , then AB = AIB* . By the S.S.S. Postulate we have

AAPB 2 AA'P'B' . - Therefore /P 2 /pi . Since rnA% = m /P and rnAIBl = m /PI ,

@b - this means that = AtB1 , which was to be proved. The converse is also valid and the proof is very similar.

THEOREM 12-9. I n the same c i r c l e o r i n congruent c i r c l e s , i f

two arcs a re congruent, then so a r e the associated chords. - That is , re fer r ing t o the f igure above, i f A t Bl , -a- then AB = A'B' . If the major a rcs a r e known t o be congruent,

then the same conclusion holds.

We now r e l a t e measure of other types of angles t o measures of intercepted arcs . The f igures below show the types of angles we consider.

Ãˆ

Figure c

Figure b

^ Figure d

In Figure a the one ray i s contained i n a tangent and the other ray contains a chord. Describe the rays i n Figure b,

i n Figure c , i n Figure d.

DEFINITION. I f the vertex of an angle i s on a c i r c l e and one of i t s s ides i s contained i n a tangent, and i t s other s ide contains a chord, then the angle i s cal led a tangent-chord angle.

DEFINITIONS. If the vertex of an angle is an exterior point of a circle and its sides are contained in two

secants, or two tangents, or a secant and a tangent, then it is called respectively a secant-secant angle, or a tangent-tangent angle or a secant-tangent angle.

THEOREM 12-10. The measure of a tangent-chord angle is one-half the measure of its intercepted arc.

Proof: A

Given: Circle with center P . 4--D QS is a tangent at Q . - QR is a chord. 6 is the intercepted arc of

/RQS . 1 - To prove: mFQS = pQR .

We consider three cases. Q

+ Case (1). P is in QR . Case (2). P is an exterior point of /RQS . Case (3). P is an interior point of /RQS .

+ Case (1). P is on QR . Then by Corollary 12-4-1

1 - /RPQ is a right angle. Since = 180 , m ~ R P Q = FQR .

Case (2). P Is an exterior point of /RQS . Consider diameter ^T . By

the Betweenness-Angles Theorem

m /RQS = m /TQS - m /TQR . 1 1 - m /TOS = ~ 1 8 0 , m /TQR = ~ I T R .

1 1 - Therefore m /RW = -^(180 - m z ) = p R S .

Case (3) is left as a problem.

THEOREM 12-11. The measure of an angle whose vertex is in the

interior of a circle and whose sides are contained in two secants, is one-half the sum of the measures of the intercepted arcs.

Proof:

Given: A circle with secants - - AB and CD intersecting at E .

To prove: rn /DEB = $(m% + m~?) ,

The remainder of the proof is left as a problem.

(~int: m /DEB = m LEAD + m LADE) .

THEOREM 12-12. The measure of a secant-secant angle, or a tangent-tangent angle or a secant-tangent angle is one- half the difference between the measures of the intercepted arcs.

The proof of this theorem for a secant-secant angle should

suggest the proofs for the remaining two angles .

Let the secants be as shown in the diagram.

The rest of the proof is easily

completed by noting that 1- m /ABC = ?AC (Why? ) and 1 - m /BCD = p B D .

Problem Set 12-4b -- 1. Consider the points in the following diagrams to be located

as the figures suggest. The degree measures indicated are

assigned to the arcs. R<

Q v

M

R 30 A

90

s 0

Figure (1) Figure (2) Figure (3)

(a) Match the numbers (1) through (5) from the diagrams with the appropriate angle-name selected from

tangent-tangent angle, tangent-secant angle, secant-

secant angle, tangent-chord angle, central angle,

inscribed angle.

(b) Give the measures of each indicated angle.

(1) Figure (l), m /AVB =

(2) Figure ( 2 ) , m /RVS =

(3) Figure ( 3 ) , m /AVB =

(4) Figure ( 4 ) , m /AVB =

(5 ) Figure (51, m /TvQ =

(6) Figure ( 6 ) , m /AvB =

12-4

2. Find the measures of each indicated par t . The a rcs have degree measures a s marked.

P i s the center of the c i r c l e . E i s i n the i n t e r i o r of

the c i r c l e . F Is i n the ex te r io r of the c i r c l e . S, T, + +

B, M, C, R a re on the c i r c l e . A S and AM a r e tangents - t o the c i r c l e a t S and M respectively. TR i s a diameter.

Prove Theorem 12-9: I n the same c i r c l e o r i n congruent c i r c l e s , i f two a rcs a r e congruent, then so a r e the associated chords.

I n the f igure A F = BH . Prove : - - - ( a ) FB = AH

( b ) ABMF 2' A A M H

12-4

5. ABCD i s a square. E i s any point of DC , as shown i n t h i s f igure . Prove t h a t * + EA and EB t r i s e c t /DEC .

I n the f igure, A, B, C , D - a re on the c i r c l e and EF i s tangent t o the c i r c l e a t A . Complete the following s t a t e - ment s :

(d) LEAD i s supplementary t o . ( e ) /DAB Is supplementary t o . ( f ) /ABC i s supplementary t o .

2 - (s) /ME 2r (h) /DBA i s supplementary t o . (1) LADE i s supplementary t o . ( j ) /DAC 2- - 4--b 7. I n the f igure CP and AQ

a re tangents, "P i s a diameter of the c i r c l e . If

w

rns = 120 and i f the radius of the c i r c l e i s 3 , f ind

the length of .

w

8. Two circles are tangent, either internally or externally,

at a point H . Let u be any line through H meeting

the circles again at M and N . Prove that the tangents

at M and N are parallel.

9. Given: Tangent and

secant %?. B is the - midpoint of PR . Prove: B is equidistant - from %? and PR .

10. Prove the theorem: If two parallel lines intersect a

circle, they intercept congruent arcs. Q -

- - P P

Case I Case I1 Case I11

(One tangent one (TWO secants) (TWO tangents) secant)

2 *11. Consider the c i r c l e 0 , ( (x,y): x + y2 = 251 , and

the point ~ ( 0 , 3 ) . 4-+ (a) Find the in tersec t ions A,B of the l i n e Q,A with - the c i r c l e 0 , given QA = ((x,y): y = 3 ) .

( b ) Find the in tersec t ions C , D of the l i n e - QC = ((x,y): x = 0) , with the c i r c l e 0 .

( c ) What i s the product of the lengths of segments (%A and ? O f segments and %D ? Are these products equal? (If not, check your work.)

2 *12. Consider the c i r c l e 0 = ((x,y) : x + y2 = 25) and the

point ~ ( 8 , 3 ) . (a) Find the points A,B which a r e the in tersec t ions of -

PA which equals ((x,y): y = 3 ) , with the c i r c l e 0 . ( b ) Find the points C,D which a re the intersect ions of

4--b PC = ((x,y): y = x - 5) with the c i r c l e 0 .

( c ) Find the product PA PB and the product PC PD ,

(if these products a r e not equal, check your work.) - *13. Given AD tangent t o the

c i r c l e a t A and the 4--b

secant BD in te r sec t ing the c i r c l e a t B and C . ( a ) What i s the r e l a t i o n

between AADB and ACDA ? Why?

( b ) Why does AD (of AADB) = k CD (of ACDA) ?

Why does BD (of AADB) = k AD (of AcDA) ?

( c ) Assume CD = 6 . Express AD i n terms of k . Express BD i n terms of k . A

(d) Compare AD AD with BD CD . Would t h i s r e l a t i o n be t r u e f o r a l l values of k ?

Would it be t rue f o r every value C

assigned t o CD ?

12-5. Lengths - of Tangent and Secant Segments. -

DEFINITION. If the line 'an* is tangent to a circle - at R , then the segment W. is a tangent-segment

from Q to the circle.

THEOREM 12-13. The two tangent-segments to a circle from an external point are congruent, and form congruent angles with the line Joining the external point to the center of the circle.

Proof : -

Given: QR is tangent to the circle C at R , and ^S is tangent to C at S .

- TO prove : QR 2 QS --

/FOR ~ P Q S

By Corollary 12-4-1, APQR and APQS are right triangles, with right angles at R and S . Obviously 2 P , and - * - PR = PS because R and S are points of the circle. By the Hypotenuse-Leg Theorem, this means that

APQR 2 APQS . Therefore 2 , and 2 /PQs , which was to be proved.

- DEFINITION. If secant OR intersects a circle in A and B such that A is between Q and B , then B is

called the secant-segment from Q to the circle and is called the external secant-segment from Q to the circle.

THEOREM 12-14. The product of the length of a secant-segment from a given exterior point and the length of its external secant-segment is constant for any secant containing the given point.

Proof: Let Q be the given point and let ^5 and ^T be two secant-segments, having respective external secant- - segments QR and . By the A.A. Theorem for similar triangles we prove

ASQU - ATQR .

12-5

It follows tha t

and therefore

We prove tha t the product QS QR i s equal t o the product of the length of any secant-segment from Q and the length of i t s external secant-segment. This proves the constant t o be

QS QR

Notice tha t t h i s theorem means t h a t the product QR QS

i s determined merely by the given c i r c l e and the given external point, and Is independent of the choice of the secant. h he theorem t e l l s us t h a t any other secant gives the same product . ) This constant product i s ca l led the power of the point with respect t o the c i r c l e .

The following theorem a s s e r t s t h a t , i n the f igure below,

QR QS = ( Q T ) ~ .

THEOREM 12-15. Given a tangent-segment ^T t o a c i r c l e a t T

and a secant through Q , in tersec t ing the c i r c l e i n

points R and S . Then

The main s teps i n the proof a re as follows. You should f ind the reasons i n each case.

1. m / S = $ & . 4 . AQRT - AQTS .

The following theorem is a further variation on the preceding two; the difference is that now we are going to draw

two lines through a point in the interior of the circle. The

theorem says that in the figure below, we have

Q R 0 Q S = Q U * Q T .

You will recognize this theorem as the generalization

arrived at in Problem 5 of Problem Set 1-4.

THEOREM 12-16. If two chords of a circle intersect, the product of the lengths of the segments of one is equal to the

product of the lengths of the segments of the other.

The find the

1.

2.

main steps in the proof are as follows. You should

reason in each case.

/S 2 /T . /SQU 2 /TOR

3 . ASQU - ATQR . 4. (QS,QT) P = (QU,QR) . 5 - QS = QU QT

Problem Set 12-5

1. Complete the following statements by replacing the blanks

with appropriate words or expressions.

(a) If M is any point in the exterior of a circle,

there are tangent-segments to the circle from M and their are equal. If P Is the center of the circle,

+ then MP is the of the containing

the tangent-segments. Illustrate with a diagram. (b) If is a secant-segment, R is in the exterior

of a circle and S is the circle. If HK - is the external secant-segment and a subset of RS,

then A is - the circle and is between and . Illustrate with a diagram. - -

(c) If in a circle three chords AB , CD and intersect at X and if AX XB = 12 , then what is C X - X D ? Whatis E X - X F ?

2. Points P, R, M are in the exterior of the circles and

A, B, X, H, K, S, T are on the circles as indicated in the diagrams. R is the center of the circle in Figure a. A, B, and X are points of contact of tangents in Figures (a) and (b), respectively.

Figure a. Figure b . Figure c.

(a) Refer to Figure a. -

(1) Segments PA and are . (2) If PA = 10 , then PB = . Why?

(State the theorem.)

(3) If m /APR = 45 , then m /BPR = . Why? (State the theorem.)

(b) Refer to Figure b.

(1) The tangent-segment is . - (2) The secant-segment is . (3) The external secant-segment is

(4) R A + A B = . What theorem did you use? (5) If XR = 12 , could RA = 8 when AB = 10 ?

Justify your answer. (6) If XR = 12 , could RA = 6 when A 3 = 18 ?

Given another pair of numbers which could be the measures of and respectively.

(c) Refer to Figure c.

(1) If M H = a , H T = b , M K = x , K S = y

then a ( ) = x a ( ) . State the theorem which justifies your answer.

(2) Could MH = 3 , HT = 17 , MK = 4 , KS = 11 ?

Explain,

3. In the diagram P is the center of the circle, Q Is a point in the ex-

terior of the circle and A and B are on the - B Q circle. QB is a secant- - segment, QA is an external - secant-segment and Q A is a tangent - segment.

We want to discover the relation between Theorem 12-14

and Theorem 12-15 by noting relations in the diagram. -

(a) In what position does QB appear to have its greatest length?

(b) In what position does the external secant-segment

appear the shortest?

(c) If takes on a sequence of positions on the circle, - - changing from position QPB to position Q A , the length of appears to . The length of the external secant-segment appears to The length of the secant-segment appears to decrease

and approach the length of the . The length of the external secant-segment appears to Increase and approach the length of .

(d) Since QJ3 QA = QB* Q A * , etc. for all positions of Bl and A * , when Bl becomes A" and A *

becomes A" we would expect QB QA to equal

(Q-) * (Q-) . Thus the situation in Theorem 12-15 is what we might call the limiting case of the situation in Theorem 12-14.

4-+- - 4. AC , CE and EF are

tangent to a circle at C D E

B, D, and F respectively.

Prove: CB + EF = CE .

5. Use the data as it appears in the diagrams.

BA = 20

CE = 16 EA = 6

Figure (a) Figure (b) Figure ( c )

( a ) Use Figure ( a ) and compute DA . (b) Use Figure (b) and compute AT . ( c ) Use Figure ( c ) and compute PD . I n the f igure DC = 8 , C R = 6 , R B = 7 .

Find BA . R

A - - Secants CA and CE i n t e r s e c t the c i r c l e a t A, B, and D, E ac a s shown i n t h i s f igure. I f A B = 8 , B C = 4 , E D = 1 3 ,

f ind DC . E

- 8. I n t h i s f igure AB i s tangent B

t o the c i r c l e a t A and secant w i n t e r s e c t s the c i r c l e a t K

and W . I f AB = 6 and - WK = 5 , how long i s BK ?

9. Given a c i r c l e with in tersec t ing chords as shown and w i t h x < w . If AB = 19 , f i n d x and w .

10. Given a c i r c l e w i t h a chord of length 12 whose distance from the center i s 8, f ind the radius of the c i r c l e .

- In the figure, CD is a tangent-

segment to the circle at D and - AC is a secant-segment which C

contains the center of the circle.

If CD = 12 and CB = 4 , find the radius of the circle.

If two tangent-segments to a circle form an equilateral triangle with the chord having the points of tangency as its endpoints, find the measure of each arc of the chord.

If a common tangent of two circles meets the line of center's at a point between the centers it is called a common internal tangent. If it does not meet the line of centers at a point between the centers it is called

a common external tangent.

- - In the figure AB Is a common internal tangent and CD

is a common external tangent.

(a) In the figure above, how many common tangents are

possible? Specify how many of each kind.

(b) If the circles were externally tangent, how many

tangents of each kind?

(c) If the circles were intersecting at two points?

( d ) If the circles were internally tangent?

(e) If the circles were concentric?

12 -5

14. Given: The sides of quadrilateral CDRS are tangent circle in the figure.

Prove: SR + CD = SC + RD .

are tangent to a circle with center 0 at A and C , respectively, and m /ABC equals 120 . Prove that AB + BC = OB .

16. The radii of two circles have lengths 22 and 8 .. respectively, and the distance between their

centers is 50 . Find the length of the common external tangent-segment.

(~int: Draw a perpendicular - from Q to AP .)

17. Two circles have a common external tangent-segment 36 inches long. Their radii are 6 inches and 21 Inches respectively. Find the distance between their centers.

18. In the accompanying diagram 1, and are tangent to

a circle with center A at B and C respectively. A second circle with center A t lies in the union of /CPB and its interior. 1, and l2 are tangent to circle A1

at B1 and Cl respectively.

--+ d (a) Are PA and PA* distinct? Explain.

(b) If minor arc has a degree measure of 130 , n

what is the degree measure of minor arc BICt ?

Justify your answer.

19. Show that it is not possible for the lengths of the segments of two Intersecting chords to be four

consecutive Integers.

2 20. Consider circle C = ((x,y): (x - 1)2 + (y + 3) = 64) and lines - , A t such that

-<?= ((x,~): Y = 51 and ,dl = ((x,y): x - y = 12) . (a) Find the coordinates of P , the intersection of

1 and Â £ . (b) Find the coordinates of T , a point of intersection

of 4 with the circle. (!There is only one point in

this case; /^ is tangent to C .) (c) Find the coordinates of R and S * , two points of

Intersection of 4 1 (a secant) with the circle. (d) Find PT and square it. (e) Find PR and PSI and their product. (f) Did you expect (PT)~ and PR PSI to be equal?

Why?

879

12 -5

21. Prove: The common in te rna l tangents of two c i r c l e s mee the l i n e of centers a t the same point.

( ~ i n t : Use an ind i rec t proof. )

22. Standing on the bridge of a la rge sh ip on the ocean, the captain asked a young o f f i c e r t o determine the distance t o the horizon. The young o f f i c e r took a pencil and paper and i n a few moments came up with an answer. On the paper he had wr i t ten the formula

d = 2 6 miles.

Show t h a t t h i s formula i s approximately correct i f h i s

the height i n f e e t of the observer above the water and i f

d i s the distance i n miles t o the horizon. (~ssume the diameter of the ea r th t o be 8,000 miles. )

Review Problems

(chapter 12, Sections 1-5)

1. Consider the problems below w i t h reference t o the four s e t s C , Ll , Lp and Lo .

2 C = ((x,y) : x + y 2 = 100) ;

L, = ((x,y) : x = -10) ;

Ln = f (x,y): Y = 61 ;

4 L3 = ( ( x , ~ ) : y = Tx)

( a ) c is a with radius and center a t the point w i t h coordinates

(b) A(6, 81, B(7, -7), and C(-8, 10) are three points in the xy-plane. For each point

determine whether it is on the circle, in-the interior of the circle or In the exterior of the

circle. Show your computations. (c) Find the intersection of Ln and C . (d) Find the intersection of Ln and C . (e) Find the intersection of Lg and C .

2. Consider a sphere S with radius 10 and an xyz-coordinate system which has its origin at the center of S .

Write an equation

of S . Give the coordinates

of the points of

intersection of S with

(1) the x-axis

(2) the y-axis Y (3) the z-axis Give the intersection

of S with the xy-plane, that is,

with ((x,y,z) : z = 0) . Give the intersection of S with the xz-plane. Give the intersection of S with the yz-plane.

Given the points ~(3 , -4 ,5 f l ) , ~(3,-5,7) , ~(9,6,1) . For each point determine whether it is (1) in S , (2) in the interior of S or (3) in

the exterior of S . Write an equation of a circle in the xy-plane which has its center at (3,-2) and its radius equal to 4 . Write an equation of a sphere with center at (2,-1,3) and with radius equal to 3 .

4. From the presentation i n Sections 12-1 through 12-5, we have several s i tua t ions i n which two angles, two segments o r two a rcs a re congruent.

(a) Give 6 conditions under which 2 segments related t o a c i r c l e are congruent.

( b ) Give 3 circumstances under which an angle re la ted t o a c i r c l e i s a r i g h t angle.

( c ) Give 4 conditions under which two angles re la ted t o a c i r c l e a re congruent.

(d) Give 4 conditions under which 2 a rcs have

the same degree measure.

5. (a) How is the degree measure of the a r c i n which an angle i s inscribed re l a t ed t o the degree measure of the arc which i t in tercepts?

(b ) Explain how the r e l a t i o n between the measure of a cen t ra l angle and the degree measure of i t s associated arc might be considered a special case of the r e l a t i o n between the measure of an angle formed by two chords wiiich in te r sec t i n the i n t e r i o r of a

c i r c l e and the degree measure of i t s associated arcs .

6. For the c i r c l e centered a t 0 ,

Given: I n the f igure , the c i r c l e with center 0 has - diameter . AF 1 1 OH , m /A = 55 . Find m= and ni~> .

- 8. Given: AB i s a diameter

of the c i r c l e w i t h center C . --b XY bisec ts /AXB .

A Prove: CY 1 AB . ( ~ i n t : Find m /AXY . )

9. Indicate whether each of the following statements i s t rue or f a l s e .

( a ) If a point i s the midpoint of two chords of a c i r c l e , then the point i s the center of the c i r c l e .

(b) If the measure of one arc of a c i r c l e i s twice the measure of a second arc , then the chord of the f i r s t a rc i s l e s s than twice as long a s the chord of the second a rc .

( c ) A l i n e which b isec ts two chords of a c i r c l e i s perpendicular t o each of the chords.

(d) If the ver t ices of a quadr i la te ra l a re on a c i r c l e ,

then each two of i t s opposite angles a re supplementary.

(e ) If each of two c i r c l e s i s tangent t o a t h i r d c i r c l e , then the two c i r c l e s a re tangent t o each other.

( f ) A c i r c l e cannot contain three co l l inea r poin ts .

( g ) If a l i n e bisect6 a chord of a c i r c l e , then it

b isec ts the minor arc of t h a t chord. (h) I f PR i s a diameter of a c i r c l e and Q i s any -

point i n the i n t e r i o r of the c i r c l e not on PR , then /PQR i s obtuse.

(1) A tangent t o a c i r c l e a t the midpoint of an arc i s pa ra l l e l t o the chord of t h a t a rc .

( j ) It i s possible f o r two tangents t o the same c i r c l e

t o be perpendicular t o each other .

- 10. Given: In the figure BX is tangent to the circle

at B . AB = AC . r n 5 = 100 . Find m /C , m /ABX , and m /CBA .

11. Given: Two circles tangent

at P with common tangent -

"W*'. "AX*' is tangent to one - circle at X and AY is

tangent to the other circle

at Y . Prove: AY = AX .

12. A hole 40 inches in diameter is cut in a sheet of

plywood, and a globe 50 inches in diameter is set in

this hole. How far below the surface of the board will

the globe sink?

13. A wheel is broken so that only a portion of the rim

remains. In order to find the diameter of the wheel the following measurements are made: three points C, A, and

3 are taken on the rim so that chord 2 chord 'KS . The chords and are each 15 inches long, and the chord is 24 Inches long. Find the diameter of

the wheel.

14. Diameter AD of a circle with center C contains a point

B which lies between A and C . Prove that is

the shortest segment Joining B to the circle and

is the longest.

15. Given: C i r c l e with cen te r C ,

tangent t o t h e c i r c l e a t H . Prove: = rn /RHN . ( ~ o t e : The c i r c l e may be considered t o represent t h e - ea r th , with PQ the e a r t h ' s ax i s , ,@-IN t he angle of e leva t ion of the North S t a r , and the l a t i t u d e of a point H . )

16. Assume that the e a r t h Is a sphere of rad ius 4,000 miles. A - straight tunnel AB 200 miles long connects two po in t s A and

B on the surface , and a vent i - l a t i o n sha f t CD Is constructed a t the cen te r of t h e tunnel . What i s the length, i n miles, of t h i s s h a f t ?

17. Describe t he s e t s Indicated below.

2 2 ( b ) M = [(x,Y): ( X - 2) + (y + 4 ) = 49) .

2 (d) R = ( (x ,y ,z) : x + y2 + z2 = 25 , z = 3 ) .

2 ( e ) The i n t e r s e c t i o n of A = [ (x ,y ,z ) : x2 + y2 + z = 1)

and B = [ (x ,y ,z ) : 1x1 = l} . 2

(f) The i n t e r s e c t i o n of D = f (x,y, z ) : x + y2 + z2 = 161 2 and F = [ (x ,y ,z) : x + y2 + z2 = 8)

2 (a ) The i n t e r s e c t i o n s of T = ( ( x , ~ , z ) : x + y2 + z2 = 25) 2

and U = {(x,y ,z) : x + y2 = 9) .

18. In the figure, "AP* is tangent to the circle

at A . AP = PX = XY . If PQ = 1 and QZ = 8 , find AX .

19. Given: , and

are 120' arcs on a circle P and P is a point on . Prove: PA + PI3 = PC . (~int: Consider a parallel B to through A inter- - seating PC in R and the

circle in Q .)

20. Prove that if two circles intersect, the common

secant bisects both common tangent-segments.

12-6. The Circumference of a Circle; the Number TT . - -- - - It makes sense t o ask of someone who made a t r i p how far

he went. If he t raveled In a s t r a i g h t l i n e t h e answer would be the distance between h i s s t a r t i n g point and a r r i v a l point. I f he t raveled i n a curved path the answer would not be so easy t o give. If the path were a c i r c u l a r a rc , the degree measure of the arc i s not a sa t i s fac to ry way of describing i t s length. Can you see t h a t it i s possible f o r two a r c s t o have t h e same degree measure and have d i f fe ren t lengths? Can you a l so see t h a t it i s possible f o r two a rcs t o have d i f fe ren t degree measures and have t h e same length?

n length of A1Bl length of length of

n ( i n nun.) = 93 AnBn = 62 dp3- 62

We are going t o try t o say what we mean by the length of c i r c u l a r a rcs and t o derive ways of f inding such lengths. The subject Is discussed more thoroughly I n a branch of mathematics known as "calculus," where a l l kinds of curved paths a r e discussed. We first proceed informally, r e fe r r ing t o the physical world. We Imagine t h a t w e have made a complete c i r c u i t around a c i r c u l a r path and inquire how far w e have gone. We

call this distance the circumference of the circle and denote it by C . It seems reasonable to suppose that if we want to

measure C approximately, we can do it by inscribing a regular polygon with a large number of sides and then measuring the perimeter of the polygon. That Is, the perimeter p ought

to be a good approximation to C when n , the number of sides, is large. Putting it another way, if we decide how close we want p to be to C , we ought to be able to get p to be this close to C merely by making n large enough. We describe this situation in symbols by writing p-C , and we say that p has C as a limit.

We cannot prove this, however; and the reason why we cannot prove it is rather unexpected. The reason is that so far, we have no mathematical definition of what is meant by the circumference of a circle. (We cannot get the circum-

ference merely by adding the lengths of certain segments, the way we didto get the perimeter of a polygon, because a circle does not contain any segments. Every arc of a circle, no matter how short, is curved at least slightly.) But the remedy is

easy; we take the statement

as our definition of C , thus:

DEFINITION. The circumference of a circle is the limit of the perimeters of the inscribed regular

polygons.

We would now like to go on to define the number IT as the quotient of the circumference of a circle divided by its

diameter. But to make sure that this definition makes sense, we first need to know that the number 5 is the same for all circles, regardless of their size. Tius we need to prove the

following:

THEOREM 12-17. The quotient of the circumference divided by n b the diameter, E, is the same for all circles.

Proof: We use similar triangles. Given a circle with center Q and radius r , and another circle, with center Ql

and radius r t , we inscribe a regular n-gon in each of them. h he same value of n must be used in each circle.)

