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ISOMETRIES OF THE PLANE
Thomas W. Shilgalis, Emeritus Professor of Mathematics, Illinois State University
Summer 2015
Definition:
An isometry is a one-to-one distance-preserving transformation.
That is, if an isometry T acts on two points P and Q to give the image points P¢ and Q¢,
the distances PQ and P¢Q¢ are equal. There are four types:
(1)
(2)
(3)
(4)
Reflection across a line k, often denoted by Mk, where k is called the axis or mirror of the
reflection.
If point P is on line k, then P¢ = P.
If point P is not on line k, then k is the perpendicular bisector of segment PP¢ .
Translation with vector v, often denoted by Tv.
Vector PP¢ is equivalent to vector v.
Rotation with center A and directed angle q , often denoted by RA,q .
RA,q (A) = A and, if P ¹ A, directed angle ÐPAP¢ = q .
Glide-reflection with vector v and axis line k, is the composite (or product) of the
translation Tv and the reflection Mk, in either order, where v and k are parallel.
A common notation is G = Tv M k (or G = M k Tv).
A transformation is said to preserve a geometric property if the image points possess the
property possessed by the original points.
An isometry preserves (in addition to distance, the defining property) collinearity of points,
concurrence of lines, betweenness of points, angle measure (but not necessarily angle
sense).
Isometries are sometimes called rigid motions since the image of a figure is a congruent figure.
Some references say that two figures are congruent if there is an isometry that maps one
onto the other.
Students are encouraged to consult Internet sources for more examples than those contained here.
These may have variations in language and notation. (ICTM oral competition questions
will follow the language and notations herein.)
(Students who have access to software such as Geometer’s Sketchpad TM and WolframAlpha
may find these useful in constructing figures to illustrate definitions and explore
properties.)
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TM
SOME GIVEN FACTS ABOUT ISOMETRIES (Students may use these without proof.)
I.
The identity isometry can be thought of as either a translation with the zero vector or a
rotation with arbitrary center and angle zero degrees.
II.
The product of two isometries is an isometry. Products obey the associative law of
function composition, but they do not always commute.
III.
The inverse of an isometry is an isometry.
IV.
An isometry that fixes three non-collinear points is the identity isometry, so two
isometries that agree at three non-collinear points are the same isometry.
V.
Translations and rotations are direct or sense-preserving, whereas reflections and glidereflections are opposite or sense-reversing. So if triangle ABC is mapped to triangle
A¢B¢C ¢ under a direct isometry, the order of traversal of vertices is the same (clockwise
or counterclockwise) for both triangles. Order of traversal is reversed if the isometry is
opposite.
VI.
The product of two translations is a translation whose vector is the sum of the two given
vectors.
VII.
The product of two rotations with the same center is a rotation whose angle is the sum of
the two given angles.
VIII.
The product of two rotations with different centers is a rotation whose angle is the sum of
the two given angles (with center to be determined) unless the sum of the two given
angles is a multiple of 360° (which includes 0°); in that case the product is a translation.
IX.
The product of two reflections across parallel lines is a translation through twice the
directed distance from the first line to the second.
X.
A translation is expressible as the product of two reflections across parallel lines that are
separated by half the length of the translation vector.
XI.
The product of two reflections across intersecting lines is a rotation through twice the
directed acute angle from the first line to the second. If the lines are perpendicular, the
angle of rotation is 180° (or -180°) and the rotation is sometimes called a half-turn.
XII.
XIII.
A rotation is expressible as the product of two reflections across intersecting lines for
which the directed acute (right?) angle from first to second is half the directed angle of
the rotation. (Note: a rotation of 400° is to be thought of as one of 40°.)
The product of three reflections is a reflection if the three axes are parallel or concurrent,
and a glide-reflection otherwise.
XIV. Every isometry is either a reflection or expressible as the product of two or three
reflections. (Cartan-Dieudonne theorem)
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XV.
