1. What is the maximum number of angles of a Saccheri quadrilateral that could be congruent to each other?

Ans. Two Solution:Since a Saccheri quadrilateral has two right angles, if more than two angles were congruent, a summit angle would have to be a right angle. Since the summit angles are congruent, this would mean that all the angles were right angles giving a quadrilateral with angle sum of 2\pi which is a contradiction.

2. In hyperbolic geometry, why can there be no squares or rectangles?

Solution:Their angle sums would be 2\pi.

3. Show that for a figure such as:

if AD > BC then the measure of angle BCD > measure of angle ADC.

Solution:Extend side BC to BC', where BC' = AD. Angle BCD is an exterior angle of triangle CC'D, and so, is greater than angle CC'D. But, ADC'B is a Saccheri qualrilateral, so angle CC'D is congruent to angle C'DA. In turn, angle C'DA is greater than angle ADC, so m(angle BCD) > m(angle CC'D) = m(angle C'DA) > m(angle ADC).

4. For a Lambert quadrilateral, which is longer, a side adjacent to the acute angle or the side opposite? Prove that your answer is correct.

Ans: The adjacent side.Solution:If the sides were equal the quadrilateral would be a Saccheri quadrilateral and hence a rectangle which is impossible. If the opposite side was longer, the quadrilateral would look like the one pictured above with angle ADC a right angle. Since, by problem 3, angle BCD is greater than angle ADC, angle BCD would be an obtuse angle which is a contradiction since the fourth angle of a Lambert quadrilateral is acute. The only remaining possibility is that the adjacent side is longer.

5.What does the answer to Exercise 4 tell you about the distance between non-intersecting lines?

Solution: The shortest distance between non-intersecting lines occurs at the common perpendicular to both lines.

6. Could a Saccheri quadrilateral also be a Lambert quadrilateral? Why?

Ans: No. Solution: If a Saccheri quadrilateral was also a Lambert quadrilateral, then one of the summit angles of the Saccheri quadrilateral would be a right angle (and hence, so would the other one) contradicting the fact that the summit angles of a Saccheri quadrilateral are acute.

7. Which is longer, the base or the summit of a Saccheri quadrilateral?

Ans:The summit.

8. Prove that your answer to to Exercise 6 is correct.

Solution: The sides of a Saccheri quadrilateral are non-intersecting lines, so, by exercise 5, the shortest distance between them occurs at the common perpendicular, which in this case is the base of the Saccheri quadrilateral.

9. Is the segment joining the midpoints of the sides of a Saccheri quadrilateral perpendicular to the sides? Prove that your answer is correct.

Ans: No. Solution:If it were, then the bottom quadrilateral would be a rectangle which is impossible.

10. Suppose a triangle has a defect of 0.426º. If two angles have measurements of 37.514º and 71.689º, find the size of the third angle.

Solution:Since the defect is 0.426º, the sum of the angles is 179.574º. Thus, the third angle measures 70.371º.

11. Let a triangular region be partitioned into two triangular regions by a segment through a vertex. Compare the defect of the original triangle with the defect of the two smaller triangles formed.

Solution: The defect of the original triangle is the sum of the defects of the two smaller triangles. To see this, consider the following diagram:

The defects of the smaller triangles are \pi - m(B) - m(1) - m(3) and \pi - m(C) - m(2) - m(4). Their sum is thus, 2\pi - m(B) - m(C) - (m(1) + m(2)) - (m(3) + m(4)) = 2\pi - m(B) - m(C) - m(A) - \pi = \pi - m(A) - m(B) - m(C) = defect of the original triangle.

12. Let a triangular region be partitioned into any number of triangular regions by segments through the same vertex. State and prove a theorem about the defects of the original triangle and the defects of the subtriangles.

Solution:The defect of the original triangle is the sum of the defects of the subtriangles. Create the subtriangles one at a time, and at each step apply exercise 11.

13. Point out in the proof of Theorem 9.10 exactly where the statements first differ from correct statements in Euclidean geometry.

Solution:The first and only place in the proof where a statement differs from a correct statement in Euclidean geometry, occurs when it is claimed that the angle at A is acute (in Euclidean geometry this angle is a right angle).