Homework 27 Answers
#1 Hint: Use the defect theorem 4.8.2.
#3 Hint: Same hint as #2.
#4 Hint: Show that none of the properties hold for a Saccheri quadrilateral, which is a parallelogram.
1. Suppose that every triangle has the same defect c. Let ABC be a triangle, and suppose that point E is in the interior of BC. We know that δ(ABC) = δ(ABE) + δ(ECA). But the defect of each of these triangles is c, so c = c + c, which implies that c = 0. Because it is not possible to have a triangle with a defect of 0 in a hyperbolic geometry, then triangles in a hyperbolic geometry can't all have the same defect.
2. Let MN be the altitude of the Saccheri quadrilateral ABCD. Since the D N C altitude is perpendicular to both AB and CD, then ∠DAM, ∠AMN, and ∠MND are all right angles. Then it follows that AMND is a Lambert quadrilateral. But this implies that MN < AD. Through a similar proof, A B MN < BC. M
3. Let MN be the altitude of the Saccheri quadrilateral ABCD. Let ∠DAM, ∠AMN, and ∠MND are all right angles. Note that AMND and BMNC are Lambert quadrilaterals, so by #2 above, AM < DN and BM < CN. Also, since M is in the interior of AB and N is in the interior of CD, A * M * B and C * N * D. Then AB = AM + MB < DN + NC = CD.
4. Let ABCD be a Saccheri quadrilateral. 1. Suppose that the diagonal AC divides the quadrilateral so that ACD ≅ ACB (RAA). Then because ∠ABC is a right angle, so is the summit angle ∠ADC. But the summit angles must be acute in hyperbolic space, so this is a contradiction. A similar argument can be constructed to show that BD cannot divide the quadrilateral into two congruent triangles. 2. We know from #3 that AB < CD, so this pair of opposite sides is not congruent. 3. We know that the summit angles are acute and the base angles are right angles, so opposite angles are not congruent. D C 4. Suppose that the angles bisect each other at point E. Then DE = EB, AE = EC, and since ∠AEB and ∠CED are vertical angles, they are congruent. Then AEB ≅ E CED. But then AB = CD, a contradiction. A B