In the figure we show only one side of each n-gon, with the associated isosceles triangle. Let e and el be their lengths as shown. Now /AQB 2 /A'Q~B~ , because each of these angles has measure - Therefore, since the adjacent sides n o are proportional,

AAQB - AAtQtB1 by the S.A.S. Similarity Theorem. Therefore (e,r) = (el,rl)

P or (ne,r) ; (nef,rl) . But ne is the perimeter of the first

n-gon, and net is the perimeter of the second. We can write, (p,r) 5 (pl,ri) . Now let C and Cl be the respective

circumferences of the two circles. Then p+C and

p*+C' , by definition of circumference of a circle. It is plausible that (C,r) = (c' ,rl) . By alternation we can write

P this as (c,c~) = (r,rf) . It follows that (c,cl) = (2r,2r1) .

P P The last proportionality shows the constant of proportionality

tobe 5 . ' It is designated by T . We see that the circumference of - any circle divided by its diameter is IT .

We express the conclusion of Tneorem 12-17 in the well-known formula

"\

c=27rr.

It can be proved that TT is not a rational number, that is, it cannot be represented by an expression Ã , where p and q are integers. It can, however, be approximated as closely as we please by rational numbers. Some rational approximations to IT are

3.14 , 22 3 , - 3.1416 , 355

7 ' IT?' 3.1415926535 . (As a general rule, if there is a choice, you should leave

your answers to problems in terms of TT .)

Corollary 12-17-1. The circumference of circles are

proportional to their radii.

Problem Set 12-6 -- 22 1. Wnich is the closer approximation to TT , 3.14 or - ? 7

2. In the following problems C = circumference, r = radius and d = diameter of a circle. Find the indicated parts.

(a) r = 7 , d = 1 c =

(b) C = 36 , d = 1 r = . (c) d = 15 , c = 9 r = . (d) r = 6a , c = 1 d =

(e) r=xi/3', C = , d = . 3. C, and C2 are the circumferences of two circles with

radii rl and r2 respectively. Pill in the blanks with

the appropriate multiplier.

(a) If r1 = 3r2 , then C, = C2 . (b) If C p = 5 5 , then d2 = dl .

1 ( c ) If d2 = -d 2 1 ' then Cl = C2 . (d) If r2 = dl , then C2 = Cl .

4. Two concentric c i r c l e s have cen t ra l angles /BOA as indicated.

A ( a ) The mG ( i n degrees)

equals how many times - mBIA1 ( i n degrees) ?

(b) If OA = 20A1 , then you would expect the length of - c BA t o equal how many times the length of B t A 1 ?

[ ~ e n g t h s of a rcs w i l l be considered more prec ise ly i n Section 12-8. I

0 A (c ) If Â¥ST-[ = f , what would you expect

length of - t o equal? length of BIA1

2 2 5. Given T = ((x,y): x + y = 36)

U = ( (x ,y) : x2 + y 2 = 1.6) . ( a ) What i s the r a t i o of the circumference of T t o

the circumference of U ?

(b) What would you expect the r a t i o of the lengths of the subsets of T and U which a re i n the union of the posi t ive x-axis. the pos i t ive y-axis and Quadrant I t o be?

6. The moon i s about 240,000 miles from the ear th, and i t s path around the ea r th i s nearly c i r cu la r . Find the circumference of the c i r c l e which the moon describes every month.

7. The ear th i s about 93,000,000 miles from the sun. The

path of the ea r th around the sun i s nearly c i r cu la r . Find how f a r we t r a v e l every year " i n orb i t . " What i s our speed i n t h i s o r b i t I n miles per hour?

8. A ce r t a in t a l l person takes s teps a yard long. H e walks around a c i rcu la r pond close t o the edge taking 628 steps. What i s the approximate radius of the pond? (use 3.14 a s an approximation f o r IT . )

12 -6

9 . A regular polygon i s inscr ibed i n a c i r c l e , then another with one more s ide than the first i s inscribed, and so on endlessly, each time increasing the number of s ides by one.

( a ) What i s the l i m i t of the length of the apothem?

( b ) What i s the l i m i t of the length of a s ide?

( c ) What i s t h e l i m i t of the measure of an angle?

(d) What is the l i m i t of the perimeter of the polygon?

10. The f igure represents pa r t of a regular polygon of which - AB and a r e s ides , and R i s the center of the c i r c l e i n which the polygon i s inscribed. Copy and complete the tab le :

lumber )f s ides

11. The s ides of a regular hexagon a re each 2 u n i t s long. I f i t i s Inscribed i n a c i r c l e , f ind the radius of the c i r c l e and the apothem of the hexagon.

12. The s ide of a square i s 1 2 inches. What i s the

circumference of i t s inscribed c i r c l e ? O f i t s circumscribed c i r c l e ?

13. ( a ) One c i r c l e has a radius of 10 f e e t . A second c i r c l e has a radius one foot longer. How much longer i s the circumference of the second c i r c l e than t h a t of the f i r s t ?

(b) How much longer i s the circumference of a c i r c l e whose radius i s 1001 f e e t than t h a t of a c i r c l e whose radius i s one foot shorter?

14. A regular octagon with s ides 1 u n i t long i s inscribed i n a c i r c l e . Find the radius of the c i r c l e .

I n the f igure , square XYZW i s D

inscribed i n a c i r c l e w i t h center 0 , and square ABCD i s circumscribed about t h i s c i r c l e . The diagonals of both squares l i e - i n AC and *W . Given t h a t C a square PQRS i s formed when the midpoints P, Q, R and S

of "A3? , BY , , and a re joined, i s the perimeter of t h i s square equal to , g rea te r than,

B or l e s s than the circumference of the c i r c l e ? Let OX = 1 and j u s t i f y your answer by computation.

12-7. Area of a Circle. --- In Chapter 11 we considered areas of polygonal-regions,

defined i n terms of a basic region, the tr iangular-region,

which i s the union of a t r i ang le and i t s i n t e r i o r . In ta lk ing

about areas associated with a c i r c l e we make a similar basic def ini t ion.

DEFINITION. A circular-region i s the union of a c i r c l e and i t s i n t e r i o r .

I n speaking of "the area of a triangular-region" we found It convenient t o abbreviate t h i s phrase t o "the area of a t r iangle ." Similarly, we usual ly say " the area of a c i rc le" a s an abbreviation of "the area of a circular-reglon."

We now get a formula f o r the area of a c i r c l e . We already have a formula f o r the area of an inscribed regular n-gon; t h i s i s

1 An = P

where a i s the apothem and p i s the perimeter.

p3

I n t h i s s i t u a t i o n there a re th ree quan t i t i e s involved, each depending on n , the number of s ides . These a re p , a and A . To get a formula f o r the area of a c i r c l e , we need t o f i n d out what l i m i t s these quan t i t i e s have a s n becomes la rge .

(1) What happens - t o A . An i s always s l i g h t l y l e s s than the area A of the c i r c l e , because there a re always some regions t h a t l i e Inside the c i r c l e but outside the regular n-gon.

But the difference between An and A i s very small when n

i s very large, because when n i s very large the polygonal- region almost f i l l s up the i n t e r i o r of the c i r c l e . Thus we expect tha t

An-A . But jus t a s In the case of the circumference of the c i r c l e , t h i s

cannot be proved u n t i l we say what we mean by the area of a c i r c l e . Here also, the way out i s easy:

DEFINITION. The area of a c i r c l e i s the l i m i t of --- the areas of the inscribed regular polygons.

Thus, A n d A by def in i t ion .

( 2 ) What happens - t o - a . The apothem a i s always s l igh t ly l e s s than r , because e i t h e r l eg of a r i g h t t r i ang le i s shorter than the hypotenuse. But the difference between a and r i s very small when n i s very large. Thus, we expect tha t

( 3 ) What happens - t o - p . By de f in i t ion of C , we have

p--+c . F i t t i n g together the r e s u l t s i n (2 ) and ( 3 ) , we get

Theref ore 1 An-rC .

But we knew from (1) t h a t

An-A . Since An has only one l i m i t , it follows t h a t

1 A = w r C . Combining t h i s with the formula C = 2-nr gives

2 A = m .

Thus the formula that you have known for years finally becomes a theorem:

2 THEOREM 12-18. The area of a circle of radius r is m .

Corollary 12-18-1. The areas of two circles are proportional to the squares of their radii.

Problem Set 12-7 -- 1. Find to the nearest tenth the circumference and area of a

circle with radius

2. Find exactly (in terms of i r ) the circumference and area of a circle with radius

3. Find the circumference of a circle whose area is

4. Find the area of a circle whose circumference is

(a) 127r (b) 207r

Find the area of one face

of this iron washer if its diameter is 4 centimeters and the diameter of the

hole is 2 centimeters.

Would the area be changed if the two circles were not concentric?

The radius of the l a rge r of two c i r c l e s i s three times the radius of the smaller. Find the r a t i o of the area of the f i r s t t o tha t of the second.

The circumference of a c i r c l e and the perimeter of a square a re each equal t o 20 inches. Which has the greater area? How much g rea te r i s it?

Given a square whose s ide i s 10 inches, what i s the area between i t s circumscribed and inscribed c i r c l e s ?

An equi la tera l t r i ang le i s inscr ibed i n a c i r c l e . I f the side of the t r i ang le i s 12 inches, what i s the radius of the c i r c l e ? The circumference? The area?

The cross inside the c i r c l e

i s d iv i s ib le in to 5 squares. Find the area which i s inside the c i r c l e and outside the cross.

Given: Two concentric c i r c l e s . - AC i s a chord of the la rger and i s tangent t o the smaller a t B . Prove: The area of the r ing (annulus) i s T(BC)~ .

I n a sphere whose radius i s 10 inches, sections a re made by planes 3 inches and 5 inches from the center. Which section w i l l be the la rger?

Prove tha t your answer i s correct .

13. In the figure, A N D is a square in which E, F, G

are midpoints of AD , AC , and , respectively.

and are circular

arcs with centers E and G respectively. If the side of the square is s , find the area of the shaded portion.

In the figure, semicircles are drawn with each side of right triangle ABC as diameter. Areas of each region in the figure ar

indicated by lower-case

letters.

Prove: r + s = t . A B -

A special archery target, with which a novice can be expected to hit the bulls-eye as often as any ring, is

+ + constructed in the following way. Rays OM and PN are parallel. A circle with center 0 and radius r , equal to the distance between the rays, is drawn inter-

+ seating OM at Q . 1 M . Then a circle with center 0 and radius OA , or r, , is drawn intersecting OM in R . This process is repeated by drawing perpen-

diculars at R and at S , and circles with radii OB and OC . Note that we arbitrarily stop at four concentric circles.

(a) Find rl , rg , rg in terms of r . (b) Show that the areas of the inner circle and the

three "rings," represented by a, b, c, and d

are equal.

16. An isosceles trapezoid whose bases a r e 2 inches and 6 Inches i s circumscribed about a c i r c l e . Find the area of the portion of the trapezoid which l i e s outside the c i r c l e .

12-8. Lengths of Arcs. Areas of Sectors. -- -- Just a s we define the circumference of a c i r c l e a s the

l i m i t of the perimeters of inscribed regular polygons, so we can define the length of a c i r c u l a r a rc a s a su i tab le l i m i t .

If is an a r c of a c i r c l e with center Q , we take n - 1 points PI , P2 * . . , Pn - on so t h a t each of the

DEFINITION. The length of a rc AB i s the l i m i t of -- APl + P P + a - 0 + Pn - 1 1 2 B a s we take n la rger

and l a rge r .

Notation. We sometimes wri te " 1 ~ " t o mean "the length of S B . "

It is convenient, i n discussing lengths of arcs , t o consider an e n t i r e c i r c l e as an arc whose degree measure i s 360 . The circumference of a c i r c l e can then be considered t o be simply the length of an a r c whose degree measure i s 360 ,

We now have two types of measure f o r c i r c u l a r arcs , t h e i r degree measure and t h e i r length. There i s a simple connection between these measures, i n the case of congruent c i r c l e s , namely, t h a t the lengths of a r c s on congruent c i r c l e s a re proportional t o t h e i r degree measures. It i s possible t o prove t h i s f a c t , but the proof i s very d i f f i c u l t . We prefer t o s t a t e it as a postulate .

Postulate - 31. The lengths of a rcs in congruent c i r c l e s a r e proportional t o t h e i r degree measures.

n n I f we take AtB1 t o be a semicircle, then rnAIBt = 180 ,

n and Y A I B ' = T T ~ .

Then we may wri te ( ~ A B , Ã ˆ ~ 5 (rn~%,l80) . Clearly the constant of proport ional i ty i s 6 .

THEOREM 12-19. An arc of degree measure q contained i n a c i r c l e whose radius i s r has length L , where

This r e s u l t follows from the proportionali ty,

( ~ , m ) 5 (q,l80)

A sector of a c i r c l e i s a region bounded by two r a d i i and

an arc , l i k e th i s :

More precisely:

DEFINITIONS. If is an arc of a c i r c l e with center Q and radius r , then the union of a l l segments F , where P i s any point of , i s a sector. - AB i s the arc of the sec tor and r i s t h e radius --- of the sector. --

The following theorem i s proved Jus t l i k e Theorem 12-18.

THEOREM 12-20. The area of a sec tor i s half the product of

i t s radius and the length of i t s arc .

Combined with Theorem 12-19, we get

THEOREM 12-21. The area of a sector of radius r and arc measure q i s 0

Problem Set 12-8 -- 1. The radius of the c i r c l e w i t h

center A i s 20 . The radius of the c i r c l e with

c

center B i s 10 . mcD = 60 . mG = 60 . Is the length of CD grea ter than, equal to , o r l e s s than the length of EF ?

2. Which has the grea ter degree measure: an arc of one inch on the equator of the ea r th o r an a r c of one inch on a half do l l a r?

3 . Are the degree measures of congruent a rcs equal? Are the lengths of congruent a r c s equal?

4. Suppose 5 on one c i r c l e has a la rger degree measure than CD on another c i r c l e . Does t h i s information permit you t o conclude t h a t the length of I s grea ter than the length of 5 ? Suppose t h a t you were a l so to ld t h a t the length of i s equal t o the length of CD . Which c i r c l e has the g rea te r radius?

5. The radius of a c i r c l e i s 15 inches. What i s the length of an a rc of 60' ? of 90' ? of 72' ? of 36' ?

6. The radius of a c i r c l e i s 6 . What i s the area of a sec tor w i t h an arc of 90' ? of lo ? of 60' ? of 5b0?

7. What i s the area of a sector whose arc has degree measure 90' and a rc length 3ir ?

8. What Is the length of A% i n the c i r c l e with center 0

i f m /AOB = 60 and the area of the sec tor AOB i s 67r ?

9. I f the length of a 60' arc i s one centimeter, f ind the radius of the arc . Also f ind the length of the chord of the arc .

10. I n a c i r c l e of radius 2 , a sector has area TT . What i s the measure of i t s a rc?

2 2 11. Given S = ((x,y,z): x + y 2 + z - 2 5 )

12. Find

Describe the in tersec t ion of S and P . Compare the circumference of the c i r c l e of i n t e r - sect ion of S and P and the circumference of a great c i r c l e of S . Compare the area of the c i r c l e of in te r sec t ion of S and P and the area of a great c i r c l e of S . Compare the a r c s of the c i r c l e s i n P a r t s (b) and ( c ) such t h a t all points of the a rcs s a t i s f y the conditions t h a t x - > 0 and y - > 0 .

2 the area of the sector of the c i r c l e x + y2 = 100

which has the posi t ive x-axis and the l i n e y = x as p a r t i a l boundries and which s a t i s f i e s the condition tha t

x > o , Y > O

13. A segment of a c i r c l e is the - - region bounded by a chord and an arc of the c i r c l e . The area of a segment i s found by subtracting the area of the t r i ang le formed by the chord \ and the r a d i i t o i t s endpoints \ / from the area of the sector . I n the f igure, m /APE = 90 . I f PB = 6 , then

Area of sector PAB = 1 v - 62=97r.

1 Area of t r i ang le FAB = -y 62 = 18 . Area of segment = 9 - 18 o r approximately 10.26 ,

Find the area of the segment i f :

( a ) m /APB = 60 ; r = 12 . (b) m /APB = 120 ; r = 6 ,

( c ) m ÂPE = 45 ; r = 8 . 1 4 . If a wheel of radius 10 inches ro ta te s through an

angle of 3 6 , ( a ) how many inches does a point on the r i m of the

wheel move? (b ) how many inches does a point on a spoke 5 inches

from the center move?

15. A continuous b e l t runs around two wheels of r a d i i 6 inches and 30 inches. The centers of the wheels a re 48 inches apar t . Find the length of the be l t .

16. I n t h i s f igure ABCD i s a square whose s ide i s 8 inches. With the midpoints of the s ides of the square as centers, a rcs a r e drawn tangent t o the diagonals. Find the area enclosed by the four arcs .

Inscribed and Circumscribed Circles .

DEFINITIONS. A c i r c l e i s Inscribed i n a t r i ang le , o r - - the t r i ang le Is circumscribed about the c i r c l e , i f -- each side of the t r i ang le i s tangent t o the c i r c l e .

A c i r c l e is circumscribed about a t r i ang le , o r the -- t r i ang le i s inscribed i n the c i r c l e , i f each vertex -- of the t r i ang le l i e s on the c i r c l e .

In these figures, A ABC i s inscribed i n c i r c l e , C, and A A * B I C * Is circumscribed about c i r c l e C2. Circle Cn i s inscribed i n A A *B*C1 and c i r c l e Cl Is circumscribedabout A ABC.

The problem of finding a c i r c l e circumscribed about a given t r i ang le can be s t a t ed i n a s l i g h t l y d i f f e ren t way. If

A, B, C a re three p o i n t s , i t i s na tura l t o inquire i f there i s a c i r c l e which contains a l l three points, and i f so, how many such c i r c l e s there a re . We know t h a t no c i r c l e has three col l inear points, so we ought t o r e s t r i c t our a t t e n t i o n t o three points which a r e noncollinear. Can you see any difference between the problem of circumscribing a c i r c l e about a given t r iangle and t h a t of passing a c i r c l e through each of three given noncollinear points?

If we want t o circumscribe a c i r c l e about a given t r iangle ABC, we would first have t o loca te i t s center 0 . We s t a r t w i t h the requirements t h a t OA = OB = OC . Where do we look f o r 0 i f OA and OB a re t o be equal? If OB and OC a r e t o be equal? If OC and OA a r e t o be equal? 3y Theorem 8-28 we know t h a t the s e t of a l l points equally d i s t a n t from two points i s the perpendicular b isec tor l i n e segment joining the given points. We a r e therefore led t o f ind the perpendicular b isec tors of , and ^B" . W i l l these perpendiculars meet i n one point? I n f a c t , they do, as is proved i n Corollary 8-28-1. Since 0 i s the only point t h a t has the property that OA = OB = OC and since OA i s the only radius t h a t we can use f o r our c i r c l e we therefore conclude t h a t there i s exactly one c i r c l e t h a t circumscribes A A E C . And thus we have proved

THEOREM 12-22. A t r i a n g l e has one and only one circumscribed c i r c l e . The center of t h i s c i r c l e i s the in tersec t ion of the perpendicular b isec tors of the s ides of the t r i ang le .

We now tu rn t o c i r c l e s inscr ibed i n t r iangles . We look f o r a c i r c l e which has the three s ides as tangents. The segment from i t s center 0' t o a point of contact i s perpen- d icular t o t h i s s ide . Why? Then the length of t h i s segment i s the dis tance from O* t o the s ides . Why? Similarly, the lengths of t he perpendicular segments from 0' t o the other s ides

would be the distances t o those s ides . Our problem then i s t o f ind a point equally d i s t an t from the s ides of a t r iangle. By Theorem 8-29 we know t h a t the s e t of a l l points equally d i s t an t from the s ides of an angle i s i t s midray. Furthermore, by Corollary 8-29-1, we know that the midrays of the angles of a t r iangle meet i n one point. We may therefore f i n d the center of an inscribed c i r c l e by f inding where the three angle bisectors of the t r i ang le meet. The radius of t h e c i r c l e i s the distance from t h i s point t o a s ide. Moreover the c i r c l e i s

unique because i t s center and radius a r e unique. We have thus

proved

THEOREM 12-23. A t r i ang le has one and only one inscr ibed c i r c l e . The center of t h i s c i r c l e i s the in te r sec t ion of the midrays of the angles of the t r i ang le .

Problem Set 12-9 -- Investigate whether the center of the circumscribed c i r c l e about a given t r i ang le i s i n t h e i n t e r i o r , on, o r i n the exter ior of the t r iangle . Consider three cases: a t r i ang le whose angles a r e acute ,a t r i a n g l e with a r igh t angle, and a t r i ang le with an obtuse angle. After making drawings f o r each case s t a t e what seems t o be t r u e In each case. Then prove each statement.

2. Must the center of the inscribed c i r c l e of a given t r i ang le be i n the i n t e r i o r of the t r i ang le? Write an argument t o support your answer.

3. The center of the circumscribed c i r c l e of a c e r t a i n t r i ang le i s on one s ide and i s 12 inches from each vertex. Find the length of the median t o the longest s ide.

4. Can a c i r c l e be circumscribed about a given rectangle? Can a c i r c l e be inscr ibed i n a given rectangle? Prove your answers.

5. Can a c i r c l e be Inscribed i n a given rhombus? Prove your answer. Can a c i r c l e be circumscribed about a given rhombus?

Prove t h a t an equ i l a t e ra l t r i ang le has concentric inscr ibed and circumscribed c i r c l e s .

Prove t h a t , i f a t r i ang le has concentric inscribed and circumscribed c i r c l e s , then it i s equi la tera l .

Prove t h a t a quadr i la te ra l can be circumscribed by a c i r c l e if a p a i r of opposite angles a re supplementary. (Hint: Use the ind i rec t method.)

Can a c i r c l e be circumscribed about a given isosceles trapezoid? Prove your answer. Can a c i r c l e be inscribed i n a given i sosce les trapezoid?

The radius of the inscribed c i r c l e of a r i g h t t r i ang le i s 2 inches long. The point of contact on the hypotenuse divides it i n t o two segments whose lengths have the r a t i o 3:2 . Find the length of the hypotenuse.

Consider t r i ang le XYZ w i t h ve r t i ces X(O,O) , Y ( ~ , o ) ,

Find the midpoint and the slope of , and write an equation of the l i n e containing t h i s point and having a s i t s slope the negative reciprocal of the slope of v. That is , wri te an equation of the perpendicular b isec tor of ^Y . Similarly, obtain an equation of the perpendicular b isec tor of . Find the point of intersect ion, C , of the - respective perpendicular bisectors of XY and . Obtain an equation of the perpendicular bisector of - YZ . Do the coordinates of C obtained i n ( c ) , s a t i s f y t h i s equation? Are the three perpendicular bisectors of t r i ang le XYZ concurrent i n the point C ? (1f you have not found t h i s t o be the case,you shou l d check your work. ) Show t h a t C i s equidis tant from the ve r t i ces of t r i a n g l e XYZ . What i s the distance CX ?

Write an equation of the c i r c l e having C a s center and nor. .aining X, Y, Z . This i s sometimes cal led the "circurncircle," and C i s the "circumcenter," of the t r i ang le .

908

12. Consider, again, the triangle XYZ with vertices X(O,O) , y(8,o) , Z(5,3)

(a) Write an equation of the line containing the midpoint of and the opposite vertex Z . This line contains the median to ^? .

(b) Write an equation of the line containing the median to XZ.

(c) Write an equation of the line containing the median to YZ .

(d) Express the three equations of Parts (a), (b), and (c) above in parametric form. For the parameter k take

1 the value 3 , or 3 , and find the coordinates of the trisection points of each median which is common to all three medians. This point is sometimes called

the "centrold" of the triangle.

13. Consider, again, the triangle XYZ with vertices X(O,O) , y(8 ,o ) , 74583)

(a) Write an equation of the line which contains Z and - is perpendicular to XY . This line contains the altitude to .

(b) Write equations for the lines which contain the

altitudes to ^Z and "2? , respectively. (c) Find the coordinates of the point 0 which is common

to all three of the lines containing the altitudes. This point is sometimes called the "orthocenter" of the triangle. Is the orthocenter of triangle XYZ in the interior or the exterior of the triangle?

Sketch a triangle for which the opposite is true.

(d) It is perhaps surprising to learn that the circumcenter, the centroid, and the orthocenter, of any triangle are collinear. Use the results of Problems 11, 12, and 13 to verify that this is the case for triangle XYZ .

12-10. Summary.

The foot of the perpendicular from the center of a c i r c l e t o a l i n e i s the key t o understanding many re la t ions between c i r c l e s and tangents and between c i r c l e s and chords. Similarly

-. . t he foot of the perpendicular from t h e center of a sphere t o a plane helps t o explain the r e l a t ions between a sphere and a tangent plane and the in tersec t ion of a sphere and a plane.

We saw some in te res t ing re l a t ions between the measures of c e r t a i n angles r e l a t ed t o a c i r c l e and the measures of intercepted arcs . These re l a t ions a r e e a s i l y remembered by noting tha t i f the vertex of the angle i s an i n t e r i o r point of the c i r c l e we use one-half of t h e sum of two arc measures; i f

on the c i r c l e , one-half of an arc measure; i f an exter ior point, one-half the difference between two arc measures.

We defined the circumference of a c i r c l e , the length of an a rc , the area of a c i r c l e , and the area of a sector a s l imi t s of c e r t a i n measures r e l a t ed t o regular polygons. This enabled us t o make plausible the formulas used f o r measuring circumferences, a rc lengths, a reas of c i r c l e s and areas of sectors.

Review Problems

(chapter 12)

1. If C i s the circumference of a c i r c l e and r i s i t s c radius, what i s t h e value of Ã ?

2. Define (a ) the area of a c i r c l e .

(b) t h e length of an a r c of a c i r c l e .

3 . If the circumference of a c i r c l e i s 1 2 inches, the length of i t s radius w i l l l i e between what two consecutive integers?

4. I f the diameter of two c i r c l e s C and C' a re d and 2d

respect ively and C makes 10 revolutions i n going a distance K , how many revolutions w i l l C * make i n going the same distance?

5. What i s the radius of a c i r c l e i f i t s circumference i s equal t o i t s area?

6. If t h e radius of one c i r c l e i s 10 times the radius of another, give the r a t i o of

( a ) t h e i r diameters. ( b ) t h e i r circumferences. ( c ) t h e i r areas .

7. If a regular hexagon i s inscr ibed i n a c i r c l e of radius 5 , what i s the length of each s ide? What i s the length of the arc of each side?

8. Show tha t the area of a c i r c l e i s given by the formula 1 2 A = -~r~rd , where d i s the diameter of the c i r c l e .

9. ( a ) If both a square and a regular octagon a r e inscr ibed i n the same c i r c l e , which has the grea ter apothem? the g rea te r perimeter?

(b) Answer the same questions f o r circumscribed f igures .

10. From what formula r e l a t i n g t o regular polygons i s the formula f o r the area of a c i r c l e derived?

11. A wheel has a 20 inch diameter. How many revolutions w i l l it make I n going 100 f e e t ?

12 . The angle of a sector i s 10' and i t s radius i s 12 inches. Find the area of the sector and t h e length of i t s a rc .

