Isometries - Millersville University of Pennsylvaniasites.millersville.edu/rumble/Math.355/Book/Chapter 1.pdf · Tbe the unique point in the plane such that (T)=S=The function inverse

Chapter 1

Isometries

The first three chapters of this book are dedicated to the study of isometriesand their properties. Isometries, which are distance-preserving transformationsfrom the plane to itself, appear as reflections, translations, glide reflections, androtations. The proof of this profound and remarkable fact will result from ourwork in this and the next two chapters.

1.1 Transformations of the PlaneWe denote points (resp. lines) in R2 by upper (resp. lower) case letters such as

(resp. ) Lower case Greek letters such as denotefunctions.

Definition 1 A transformation of the plane is a function : R2 R2 with

domain R2

Definition 2 A transformation : R2 R2 is injective (or one-to-one) if and

only if the images of distinct points under are distinct, i.e., if and aredistinct points in R2 then ( ) 6= ( ).

Example 3 The transformation³£ ¤´

=£ 2¤

fails to be injective because¡£11

¤¢=

¡£11

¤¢while

£11

¤ 6= £11¤


=£+22

¤is injective as the following

argument (using the contrapositive of the implication in the definition) shows:Suppose that

¡£ ¤¢=

¡£ ¤¢Then

μ· ¸¶=

·+ 2

2

¸=

·+ 2

2

¸=

μ· ¸¶

1

2 CHAPTER 1. ISOMETRIES

and by equating and coordinates we obtain

+ 2 = + 22 = 2

A simple calculation now shows that = and = so that£ ¤

=£ ¤.

Definition 5 A transformation : R2 R2 is surjective (or onto) if and only

if every point lies in the image of i.e., given any point R2 there is some

point R2 such that ( ) =


=£ 2¤

discussed in Example 3 fails to

be surjective because there is no point£ ¤

R2 for which

³£ ¤´=£11

¤


=£+22

¤discussed in Example 4 is

surjective as the following argument shows: Let =£ ¤

be any point in the

plane. Are there choices for and such that³£ ¤´

=£ ¤? Equivalently,

does£+22

¤=£ ¤

for appropriate choices of and ? The answer is yes if thesystem

+ 2 =2 =

has a solution, which indeed it does since the determinant

¯̄̄¯ 1 22 1

¯̄̄¯ = 1 4 =

5 6= 0 By solving simultaneously for and in terms of and we find that

= 15 + 2

5

= 25

15

Therefore³£ 1

5 + 25

25

15

¤´=£ ¤

and is surjective by definition.

Definition 8 A bijective transformation is both injective and surjective.

Example 9 The transformation discussed in Examples 4 and 7 is bijective;the transformation discussed in Examples 3 and 6 is not.

Distance-preserving transformations of the plane are fundamentally impor-tant in this course. The discussion below uses the following notation: If andare distinct points, the symbols and denote distance be-

tween and the line segment connecting and the line through andand the circle centered at through respectively.

1.1. TRANSFORMATIONS OF THE PLANE 3

Definition 10 An isometry is a distance-preserving transformation : R2

R2 i.e., for all R

2 if 0 = ( ) and 0 = ( ) then = 0 0

Example 11 The identity transformation : R2 R2 defined by ( ) = is

an isometry

Proposition 12 Isometries are injective.

Proof. Let be an isometry and let and be distinct points. Then0 = ( ) and 0 = ( ) are distinct since 0 0 = 0

Isometries are also surjective, but the proof requires the following lemma:

Lemma 13 Three concurrent circles with non-collinear centers have a uniquepoint of concurrency.

Proof. We prove the contrapositive. Suppose that circles andare concurrent at two distinct points and Since all three circles share chord

their centers and are collinear since they lie on the perpendicularbisector of .

Theorem 14 Isometries are surjective.

Proof. Given an isometry and an arbitrary point show there exists apoint such that ( ) = If ( ) = then = and we’re done, soassume that = 0 = ( ) 6= Then 0 = ( ) lies on circle since

= 0 0 = 0 ( is an isometry). Again, if 0 = then = andwe’re done, so assume that 0 6= . Choose a point and notethat 4 and 4 0 0 0 are equilateral ( is an isometry). Then 0 = ( )lies on since = = 0 0 = 0. Again, if 0 = then = andwe’re done, so assume that 0 6= and consider 4 0 0 whose vertices lie on

By Lemma 13, is the unique point of concurrency for circles 0 0 and0 (see Figure 1.1). Choose a point on such that 4 = 4 0 0

with = 0 0 . I claim that 0 = ( ) = Since 0 and lieon and is an isometry, 0 = = 0 0 and 0 lies on 0 Sincecorresponding parts of congruent triangles are congruent (CPCTC) and is anisometry, 0 = = 0 0 and 0 lies on 0 Since CPCTC and is anisometry, 0 = = 0 0 and 0 lies on 0 Therefore 0 is a point ofconcurrency for 0 0 and 0 We conclude that 0 = by uniqueness.


D

C'

C

A B=A'

B'

Figure 1.1.

Definition 15 Let be an isometry, let be any point in the plane, and letbe the unique point in the plane such that ( ) = The function inverse todenoted by 1 is defined by 1 ( ) =

Proposition 16 Let and be an isometries.

1. The composition is an isometry.

2. = = i.e., the identity transformation acts as an identityelement.

3. 1 is an isometry.

Proof. The proof is left as an exercise for the reader.

As we shall see, the isometries include reflections in a fixed line, rotationsabout a fixed point, translations, and glide reflections. We begin our study ofisometries with a look at the reflections. One of the fundamental results of thiscourse is that every isometry can be written as a composition of three of fewerreflections, but that’s getting ahead of the story. We conclude this section withone additional fact about isometries.

