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MATH 3181 001 SPRING 1999 HOMEWORK ASSIGNMENT #6 SOLUTIONS 1. (Exercises 9.1.22 — 9.1.27, page 369) It can be proved that each of the statements in Exercise 22–27 is equivalent to Euclid’s fifth postulate. Rewrite each sentence, using the negation of the conclusion, to give a statement assume to be true in non-Euclidean geometry. 9.1.22: If a straight line intersects one of two parallel lines, then it will always intersect the other. If a straight line intersects one of two parallel lines, then it is not necessary that it will intersect the other. 9.1.23: Straight lines parallel to the same straight line are always parallel to one another. There exist two intersecting straight lines both of which are parallel to the same line.. 9.1.24: There exists one triangle for which the sum of the measures of the angles is π radians. For every triangle the sum of the measures of the angles is less than π radians. 9.1.25: There exists a pair of similar but non-congruent triangles. Every pair of similar triangles is congruent. 9.1.26: There exists a pair of straight lines the same distance apart at every point. For every pair of straight lines, the set of points equidistant from each other contains at most 2 points. 9.1.27: It is always possible to pass a circle through three non-collinear points. There exist three non-collinear points through which no circle passes. 2. (Exercise 9.2.4, page 374) Without using an axiom of parallelism, prove that if a transversal of two lines makes the alternate angles congruent, then the two given lines do not intersect. Let m and n be the two lines cut by the A Q transversal, k. Let A and B be the points of n intersection of k and n and k and m, respectively. We are told that the angle at A and M the angle at B are congruent. Let M be the midpoint of AB. Let P be the foot of M on m m B P and let Q be the point of intersection of n and uuuuv the ray PM . Then we have than AM ≅ BM , RMBP ≅ RMAQ (given), and RBMP ≅ RAMQ by vertical angles, since P, M, and Q are ©David Royster Hyperbolic Geometry FOR CLASSROOM USE ONLY MATH 3181 001 HOMEWORK ASSIGNMENT #6 SOLUTIONS SPRING 1999 collinear. Thus, ∆BMP ≅ ∆AMQ , and it follows that RBPM ≅ RAQM . This makes RAQM a right angle. Hence, m and n have a common perpendicular. By our previous theorem, they must be non-intersecting. 3. Let ¨ABDC be a Saccheri quadrilateral, so D that RA and RB are right angles and AD ≅ BC . Prove that RC ≅ RD . See Figure 1. C Since RA and RB are right angles, AD ≅ BC , and AB ≅ AB , we have that ∆ADB ≅ ∆BCA . Thus, AC ≅ BD . Then, since CD ≅ CD , it A follows that ∆ADC ≅ ∆BCD . Then, it follows that RBCD ≅ RADC . B 4. This problem has five parts. We first construct the Saccheri quadrilateral associated with a given triangle, and then apply this construction to triangles in the hyperbolic plane. (a) Given ∆ABC , let I, J, and K be the midpoints of BC, CA, and AB, respectively. Drop B K A perpendiculars AD, BE, and CF from the vertices sur to IJ . Prove that AD ≅ CF ≅ BE , and, hence that ¨EDAB is a Saccheri quadrilateral. Why F should this quadrilateral have the same area as E I J D the triangle? See Figure 2. We know that BI ≅ CI , RBIE ≅ RCIF , and RBEI ≅ RCFI . Thus, ∆BIE ≅ ∆CIF by AAS. This means that BE ≅ CF . Similarly, AD ≅ CF C using an identical argument. By transitivity it follows BE ≅ AD . Therefore, by definition, ¨EDAB is a Saccheri quadrilateral. sur IJ . What does this (b) Prove that the perpendicular bisector of AB is also perpendicular to suur sur say about the lines