Isometries are often given by a pair of linear equations:
(i)
(ii)
(iii)
Translation with vector v = < a, b > :
x¢ = x + a
y¢ = y + b
x¢ = (x - a)cosq - (y - b)sin q + a
y¢ = (x - a)sin q + (y - b)cos q + b
Note that in each of these the determinant of the coefficient matrix is 1.
1 0
cosq -sin q
For a translation it is
; for a rotation it is
.
0 1
sin q cosq
Rotation with center (a, b) and angle
q:
For reflections and glide-reflections the form is less definitive.
x¢ = x cos a + ysin a + h
For each the form is
.
y¢ = x sin a - y cos a + k
æ cos a sin a ö
Here the determinant of coefficients is -1 ç
÷ and the
è sin a -cos a ø
isometry is a reflection if there is a line of fixed points and a
glide-reflection if there are no fixed points.
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ISOMETRIES – EXAMPLES followed by EXERCISES
Example 1:
In the following figure triangle ABC, where A = (4,-1), B = (8,1), C = (9, 4), is
separately subjected to
(i)
(ii)
(iii)
(iv)
the translation with vector PQ = < 4,6 > to give triangle DEF.
the reflection with axis VW which has slope 3/2 to give triangle GHI.
the rotation with center V(-2, -3) and angle 100° to give triangle JKL.
the glide-reflection which is the product of (i) and (ii) in either order,
yielding triangle MNO.
Figure 1
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Example 2:
Using the same original triangle ABC we show the composition of (first) the rotation and
(second) the translation from example 1 to give triangle STU. This composite isometry is
a rotation with angle 100° and center to be determined. To find the center, called Z, we
proceed as follows: the product must map A to S, so the center Z must lie on the
perpendicular bisector of AS , as well as the perpendicular bisectors of BT and CU.
Two of these will suffice to locate Z as shown at (-2.517, 1.678).
This, of course, is a solution by software (GSP).
Figure 2
Now let’s use Property XV to solve this problem differently (analytically).
x¢ = (x + 2)cos(100) - (y + 3)sin(100) - 2
Part (ii) gives these equations for the rotation:
.
y¢ = (x + 2)sin(100)+ (y + 3)cos(100) - 3
The translation is given by the equations x¢ = x + 4, y¢ = y + 6.
Since the rotation is done first, the composite isometry is given by
x¢ = (x + 2)cos(100) - (y + 3)sin(100) - 2 + 4
.
y¢ = (x + 2)sin(100)+ (y + 3)cos(100) - 3+ 6
The center Z is the fixed point, that is, x¢ = x, y¢ = y, so we solve the equations
x = (x + 2)cos(100) - (y + 3)sin(100) - 2 + 4
on a calculator to give x = -2.517, y = 1.678.
y = (x + 2)sin(100) + (y + 3)cos(100) - 3+ 6
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Here are some more (and easier) examples:
Example 3:
Analyze the composite of these four reflections across (first) the line x = 1, (second) the
line x = 3, (third) the line x = 10, and (fourth) the line x = 7.
Solution: From Property VI the product of the first two reflections is the translation with
vector v1 =< 4, 0 > . The product of the third and fourth reflections is the translation with
vector v2 =< -6,0 > . By Property VI the product of all four is the translation with vector
w =< -2, 0 > .
Example 4:
Let P = (8, 2), Q = (16, 8), R = (12,11). If reflection across the line y = x maps triangle
PQR onto triangle P¢Q¢R¢ , find the area of triangle P¢Q¢R¢.
Solution: Not much to do here since the image and original triangles are congruent. The
original is a right triangle with legs 10 and 5 and area 25, so 25 is also the area of the
image triangle.
Example 5:
The translation with vector v =< 0, 4 > is followed by the rotation about the origin
through 90° (positive means counter-clockwise). The product is a rotation of 90° about
some center C. Find the coordinates of C.
