Overview of Section 5.1 **Reflections, Translations, and Rotations**

Definitions:

Mapping -- a function that assigns elements of a domain to elements of a range.

Image -- the elements of a range

Preimage -- the elements of a domain

Reflection--A reflection in a line

lis a correspondence that pairs each pointPin the plane and not on linelwith a pointP' such thatlis the perpendicular bisector ofPP'. If P is on linel, thenPis paired with itself. That isP' = P.Notation: will designate a reflection (think 'mirror') of the plane in line L. (I will use Ml or

as substitute symbols.)MkFrom the definition, if we have a point B in the plane we can find its preimage A -- that is, given a line

l,(A) = B. Because every point in the plane has a preimage, the rangeMlis the whole plane and the mapping isMlonto. Thereforemaps the plane onto the entire plane.MlAlso, if A ≠ B then

(A) ≠Ml(B). SoMlis a one-to-one mapping.MlSo a reflection is a transformation of the plane.

Transformation-- a one-to-one onto mapping of the plane to the plane.

Pis a point in the plane.T(P)is the image of pointPunder the transformation.T(P) = P'is an alternative notation. P is the preimage of T(P).A transformation is a function whose range is the same as the domain.

Identity-- I is the identify function for the plane if I(P) = P for all P in the plane

Composition of transformations- We writeT1withT2-T2 º T1meaning that first T1 acts on ap point, and then T2 acts on its image. We write (T2 º T1) = T2(T1(P) for all P in the plane.

Inverse of-- (Q) = P if and only if Q =T(P).TIf follows that

º=T=T ºI

Isometry-- A transformation that preserves distance. An isometry is a transformation of the plane such that for every two points A and B the distance between A and B equals the distance between T(A) and T(B) . Or if we use the notation that A' is the image of A and B' is the image of B, the transformation is an isometry if and only if AB = A'B'. In some contexts, d(P,Q) denotes the distance between two points P and Q in the plane. For an isometry d(T(A), T(B)) = d(A, B). We also call an isometry arigid motion.

Translation-- A translation is a transformation of the plane from A to B that assigns every point in the plane point P'. If P is ont on line AB, then P' is the point in the plane for which ABPP' is a parallelogram. If P is on line AB, the P' is the point P' for which ABPP' is a degenerate parallelogram. AB is adirected segmentor avector. That is, it has both length and direction.

Rotation-- If O is any point in the plane and is a real number, then the rotation about O as a center through , denoted by , if a function from the plane to the plane that maps O onto itself and any other point P onto point P' such that OP = OP' and .

Half Turn-- A rotation through 180 degrees is a half turn. We denote with . Note

Fixed Point-- A point is invariant if it the image of itself.

Symmetry of a Figure-- Given a figureS, an isometryTis a symmetry of the figure isT(S) = S.A symmetry is simply a case where a figure remains unchanged under a transformation. Notice that symmetry is about the figure, not the individual points.

Example 5.2 lists the symmetries of a rectangle

Reflections in the lines parallel to the sides through the center

A rotation about the center

The identity transformation

Orientation-- Clockwise orientation and counterclockwise orientation.Figure orientation is

Invariant Under a Transformationif for all triples of noncollinear points A, B, C on the figure the orientation of triangle ABC is the same as that of the orientation of image triangle A'B'C', it is calleddirecttransformation; otherwise it isoppositetransformation.

Glide ReflectionThe Glide Reflection is an isometry because it is defined as the composition of two isometries: º

, where P and Q are points on lineMlor a vector parallel to linel. An issue, of course, is whether this composition is equivalent to some existing isometry -- a reflection, rotation, or translation.lAssume P ≠ Q. The resulting glide reflection has opposite orientation so it can not be a rotation or a translation. Therefore all that remains is to consider if there is a reflection through some line other than

for which A"B"C"D" is the image of ABCD. However, a reflection has all of the points on the line of reflection as fixed points whereas a glide reflection would have no fixed points. Therefore it is a new isometry.l

NOW SOLVE THIS 5.11. Show that a composition of two isometries is an isometry.

Proof: LetandT1be isometries.T2(A) = A' andT1(B) = B'. Therefore by isometryT1, AB = A'B'T1

(A) = A'' andT2(B) = B''. Therefore by isometryT2, AB = A''B''T2

ºT2T1 = T2(=T1(A))(A') = A'''.T2ºT2T1 = T2(=T1(A))(B') = B'''. Therefore under composition AB = A'''B''' and distance is preserved.T22. Show that the inverse of an isometry is an isometry.