13. Prove t h a t the area of an equ i l a t e ra l t r i a n g l e circumscribed about a c i r c l e i s four times the a rea of an equ i l a t e ra l t r i ang le inscribed i n the c i r c l e .

1 4 . I n the f igure , %?is tangent + t o the c i r c l e at D and CA

E

bisec ts /BCD . I f & = 88 and rnm = 62 , f ind the measure of each a rc and each angle indicated i n the f igure .

F

The dis tance between the centers of two c i r c l e s having r a d i i of 7 and 9 i s 20 . Find the

length of the common in te rna l tangent-segment.

Given Inscribed quadr i la te ra l ABCD with diagonals i n t e r - sect ing a t P *

Prove :

( a ) AAPD - ABPC . (b) AP PC = PD PB .

2 Given the c i r c l e C = [(x,y): x + y2 = 16) . ( a ) Are points A(4,0) , ~ ( 0 , 4 ) on the c i r c l e ?

Find the slope of (b) Find the midpoint of 7?B . ( c ) Write an equation of the l i n e containing the -

midpoint of AB and perpendicular t o . (d) Does the l i n e of Part ( c ) contain the or igin?

What theorem does t h i s i l l u s t r a t e ?

( e ) Find the point of in tersec t ion of the perpendicular - bisector of AB with C . T h i s point i s the - of AB .

Given: I n the f igure, P

i s the center of the c i r c l e , and m /AEP = m /DEP .

REVIEW PROBLEMS

Chapters 10-12

Write (+) i f the statement i s t r u e and (0) i f it i s f a l s e . Be able t o explain why you mark each statement t rue o r f a l s e .

There can be a region which i s completely surrounded by a polygonal-region and which does not contain a point of the polygonal-region.

I f a polygon i s equi la tera1 , i t must be a regular polygon.

No polygon can be a convex s e t , but some polygons can

enclose a convex s e t .

Each i n t e r i o r angle of a regular pentagon has measure 72 . Every polygonal-region I s e i t h e r a t r iangular-region o r the union of two o r more coplanar tr iangular-regions.

The sum of the measures of the face angles of a polyhedral angle can equal 360 . The number of the diagonals from a given vertex of a convex polygon i s equal t o the number of s ides of the

polygon.

An ex te r io r angle of a regular polygon i s congruent t o the cen t ra l angle of the polygon.

The sum of the measures of the i n t e r i o r angles of a convex polygon of n s ides i s (n -2)18o . The sum of the measures of the e x t e r i o r angles of a convex polygon, considering one a t each vertex, i s equal t o the sum of the measures of four r i g h t angles.

If a l i n e i n t e r s e c t s a c i r c l e i n one point, it In te r sec t s the c i r c l e i n two points.

It i s poss ib le f o r two t r i ang les t o be congruent and t o be the boundaries of t r iangular regions with d i f fe ren t areas.

The area of a r igh t t r iangular region i s one-half the product of the length of the hypotenuse and the length of the shor ter leg.

14. The area of the i n t e r i o r of a parallelogram is the product of the lengths of any two consecutive s ides .

15. If ~ ( 0 , 0 ) and ~ ( o , 6 ) a r e the endpoints of a diameter o f a c i r c l e , then ~ ( 3 . 3 ) i s a point on the c i r c l e .

16. If two parallelograms have congruent a l t i t u d e s , the areas of t h e i r i n t e r i o r s a r e proportional t o the lengths of t h e i r bases. . .

17. If a plane i n t e r s e c t s a sphere i n a t l e a s t two points, . . t he in te r sec t ion i s a l i n e .

18. If a sphere and a c i r c l e have the same center and i f they in te r sec t , thei5"'the intersecti 'on i s a c i r c l e .

19. I f a l i n e i s tangent t o a c i r c l e , it i s perpendicular t o the plane of the c i r c l e .

20. A l i n e which i s perpendicular t o and b i sec t s a chord of a c i r c l e contains the center of the c i r c l e .

2 21. the s e t ((x,y,z): x + y2 + z2 = 9) i s a sphere with center a t the o r ig in and w i t h a radius equal t o 9 .

Ã 2 22. If C = [ (x, y): x + y2 = 25) , then the l i n e

[(x,y): y = 5 ) i s tangent t o C . 23. The distance between the point A(1,2,3) and the point

~ ( 0 , - 4 , - 1 ) i s </53" . 24. Any quadr i la te ra l can be inscr ibed i n some c i r c l e .

25. If the l a t e r a l edge of a prism i s congruent t o the a l t i t u d e of the prism, the prism i s a r i g h t prism.

26. Any two grea t c i r c l e s of a sphere in te r sec t .

27. Two tangent c i r c l e s a r e ex terna l ly tangent only i f t h e i r centers l i e on opposite s ides of each common tangent l ine .

28. The point of tangency of two tangent c i r c l e s i s col l inear wi th the centers of the two c i r c l e s .

29. The areas of two similar polygons a re proportional t o the squares of the lengths of any two corresponding sides.

30. The apothem of a regular hexagon of side s i s equal t o the a l t i t u d e of an equ i l a t e ra l t r i ang le of s ide s .

The radius of a c i r c l e i s congruent t o the median on the hypotenuse of a r i g h t t r i ang le inscribed i n the c i r c l e .

I n a c i r c l e of radius 12 , an inscr ibed angle of 135' in tercepts an a r c of measure 9-n- . A l i n e can in te r sec t a sphere i n exact ly one point.

Two planes tangent t o the same sphere must i n t e r s e c t . . . .- If two chords in te r sec t within a c i r c l e , the difference

i n the lengths of the segments of one chord i s equal t o the difference i n the lengths of the segments of the other .

If the l a t e r a l edge of a paral le lepiped whose base i s a !' ' .

square i s congruent t o a s ide of the square base, then the parallelepiped Is a cube.

I f a l i n e in te r sec t s the ex te r io r of a sphere, then It must in te r sec t the sphere.

' ^'1

Concentric c i r c l e s have concurrent diameters.

A sphere has radius 5 . I f a plane 3 u n i t s from the center in te r sec t s the sphere i n a c i r c l e , the radius of t h i s c i r c l e Is 4 . The plane [(x,y,z) : z = 7) i n t e r s e c t s the sphere

2 2 ((x.y,s): x + y2 + z = 16) i n a c i r c l e . 2 2 Point (2.3) l i e s on the c i r c l e ( (x ,y ) : x + y = 5) .

The product of a secant-segment and i t s external secant- segment i s constant f o r any given c i r c l e and ex te r io r

point.

A pyramid i s a regular pyramid i f the foot of the perpendicular from i t s vertex i s one of the ve r t i ces of the case.

1 The formula A = p p i s the formula f o r the l a t e r a l a rea of any pyramid.

The degree measure of a minor a rc i s the same a s the measure of the cen t ra l angle which in tercepts i t .

A l l a r c s with the same degree measure have t h e same length i f they a re a l l of the same c i r c l e .

Every c i r c l e forms with i t s i n t e r i o r a polygonal-region.

The measure of a tangent-chord angle i s one-half the measure of i t s intercepted arc .

2 An arc whose degree measure i s 120 has length -XT i f the length of i t s radius i s one.

2 Line y = 2x i n t e r s e c t s the c i r c l e x + y2 = 100 in the points with coordinates ( 2 -/5 , 4 i /^ ) and

(-2 1/5 9 -4^) 2

(x - 2) + (y + 5)2 = 25 i s a c i r c l e with center a t

(295)

I f the measure of the radius of a c i r c l e i s one, i t s circumference i s 27 and i t s area i s TT . The median of an i sosce les t r i ang le i s pa ra l l e l t o the base.

If two chords of a c i r c l e b isec t each other , then each must be a diameter of the c i r c l e .

The quadr i la te ra l w i t h ve r t i ces (0,0) , (4,0) , (4,4) , (0,4) i s a square.

In a sphere the planes of great c i r c l e s a re pa ra l l e l .

A unit-square must have a measure of one inch f o r each s ide.

Angles inscr ibed i n the same a r c a re congruent.

The area of a c i r c l e i s smaller than the area of any regular polygon i n which the c i r c l e i s inscribed.

The area of a trapezoidal-region i s the product of the lengths of i t s a l t i t u d e and i t s median.

A rectangular prism whose bases a re squares, and each of whose l a t e r a l edges i s twice as long a s the s ide of the base, has a t o t a l surface area which i s ten times the

area of a base. 916

If two triangles have bases of the same length and altitudes of the same length, they have the same area and are similar polygons.

If a circle is inscribed in a parallelogram, then the parallelogram must be equilateral.

If the degree measure of one arc of a circle is twice the degree measure of a second arc, then the chord

associated with the first arc is twice as long as the chord associated with the second arc.

If a set of points is a quadrilateral, then the points are coplanar.

It is possible for the incenter, the circumcenter', and the orthocenter of a triangle to be the center of the same circle.

The square of the length of a tangent-segment from a given exterior point is equal to the product of the lengths of any secant-segment from that point and the length of its external segment.

The points of a circle are said to be collinear.

If the circumference of a circle is 12-n- , then the area is 1 4 4 ~

The area of a sector of radius r whose arc has a degree 2 measure q is rr

The point which is equally distant from all three sides of a triangle, is the Intersection of the midrays of the angles of the triangle.

The circumference of a circle is the limit of the perimeters of the circumscribed regular polygons.

If two parallelograms have the lengths of the base and altitude of one proportional to the lengths of the base and altitude of the other, the parallelograms are similar.

A quadrilateral whose vertices have xy-coordinates (0,0) , (3,0) , (4,2) , (1,2) is a parallelogram.

All bisectors of the interior angles of a regular polygon

intersect In a single point.

The perpendicular bisectors of the sides of a triangle are concurrent.

A triangle has one and only one circumscribed circle.

The center of a circle circumscribed about a right triangle is not in the interior of the triangle.

The least number of faces a polyhedron can have is three.

If the perimeter of a regular hexagon is p , then the radius of the inscribei ci~cle is ,q . There are only five types of regular polyhedrons.

The point (4,-3) is a point of a circle whose center is the point (0,0) and whose radius is 4 . The radii of two circles are 3 and 5 ; the radius of a circle whose area is equal to the sum of the areas of these circles is 1/3 . The face angles of a trihedral angle may have measures 120, 2.5, and 9 0 .

86. A regular polyhedron having as many faces as there are months in the year is called a dodecahedron.

2

87. If AB = C D , then (A,B) A (cTD) . 88. The addition of directed line segments is commutative.

89. Vector addition is commutative.

90. The vector [6,31 determines a unique directed line segment.

91. The vector [6,31 is the additive inverse of [-6,-31 . 92. If a"# ? , then la cannot equal 121 .

& - A & A A A A

93. If a + b = b + a , we can conclude a - b = b - a . 94 The scalar product of two vectors is commutative.

95. The scalar product of two vectors is not zero if the vectors are perpendicular.

A 96. The o r ig in of (A,B) i s A .

2

97. The terminusof (A,B) i s B . A A 3 A 2 A

98. If (A,B) + ( c ,D) = (A,x) and (c,D) + (A,B) = (c,Y) 2 2

then ( A , x ) A (c,Y) . 2

99. I f (A,B) = (a) then A = C . 2

100. If (A,B) is not equivalent t o (c ,D) , then the vectors

they represent cannot be equal.

Appendix V

HOW TO DRAW PICTURES OF SPACE FIGURES

Simple Drawing.

A course in mechanical drawing is concerned with precise

representation of physical objects seen from different positions

in space. In geometry we are concerned with drawing only to the

extent that we use sketches to help us do mathematical thinking.

There is no one correct way to draw pictures in geometry, but

there are some techniques helpful enough to be in rather general

use. Here, for example, is a technically

correct drawing of an ordinary pyramid,

for a person can argue that he is looking

at the pyramid from directly above. But

careful ruler drawing is not as helpful

as this very crude free-hand sketch. The

first drawing does not suggest 3-space;

the second one does.

The first part of this discussion offers suggestions for

simple ways to draw 3-space figures. The second part introduces

the more elaborate technique of drawing from perspective. The

difference between the two approaches is suggested by these two

drawings of a rectangular box.

In the first drawing the base is shown by an easy-to-draw parallelogram. In the second drawing, the front base edge and the

back base edge are parallel, but the back base edge is drawn

shorter under the belief that the shorter length will suggest I1 more remote".

No matter how a rectangular box i s drawn, some sac r i f i ces must be made. A l l angles of a rectangular s o l i d a r e r ight angles,

- but i n each of the drawings shown on the previous page two- t h i r d s of the angles do not come close t o indicat ing ninety degrees when measured with a protractor . We a r e wi l l ing t o give up the drawing of r i g h t angles tha t look l i k e right angles i n order t h a t we make the f igure a s a whole more suggestive.

You already know t h a t a plane i s generally pictured by a parallelogram.

It seems reasonable t o draw a horizontal plane i n e i t h e r LIZ7 r 3 of the ways shown, and t o draw a v e r t i c a l plane l i k e t h i s .

0 I f we want t o indicate two p a r a l l e l planes, however, we can not be e f fec t ive i f we just draw any two "horizontal" planes. Notice how the drawing t o the r i g h t below improves upon the one t o the l e f t . Perhaps you prefer s t i l l another kind of drawing.

Various devices are used t o indica te t h a t one part of a f igure passes behind another par t . Sometimes a hidden part i s simply omitted, sometimes i t i s indicated by dotted l ines . Thus, a l i n e piercing a plane may be drawn i n e i t h e r of the two ways:

Two intersecting planes are illustrated by each of these

drawings.

The second is better than the first because the line of intersec-

tion is shown and parts concealed from view are dotted* The

third and fourth drawings are better yet because the line of

intersection is visually tied in with plane @ as well as planeQ by the use of parallel lines in the drawing.

Here is a drawing which has the advantage

of simplicity and the disadvantage of

suggesting one plane and one halfplane*

In any case a line of intersection is a particularly important

part of a figure.

Suppose that we wish to draw two intersecting planes each

perpendicular to a third plane* An effyctive procedure is shown $

by this step-by-step development.

Notice how the last two planes drawn are built on the line of

intersection. A complete drawing showing all the hidden lines is

just too involved to handle pleasantly. The picture below is

much more suggestive*

A dime, from different angles, looks like this:

Neither the first nor the last is a good picture of a circle in 3-space. .Either of the others 1 s satisfactory. The thinner oval

is perhaps better to use to represent the base of a conem

Certainly nobody should expect us to interpret the figure shown

A few additional drawings, with verbal descriptions, are shown.

A line parallel to a planem

A cylinder cut by a plane parallel to the base.

A cylinder cut by a plane not parallel to the base.

A pyramid cut by a plane pa ra l l e l t o the base.

It is important t o remember tha t a drawing i s not an end i n i t s e l f but simply an a i d t o our understanding of the geometrical s i tuat ion. We should choose the kind of p ic ture t h a t w i l l serve us best f o r t h i s purpose3 and one person's choice may be different from another.

Perspective-

The rays a b , c d , e f i n the left-hand f igure below suggest coplanar l i n e s in tersec t ing a t V ; t he corresponding rays i n the right-hand f igure suggest p a r a l l e l l i n e s i n a three-dimensional drawing. Think of a ra l l road t r ack and t e l e - phone poles as you look a t the right-hand figure.

a c

0 d

The right-hand f igure suggests ce r t a in pr inc ip les which a r e use- f u l i n making perspective drawings-

(1) A s e t of p a r a l l e l l i n e s which recede from the viewer a re drawn a s concurrent rays; f o r example, rays a b c d e f.

The point on the drawing where the rays meet is known as t h e "vanishing point" .

( 2 ) Congruent segments a re drawn smaller when they a re fu r the r from the viewer. (Find examples i n the drawing.)

( 3 ) Para l l e l l$nes which a re perpendicular t o the l i n e of

s ight of the viewer a re shown as p a r a l l e l l i n e s In the drawing. (Find examples i n the d~awing.)

A person does not need much a r t i s t i c a b i l i t y t o make use of these three pr inc ip les .

The s teps t o follow i n sketching a rectangular so l id are shown below.

1

maw the f ront face as a ~ e c ~ a n g l e .

Select a vanishing point and a:>aw segments from it t o the ver t ices . O m i t segments t h a t cannot be seen.

L ,

maw edges p a r a l l e l t o those of the f ront face. F ina l ly erase l i n e s of perspectfve.

Under t h i s technique a s ingle horizontal plane can be drawn

a s the top face of t h e s o l l d shown above.

A s lngle v e r t i c a l plane can be represented by the front face or the right-hand face of the so l id .

After this brief account of two approaches to the drawing of figures in 3-space we should once again recognize the fact that there Is no one correct way to picture geometric ideas. However,

the more "real" we want our picture to appear, the more attention we should pay to perspective. Such an artist as Leonard0 da Vinci paid great attention to perspective. Most of us find this done for us when we use ordinary cameras.

I t See some books on drawing or look up perspective" In an encyclopedia if you are Interested in a detailed treatment.

Aopend ix V I I

SURFACE AREA A N D VOLUME

I n t r o d u c t i o n .

I n y o u r s t u d y o f i n f o r m a l g e o m e t r y you l e a r n e d f o r m u l a s f o r

f i n s i n g s u r f a c e a r e a s and volumes o f f a m i l i a r f i g u r e s . You may

r e c a l l t h a t a s o h e r e o f r a d i u s r h a s vo lume ^ 3 3 ~ ; , f o r i n -

s t a n c e , and t h a t i t s s u r f a c e a r e a i s g i v e n b y ^-frr . T h i s

c o u r s e i n g e o m e t r y t h a t you h a v e j u s t c o m p l e t e d c o v e r e d f o r m a l l y mos t o f t h e o t h e r t o p i c s t h a t you met i n i n f o r m a l g e o m e t r y ,

and you may wonder why t o p i c s o f s u r f a c e a r e a and volume were o m i t t e d . The r e a s o n h a s t o d o w i t h a b r a n c h o f h i g h e r mathema-

I I t i c s known a s " i n t e g r a l c a l c u l u s . U n t i l t h e i n t e g r a l c a l c u l u s was i n v e n t e d , i n t h e s e v e n t e e n t h c e n t u r y , t h e s t u d y o f a r e a

and volume was s k e t c h y . T h e r e w e r e n o s a t i s f a c t o r y d e f i n i t i o n s

o f a r e a and volume and no s y s t e m a t i c ways o f f i n d i n g them.

The s u b j e c t c o n s i s t e d o f t h e d i s c o v e r y and s t u d y o f f o r m u l a s

f o r t h e a r e a s and vo lumes o f i n d i v i d u a l f i g u r e s . Moreove r

t h e d e r i v a t i o n o f t h e s e f o r m u l a s , i n p r e - c a l c u l u s g e o m e t r y , a r e a l m o s t w i t h o u t e x c e p t i o n e i t h e r v e r y I o n s and d i f f i c u l t

t o f o l l o w , o r l o g i c a l l y unsound . It seemed u n f a i r t o i n f l i c t t h i s k i n a o f s t u d y on t h e h i g h s c h o o l s t u d e n t when ( a ) t h e r e

a r e p a r t s o f g e o m e t r y f r o m wh ich h e c o u l d p r o f i t more and ( b ) h e w i l l see a s u i t a b l e d e v e l o p m e n t o f t h i s s u b j e c t when

( a n d i f ) he s t u d i e s c a l c u l u s .

S u r f a c e Area . - We s h a l l d i s c u s s s u r f a c e a r e a i n a n i n f o r m a l way, more a s

t h o u g h we were t a l k i n g a b o u t p h y s i c a l o b j e c t s t h a n m a t h e m a t i c a l

o n e s .

I t i s e a s y t o f i n d t h e s u r f a c e a r e a o f s o l i d s s u c h a s

p r i s m s and p y r a m i d s , b e c a u s e t h e i r s u r f a c e s a r e composed o f p l a n e f i g u r e s , n a m e l y p o l y g o n s . T h e r e a r e a few o t h e r t y p e s

o f s o l i d s , whose s u r f a c e s a r e n o t made u p o f p l a n e f i g u r e s ,

whose s u r f a c e a r e a c a n b e f o u n d by f i n d i n g a n e q u i v a l e n t p l a n e

f i g u r e .

T h e c y l i n d e r and t h e c o n e a r e s u c h f i g u r e s . If you imag ine

s l i t t i n g a c y l i n d e r and u n r o l l i n g i t s s u r f a c e o n t o a p l a n e ,

what f i g u r e d o you t h i n k i s o b t a i n e d ?

Can you see t h a t I t i s a r e c t a n g l e whose a l t i t u d e i s t h e a l -

t i t u d e o f t h e c y l i n d e r and whose b a s e i s t h e c i r c u m f e r e n c e o f t h e b a s e o f t h e c y l i n d e r ? I f r i s t h e r a d i u s o f t h e b a s e o f

a c y l i n d e r and h i s its h e i g h t , t h e n t h e s u r f a c e a r e a o f t h e

c y l i n d e r ( c a l l e d i t s l a t e r a l s u r f a c e ) i s e q u a l t o t h e a r e a

o f t h e r e c t a n g l e o b t a i n e d by u n r o l l i n g t h e c y l i n d e r . Thus , i t s l a t e r a l s u r f a c e i s 21rrh.

Cones c a n b e t r e a t e d I n a s i m i l a r way. If a cone i s s l i t

p r o p e r l y , i t c a n b e u n r o l l e d t o l i e f l a t .

Can you see t h a t t h e u n r o l l e d c o n e i s a s e c t o r o f a c i r c l e ?

T h i s p r o c e d u r e r e d u c e s t h e problem o f f i n d i n g t h e s u r f a c e a r e a

of a c o n e to t h a t o f f i n d i n g t h e a r e a o f a s e c t o r .

T h i s l a t t e r p r o b l e m h a s a l r e a d y b e e n s o l v e d �Ã„ he or e 1 2 - 2 0 ) .

A l l we need t o know i n a n y g i v e n c a s e i s t h e r a d i u s of t h e

c i r c l e and t h e l e n g t h o f t h e i n t e r c e p t e d a r c . I n t h e c a s e o f a s e c t o r o b t a i n e d by u n r o l l i n g a c o n e , t h e r a d i u s o f t h e

s e c t o r is t h e s l a n t h e i g h t o f t h e c o n e and t h e l e n g t h o f t h e i n t e r c e p t e d a r c i s t h e c i r c u m f e r e n c e o f t h e b a s e o f t h e c o n e .

T h e r e f o r e , t h e f o r m u l a f o r t h e l a t e r a l s u r f a c e of a c o n e i s J! C , whe re is i t s s l a n t h e i g h t and C i s t h e c i r c u m - 2

f e r e n c e o f i t s b a s e . A n o t h e r i m p o r t a n t s u r f a c e f o r w h i c h 2 t h e r e i s a n a r e a f o r m u l a is t h e s p h e r e . The f o r m u l a i s h r ,

where r i s t h e r a d i u s of t h e s p h e r e .

I t i s n a t u r a l t o t r y t o d e r i v e t h i s f o r m u l a by s l i t t i n g t h e s p h e r e and u n r o l l i n g i t o n t o some p l a n e f i g u r e . However,

i t h a s been p roved i n h i g h e r m a t h e m a t i c s t h a t i t i s i m p o s s i b l e

t o f l a t t e n o u t t h e s p h e r e i n t h i s way. (Can you see a n y con-

n e c t i o n be tween t h i s s t a t e m e n t and t h e f a c t t h a t maps o f l a r g e p o r t i o n s of t h e e a r t h h a v e t o d i s t o r t t h e s h a p e s o f t h e r e g i o n s

wh ich t h e y d e p i c t ' : ' )

T h e r e i s a s i m p l e e x p e r i m e n t w h i c h you c a n p e r f o r m t o tes t

t h e f o r m u l a 4 i r 2 f o r t h e s u r f a c e a r e a o f a s p h e r e . Wind a s t r i n g a r o u n d a s p h e r e , i n a s p i r a l , u n t i l a n e n t i r e h e m i s p h e r e

i s c o v e r e d , and m e a s u r e t h e l e n g t h of t h e s t r i n g r e q u i r e d . Then

wind a s t r i n g ( o f t h e same d i a m e t e r ) i n a p l a n a r s p i r a l u n t i l

i t j u s t c o v e r s t h e i n s i d e o f a c i r c l e o f t h e same r a d i u s a s t h e g i v e n s p h e r e , and measure t h e l e n g t h o f t h e s t r i n g r e q u i r e d .

If y o u r work i s a c c u r a t e t h e l e n g t h o f s t r i n g r e q u i r e d t o c o v e r t h e h e m i s p h e r e w i l l be t w i c e t h e l e n g t h r e q u i r e d t o c o v e r t h e

2 c i r c l e . S i n c e t h e a r e a o f t h e c i r c l e i s irr , a hemisphere 2 s h o u l d have s u r f a c e a r e a 27rr , and t h e whole s p h e r e s h o u l d

2 h a v e s u r f a c e a r e a 4 ~ r r . A more s o p h i s t i c a t e d a p p r o a c h t o d e r i v i n g t h e f o r m u l a f o r

t h e s u r f a c e a r e a o f a s p h e r e i s t o a p p r o x i m a t e t h e s u r f a c e o f a s p h e r e by r e v o l v i n g s u i t a b l e c h o r d s a r o u n d a d i a m e t e r , a s shown below.

T h i s makes i t p o s s i b l e t o f i n d t h e s u r f a c e o f a s p h e r e , ap- p r o x i m a t e l y , i n terms o f s u r f a c e s o f p o r t i o n s o f c o n e s . A

l i m i t i n g p r o c e s s t h e n y i e l d s t h e f o r m u l a Wr2 for- t h e s u r f a c e a r e a o f a s p h e r e .

P o s s i b l e D e f i n i t i o n s - o f Volume.

It i s a s t r a n g e f a c t t h a t m a t h e m a t i c i a n s d i s c o v e r e d f o r - mulas f o r t h e volume o f many f i g u r e s l o n g b e f o r e t h e y knew

what volume was, o r a t l e a s t b e f o r e t h e y had a f o r m a l d e f i n i - t i o n o f volume. One way o f u n d e r s t a n d i n g t h i s i s t o o b s e r v e

t h a t t h e y had some g e n e r a l n o t i o n s a s t o what s h o u l d b e t r u e a b o u t volumes, which , i n t h e c a s e of some of t h e s i m ~ l e r f i g -

u r e s , were s u f f i c i e n t t o l e a d t o d e f i n i t e f o r m u l a s .

L e t u s draw up a l i s t o f some r e q u i r e m e n t s t h a t are

r e a s o n a b l e t o impose on a n y p o s s i b l e d e f i n i t i o n o f volume.

1. It i s r e a s o n a b l e t o e x p e c t t h a t t h e volume o f a s o l i d s h o u l d be a n o n - n e g a t i v e r e a l number.

2. I t i s r e a s o n a b l e t o e x p e c t t h a t i f a s o l i d i s p a r t i t i o n e d i n t o s e v e r a l p a r t s and t h a t i f e a c h of t h e s o l i d s i n v o l v e d h a s a volume, t h e n t h e volume of t h e o r i g i n a l s o l i d s h o u l d e q u a l t h e sum of t h e volumes of t h e p a r t s .

