transform_geom_II - Transformational Geomegy Part II...

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Unformatted text preview: Transformational Geomegy --- Part II Theorems (NIB) 5.2 — 5.4 below fully characterize isometries of the Euclidean Plane. They show that every isometry of the Euclidean plane is one of the following: (1) areflection, (2) a composition of two reflections, or (3) a composition of three reflections . They also show that every isometry of the Euclidean plane is one of the following: (1) a translation, (2) a rotation, (3) a reflection , or (4) a glide reflection. Theorems (NIB) 5.2 and 5.3 are proved in Neutral Geometry so that they may also be used in the study of Hyperbolic Geometry. Theorem g NIB) 5.2 (The Congment Triangles Isomet_ry Theorem): Suppose that points A, B and C, and points D, E, and F are sets of three non-collinear points each in Neutral Geometry such that AABC 2 ADEF. Then, there is an isometry t which maps AABC onto ADEF, that is, there is an isometry t such that t(A) = D, t(B) = E, and t(C) = F , and this transformation t can be chosen D = t(A) such that at least one of the following is true: 3 t pl: _ ac) 1) t is a reflection about a mirror line, /\E= “B, 2) t is a composition of two reflections, 3) t is a composition of three reflections. c Proof of Theorem 5.2 The Con ent Trian les Isome Theorem : The proof is accomplished by constructing the isometry t as a composition of reflections in a step-by-step manner, considering all cases of the possible placement of the target triangle ADEF as it relates to the position of the source triangle AABC . The first case to consider is the possibility that AABC and ADEF are names of the same triangle. IfA=D andB=E andC=F, B E then we can choose t to be t = the identity transformation where t(X) = X, for every point X inthe plane. Then, t is an isometry such that t(A)=A=D,t(B)=B=E, andt(C)=C=F. In this case, t is the composition of two reflections, where both A = D reflections have the same mirror line B . _ Forexample,ifwelet line e = 71; , then t = Re 0 Re. LI" 6 C=F Thus, we can assume that one of the points A, B, or C is distinct from D, E, or F, respectively. Relabelling the points (if necessary), we assume that A 9e D . Let line 8 be the perpendicular bisector of segment E . Let transformation t1 = R e . Since the mirror line e is the perpendicular bisector of 213 , t1 (A) = R e (A) = D , by definition of reflections. Thus, A’ = t1(A) = D. LetB’= t1(B) and C’= t1(C). Now, t1 may or may not be the complete transformation t we are seeking. It depends on the positions of E and F of the target triangle ADEF . lfB'= E and C’= F, E=B'=t(3) then ADEF is positioned o = A' = t(A) so that it is the mirror image AABC and F = c' = t(C) we can choose t = t1, Mirror lin e and we have proven the theorem A = th , perpendicular bisector ofzfi statement in the case that B’ = E and C’ = F , where tis the reflection Re and line fl is the perpendicular bisector of segment 25 . Now, stillassumingthat t1 = Re, with A’ = t1(A), B' = t1(B) and C’= t1(C) andwith A 75 D but with A’ = D, weknowthat B'= E or B'96 E. We assume first that B ’ = E . In this case , we can assume that C ’ 7i F, because the theorem statement was just proved to be true if C ’ = F . Thus,inthis case,weareassumingthat A’ = D , B’ = E, and C’ 75 F. Let line 11 be the perpendicular bisector of the segment 5’? . Let t 2 = Rn . Thus, t2( C') = RI1 (CI) = F , by the definition of reflection. We continue the construction of the isometry t by following the reflection t1 by the reflection t2. Thus, we consider the transformation