Proof:Given T and isometry. This means T(A) = A' and T(B) = B' with AB = A'B'

The inverse of T, , maps A' to A and B' to B, so A'B' maps to AB. But A'B' = AB and so the inverse is an isometry.

3. What kind of a single isometry is

?Ml º Ml

Proof:This notation is for the composition of a reflection with itself.

maps P to P' so that lineMlis the perpendicular bisector of PP'l.

maps P' to a point P'' such that lineMl(Ml)is the perpendicular bisector of P'P''. Assume P'' ≠ P the we have a triangle PP'P'' with two adjacent side perpendicular to the same line. This is a contradiction and so P'' = P. Thereforel=Ml º MlI, the identity transformation.4. What kind of isometry is ?

The notation is that this is the inverse of a reflection in a line

l.From the definition of an inverse, (Q) = P if and only if Q = Ml(P)

If (Q) = P then line

lis the perpendicular bisector of QP. Because linelis the perpendicular bisector of QP, then Q is the image of P under Ml.If Ml(P) = Q, then line

lis the perpendicular bisector of QP. Because linelis the perpendicular bisector of QP, the P is the image of Q under

NOW SOLVE THIS 5.2

1. Show that a translation is an isometry.

Proof: Under a translation points P and Q are mapped by vector AB to points P' and Q'. ABP'P is a parallelogram with AB = P'P and ABQ'Q is a parallelogram with AB = Q'Q. Therefore PP'Q'Q is a parallelogram and under a translation and PQ = P'Q'. Therefore a translation is an isometry.2. What is the image point P under

?Answer: P

3. Based on the answer to part (2) what kind of mapping is ?

Answer: The identity.

4. Find

Given an translation a point P is mapped to P' so that ABP'P is a parallelogram.

would map P' to P and a parallelogram with the vector DE forming a parallelogram DEPP'.

But ABDE would also be a parallelogram with DE parallel to AB but in opposite directions. Therefore

=

NOW SOLVE THIS 5.3

1. In problem 1 you are given a context under which you were to consider transformation, Half Turns, about three points A, B, and C and follow the location of a point R and its subsequent images. The problem is

HA(R) = R1

HB(R1) = R2

HC(R2) = R3

HA(R3) = R3

HB(R4) = R3

HC(R5) = ?

Note that this is a composition of transformations (HC º HB º HA º HC º HB º HA)(R) = ? Does the order of the compositions matter for half turns?

This certainly asks for proof even though a GSP implementation quickly confirms that after the sequence of half turns of the images of R, the final jump return to original location.

Consider (HA º HA)(R). This appears to be the identity transformation, return R to its original site.

Does (HC º HB º HA º HC º HB º HA)(R) = (HC º HC º HB º HB º HA º HA)(R) for half turns?

ADDITIONAL EXPLORATIONS

The sequence of half-turns around a triangle leads back to the starting point. In class, it was asked if this would work (return to the starting point after some sequence of half-turns) for other figures. Allyson Hallman has done some exploration for a 4-sided figure and a 5 sided figure.

See the GSP File from Allyson.Clearly it works in n = 1. For n = 2, it appears not to work. The successive image points for a segment AB lie on two lines parallel to AB but 'spiraling' progressively larger. (construct it and see).

Allyson has the hypothesis that it works for n-sided figures where n is odd but not for n-sided figures where n is even.

If we begin with a 4-sided figure that is a square, a rectangle, a rhombus, or a parallelogram, ABCD, then for any starting point, successive have turns about ABCD will trace a path where A, B, C, and D are each midpoints of the segment. However, we know for a general quadrilateral, the midpoints of the sides will define a parallelogram (or in special cases a square, rhombus, or rectangle).

Now, if S is mapped by HA, that image mapped by HB, that image by HC, and that image mapped by HD we return to S.

Other explorations are needed.

If we have a 6-sided polygon . . . the path may be finite if the opposite sides of the hexagon are parallel. . . Test it.

2. A half-turn can be defined without reference to a rotation. Write such a definition.

Consider perpendicular lines l and m intersecting at O. Then

is a half turn about point O.Mm º Ml

3. Prove that .

Proof: Given a half-turn that maps P to P'. That is AP = AP' and the measure of angle PAP' = πmaps P' to some point P'' . Thus AP' = AP'' and the measure of angle P'AP'' is π.