Solution: Choose two easy points P and Q and find their images P¢¢ and Q¢¢ and then
find the intersection of the perpendicular bisectors of PP¢¢ and QQ¢¢. We find from a
sketch:
and
The
perpendicular bisector of PP¢¢ passes through (0, 2) and has slope 2 and the equation
y = 2x + 2 . The perpendicular bisector of QQ¢¢ passes through (-2, 0) and is vertical, so
its equation is x = -2. The intersection is C = (-2, -2).
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Example 6:
A certain isometry S is the composite of three reflections: (first) across the x-axis,
(second) across the y-axis, and (third) across the line x + y = 4.
By Property XIII the product is a glide-reflection.
Find the axis of the reflection part and the vector of the glide (translation) part.
Solution: Refer to Figure 1 and notice that the midpoints of AM , BN and CO all lie on
the axis of the reflection ( VW ) part of the glide-reflection (iv). This is useful in our
solution. We choose two easy points and find their ultimate images. Under the three
given reflections we find
and the midpoint of
and the midpoint of
PP¢¢¢ is M(2, 2). Also,
QQ¢¢¢ is N(4, 4). The line MN , whose equation is y = x, is the axis of the reflection part
of the glide-reflection S. An easy way to find the glide vector is to choose a point on
MN and find its ultimate image. Fortunately, we already have one; it’s P(0,0). Since
P¢¢¢ = (4, 4), the glide vector is w =< 4, 4 > .
Example 7:
What type of isometry has the equations x¢ =
Give details.
2
1
1
2
x+
y, y¢ =
xy?
5
5
5
5
Solution: The equations are of type (iii) in Property XV, with a = b = 0 and
2
1
cos a =
, sin a =
. The point (0, 0) is an obvious fixed point, so the isometry is a
5
5
reflection since a glide-reflection has no fixed points. Solving x¢ = x, y¢ = y gives the
line of fixed points (axis of the reflection) whose equation is y = ( 5 - 2)x.
Oral competition note:
Examples 3 through 7 could be made more complicated, but not conceptually more
difficult, by using messier given information. Since time is short, there is no point in
doing that, so the contest question writer(s) will strive to focus on concepts and minimize
drudgery. Not all of the ensuing examples and exercises will follow this guideline, so the
object here is to illustrate ideas and techniques that provide practice and should make
contest day easier than the training period, apart from the usual pressure, of course.
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Example 8:
x¢ = x cos a + ysin a + h
are not terribly helpful if, for example,
y¢ = x sin a - y cos a + k
we want equations for the reflection across the line y = 2x + 3.
Equations XV(iii):
In this example we derive equations for reflection across the line y = mx + b. First note
that (0,b) and (-b / m,0) lie on the axis of the reflection and thus are fixed points.
These facts yield the equations:
0 = bsin a + h
b = -b cos a + k
and
-b / m = (-b / m)cos a + h
0 = (-b / m)sin a + k
, so
h = -b sin a = (b / m)(1+ cos a )
k = b(1+ cos a ) = (b / m)sin a
.
Next note that a point not on the axis is reflected across the axis and that the axis is the
perpendicular bisector of the segment joining such a point and its image.
æhö
k-0
1
h
k
= - , or k = - , and = mç ÷ + b , or
For example,
and
h-0
m
m
è2ø
2
k = mh + 2b .
2bm
2b
These last two equations give h = .
2 and k =
1+ m
1+ m 2
2b
.
Then 0 = (-b / m)sin a + k yields the result: sin a =
1+ m 2
1- m 2
Also, -b sin a = (b / m)(1+ cos a ) gives cos a =
.
1+ m 2
This completes the derivation and yields the equations for the reflection in terms of the
given parts m and b:
æ1- m 2 ö æ 2m ö
2bm
x¢ = ç
2 ÷x + ç
2 ÷y 2
è 1+ m ø è 1+ m ø 1+ m
æ 2m ö æ1- m 2 ö
2b
y¢ = ç
2 ÷x - ç
2 ÷y +
è1+ m ø è 1+ m ø 1+ m 2
. (File these for future use.)