3 . It is r e a s o n a b l e t o e x p e c t t h a t c o n g r u e n t s o l i d s s h o u l d h a v e

equa 1 volumes ,

4. It i s a n i m p o r t a n t f a c t a b o u t volume (which may o r may n o t

seem r e a s o n a b l e ) t h a t i f s o l i d S is s i m i l a r t o s o l i d S t , and i f t h e p r o p o r t i o n a l i t y f a c t o r i s k, t h e n t h e volume

o f s o l i d S t i s k3 times t h e volume o f s o l i d S. F o r i n s t a n c e , c o n s i d e r two c u b e s , S and St , s u c h t h a t t h e e d g e of S t i s t w i c e a s l o n g a s t h e e d g e o f 3 . Then S

i s s i m i l a r t o 31 and t h e p r o p o r t i o n a l i t y f a c t o r i s 2.

N o t i c e t h a t S t c a n b e ~ a r t i t i o n e d i n t o e i g h t c u b e s e a c h c o n g r u e n t t o S. Is i t n o t r e a s o n a b l e t o e x p e c t t h a t t h e

volume o f S t s h o u l d b e 2 times t h e volume o f S?

5. I t i s r e a s o n a b l e t o r e q u i r e t h a t t h e volume of a r e c t a n g u - l a r p a r a l l e l e p i p e d s h o u l d b e t h e p r o d u c t o f i t s a l t i t u d e by t h e a r e a o f i t s b a s e .

Cavalier! ' s P r i n c i p l e .

Even though we have n o t been a b l e t o d e f i n e " volume", t h e r e a r e some s i t u a t i o n s i n which we c a n r e a s o n a b l y s a y t h a t

two s o l i d s have t h e same volume. We a r e g o i n g now t o i l l u s - t r a t e o n e i m p o r t a n t c a s e o f t h i s s o r t . I t w i l l h e l p u s unaer -

s t a n d t h e c a s e i n q u e s t i o n i f we first t h i n k of a p h y s i c a l model. We c a n make a n a p p r o x i m a t e model o f a s q u a r e pyramid by f o r m i n g a s t a c k of t h i n c a r d s , c u t t o t h e p r o p e r s i z e , l i k e t h i s :

The f i g u r e on t h e l e f t r e p r e s e n t s t h e e x a c t pyramid, and t h e

f i g u r e on t h e r i g h t i s t h e a p p r o x i m a t e model made f rom c a r d s .

Now, s u p p o s e we d r i l l a narrow h o l e i n t h e model, f rom t h e

t o p t o some p o i n t of the b a s e , and i n s e r t a t h i n rod s o t h a t i t g o e s t h r o u g h e v e r y c a r d i n t h e model. T h i s e n a b l e s u s t o

t i l t t h e rod I n a n y way we want , k e e p i n g i t s bo t tom end f i x e d on t h e b a s e . Such t i l t i n g c h a n g e s t h e s h a p e o f t h e model, b u t n o t i t s volume. The r e a s o n i s t h a t i t s volume i s s i m p l y t h e t o t a l volume o f t h e c a r d s ; and t h i s t o t a l volume d o e s n o t change a s t h e c a r d s s l i d e a l o n g e a c h o t h e r .

The same p r i n c i p l e a p p l i e s more g e n e r a l l y . Suppose we h a v e two s o l i d s w i t h b a s e s i n a p l a n e which w e s h a l l t h i n k o f a s h o r i z o n t a l . If a l l h o r i z o n t a l c r o s s - s e c t i o n s o f t h e two s o l i d s a t t h e same l e v e l have t h e same a r e a , t h e n t h e two s o l i d s have t h e same volume. To see t h i s , o b s e r v e t h a t i f we make a c a r d model o f e a c h o f t h e s o l i d s , t h e n e a c h c a r d i n t h e f i r s t

model h a s e x a c t l y t h e same volume a s t h e c o r r e s p o n d i n g c a r d i n t h e second model .

T h e r e f o r e , t h e volumes o f t h e two m o d e l s o u g h t t o b e t h e

same. The a p p r o x i m a t i o n g i v e n by t h e m o d e l s i s a s c l o s e a s we p l e a s e , if o n l y t h e c a r d s a r e t h i n enough . T h e r e f o r e , t h e

vo lumes

of t h e two s o l i d s t h a t we s t a r t e d w i t h o u g h t t o be t h e same.

The p r i n c i p l e i n v o l v e d h e r e i s c a l l e d C a v a l i e r i t s P r i n c i p l e .

We h a v e n o t p roved i t ; we h a v e m e r e l y b e e n e x p l a i n i n g why i t i s

r e a s o n a b l e . L e t u s s t a t e i t e x p l i c i t l y .

C a v a l i e r i l s P r i n c i p l e : G iven two s o l i d s and a p l a n e .

If , f o r e v e r y p l a n e w h i c h i n t e r s e c t s t h e s o l i d s and

i s p a r a l l e l t o t h e g i v e n p l a n e , t h e two i n t e r s e c t i o n s

h a v e e q u a l a r e a s , t h e n t h e two s o l i d s h a v e e q u a l

volumes.

C a v a l i e r i l s P r i n c i p l e c a n b e used a s a k e y t o t h e c a l c u l a -

t i o n o f vo lumes , a s we s h a l l see i n t h e n e x t s e c t i o n .

P r i s m s - and Pyramids .

The f o r m u l a f o r t h e volume o f a r e c t a n g u l a r p a r a l l e l e p i p e d

a l s o a p p l i e s t o g e n e r a l p a r a l l e l e p i p e d s .

T h i s c a n be s e e n by u s i n g C a v a l i e r l l s P r i - n c i p l e t o compare t h e vo lume o f a p a r a l l e l e p i p e d w i t h t h a t o f t h e a p p r o p r i a t e r e c - ' - . angu la r p a r a l l e l e p i p e a .

T h e volume o f a n y p r i s m i s t h e p r o d u c t o f i t s a l t i t u d e and

t h e a r e a o f i t s b a s e .

T h e volume o f a pyramid i s g i v e n b y hB w h e r e h i s i t s -3 a l t i t u d e a n d B i s t h e a r e a o f i t s base. N o t i c e t h e o c c u r -

1 r e n c e of t h e f a c t o r -7 i n t h i s f o r m u l a . P e r h a p s i t r e m i n d s 1 1 you o f t h e f a c t o r - w h i c h o c c u r s i n t h e f o r m u l a 2 hb f o r 2

t h e a r e a o f a t r i a n g l e . T h e s e f a c t o r s a r e i n d e e d a n a l o g o u s , 2nd

we now t r y t o show how. I n t h e d e r i v a t i o n of t h e f o r m u l a f o r

t h e a r e a o f a t r i a n g l e o f a l t i t u d e h and b a s e b a n

a u x i l i a r y p a r a l l e l o g r a m was i n t r o d u c e d , a l s o h a v i n g a l t i t u d e

h and b a s e b, a n d w h i c h c o u l d be d i s s e c t e d i n t o two t r i a n g l e s ,

e a c h c o n g r u e n t t o t h e o r i g i n a l o n e . S i n c e t h e p a r a l l e l o g r a m 1 was known t o h a v e a r e a hb, t h e f o r m u l a 7 h b was r e a d i l y a e -

duced . I n t h e c a s e o f a t r i a n g u l a r pyramid w i t h a l t i t u d e h

and b a s e a r e a B, t h e a u x i l i a r y f i g u r e i s a t r i a n g u l a r p r i s m

w i t h t h e same b a s e and a l t i t u d e , and w h i c h c a n be d i s s e c t e a

i n t o t h r e e t r i a n g u l a r p y r a m i d s , e a c h h a v i n g t h e same volume a s

t h e g i v e n one . S i n c e t h e p r i s m h a s volume hB, t h e f o r m u l a

hB i s r e a d i l y d e d u c e d . 3 The f o r m u l a f o r the volume o f a n y p y r a m i d , n o t n e c e s s a r i l y

hB. It can be d e r i v e d by d i s s e c t i n g t h e t r iangular , I s also - 3 g i v e n p y r a m i d i n t o t r i a n g u l a r p y r a m i d s a n d o b s e r v i n g tha t t h e

volume o f the o r i g i n a l f igure i s the sum o f t h e v o l u m e s of the

a u x i l i a r y p y r a m i d s .

C y l i n d e r s . , - - Cones a n d S p h e r e s .

One way of f i n d i n g t h e volume o f a c y l i n d e r i s t o i n t r o d u c e

a s u i t a b l e p r i s m and u s e C a v a l i e r i t s P r i n c i p l e .

-

-area of cross-section is B~

By r e f e r r i n g t o a p r i s m h a v i n g t h e same a l t i t u d e h and c r o s s - s e c t i o n a l a r e a B a s o u r c y l i n d e r , w e see t h a t t h e volume o f

t h e cyl inder i s hB.

One way o f f i n d i n g c h e volume o f a c o n e i s t o i n t r o d u c e

2 s u i t a b l e pyramid and u s e C a v a l i e r i l s P r i n c i p l e .

By r e f e r r i n g t o a pyramid h a v i n g t h e same a l t i t u d e h a s o u r

c o n e , and e q u a l c r o s s - s e c t i o n a l a r e a a t c o r r e s p o n d i n g l e v e l s , 1 we c a n i n f e r t h a t t h e volume o f t h e c o n e i s -3 KB, where B

i s t h e a r e a o f i t s b a s e .

A n o t h e r i m p o r t a n t s o l i d f o r w h i c h t h e r e i s a volume f o r m u l a i s t h e i n t e r i o r o f a s p h e r e . T h e volume o f s u c h a r e g i o n i s 4 3 -5- Trr , w h e r e r i s t h e r a d i u s o f t h e s p h e r e . L e t u s s e e i f

we c a n f i n d some j u s t i f i c a t i o n f o r t h i s s t a t e m e n t .

If p l a n e s a r e drawn t h r o u g h t h e c e n t e r o f t h e s p h e r e , t h e

s p h e r e i s p a r t i t i o n e d i n t o s o l i d s wh ich a r e v e r y much l i k e

p y r a m i d s . T h e s e s o l i d s h a v e c u r v e d b a s e s , s o t h e f o r m u l a w e

h a v e f o r t h e volume o f a pyramid o u g h t n o t b e used f o r f i n d i n g

t h e i r vo lumes . However , i f enough p l a n e s a r e drawn, t h e b a s e

o f a n y o n e o f t h e s e s o l i d s i s a l m o s t f l a t , i t s a l t i t u d e i s

a l m o s t e q u a l t o t h e r a d i u s of t h e s p h e r e , s o it i s n o t u n r e a s o n -

a b l e t o b e l i e v e t h a t i t s volume i s g i v e n b y

w h e r e B i s the s u r f a c e area of i t s b a s e . T h e r e f o r e , t h e t o t a l

vo lume o f t h e s p h e r e a p p e a r s t o b e t h e sum o f a l l these vo lumes . 1

A l i t t l e algebraic m a n i p u l a t i o n shows t ha t t h e i r sum i s J.

7 times the sum o f a l l the areas B. Thus , t he volume of the

1 s p h e r e a p p e a r s t o b e -rS, where S i s t h e s u r f a c e area o f t h e

s p h e r e .

'r 3 S i n c e S is krr * , i t t h e r e f o r e a p p e a r s t h a t Â¥n- i s

the co r r ec t f o r m u l a .

Append ix V I I I

HOW ERATOSTHENES MEASURED THE EARTH

The c i r c u m f e r e n c e o f t h e e a r t h , a t t h e e q u a t o r , i s a b o u t

40 ,000 k i l o m e t e r s , o r a b o u t 2 4 , 9 0 0 miles. C h r i s t o p h e r Colurn-

b u s a p p e a r s t o h a v e t h o u g h t t h a t t h e e a r t h was much s m a l l e r

t h a n t h i s . A t a n y r a t e , t h e West I n d i e s g o t t h e i r name, be-

c a u s e when Columbus r e a c h e d them, h e t h o u g h t t h a t h e was

a l r e a d y i n I n d i a . H i s m a r g i n o f e r r o r , t h e r e f o r e , was

somewhat g r e a t e r t h a n t h e w i d t h of t h e P a c i f i c Ocean.

I n t h e t h i r d c e n t u r y B.C. , however , t h e c i r c u m f e r e n c e of

t h e e a r t h was m e a s u r e d , by a G r e e k m a t h e m a t i c i a n , w i t h a n e r r o r o f o n l y o n e o r two p e r c e n t . The man was E r a t o s t h e n e s ,

and h i s method was a s f o l l o w s :

It was o b s e r v e d t h a t a t A s s u a n o n t h e Nile, a t noon on

;he Summer S o l s t i c e , t h e s u n was e x a c t l y o v e r h e a d . T h a t i s ,

a t noon o f t h i s p a r t i c u l a r d a y , a v e r t i c a l p o l e c a s t n o

shadow a t a l l , and t h e b o t t o m o f a d e e p w e l l was c o m p l e t e l y

l i t up.

I n t h e f i g u r e , C i s t h e c e n t e r of t h e e a r t h . A t noon on

t h e Summer S o l s t i c e , i n A l e x a n d r i a , E r a t o s t h e n e s measu red t h e

a n g l e marked - a on t h e f i g u r e , t h a t is , t h e a n g l e be tween a

v e r t i c a l p o l e and t h e l i n e o f i t s shadow. He f o u n d t h a t t h i s 0 1 a n g l e was a b o u t 7 1 2 ' , o r a b o u t - o f a c o m p l e t e c i r c u m f e r -

30 e n c e .

Now, t h e s u n i s r a y s , o b s e r v e d on e a r t h , a r e v e r y c l o s e t o b e i n g p a r a l l e l . Assuming t h a t t h e y a r e a c t u a l l y p a r a l l e l , i t

f o l l o w s when t h e l i n e s L-, and L2 i n t h e f i g u r e a r e c u t by

a t r a n s v e r s a l , a l t e r n a t e i n t e r i o r a n g l e s a r e c o n g r u e n t .

T h e r e f o r e , / a / b . T h e r e f o r e , t h e d i s t a n c e f r o m Assuan t o 1 A l e x a n d r i a mus t be a b o u t - 5o o f t h e c i r c u m f e r e n c e o f t h e e a r t h .

T h e d i s t a n c e f r o m Assuan t o A l e x a n d r i a was known t o b e

a b o u t 5,000 Greek s t a o i a . ( A s t a d i u m was a n a n c i e n t u n i t o f d i s t a n c e . ) E r a t o s t h e n e s c o n c l u d e d t h a t t h e c i r c u m f e r e n c e

o f t h e e a r t h mus t be a b o u t 2 3 0 , 0 0 0 s t a a i a . C o n v e r t i n g t o

miles, a c c o r d i n g t o what a n c i e n t s o u r c e s t e l l u s a b o u t what

E r a t o s t h e n e s mean t b y a s t a d i u m , we g e t 24,662 miles.

Thus , E r a t o s t h e n e s 1 e r r o r was w e l l u n d e r two p e r c e n t .

L a t e r , h e changed h i s e s t i m a t e t o a n e v e n c l o s e r one, 232 ,000

s t a d i a , b u t nobody seems t o know on wha t b a s i s h e made t h e

c h a n g e . On t h e b a s i s o f t h e e v i d e n c e , some h i s t o r i a n s b e l i e v e

t h a t h e was n o t o n l y v e r y c l e v e r and v e r y c a r e f u l , b u t a l s o

v e r y l u c k y .

Append ix I X

RIGID MOTION

The G e n e r a l I d e a of a R i g i d Mot ion . - --- I n C h a p t e r s 5 and 1 2 we h a v e d e f i n e d c o n g r u e n c e i n

a number o f d i f f e r e n t ways, d e a l i n g w i t h v a r i o u s k i n d s of

f i g u r e s . The c o m p l e t e l i s t l o o k s l i k e t h i s :

( 1 ) CD i f t h e two s e g m e n t s h a v e t h e same l e n g t h ,

t h a t IS, if AB = CD.

( 2 ) LA 0 i f t h e two a n g l e s h a v e t h e same m e a s u r e , t n a t is, i f m LA = m LB.

( 3 ) AABC 3 A DEF i f , u n d e r t h e c o r r e s p o n d e n c e ABC-DEF, e v e r y two c o r r e s p o n d i n g s i d e s a r e c o n g r u e n t

and e v e r y two c o r r e s p o n d i n g a n g l e s a r e c o n g r u e n t .

( 4 ) Two c i r c l e s a r e c o n g r u e n t i f t h e y have t h e same

r a d i u s . - - (5 ) Two c i r c u l a r a r c s AB and CD a r e c o n g r u e n t i f

t h e c i r c l e s t h a t c o n t a i n them a r e c o n g r u e n t a n d t h e two a r c s h a v e t h e same degree m e a s u r e .

T h e i n t u i t i v e i d e a o f c o n g r u e n c e i s t h e same i n a l l f i v e o f t h e s e c a s e s . I n e a c h c a s e , two c a r d b o a r d f i g u r e s a r e c o n g r u e n t i f o n e o f them c a n be moved s o a s t o c o i n c i d e

w i t h t h e o t h e r .

A t t h e b e g i n n i n g o f o u r s t u d y of c o n g r u e n c e , t h e scheme used i n C h a p t e r s :? and 1 2 i s t h e e a s i e s t and p r o b a b l y t h e b e s t . I t i s a p i t y , however , t o h a v e f i v e d i f f e r e n t

s p e c i a l ways of d e s c r i b i n g t h e same b a s i c i d e a i n f i v e s p e c i a l c a s e s . And, i n a way, i t i s a p i t y f o r t h i s b a s i c i d e a t o b e

l i m i t e d t o t h e s e f i v e s p e c i a l c a s e s .

F o r example , a s a m a t t e r o f common s e n s e , i t i s p l a i n t h a t two

s q u a r e s , e a c h o f e d g e 1, mus t b e c o n g r u e n t i n some v a l i d s e n s e :

T h e same o u g h t t o b e t r u e f o r p a r a l l e l o g r a m s , i f c o r r e s p o n d i n g s i d e s and a n g l e s a r e c o n g r u e n t , l i k e t h i s :

I t i s p l a i n , however , t h a t n o n e o f o u r f i v e s p e c i a l d e f i n i t i o n s o f c o n g r u e n c e a p p l i e s t o e i t h e r o f t h e s e c a s e s .

I n L h i s a p w e n a i x , we s h a l l e x p l a i n t n e i a e a o f r i g i d m o t i o n . T h i s i d e a i s d e f i n e d i n e x a c t l y t h e same way, r e g a r d -

less of t h e t y p e o f f i g u r e t o wh ich we happen t o b e a p p l y i n g i t . We s h a l l show t h a t f o r s e g m e n t s , a n g l e s , t r i a n g l e s , c i r c l e s

and a r c s i t means e x a c t l y t h e same t h i n g a s c o n g r u e n c e . F i n a l l y ,

we w i l l show t h a t t h e s q u a r e s and p a r a l l e l o g r a m s i n t h e f i g u r e s

a b o v e c a n be made t o c o i n c i d e by r i g i d m o t i o n . Thus , f i r s t ,

t h e i d e a o f c o n g r u e n c e w i l l b e u n i f i e d , and s e c o n d , t h e r a n g e

of i t s a p p l i c a t i o n w i l l b e e x t e n d e d .

B e f o r e we g i v e t h e g e n e r a l d e f i n i t i o n o f a r i g i a m o t i o n ,

l e t u s l o o k a t some s i m p l e e x a m p l e s . C o n s i d e r two o p p o s i t e

s i d e s o f a r e c t a n g l e , l i k e t h i s :

The v e r t i c a l s i d e s a r e d o t t e d , b e c a u s e we w i l l n o t b e e s p e c i a l l y

c o n c e r n e d w i t h them. F o r e a c h p o i n t P, Q, . . . , o f t h e t o p e d g e l e t u s d r o p a p e r p e n d i c u l a r t o t h e b o t t o m e d g e ; and l e t

t h e f o o t o f t h e p e r p e n d i c u l a r b e P 1 , Q f . . . . Under t h i s

p r o c e d u r e , Lo e a c h p o i n t o f t h e t o p e d g e t h e r e c o r r e s p o n d s

e x a c t l y o n e p o i n t o f t h e b o t t o m edge . And c o n v e r s e l y ,

t o e a c h p o i n t o f t h e b o t t o m e d g e t h e r e c o r r e s p o n d s e x a c t l y

o n e p o i n t o f t h e t o p e d g e . We c a n ' t wri te down a l l o f t h e m a t c h i n g p a i r s

i n f i n i t e l y many e x p l a i n i n g wha t

i s wha t w e h a v e

p a i r

and e x p l a i n t h e

P-Pf , Q-Qt , . . ., b e c a u s e t h e r e a r e

of them. We c a n , however , g i v e a g e n e r a l r u l e ,

i s t o c o r r e s p o n d t o w h a t ; and i n f a c t , t h i s

done . U s u a l l y , we w i l l write down a t y p i c a l

P<Ã‘>P1 r u l e by w h i c h t h e p a i r s a r e t o b e fo rmed .

N o t i c e t h a t t h e i d e a o f a o n e - t o - o n e c o r r e s p o n d e n c e i s

e x a c t l y t h e same i n t h i s c a s e a s i t was when w e were u s i n g i t

f o r t r i a n g l e s i n C h a p t e r 5. T h e o n l y d i f f e r e n c e i s t h a t i f we a r e m a t c h i n g up t h e v e r t i c e s o f two t r i a n g l e s , w e c a n wri te down a l l o f t h e m a t c h i n g p a i r s , b e c a u s e t h e r e a r e o n l y three

o f them. (ABC-DEF means t h a t A-D, B S E and C-F.) A t p r e s e n t , we a r e t a l k i n g a b o u t e x a c t l y t h e same s o r t o f

t h i n g s , o n l y t h e r e a r e t o o many of them t o wri te down.

I t i s v e r y e a s y t o c h e c k t h a t i f P and Q a r e a n y

two p o i n t s o f t h e t o p e d g e , and Pt a n d Ql a r e t h e c o r r e s -

wonaing p o i n t s o f t h e b o t t o m e d g e , t h e n

- T h i s i s t r u e b e c a u s e t h e s e g m e n t s PQ and P t Q 1 a r e o p p o s i t e

s i d e s o f a r e c t a n s l e . We e x p r e s s t h i s f a c t by s a y i n g t h a t

c o r r e s p o n d e n c e P<Ã‘>P p r e s e r v e s d i s t a n c e s .

The c o r r e s p o n d e n c e t h a t we h a v e j u s t set u p i s o u r f i r s t

and s i m p l e s t example o f - a r i g i a m o t i o n . To b e e x a c t :

D e f i n i t i o n : G iven two f i g u r e s F and F 1 , a r i g i d

n o t i o n be tween F and F l i s a o n e - t o - o n e c o r r e s p o n d e n c e

be tween t h e p o i n t s o f F and t h e p o i n t s o f F l , p r e s e r v i n g

d i s t a n c e s .

If t h e c o r r e s p o n d e n c e P-P1 i s a r i g i d mot ion between F and F i , t h e n we s h a l l write

F eg P i . T h i s n o t a t i o n i s l i k e t h e n o t a t i o n AABC 2 AAiB 'Ci f o r c o n g r u e n c e s between t r i a n g l e s . We c a n r e a d F w F1 a s "I?

i s o m e t r i c t o FI .I1 s so metric" means " e q u a l measure.")

Problem S e t I X - 1 -- C o n s i d e r t r i a n g l e s AABC and AA'BtCl , and suppose t h a t A ABC s A AIBiCi . L e t F be t h e se t c o n s i s t i n g o f t h e v e r t i c e s o f t h e f irst t r i a n g l e , and l e t F ' be t h e set c o n s i s t i n g o f

t h e v e r t i c e s o f t h e second t r i a n g l e . Show t h a t t h e r e i s

a r i g i d mot ion

F w F 1 .

L e t F b e t h e set c o n s i s t i n g o f t h e v e r t i c e s of a s q u a r e o f edge 1, and l e t F i be t h e set c o n s i s t i n g o f t h e v e r t i c e s o f a n o t h e r s q u a r e o f e d g e 1, a s i n t h e f i g u r e a t t h e b e g i n n i n g o f t h i s Appendix. Show t h a t t h e r e i s

a r i g i d mot ion F w F ' .

( F i r s t , you h a v e t o e x p l a i n wha t c o r r e s p o n d s t o what , and second you have t o v e r i f y t h a t d i s t a n c e s a r e p r e -

s e r v e d . )

Do t h e same f o r t h e v e r t i c e s o f t h e two p a r a l l e l o g r a m s i n

t h e f i g u r e a t t h e s t a r t o f t h i s Appendix.

Show t h a t i f F c o n s i s t s o f t h r e e c o l l i n e a r p o i n t s , and F i c o n s i s t s o f t h r e e n o n - c o l l i n e a r p o i n t s , t h e n t h e r e

i s no r i g i d m o t i o n between F and F l . (What you will have t o d o i s t o assume t h a t s u c h a r i g i d mot ion e x i s t s , and t h e n show t h a t t h i s a s s u m p t i o n l e a d s t o a con t rad ic -

t i o n . )

Show t h a t t h e r e i s n e v e r a r i g i d mot ion between two segments . o f d i f f e r e n t l e n g t h s .

Show t h a t there is n e v e r a r i g i d m o t i o n between a l i n e and a n a n g l e . ( ~ i n t : Apply Problem 4.)

7. Show t h a t g i v e n a n y two r a y s , t h e r e is a r i g i d m o t i o n be -

tween them. ( ~ i n t : Use t h e P o i n t P l o t t i n g Theorem.)

3. Show t h a t t h e r e i s n e v e r a r i g i d m o t i o n be tween two c i r c l e s

o f d i f f e r e n t r a d i u s .

R i g i d Mot ion - o f Segmen t s .

Theorem I X - 1 . I f AB = CD, t h e n t h e r e i s a r i g i d m o t i o n

P r o o f : F i r s t , we need t o set u p a c o r r e s p o n d e n c e P < Ã ‘ > P - between AB and m. Then, w e need t o c h e c k t h a t d i s t a n c e s

a r e p r e s e r v e d .

By t h e R u l e r P o s t u l a t e , t h e p o i n t s o f t h e l i n e %? c a n

be g i v e n c o o r d i n a t e s i n s u c h a way t h a t A h a s c o o r d i n a t e

z e r o and B t h e p o s i t i v e c o o r d i n a t e P.B.

I n t h e f i g u r e , we have shown t y p i c a l p o i n t s P, Q w i t h

t h e i r c o o r d i n a t e s x and y.

I n t h e same way, t h e p o i n t s o f CD c a n b e g i v e n c o o r a i n -

a tes:

N o t i c e t h a t D h a s t h e c o o r d i n a t e AB, b e c a u s e CD = AB.