Clearly P, P' and P'' lie on a circle with center at A. Assume P ≠ P'' Then PP' and P'P'' are both diameters and we have a contradiction.

Therefore

Now Solve This 5.4

List all of the symmetries for the following figures. Recall that a symmetry of a figure is a transformation that maps the figure onto itself

1. Equilateral triangle:

I, RG,120, RG,240, Md, Me,Mf

2. A square

I, RC,90 , RC,180 , RC,270 , Mh, Mv,Md1,Md2

3. A regular pentagon

10 of them: I, Rotations of 72, 144, 216, and 288 degrees, 5 reflections about lines through a vertex perpendicular to the opposite side.

4. A regular hexagon

12 of them: I, Rotations of 60, 120, 180, 240, and 300 degrees, 3 reflections about diagonals, 3 reflections about lines perpendicular to opposite sides.

5. A circle

An infinite number.

6. A segment

2 of them. I and a reflection about the perpendicular bisector.

7. A straight line

An infinite number

Now Solve This 5.5

1. Use the concept of orientation to prove that a rectangle has only four symmetries. In Example 5.2 we identified the four symmetries as

andMl, Mk, HO.IHow does finding the following identities contribute to a proof that there are only four symmetries of the rectangle? Essentially we are testing all the possibilities of composition of two of these four symmetries and showing that any composition leads to one of the existing four symmetries.

Show

.Mk º Ml = Ml º Mk = HOSee Prob 2 of NST 5.3. A half turn is the composition of two perpendicular reflections.

Find

andMl º HOHO º MlEach of them is equal to

Mk.Further,

Mk º HO = HO º Mk = MlShow

= I (see Prob 3 of NST 5.1)

= I (See Prob 3 of NSF 5.1)

= I

2. In the Cartesian coordinate plane, let

Mkbe the reflection in the x-axis andMlthe reflection in the y-axis andHObe a half-turn about the origin. Showa.

Mk(x, y) = (x, -y)b.

Ml(x, y) = (-x, y)c.

HO(x,y) = (-x, -y)3. Assume that the lines k and l in example 5.2 are the x-axis and the y-axis, respectively, and notice that for all points (a,b):

In a similar way show that

(a)

Mk º Ml = HO(b)

Ml º HO = Mk(c)

HO º Ml = Mk(d) = = I

Now Solve This 5.6

1. Consider the glide reflection º , P ≠ Q. Choose the line

to be the x-axis, and show that the image of any pointlto under the glide reflection is(x,y)for some fixed, non-zero number(x+a, -y).aLet

be atPand(0,0)be atQ. Take a point(a,0)atK. Under the reflection(x,y),Mlis mapped toK. SinceK'bisectsl', the coordinates ofKKwill beK'. Now the translation of(x,-y)toK'moves along a line parallel toK", the x-axis a distance oflto form a parallelograma. The coordinates ofPQK"K'will therefore beK". Thus .(x+a,-y)

2. Using the result form part 1, prove that a glide reflection in which the translation is not by a zero vector has no fixed point.

Proof: Assume thatis a fixed point. Then by a glide reflection(r,s)maps to(r,s). But by part 1(r,s)maps to(r,s). This would mean that(r+a,-s)andr = r + a. The first would mean thats = -sbut that contradicts a nonzero vector. Therefore there is no fixed point in the glide reflection with a translation having a nonzero vector.a = 0

3. Prove that if º

is a glide reflection, then ºMl=Mlº .MlAs we did in part 1, choose the line

to be the x-axis. We will examine the compositionlº .MlLet

be atPand(0,0)be atQ. Take a point(a,0)atK. Under the translation ,(x,y)is moved toKalong a vector of lengthK'parallel toato form a parallelograml. The coordinates ofPKK'Q' will beK. Then under a reflection(x+a, y),Mlis mapped toK'so that the segmentK''' is bisected by the lineK'K'. The coordinates oflwill beK"since(x+a,-y)is the x-axis. Therefore the compositionlº is the same glide reflection as ºMl,Ml

4. Prove that a glide reflection composed with itself is a translation.

Let be the first glide reflection. In composition,

is mapped to(x+a, -y)=(x+a+a, -(-y))which is a translation along a vector parallel to line(x+2a, y)with a vector of lengthl.2a5.

6.