3. Explain why a line in hyperbolic geometry must contain two distinct ideal points.

Solution: Let l be a line and P a point not on this line. If l had only one ideal point, then the two parallel rays through P would both contain this ideal point. But since they already have point P in common, they would have to be the same line, a contradiction to the characteristic postulate of hyperbolic geometry.

4. Sketch three omega triangles, all with the same ideal vertex and each two of which also have an ordinary vertex in common.

Solution:

5. Sketch a three-sided figure with two ideal vertices.

Solution:

6. Sketch a three-sided figure with three ideal vertices.

Solution:

7.Name all the omega triangles shown in Figure 9.12

Solution: AB\Omega, AC\Omega, BC\Omega, CD\Omega, CE\Omega, AF\Omega, FB\Omega (where F is the unlabelled point on C\Omega ).

8. Prove that a line intersecting a side of an omega triangle at a point other than a vertex intersects a second side.

Solution:Let AB\Omega be an omega triangle and let M be a point of AB, and C a point of A\Omega.

9. Prove that the sum of the measures of the two angles at ordinary vertices of an omega triangle is less than \pi. (Hint: Use Thm. 9.4)

Solution: Let AB\Omega be an omega triangle with ordinary interior angles at A and B, and let C be the exterior angle at A obtained by extending the line AB. Notice that the angle at A and C are supplementary. If the sum of the angles at A and B is greater than or equal to \pi, then C would have to be smaller than or equal to the angle at B. This contradicts Thm. 9.4, which says that the exterior angle C is greater than the angle at B.

10. Prove Thm. 9.6. (Hint: Use Thm. 9.5)

Solution:Assume that AB\Omega and A'B'\Omega are two omega triangles with the angles at A and A', and B and B' congruent. If AB is longer than A'B', we can find a point A'' on AB so that A''B is congruent to A'B'. Consider the omega triangle A''B\Omega. By Thm. 9.5 this omega triangle is congruent to A'B'\Omega, so the angle at A'' is congruent to the angle at A', and hence, the angle at A. This angle however, is an exterior angle of the omega triangle AA''\Omega, and by Thm. 9.4 it must be larger than the angle at A. This contradiction shows that AB must be congruent to A'B', so by Thm 9.5 the two original omega triangles are congruent.

11. Prove that the angle of parallelism is constant for a given distance

Solution:Let AB\Omega and A'B'\Omega be two omega triangles with the angles at B and B' both right angles and sides AB and A'B' congruent. By Thm. 9.5, these are congruent omega triangles, so the angles at A and A' are also congruent. However, these are the angles of parallelism for the lines B\Omega and B'\Omega at the points A and A' respectively. Thus, the angle of parallelism for the distance AB must be constant.

12. As the distance increases, does the angle of parallelism increase or decrease? Prove that your answer is correct.

Ans:Decrease. Solution: Let l be a line, AB be a perpendicular to l at the point B, and C a point on AB such that A is between C and B. Consider the lines A\Omega and C\Omega that are parallel to l on the same side of AB. AB\Omega and AC\Omega are both omega triangles. The angle of parallelism at A to l, is an exterior angle of the omega triangle AC\Omega. By Thm. 9.4, this angle is greater than the angle of parallelism at C to l, thus, the angle of parallelism decreases as the distance from l increases.

13. Prove that if the two angles at ordinary vertices of an omega triangle are congruent, then the line from the ideal vertex to the midpoint of the opposite side is perpendicular to that side.

Solution:Let AB\Omega be an omega triangle and let M be the midpoint of AB. Since the angles at A and B are congruent and the sides AM and MB are congruent, the omega triangles AM\Omega and BM\Omega are congruent by Thm. 9.5. Thus, the two angles at M are congruent and since they are also supplementary, they must both be right angles. That is, M\Omega is perpendicular to AB.

14. Prove the converse of the statement in Exercise 13.

Solution:Let AB\Omega be an omega triangle and let M be the midpoint of AB. If M\Omega is perpendicular to AB, then the omega triangles AM\Omega and BM\Omega have a pair of angles (the right angles) congruent and the ordinary side congruent. Thus, by Thm. 9.5 these are congruent omega triangles, and so, the angles at A and B are congruent.