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Example 9:
In this example we use the fact (Property XII) that a rotation is expressible (factorable) as
a product of two reflections across lines passing through the center of the rotation and
forming an angle half the size of the rotation’s angle.
We use this to illustrate Property VIII: the product of two rotations with different centers
is a rotation whose angle is the sum of the two given angles. We take the case where the
sum of the two angles, first 60°, then -150°, is not a multiple of 360°.
A is the first center and B the second in the figure.
The rotation RA,60 is factorable into reflections across lines p and q if the acute angle from
p to q is 30°. In the figure we make a clever choice for line q, It’s AB. We also use AB
as the axis of the first reflection that makes up the second rotation RB,-150. We then must
use line r as shown because the acute angle from q to r must be - 75° by Property XII.
Why is the choice to use AB twice a clever one? Here’s why. Composition of functions
is associative (but not usually commutative) and we have:
RB,-150 RA,60 = (M r M q )(M q M p ) = M r (M q M q )M p = M r (I )M p = M r M p .
Thus the product of the rotations has been shown to be the product of the reflections
across p, then r, and the acute angle from the former to the latter is -45° from elementary
geometry. The composite is thus a rotation about C through -90°.
Figure 3
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Example 10:
Property XIII says that the product of three reflections is a glide-reflection when the axes
determine a triangle or when two of them are parallel but the other is not parallel to them.
Let’s see how to justify the first of these assertions.
Let the axes, in order, be L1 , L2 , L 3 . The first two reflections are equivalent to a rotation
about the common point G. We replace those first two axes by two others L4 and L5
(dashed) that form the same acute angle and so that L5 ^ L 3 .
Now replace L5 and L 3 by L6 and L 7 (dotted) so that L6 ^ L 7 at H and L6 || L 4 .
We then have M 3 ( M 2 M 1 ) = M 3 ( M 5 M 4 ) = ( M 3 M 5 ) M 4 = ( M 7 M 6 ) M 4 = M 7 ( M 6 M 4 ).
This exhibits the original product as the product of reflections across the parallel lines L 4
and L 6 followed by reflection across line L 7 which is parallel to L 4 and L 6.
The product M 6 M 4 is a translation, so the net result is a translation followed by a
reflection across a line parallel to the translation vector, i.e., a glide-reflection.
Figure 4
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Example 11:
Analyze the product of a translation and a rotation.
We show the case in which the rotation is first and the translation is second.
The other direction is left as an exercise.
Let the rotation be RA,q , shown in Figure 5 with q = 110 for illustrative purposes.
It is expressible as the product of reflections across two lines making an angle q / 2 at A.
The translation is Tv and its vector v is shown as horizontal. The translation is
expressible as the product of two reflections across, respectively, vertical lines L2 and
L3, for which the directed distance AB from the first to the second is v / 2 .
We make a clever choice: L2 is the axis of both the second reflection of the rotation and
the first reflection of the translation. We then have Tv RA,q = ( M 3 M 2 )( M 2 M 1 )
= M 3 ( M 2 M 2 ) M 1 = M 3 ( I ) M 1 = M 3 M 1. This final product is the product of reflections
across L1 then L3, which make an angle of q / 2 at C. This product is thus RC,q .
Figure 5
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Example 12:
x 2 y2
+ = 1 under the 60° rotation about (0, 0).
4 9
1
3
x¢ = x y
2
2
Solution: From Property XV(ii) the equations of this rotation are
.
3
1
y¢ =
x+ y
2
2
To find an equation of the image curve, solve these equations for x and y and substitute in
the equation of the given ellipse. We get x = x¢ / 2 + 3y¢ / 2, y = - 3x¢ / 2 + y¢ / 2 .
2
2
Substitution reduces to 21x¢ +10 3x¢y¢ + 31y¢ = 144, an equation whose graph is not
recognizable as it is, but we know that an isometry maps every figure onto a congruent
figure, so this is an equation of an ellipse. The image is shown dashed in Figure 6.