Definition 17 A collineation is a transformation that maps lines to lines.

Proposition 18 Isometries are collineations.

1.1. TRANSFORMATIONS OF THE PLANE 5

Proof. The proof is left as an exercise for the reader.

Exercises1. Which of the following transformations are injective? Which are surjec-tive? Which are bijective?

³£ ¤´=£ 3¤ ³£ ¤´

=£cossin

¤ ³£ ¤´=£ 3 ¤

³£ ¤´=£23

¤ ³£ ¤´=£+3

¤ ³£ ¤´=£3+2

¤³£ ¤´

=£ 3 ¤ ³£ ¤´

=£ ¤ ³£ ¤´

=£+23

¤

2. Prove that the composition of transformations is a transformation.

3. Prove that the composition of isometries is an isometry.

4. Prove that the composition of functions is associative, i.e., if arefunctions, then ( ) = ( ) (Hint: Show that [ ( )] ( ) =[( ) ] ( ) for every element in the domain.)

5. Prove that the identity transformation is an identity element for the setof all transformations with respect to composition, i.e., if is a transfor-mation, then = =

6. Prove that the inverse of a bijective transformation is a bijective transfor-mation. (Remark: Exercises 2,4,5 and 6 show that the set of all bijectivetransformations is a group under composition.)

7. Prove that the inverse of an isometry is an isometry (Remark: Exercises3,4,5 and 7 show that the set of all isometries is a group under composi-tion.)

8. Let and be bijective transformations. Prove that ( ) 1 = 1

1 i.e., the inverse of a composition is the composition of the inversesin reverse order.

9. Prove Proposition 18: Isometries are collineations.

10. Let be the line with equation + + = 0. Show that each of thefollowing transformations is a collineation by finding the equation of theimage line 0 = ( )

a.³£ ¤´

=£ ¤

.


b.³£ ¤´

=£ ¤

c.³£ ¤´

=£ ¤

d.³£ ¤´

=£2

2

¤

11. Which of the transformations in Exercise 1 are collineations? For eachcollineation in Exercise 1, find the equation of the image of the line withequation + + = 0.

12. Consider the collineation³£ ¤´

=£3¤and the line 0 whose equation

is 3 + 2 = 0 Find the equation of the line such that ( ) = 0

13. Find an example of a bijective transformation that is not a collineation.

1.2 Reflections

Definition 19 A point is a fixed point for a transformation if and only if( ) =

Definition 20 Let be a line. The reflection in line is the transformation: R2 R

2 that satisfies the following two conditions:

1. Each point is a fixed point for and

2. If and 0 = ( ) then is the perpendicular bisector of 0

Theorem 21 Reflections are isometries.

Proof. Let be any line and let be the reflection in line Let andbe distinct points in the plane and let 0 = ( ) and 0 = ( ) We shall

consider several cases depending upon the position of and relative to

Case 1: Suppose that and lie on Then 0 = and 0 = so 0 0 =as required.

Case 2: Suppose that lies on and lies o of Then = 0 and is thebisector of 0

Subcase 2a: If then 0 lies on and is the midpoint of 0 so that= 0 = 0 0 as required.

Subcase 2b: Otherwise, let be the point of intersection of with 0 Observe

1.2. REFLECTIONS 7

that 4 = 4 0 by where the angles considered here are the rightangles (see Figure 1.2).

'

Figure 1.2.

Since corresponding parts of congruent triangles are congruent ( ) wehave = 0 = 0 0 as required.

Case 3: Suppose that both and lie o and on the same side of

Subcase 3a: If , let be the point of intersection of with Ifthen = = 0 0 = 0 0 and similarly if

Subcase 3b: Otherwise, let be the point of intersection of with 0 and letbe the point of intersection of with 0 Then 4 = 4 0 by

so that = 0 by (see Figure 1.3)

'

'

Figure 1.3.

Now 0 and 0 so 0 k 0. Lines and 0 are transversals

so = = 0 = 0 0 Since = 0 we have 4 =4 0 0 from which it follows that = 0 0 by .

Case 4: Suppose that both and lie o and on opposite sides of (Proofsof these subcases are left as exercises for the reader.)

Subcase 4a: If let be the point of intersection of with

Subcase 4b: Otherwise, let be the point of intersection of with 0, let

be the point of intersection of with 0 and let be the point of intersectionof with


When R2 comes equipped with a Cartesian system of coordinates, one canuse analytic geometry to calculate the coordinates of 0 = ( ) from theequation of and the coordinates of . We now derive the formulas (calledequations of ) for doing this. Let be a line with equation + + = 0 with2 + 2 0, and consider points =

£ ¤and 0 =

£ 00¤such that 0 = ( )

Assume for the moment that is o By definition, 0 So if neithernor 0 is vertical, the product of their respective slopes is -1, i.e.,

·00 = 1

or 00 =

Cross-multiplying gives( 0 ) = ( 0 ) (1.1)

Note that equation (1.1) holds when is vertical or horizontal as well. If isvertical, its equation is + = 0 in which case = 1 and = 0 But reflectionin a vertical line preserves the -coordinate so that = 0 On the other hand,if is horizontal its equation is + = 0 in which case = 0 and = 1 Butreflection in a horizontal line preserves the -coordinate so that = 0 But thisis exactly what equation (1.1) gives in either case. Now the midpoint ofand 0 has coordinates

=

· + 02+ 02

¸

Since lies on , its coordinates satisfy + + = 0 which is the equationof line Therefore μ

