1. Explain why two distinct lines cannot have more than one common perpendicular.

Solution: If they had more than one common perpendicular there would exist a rectangle formed by two common perpendiculars and the original two lines.

2. Where are two non-intersecting lines closest together?

Solution:At their common perpendicular (see exercise 5 in homework #25).

3. How many gamma points are on each line in hyperbolic geometry?

Ans: An infinite number. Solution:Given a line l and a point P not on this line, there are an infinite number of lines through P that do not intersect l. Each of these has a gamma point in common with l. These gamma points are all distinct because the lines that determine them already have the point P in common.

4. Do two lines in hyperbolic geometry always diverge from their point of intersection?

Ans: No.Solution: Two lines meeting at a gamma point converge towards their common perpendicular, so they must diverge from their point of intersection. If gamma points are not considered, the answer is yes.

5.Prove that two parallel lines converge continuously in the direction of parallelism.

Solution: Consider the diagram below:

AB and DC are parallel. Since the angle of parallelism at C is acute, the internal angle at C is obtuse and thus greater than the angle of parallelism at D. Since the summit angles are not congruent, the quadrilateral ABCD can not be a Saccheri quadrilateral and so, AD is not congruent to BC. If BC was larger than AD, then by exercise 3 of homework #25, the angle of parallelism at D would be larger than the obtuse internal angle at C, a contradiction. So, BC is smaller than AD and the parallel lines converge continuously in the direction of parallelism.

6. Suppose a certain triangle in hyperbolic geometry has a defect of 0.00042 radians. In the formula for the area of a triangle, what value should be chosen for the constant k if the given triangle is to have a unit area?

Ans: 1/0.00042 = 2,380.952380...

7. Prove that two polygonal regions equivalent to the same polygonal region are equivalent to each other.

Solution:Let A and B be two polygonal regions that are equivalent to the same polygonal region C. A and C can be partitioned into the same number of pairs of congruent triangles, as can B and C. Consider superimposing the two partitions of C induced by A and B. This partitions C into polygonal regions. Divide each of these small regions into triangles. Each of the partitions of A and B can then be replaced by the finer partitions of C. The number of small triangles in these finer partitions are equal, since they are both equal to the number of small triangles in the finer partition of C. As the triangles used are congruent in pairs in the partitions of A and B, A and B are equivalent.

8. Prove that a quadrilateral exists with four congruent angles.

Solution: Take any Saccheri quadrilateral and construct on its base a congruent Saccheri quadrilateral in the opposite direction (see diagram). The sides of the large quadrilateral are lines since the angles at the midpoint are supplementary (two right angles). All four angles are of the large quadrilateral are congruent.

9. For the quadrilateral in Exercise 8, are the four sides congruent? Prove that your answer is correct.

Ans: Not in general, but they can be made so. Solution: To obtain a quadrilateral with four congruent sides and four congruent angles, we would start with a construction of a special Lambert quadrilateral, one where the two sides at the acute angle are congruent. This can be constructed as follows: At a point P on a given line, construct the perpendicular to that line. Take a point Q on this perpendicular and construct the perpendicular to that line at Q. Now construct the angle bisector at P and let the point where it intersects this second perpendicular be R (if this angle bisector does not meet that line, start again with Q choosen closer to P). On the original line, find point S on the same side of the first perpendicular as R so that PS = PQ. Join S to R. Triangle PSR is congruent to triangle PQR by SAS. Thus the angle at R is a right angle, so PQRS is a Lambert quadrilateral, and QR = RS. Now, double this Lambert quadrilateral to get a Saccheri quadrilateral whose sides are half the length of the summit. Double the Saccheri quadrilateral (as above) to get a quadrilateral with four congruent sides and four congruent angles.

10. Prove that the segment joining the midpoints of two sides of a triangle has a length less than half the length of the third side.

Solution:In the construction of the Saccheri quadrilateral equivalent to the original triangle, due to the congruent triangles used to show that the quadrilateral is equivalent to the original triangle, it can be seen that the base of the Saccheri quadrilateral is exactly twice the length of the line segment joining the midpoints of the sides of the triangle. Since the summit of the Saccheri quadrilateral is the third side of the original triangle, and we know that the summit of a Saccheri quadrilateral is longer than its base (exercise 7 of Homework #25), the line segment joining the midpoints must be less than half of the length of the third side.

11. In Figure 9.29, prove that BCGF and IJNM are Saccheri quadrilaterals.

Solution:The proof is the same for both, so we shall only do one. Draw the perpendicular from A to line ED and let the point of intersection be labeled H. Triangles CGE and AHE are congruent by AAS as are triangles AHD and DFB. By the first pair of congruent triangles, CG is congruent to AH. By the second pair, FB is congruent to AH, thus CG is congruent to FB and so BCGF is a Saccheri quadrilateral.

12. Prove that triangle ABC, Figure 9.29, is equivalent to Saccheri quadrilateral BCGF.

Solution:Using the same construction as in exercise 11 and adding the line segment BE, we see that triangle ABC is partioned into the four triangles, AEH, ADH, CEB and BDE. The Saccheri quadrilateral BCGF is partitioned into the four triangles, CGE, CEB, BDE and BDF. Since triangles AEH and CGE are congruent and ADH and BDF are congruent (see exercise 11), both regions are partitioned into the same number of pairs of congruent triangles (CEB and BDE are congruent to themselves), and so, are equivalent.

13. Why, in Figure 9.29, do the two Saccheri quadrilaterals have congruent summit angles?

Solution:The sum of the measures of the summit angles is equal to the sum of the measures of the three angles of the original triangles.