Math 305 Final Exam Key

Math 305 Final Exam Key Instructions 1. No assistance may be received from anyone on this take-home exam. 2. Do NOT write your answers on these sheets...
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Math 305 Final Exam Key Instructions 1. No assistance may be received from anyone on this take-home exam. 2. Do NOT write your answers on these sheets. Nothing written on the test papers will be graded. 3. Please begin each section of questions on a new sheet of paper. 4. Do not write problems side by side. 5. Do not staple test papers. 6. Limited credit will be given for incomplete or incorrect justification. Questions 1. Conjecture (5 each) For the following problems use any tools necessary (e.g., Geogebra with Poincare Disc model) to experiment. Provide your method of experiment and guess about the result. For this section you are not required to provide a proof or any reasons. (a) In hyperbolic geometry do two perpendicular lines always have a single line that is sensed parallel to both lines? Construct a pair of perpendicular lines. Construct another line that shares an omega point with one of the lines. See if the other point can be moved to make the line a sensed parallel of the second line. This works for all cases. Note that the sensed parallel can be defined by two omega points (one from each of the lines). (b) Is the orthogonality important to the result above? Change the experiment to start with non-perpendicular lines. The same result holds. Orthogonality was not important. (c) What is the smallest the angle of parallelism can be? Construct a line. Construct a line perpendicular to that line. Select a point on the perpendicular and construct a line containing it and the omega point of the original line. Measure the angle as the point is moved. The angle varies in (0, π/2). There is no smallest, but it tends toward 0. (d) Triangles do not have an angle sum of π in hyperbolic geometry. Find some pattern to the difference between π and the angle sum of equilateral triangles. Construct a triangle. Measure the three angles. Add these angles. As the triangle’s size is increased the difference between π and the angle sum is decreased.

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2. Triangulate (5 each) See Figure 1. Use the following additional information. l1 ⊥ l2 . bi are angle bisectors of angles formed by l1 and l2 . Ωi are omega points. pi are sensed parallels to l1 and l2 . AB, BC, CD, and DA are line segments. E, F, G, and H are the intersections of l1 and l2 with the line segments. Suppose |OA| = |OB| = |OC| = |OD|. (a) Prove that rays b1 and b3 are on the same line. m6 AOH + m6 HOD + m6 DOG + m6 GOC = π because definition of perpendicular and angle bisector. Thus b1 and b3 are a line. (b) Prove that rays b2 and b4 are on the same line. The argument above works here as well. (c) Identify all already drawn triangles that are congruent. 4AOB ∼ = ∼ 4BOC = 4COD ∼ = 4DOA 4AOE ∼ = 4EOB ∼ = 4BOF ∼ = 4F OC ∼ = 4COG ∼ = 4GOD ∼ = 4DOH ∼ = 4HOA 4ABC ∼ = 4BCD ∼ = 4CDA ∼ = 4DAB.

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(d) Prove the congruences. OA ∼ = OA by reflexivity. OB ∼ = OD is given. m6 AOE + m6 EOB = π = m6 AOH + m6 HOD by definition of perpendicular and angle bisector. Thus 4AOB ∼ = AOD by side-angle-side. ∼ ∼ Similarly 4AOD = 4DOC = 4COB. By CPCTC 6 OAE ∼ = 6 OAH ∼ = 6 OCG ∼ = 6 OCF and 6 ODH ∼ = 6 ODG ∼ = 6 OBF ∼ = 6 OBE. ∼ ∼ 6 6 OA = OA by reflexivity. AOE = AOH by definition of angle bisector. Thus 4OAE ∼ = 4OAH. Similarly 4OHD ∼ = 4OGD, 4OCG ∼ = 4OF C, 4OF B ∼ = 4OEB. Also, OA ∼ = OB is given. OE ∼ = OE by reflexivity. 6 AOE ∼ = 6 BOE by definition of perpendicular and ∼ angle bisector. Thus 4AOE = 4BOE by side-angle-side. Similarly 4BOF ∼ = 4COF, 4COG ∼ = 4DOG, and 4DOH ∼ = 4AOH. ∼ ∼ ∼ Also, by CPCTC AB = BC = CD = DA. Further |AB| = |BO| + |OA|. Similarly |CD| = |DO| + |OC|. By given the parts are equal so also AB ∼ = CD. Thus by SSS 4ABC ∼ = 4BCD. ∼ ∼ Similarly 4BCD = 4CDA = 4DAB. Combining these with the previous set proves all the small triangles are congruent. (e) Label all congruent angles and all congruent line segments on the Figure 1. See figure. (f) Label all the right angles in Figure 1. See figure. (g) Prove that these angles are right angles (unless that is given above). By CPCTC 6 AEO ∼ = 6 BEO. Also m6 AEO + m6 BEO = π, because this is a line. Thus m6 AEO = m6 BEO = π/2. Similarly m6 AHO

=

m6 DHO

=

m6

DGO

=

m6

CGO

=

m6 CF O

=

m6 BF O

=

π/2.

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3. Non-Euclidean (5 each) See Figure 1. Use the same information as above and everything your prove above. (a) Identify all omega triangles that are congruent. Based on congruencies above the following are congruent.

4AOΩ2 4BOΩ2 4BOΩ3 4COΩ3

∼ = ∼ = ∼ = ∼ =

4DOΩ4

∼ = ∼ =

4DOΩ1

∼ =

4COΩ4

4AOΩ1 . 4AEΩ2 4BEΩ2 4BF Ω3 4CF Ω3

∼ = ∼ = ∼ = ∼ =

4DGΩ4

∼ = ∼ =

4DHΩ1

∼ =

4CGΩ4

4AHΩ1

4BCΩ3

∼ = ∼ =

4CDΩ4

∼ =

4ABΩ2

4DAΩ1 . (b) Prove the congruences. The first set are congruent by side-angle using given congruences of sides and angle bisector of congruent (orthogonal) angles. From previous and vertical angles (or supplementary angle of line) 6 AEΩ2 and similar are right angles. Thus AEΩ2 ∼ = BEΩ2 . The others are similarly congruent. By CPCΩ TC 6 EAΩ2 ∼ = EBΩ2 . The other pairs are likewise congruent. Thus 4ABΩ2 ∼ = 4BCΩ3 . The other pairs are similarly congruent. → , the angle bisector of l1 and l2 , (c) Let P be a point on p1 to the left or right of A. Draw OP . Prove that OA must intersect p1 . WoLoG suppose P is to the right of A. Thus AOΩ2 is an omega triangle. Thus by Pasch’s axiom for omega triangles the angle bisector must intersect p1 .

Final Exam

4. Transformations (5 each) (a) Identify all symmetries of Figure 1. The figure has reflectional symmetry over l1 , l2 and the angle bisectors. These reflections produce rotations of degree four (angle π/2 about O). (b) Prove the symmetries you identified. The triangles on opposing sides of the lines of reflection have been shown to be congruent. The rotations are compositions of the reflections.

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Ω2

b2

b1

l1

B

E

s2

p2

A p1

O

s1 l2

F

Ω3

Ω1

H s3

p3

p4 C

G

s4

b3

D b4

Ω4

Figure 1: Hyperbolic Figure

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