t2 0 t1. W6WlllWI‘lt6 A”: (tzo t1)(A) , B”: (tzo t1)(B) , CH: (tzo NOW, A”: t2(t1(A)) = t2(D) and B”: t2(t1(13)) = t203)- Weshowthat t2(D) = D and that t2(E) = E byshowingthatD and E are points on the mirror line n. 25 _ 5? since AABC .2. ADEF , by assumption. I2 ll :15 ; A'C’ _ DEsince t1 isanisometry. D is equidistant from C l and F , and so, by Theorem 3.2.8, D is a point on line 11, the perpendicular bisector of C'F . Mirror Line n = the perpendiculaL bisector of C'F « rpendicular__ ' isector of AD Similarly, F5 2 5 since AABC z ADEF , by assumption. EE B'C’ _ E—C' since t1 is an isometry. III II E is equidistant from C ’ and F , and so, by Theorem 3.2.8, B is a point on line 11, the perpendicular bisector of C’F . Therefore, D and E are fixed points of the reflection t2 = Rn since both points are on the mirror line n. Therefore, t; (D) = D and t; (E) = E . Now, A”= (tzot1)(A) = t2(t1(A)) = t2(A’) = t2(D) = D. E. Also, B”: (tzot1)(B) = t2(t1(B)) = t2(B’) = mm = Since the mirror line 11 is the perpendicular bisector of C'F, t2(C’) = Rn(C’) = F. .-.c”= (t20t1)(C) = t2(t1(C)) = t2(C') = F. Thus, wecanchoose t = (t2 0 t1). .'.t(A) = D, t(B) = E, and t(C) = F. Thus, we have proved the theorem statement in the case that A’=D, B’= E and CliF and the isometry t is the composition of two reflections: Re followed by Rn. Recall that we have assumed that A at D and, by definition of t1, A] = D. We have proved the theorem statement under these conditions in all cases for which B’ = E. Thus, weassumethat A’ = D and B’ i E. Let line m be the perpendicular bisector of the segment E5 . Let t3 = RIn . Thus, t3( B ’ ) = Rm (B ’) = E , by the definition of reflection. We continue the construction of the isometry t by following the reflection t1 by the reflection t3. Thus, we consider the transformation t3 0 t1. Wewillwrite A”: (t30 t1)(A) , B”: (tgo t1)(B) , C”= (t3o t1)(C). NOW, A”: t3(t1 = t3(D) and CH: t3(t1 = t3 We show that t3 (D ) = D and by showing that D is a point on the mirror line m. . B 5 DE since AABC ; ADEF, by assumption. AB .=. A’ ' E DB' since t1 is an Perpendicular; isometry. bisector of B'E . F D is equidistant from B ’ and E , and so, by Theorem 3.2.8, C D is a point on line m, the perpendicular bisector of B71? . Therefore, D is a fixed point of the reflection t 3 = Rm since D is on the mirror line n. Therefore, A" = t3 (A’) = t3 (D) = D and, bydeflnitionof t3 , B" = t3 (B’) = E. Now, either C”= t3(C’) = F or C”= t3(C/) 75 F . If C” = t3 (C’) = F , then wecanchoose t = (t3 0 t1) becauseofthe following calculations MA) = A” = (tsot1)(A) = t3(t1(A)) = t3(D) = D, t(B) = B” = (t30t1)(B) = t3(t1(B)) = ts(B') = E, t(C) = C” = (t30t1)(C) = t3(t1(C)) = t3(C’) = F, provethat the theorem statement is true in this case, and the isometry t Perpendicular is the composition of “sector " ' two reflections : - dicular t = RmOR‘ ' A rof-AB. This leaves one final case to consider and that is the case in which C” ¢ F , where C" = (t3 0 t1)(C) , which is illustrated in the figure to the right : A]: = D, I t1(B) i E, / t1(C) 9 // = B C A” = t3(AI) = D, B E , C N = t3(C/) i This is the situation for which the desired isometry t is a composition of 3 reflections. Let line k be the perpendicular bisector of the segment W . Let t4 = Rk . Thus, t4( C H) = Rn (C H) = F , by the definition of reflection. We continue the construction of the isometry t by following the composition of reflections t 3 o t 1 by the reflection t4. Thus, let isometry t be t= t4ot3ot1. B /$ t(C) = (t40t30t1XC) = t4(t3(t1(c))) = t4(t3(C/))- t(C) = t4(C”) = F. e k = the perpendicular __ bisector of