+ 0

2

¶+

μ+ 0

2

¶+ = 0 (1.2)

Now rewrite equations (1.1) and (1.2) to obtain the following system of linearequations in 0 and 0 : ½ 0 0 =

0 + 0 =

where = and = 2 Write this system in matrix form as· ¸ · 0

0

¸=

· ¸(1.3)

Since 2 + 2 0 the coe cient matrix is invertible, we may solve for£ 0

0¤and

obtain· 0

0

¸=

· ¸ 1 · ¸=

12 + 2

· ¸ · ¸=

12 + 2

·+

¸

1.2. REFLECTIONS 9

Substituting for and gives

0 =1

2 + 2[ ( ) + ( 2 )] (1.4)

=1

2 + 2

¡2 2 2

¢

=1

2 + 2

¡2 + ( 2 2 ) 2 2 2

¢

=22 + 2

( + + )

and similarly

0 =2

2 + 2( + + ) (1.5)

Finally, if =£ ¤

is on then + + = 0 and equations (1.4) and (1.5)reduce to

0 =0 =

in which case is a fixed point as required by the definition of We haveproved:

Theorem 22 Let be a line with equation + + = 0 where 2+ 2 0The equations of (the reflection in line ) are:

0 = 22+ 2 ( + + )

0 = 22+ 2 ( + + )

(1.6)

Remark 23 The equations of are not to be confused with the equation ofline

Exercises1. Words such as MOM and RADAR that spell the same backward and forward,are called palindromes.

a. When reflected in their vertical midlines, MOM remains MOM but the R’sand D in RADAR appear backward. Find at least five other palindromes likeMOM that are preserved under reflection in their vertical midlines.

b. When reflected in their horizontal midlines, MOM becomes WOW, but BOBremains BOB. Find at least five other words like BOB that are preservedunder reflection in their horizontal midlines.


2. What capital letters could be cut out of paper and given a single fold toproduce the figure below?

3. The diagram below shows a par 2 hole on a miniature golf course. Usea MIRA to construct the path the ball must follow to score a hole-in-oneafter banking the ball o

a. wall

b. wall

c. walls and

d. walls and

4. Two cities, located at points and in the diagram below, need to pipewater from the river, represented by line . City is 2 miles north of theriver; city is 10 miles downstream from and 3 miles north of the river.The State will build one pumping station along the river.

a. Use a MIRA to locate the point along the river at which the pumpingstation should be built so that the minimum amount of pipe is used to

1.2. REFLECTIONS 11

connect city to and city to .

10 miles

2 miles

3 miles

b. Having located point , prove that if is any point on distinct fromthen + + .

5. Given two parallel lines and in the diagram below, use a MIRA toconstruct the path of a ray of light issuing from and passing throughafter being reflected exactly twice in and once in

6. Use the fact that the path followed by a the ray of light traveling frompoint to point is the line segment to prove the following statement:When a ray of light is reflected by a flat mirror, the angle of incidenceequals the angle of reflection.

7. A ray of light is reflected by two perpendicular flat mirrors. Prove thatthe emerging ray is parallel to the initial incoming ray as indicated in thediagram below.


8. The Smiths, who range in height from 170 cm to 182 cm, wish to purchasea flat wall mirror that allows each Smith to view the full length of his orher image. Use the fact that each Smith’s eyes are 10 cm below the topof his or her head to determine the minimum length of such a mirror.

9. Graph the line with equation + 2 6 = 0 on graph paper. Plotthe point =

£53

¤and use a MIRA to locate and mark its image 0

Visually estimate the coordinates£ 0

0¤of the image point 0 and record

your estimates. Using formulas (1.6), write down the equations for thereflection and use them to compute the coordinates of the image point0 Compare these analytical calculations with your visual estimates of

the coordinates.

10. Fill in the missing entry in each row of the following table:

Equation of Point ( )

= 0£ ¤

= 0£ ¤

=£23

¤

=£ ¤

= 2£23

¤

= 3£41

¤

Equation of Point ( )

= 3£ ¤£53

¤ £83

¤£03

¤ £30

¤£ ¤ £ ¤

= 2£43

¤

2 = 3 + 5£ ¤

11. Horizontal lines and in the diagram have respective equations = 0and = 5

a. Use a MIRA to construct the shortest path from point£03

¤to point

£161

¤that first touches and then

1.3. TRANSLATIONS 13

b. Determine the coordinates of the point on and the point on touchedby the path constructed in part a.

c. Find the length of the path from to constructed in part a.

12. For each of the following pairs of points and 0, determine the equationof the reflecting line such that 0 = ( )

a.£11

¤ 0£ 11

¤b.

£26

¤ 0£48

¤

13. The equation of line is = 2 5 Find the coordinates of the imagesof£00

¤ £13

¤ £21

¤and

£24

¤under reflection in line

14. The equation of line is 2 + 3 = 0 Find the coordinates of theimages of

£00

¤ £41

¤ £35

¤and

£36

¤under reflection in line

15. The equation of line is 2 + 3 + 4 = 0; the equation of line is2 + 3 = 0 Find the equation of the line = ( )

16. Prove subcases 4a and 4b in the proof of Theorem 21.

17. For any line prove that 1 =

1.3 Translations

A translation of the plane is an isometry whose e ect is the same as sliding theplane in a direction parallel to some line for some finite distance.

Definition 24 Let and be points. The translation from to is thetransformation : R2 R

2 with the following properties:

1. = ( )

2. If = , then =

3. If 6= let be any point on and let be any point o ; let0 = ( ) and let 0 = ( ). Then quadrilaterals ¤ 0 and¤ 0 0 are parallelograms (see Figure 1.4).


