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 PN 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 PN 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 AN, BN, and CN , then: 1. If A, B, and C are collinear, then AN, BN, and CN are also collinear. 2. If A*B*C, then AN*BN*CN. 3. If A, B, and C are non-collinear, mpABC = mpANBNCN ~ 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 ANBN + BNCN = ANCN. By the triangle inequality, the only way AN, BN and CN could fail to be collinear is if ANBN + BNCN > ANCN, which we have shown is not true. In addition, we also conclude AN*BN*CN, which proves (2). For (3), note that AB = ANBN and AC = ANCN and BC = BNC N so by SSS ªABC is congruent to ªANBNCN. By CPCF, the corresponding angles have equal measure. 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 AN*XN*BN so XN is on segment . Now f–1 is an isometry as well, so if AN*Y*BN, , 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. Definition: Let l be a line. Define the reflection over line l as a N function sl that assigns to each point P a point P defined as follows: 1. If P is on l, then sl(P) = P. 2. Otherwise, drop a perpendicular from P to l, with foot F. Find a point PN on the other side of l from P that lies N on the perpendicular and such that PF = FP . Let sl(P) = PN. Note that sl has the property that l is the perpendicular bisector of every segment formed by a point and its image under the function. Theorem: A reflection is an isometry. ~ We show that for any two points P and Q, PQ = PNQN. We consider cases: Case 1: If P and Q are both on l they are fixed so P = PN and Q = QN and the result is trivial. Case 2: P is on l and Q is not. P is fixed so P = PN. Then ÎQFP QNFP by SAS, and CPCF gives QP = QNP = QNPN. 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 FQNPNG. By CPCF, PQ = PNQN. 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, FQPNG FQNPG. By CPCF, PQN = PNQ and pQPNP pQNPPN. Then SAS gives us ÎQNPPN ÎQPNP and CPCF gives us PQ = PNQN. 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 to the unique line containing A and B, namely, itself. Now if P is a point of , we need to show that . PN is on and in fact APN = ANPN = AP and BPN =BNPN = BP since f is an isometry. By the lemma above, P = PN, so P is a fixed point for f. 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 ªABC is fixed. Now suppose P is a point not on ªABC. By picking a point D in the interior of any side of ªABC, 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, . Theorem: Given two congruent triangles , there is a unique isometry that maps ÎABC onto ÎDEF. ~ 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 sl be the reflection across line l. This maps A onto D. We now consider , where . Stage 2: If sl mapped B onto E, proceed to Stage 3. Otherwise, let m be the perpendicular bisector of . We know that ED = BA = BNAN = BND, so BN and E are equidistant from D. Thus m will contain N the point D (=A ) and so D will be a fixed point for the reflection s.m Reflect over line m. This will map BN onto E and leave A mapped onto D. N Stage 3. If smm mapped C onto F, we are done. The mapping ssl is the desired mapping. Otherwise. AO = D and BO = 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 = BOCO = ECO and DF = AC = AOCO = DCO so CO and F are equidistant from both D and E. Thus is the perpendicular O O bisector of . Thus sn will map C onto F, while leaving A = D O and B = E fixed. The mapping ssnmsl (the composition of sl followed by smn followed by s ) is thus the desired mapping that takes onto . 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.