I t i s now p l a i n wha t r u l e we s h o u l d u s e t o s e t up t h e

c o r r e s p o n d e n c e

P-Pt - -

between t h e p o i n t s o f AB and t h e p o i n t s o f CD. The r u l e

i s t h a t P c o r r e s p o n d s - t o P t - if P - and P t h a v e t h e same --- c o o r d i n a t e . ( i n p a r t i c u l a r , A-C b e c a u s e A and C h a v e

c o o r d i n a t e z e r o , and B-D b e c a u s e B and D h a v e c o o r -

d i n a t e A B . )

I t i s e a s y t o see t h a t t h i s c o r r e s p o n d e n c e i s a r i g i d mot ion . If P-PI and Q-Qt , and t h e c o o r d i n a t e s a r e x and y , a s i n t h e f i g u r e , t h e n PQ = P ' Q ' , b e c a u s e

PQ = l y - x = P ' Q f . We t h e r e f o r e have a r i g i d mot ion - -

AB a CD,

and t h e t h e o r e m i s proved .

N o t i c e t h a t t h i s r i g i d m o t i o n between t h e two segments i s

c o m p l e t e l y d e s c r i b e d i f we e x p l a i n how t h e e n d - p o i n t s a r e t o

be maccned up. We t h e r e f o r e w i l l c a l l i t t h e r i g i d mot ion - induced & - t h e c o r r e s p o n d e n c e

A-C

B-<Ã‘ D. - Theorem IX-2. If t h e r e i s a r i g i d mot ion AB w ^H between -

two segments , t h e n AB = CD.

The p r o o f i s e a s y . h his theorem was Problem 5 i n t h e

p r e v i o u s Problem S e t . )

Problem S e t IX-2

Show t h a t t h e r e is a n o t h e r r i g i d mot ion between t h e con- - g r u e n t segments AB and my Induced by t h e c o r r e s p o n -

d e n c e A < Ã ‘ Ã

B<Ã‘>C

Show t h a t t h e r e a r e two r i g i d m o t i o n s between a segment and i tself . (one o f t h e s e , o f c o u r s e , i f t h e i d e n t i t y

c o r r e s p o n d e n c e P < Ã ‘ > P ~ u n d e r which e v e r y p o i n t c o r r e s - ponds t o i t s e l f ; t h i s i s a r i g i d mot ion b e c a u s e PQ = PQ

f o r e v e r y P and Q.)

R i g i d Motion of Rays, Anfqles - and T r i a n g l e s .

+ + Theorem I X - 3 . Given a n y two r a y s AB ana CD, t h e r e i s a r i g i d m o t i o n + +

AB W CD.

The p r o o f o f t h i s theorem i s q u i t e s i m i l a r t o t h a t of Theorem IX-1, and t h e d e t a i l s a r e l e f t t o t h e r e a d e r .

Theorem IX-4. I f L ABC Â¥ LDEF, t h e n t h e r e i s a r i g i d

mot i o n

LABC Ã‡LDE between t h e s e two a n g l e s .

P r o o f : We know t h a t t h e r e a r e r i g i d m o t i o n s

and

be tween c h e r a y s wh ich f o r m t h e s i d e s o f t h e two a n g l e s .

L e t u s a g r e e t h a t two p o i n t s P and Pt ( o r Q and Q I ) a r e

t o c o r r e s p o n d t o o n e a n o t h e r i f t h e y c o r r e s p o n d u n d e r o n e o f

t h e s e two r i g i d m o t i o n s . T h i s g i v e s u s a o n e - t o - o n e c o r r e s -

pondence be tween t h e two a n g l e s . What we need t o show i s t h a t

t h i s c o r r e s p o n d e n c e p r e s e r v e s d i s t a n c e s .

S u p p o s e t h a t we h a v e g i v e n two p o i n t s P, Q o f ABC

and t h e c o r r e s p o n d i n g p o i n t s P I , Q 1 o f DEF. If P a n d

Q a r e o n t h e same s i d e o f ABC, t h e n o b v i o u s l y

P t Q 1 = PQ,

b e c a u s e d i s t a n c e s a r e p r e s e r v e d o n e a c h o f t h e r ay s t h a t f o r m

LABC. Suppose , t h e n , t h a t P and Q a r e o n d i f f e r e n t s i d e s

o f L ABC, s o t h a t Pt and Q 1 a r e o n d i f f e r e n t sides o f

LDEF, l i k e t h i s :

By t h e S.A.S. P o s t u l a t e , we h a v e

T h e r e f o r e PQ = P ' Q l , w h i c h was t o b e p r o v e d . N e x t , we need t o p r o v e t h e a n a l o g o u s t h e o r e m f o r t r i a n g l e s :

Theorem - IX-5. If

AABC A A I B ' C * , t h e n t h e r e i s a r i g i d m o t i o n

AABC N A A I B ' C 1 , u n d e r w h i c h A , B and C c o r r e s p o n d t o A 1 , B t , and C 1 ,

P r o o f : F i r s t , we s h a l l se t up a o n e - t o - o n e c o r r e s p o n d e n c e b e t w e e n t h e p o i n t s o f A ABC and t h e p o i n t s o f A A t B I C 1 . We

h a v e g i v e n a o n e - t o - o n e c o r r e s p o n d e n c e

ABC<Ã‘^- ' B t C 1 f o r t h e v e r t i c e s . By Theorem V I I I - 1 ">;is g i v e s u s t h e i n -

d u c e d r i g i d m o t i o n s A B w m ,

and

between t h e s i d e s o f t h e t r i a n g l e s . T h e s e t h r e e r i g i d m o t i o n s , t a k e n t o g e t h e r , g i v e u s a o n e - t o - o n e c o r r e s p o n d e n c e P-P1

between t h e p o i n t s o f t h e two t r i g n a l e s . We need t o show t h a t

t h i s c o r r e s p o n d e n c e p r e s e r v e s d i s t a n c e s .

If P a n d Q a r e o n t h e same s i d e o f t h e t r i a n g l e , t h e n we know a l r e a d y t h a t

P'Q1 = PQ.

S u p p o s e , t h e n , t h a t P and Q a r e o n d i f f e r e n t s i d e s , say, - AB a n d E, l i k e

B

t h i s :

B'

We know t h a t

AP = A l p 1 , because Ã A'B' i s a rigid mot ion . F o r t h e same r e a s o n ,

AQ = AfQ'y and LA = L At, b e c a u s e AABC 2! A AlB@C*. By t h e S.A.S.

P o s t u l a t e , A PA& 2 APrA'Q* .

T h e r e f o r e , PQ = P'Q1 ,

which was t o b e p roved .

N o t i c e t h a t w h i l e t h e f i g u r e d o e s n o t show t h e c a s e P = B, t h e p r o o f t a k e s c a r e o f t h i s c a s e . The p r o o f i s more

i m p o r t a n t t h a n t h e figure, anyway.

Problem -- S e t I X - 3

1. L e t ABC +A'BICt

be a r i g i d mot ion , and s u p p o s e t h a t A , By and C

a r e c o l l i n e a r . Show t h a t i f B is between A and C ,

t h e n Bl i s between A * and C 1 .

2. Given a r i g i d mot ion F w F ' .

L e t A and B be p o i n t s o f F, and s u p p o s e t h a t F -

c o n t a i n s t h e segment AB. Show t h a t F' c o n t a i n s t h e

segments A I B t .

3 . Given a r i g i d mot ion F as F ' .

Show t h a t i f F is convex, t h e n s o a l s o is Ff.

4. Given a r i g i d m o t i o n F w F l . Show t h a t i f F i s a r a y , t h e n s o a l s o i s F 1 .

5. Show t h a t t h e r e is no r i g i d mot ion between a segment and

a c i r c u l a r a r c ( n o m a t t e r how s h o r t b o t h o f them may be ) .

R i d Mot ion o f C i r c l e s a n d A r c s . - -- Theorem IX-6. L e t C and C t be c i r c l e s o f t h e same r a d i u s - r. Then, t h e r e i s a r i g i d m o t i o n

b e t w e e n C and C t .

P r o o f : L e t t h e c e n t e r s o f t h e c i r c l e s b e P and P I .

L e t be a d i a m e t e r o f t h e f i r s t c i r c l e , and l e t A'B' be a d i a m e t e r o f t h e s e c o n d . L e t H and H2 b e t h e

<!-Ã h a l f - p l a n e s d e t e r m i n e d b y t h e l i n e AB; and l e t H I , and

- a H t 2 b e t h e h a l f - p l a n e s d e t e r m i n e d b y t h e l i n e A'B1.

We c a n now set up o u r o n e - t o - o n e c o r r e s p o n d e n c e Q-Q1,

i n t h e f o l l o w i n g way: . ( l ) L e t A t a n d B' c o r r e s p o n d t o A and B, r e s p e c t i v e l y . ( 2 ) If Qi i s a p o i n t o f C ,

l y i n g i n Hi, l e t Q t l be t h e p o i n t o f C t , l y i n g i n

H t l, s u c h t h a t

( 3 ) 1f Qg i s a p o i n t o f C , l y i n g I n Hz, l e t Q,12 b e

t h e p o i n t of C2, l y i n g i n H t g , s u c h t h a t

We need t o c h e c k t h a t this c o r r e s p o n d e n c e p r e s e r v e s d i s -

t a n c e s .

Thus , f o r e v e r y two p o i n t s Q, R o f C , w e mus t h a v e

Q 'Rt = QR.

If Q and R a r e t h e e n d - p o i n t s o f a d i a m e t e r , t h e n s o a r e

Q 1 and R 1 , and Q'RI = QR = 2r. O t h e r w i s e , we a l w a y s

h a v e AQPR 'Z A Q I P I R t , s o t h a t QtRt = Q R . ( p r o o f ? T h e r e

a r e two c a s e s t o c o n s i d e r , a c c o r d i n g a s B i s i n t h e i n t e r i o r

o r t h e e x t e r i o r o f L Q P R . )

You s h o u l d p r o v e t h e f o l l o w i n g two t h e o r e m s f o r y o u r s e l f .

They a r e n o t h a r d , o n c e we h a v e g o n e t h i s f a r .

Theorem IX-7. L e t C and C t b e c i r c l e s w i t h t h e same

r a d i u s , a s i n Theorem IX-6.

L e t /XPB and /XIPIB* be c o n g r u e n t c e n t r a l a n g l e s o f C

and C1. r e s p e c t i v e l y .

Then a r i g i d mot ion C Ã C t c a n be chosen i n such a way - n t h a t B-Bl, X<Ã‘Ãˆ-X and BX U BIX'.

Theorem IX-8. Given a n y two c o n g r u e n t a r c s , there i s a r i g i d m o t i o n between them. The p r o o f i s l e f t t o t h e r e a d e r .

R e f l e c t i o n s .

The d e f i n i t i o n o f r i g i d mot ion g i v e n i n S e c t i o n V I I I - 1 i s

a p e r f e c t l y good m a t h e m a t i c a l d e f i n i t i o n , b u t we might

c l a i m t h a t f r o m an i n t u i t i v e v i e w p o i n t i t d o e s n o t convey a n y i d e a o f "mot ion" . We w i l l d e v o t e t h i s s e c t i o n t o showing how a p l a n e f i g u r e c a n b e "moved" i n t o c o i n c i d e n c e w i t h any i s o m e t r i c f i g u r e i n t h e same p l a n e .

Throughout t h i s s e c t i o n a l l f i g u r e s w i l l be c o n s i d e r e d a s l y i n g i n a f i x e d p l a n e .

D e f i n i t i o n s . A one- to -one c o r r e s p o n d e n c e between two

f i g u r e s i s a r e f l e c t i o n i f t h e r e i s a l i n e L, such t h a t f o r a n y p a i r o f c o r r e s p o n d i n g p o i n t s P and P I , e i t h e r ( 1 ) P = PI

and l i e s on L o r ( 2 ) L i s t h e p e r p e n d i c u l a r b i s e c t o r of - PP' . L is c a l l e d t h e -- a x i s o f r e f l e c t i o n , and e a c h f i g u r e i s s a i d t o be t h e r e f l e c t i o n , o r t h e image, o f t h e o t h e r f i g u r e i n L.

I n t h e ~ i c t u r e

t l o n s of s i m p l e f

s below a r e shown some e x a m p l e s of r e f l e c -

i g u r e s . P B

A

A A'

v

L A' Theorem IX-9. A r e f l e c t i o n i s a r i g i d m o t i o n .

P r o o f : We must show t h a t i f P and 0. a r e a n y two p o i n t s ,

and P1 and Q1 t h e i r i m a g e s i n a l i n e L, t h e n PQ = P I Q ' .

T h e r e a r e f o u r c a s e s t o c o n s i d e r .

Case (1) Case (2 ) Case ( 3 ) Case ( 4 )

C a s e 1. P and Q a r e o n t h e same s i d e of L. L e t PP' - i n t e r s e c t L a t A and QQ1 i n t e r s e c t L a t B. By t h e

d e f i n i t i o n o f r e f l e c t i o n PP* 1 L and PA = P'A, and - QQt 1 L and QB = Q'B. Hence APAB A PIAB, a n d PE = PrB,

/. PBA Z /. P'BA. By s u b t r a c t i o n , LPN L?BQt . We t h e n h a v e

(by S.A.S.) APBQ A P*BQ1, and s o PQ = P l Q * .

Case 2. The p r o o f i s t h e same, e x c e p t t h a t i n p r o v i n g LPBQ 2 LPBQ* we add a n g l e m e a s u r e s i n s t e a d o f s u b t r a c t i n g .

Case 3. Q. i s on L. Then Q = Q l and PQ = P I Q 1 - s i n c e Q i s o n t h e p e r p e n d i c u l a r b i s e c t o r o f P P * . The c a s e

P on L and Q n o t o n L i s j u s t t h e same.

C a s e 4. P and Q b o t h on L. S i n c e P = P* and C = Q* we c e r t a i n l y h a v e PQ = P * Q t -

S t a r t i n g w i t h a f i g u r e F we can r e f l e c t i t i n some l i n e

t o g e t a f i g u r e F , F l c a n be r e f l e c t e d i n some l i n e t o g e t

a f i g u r e Fy, and s o on. I f w e end up w i t h a f i g u r e Fl

a f z e r n s u c h s t e p s we s h a l l s a y t h a t F h a s been c a r r i e d

i n t o F * by a c h a i n o f n r e f l e c t i o n s . -

C o r o l i a r y IX-9-1, A c h a i n o f r e f l e c t i o n s c a r r y i n g F i n t o

F * d e t e r m i n e s a r i g i d mot ion between F and F 1 .

2oming b a c k t o o u r o p e n i n g d i s c u s s i o n i n t h i s s e c t i o n , a

r e j e c t i o n c a n b e t h o u g h t o f a s a p h y s i c a l mot ion , o b t a i n e d

(22 r o t a t i n g t h e whole p l a n e t h r o u g h 180' a b o u t t h e a x i s o f

i n f l e c t i o n . The a b o v e c o r o l l a r y s a y s t h a t a c e r t a i n t y p e o f r i g i d mot ion , namely, t h o s e o b t a i n a b l e a s a c n a i n o f r e f l e c -

-. ,

t i o n s , can b e g i v e n i n a p h y s i c a l i n t e r p r e t a t i o n . What we

s h a l l now show i s t h a t e v e r y r i g i a m o t i o n - i f o f t h i s t y p e .

The p r o o f w i l l be g i v e n i n two s t a g e s , t h e f i r s t s t a g e i n -

v o l v i n g o n l y a v e r y s i m p l e f i g u r e . F o r c o n v e n i e n c e we w i l l

u s e t h e n o t a t i o n F 1 F t i f F and F * a r e r e f l e c t i o n s o f

e a c h o t h e r i n some a x i s .

Theorem I X - 1 0 . L e t A , B, C , A * , B * , C * b e s i x p o i n t s such t h a t A B = A * B t , A C = A t C t , B C = B * C 1 . Then t h e r e i s a c h a i n o f a t most t h r e e r e f l e c t i o n s t h a t c a r r i e s A , B, C i n t o

A ' , B ' , C * .

P r o o f :

S t e p 1.

- L e t L2 b e t h e p e r p e n d i c u l a r b i s e c t o r of A A 1 , and l e t B2

and C2 b e t h e r e f l e c t i o n s o f B1 and Cl i n L2. Then

t . , B2,C2 1 A 1 , B i , C t .

- L e t I,, be t h e p e r p e n d i c u l a r b i s e c t o r o f BBo. S i n c e

AB = AfB1 and s i n c e by Theorem IX-9, A'B1 = AB2, i t f o l l o w s

t h a t AB = AB2. T h e r e f o r e A l i es o n Ll and i s i t s own

image i n t h e r e f l e c t i o n i n L. Thus, t h e image of A , By, C2 i n Ll i s A , B, Cl .

S t e p 3.

By a r g u m e n t s s i m i l a r t o t h e o n e a b o v e , w e see t h a t AC = AC <Ã‘

and B C = B C l . Hence, AB is t h e p e r p e n d i c u l a r b i s e c t o r of

1 - ec.' and t h e image A , B, C i i n 3 i s A , B, C.

We t h u s have ,

a s was d e s i r e d .

Any o n e o r two of t h e t h r e e s t e p s may be u n n e c e s s a r y If t h e p a i r o f p o i n t s we a r e w o r k i n g o n ( A i n s t e p 1, B i n s t e p 2 ,

C i n s t e p 3) happen t o c o i n c i d e .

We a r e now r e a d y f o r t h e f i n a l s t a g e o f t h e p r o o f .

Theorem I X - 1 1 . Any r i g i d m o t i o n i s t h e r e s u l t o f a c h a i n of a t mos t t h r e e r e f l e c t i o n s .

P r o o f : We a r e g i v e n a r i g i d m o t i o n F Ã Ft. L e t A, B, C be t h r ee n o n - c o l l i n e a r p o i n t s i n F, and A', B 1 , C l

t h e c o r r e s p o n d i n g p o i n t s i n F 1 .

(1f a l l p o i n t s o f F a r e c o l l i n e a r a s e p a r a t e , b u t s i m p l e r , p r o o f i s n e e d e d . The d e t a i l s o f t h i s a r e l e f t t o t h e s t u d e n t . )

By Theorem IX-10, we c a n p a s s f r o m A t , Bl, C 1 t o

A , B, C b y a c h a i n o f a t mos t t h r e e r e f l e c t i o n s . By c o r o l l a r y IX-9-1, t h i s c h a i n d e t e r m i n e s a r i g i d m o t i o n F f Ã Fit , and by t h e c o n s t r u c t i o n o f t h e r e f l e c t i o n s w e h a v e A" = A ,

B1I = B and C l l = C . S c h e m a t i c a l l y t h e s i t u a t i o n i s some- t h i n g l i k e t h i s :

We s h a l l show t h a t f o r e v e r y p o i n t P a f F, w e h a v e PI1 = P. T h i s w i l l show t h a t F l l c o i n c i d e s w i t h F, and

t h a t t h e g i v e n r i g i d m o t i o n F w F 1 is i d e n t i c a l w i t h t h e o n e d e t e r m i n e d b y t h e c h a i n o f r e f l e c t i o n s .

L e t u s c o n s i d e r , t h e n , a n y p o i n t P of F, i t s c o r r e s p o n d -

i n g p o i n t P 1 i n F * d e t e r m i n e d by t h e r i g i d m o t i o n F Ã P 1 , a n d t h e p o i n t P t l i n F w d e t e r m i n e d f r o m PT by t h e

c h a i n of r e f l e c t i o n s . We r e c a l l t h a t A m = A , Bt1 = B,

C " = c .

S i n c e a l l o u r r e l a t i o n s h i p s a r e r i g i d m o t i o n s , we h a v e AP" = A t P 1 = AP. Similarly, BP" = BP and CP" = CP. From

t h e f irst two o f t h e s e , and AB = AB, we g e t t h a t A ABP AABP", a n d s o LBAP = LBAP" . I f P and P"

<Ã‘ a r e on t h e same s i d e o f AB t h e n b y t h e A n g l e C o n s t r u c t i o n Theorem 3 = AP", and s i n c e AP = AP" i t f o l l o w s from t h e

P o i n t P l o t t i n g Theorem t h a t P = P", w h i c h i s w h a t we wan ted t o p r o v e .

S u p p o s e t h e n t h a t P and P" l i e o n o p p o s i t e s ides o f

S i n c e PA = P"A and PB = P"B i t fo l lows t h a t A and B

l i e o n t h e p e r p e n d i c u l a r b i s e c t o r o f "PP"". S i n c e PC = P"C,

C a l s o l i e s o n t h i s l i n e , c o n t r a r y t o t h e c h o i c e o f A , B, and C a s n o n - c o l l i n e a r . Hence, t h i s c a s e d o e s n o t a r i s e , and we

a r e l e f t w i t h P = P , t h u s p r o v i n g t h e theo rem.

Problem S e t I X - 5 -- 1. I n e a c h o f t h e f o l l o w i n g c o n s t r u c t , w i t h any i n s t r u m e n t s

you f i n d c o n v e n i e n t , t h e image o f t h e g i v e n f i g u r e i n t h e

l i n e L.

2. F i n d a c h a i n o f t h r e e o r fewer r e f l e c t i o n s t h a t w i l l c a r r y ABCD i n t o A t B t C I D t .

3. a . C a r r y AABC t h r o u g h t h e c h a i n o f f o u r r e f l e c t i o n s i n

t h e a x e s Ll, L2, Lg, L4.

b. F i n d a s h o r t e r c h a i n t h a t w i l l g i v e t h e same r i g i d rno t i o n .

D e f i n i t i o n s : A f i g u r e i s symmet r i c i f i t i s i t s own image i n some a x i s . Such a n a x i s i s c a l l e d a n a x i s o f symmetry o f t h e -- f i g u r e .

Show t h a t a n i s o s c e l e s t r i a n g l e i s symmet r i c . What i s t h e a x i s ?

A f i g u r e may have more t h a n o n e a x i s o f symmetry. How many d o e a c h o f t h e f o l l o w i n g f i g u r e s have?

a . A rhombus.

b. A r e c t a n g l e .

c . A s q u a r e .

d . An e q u i l a t e r a l t r i a n g l e .

e. A c i r c l e .

The r i g i d mot ion d e f i n e d by a c h a i n o f two r e f l e c t i o n s i n

p a r a l l e l a x e s h a s t h e p r o p e r t y t h a t i f P-PI t h e n - PPI h a s a f i x e d l e n g t h ( t w i c e t h e d i s t a n c e be tween t h e a x e s ) and d i r e c t i o n ( p e r p e n d i c u l a r t o t h e a x e s ) . P rove t h i s . Such a mot ion i s c a l l e d a t r a n s l a t i o n .

The r i g i d mot ion d e f i n e d by a c h a i n o f two r e f l e c t i o n s i n

a x e s which i n t e r s e c t a t Q h a s t h e p r o p e r t y t h a t i f

P < Ã ‘ > P l t h e n Z. PQP' h a s a f ixed m e a s u r e ( t w i c e t h e

measure o f t h e a c u t e a n g l e be tween t h e a x e s ) . P r o v e t h i s .

Show how by u s i n g t h e r e s u l t s o f P rob lems 6 and 7 t h e

Fundamenta l Theorem I X - 1 1 c a n be r e s t a t e d i n t h e f o l l o w i n g form:

Any r ig id mot ion i n a p l a n e i s e i t h e r a r e f l e c t i o n , a t r a n s - l a t i o n , a r o t a t i o n , a t r a n s l a t i o n f o l l o w e d by a r e f l e c t i o n , o r a r o t a t i o n f o l l o w e d by a r e f l e c t i o n .

Appendix X

TRIGONOMETRY

T r i g o n o m e t r i c F u n c t i o n s .

The e l e m e n t a r y s t u d y o f t r i g o n o m e t r y i s based on t h e

f o l l o w i n g theorem.

Theorem X-1. If an a c u t e a n g l e o f one r i g h t t r i a n g l e i s

c o n g r u e n t t o a n a c u t e a n g l e o f a n o t h e r r i g h t t r i a n g l e , t h e n

t h e two t r i a n g l e s a r e s i m i l a r .

P r o o f : I n A ABC and A A I B t C f l e t L C and L C * b e r i g h t a n g l e s and l e t m L A = m L A'. Then A ABC - A A ~ B ~ C !

by A . A . S i m i l a r i t y Theorem 7-6.

We a p p l y t h i s theorem a s f o l l o w s : L e t r be any

number be tween 0 and 90, and l e t A ABC be a r i g h t

t r i a n g l e w i t h m L C = 90 and ML A = r. For c o n v e n i e n c e

s e t

A B = c , A C = b , BC=a. 2 2 h he P y t h a g o r e a n Theorem t h e n t e l l s u s t h a t c = a + b .)

If w e c o n s i d e r a n o t h e r s u c h t r i a n g l e A A*B7Ct w i t h

m LC' = 90 and m ^- A * = r, w e get three c o r r e s p o n d i n g

numbers a * , b l , c t , which would generally b e d i f f e r e n t from a , b y c . However, we h a v e ( a , c ) - ( a t , c l ) . By a l t e r n a t i o n

we c a n t h e n wri te ( a , a * ) - ( c , c 7 ) and i t s c o n s t a n t of a p r o p o r t i o n a l i t y i s 5.

a does n o t depend on t h e p a r t i c u l a r Thus , t h e number

t r i a n g l e w e u s e , b u t o n l y on t h e measure r o f t h e a c u t e

a n g l e . The v a l u e of t h i s number i s c a l l e d t h e s i n e af ru, w r i t t e n s i n r' f o r s h o r t . The r e a s o n w e s p e c i f y t h a t w e a r e i

I u s i n g d e g r e e measure Is t h a t i n more advanced a s u e c t s of

t r i g o n o m e t r y a d i f f e r e n t measure o f a n g l e , r a d i a n measure ,

i s common. â‚

L e t u s see what we c a n s a y

a b o u t s i n 30'. We know f rom Theorem 11-9 t h a t i n t h i s c a s e a

1 i f c = 1, t h e n a = - y . Hence, a 1 - = - s i n 3 0 ' = ~ Ã£ A

b c

It i s e v i d e n t t h a t t h e number b - c a n be t r e a t e d i n t h e same way a s c

- b a The number Ã i s c a l l e d t h e c o s i n e - 0 o f r , w r i t t e n c o s rO. From t h e n 1 P h y t h a g o r e a n Theorem, we s e e t h a t i f a = 75- and c = 1, t h e n

b =A. Hence, c o s 30' =/I. O f t h e f o u r o t h e r p o s s i b l e q u o t i e n t s o f t h e t h r e e s i d e s

a o f t h e t r i a n g l e , we s h a l l u s e o n l y one, r. T h i s i s c a l l e d u

t h e t a n g e n t o f rO, w r i t t e n t a n rO. We see t h a t 1 t a n 3 0 = 8 . h his u s e o f t h e word " t a n g e n t " h a s o n l y a n

u n i m p o r t a n t h i s t o r i c a l c o n n e c t i o n w i t h i t s u s e w i t h r e l a t i o n t o a l i n e and a c i r c l e . )

T h e s e t h r e e q u a n t i t i e s a r e c a l l e d t r i g o n o m e t r i c f u n c t i o n s .

