Transformations and Isometries
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1 Transformations and Isometries Definition: A transformation in absolute geometry is a function f that associates with each point P in the plane some other point P in the plane such that (1) f is one-to-one (that is, if for any two points P and Q, then P = Q). and (2) f is onto (that is, if Q is any point in the plane, then there is a point P such that ). We will often denote f(p) by P Definition: A transformation of the plane is said to have a point A as a fixed point iff f(a) = A. Definition: A transformation of the plane is said to be the identity mapping if every point of the plane is a fixed point. The identity mapping is denoted by e. Thus e(p) = P for all points P of the plane. Definition: An isometry is a transformation of the plane that preserves distances; that is, if P and Q are two points, then. Theorem: An isometry preserves collinearity, betweenness, and angles. That is, if A, B, and C are three points in the plane and their images under an isometry are A, B, and C, then: 1. If A, B, and C are collinear, then A, B, and C are also collinear. 2. If A*B*C, then A *B *C. 3. If A, B, and C are non-collinear, m ABC = m A B C For (1), suppose A, B, and C are collinear. One of the points must be between the others; WLOG, suppose A*B*C. Then AB + BC = AC, so because we have an isometry A B + B C = A C. By the triangle inequality, the only way A, B and C could fail to be collinear is if A B + B C > A C, which we have shown is not true. In addition, we also conclude A *B *C, which proves (2). For (3), note that AB = A B and AC = A C and BC = B C so by SSS ABC is congruent to A B C. By CPCF, the corresponding angles have equal measure.
2 Theorem: The inverse of an isometry is an isometry. Let f be an isometry and. Let R and S be points of the plane. Let, so. Then: thus preserves distances. Theorem: The composition of two isometries is an isometry. Choose points A and B, and isometries f and g. Then and. Theorem: In addition to preserving lines, angles, betweenness, angles, and distances, an isometry f also preserves: a. Segments: for points A and B, (and ) b. Betweenness of rays: If, then. c. Triangles: for noncollinear A, B, and C, is a triangle and. d. Circles: If is a circle with radius r and center O, then is a circle with radius r and center. For (a), if X is a point such that A*X*B, then A *X *B so X is on 1 segment. Now f is an isometry as well, so if A *Y*B,, so, so for every Y on segment there is an X (namely ) on such that. Statement (b) follows from preservation of angle measures. Statement (c) follows from SSS. Statement (d) is almost immediate from the definitions of isometry and circle.
3 Definition: Let l be a line. Define the reflection over line l as a function s that assigns to each point P a point P defined as follows: l 1. If P is on l, then s l(p) = P. 2. Otherwise, drop a perpendicular from P to l, with foot F. Find a point P on the other side of l from P that lies on the perpendicular and such that PF = FP. Let s l(p) = P. l Note that s has the property that l is the perpendicular bisector of every segment formed by a point and its image under the function.
4 Theorem: A reflection is an isometry. We show that for any two points P and Q, PQ = P Q. We consider cases: Case 1: If P and Q are both on l they are fixed so P = P and Q = Q and the result is trivial. Case 2: P is on l and Q is not. P is fixed so P = P. Then QFP Q FP by SAS, and CPCF gives QP = Q P = Q P. Case 3: Neither P nor Q are on l, but they are on the same side of l. Let F and G be the feet of the perpendiculars to l from Q and P, respectively. By SASAS, FQPG FQ P G. By CPCF, PQ = P Q. Case 4: Neither P nor Q are on l, and are on opposite sides of l. As in Case 3, let F and G be the feet of the perpendiculars to l from Q and P, respectively. Again by SASAS, FQP G FQ PG. By CPCF, PQ = P Q and QP P Q PP. Then SAS gives us Q PP QP P and CPCF gives us PQ = P Q.
5 We are now going to prove an important result about isometries: Every isometry is defined by what it does to three noncollinear points. We make use of the ideas of fixed points and the identity transformation. We need a couple of nice lemmas first: Lemma: If P and Q are points on a line such that AP = AQ and BP = BQ, then P = Q. Use the ruler postulate to establish a coordinate system. Using the notation P[x} to denote that the coordinate of point P is x. Now let A[0], B[b] with b > 0, P[p], and Q[q]. Then, so p = ±q. If p = q, then P = Q by the ruler postulate and we are done. Otherwise, suppose q = -p. Now so (p - b) = ± (q - b). If (p - b) = (q - b), then p = q, and this combined with q = -p gives us p = -p so p = 0, and P = A = Q. If (p - b) = -(q - b), then p - b = - q + b. This, together with q = -p, gives us -b = b, or b = 0, a contradiction. Thus in all possible cases, P = Q. Lemma: If an isometry f fixes two points A and B, it also fixes every point on. Let P be on. Because f is an isometry it will map to another line, and since A and B are fixed it will in fact map unique line containing A and B, namely, itself. Now if P is a point of, we need to show that. P is on to the and in fact AP = A P = AP and BP =B P = BP since f is an isometry. By the lemma above, P = P, so P is a fixed point for f.
6 Theorem: The only isometry to have three noncollinear fixed points is the identity. Let A, B, and C be noncollinear, and suppose f has A, B, and C as fixed points. That is,. We will show that for any other point P of the plane P is also a fixed point. We note first that by the previous lemma, every point on the triangle ABC is fixed. Now suppose P is a point not on ABC. By picking a point D in the interior of any side of ABC, we can guarantee that intersects the triangle exactly twice (either it contains a vertex, or PASCH guarantees that is crosses another side); call the other intersection point E. Since f fixes every point on the triangle, in particular points E and D, it fixes all of, including point P. Thus f fixes every point, and must be the identity. Theorem: If f and g are isometries and A, B, and C are noncollinear points such that, then f = g. If, then. In other words, is an isometry that fixes three noncollinear points. By the previous theorem, must be the identity, so for all points P. But this implies that for all P,.
7 Theorem: Given two congruent triangles a unique isometry that maps ABC onto DEF., there is First, note that it is sufficient to show that there is a unique isometry that maps A to D, B to E, and C to F, because segments map to segments. We build the isometry from reflections. We proceed in stages: Stage 1: Let l be the perpendicular bisector of and let s l be the reflection across line l. This maps A onto D. We now consider, where.
8 Stage 2: If s l mapped B onto E, proceed to Stage 3. Otherwise, let m be the perpendicular bisector of. We know that ED = BA = B A = B D, so B and E are equidistant from D. Thus m will contain the point D (=A ) and so D will be a fixed point for the reflection s m. Reflect over line m. This will map B onto E and leave A mapped onto D. Stage 3. If s mapped C onto F, we are done. The mapping s s is the m desired mapping. Otherwise. A = D and B = E, so has been mapped onto. Let n be the line. We know that a reflection through n will leave fixed. Moreover, we know that EF = BC = B C = EC and DF = AC = A C = DC so C and F are equidistant from both D and E. Thus is the perpendicular bisector of. Thus s n will map C onto F, while leaving A = D and B = E fixed. The mapping snss m l(the composition of s lfollowed by s m followed by s n) is thus the desired mapping that takes onto. m l
9 Theorem: Every isometry is the product of at most three reflections. Let f be an isometry and A, B, and C be noncollinear points. Let. Because f is an isometry and preserves angles and distances,. By the previous theorem, we know that is also mapped to by a product of at most three reflections. Because this product of reflections maps the three noncollinear points A, B, and C to the same points as f, f is equivalent to the product of reflections.
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