Mathematics 138
Final Problem Set due at the final exam
Since the lowest problem-set grade will be ignored, you may not need to hand this in, but it is crucial preparation for the final exam! 
The precise mechanism for ignoring the lowest grade is that whatever assignment is the most points short of a perfect score will be converted to a perfect score.

1a. Construct the matrix for a reflection in an arbitrary line of the parallel pencil through the point x=1, y = 0 (and z = 1).  Use a parameter r to distinguish the lines of the pencil, so that r = 0 for the horizontal line and r =1 for the line through x = 0, y = 1.  The resulting matrix will look somewhat like the one on p. 164 of Ryan.  You can use Mathematica to construct the matrix.
b. Construct the matrix Horo(h) for a horolation with parameter h that holds x = 1, y = 0 fixed and that moves the point (0,0) up rather than down.  You can use Mathematica to derive this matrix from the one in part a.  Using Mathematica, check that Horo(h1)Horo(h2) = Horo(h1+h2).

2. Let D, E, and F be the midpoints of the sides opposite vertices A, B, C of a triangle.  Suppose that the angles and A, B, and C are a, b, and g respectively, and that the sum of the three angles of the triangle DEF is equal to the sum of the angles of ABC (which is not true in hyperbolic geometry.) Prove that a+b+g = p.

3. Ryan, problem 35 on page 181. The theorem to prove is that the product of three reflections in lines that do not all belong to the same pencil is a nontrivial glide reflection. You may recall that the corresponding theorem was easy to prove in spherical geometry, where any two lines intersect, but was a bit more difficult in Euclidean geometry because of the special case where
two of the three lines were parallel. In hyperbolic geometry there are numerous different cases; for example, all three intersection points might be lightlike.  The secret of an effecient proof is to start with the easy case where two of the lines intersect in a timelike point, then reduce all the other cases to that case.

4. Prove that all horocycles are congruent.
5. Consider a hyperbolic right triangle with sides a and b and hypotenuse c.  Under what circumstances is the circumscribed circle (through the three vertices) a horocycle?  A circle? An equidistant curve?  Sketch a diagram for each case.

6. By considering the limit of a regular polygon with n sides, show that the area of a hyperbolic circle of radius r is 4p sinh2 (r/2)

7. Return to the hyperbolic landowner from problem 8 of the previous problem set. He had seven children and decided to leave each of them an equilateral triangle of land so that the triangles fit neatly together to form a hyperbolic heptagon. 
a. An eighth child is born, and the landowner must redraft his will so that the plots of land received by the children now form an octagon of equilateral triangles.  By what fraction does the sum of the areas of all the plots of land increase as a result of this revision?
b. The eight ungrateful children complain so much that their father buys more land so that he can create eight congruent doubly asymptotic triangles for the children.  By what fraction does the sum of the areas of all the plots of land increase (above the answer to part a) as a result of this revision?
8. Suppose that a singly asymptotic hyperbolic triangle has one right angle and one 30 degree angle.
a. Using the angle of parallelism formula, draw an accurate representation of this triangle in the Klein model with the right angle at the center of a circle of radius 1.  In this case, what is Euclidean length of the side joining the two timelike vertices, and what is the Euclidean measure of the angle between the lines that meet at the 30 degree angle?
b. Draw an accurate representation of this triangle in the Klein model with the 30 degree angle at the center of a circle of radius 1. This is much easier, because when hyperbolic lines meet at the center, the Euclidean and hyperbolic angles are equal.