'

'

Figure 1.4.

When 6= one can think of a translation in the following way: If is anypoint and is the line through parallel to then = ( ) is the pointon whose directed distance from isIn the discussion below, the symbol denotes the ray from through

Definition 25 Let and be distinct points. The vector PQ is the quantity

with magnitude and direction . The vector PP called the zero vector,has magnitude 0 and arbitrary direction. If =

£ ¤and =

£ ¤, the quantities

and are called the -component and -component of PQ respectively.

Since a vector P =£ ¤

is uniquely determined by its components, we identifythe vector P with the point =

£ ¤and write P =

Definition 26 The vector sum of PQ =£ ¤

and RS =£ ¤

is the vector

PQ+RS =

·+

+

¸

Definition 27 Let and be points. The translation by vector PQ is thetransformation PQ : R

2R2 defined by

PQ ( ) = R+PQ

Vector PQ is called the vector of PQ

The next theorem relates the two formulations of translation defined above;each has its advantages.

Theorem 28 = PQ for all points and


Proof. If = , then = = PQ. So assume that 6= If is anypoint on and is any point o let 0 = ( ) and 0 = ( )Then ¤ 0 and ¤ 0 are parallelograms by definition, in which caseAA0 = BB0= PQ Therefore, PQ ( ) = A+PQ = A+AA0= A0= 0 =

( ) and ( ) = B+PQ = B+BB0 = B0 = 0 = ( )

Corollary 29 If PQ( ) = then PQ = RS

Proof. If = then PQ = and = ; hence RS = If 6= thenPQ ( ) = = ( ) by Theorem 28. But ¤ is a parallelogram bydefinition of , hence PQ = RS and PQ = RS

Corollary 29 tells us that a translation is uniquely determined by any pointand its image. Consequently, we shall often refer to a general translationwithout specific reference to a point and its image or to a vector PQWhen we need the vector of for example, we simply evaluate at any point£ ¤

and obtain the image point£ 0

0¤; the desired components are 0 and 0

Furthermore, given PQ =£ ¤

=£ 0

0¤we immediately obtain the equations of

the translation by vector PQ:

Proposition 30 Let PQ =£ ¤

The equations for the translation PQ are

0 = +0 = +

(1.7)

Example 31 Let =£45

¤and =

£13

¤Then PQ =

£52

¤and the equations

for PQ are0 = 50 = 2

In particular, PQ

³£75

¤´=£27

¤

Theorem 32 Translations are isometries.

Proof. Let be a translation, let and be arbitrary points, let 0 = ( )and 0 = ( ) If = then 0 = 0 and = 0 = 0 0 So assumethat 6= Then ¤ 0 0 is a parallelogram by definition, in which case

= 0 0 as required. The case of collinear points 0 0 is left as anexercise for the reader.Although function composition is not commutative in general, the composi-

tion of translations is commutative. Intuitively, this says that you will arrive atthe same destination either by a move through directed distance 1 parallel toline 1 followed by a move through directed distance 2 parallel to line 2 or by


a move through directed distance 2 parallel to 2 followed by a move throughdirected distance 1 parallel to 1 The paths to your destination follow the tworoutes along the edges a parallelogram from one vertex to its diagonal opposite.This fact is the second part of the next proposition, whose proof is left as anexercise.

Proposition 33 Let and be arbitrary points.

1. The composition of translations is a translation. In fact,

RS PQ = PQ+RS

2. The composition of translations commutative:

RS PQ = PQ RS

Exercises

1. A river with parallel banks and is to be spanned by a bridge at rightangles to and .

a. Using an overhead transparency to perform a translation and a MIRA,locate the bridge that minimizes the distance from city to city .

b. Let denote the bridge at right angles to and constructed in parta. Prove that if is any other bridge spanning river distinct from andparallel to then + + + + .


2. Let be the translation such that£13

¤=£52

¤a. Find the vector of

b. Find the equations of

c. Find£00

¤ £37

¤and

£52

¤.

d. Find and such that£ ¤

=£00

¤

3. Let be the translation such that£46

¤=£710

¤a. Find the vector of


c. Find£00

¤ £12

¤and

£34

¤.

d. Find and such that£ ¤

=£00

¤

4. Let =£41

¤and =

£35

¤.

a. Find the vector of


c. Find£36

¤ £12

¤and

£34

¤.

d. Let be the line with equation 2 + 3 + 4 = 0 Find the equation ofthe line 0 = ( )

5. Complete the proof of the statement in Theorem 32: Let be a translation,let and be distinct points, let 0 = ( ) and let 0 = ( ) If 0

and 0 are collinear, prove that = 0 0

6. Prove that the composition of translations is a translation.

7. Prove the composition of translations is commutative.

8. Let be a line containing distinct points and Prove that ( ) =

9. Let and be points. Prove that 1AB = BA

10. Let and be the lines with respective equations + 2 = 0 and+ + 8 = 0

a. Compose the equations of and and show that the compositionis a translation .

b. Compare the norm of the vector of with the distance between and.


1.4 Glide Reflections

In this section we introduce glide reflections, which are certain compositionsof reflections with translations. In exercise 3 of Section 1 above, you provedthat the composition of isometries is an isometry. Thus glide reflections areautomatically isometries. Before we begin, let’s consider an example of reflectioncomposed with a translation that fails to be a glide reflection in the sense definedbelow.

