Overview of Section 5.1 Reflections, Translations, and Rotations
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
A reflection in a line l is a correspondence that pairs each point P in the plane and not on line l with a point P' such that l is the perpendicular bisector of PP'. If P is on line l, then P is paired with itself. That is P' = P.
Notation: will designate a reflection (think 'mirror') of the plane in line L. (I will use Ml or Mk as substitute symbols.)
From the definition, if we have a point B in the plane we can find its preimage A -- that is, given a line l, Ml(A) = B. Because every point in the plane has a preimage, the range Ml is the whole plane and the mapping is onto. Therefore Ml maps the plane onto the entire plane.
Also, if A ≠ B then Ml(A) ≠ Ml(B). So Ml is a one-to-one mapping.
So a reflection is a transformation of the plane.
Transformation -- a one-to-one onto mapping of the plane to the plane.
P is a point in the plane. T(P) is the image of point P under 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 T1 with T2 - - We write T2 º T1 meaning 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 T -- (Q) = P if and only if Q = T(P).
If follows that º T = T º = I
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 a rigid 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 a directed segment or a vector. 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 figure S, an isometry T is a symmetry of the figure is T(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 Transformation if 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 called direct transformation; otherwise it is opposite transformation.
The Glide Reflection is an isometry because it is defined as the composition of two isometries: º Ml, where P and Q are points on line l or a vector parallel to line l. An issue, of course, is whether this composition is equivalent to some existing isometry -- a reflection, rotation, or translation.
Assume 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 l 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.
NOW SOLVE THIS 5.1
1. Show that a composition of two isometries is an isometry.
Proof: Let T1 and T2 be isometries. T1(A) = A' and T1(B) = B'. Therefore by isometry T1, AB = A'B'
T2(A) = A'' and T2(B) = B''. Therefore by isometry T2, AB = A''B''
T2 º T1 = T2(T1(A)) = T2(A') = A'''. T2 º T1 = T2(T1(A)) = T2(B') = B'''. Therefore under composition AB = A'''B''' and distance is preserved.
2. Show that the inverse of an isometry is an isometry.
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 ?
This notation is for the composition of a reflection with itself. Ml maps P to P' so that line l is the perpendicular bisector of PP'.
Ml(Ml) maps P' to a point P'' such that line l 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. Therefore Ml º Ml = I , 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 l is the perpendicular bisector of QP. Because line l is the perpendicular bisector of QP, then Q is the image of P under Ml.
If Ml(P) = Q, then line l is the perpendicular bisector of QP. Because line l is 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 ?
3. Based on the answer to part (2) what kind of mapping is ?
Answer: The identity.
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?
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.
Open GSP file for this one.
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 Mm º Ml is a half turn about point O.
3. Prove that .
Proof: Given a half-turn that maps P to P'. That is AP = AP' and the measure of angle PAP' = π
maps 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.
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 Ml, Mk, HO and I.
How 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 = HO.
See Prob 2 of NST 5.3. A half turn is the composition of two perpendicular reflections.
Find Ml º HO and HO º Ml
Each of them is equal to Mk.
Further, Mk º HO = HO º Mk = Ml
= I (see Prob 3 of NST 5.1)
= I (See Prob 3 of NSF 5.1)
2. In the Cartesian coordinate plane, let Mk be the reflection in the x-axis and Ml the reflection in the y-axis and HO be a half-turn about the origin. Show
a. 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 l to be the x-axis, and show that the image of any point (x,y) to under the glide reflection is (x+a, -y) for some fixed, non-zero number a.
Let P be at (0,0) and Q be at (a,0). Take a point K at (x,y). Under the reflection Ml, K is mapped to K'. Since l bisects KK', the coordinates of K' will be (x,-y). Now the translation of K' to K" moves along a line parallel to l, the x-axis a distance of a to form a parallelogram PQK"K'. The coordinates of K" will therefore be (x+a,-y). Thus .
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 that (r,s) is a fixed point. Then by a glide reflection (r,s) maps to (r,s). But by part 1 (r,s) maps to (r+a,-s). This would mean that r = r + a and s = -s. The first would mean that a = 0 but that contradicts a nonzero vector. Therefore there is no fixed point in the glide reflection with a translation having a nonzero vector.
3. Prove that if º Ml is a glide reflection, then º Ml = Ml º .
As we did in part 1, choose the line l to be the x-axis. We will examine the composition Ml º .
Let P be at (0,0) and Q be at (a,0). Take a point K at (x,y). Under the translation , K is moved to K' along a vector of length a parallel to l to form a parallelogram PKK'Q. The coordinates of K' will be (x+a, y). Then under a reflection Ml, K' is mapped to K'' so that the segment K'K'' is bisected by the line l. The coordinates of K" will be (x+a,-y) since l is the x-axis. Therefore the composition Ml º is the same glide reflection as º Ml,
4. Prove that a glide reflection composed with itself is a translation.
Let be the first glide reflection. In composition, (x+a, -y) is mapped to (x+a+a, -(-y)) = (x+2a, y) which is a translation along a vector parallel to line l with a vector of length 2a.
Problem Set 5.1