Problem Set X-1 --

1. I n e a c h of t h e f o l l o w i n g g i v e t h e r e q u i r e d i n f o r m a t i o n i n terms o f t h e i n d i c a t e d l e n g t h s of t h e s i d e s .

b. s i n rO = ?, c o s P = ?, t a n P = ?.

2. I n each o f t h e fo l lowing , f i n d t h e c o r r e c t numer ica l value f o r x.

a . c o s P = x .

b. t a n a' = x.

f 0 f 0 3. Find: s i n 00 , c o s 60Â° t a n 00 . 4. Find: s i n 4 5 ' cos Â¡ t a n 45O.

5. By making c a r e f u l drawings w i t h r u l e r and p r o t r a c t o r de te rmine by measuring

a . s i n 20Â° c o s 20Â° t a n 20';

b. s i n % ' , cos53 ' , t an53 ' .

T r i g o n o m e t r i c T a b l e s - and A p p l i c a t i o n s .

Al though t h e t r i g o n o m e t r i c f u n c t i o n s c a n b e computed e x a c t l y f o r a few a n g l e s , s u c h a s 3 0 , 60' and 4 5 , i n

most c a s e s , we have t o be c o n t e n t w i t h a p p r o x i m a t e v a l u e s . T h e s e c a n be worked o u t by v a r i o u s advanced methods and

a t t h e end o f t h i s Appendix, we g i v e a t a b l e o f t h e v a l u e s of t h e three t r i g o n o m e t r i c f u n c t i o n s c o r r e c t t o three dec imal p l a c e s .

Hav ing a " t r i g t a b l e " , and a d e v i c e f o r m e a s u r i n g a n g l e s , s u c h a s a s u r v e y o r ' s t r a n s i t ( o r s t r i n g s and a p r o t r a c t o r ) one

c a n s o l v e v a r i o u s p r a c t i c a l problems.

Example s. From a p o i n t 100 feet f r o m t h e b a s e of a f l a g p o l e t h e a n g l e between t h e h o r i z o n t a l and a l i n e t o t h e t o p o f t h e p o l e i s found t o be 23'. L e t x be t h e h e i g h t o f t h e

p o l e . Then

Hence, x = 42.5 fee t . An a n g l e l i k e t h e o n e used I n t h i s exzmple i s f r e q u e n t l y c a l l e d t h e a n g l e - o f e l e v a t i o n o f t h e

o b j e c t .

Example G. I n a c i r c l e o f radius 6 cm. a chord - AB h a s l e n g t h 10 cm. What i s t h e measure o f a n a n g l e i n s c r i b e d I n t h e m a j o r a r c s? We have AC = 8,

AQ = 1 5 Â¥ 10 = 5. Hence, s i n (/AcQ = 8 = .625,

m L ACQ = 3g0, m(minor a r c AT) = m LACB =

2(m L A C Q ) = 78'. 1 -

Hence, m LAPB = p ( a r c AB) =

39' t o t h e n e a r e s t d e g r e e .

Problem S e t X-2. -- 1. From t h e t a b l e f i n d : s i n 1 7 O , c o s 46O, t a n 82O,

c o s 3 3 O , s i n 60'. Does t h e l a s t v a l u e a g r e e w i t h t h e

o n e found I n Problem 3 of Set X-l?

2 . From t h e t a b l e f i n d x t o t h e n e a r e s t d e g r e e i n e a c h

o f t h e f o l l o w i n g c a s e s :

s i n x = .413, t a n x = 2, c o s x = 1 3 '

3. A h i k e r c l i m b s f o r a h a l f m i l e up a s l o p e whose i n c l i n a - t i o n i s 17'. How much a l t i t u d e d o e s h e g a i n ?

4 . When a s i x - f o o t p o l e c a s t s a f o u r - f o o t shadow, wha t Is

t h e a n g l e o f e l e v a t i o n o f t h e s u n ?

5 An i s o s c e l e s t r i a n g l e h a s a b a s e o f 6 i n c h e s and a n o p p o s i t e a n g l e o f 30'. F i n d :

a . The a l t i t u d e of t h e t r i a n g l e .

b . The l e n g t h s o f t h e a l t i t u d e s t o t h e e q u a l s i d e s .

c . The a n g l e s t h e s e a l t i t u d e s make w i t h t h e b a s e .

d . The p o i n t o f i n t e r s e c t i o n o f t h e a l t i t u d e s .

6 . A r e g u l a r d e c a g o n ( 1 0 s i d e s ) i s i n s c r i b e d i n a

c i r c l e of r a d i u s 12 . F i n d t h e l e n g t h o f a s i d e , t h e apothem, and t h e a r e a o f t h e d e c a g o n .

7. G i v e n , m L P . = ~ 6 ' ~ ~ tnLCBD = 42', D

R e l a t i o n s Among t h e T r i ~ o n o m e t r i c F u n c t i o n s . - Theorem X-2. F o r a n y a c u t e L A ; s i n A < 1, c o s A < 1. -

P r o o f : I n t h e r i g h t t r i a n g l e AABC o f S e c t i o n X-1,

a < c and b < c.' D i v i d i n g each of t h e s e i n e q u a l i t i e s by

c g i v e s

wh ich i s wha t we wan ted t o p r o v e .

Theorem X-3 . F o r a n y a c u t e a n g l e A , - s i n A 2 2 - = t a n A , and ( s i n A ) + ( c o s A ) = 1. c o s A

Proof : a -

s i n A - = - = tan^. c o s A - b

2 b2 s i n A ) ~ + ( c o s A ) = a ^ +

c c

Theorem - X-4. If A and 3 a r e complementary a c u t e a n g l e s , t h e n s i n A = c o s B, c o s A = s i n B, and

t a n A = 1 . t a n B B

x

A Y

P r o o f : I n t h e n o t a t i o n o f t h e f i g u r e , w e have x s i n A = - = c o s B, z

c o s A = -^ = s i n B, z

x 1 t a n A = - = - = - 1 y v t a n B* x

Problem Set X - 3 -- Do t h e f o l l o w i n g problems w i t h o u t u s i n g t h e t a b l e s .

1. If s i n A = 1 what; i s t h e v a l u e o f c o s A ? What i s t h e v a l u e o f t a n A? ( u s e Theorem X-3.)

2 . W i t h r u l e r and compass c o n s t r u c t LA, i f poss ib le , i n

each of t h e fo l lowing. You a r e allowed t o use t h e r e s u l t s of e a r l i e r p a r t s t o s i m p l i f y l a t e r on s.

a . cos A = .8.

\

Solu t ion : Take AC any convenient segment and c o n s t r u c t + c o n s t r u c t a n a r c CQ 1 E. W i t h c e n t e r A and r a d i u s ~g +

i n t e r s e c t i n g CQ a t B. Then cos( LBAC)= .8.

s i n A = -8 .

s i n A = .7.

2 t a n A - - 3- 3 t a n A = F.

Table of Trigonometric Ratios

Tan- Angle S ine Cosine gent

Tan- Angle Sine Cosine gent

Solutions t o Appendix

Problem Set X - 1 --

Problem S e t X-2 --

43*, 23'. 17', 24O, 63'' 71'.

s i n 17'

t a n x =

m L A =

- - x x = .292 2640 = 771 feet. 5260 2

0 = 1.5. x = 56'.

30, m L B = rn L C = 75'.

- - AD t a n C . AD = 3.732 3 = 11.196. CD - - CE - sin B. CE = .966-6 = 5.796. CB - rn L E C B = 90' - rn L B = l 5 O .

- - DF tan 15'. DP = .268*3 = .804. CD -

B C

b s i n 18' = - 12 '

a = 11.41.

1 a r e a = - 10 7.42 11.41 = 423. 2

HP. ^ tan 26' = Ã‘ .AC = 92.2, P,B = 42.2.

45 s i n 26' = m, AD = 103. -

Problem Set, X-3 -- 2 1 2 (sin f i 1 2 + ( c o s A ) = 1, - 3 + (cos A ) = 1,

,

s i n A - t a n A = - - 3 - 1 - - cos A 2,/2 2,/?

7 3 - ( c ) is impossible.

( d ) A here i s congruent to B of part ( a ) .

(g) A here is the complement of the A of p a r t ( f ) .

Appendix X I

VECTORS I N SPACE

I n C h a p t e r 10 o u r work d e a l t s o l e l y w i t h v e c t o r s i n a p l a n e . However, a l l t h a t we d i d t h e r e c a n b e ex tended t o s p a c e , and

i n t h i s a p p e n d i x , we summarize, w i t h o u t p r o o f , t h e e x t e n - s i o n s o f o u r r e s u l t s t o t h r e e d i m e n s i o n s . ( ~ e f i n i t i o n s , n o t a - t i o n s , and theorems which a r e n o t r e p e a t e d h e r e a r e t h e same i n two and t h r e e d i m e n s i o n s . )

D e f i n i t i o n s . A v e c t o r i n s p a c e i s a n o r d e r e d t r i p l e o f r e a l numbers [ a , b , c ] . The numbers a , b y c a r e c a l l e d t h e components o f t h e v e c t o r .

D e f i n i t i o n . The o r d e r e d t r i p l e [O, 0, 01 i s c a l l e d t h e z e r o v e c t o r . -

a D e f i n i t i o n . I f u = [ a , b, c ] and h i s a s c a l a r , t h e v e c t o r [ h a , hb, h e ] i s c a l l e d t h e p r o d u c t - o f

A t h e v e c t o r u and t h e s c a l a r &. - - -- D e f i n i t i o n . I f 1?= [ a , b , c ] , t h e v e c t o r [ - a , -b, - c ] i s c a l l e d t h e o p p o s i t e o f u.

A D e f i n i t i o n . I f u = [ a , b, c ] t h e number 2

A i s c a l l e d t h e magni tude o r l e n g t h o f u.

D e f i n i t i o n . Two v e c t o r s a r e e q u a l i f and o n l y i f

t h e y have t h e same components.

2 2

D e f i n i t i o n . I f u = [ a , b, c ] and v = [ d , e, f ] , t h e v e c t o r [ a + d , b + e, c + f ] i s c a l l e d t h e

2

sum o f T? and -v. -

P r o p e r t i e s 1-11 h o l d e q u a l l y well f o r v e c t o r s i n a p l a n e and v e c t o r s i n s p a c e .

D e f i n i t i o n . T h r e e v e c t o r s G, CD, EF a r e s a i d t o <Â¥ <Ã‘ b e c o p l a n a r i f t h e r e exis ts a p l a n e t o which AB, CE, -

E F a r e a l l p a r a l l e l .

2 2 Theorem X I - 1 . If e, v, w a r e t h r e e non-ze ro and non- - c o p l a n a r v e c t o r s , and i f z i s a n y v e c t o r , t h e n t h e r e e x i s t s c a l a r s p, q, r s u c h t h a t

a A Theorem - X I - 2 . If u, v, w a r e t h r e e non-zero and non- c o p l a n a r v e c t o r s , and i f p l , ql , rl, p2, qg, r a r e s c a l a r s

2 s u c h t h a t

a 2

D e f i n i t i o n . If u = [ a , b, c ] and v = [ d , e, f ] ,

t h e number ad + b e + c f i s c a l l e d t h e s c a l a r p r o d u c t A

of u and v.

The p r o p e r t i e s o f t h e s c a l a r p r o d u c t a r e t h e same i n two and t h r e e d i m e n s i o n s .

Theorem - XI-3. Two non-ze ro v e c t o r s a r e p e r p e n d i c u l a r i f

and o n l y i f t h e i r s c a l a r p r o d u c t i s z e r o .

A

Theorem - XI-&. If and v a r e non-ze ro v e c t o r s , t h e a b s o l u t e v a l u e o f t h e i r s c a l a r p r o d u c t i s e q u a l t o

2

( a ) The l e n g t h o f u m u l t i p l i e d by t h e l e n g t h o f A A

t h e p r o j e c t i o n of v on u; o r , e q u a l l y well, 2

( b ) The l e n g t h of v m u l t i p l i e d by t h e l e n g t h of t h e a A

p r o j e c t i o n of u on v.

APPENDIX X I 1

APPLICATIONS O F GEOMETRIC THEORY

t o t h e

USE OF STRAIGHTEDGE AND COMPASSES

i n

DRAWING PICTURES OF PLANE FIGURES

I n your s tudy of informal geometry i n previous mathematics

courses you may have l ea rned how t o use a s t r a i g h t e d g e and

compasses i n drawing p i c t u r e s o f plane geometric f i g u r e s . The

p i c t u r e s below sugges t some of t h e b a s i c o p e r a t i o n s i n t h e use

of t h e s e t o o l s . The comments g i v e some i n d i c a t i o n of t h e

r e l a t e d theory f o r t h e p lane which may be a p p l i e d t o "prove t h e

accuracy" of t h e p i c t u r e . + -

1. Given a r a y AB and a segment C D , t o draw t h e p o i n t E -+ - - -

on AB such t h a t AE = CD .

(The Point P l o t t i n g he or ern)

2. Given two d i s t i n c t p o i n t s , t o draw t h e l i n e which c o n t a i n s

them.

( P o s t u l a t e 3 )

3. Given two d i s t i n c t p o i n t s A and B , t o draw t h e c i r c l e

with c e n t e r A and r a d i u s AB .

( ~ e f init ion of c i r c l e )

975

4 . Given a segment and a p o i n t , t o draw t h e c i r c l e wi th

c e n t e r a t t h e p o i n t and r a d i u s equal t o t h e measure of

t h e segment.

( ~ e f i n i t i o n of c i r c l e ) + - 5. Given a r a y AB and a segment CD , t o draw t h e po in t E - -

on AB such t h a t 'm = 3 CD ,

( p o i n t P l o t t i n g he or em)

6 . To draw t h e midpoint of a g iven segment.

(An a p p l i c a t i o n of t h e Tr iang le Congruence ~ o s t u l a t e s )

7. To draw t h e pe rpend icu la r b i s e c t o r of a g iven segment.

To draw t h e b i s e c t o r of a g iven ang le .

(An a p p l i c a t i o n of t h e S.S.S. Congruence p o s t u l a t e )

To draw a l i n e perpendicular t o a given l i n e a t a g iven

po in t on i t . 4

( ~ n a p p l i c a t i o n of t h e S.S.S. Congruence p o s t u l a t e )

Given a l i n e and a po in t n o t on i t , t o draw t h e f o o t of

t h e perpendicular from t h e p o i n t t o the l i n e .

(An a p p l i c a t i o n of the S.S.S. Congruence p o s t u l a t e )

Given a l i n e and a po in t n o t on it, t o draw t h e l i n e which

- 12. Given an angle /ABC , a ray DE and a point G not on - DE , draw the ray DF such that F and G are on the

same side of '*Sf' and /FDE 2 /ABC .

( ~ n application of the S.S.S. Congruence postulate) - - -

13. Given three segments PP* , QQ* , RR* , such that the sum of the lengths of every pair of them exceeds the length of the third segment, draw a triangle AABC such that

AB = PP1 , BC = QQ* , AC = RR* .

(Some algebra involving coordinates and equations may be

used to show that the circles with centers at A and B , with AB = PP* , and radii QQ* and RR* , respectively, - intersect in two points, one on each side of AB . )

14. Given two segments and an angle, to draw a triangle having

two sides and an included angle congruent to the given

segments and angle.

(An application Involving some of the incidence postulates

and (12) above.)

15. Given a l i n e 1 a po in t P not on It, t o draw t h e l i n e through t he given po in t and p a r a l l e l t o t h e given l i n e .

( ~ n a p p l i c a t i o n involving (12) above. )

Problem Se t XI1 -- The fo l lowing are e x e r c i s e s I n t h e use of s t r a i g h t e d g e and compasses:

Given a c i r c l e wi th c e n t e r M and chord no t conta in- i n g M , t o draw a chord through M and perpendicular t o RS . The chord t h e chord . The chord i s a o f t h e c i r c l e .

Given a c i r c l e and a r a d i u s which j o i n s i t s c e n t e r t o a p o i n t P on it, t o draw a l i n e perpendicular t o t h e r a d i u s a t P . Given a c i r c l e wi th c e n t e r P and chord n o t conta in- - - i n g P , t o draw the cnord AC perpendicular t o AB , and a l s o t o draw t h e cnord B5" . The chord t h e p o i n t P . Given a c i r c l e wi th c e n t e r P and chord CD not - c o n t a i n i n g P , t o draw t h e perpendicular b i s e c t o r of CD . This b i s e c t o r c o n t a i n s t h e p o i n t

- Given a c i r c l e wi th c e n t e r R and chords MA and AC , t o draw t h e i n t e r s e c t i o n o f t h e perpendicular b i s e c t o r s - of MA and AC . This i n t e r s e c t i o n i s

Given t h r e e n o n c o l l i n e a r p o i n t s A, B, C , t o draw a c i r c l e which c o n t a i n s A , B, and C . How many d i s t i n c t c i r c l e s c o n t a i n t h e s e t h r e e p o i n t s ?

Given a t r i a n g l e ABC , t o draw t h e t r i a n g l e whose v e r t i c e s are t h e midpoints of AABC . Given a convex q u a d r i l a t e r a l ABCD t o d r a w t h e q u a d r i l a t e r a l AIBICID1 where At, B1, C t , Dl a r e t h e - - - - midpoints of A B , BC , CD , DA , r e s p e c t i v e l y .

- Given a l i n e segment AB , t o draw t h e po in t C on

such t h a t AC = 2 CE .

10. Given two distinct points A and B , to draw a point C

such that AC = ifS AB . 11. Given two distinct points A and B , to draw an angle

/CBA such that m /CBA = 60 . 12. Given two distinct points A and B , to draw a regular

hexagon ABCDEF . 13. Given three noncollinear points, to draw a parallelogram

having the given points as three of its vertices.

14. Given a triangle, to draw its medians. The medians

intersect in the centroid of the triangle.

15. Given a triangle, to draw its angle bisectors. These

angle bisectors intersect in the incenter of the triangle.

16. Given a triangle, to draw its altitudes. These altitudes

intersect in the orthocenter of the triangle.

17. Given a triangle, to draw the perpendicular bisectors of

its sides. These bisectors intersect In the circumcenter

of the triangle.

18. Given a triangle, to draw its incircle.

19. Given a triangle ABC , to draw a triangle DEF similar 2 to ABC with constant of proportionality 3 .

20. Given three noncollinear points, to draw an equilateral triangle containing the three points.

The Meaning and Use of Symbols - ---

General .

= A = B can be read a s " A e q u a l s B", " A i s e q u a l t o B", " A equa l B" ( a s i n " ~ e t A = B"), and p o s s i b l y o t h e r ways appea r s . However, we should n o t u s e t h e

symbol, = , i n such forms a s A and B a r e = I 1 - , i t s

p r o p e r u s e i s between - two e x p r e s s i o n s . If two e x p r e s s i o n s a r e connected by It=" i t i s t o be unders tood t h a t t h e s e

two e x p r e s s i o n s a r e names o f t h e same ma thema t i ca l ob- j e c t , i n o u r c a s e e i t he r a r e a l number o r a p o i n t set.

# " ~ o t e q u a l to" . A # B means t h a t A and B do - n o t r e p r e s e n t t h e same o b j e c t . The same v a r i a t i o n s and

c a u t i o n s a p p l y t o t h e u s e o f # a s t o t h e u s e o f = . { a : p r o p e r t y ) . The set o f a l l e l emen t s a e a c h o f which

has t h e s t a t e d p r o p e r t y . h he " s e t - b u i l d e r " n o t a t i o n ) .

Algebra ic . + . - + These f a m i l i a r a l g e b r a i c symbols f o r o p e r a t -

ing w i t h r e a l numbers need no comment. The b a s i c p o s t u l a t e s about them a r e p r e s e n t e d i n Appendix 11.

< > < > Like = , these can be r ead I n v a r i o u s ways i n - - sen tences , and A < B may s t a n d f o r t h e u n d e r l i n e d p a r t of "1f A i s less than B" , " ~ e t A b e less t h a n B" , " A less than B impl ies1 ' , e t c . S i m i l a r l y f o r t h e

o t h e r t h r e e symbols, r ead " g r e a t e r than", "less t h a n o r I t e q u a l t o ' , g r e a t e r than o r e q u a l to". These i n e q u a l l -

t ies app ly on ly t o r e a l numbers. T h e i r p r o p e r t i e s a r e mentioned b r i e f l y i n S e c t i o n 3-2, and i n more d e t a i l i n

S e c t i o n 3-3.

A I "Absolute va lue of A " . Discussed i n S e c t i o n s 3-2 and

8-3.

(x , y ) The c o o r d i n a t e s of a p o i n t ( a n o rde red p a i r o f r e a l

numbers); a l s o used a s t h e name of t h e p o i n t . 11 = P r o p o r t i o n a l i t y . " ( a , b, c ) = (d , e, f ) " i s read a , b , ctl

P P a r e p r o p o r t i o n a l t o I'd, e , f " .

G e o m e t r i c .

P o i n t S e t s . A s i n g l e l e t t e r may s t a n d f o r a n y s u i t a b l e

d e s c r i b e d p o i n t se t . Thus , w e may s p e a k o f a p o i n t P,

a l i n e m, a h a l f p l a n e w, a c i r c l e C , a n a n g l e x,

a s e g m e n t b, e t c . <-> AB T h e l i n e c o n t a i n i n g t h e two p o i n t s A and B. - AB The s e g m e n t h a v i n g A a n d B a s e n d p o i n t s . 4 AB. The r a y w i t h A a s I t s e n d p o i n t and c o n t a i n i n g p o i n t B.

+ 4 LABC T h e a n g l e h a v i n g B a s v e r t e x and BA and BC a s s i d e s .

AABC T h e t r i a n g l e h a v i n g A , B, C a s v e r t i c e s .

LABC is a r i g h t a n g l e . <Ã‘

LA-BC-D. The d i h e d r a l a n g l e h a v i n g l i n e BC a s e d g e and

w i t h f a c e s c o n t a i n i n g A and D. 2

(A,B) D i r e c t e d s e g m e n t whose o r i g i n i s A and whose

t e r m i n u s i s B. Read " d i r e c t e d segmen t AB" . 2 A u T h e v e c t o r u, r e a d " u v e c t o r 1 ' o r " t h e v e c t o r u".

[ a , b ] T h e v e c t o r whose componen t s a r e a and b . Note

t h a t a and b f o r m a n o r d e r e d ' p a i r and i n g e n e r a l

[a,b] i s n o t t h e same a s [ b , a ] . A

The m a g n i t u d e , o r l e n g t h , o f u A

PQ T h e v e c t o r whose c o m p o n e n t s a r e t h e same a s t h o s e of A

d i r e c t e d s e g m e n t ( P , Q ) . Read "PQ v e c t o r " o r " t h e

v e c t o r PQ". 2

0 T h e z e r o v e c t o r ; t h a t i s t h e v e c t o r whose componen t s

a r e b o t h z e r o . A A A A u - v T h e s c a l a r p r o d u c t o f t h e v e c t o r s u a n d v. Read

" u d o t v".

^-v-Q1~2.. . Q The p o l y h e d r a l a n g l e whose v e r t e x i s V and -+ + +

whose e d g e s a r e VQ,, VQ2, ..., ^n - AB An a r c of a c i r c l e w i t h e n d p o i n t s A and B. Read

II a r c AB" . A% The a r c o f a c i r c l e w i t h e n d p o i n t s A and B and

c o n t a i n i n g t h e p o i n t X. Read " a r c AXB" . - mAXB The d e g r e e measure o f t h e a r c whose e n d p o i n t s a r e A

and B and which i n c l u d e s t h e p o i n t X.

Tf The g r e e k l e t t e r p i , used h e r e and many o t h e r p l a c e s

t o d e n o t e t h e q u o t i e n t o f t h e c i r c u m f e r e n c e o f a c i r c l e d i v i d e d b y i t s d i a m e t e r . The number TT i s t h e

same f o r a l l c i r c l e s .

R e a l Numbers. - AB The p o s i t i v e number which i s t h e d i s t a n c e between t h e

two p o i n t s A and B, and a l s o t h e l e n g t h o f t h e - segment AB.

m LABC. The r e a l number be tween 0 and 180 which i s t h e d e g r e e measure o f L ABC.

- re relative t o (A, A ' ) ) The m e a s u r e o f t h e segment PQ

w i t h r e s p e c t t o t h e u n i t - p a i r ( A , A']. -

m m The s l o p e o f segment AB.

-+ m 8 The s l o p e o f r a y AB.

<-> *-,Ã The s l o p e o f l i n e AB.

R e l a t i o n s .

a w b , a i s matched w i t h b.

fv - Congruence . A ^ B i s r e a d " A i s c o n g r u e n t t o B" , b u t w i t h t h e same p o s s i b l e v a r i a t i o n s and r e s t r i c t i o n s a s A = B. I n t h e t e x t A and B may be segments , a n g l e s , o r t r i a n g l e s .

1 P e r p e n d i c u l a r . A 1 B i s r e a d " A i s p e r p e n d i c u l a r t o B", b u t w i t h t h e same comment a s f o r S. A and

B may b e e i t h e r two l i n e s , o r s u b s e t s o f l i n e s ( r a y s o r

s e g m e n t s ) .

F w F 1 F i s i s o m e t r i c t o F 1 . ( ~ p p e n d i x IX),.

1 ' P a r a l l e l i s m . We r e a d " p 1 1 q" a s " l i n e p i s p a r a l l e l t o l i n e q". - S i m i l a r i t y . We r e a d " A ABC - A DEF" a s " t r i a n g l e

ABC Is s i m i l a r t o t r i a n g l e DEF".

- - E q u i v a l e n c e f o r d i r e c t e d - segments ( A , B ) A ( C , D) i s r e a d " d i r e c t e d segment ( A , B ) i s e q u i v a l e n t t o d i r e c t e d

A

segment (c, D ) " .

The Greek Alphabet

a lpha

b e t a

gamma

del ta

e p s i l o n

z e t a

e t a

t h e t a

iota

kappa

lambda

mu

nu

x i

omicron

p i

rho

sigma

t a u

u p s i l o n

phi

c h i

p s i

omega

LIST OF CHAPTERS AND POSTULATES

CHAPTER 8. Coordinates i n a Plane

CHAPTER 9. Perpend icu la r i ty , P a r a l l e l i s m , and

Coordinates i n Space

POSTULATE 24. There i s a unique p lane which c o n t a i n s a g iven

p o i n t and i s pe rpend icu la r t o a g iven l i n e .

POSTULATE 25. Two l i n e s which a r e pe rpend icu la r t o t h e same

plane a r e p a r a l l e l .

CHAPTER 10. Direc ted Segments and Vectors

CHAPTER 11. Polygons and Polyhedrons

POSTULATE 26. If R is any g iven polygonal-region, t h e r e i s a correspondence which a s s o c i a t e s t o each

polygonal-region i n space a unique p o s i t i v e

number, such t h a t t h e number ass igned t o t h e

g iven polygonal-region R i s one.