IJ and AB . From a theorem in class we know that the line perpendicular to the summit of Saccheri quadrilateral at its midpoint isralso perpendicular to the base of the Saccheri quadrilateral. suu sur Thus, we have that IJ and AB are hyperparallel. (c) Prove that IJ ≅ 12 ED . Conclude that in hyperbolic geometry IJ < 12 AB . From the triangles in part (a) we have that EI ≅ FI and FJ ≅ JD . Then, EI + IF + FJ + JD = ED 2 IF + 2 FJ = ED 2 IJ = ED 1 So, IJ ≅ 2 ED . Since the summit of a Saccheri quadrilateral is longer than its base, We know that AB > ED, so IJ < 12 AB . ©David Royster Hyperbolic Geometry FOR CLASSROOM USE ONLY MATH 3181 001 HOMEWORK ASSIGNMENT #6 SOLUTIONS SPRING 1999 (d) Suppose, in this part, that RC is a right angle. Prove that the Pythagorean Theorem doe NOT hold in hyperbolic geometry. (If it were valid, what would applying to the triangles ∆ABC and ∆JIC imply?) Assume that the Pythagorean Theorem does hold, then in ∆ABC we have AC 2 + BC 2 = AB 2 , and in ∆JIC we have, IC 2 + JC 2 = IJ 2 . Since I is the midpoint of AC and J is the midpoint of BC, the first equation gives us AB 2 = (2 IC ) 2 + (2 JC ) 2 = 4( IC 2 + JC 2 ) AB 2 = 4 IJ 2 Thus, from the last equation we get that 2IJ ≅ AB , which we just showed to be impossible. Thus, the Pythagorean Theorem does not hold in hyperbolic geometry. (e) In this part, assume that ∆ABC is an isosceles triangle with AC ≅ BC . Prove that K, F, and C are collinear but F is not the midpoint of CK. Since AC ≅ BC and I and J are the midpoints, it follows that IC ≅ JC . F is the foot of C on ED, so by Hypotenuse-Leg, we have that RICF ≅ RJCF . Let X uuur denote the point at which the ray CF intersects AB. It must intersect AB by the Crossbar Theorem. Then by SAS, ∆BCX ≅ ∆ACX . Hence X is the midpoint of AB, and X = K . Thus, K, F, and C are collinear. K B A F E I J D If F is the midpoint of CK, then KF ≅ FC ≅ BE ≅ AD suur C sur and the lines AB and IJ have three equidistant points. This contradicts the theorem that we proved in class. Thus, F cannot be the midpoint of CK. 5. Let ∆ABC be any triangle, and let L, M, and N be the midpoints of BC, AB, and AC, respectively. A (a) Prove that ∆AMN is not similar to ∆ABC . (HINT: Otherwise the defect of ¨MBCN would be 0.) N D M Assume that ∆AMN is similar to ∆ABC . Then we have that RAMN ≅ RABC and RANM ≅ RACB . Since RAMN + RBMN = 180° and RANM + RCNM = 180° , it B L C follows that RCBM + RBMN + RMNC + RNCB = 360° . Thus, ¨MBCN has defect 0, contradicting the fact that no quadrilateral in hyperbolic geometry has defect 0. Thus the two triangles cannot be similar. ©David Royster Hyperbolic Geometry FOR CLASSROOM USE ONLY MATH 3181 001 HOMEWORK ASSIGNMENT #6 SOLUTIONS SPRING 1999 (b) Prove that MN is not congruent to BL by assuming the contrary and deducing that ∆ABC has defect 0. (HINT: Choose D such that N is between M and D and ND ≅ MN .) Assume that MN ≅ BL . Choose D such that N is between M and D and ND ≅ MN . Since N is the midpoint of AC and MNA DNC , by vertical angles, it follows that ∆AMN ≅ ∆CDN . Therefore, ≅ ≅ RAMN ≅ RCDN and RMAN ≅ RDCN . Now, we have that MD ≅ BC , CD ≅ MB , and CM ≅ CM . Thus, by SSS, ∆MBC ≅ ∆CDM and RB = RMBC ≅ RCDM ≅ RAMN . Likewise, RBMC ≅ MCD . Now, RBMC + RCMD + RDMA = 180° RBMC + RCMD + RB = 180° RBMC + RBCM + RB = 180° RMCD + RBCM + RB = 180° RBCA + RNCD + RB = 180° RA + RB + RC = 180° Thus, the defect of ∆ABC is 0. This contradiction proves that MN is not congruent to BL. ©David Royster Hyperbolic Geometry FOR CLASSROOM USE ONLY

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