Find the image of the ellipse
Figure 6
Continuing, suppose that we are presented with a second degree equation such as
21x 2 +10 3xy + 31y2 = 144, whose graph is a conic section, possibly degenerate. The
(
)
indicator of this conic is B - 4AC = 10 3 - 4 · 21· 31 < 0, so the conic is an ellipse.
In trigonometry we learn that a rotation of axes will put the equation into a standard
form. The angle q through which the axes should be rotated is given by the equation
B
10 3
tan 2q =
, which in this case is
= - 3. Then 2q = 120 and q = 60. Now let
A-C
21- 31
1
3
3
1
u 2 v2
2
2
+ = 1.
x= uv and y =
u + v. Substitution yields 36u +16v = 144, or
4 9
2
2
2
2
2
12
2
EXERCISES
Exercise 1:
Example 1 is too messy to be an oral competition problem, but for practice try doing a
modification of it by performing the translation first and then the rotation. Make a sketch
before looking at the next figure.
Figure 7
The perpendicular bisectors of AA¢¢ and BB¢¢ meet at the center Y of this composite rotation.
Using the second method, Property XV gives these equations for the composite:
x¢ = ((x + 4) + 2)cos(100) - ((y + 6) + 3)sin(100) - 2
y¢ = ((x + 4)+ 2)sin(100)+ ((y + 6)+ 3)cos(100) - 3
The fixed point (center of the composite) is found by solving the equations:
x = ((x + 4) + 2)cos(100) - ((y + 6) + 3)sin(100) - 2
to get x = -6.517, y = -4.322.
y = ((x + 4)+ 2)sin(100)+ ((y + 6)+ 3)cos(100) - 3
Exercise 2:
The derivation for equations (Example 8) of a reflection involved m in a denominator but
the final equations do not make m = 0 an issue. Use 0 for m in these and write the
resulting equations and verify that they do indeed represent a reflection across the
horizontal line y = b.
Answer: x¢ = x, y¢ = -y + 2b.
Exercise 3:
The derivation for equations (Example 8) of a reflection does not apply to vertical lines
either. Use the definition of reflection to verify the following equations for reflection
across a vertical line x = c: x¢ = -x + 2c, y¢ = y .
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Exercise 4:
Write equations for each of these reflections – first across y = 2x + 3 and then across
y = 2x -1. Use these to verify that (Property IX) the composite is the translation
perpendicular to these two parallel lines and through twice the distance from the first to
the second.
Answer: First reflection: x¢ = (-3/ 5)x + (4 / 5)y -12 / 5, y¢ = (4 / 5)x + (3/ 5)y + 6 / 5
Second reflection: x¢ = (-3/ 5)x + (4 / 5)y + 4 / 5, y¢ = (4 / 5)x + (3/ 5)y - 2 / 5
The product has the equations: x¢ = x -16 / 5, y¢ = y + 8 / 5. This is a translation.
8 5
The translation vector has slope -1/2 and length
. The distance from (0, 3) (on the
5
2 · 0 -1· 3 -1
4 4 5
=
=
.
first line) to the second line is
2
2
5
5
2 + (-1)
Exercise 5:
Make up a glide-reflection whose reflection axis is y = 2x + 3 and verify that it has no
fixed points.
x¢ = (-3/ 5)x + (4 / 5)y + 8 / 5
One answer: Using the vector < 4, 8 >, the equations are
.
y¢ = (4 / 5)x + (3/ 5)y + 46 / 5
(Notice that it is easier to do the translation after the reflection.)
x = (-3/ 5)x + (4 / 5)y + 8 / 5
Trying to solve
for a fixed point, we get the system
y = (4 / 5)x + (3/ 5)y + 46 / 5
(8 / 5)x - (4 / 5)y = 8 / 5
, which has no solution since 46 / 5 ¹ (-1/ 2)(8 / 5).
(-4 / 5)x + (2 / 5)y = 46 / 5
Exercise 6:
Illustrate Example 9 analytically by choosing A to be (0, 0) and B to be (10, 0) and find
the coordinates of C from the equations of the composite rotation.