POSTULATE 27. Suppose t h a t t h e polygonal-region R i s t h e

union of two polygonal-regions R and R2

such t h a t t h e i n t e r s e c t i o n of R, and R2 is

contained i n a union of a f i n i t e number o f

segments. Then, r e l a t i v e t o a g iven un i t - a rea ,

t h e a r e a of R i s t h e sum of t h e a r e a s of R.,

and Rg . POSTULATE 28. If two t r i a n g l e s a r e congruent , then t h e

r e s p e c t i v e t r i a n g u l a r - r e g i o n s c o n s i s t i n g of t h e

t r i a n g l e s and t h e i r i n t e r i o r s have t h e same

a r e a r e l a t i v e t o any g iven u n i t - a r e a .

POSTULATE 29. Given a u n i t - p a i r f o r measuring d i s t a n c e , t h e

a r e a of a r e c t a n g l e r e l a t i v e t o a un i t - square

i s t h e product of t h e measures ( r e l a t i v e t o t h e

g iven u n i t - p a i r ) of any two consecut ive s i d e s

of t h e r e c t a n g l e .

CHAPTER 12. C i r c l e s and Spheres

POSTULATE 30. If and a r e a r c s of t h e same c i r c l e having only t h e p o i n t B i n common, and i f - t h e i r union i s a n a r c AC , then m E + m% =

POSTULATE 31. The l e n g t h s of a r c s i n congruent c i r c l e s a r e p r o p o r t i o n a l t o t h e i r degree measures.

LIST OF THEOREMS AND COROLLARIES

THEOREM 8 - 1 .

THEOREM 8-2.

THEOREM 8 -3.

THEOREM 8-4.

THEOREM 8-5.

THEOREM 8 -6.

THEOREM 8-7.

THEOREM 8-8.

If P and Q a r e p o i n t s on t h e same v e r t i c a l l i n e , then PQ = lyp - yQl . If P and Q a r e p o i n t s on t h e same horizon- t a l l i n e , then PQ = 1xp - x i . Every v e r t i c a l l i n e i s pe rpend icu la r t o every h o r i z o n t a l l i n e .

If Pl(x1,yl) and pg(x2,yg) a r e two p o i n t s

2 + ('2 - yl)

I f P and 0, a r e two p o i n t s i n t h e same v e r t i c a l l i n e , then t he midpoint M o f

If P and Q a r e two p o i n t s on t h e same h o r i z o n t a l l i n e , then t h e midpoint M of - PQ Is t h e p o i n t

If P and Q a r e d i s t i n c t p o i n t s on a l i n e which i s n e i t h e r v e r t i c a l n o r h o r i z o n t a l , t hen t h e midpoint M of i s t h e p o i n t

I f P = ( x , y l ) and Q = (x2,y2) a r e any two d i s t i n c t p o i n t s i n a plane, t h e n t h e

midpoint M of i s t h e po in t

THEOREM 8-9.

THEOREM 8-10.

THEOREM 8-11.

THEOREM 8-12.

THEOREM 8-13.

THEOREM 8-14.

THEOREM 8-15.

COROLLARY 8-15.

Let a be any r e a l number. Then t h e s e t of

a l l p o i n t s i n t h e xy-plane each of which has

x-coordinate a i s a v e r t i c a l l i n e .

Let b be any r e a l number. The s e t of a l l

p o i n t s i n t h e xy-plane wi th y-coordinate b

i s a h o r i z o n t a l l i n e .

If ~ ~ ( x ~ , y ~ ) and P2(x2,y2) a r e any two

p o i n t s , then

PIP2 = ( ( x , y ) : x = x + k ( x 2 - x,), y =

Yl + k(y2 - y l ) , k i s r e a l )

If a , b , c , d a r e r e a l numbers such t h a t b

and d a r e not both zero and i f

S = [ ( x , y ) : x = a + bk, y = c + dk, k i s r e a l ) , then S i s a l i n e .

The s l o p e of a non-ver t i ca l l i n e p i s

2 - '1 , where I s any segment of p X- - X-

I f p i s t h e l i n e through ( x , , ~ , ) with , then s lope m = -

g

1. p = ( ( x , y ) : x = x + kg , Y = y1 + kf

k i s r e a l ) and

2. p = { ( x , y ) : x = x, + k , Y = yl + km , k i s r e a l ) .

Two n o n - v e r t i c a l l i n e s a r e p a r a l l e l i f and

only i f t h e i r s l o p e s a r e equal .

Three p o i n t s A , B, C a r e c o l l i n e a r i f and

only if m = m - , o r they l i e on a v e r t i c a l

l i n e . BC

THEOREM 8-16. ~f P = ( x , y 1 ) and Q = (x2,y2) and If p a

i s a n o b l i q u e l i n e , t h e n

COROLLARY 8-16-2. If p i s t h e l i n e which p a s s e s th rough

THEOREM 8 -17.

THEOREM 8-18.

THEOREM 8-19.

THEOREM 8-20.

THEOREM 8-21.

THEOREM 8-22.

THEOREM 8-23.

p(x1,yl) w i t h s l o p e m , t h e n

P = l ( x , y ) : Y - Y1 = m(x - x l ) L

Two n o n - v e r t i c a l l i n e s are p e r p e n d i c u l a r i f

and on ly i f t h e product of t h e i r s l o p e s i s -1 . A q u a d r i l a t e r a l i s a p a r a l l e l o g r a m i f each

o f i t s s i d e s i s congruent t o t h e s ide o p p o s i t e

i t .

A q u a d r i l a t e r a l i s a p a r a l l e l o g r a m i f and o n l y

i f each a n g l e i s congruent t o t h e a n g l e

o p p o s i t e i t .

A q u a d r i l a t e r a l i s a r e c t a n g l e i f and o n l y

i f It i s e u u i a n g u l a r .

A q u a d r i l a t e r a l i s a rhombus i f and on ly i f

i t i s e q u i l a t e r a l .

A l i n e segment which j o i n s t h e midpo in t s o f

two s i d e s of a t r i a n g l e i s p a r a l l e l t o t h e

t h i r d s i d e and i t s l e n g t h i s h a l f t h e l e n g t h

o f t h e t h i r d s i d e .

Given q u a d r i l a t e r a l ABCD w i t h A = (0,O) , B = ( a , 0 ) , D = ( b , c ) , t h e n ABCD i s a

p a r a l l e l o g r a m i s and on ly i f C = ( a + b , c ) . COROLLARY 8-23-1. If t h e c o o r d i n a t e s o f t h e v e r t i c e s o f a

p a r a l l e l o g r a m are A = ( 0 , 0 ) , B = ( a , 0 ) , C = ( a + b , c ) , and D = ( b , c ) , t h e n t h e

p a r a l l e l o g r a m i s a r e c t a n g l e i f and o n l y i f

b = O .

COROLLARY 8-23-2. If t h e coord ina tes of t h e v e r t i c e s of a

THEOREM 8-24.

THEORME 8-25.

THEOREM 8-26.

THEOREM 8-27.

THEOREM 8-28.

para l le logram a r e A = ( 0 , 0 ) , B = ( a , 0 ) , C = ( a + b , c ) and D = ( b , c ) where a > 0,

then t h e i s a rhombus i f and

A q u a d r i l a t e r a l i s a para l le logram i f and only

i f t h e d iagona l s b i s e c t each o t h e r .

A para l le logram i s a r e c t a n g l e i f and only i f t h e d iagona l s a r e congruent.

A para l le logram i s a rhombus i f and only i f

t h e d iagona l s a r e perpendicular .

A para l le logram i s a rhombus i f and only i f

a d iagonal b i s e c t s one of i t s ang les .

The s e t of a l l p o i n t s i n a p lane which a r e e q u i d i s t a n t from two given p o i n t s i n t h e p lane i s t h e pe rpend icu la r b i s e c t o r of t h e

segment Jo in ing trle g iven p o i n t s .

COROLLARY 8-28-1. The pe rpend icu la r b i s e c t o r s of t h e s i d e s of a t r i a n g l e a r e concurrent a t a p o i n t equi - d i s t a n t from t h e v e r t i c e s of t h e t r i a n g l e .

THEOREM 8 -29. The set of a l l p o i n t s i n t h e i n t e r i o r of a n

ang le which a r e e q u i d i s t a n t from t h e l i n e s which c o n t a i n t h e s i d e s of t h e angle Is t h e

i n t e r i o r of t h e mldray of t h e ang le .

COROLLARY 8-29-1. The l i n e s which c o n t a i n t h e a n g l e b i s e c t o r s

THEOREM 9-1.

THEOREM 9-2.

THEOREM 9-3 .

THEOREM 9-4.

THEOREM 9-5.

THEOREM 9-6.

THEOREM 9-7.

THEOREM 9-8.

THEOREM 9-9.

of t h e a n g l e s of a t r i a n g l e are c o n c u r r e n t

a t a p o i n t e q u i d i s t a n t from t h e s i d e s o f t h e

t r i a n g l e .

The p l a n e which i s p e r p e n d i c u l a r t o a g i v e n

l i n e a t a p o i n t c o n t a i n s eve ry l i n e which i s

p e r p e n d i c u l a r t o t h e g iven l i n e a t t h a t

p o i n t .

I f a l i n e i s p e r p e n d i c u l a r t o each of two

i n t e r s e c t i n g l i n e s a t t h e i r p o i n t of i n t e r -

s e c t i o n , it i s p e r p e n d i c u l a r t o t h e p l a n e

de te rmined by t h e two l i n e s .

There i s a qu ique l i n e which i s p e r p e n d i c u l a r

t o a g iven p l a n e a t a g i v e n p o i n t I n t h e p l a n e .

I f a p l ane i n t e r s e c t s one o f two d i s t i n c t

p a r a l l e l l i n e s i n a p o i n t , i t i n t e r s e c t s t h e

o t h e r l i n e i n a p o i n t a l s o .

I f ap l ane i s p a r a l l e l t o one of two p a r a l l e l

l i n e s , i t i s a l s o p a r a l l e l t o t h e o t h e r .

I f a p l a n e i n t e r s e c t s each of two d i s t i n c t

p a r a l l e l p l a n e s , t h e i n t e r s e c t i o n s a r e two

d i s t i n c t p a r a l l e l l i n e s .

I f a l i n e i n t e r s e c t s one of two d i s t i n c t

p a r a l l e l p l a n e s i n a s i n g l e p o i n t , i t i n t e r s e c t s t h e o t h e r p l ane i n a s i n g l e p o i n t a l s o .

If a l i n e Is p a r a l l e l t o one of two p a r a l l e l

p l a n e s , i t Is p a r a l l e l t o t h e o t h e r a l s o .

Two p l a n e s which a r e p e r p e n d i c u l a r t o t h e same

l i n e are p a r a l l e l .

THEOREM 9-10.

THEOREM 9-11.

THEOREM 9-12.

THEOREM 9-13.

THEOREM 9-14.

THEOREM 9-15.

THEOREM 9 -16.

THEOREM 9-17.

THEOREM 9-18.

THEOREM 9-19.

If a l i n e i s p e r p e n d i c u l a r t o one of two

d i s t i n c t p a r a l l e l p l a n e s i t i s p e r p e n d i c u l a r

t o t h e o t h e r a l s o .

I f a p l a n e i s p e r p e n d i c u l a r t o one o f two

d i s t i n c t p a r a l l e l l i n e s , it i s p e r p e n d i c u l a r

t o t h e o t h e r l i n e a l s o .

I f two l i n e s a r e p a r a l l e l t o a t h i r d

l i n e , t h e y are p a r a l l e l t o each o t h e r .

Given a p l a n e and a p o i n t n o t i n t h e p l ane ,

t h e r e i s a unique l i n e which p a s s e s th rough

t h e p o i n t and i s p e r p e n d i c u l a r t o t h e p lane .

There i s a unique p l a n e p a r a l l e l t o a g iven

p l a n e th rough a g i v e n p o i n t .

I f two p l a n e s a r e each p a r a l l e l t o a t h i r d

p l a n e , t h e y are p a r a l l e l t o each o t h e r .

The s h o r t e s t segment j o i n i n g a p o i n t t o a

p l ane n o t c o n t a i n i n g t h e p o i n t i s t h e segment

p e r p e n d i c u l a r t o t h e g i v e n p l ane .

A l l segments which a r e p e r p e n d i c u l a r t o each

of two d i s t i n c t p a r a l l e l p l anes and have t h e i r

e n d p o i n t s i n t h e p l a n e s have t h e same l e n g t h .

The s e t of a l l p o i n t s which are e q u i d i s t a n t

from t h e e n d p o i n t s o f a g i v e n segment i s t h e

p l a n e which c o n t a i n s t h e midpoint o f t h e

segment and i s p e r p e n d i c u l a r t o t h e l i n e which

c o n t a i n s t h e segment.

Any two p l a n e a n g l e s o f a d i h e d r a l a n g l e a r e

congruent .

THEOREM 9-20.

THEOREM 9-21.

THEOREM 9-22.

THEOREM 9-23.

THEOREM 9-24.

THEOREM 9 -25.

I f a l i n e i s p e r p e n d i c u l a r t o a p l ane ,

t h e n any p l ane c o n t a i n i n g t h i s l i n e i s

p e r p e n d i c u l a r t o t h e g i v e n p l ane .

If two p l a n e s a r e p e r p e n d i c u l a r , t h e n any

l i n e i n one of t h e p l a n e s which i s pe rpend i -

c u l a r t o t h e i r l i n e o f i n t e r s e c t i o n i s

p e r p e n d i c u l a r t o t h e o t h e r p l a n e .

If two p l a n e s a r e p e r p e n d i c u l a r , t h e n any l i n e

p e r p e n d i c u l a r t o one of t h e p l a n e s a t a p o i n t

on t h e i r l i n e of i n t e r s e c t i o n l i e s i n t h e

o t h e r p l ane .

I f two i n t e r s e c t i n g p l a n e s are e a c h pe rpend i -

c u l a r t o a t h i r d p l ane , t h e n t h e i r l i n e o f

i n t e r s e c t i o n i s p e r p e n d i c u l a r t o t h i s p l ane .

If Pl and P2 a r e p o i n t s on a l i n e p a r a l l e l

t o t h e x - a x i s , t h e n PIPn = lx, - x21 , where

x-, and x2 a r e t h e x - c o o r d i n a t e s o f Pl and

Pp , r e s p e c t i v e l y .

I f P and P2 are p o i n t s on a l i n e p a r a l l e l

t o t h e y - a x i s , t h e n PlP2 = lyl - y21 , where

yl and y2 a r e t h e y - c o o r d i n a t e s of Pl and

P2 , r e s p e c t i v e l y .

THEOREM 9-26. If P and 5 a r e p o i n t s on a l i n e p a r a l l e l

t o t h e z - ax i s , t h e n PIP2 = Izl- z21 , where

z and z a r e t h e z - coord ina t e s of Pl and 1 2

P r e s p e c t i v e l y . 2

THEOREM 9-27. The d i s t a n c e between t h e p o i n t s Pl(xl ,yl ,z,)

and P ~ ( X ~ , Y ~ , Z ~ ) i s g i v e n by

1

THEOREM 9-28. If P ~ ( ~ ~ , Y ~ , ~ ~ ) and P2(x29~29z2) are any two d i s t i n c t p o i n t s , t h e n f o r every va lue

of k t h e p o i n t whose coord ina tes a r e

x = x, + k(x2 - x )

Y = Y, + k(y2 - Y,) z = Z , + k ( z 2 - z,)

l i e s o n x and, conversely, t o every poicit

on 1?.,?"" t h e r e corresponds a unique value

o f k such t h a t t h e s e equa t ions g ive t h e

coord ina tes of t h e p o i n t .

THEOREM 9-29. Every p lane has a n equa t ion of t h e form ax + by + cz = d , where one o r more of t h e

numbers a , b, c i s d i f f e r e n t from zero; and

every equa t ion of t h i s form i s a n equat ion of

a p lane .

THEOREM 10-1. There i s one and only one d i r e c t e d segment which i s e q u i v a l e n t t o a g iven d i r e c t e d

segment and has i t s o r i g i n a t a g iven p o i n t . A A

THEOREM 10-2. Two d i r e c t e d segments (p1,p2) and (p3,P[0

are e q u i v a l e n t i f and only i f they have t h e

same components . THEOREM 10-3. I f P and Pp have coord ina tes (x,,yl)

and (xq,yn) , r e s p e c t i v e l y , t h e l eng th of c r A

any d i r e c t e d segment equ iva len t t o (p1,pp)

THEOREM 10-4. If t h e coord ina tes of Pi and P2 a r e

(xl,y,) and (x2,y2) , r e s p e c t i v e l y , then

t h e components of t h e d i r e c t e d segment L

(p1,P3) which Is k times t h e d i r e c t e d L

segment (p1,p2) a r e k(x2 - xl) and

THEOREM 10-5.

THEOREM 10-6.

THEOREM 10-7.

THEOREM 10-8 .

THEOREM 10-9.

THEOREM 10-10.

THEOREM 10-11.

THEOREM 10-12.

THEOREM 10 -13.

A - The components of (Pl,P2) + (P3,P4) a r e

t h e sums of t h e corresponding components

a r e equ iva len t d i r e c t e d segments.

2 A I f PIP2 and P3P4 a r e p a r a l l e l v e c t o r s ,

then and a r e p a r a l l e l .

A I f and v a r e p a r a l l e l vec to r s , then

A v = kt7

where k = EL.

$ 1

The sum of t h e v e c t o r PlP2 and t h e v e c t o r A A P,P4 i s t h e v e c t o r P X where X i s t h e

1 A A

unique po in t such t h a t P2X = 3 - 4 2 2

I f OA and OB a r e two non-zero v e c t o r s A

which a r e not p a r a l l e l and i f OP i s any

v e c t o r i n t h e plane OAB , then t h e r e e x i s t

s c a l a r s h and k such t h a t

A I f "$ and v a r e non-zero, n o n - p a r a l l e l

vec to r s , and i f x, y , z , w a r e s c a l a r s such

t h a t

xu + yv = z u + wv , then

x = z a n d y = w .

The midpoints of t h e s i d e s of any q u a d r i l a t e r a l

a r e t h e v e r t i c e s of a para l le logram.

THEOREM 10 -14.

THEOREM 10-15.

THEOREM 10-16.

THEOREM 11-1.

COROLLARY 11-1-1.

THEOREM 11-2.

COROLLARY 11-2-1.

THEOREM 11-3.

THEOREM 11 -4.

COROLLARY 11-4-1.

The segment j o i n i n g t h e midpoints of two

s i d e s of a t r i a n g l e i s p a r a l l e l t o t h e t h i r d

s i d e and t h e l eng th of t h e segment i s one

h a l f t h e l eng th of t h e t h i r d s i d e .

A q u a d r i l a t e r a l i s a para l le logram i f and

only i f i t s d iagona l s b i s e c t each o t h e r .

Two non-zero vec to r s a r e perpendicular i f

and only i f t h e sum of t h e products of' t h e i r

r e s p e c t i v e components i s zero.

The sum of t h e measures of t h e ang les of a convex polygon of n s i d e s i s ( n - 2)

180 . The measure of each ang le of a r e g u l a r

polygon of n s i d e s i s

( n - 2)180 360 , o r 180 - - . n n

For any convex polygon of n s i d e s , t h e

sum of t h e measures of e x t e r i o r angles ,

one a t each v e r t e x of t h e polygon, i s 360 . The measure of each e x t e r i o r angle of a

360 r e g u l a r polygon of n s i d e s i s - n

The area of a r i g h t t r i a n g l e i s one h a l f

t h e product of t h e l e n g t h s of i t s two l egs .

The a r e a of a t r i a n g l e i s one-half t h e

product of any base and t h e a l t i t u d e t o t h a t

base.

The a r e a A of an e q u i l a t e r a l t r i a n g l e

whose s i d e has l eng th s i s given by:

THEOREM 11-5. The a r e a of a rhombus i s one h a l f t h e product of t h e l e n g t h s of t h e d iagona l s .

COROLLARY 11-5-1. The a r e a A of t h e square whose d iagona l

has l eng th d i s glven by

THEOREM 11-6. The a r e a of a para l le logram i s t h e product of any base and t h e a l t i t u d e t o t h a t base.

THEOREM 11-7. The a r e a of a t r a p e z o i d i s one-half t h e product of i t s a l t i t u d e and t h e sum of i t s

bases.

COROLLARY 11-7-1. The a r e a of a t r apezo id i s equal t o t h e product of i t s a l t i t u d e and t h e l eng th of i t s median.

THEOREM 11-8. Consider a s e t of two o r more t r i a n g l e s .

(a ) If t h e bases of a l l t h e t r i a n g l e s a r e equal , then t h e a r e a s o f t h e t r i a n g l e s a r e p r o p o r t i o n a l t o t h e corresponding a l t i t u d e s .

( b ) If t h e a l t i t u d e s of a l l t h e t r i a n g l e s a r e equal , then t h e a r e a s of t h e t r i a n g l e s a r e p r o p o r t i o n a l t o t h e

corresponding bases .

( c ) If t h e a r e a s of a l l t h e t r i a n g l e s a r e equal , t h e n t h e bases of t h e t r i a n g l e s a r e i n v e r s e l y p r o p o r t i o n a l t o t h e corresponding a l t i t u d e s .

THEOREM 11-9. Cons ider a set of two o r more pa ra l l e log rams .

( a ) If t h e bases of a l l t h e p a r a l l e l o g r a m s

a r e e q u a l , t h e n t h e a r e a s of t h e p a r a l l e l -

ograms a r e p r o p o r t i o n a l t o t h e c o r r e s -

ponding a l t i t u d e s .

( b ) I f t h e a l t i t u d e s of a l l t h e pa ra l l e log rams

a r e e q u a l , t h e n t h e a r e a s o f t he p a r a l l e l -

ograms are p r o p o r t i o n a l t o t h e c o r r e s -

ponding bases .

( c ) If t h e a r e a s o f a l l t h e pa ra l l e log rams

a r e e q u a l , t h e n t h e bases o f t h e

p a r a l l e l o g r a m s a r e I n v e r s e l y p r o p o r t i o n a l

t o t h e co r r e spond ing a l t i t u d e s .

THEOREM 11-10. Every s i m i l a r i t y between t r i a n g l e s h a s t h e

p r o p e r t y t h a t t h e measures of t h e t h r e e s i d e s

and any a l t i t u d e o f t h e one t r i a n g l e a r e

p r o p o r t i o n a l t o t h e measures of t h e c o r r e s -

ponding s i d e s and t h e cor responding a l t i t u d e

o f t h e o t h e r t r i a n g l e .

THEOREM 11-11.

THEOREM 11-12.

THEOREM 11-13.

Every s i m i l a r i t y between t r i a n g l e s has t h e

p r o p e r t y t h a t t h e a r e a s o f t h e t r i a n g l e s a r e

p r o p o r t i o n a l t o t h e s q u a r e s of t h e l e n g t h s

o f any p a i r of' cor responding s i d e s .

Every s i m i l a r i t y between convex polygons

w i t h n s i d e s has t h e p r o p e r t y t h a t t h e

l e n g t h s o f t h e n s i d e s and t h e p e r i m e t e r

o f one polygon a r e p r o p o r t i o n a l t o t h e l e n g t h s 1 of t h e co r r e spond ing s i d e s and t h e p e r i m e t e r '1 of t h e o t h e r polygon.

Every s i m i l a r i t y between convex polygons

w i t h n s i d e s has t h e p r o p e r t y t h a t t h e a r e a s

of t h e po lygona l - r eg ions ( c o n s i s t i n g o f t h e

polygons and t h e i r i n t e r i o r s , r e s p e c t i v e l y )

a r e p r o p o r t i o n a l t o t h e s q u a r e s of t h e

l e n g t h s of any p a i r o f co r r e spond ing s i d e s .

THEOREM 11-14.

THEOREM 11-15

THEOREM 11 -16.

THEOREM 11-17.

THEOREM 11-18.

THEOREM 11 -19.

THEOREM 11-20.

COROLLARY 11-20-1.

THEOREM 11-21.

The b i s e c t o r s of t h e i n t e r i o r a n g l e s of a r e g u l a r convex polygon of n s i d e s i n t e r s e c t

a t a p o i n t .

Every c e n t r a l t r i a n g l e of a r e g u l a r polygon

i s i s o s c e l e s and is congruent t o every o t h e r

c e n t r a l t r i a n g l e .

The a r e a of a r e g u l a r polygon I s one-half t h e

product of t h e apothem and t h e pe r imete r of

t h e polygon.

The sum of the measures of any two face

ang les of a t r i h e d r a l ang le i s g r e a t e r than

t h e measure of t h e t h i r d .face ang le .

The sum of t h e measures of a l l t h e f a c e ang les

of any polyhedra l ang le i s l e s s than 360 . There a r e no more than f i v e fcy-ses of r e g u l a r

polyhedrons.

The l a t e r a l a r e a of a prism i s e q u a l t o t h e

product of t h e l e n g t h of a l a t e r a l edge and

t h e pe r imete r of a r i g h t - sec t ion .

The lateral a r e a of a r i g h t prism i s t h e

product of t h e l eng th of a l a t e r a l edge and

t h e per imeter of a base.

Let a t r i a n g u l a r pyramid be given.

( a ) Every c r o s s - s e c t i o n of t h e pyramid i s

a t r i a n g l e s i m i l a r t o t h e boundary of

t h e base.

( b ) I f t h e d i s t a n c e from trie v e r t e x of t h e

pyramid t o t h e plane con ta in ing t h e

c r o s s - s e c t i o n i s k and i f t h e z l t i t u d e

of t h e pyramid i s h , then t h e a r e a of

t h e c r o s s - s e c t i o n and t h e a r e a of t h e

base a r e p r o p o r t i o n a l t o t h e numbers

k2 and h2 . 1003 ,CEhSREL - CSMP WAR^

., 103 S. WASHINGTON ST. ,-NDALE, ILL. 62901

THEOREM 12-1. The i n t e r s e c t i o n of a sphere wi th a plane through i t s c e n t e r i s a c i r c l e whose c e n t e r

and r a d i u s a r e t h e same a s those of t h e sphere .

THEOREM 12-2. The r a d i i of a c i r c l e o r congruent c i r c l e s , o r of a sphere o r congruent spheres , a r e congruent .

THEOREM 12-3.

THEOREM 12-4.

COROLLARY 12-4-1.

COROLLARY 12 -4 -2.

COROLLARY 12-4-3.

The d iameters c i r c l e o r congruent o r of a sphere o r congruent spheres , a r e congruent.

Given a l i n e A! and a c i r c l e C i n t h e same plane . Let P be t h e c e n t e r of t h e c i r c l e , and l e t F be t h e f o o t of t h e perpendicular from P t o t h e l i n e . (1) Every p o i n t of ^? i s o u t s i d e C i f and

only i f F i s o u t s i d e C . ( 2 ) ^Â Is a tangent t o C i f and only i f

F i s on C . ( 3 ) A? i s a secan t of C i f and only i f

F i s i n s i d e C . Given a c i r c l e and a coplanar l i n e , t h e l i n e i s a tangent t o t h e c i r c l e i f and only i f it

i s pe rpend icu la r t o a r a d i u s of t h e c i r c l e a t i t s o u t e r end.

