You may use pencil, pens, protractor, ruler, compass, scissors, calculator, book or notes on this test. You may not use a cell phone or any device with the ability to communicate outside the room. Please take your time on each problem and show your work in the space provided. If you get stuck on a problem, move on to the next and return when you have time.
Discussion of Proof: The goal of the proof is to find an isometry that takes triangle ABC to triangle DEF. We could start by using axioms 6 and 7 to "move" side BC to side EF, but then we'd have to worry about which side of line EF the image of A was on. Instead we use the definition of congruence and the fact that we some angles are congruent to find an isometry h which takes angle B to angle E and point C to point F.
Proof: By the definition of congruence, there is some isometry f for which f(B) = E and which sends angle ABC to angle DEF.
If f(C) = C' is on ray EF, then C' = F because |BC| = |EF|; let h=f.
This next step is incorrect, in several ways. If f(C) = C' is on ray ED, then use axiom 6 and the fact that |BC| = |EC'| = |EF| to find an isometry g which fixes B and moves C' to F. Let h = g ° f.
We now have an isometry h for which h(B) = E, h(C) = F, and h(A) = A' lies on ray ED. If A' = D then h is an isometry that takes triangle ABC to triangle DEF, proving the two triangles are congruent and completing our proof.
Is it possible that h(A) = A' is not equal to D? No: we know that angle EFD is congruent to angle BCA (by hypothesis) and that angle BCA is congruent to angle EFA' (because the isometry h sends angle BCA to angle EFA') so angle EFD is congruent to angle EFA'. Hence ray FD is the same as ray FA'.
The point of intersection of rays ED and FD is D. Point A' also lies on the intersection of rays ED and FD, so A' = D. (Note that we're assuming here that two lines in the plane can intersect in only one point. This is similar to the lemma about intersection of circles that we used to prove side-side-side.)
We have proven the existence of an isometry that maps triangle ABC to triangle DEF, so the two triangles are congruent.//
Look at figure 3.1(b) on page 71. Follow the steps used to construct that figure, then construct the perpendicular bisectors of segments OS and SQ. The points of intersection of the perpendicular bisectors with circle CP(|OP|) together with the vertices of the square form the vertices of a regular octagon.
Yes it is true. Theorem 6.3.3 on page 121 of the text says that if two hyperbolic triangles are similar then they are also congruent. The side length ratios discussed in Corollary 1.7.4 are all equal to 1.
a) (10 pts) Given the points A and B below find two points C and D so
that ABCD is a rhombus, or explain why that is not possible.
There is more than one solution to this problem. One of the simpler solutions is to let point C be six blocks east of B, and let D be six blocks east of A.
b) (20 pts) Is it true that if ABCD is a rhombus there is only one other rhombus in the taxicab plane that shares edge AB? Justify your answer. (In other words, can there be more than one rhombus on the "same side" of edge AB?)
No, it is not true that there are only two rhombi with edge length six that include edge AB.
Let C' be a point four blocks east and two blocks north of B. Let D' be a point four blocks east and two blocks north of A. Then ABCD (from (a)) and ABC'D' are both rhombi with edge AB "on the same side" of edge AB.