MATH A305 Introduction to Geometries Exam 3 Key

Instructions

1. Do NOT write your answers on these sheets. Nothing written on the test papers will be graded. 2. Please begin each section of questions on a new sheet of paper. 3. Do not write problems side by side.

4. Do not staple test papers. 5. Limited credit will be given for incomplete or incorrect justification.

Questions

1. Quick Facts (2 each) For each of the following answer the question and state the axiom or theorem used. (a) Let ` and n be sensed parallels. Let P be on ` and Q be the foot of the perpendicular from P to n. Let M be the midpoint of P Q. Let m be the sensed parallel to ` at M. Are the omega triangles PMΩ and QMΩ congruent? No. One has a right angle, the other two cannot (angle or parallelism is less than a right angle). (b) Let P be a point on `. Let c be a line orthogonal to ` at P. Let Q and R be points on opposite sides of c such that PQ =∼ PR. Let m be the sensed parallel to ` through Q and n be the sensed parallel to ` through R. Are the omega triangles QP Ω and RP Ω congruent? Yes. Angle side congruency of omega triangles. (c) Let ABΩ be an omega triangle. If ` intersects side AΩ must it intersect one of the other sides? Yes. Because AΩ is the sensed parallel at all points on it, any line cutting through must intersect the other line. (d) Let ABΩ be an omega triangle. If ` intersects side AB must it intersect one of the other sides? No. Through each point of AB there is a line that is a sensed parallel. (e) Why can’t a rectangle exist in hyperbolic geometry? The rectangle meets the conditions of the Saccheri quadrilateral which is proven to have upper angles less than right angles.

1 Exam 3: Non-Euclidean 2

2. Proof Prove the following about Figure 1. Note mL and mR are the left and right, respectively, sensed parallel to l through P. ∼ (a) 4PQLR = 4PQRR. ∼ ∼ ∼ Note PR = PR by reflexivity. PQL = PQR is given. 6 RPQL = 6 RPQR because they are angles of ∼ parallelism (sensed parallels is given). Thus by side-angle-side 4RPQL = RPQR. ∼ (b) 4QLFLR = 4QRFRR. ∼ ∼ Note QLR = QRR from previous proof by CPCTC. Note 6 PRQL = 6 PRQR from previous proof by ∼ CPCTC. Thus 6 QLRFL = 6 QRRFR by subtraction of congruent angles from given right angles. Note ∼ ∼ 6 QLFLR = 6 QRFRR is given. Thus by angle-angle side 4QLFLR = 4QRFRR. ←→ ←→ ←→ (c) QLFL, PR, and QRFR are all parallel. Each of these pairs of lines are at right angles to a shared transversal. This is greater than the angle of parallelism and less than the opposing angle of parallelism. Exam 3: Non-Euclidean 3

3. Symmetry (Euclidean)

(a) Prove that the diagonals of a square meet at a point half way from each opposing set of corners. Consider square EF GH. Note G and E are on opposite sides of FH so by the plane separation postulate FH and EG intersect. Call this point I. Note EF =∼ GF =∼ EH =∼ GH by definition of square. Also 6 F GH =∼ 6 FEH because they are both right. Thus by SAS 4FEH =∼ 4F GH. Thus 6 HFG =∼ 6 HFE. Also 6 HFG =∼ 6 FHG =∼ 6 HFE =∼ 6 FHE because these are isosceles triangles. Similarly 6 GEF =∼ 6 EGF =∼ 6 EGH =∼ 6 GEH. Thus 4IHG =∼ 4IGF =∼ 4IFE =∼ 4IEH by angle-side-angle. By CPCTC IH =∼ IF and IE =∼ IG meaning that I is half way from each opposing set of corners. (b) Prove that the line segments connecting midpoints of opposing sides of a square are orthogonal to both (opposing) sides. Consider square EF GH. Let J be the midpoint of FE, K be the midpoint of EH, L be the midpoint of HG, and M be the midpoint of GF . Note GL =∼ HL (midpoint) and FG =∼ EH (square). Also 6 F GL =∼ 6 EHL (square). Thus 4EHL =∼ 4F GL. LF =∼ LE by CPCTC. JF =∼ JE (midpoint). 6 LF G =∼ 6 LEH by CPCTC. 6 GF E =∼ 6 HEF (square). So by angle sums 6 LF J =∼ 6 LEJ. Thus by side-angle-side 4LJF =∼ 4LJE. By CPCTC 6 LJF =∼ 6 LJE and they are supplementary, so they are right angles. Proofs for the other pairs are similar. (c) Prove that Figure 2 has four fold rotational symmetry. First we show that the lines connecting E, F, G, H to the center are at right angles. This was done above. Second we show that the lines connecting A, B, C, D to the center are at right angles. Note AE =∼ AF implies 6 AF J =∼ 6 AEJ by the isosceles triangle theorem. Further 4AJF =∼ 4AJE by side-angle-side. Thus 6 AJF is a right angle as is 6 FJL from above. This means A−J −L−C. Similarly B −M −K −D. AC ⊥ FE and FE k BD thus AC ⊥ BD. Thus the two sets of four vertices are at right angles from each other. This matches the definition of rotational symmetry. Congruency of the triangles ensures the points in between the vertices also remain constant under the rotations. Exam 3: Non-Euclidean 4

P m L mR QL QR

l R F R F L

Figure 1: Quadrilaterals

A

F E

B D

G H

C

Figure 2: Rotational Symmetry