Example 34 Consider the composition PQ where PQ is the translationwith vector PQ =

£20

¤and is reflection in the -axis. The equations for PQ

are 0 = + 2 and 0 = ; the equations for are 0 = and 0 = Toobtain the equations for the composition PQ compose equations as follows:0 = ( + 2) = 2 and 0 = As the reader can easily check, these areexactly the equations for reflection in line : + 1 = 0 Thus PQ =

Note that the composition = PQ in Example 34 “fixes” every lineparallel to the translation vector. This distinguishes from a glide reflection,which fixes exactly one line parallel to the translation vector.

Definition 35 A transformation fixes a set if and only if ( ) = Atransformation fixes a set pointwise if and only if ( ) = each point

Example 36 Let be a line. The reflection fixes pointwise. If PQ is anon-zero vector parallel to the translation PQ fixes but not pointwise. Thecomposition PQ fixes line but not lines distinct from and parallel to

Definition 37 A transformation : R2 R2 is a glide reflection if and only

if there exists a non-identity translation and a reflection such that

1. = and

2. ( ) =

The vector of is called the glide vector of ; the line is called the axis of .

In your mind, picture the footprints you leave when walking in the sand.Imagine that your footprint pattern extends infinitely far in either direction.Imagine a line positioned midway between your left and right footprints. Inyour mind, slide the entire pattern one-half step in a direction parallel to thenreflect in line The image pattern exactly superimposes on the original pattern.This transformation is an example of a glide reflection with axis (see Figure1.5).

1.4. GLIDE REFLECTIONS 19

Figure 1.5: Footprints fixed by a glide reflection.

As mentioned above, the following proposition is an immediate consequenceof exercise 3 in Section 1:

Proposition 38 Glide reflections are isometries.

Alternatively, we can think of a glide reflection with axis as a reflection inline followed by a translation parallel to

Proposition 39 If is a non-identity translation fixing line then =

Proof. Left as an exercise for the reader.

Since a glide reflection is the composition of a reflection in some line witha non-identity translation whose vector is parallel to , the equations of a glidereflection are easy to obtain.

Proposition 40 Let be a glide reflection with axis + + = 0 and glidevector

£ ¤where + = 0 Then the equations of are given by:

0 = 22+ 2 ( + + ) +

0 = 22+ 2 ( + + ) +

Proof. Composing the equations for a reflection given in Theorem 22 withthose for a translation given in Proposition 30 gives the result.

Example 41 Let be the line with equation 3 4 + 1 = 0; let be thetranslation with vector

£43

¤Since + = (3) (4) + ( 4) (3) = 0 the vector of

is parallel to and = is a glide reflection whose equations are

0 = 125 (7 + 24 + 94)

0 = 125 (24 7 + 83)

Let =£019

¤and 0 =

£222

¤=

¡£019

¤¢Then =

£11172

¤which is the midpoint

of and 0 lies on This is no accident, as the next proposition shows.

Proposition 42 Let be a glide reflection with axis

1. has no fixed points


2. The midpoint of point and its image ( ) lies on

3. interchanges the halfplanes of

4. fixes exactly one line, its axis

Proof. Let = be a glide reflection and let be any point. Then6= and ( ) = Let = ( ) and 0 = ( ) ; then k If lies on

then 0 = is a point on distinct from , and the midpoint of and 0 lieson But if lies o and 0 lie on opposite sides of and interchangesthe halfplanes of its axis Furthermore, 0 intersects at some point (seeFigure 1.6).

Figure 1.6.

Let be the foot of the perpendicular from to . To show that isthe midpoint of 0 it su ces to prove that 4 0 = 4 in which case

= 0 since By definition of a reflection, is the perpendicularbisector of 0; hence = 0 is the midpoint of 0. Then =and = 0 since these angles are corresponding. Also 0 and

are right angles so 4 0 = 4 by Finally, to prove part(d), let be a line fixed by and let be a point on Then 0 = ( ) lies

on and = 0 Now by part (2), the midpoint of and 0 lies onSimilarly, if 0 = ( ) then = 0 and the midpoint of and 0 lieson Thus and = = 0 =

Exercises

1. A glide reflection maps 4 onto 4 0 0 0 in the diagram below.Use a MIRA to construct the axis and glide vector of .

1.5. HALFTURNS 21

A'

B'

C'

C

B

A

2. Let be the line with equation 2 +3 = 0; let =£41

¤and =

£81

¤a. Prove that = PQ is a glide reflection and find the equations of

b. Find the image of£12

¤ £25

¤and

£32

¤under

c. Write the equations for the composite transformation PQ andcompare with the equations found in part a.

3. Prove Proposition 39: Let be a non-identity translation that fixes lineThen =

4. If = is a glide reflection with axis , prove thata. 1 = 1

b. 1 is a glide reflection with axis

5. Let be a glide reflection with axis and let be a translation that fixesUse Exercise 3 to prove that =

6. Let be a glide reflection with axis and glide vector v Given any pointon construct a point such that is the midpoint of and ( )

1.5 Halfturns

In this section we consider halfturns, which have the same e ect as 180 rotationsof the plane about some fixed point. Halfturns play an important role in ourtheoretical discussion to follow.


Definition 43 Let be a point. A transformation : R2 R2 is a halfturn

about center if and only if each point and its image 0 = ( ) are relatedin one of the following ways:

1. If = then = 0.

2. If 6= then is the midpoint of 0

Condition (1) implies that a halfturn fixes its center of rotation.

P'

Figure 1.7: A halfturn about

Remark 44 Some authors refer to the halfturn about point as “the reflectionthrough point ” In this course we use the “halfturn” terminology exclusively.

Theorem 45 Halfturns are isometries.