Solution: The first rotation maps (x, y) to (x / 2 - 3y / 2, 3x / 2 + y / 2) = ( x¢, y¢).
The second maps ( x¢, y¢) to (y + 5 3 +10, 5 - x) = ( x¢¢, y¢¢) . The composite isometry thus
has the equations x¢¢ = y + 5 3 +10, y¢¢ = -x + 5. These identify the sine of the composite
rotation as -1 and the cosine as 0, making a rotation of - 90°.
The center is found by solving the system x = y + 5 3 +10, y = -x + 5.
5 3 +15
-5 3 - 5
» 11.83, y =
» -6.83.
A calculator gives x =
2
2
Superimposing a coordinate system on Figure 3 makes this solution reasonable.
14
Exercise 7:
A certain isometry is the product of three reflections across the parallel lines x = 0, then x
= 5, then x = 12. Use the method of Example 9 above, with a clever replacement of the
first two lines, to show that the product is a reflection across x = 7. Then use the result of
Exercise 3 to verify this.
Solution: Replace the first two lines by x = 7 and x = 12, respectively, since both
products are the translation 10 units to the right. The two applications of the reflection
across x = 12 cancel each other and we are left with just the reflection across x = 7.
Exercise 8:
Do four reflections across the axes x = 5, y = 6, x = -8, y = -10 in the order given.
What is the product isometry?
Solution: The product of the first two is a half-turn about (5, 6). The product of the third
and fourth is a half-turn about (-8, -10). The product of these two half-turns is a
translation since the sum of their angles is 360°. An easy point to use to find the
translation vector is P(5, 6), which is fixed by the first half-turn and mapped by the
second half-turn to P¢(-21,-26), so the vector is < -21- 5, - 26 - 6 > = < -26,-32 > .
Exercise 9: The letters below are in the font Geneva. Some of them have one line of symmetry,
some have two, some have none, and some have point symmetry. The latter are invariant
under a half-turn.
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Which ones are in the first set? The second? The third? The fourth?
True or false: A letter in the second of these sets must be in the fourth set as well.
A letter in the fourth set must also be in the second set.
Answers: ABCDEKMTUVWY, HIOX, FGJLNPQRSZ, HINOSXZ. True. False.
Exercise 10:
The period of an isometry T, if it exists, is the smallest positive integer n for which
T n = I, the identity isometry. For example, the period of RA,90° is 4 since four
applications on a point P gets back to P. Discuss the possibilities.
Answer: The identity has period 1. No translation (other than I) has a period. Every
reflection has period 2. No glide-reflection has a period. (Note that if the glide-vector is
0 it would not be called a glide-reflection.) If a rotation whose angle q is a positive
reduced rational number p/q, then the rotation has a period that can be computed. For
example, if q = (31/ 7)° , the period is 2520 since (31/ 7)(2520) = 11160 = 31· 360 . If
q = (72 / 7)°, the period is 35 since (72 / 7)(35) = 1· 360 . But if q = 137° , the equation
n 137 = 360m , where m is a positive integer, has no integer solution for n.
15
Exercise 11:
What can you say about the product of 13 reflections?
Answer: All you can say, without knowing more about the axes of the reflections, is that
the product is sense-reversing, hence it is either a reflection or a glide-reflection.
Exercise 12:
Five reflections are composed, with axes in order the lines x = 0, x + y = 6,
y = 6, y = x + 2, y = x + 8. Is the product a reflection or a glide-reflection? Give details.
Solution: The first three have (0, 6) in common, so the product is a reflection across
some line through that point. Choose another point, say (0, 0), and find its image. It’s
(6, 6) and the midpoint is (3, 3), so the axis of the product of the first three is x + y = 6.
The fourth and fifth axes are both perpendicular to that line, so the product of all five is
the glide-reflection with axis x + y = 6 and glide vector < -6 2, 6 2 > .