C"F . H II t4° t3°t1 ltremainsonlytoshowthatt(A) = D and t(B) = E. Now,t(A) = t4(A”) = t4(D) and t(B) = t4(B”) = t4(E). Once we have shown that D andE are fixed points of t4, we can prove that t (A) = D and t(B) = E. E a 5 since AABC ; ADEF, by assumption. Also, A ; A"C” 2 DC" since t3 0 t1 is anisometry. D is equidistant from C” and F . D is apoint on line k, the perpendicular bisector of 57}: . by Theorem 3.2.8 . E E E15 since AABC _=. ADEF, by assumption. E5 ; B"C" EC" since t 3 0 t1 is anisometry. Ill E is equidistant from C” and F. E is a point on line k, the perpendicular bisector of C"F , by Theorem 3.2.8 . A Since D and E are on line k, the perpendicular bisector of 571? , t4(D) = Rk(D) = D and t4(E) = Rk(E) = E. Thus,t(A) = t4(A”) = t4(D) = D and t(B) = t4(B”) = t4(E) = E. Recallthatt(C) = t4(C”) = F. Thus, when t is the isometry t = t40t30t1, o t(A) = D, B W t(B) = E,and t(C) = E, and t maps AABC onto ADEF. Therefore, the theorem statement is true in this case because t is the composition of 3 reflections: t RkORmORe. Therefore, the theorem is true in general because it has been proved to be true in all possible cases. Q E D Theorem (MB) 5.3 (The Isometries from Reflections Theorem): Given any isomtery t of the Neutral plane, then at least one of the following is true: 1) t is a reflection about a mirror line, 2) t is a composition of two reflections, 3) t is a composition of three reflections. Proof: Let t be any isometry of the Neutral plane. Write X’ = t(X) for all points X. Let A, B, and C be three non—collinear points. Then, AABC 2 AA'B'C' by SSS, since t is an isometry. By the Congruent Triangles Isometry Theorem (Theorem (NIB) 5.2), there is an isometry to such that to(A) = A' , t0(B) = B’, and t0(C) = C’, and to is the composition of one, two, or three reflections in the plane. Since t(A) = A’ = t0(A), t(B) = B’ = t0(B), and t(C) = C’ = tom), the isometries t and to agree at three non-collinear points in the plane. Therefore, by the Three Points Agreement Theorem (Theorem 5.3.12) , as transformations, t = to . Since to is the composition of one, two, or three reflections in the plane, t = to is also the composition of one, two, or three reflections in the plane . QED Theorem @132 5.4 (The Four Euclidean Isomefl Types Theorem): Every isometry of the Euclidean Plane is one of the following types: 1) a translation TPQ for some vector PQ, 2) areflection Ra about amirror line e. 3) a rotation R 13,9 about a point P and through an angle 6 , or ————) 4) a glide reflection GL PQ’ e for some mirror line Z and some vector PQ parallel to line (3 . Proof: (Incomplete) Let t be any isometry. By the lsometries from Reflections Theorem (Theorem (NIB) 5.3), at least one of the following is true: 1) t is a reflection about a mirror line, 2) t is a composition of two reflections, or 3) t is a composition of three reflections. As the work in a previous Lab showed, such reflections and their compositions result in transformations of the plane of only four types: translations, reflections, rotations and glide reflections. The proof of this theorem is incomplete insofar as this last claim has only been conjectured during that previous Lab and a proof of that claim is not provided here. QED (almost) 10 ...
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This note was uploaded on 11/02/2010 for the course MATH modern geo taught by Professor Shirley during the Spring '10 term at University of Texas.

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transform_geom_II - Transformational Geomegy Part II...

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