A diameter of a c i r c l e b i s e c t s a non-diameter chord of t h e c i r c l e i f and only i f i t i s perpend icu la r t o t h e chord.

I n t h e p lane of a c i r c l e , t h e perpendicular b i s e c t o r of a chord con ta ins t h e c e n t e r of t h e c i r c l e .

COROLLARY 12-4-4. If a l i n e i n t h e p lane of a c i r c l e i n t e r s e c t s

t n e I n t e r i o r of t h e c i r c l e , t h e n it I n t e r s e c t s

t h e c i r c l e i n e x a c t l y two p o i n t s .

THEOREM 12-5.

THEOREM 12-6.

Chords of congruent c i r c l e s a r e congruent i f

and only i f they a r e e q u i d i s t a n t from t h e

c e n t e r s .

Given a p lane and a sphere S wi th

c e n t e r P . Le t F be t h e f o o t of t h e

pe rpend icu la r from P t o 772. . 1. Every p o i n t of mis o u t s i d e S i f and

only if F i s o u t s i d e S . 2. '972. i s t angen t t o S i f and only i f F

i s on S . 3. %! i n t e r s e c t s S i n a c i r c l e wi th

c e n t e r F i f and only i f F i s i n s i d e

s .

COROLLARY 12-6-1. A plane i s t angen t t o a sphere i f and only

i f it i s pe rpend icu la r t o a r a d i u s a t i t s o u t e r endpoint .

COROLLARY 12-6-2. A perpend icu la r from t h e c e n t e r of a- sphere

t o a chord of t h e sphere b i s e c t s t h e chord.

COROLLARY 12-6-3. The segment jo in ing t h e c e n t e r of a sphere

t o t h e midpoint of a chord i s perpendicular

t o t h e chord.

THEOREM 12-7. The measure of an I n s c r i b e d ang le i s h a l f t h e

measure of i t s i n t e r c e p t e d a r c .

COROLLARY 12-7-1. An ang le i n s c r i b e d i n a s e m i c i r c l e i s a r i g h t

angle .

COROLLARY 12-7-2. Angles i n s c r i b e d i n t h e same a r c a r e congruent.

COROLLARY 12-7-3. Congruent ang les i n s c r i b e d i n congruent c i r -

c l e s i n t e r c e p t congruent a r c s .

THEOREM 12-8.

THEOREM 12-9.

THEOREM 12-10.

THEOREM 12-11.

THEOREM 12-12.

THEOREM 12-13.

THEOREM 12-14,

THEOREM 12-15.

THEOREM 12-16.

I n t h e same c i r c l e o r i n congruent c i r c l e s ,

i f two chords, not diameters , a r e congruent,

then s o a r e t h e a s s o c i a t e d minor a r c s .

I n t h e same c i r c l e o r i n congruent c i r c l e s ,

i f two a r c s a r e congruent, then so a r e t h e

a s s o c i a t e d chords.

The measure of a tangent-chord ang le i s one-

h a l f t h e measure of i t s i n t e r c e p t e d a r c .

The measure OF a n ang le whose v e r t e x i s i n

t h e i n t e r i o r of a c i r c l e and whose s i d e s

a r e contained i n two s e c a n t s , i s one-half

t h e sum of t h e measures of t h e i n t e r c e p t e d

a r c s .

The measure of a secant -secant angle, o r a

tangent - tangent ang le o r a secant - tangent

ang le i s one-half t h e d i f f e r e n c e between t h e

measures of t h e i n t e r c e p t e d a r c s .

The two tangent-segments t o a c i r c l e from

an e x t e r n a l po in t a r e congruent, and form

congruent ang les wi th t h e l i n e jo in ing t h e

e x t e r n a l p o i n t t o t h e c e n t e r of t h e c i r c l e .

The product of t h e l eng th of a secant-segment

from a g iven e x t e r i o r po in t and t h e l eng th of

i t s e x t e r n a l secant-segment i s cons tant f o r

any secan t con ta in ing t h e given po in t .

Given a tangent-segment ^T t o a c i r c l e a t

T and a secan t through Q , i n t e r s e c t i n g t h e

c i r c l e I n p o i n t s R and S . Then

QR as = (QT)' . I f two chords of a c i r c l e i n t e r s e c t , t h e

product of the l e n g t h s of t h e segments of

one i s equal. t o t h e product of t h e l eng ths

of t h e segments of t h e o t h e r .

THEOREM 12-17. The q u o t i e n t of t h e circumference d iv ided by i s t h e same f o r a l l t h e d iameter , -w- ,

c i r c l e s .

COROLLARY 12-17-1. The circumference of c i r c l e s a r e p r o p o r t i o n a l t o t h e i r r a d i i .

THEOREM 12-18. 2 The a r e a of a c i r c l e of r a d i u s r i s irr .

COROLLARY 12-18-1. The a r e a s of two c i r c l e s a r e p r o p o r t i o n a l

t o t h e squares of t h e i r r a d i i .

THEOREM 12-19. A n a r c of degree measure q conta ined i n a

c i r c l e whose r a d i u s i s r has l e n g t h L , where

THEOREM 12-20. The a r e a of a s e c t o r i s h a l f t h e product of i t s r a d i u s and t h e l eng th of i t s a r c .

THEOREM 12-21. The a r e a of a s e c t o r of r a d i u s r and a r c

measure q i s 2

Â¥n-I . q . ^ THEOREM 12-22. A t r i a n g l e has one and only one circumscribed

c i r c l e . The c e n t e r o? t h i s c i r c l e i s t h e

i n t e r s e c t i o n of t h e perpendicular b i s e c t o r s

of t h e s i d e s of t h e t r i a n g l e .

THEOREM 12-23. A t r i a n g l e has one and only one i n s c r i b e d

c i r c l e . The c e n t e r of t h i s c i r c l e i s t h e

i n t e r s e c t i o n of t h e midrays of t h e ang les

of t h e t r i a n g l e .

For precl-~ely defined geometric terms the reference is to tho formal definition. For other terms the reference is to an informal definition or to tcie most prominent discussion.

A.A. similarity theorem, 413 absolute value, 81 ex., 519 acute angle, 182 addition property,

of equality, 235 of order, 63 of proportionality, 398

adjacent angles, 181 alternate Interior angles, 319 alternation property of

proportion, 399 altitude

of a parallelopam, 7 8 of a prism, 799 of a pyramid, 803 of a trapezold, 523 of a triangle, 427

"and1', 534 angle ( 13 1 , 145

acute, 182 adjacent, 181 alternate Interior, 319 bisector of, 167 central, 847 complementary 18 9 congruent, 184 consecutive Interior, 319 of a convex oolygon, 213 corresponding, 319 dihedral, 215 exterior of a polygon, 758 exterior of a triangle, 292 face, 789 inscribed. 851 - - intercepts an arc, 852 interior of a polygon, 738 interior of a triangle, 292 linear pair of, 179 measure of, 154 obtuse, 182 plane ansle of a dihedral

angle, 634 polyhedral, 788 of a quadrilateral, 204 reflex, 143 right, 182 secant-secant, 862 secant-tangent , 862 sloe of, 145 " 8trai6ht I' , 144 supplementary, 189 tansent-ciiord, 861

angle(s), (con't.) tangent-tangent, 862 of a triangle, 201 trihedral, 785 vertex of, 145 vertical, 194 II zero", 144

angle construction theorem, 160 antiparallel rays, 355 apothem of regular polygon, 775 arc (a)

congruent, 855 degree measure of, 849 endpoints of, 848 intercepted, 852 lengtn of, 900 major, 848 minor, 848 of a sector, 901 semicircle, 848

area of a circle, 855 of equilateral triangle, 755 lateral, of a prism 799 of a parailelowm, 756 of polygonal regions, 744 of a rectangle, 748 of a regular polygon, 786 of a rhombus, 755 of a sector of a circle, 501 of a square, 749, 756 of a trapezoid, 758 of triangles, 753

area relations of congruent triangles, 747 of parallelograms, 763 of triangles, 767

A.S.A. postulate, 252 axl~ms, 10 base (8)

of isosceles triangle, 277 of a parallelo&raa, 748 of a prism, 796 of a pyramid, 803 of a trapezold, 593

base angles of isosceles triangle, 277 a f a trapezoid, 593

betweenneas for saints, Â£ for rays, 165

betweenness-addition components theorem, 240 of directed segments, 690

betweenness-angles theorem, 166 of vectors, 703 composite condition, 533 concentric. 820

betweenness-co6rdinates theorem, 109

betweenness-distance theorem, 117

bisection of an ansle, 167 of a segment, 92

boundary of a. polygonal region, 732

center of a 'circle, 819 of gravity, 771 ex. of a regular polygon, 779 of a sphere, 820

central angle of a circle, 847 central triangle of a

regular polygon, 779 chess, 34 chord, 821 circle (s 1, 819

area if. 895 area o f sector of, 901 center of, 819 central angle of, 847 chord of, 821 clrc--:mf erence of, 888 circumscribed., :.05 concentric, 820 c angruent , 822 diameter of, 821 exterior of, 829 great, 821 Inscribed, 905 interior of, 829 major arc of, 848 minor $ire of, 848 power ">f a point with

respect to, 871 radius of, 819, 821 secant of, 821 sect-ir of, SO1 segment of, 503 tangent of, 830 tangent externally, 835 tangent internally, 835

circular-region, 8 9 circumference ;f a circle, 888 circumscribed circles, 905 circ3mscribed triangle, 905 collinear, 40

in tnat order, 51

conclusion; 10 concurrent lines, 597 concurrent rays, 597

in t a t order, 166 concurrent segments, 597 conditionale, 244 congruence between two convex

polygons, 460 con&ruence between two

triangles, 228 congruent

angles, 184 arcs, 855 chords, 834 circles, 822 polygons, 405 segments, 115 spheres, 822 triangles, 229

consecutive interior an&les, 319 constant of proportionality, 393 contrapoeitlve, 328

property of, 329 converse, 280

of Pythagorean theorem, 434 convex polygon(s), 211

angles of, 213 consecutive angles of, 213 diagonals of, 212 interior af, 212

convex polyheciron, 884 convex set of points, 134 coordinate jlanes, 643 coordinate of a point, 76 coordl-rmte system, 76 in a plane, 509 in apace, 641 on a line, 505 origin of, 76 unit point of, 76

coordinates of a point in a plane, 511 In space, 646

coplanar, 44 correspondence, one-to-one, 30 between triangles, 227

corresponding angles, 319 counter-exaiaole, 5 counting numbers, 55

common external tangent, 877 ex.Cros3-section of a prism, 738 common Internal tangent, 77 ex.cube, 757 complement, 189 decagon, 210

decahedron, 784 ded-ictive reasoning, 10 definitions, 15

circular, 34 complete form, 241 formal, 15 If and only if form, 242 in proofs, 241

degree, 154 degree meaaure of an arc, 849 diagonals of a convex

polygon, 212 diameter, 821 dihedral angle (a), 215

edge of, 215 face of, 215 measure of, 635 plane angle of, 634 right, 635 vertical, 216

directed se ent ( a ) , 684 equal, 685 equivalent, 686

properties of, 687 length of, 685 opposite of, 693 product with a number, 693 subtraction of, 701 ex. sun of, 697 x-companent of, 650 y-component of, 690

displacement, 683 distance,

between a point and a line, 376

between a point and a plane, 628

between two parallel lines, 354

between two points, 522, 655 measure of, 70

distance formula, in a plane, 522 in space, 655

dodecagon, 210 dodecahedron 784 'dot product', 718 edz,e of

Halfpl&ne, 138 p ~lygonal-region, 734 ex. polyhedral ansle, 788 polyhedron, 783

empty set, 25 endpoints of an arc, 848 equal directed segments, 685 equal vectors, 704

equation (8 ) , 538 equivalent, 539 intercept form, 571 ex. parametric, 546, 658 of a plane, 663 point-slope form, 569 slope-intercept form, 571 ex. two-point form, 569

equlaxigular triangle, 277 equilateral triangle, 277

area ~ f , 755 equivalent directed

segents, 686 equivalent equations, 539 Euier'a theorem, 734 ex. exterior

of an angle, 176 of a circle, 829 of a sphere, 841 of a triangle, 203

exterior angle of polygon, 738 exterior angle of triangle, 292 external secant segment, 870 externally tangent circles, 835 face angle of a

polydehral angle, 789 faces

of a polygonal-region, 734 ex. of a polyhedral ansle, 789 of a polyhedron, 783

foot of a perpendicular, 294 frustum 3f a pyramid, 805 geometrical applications

of vectors, 714 grad, 153 sraph, 514 great circle of a sphere, 821 greater than, 57 halfline, 137 half plane, 138

edge of, 138 halfspace, 139 heptagon, 210 heptahedron, 784 hexagon, 210 hexahedron, 784 horizontal lines, 510 hypo tenuse, 366 hypotenuse-leg theorem, 367 hypothesis, 10 icosahedron, 784 identity correspondence, 3.5 ex. if and only If form, 242 If-then form, 17 inciaence relations, 35

points and lines, 36 points, lines, ana planes, 42

indirect method of proof, 325 indirect reasoning, 12 inductive reasoning, 5 inequalities, 558

In the same order, 370 initial point, 684 in~cribed angle, 851 inscribed circle, 905 inscribed triangle, 905 Integers, 56

negative, 56 positive, 55

intercept form of a linear equation, 571

Intercepted arc, 852 interior

of an angle, 175 of a circle, 829 of a convex polygon, 212 of a polygonal-region, 732 of a ray, 90 of a segment, 90 of a sphere, 841 of a triangle, 202

interior angle of a polygon, 738 of a triangle, 292

Internally tangent circles, 835 intersect, 27 Intersection of Bets, 24, 534 inversely proportional, 766 inversion property of

proportion, 399 isosceles trapezoid, 593 laosceles triangle, 277

base of, 277 base angles af, 277 theorem, 275 vertex of, 277

lateral area of a priam, 799 lateral edge of a prism, 797 lateral face of a prism, 797 lateral surface of a prism, 797 leg of a right triangle, 366 leg of & trapezoid, 593 length of an arc, 900 length of a segment, 114 length of a vector, 704 less than, 64 limit, 888 line ( a )

concurrent, 597 coordinate system on, 505 horizontal, 510 intercept f o m of, 571 ex. opposite sides of, 138

parallel to a plane, 617 parametric equations of, 546 perpendicular, 183 perpendicular to a plane, 610 point-slope form of, 569 projection of a point on, 428 projection of a

segment on, 428 representation of, 35 skew, 316 slope of, 556 slope-intercept

form of, 571 ex. transversal, 317 two-point f ~ r m of, 569 undefined, 33 vertical, 510

linear pair, 179 Lobachevskian geometry, 340 locus, 538 logical equivalence, 329 logical system, 2 magnitude of a vector, 704 major arc, 948 meaaure of an angle, 154 measure of an arc, 849 measure of a dihedral angle, 635 measure of distance, 70 median of a traêzoid, 593, 758 median of a triangle, 285 midpoint of a

aegaient, 91, 526, 550 midray, 167 mil, 153 minor arc. 8443 multiplication property

of equality, 236 of order, 63

nonagon, 210 nonatiedron, 784 non-2ucliSean ~eometries, 339 null set, 29 ex. nuaoers

counting, 55 Inequality of, 57 integers, 56 irrational. 56 natural, 55 - negative, 63 order properties, 63 positive, 63 rational 56 real, 56

obtuse angle, 182 octagon, 210 octahedron, 784

one-to-one correspondence, 30

II between triangles, 228

or", 534 order,

for real numbers, 63 of colllnear points, 91

ordered pair, 511 ordered triple, 646 origin, 76, 510, 684 oriRin and nit point

theorem. 78 outer end of radius, 821 parallel lines, 316, 343

distance between, 354 properties of, 347

parallel postulate, 339 parallel rays, 355 parallel segments, 350 parallel vectors, 705 parallelepiped, 757

rectangular, 797 parallelism, 315

of a line to a plane, 617 of two planes, 617

parallelogram, 351 altitude of, 748 area of, 756 base of, 748 properties of, 603

paraneter, 546 parametric equations

in a plane, 546 in space, 658

pentagon, 210 pentahedron, 784 perpendicular , 18 3

foot of, 294 lines, 183 planes, 535 sets, 183 vectors, 717

perpendicularity of a line and a plane, 610

p i , IT , 885 plane [B 1 ,

coordinate system in, 509 equation of, 663 parallel to a line, 617 parallel to another

plane, 617 peroendicular, 635

plot. 514 point ( a )

line coordinate of, 76 plane coordinates of, 511 apace coordinates of, 646 dlatance between, 522, 655 power of with respect to

a circle, 871 representation of, 35 undefined, 33

point plotting theorem, 116 point-slope form of a

linear equation, 569 point of tangency

of a circle, 830 of a sphere, 842

P O ~ Y &on ( 8 1 angles of, 213 congruence between, 405 consecutive sides of, 210 consecutive verlcen of, 210 convex, 211 regular. 297, 779 sides of, 209 eimllar, 403 vertex of, 209

polygonal-reglon(s) , 730 area of, 744 boundary of, 732 edses of, 734 faces of, 734 interior of, 732 vertices of, 734

polyhedral anglers), 788 edge of, 788 face of, 789 face angle of, 789 vertex of, 788

polyhedron(s), 783 convex, 784 edge of, 783 face of, 783 regular, 784 section of, 784 vertex of, 783

postulate(a), 10 of Usebra, 57 of conguence,

A.S.A., 252 S.A.S., 251 5.3.3.. 253 - -

peroendlcular to a line, 510 of incidence, 35, 36, 42 representation of, 44 interior of an angle, 174 tangent, 842 parallel, 339 undefined, 33 plane separation, 1%

plane angle of a dlhe&ral proportional segaents, 412 angle, 534 protractor, 159

plane separation postulate, 138 ruler, 77

power of a point with resgec t t o a c i r c l e , 871

prime number, 5 ? r l ~ r n i s ) , 756

a l t i t u d e o f , 799 base of , 796 cross-sect ion o f , 798 l a t e r a l a rea o f , 759 l a t e r a l eiige o f , 797 l a t e r a l f ace o f , 797 l a t e r a l surface o f , 797 rec tangula r , 796 r i s h t , 797 r ig:- t-sect ion o f , 798 t o t a l a r ea o f , 79Y t r i a n g u l a r , 796

pr ismat ic surface , 797 product property of

proport ion, 399 project ion,

of a point i n t o a ?lane, 630 of a point on a l i n e , 428 of a segment on a l i n e , 428 of a s e t of po in t s I n t o a

olane, 631 of a vec tor , 720

proof, 17 f i nd ing of , 271 I n d i r e c t method, 325 paragrash form, 276 two column form, 244, 276 us ing d e f i n i t i o n s In, 241 w r i t i n g o f , 260

~ r o o e r t l e s of - - congruence, 233

f o r angles , 235 f o r s e m e n t s . 234 f o r t r i a n g l e s , 235

d i r ec t ed sezmenta. 687 equa l i t y , 253, 235 order , 63 p a r a l l e l l i n e s , 347 p a r a l l e l planes, 627 p a r a l l e l ~ ~ r a m s , 503 proport ion, 359 propor t lona l l ty , 357 rec tang les , 503 rhombuses, 603 s c a l a r products, 719 a imilar convex polygons, 406 squ i res , 603 trapezoids, 503 vec tors , 707

property of t he c o n t r a ~ o s l t i v e . 329

399 - prope r t i e s o f , 359

propor t ional , 393

propor t ional segments pos tu la te , 412

propor t iona l i ty , 392 inverse , 766 p rope r t i e s o f , 397

p ro t r ac to r , 151 p ro t r ac to r pos tu la te , 159 pyramid ( 8 ) , 803

a l t i t u d e o f , 803 base o f , 803 frustum o f , 805 regula r , 804 s l a n t neight o f , 805 ver tex o f , 803

Pythagorean theorem, 435 quadrants, 513 q u a d r i l a t e r a l s ) , 204

opposi te s i de s of , 213 opposi te v e r t i c e s o f , 213 s i d e s of , 204 v e r t i c e s o f , 204

radian, 153 rad ius

of a c i r c l e , 819, 821 ou t e r end of , 821, of a r egu la r polygon, 779 of a s e c t o r of a c i r c l e , 501

r a y b ) , 84 a n t i p a r a l l e l , 355 concurrent, 597 coordinate o f , 159 endpoint o f , 84 i n i t i a l , 143 I n t e r i o r o f , 90 opposite, 85 ordered p a i r o f , 143 p a r a l l e l , 355 s lope o f , 556 terminal, 143

ray-coordinate ayatern, 159 r e a l numoers, 56 reasoning,

deauc t i v e , 10 i n d i r e c t , 12 induct ive , 5

rec tang le , 578 a rea o f , 748 proper t i es o f , 503

rec tangula r para l le lep iped , 797 rec tangula r prism, 796 r e f l e x angle , 143 r e f l ex ive property

of congruence, f o r a n ~ l e s , 235 f a r sesaents , 234 f o r t r i a n g l e s , 235

of equa l i ty , 233 of equivalent d i r ec t ed

segments, 687

reflexive property, (con' t . ) of parallel lines, 347 of parallel planes, 627 of proportionality, 397 of similar convex

polygons, 406 regular polygon (a ) , 297

apothem of, '(79 area of, 780 center of, 779 central triangle of, 779 radius of, 779

regular polyhedron, 784 regular pyramid, 804 resultant. 709 ex. rhombus, 578

area of, 755 properties of, 603

Rieaannlan geometry, 340 right angle, 182 right dihedral angle, 635 right priam, 797 right section of a prism, 798 right triangle, 366 rotation, 143 ruler postulate, 77 S.A.A. theorem, 361 S.A.S. postulate, 251 3.A.S. similarity theorem, 421 scalar(s), 683, 703 scalar product, 718

properties of, 719 secant, 821 secant-secant angle, 862 secant-segment, 870

external, 870 section of polyhedron, 784 sector of a circle, 901

arc of, 901 area of, 901 radius of, 901

segment (s ) , 86 of a circle, 903 ex. concurrent, 597 congruent, 115 directed, 584 endpoints of, 85 interior of, 90 lensth of, 114 midpoint of, 91, 526, 550 parallel, 350 slope of, 554 tangent, 862

semicircle, 848 separation,

by a llne, 138 by a plane, 139 by a oolnt, 133

set(&), 19 convex, 134 elements of, 19 empty, 26 equality of, 20 intersection of, 24, 534 null, 29 ex. of real numoers. 55 union of, 25, 534- -

set-builder notation, 531 side,

of an angle, 145 of a llne, 138 of a plane, 139 of a quadrilateral, 204 of a trian~le, 201

similar polyams, 403 properties of, 405

skew lines, 316 slant height of a pyramid, 804 slope,

of a segment, 554 of a non-vertical line, 556 of a non-vertical ray, 556

slope-intercept form of a linear equation, 571

space, 36 coordinate system in, 641

sphere(s), 820 center of, 820 chord of, 821 concentric, 820 congruent, 822 diameter of, 821 exterior of, 841 great circle of, 821 interior of, 841 radius of, 820, 821 secant of, 821 tangent to, 842

square, 578 area of, 749, 756 properties of, 603

S . 3 . 3 . postulate, 253 5 .3 -5 . similarity theoreai, 420 "straight angle", 144 subset, 22

proper, 25 ex. substitution property, 233 supplement, 189 supplement tneorem, 189 symmetric property

of conrfruence, for angles, 235 for segments, 234 for triangles, 235

of equality, 233 of equivalent directed

segments, 687

symmetric property, (con' t. ) of parallel lines, 347 of parallel planes, 627 of proportionality, 397 of similar convex

oolygons, 406 tangent to a circle, 830

common external, 877 ex. common internal, 877 ex.

tangent-chord, angle, 861 tangent circles, 835 tangent plane, 842 tangent-secant angle, 862 tangent-segment, 869 tangent-tangent an terminal point, 68 terminus, 584

Pe* 852

tetrahedron, 784 theorem(s), 10

A.A. similarity, 422 angle construction, 160 betweenness-angle, 166 betweenness-addltlon,

for points, 240 for rays, 240

betweenness-coordinate, 109 betweenness-distance, 117 hypotenuse-leg, 367 Isoceles triangle, 275 oriffin and unit point, 78 point-plotting, 116 Pythagorean, 433

converse of, 434 3.A.A.. 351 3 . A . S . similarity, 421 3.3.3. similarity, 420 su??lernent, 185 triangle inequality, 378 two-coordinate system, 103 two-point, 108

total area of a prism, 7 s transitive property

of congruence, for angles, 235 for segments, 234 for triangles, 235

of equality, 234 of equivalent directed

sesrenta, 697 of oraer, 63 of parallel lines, 347 of oarallel planes, 627 of proportionality, 398 of simllar convex

polygons, 406 transversal, 317

trapezoid, 553 altitude of, 593 area of, 758 base of, 593 base anglee of, 553 isoceles, 593 legs of, 593 median of, 593, 758 properties of, 603

triangle(a), 201 altitude of, 427 angles of, 201 area of, 753 circumscribed, Â£0 congruent , 22 9 equiangular, 277 equilateral, 277 exterior of, 203 exterior angle of, 252 inscribed., 505 interior of, 202 interior angle of, 292 Isosceles, 277 median of, 289 right, 366 sides of, 201 vertic-es of , 201

triangle inequality tneorem, 378 triangular prism, 756 triangular-region, 730 trihedral angle, 785 two-coordinate system

theorem, 103 two-point form of a linear

equation, 569 two-point tneoren, 108 unequal in t n e same oraer, 370 union of sets, 25, 534 unit, 70 unit area, 744 unit-pair, 70 unit-point , 76 unit square, 748 vector(s), 700, 703

components of, 703 equal, 704 geometrical applications, 714 lengtn of, 704 magnitude of, 704 parallel, 705 perpendicular, 717 product with scalar, 704 projection of, 718 properties of, 707 scalar product of, 718 subtraction of, 705 sum of, 705 zero, 704

v e r t e x of a polygonal - reg ion , 734 ex . of a po lyhedra l a n ~ l e , 788 o f a ;ôlyhedran, 783 of a pyramid, 803

v e r t e x a r i s l e of a n i s o s c e l e s t r i a n g l e , 277

vertical angles, 154 v e r t i c a l l i n e s , 510 x -ax i s , 510 x-coa?onant of a d i r e c t e d

segment, 650 x-coordinate, 5-1, 645 xy-plane, 510, 543 xz-plane, 643 y-axis, 510 y-coaoonent o f a d i r e c t e d

segment, 590 y-coord ina t e , 511, 645 yz-p lane , 643 z - c o o r a l n a t e , 546 zero-ray , 155 zero v e c t o r , 704

Geometry with Coordinates - the CSMP Preservation Project

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