Exercise 13:
Change Exercise 12 so that the fourth and fifth axes are y = 0 and y = -5.
Solution: We now have three reflections across lines (x + y = 6, y = 0, y = -5) that are
neither parallel nor concurrent, so the net result is a glide-reflection. Choose two points
P and Q and find their eventual images P¢ and Q¢. The midpoints M and N of PP¢ and
QQ¢ will determine the axis of the reflection part. We choose P(0,6) and Q(6,0) and
find P¢(0,-4) and Q¢(6,-10) and M (0,1) and N(6,-5). The axis of the reflection part
is thus x + y = 1. Since M ¢ = (5,-4), the glide-vector is MM ¢ = < 5,-5 > .
Exercise 14:
Derive equations for the glide-reflection in Exercise 13.
Solution: We could start from scratch and write the equations for all five of the given
reflections and do a lot of algebra. Instead, let’s take the result of Exercise 13 and write
equations for the reflection across x + y = 6 and follow it by the translation with vector
< 5,-5 > . For the reflection we use the equations derived in Example 8:
æ1- m 2 ö æ 2m ö
2bm
x¢ = ç
2 ÷x + ç
2 ÷y 2
x¢ = -y + 6
è 1+ m ø è 1+ m ø 1+ m
,
with
m
=
-1
and
b
=
6.
We
get:
.
y¢ = -x + 6
æ 2m ö æ 1- m 2 ö
2b
y¢ = ç
÷x - ç
÷y +
è1+ m 2 ø è 1+ m 2 ø 1+ m 2
The translation is given by x¢ = x + 5, y¢ = y - 5, so the net result is
x¢ = -y +11, y¢ = -x +1 .
16
Exercise 15:
A rotation of 90° about A(-1, 0) is followed by a rotation of 180° about B(1, 0). The first
rotation is applied again after that. Analyze the composite of these three rotations.
Solution: The sum of the four angles is 360°, so the composite is a translation (Property
VIII), so all we need to find the vector of the translation is one point and its image. A is a
good choice. We have
, so
v = AA¢¢¢ = < 0, 4 > .
Exercise 16:
Verify the other part of Property XIII in the case where the first two axes are parallel and
the third crosses them as in Figure 8 below. Provide an explanation, using the notation of
the figure, as to why the product M 3 M 2 M 1 is a glide-reflection.
Solution: The solid lines are the given ones, with the first two parallel. The last two are
replaced by L4 and L5 (dashed), which make the same acute directed angle as do L2 and
L3, approximately 70° and L4 is chosen to be perpendicular to L1.
Then ( M 3 M 2 ) M 1 = ( M 5 M 4 ) M 1 = M 5 ( M 4 M 1 ). Then L1 and L4 are replaced by L6 and
L7 (dotted), which are perpendicular just like their predecessors.
Then M 5 ( M 4 M 1 ) = M 5 ( M 7 M 6 ) = ( M 5 M 7 ) M 6. But this last product is the same as
( M 7 M 5 ) M 6 since both
M 5 M 7 and M 7 M 5 are the half-turn about H. Thus the original
product is expressible as M 7 ( M 5 M 6 ), and this is the translation through twice the
distance from L6 to L5 , followed by reflection across L7, i.e., a glide-reflection
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Figure 8
Exercise 17:
Verify Property IX, which says that two reflections across parallel lines are equivalent to
a translation through twice the directed distance from the first line to the second. Do this
by considering (i) a point on the first line, then (ii) a point between the lines, and third
(iii) a point outside the region between the two lines. Do the same for Property VII about
two rotations with the same center.
Solution: Not provided since these are easy.
Exercise 18:
Use an approach similar to Example 11 for the case in which the translation is first.
Solution: See Figure 8 and go from there to write the appropriate equation for RA,q Tv in
terms of reflections.
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Figure 9
Exercise 19:
On page 1 it is stated that an isometry preserves collinearity of points. That is, if three
points are collinear, their images under an isometry are also collinear. Give an indirect
proof of this assertion.