Proof. Let be a halfturn about a point let and be distinctpoints, let 0 = ( ) and let 0 = ( ) Suppose that either =or = If = then 0 = by condition (1) in the definition. But

= 0 by condition (2) hence = = 0 = 0 0 as desired,and similarly for = On the other hand, suppose that is distinct fromboth and . By condition (2), = 0 and = 0 Furthermore,

= 0 0, since vertical angles are congruent, and it follows bythat 4 = 4 0 0 (see Figure 1.8). Therefore = 0 0 since

'

'

Figure 1.8.

1.5. HALFTURNS 23

We now derive the equations of a halfturn about the point =£ ¤

Let

=£ ¤

and 0 =£ 0

0¤where 0 = ( ) If 6= then by definition is

the midpoint and 0 so that

=+ 0

2and =

+ 0

2

These equations simplify to

0 = 20 = 2 (1.8)

On the other hand, evaluating the equations in (1.8) at the point =£ ¤

gives

0 =0 =

This says that the center of rotation is fixed by the transformation whoseequations are given in (1.8), which proves:

Theorem 46 Let =£ ¤

be a point in the plane. The equations of the halfturnare given by

0 = 20 = 2

Example 47 Let denote the origin; the equations for the halfturn about theorigin are

0 =0 =

Recall that a function : R R is odd if and only if ( ) = ( ) Anexample of such a function is ( ) = sin( ) Let be odd and consider a point=£( )

¤on the graph of The image of under the halfturn is

μ·( )

¸¶=

·( )

¸=

·( )

¸

which is also a point on the graph of Thus fixes the graph of

Exercises


1. People in distress on a deserted island sometimes write SOS in the sand.

a. Why is this signal particularly e ective when viewed from searchingaircraft?

b. The word SWIMS, like SOS, reads the same after a halfturn about itscentroid. Find at least three other words that are preserved under ahalfturn about their centroids.

2. Try it! Plot the graph of = sin( ) on graph paper, pierce the graphpaper at the origin with your compass point and push the compass pointinto your writing surface. This provides a point around which you canrotate your graph paper. Now physically rotate your graph paper 180and observe that the graph of = sin( ) is fixed by in the sense definedabove.

3. The parallelogram “0” in the figure below is mapped to each of the othereight parallelograms by a reflection, a translation, a glide reflection or ahalfturn. Indicate which of these apply in each case (more than one mayapply in some cases).

4. Find the coordinates for the center of the halfturn whose equations are0 = + 3 and 0 = 8

5. Let =£23

¤; let be the line with equation 5 + 7 = 0

a. Find the equations of

b. Find the image of£12

¤ £25

¤and under

c. Find the equation of the line 0 = ( ) On graph paper, plot pointand draw lines and 0

6. Repeat Exercise 5 with =£32

¤

1.6. PROPERTIES OF TRANSLATIONS AND HALFTURNS 25

7. =£32

¤and =

£57

¤Find the equations of the composition

and inspect them carefully. These are the equations of an isometry wediscussed earlier in the course. Can you identify which?

8. In the diagram below, use a MIRA to find a line through intersectingcircles and in chords of equal length.

P

BA

9. Let =£ ¤

and =£ ¤

be distinct points. Find the equations ofand prove that the composition is a translation . Find the vectorof

10. For any point prove that 1 =

11. Let and be the lines with respective equations + 2 = 0 and+ 8 = 0

a. Compose the equations of and and show that the compositionis a halfturn .

b. Find the center of this halfturn and the coordinates of the point= What do you observe?

12. Prove that if = then =

1.6 Properties of Translations and HalfturnsIn this section we study some important properties of translations and halfturnsand some fascinating relationships between them.

Definition 48 A collineation is a dilatation if and only if ( )k for everyline

Theorem 49 Translations are dilatations.


Proof. Let be a translation and let be a line. Let and be distinctpoints on line ; let 0 = ( ) and 0 = ( ). Then ( ) = 0 0 since is acollineation by Proposition 18 and = AA0 = BB0 by Corollary 29. If0 and 0 are collinear, then = 0 0 If 0 and 0 are non-collinear,then lies o 0 and ¤ 0 0 is a parallelogram by definition of translation,and it follows that k 0 0 (see Figure 1.9).

' = ( )PQ

' = ( )PQ

Figure 1.9.

Theorem 50 A halfturn is a dilatation.


Theorem 51 If and are distinct points, then PQ fixes every line parallelto

Proof. Let be any line parallel to ; let be any point on line andlet 0 = PQ ( ) If 6= then ¤ 0 is a parallelogram by definition

of translation and k 0. Therefore 0 lies on and is fixed by PQ Thecase with = is left as an exercise for the reader.

Definition 52 A non-identity transformation is an involution if and only if2 =

Note that an involution has the property that 1 =

Proposition 53 A halfturn is an involution.


Proof. Let be a point and consider the halfturn . Then 2 ( ) =( ( )) = ( ) = If 6= let 0 = ( ); then is the midpoint

of and 0 Therefore 2 ( ) = ( ( )) = ( 0) = so that 2 =as claimed.

Proposition 54 A reflection is an involution.


Proposition 55 A line is fixed by the halfturn if and only if lies on

Proof. Let be a halfturn and let be a line. If lies on consider apoint on distinct from . Let 0 = ( ); by definition, is the midpointof 0 Hence 0 is on and is fixed by Conversely, suppose that lieso and consider any point on Then 0 = ( ) lies o since otherwisethe midpoint of 0, which is would lie on Therefore is not fixed by

The next theorem highlights a close relationship between halfturns and trans-lations.