Answer: Let A, B,C be collinear, with B between A and C. Then AB + BC = AC.
Since an isometry preserves distance, A¢B¢ + B¢C¢ = A¢C ¢. If A¢, B¢, C ¢were not collinear,
the triangle inequality would give A¢B¢ + B¢C ¢ > A¢C ¢, a contradiction.
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Exercise 20:
In Figure 10 triangles ABC and DEF are congruent (by SSS). Construct an isometry that
maps the first triangle onto the second.
C(0, 4)
A(0, 0)
B(3, 0)
D(11, 0)
F(7 0)
E(11, -3)
Figure 10
Solution:
There are several ways to proceed. One is to translate triangle ABC through the vector
AD to get triangle DB¢C ¢ (dashed in Figure 10*). Next, rotate triangle DB¢C ¢ 90°
clockwise about D to get triangle DEC ¢¢. Then reflect triangle DEC ¢¢ across line DE to
get triangle DEF as desired. The three isometries used are two direct and one opposite,
so the product is opposite. A freehand sketch is enough to convince you that this
opposite isometry is not a reflection, so it is a glide-reflection. The next exercise asks for
details about this glide-reflection.
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C'(11, 4)
C(0, 4)
A(0, 0)
B(3, 0)
D(11, 0)
F(7 0)
B'(14, 0)
C''(15, 0)
E(11, -3)
Figure 10*
Exercise 21:
Find the axis and glide-vector of the glide-reflection in Exercise 20.
Solution: Refer to Example 6. We use the fact that the midpoints of AD, BE and CF lie
on the axis of the reflection. Call the first two midpoints M(5.5, 0) and N(7, -1.5).
An equation of MN is y = -x + 5.5. The glide-vector is A¢D = < 5.5, - 5.5 > .
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Figure 11
Exercise 22: Analyze the following isometries, giving all important information about them.
(i)
(iii)
Answers:
x¢ = y + 4
y¢ = -x -10
(ii)
x¢ = x + 3
y¢ = y - 7
(iv)
3
4
16
x- y+
5
5
5
4
3
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y¢ = - x - y +
5
5
5
x¢ =
8
15
x + y -16
17
17
15
8
y¢ = x - y + 2
17
17
x¢ =
(i) is a rotation of - 90° about the fixed point (-3, -7).
(ii) is the reflection across y = (-1/ 2)x + 4 . (Solve for fixed points.)
(iii) is the translation with vector < 3,-7 >.
(iv) is the glide-reflection with axis y = (3/ 5)x + 29 / 5 and glide-vector
< -185 /17,-111/17 > » < -10.88,-6.53 > .
Use P(0, 0) and Q(17,17) to get P¢(-16, 2) and Q¢(7,9) and midpoints
M (-8,1) and N(12,13) and then M ¢(-321/17,-94 /17). See Figure 12.
Figure 12
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Exercise 23:
Refer to Example 12 to analyze the conic section given by
16x 2 + 24 xy + 9y 2 + 60x - 80y = 0 by performing a rotation of axes to simplify the
equation into one that is recognizable.
Solution:
Following Example 12, the angle q through which the axes should be rotated satisfies
B
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24
tan 2q =
=
= . Then 2q » 73.74° and q » 36.87°, from which
A - C 16 - 9 7
cosq = 4 / 5 and sinq = 3/ 5 . We then let x = (4 / 5)u - (3/ 5)v and y = (3/ 5)u + (4 / 5)v
and substitute to get what reduces to v = u 2 / 4 , whose graph is the dashed parabola in
Figure 13. (The solid parabola, with the equation y = x 2 / 4, was rotated to create the
figure and get the dashed parabola.) The listed coordinates of P, P’ and M are relative to
the x and y axes. The distances PR and PQ are shown as they should be since P’ is the
rotated image of P.
Figure 13
Example 12 and Exercise 23 give insight into the connection between rotation as an
isometry and rotation of axes to identify the nature of a “tilted” conic section.
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