Theorem 56 The composition of two halfturns with distinct centers is a trans-lation. Furthermore, if is the midpoint of and , then

= AC =

Proof. Let =£

1

2

¤=£

1

2

¤and =

£1

2

¤Since is the midpoint of

and we have = 2 = 2 so that

AC =

·2 ( 1 1)

2 ( 2 2)

¸=

·2 ( 1 1)

2 ( 2 2)

¸

and the equations for AC can be written as

0 = + 2( 1 1) = + 2 ( 1 1)0 = + 2( 2 2) = + 2 ( 2 2)

Now

( )

μ· ¸¶=

μ μ· ¸¶¶=

μ·2 1

2 2

¸¶(1.9)

=

·2 1 (2 1 )

2 2 (2 2 )

¸=

·+ 2( 1 1)

+ 2( 2 2)

¸

= AC

μ· ¸¶


A similar calculation gives

( )

μ· ¸¶=

·+ 2( 1 1)

+ 2( 2 2)

¸= AC

μ· ¸¶

completing the proof (see Figure 1.10).

Figure 1.10: A composition of two halfturns

Remark 57 Since = 2 the calculations in (1.9) tell us that the compo-sition of two halfturns with 6= is a translation through twice thedirected distance from to

Theorem 58 The composition of three halfturns with distinct centers andis a halfturn. Furthermore, if the point satisfies AB = DC then

= (1.10)

In particular, if and are non-collinear, then ¤ is a parallelogram.

Proof. Given =£

1

2

¤=£

1

2

¤and =

£1

2

¤note that

AB =

·1 1

2 2

¸=

·1 ( 1 1 + 1)

2 ( 2 2 + 2)

¸

and the point =£

1 1+ 1

2 2+ 2

¤satisfies AB = DC Thus if and are

non-collinear, then ¤ is a parallelogram. We must show that equation(1.10) holds for =

£1 1+ 1

2 2+ 2

¤. From the calculation in (1.9) above we have

( )

μ· ¸¶=

·+ 2( 1 1)

+ 2( 2 2)

¸


Applying to both sides gives

( )

μ· ¸¶=

μ·+ 2( 1 1)

+ 2( 2 2)

¸¶=

·2 1 ( + 2( 1 1))

2 2 ( + 2( 2 2))

¸

=

·2( 1 1 + 1)

2( 2 2 + 2)

¸

which is a halfturn about the point =£

1 1+ 1

2 2+ 2

¤see Figure 1.11).

Figure 1.11.

Note that Figure 1.11 shows us how to construct the center of the halfturn=

Theorem 59 Given any three of the (not necessarily distinct) pointsand the fourth point is uniquely determined by the equation

AB = (1.11)

Proof. First, given equation (1.11) we have

= ( ) ( ) (1.12)

Thus is determined by and Apply to both sides of (1.12)and obtain

( ) ( ) = ( ) ( ) =

Thus is determined by and Next, compose both sides of equation(1.11) with and obtain

AB = =

Thus given and the fourth point is the midpoint of and 0, where

0 = ( ) = ( AB ) ( ) = AB ( )


Finally apply to both sides of equation (1.11) and obtain

AB = =

Thus given and the fourth point is the midpoint of and 0, where

0 = ( ) = ( AB) ( ) = ( )

Theorem 60 For any three points and

=

Proof. By Theorem 59, given there is a point such that

=

Sequentially composing and with both sides on the left gives

=

=

=

Composing both sides with on the right gives

=

Therefore = =

Remark 61 Halfturns do not commute in general. In fact, if =then QP = PQ which holds if and only if = Thus the only halfturn thatcommutes with is itself.

Exercises1. Find all values for and such that

³£ ¤´=£ ¤

is an involution.

2. Complete the proof of Theorem 51: Let and be distinct points; letbe a point on Then 0 = PQ ( ) is a point on

3. Let and be distinct points. Prove that AB BA = AB( )


4. Use Theorem 60 to prove that any two translations commute.

5. In the figure below, sketch points such that

a. =

b. AC =

c. BC AB EA ( ) =

6. Let be a translation; let be a point. Let be the midpoint of and( )

a. Show that is the halfturn with center

b. Show that is the halfturn with center = 1 ( )

7. Prove Theorem 50: A halfturn is a dilatation.

8. Prove Proposition 54: A reflection is an involution.

9. If coin in the figure below is rolled around coin until coin is di-rectly under coin will the head on coin right-side up or up-side down?


1.7 General RotationsHalfturns are 180 rotations. In this section we define general rotations, derivetheir equations, and prove some fundamental properties. We begin with a dis-cussion of angle measure. Assume that the sine and cosine functions are definedwith respect to degree measure.

Definition 62 Given R, let =£cossin

¤and =

£cossin

¤. The directed

angle from to is the angle with initial side terminal side andmeasure A directed angle is positively directed if its measure is positive;it is negatively directed if its measure is negative.

Positively directed angles are measured counterclockwise; negatively directedangles are measured clockwise.

Definition 63 Two real numbers and are congruent mod360 if and onlyif = 360 for some Z The set = { + 360 | Z} is called thecongruence class of and we define ( ) = ( ) and + = ( + )

Each congruence class contains exactly one real number in the range 0360 For example, 10 370 and 60 300 This allows us to compare

congruence classes.

Definition 64 Given congruence classes and say that if andonly if there exist representatives and such that 0360

Consequently, 370 300 since 10 and 60 are the unique class representativesin the range 0 360

Definition 65 Let and be distinct points. The (undirected) angle from

to , denoted by is the angle with initial side and terminalside . The measure of , denoted by is the congruence classdefined as follows: Let 0 = CO ( ) and 0 = CO ( ) ; choose and so

that£cossin

¤lies on 0 and

£cossin

¤lies on 0. Define = ( )

(see Figure 1.12)

'

'Figure 1.12: = 0 0

1.7. GENERAL ROTATIONS 33

Given distinct points and there is a unique real number [0 360)such that = Note that a congruence class of angles has undefined“direction” since it contains both positive and negative directed angle measures.Nevertheless, one must be pay careful attention to the order of and since

=

Definition 66 Let be a point and let R. The rotation about throughangle is the transformation : R2 R

2 with the following properties:

1. ( ) = .

2. If 6= and 0 = ( ) then 0 = and 0 = (seeFigure 1.13).

' = ( )

Figure 1.13.

Theorem 67 A rotation is an isometry.

Proof. Let be a rotation. Let and be points with anddistinct and let 0 = ( ) and 0 = ( ) If = then by definition,

= = 0 = 0 0 So assume that and are all distinct. Ifand are non-collinear, then 4 = 4 0 0 by and = 0 0

since If and are collinear with between and then= = 0 0 = 0 0 since = 0 and = 0 by

definition, and similarly for between and But if is between andthen is between 0 and 0since 0 = 0 = Therefore= + = 0 + 0 = 0 0

We begin our derivation of the equations of a general rotation by consideringthe special case of rotations about the origin . From linear algebra we know


that a rotation about the origin is a linear transformation, i.e., given vectors£ ¤

and£ ¤

μ · ¸+

· ¸¶=

μ· ¸¶+

μ· ¸¶

for all R Let 1 =£10

¤and 2 =

£01

¤; let 0

1 = ( 1) and 02 =

( 2) Then

μ· ¸¶=

μ ·1

0

¸+

·0

1

¸¶

= ( 1) + ( 2)

= 01 +

02

and we see that is completely determined by its action on 1 and 2.

Since the directed angle measure from 1 to 01 is

01 =

·cos

sin

¸

Furthermore, the directed angle measure from 1 to 2 is 90 and from 2

to 02 is so the directed angle measure from 1 to 0

2 is + 90 and

02 = +90 ( 1) =

·cos ( + 90)

sin ( + 90)

¸=

·sin

cos

¸

Consequently,· 0

0

¸= 0

1 +02 =

·cos

sin

¸+

·sin

cos

¸=

·cos sin

sin + cos

¸

We have proved:

Theorem 68 Let R. The equations for are

0 = cos sin0 = sin + cos

Now rotate the plane about a general point =£ ¤

through directed angleby performing the following three operations in sequence:

1. Translate by vector CO;

2. Rotate about the origin through directed angle ;

3. Translate by vector OC


Then

= OC CO

and we obtain

· 00

¸=

¡OC CO

¢μ· ¸¶= OC

μ μ· ¸¶¶

= OC

μ·( ) cos ( ) sin

( ) sin + ( ) cos

¸¶

=

·( ) cos ( ) sin +

( ) sin + ( ) cos +

¸

In summary, we have proved:

Theorem 69 Let =£ ¤

and let R. The equations for the rotationare

0 = ( ) cos ( ) sin +0 = ( ) sin + ( ) cos +

Proposition 70 A non-identity rotation fixes exactly one point, namelyits center and fixes every circle with center .

Proof. That fixes follows from the definition. From the equationsof a rotation it is evident that = (the identity) if and only if = 360for some Z So 6= 0 by assumption. Now if is distinct from and0 = ( ) then 6= 0 since 0 = Therefore is the only point

fixed by Furthermore, let be a point on and let 0 = ( ). Thenby definition = = 0 so that 0 lies on Therefore ( ) =

Proposition 71 Let be a point and let R Then

= + =

Proof. Let be any point distinct from ; let 0 = ( ) and 00 =( 0) (see Figure 1.14).


' = ( )

’’ = ( ) = ( )'

Figure 1.14.

Then by definition, the directed angle measure from to 0 is the di-rected angle measure from 0 to 00 is and = 0 = 00 But

00 = ( 0) = ( ( )) = ( )( )

and the directed angle measure from to 00 is + Therefore= + by definition. Similarly, = +

Corollary 72 Involutory rotations are halfturns.

Proof. An involutory rotation is a non-identity rotation such that= = 0 By Proposition 71, = 2 so that 2 =

0 and (2 ) = 0 Hence there exists some Z such that 2 = 360 andconsequently = 180 Now cannot be even since 6= implies thatis not a multiple of 360. Therefore = 2 + 1 is odd and = 180 + 360 Itfollows that = 180 and = as claimed.

Exercises1. Find the coordinates of the points 30

¡£36

¤¢.

2. Let =£35

¤Find the coordinates of the point 45

¡£36

¤¢

3. Let be the line with equation 2 + 3 + 4 = 0

a. Find the equation of the line 30 ( )

b. Let =£35

¤Find the equation of the line 45 ( )

4. Let be a point and let R. Prove that 1 =


5. Let and be the lines with respective equations + 2 = 0 and= 3

a. Compose the equations of and and show that the compositionis a rotation .

b. Find the center and directed angle of rotation and ; comparewith the directed angle from to .

6. Let and be distinct points and let be the midpoint of andLet be a line such that PR ( ) = Using the fact that PR =prove that PR = PR .


Documents

Isometries - Millersville University of Pennsylvaniasites.millersville.edu/rumble/Math.355/Book/Chapter 1.pdf · Tbe the unique point in the plane such that (T)=S=The function inverse