Math 451: Euclidean and Non-Euclidean Geometry MWF 3pm, Gasson 204 Homework 8 Solutions

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1 Math 451: Euclidean and Non-Euclidean Geometry MWF 3pm, Gasson 204 Homework 8 Solutions Exercises from Chapter 2: 5.5, 5.10, 5.13, 5.14 Exercises from Chapter 3: 1.2, 1.3, 1.5 Exercise 5.5. Give an example of a (non-proper spherical quadrilateral that cannot be triangulated. Solution. Here is a picture of such a quadrilateral: B C A D Figure 1. The non-proper spherical quadrilateral A, B, C, D is shaded. Of the two spherical line segments between A and C, neither is contained in the quadrilateral pictured. The same is true of B, D, so no triangulation of the quadrilateral is possible. Exercise For each g, describe the construction of a specific polyhedron with g holes in which you can easily show that the Euler characteristic is 2 2g. Solution. We fit together g copies of the one-holed polyhedral set we discussed in class, gluing them together along faces. Here is picture for g 4: 1

2 2 Figure 2. A polyhedral set with g holes (and g 4. Each of these pieces has 16 vertices, 32 edges, and 16 faces. When g 2 copies are fit together as pictured, we compute the numbers of vertices, edges, and faces: V 12(g g + 4, E 28(g g + 4, and F 14(g 2 + 2(15 14g + 2. Thus V E + F ( g + ( g, as desired. Exercise Instead of polygons, one could tile arbitrary regions of the plane. What do you think the Euler characteristic records? Solution. It turns out that the Euler characteristic of a polygonal subset of R 2 with g holes is 1 g. Here is a family of examples: The Figure 3. A polygonal subset of the plane with g holes (and g 4. number of vertices can be computed as in Exercise 5.10: V 6(g g + 2, E 11(g g + 1, and F 4g. We conclude that V E + F ( g + (2 1 1 g.

3 3 Exercise Show that as r, the ratio of interior vertices to total vertices approaches 1. Solution. Let s indicate the number of interior vertices by, exterior vertices by V ext, interior polygons by F int, and exterior polygons by F ext. Letting D indicate the diameter of P, the union of interior polygons must contain the disk of radius R 2D. Thus we have Area(P F int π(r 2D 2. Similarly, the union of the exterior polygons is contained in the annulus with inner radius R 2D and exterior radius R + 2D, so we have Area(P F ext π ( (R + 2D 2 (R 2D 2 8πDR. Summing the number of vertices from each tile over all of the interior tiles, we end up counting each vertex according to the number of edges touching it. Since the sum of the angles formed at each vertex is 2π, this number is at most 2π divided by the minimum angle of P. Denoting this last quantity by α(p, we have n F int 2π α(p. We may do the same count for the exterior tiles, where we may use the simpler observation that each exterior vertex is counted at least once when we sum the number of vertices over the exterior tiles. Thus Putting this info together, we have n F ext V ext. n α(p π(r 2D2 n α(p (R 2D2 2π Area(P 2Area(P 8πDR V ext n Area(P 8πnD Area(P R. This means we have V ext 8πnD R Area(P 16πD nα(p (R 2D2 α(p 2Area(P which goes to zero as R. Thus we have as R, as desired. + V ext Vext R (R 2D 2

4 4 Exercise 1.2. Show that if ax 2 + ay 2 + bx + cy + d 0 has more than one solution, and a 0, then it describes a circle in R 2. Solution. Since a 0 we may divide by a, so that x 2 + y 2 + b a x + c a y + d a 0. Rearranging the terms and adding b 2 /(2a 2 and c 2 /(2a 2 to both sides, b 2 + c 2 4ad (x 2 ba b2 + x + + (y 2 ca c2 + y + 4a 2 4a 2 4a 2 ( x + b 2 ( + y + c 2. 2a 2a If b 2 + c 2 4ad 0, this equation has only one solution ( b, c 2a 2a, and if b 2 + c 2 4ad < 0 the equation has no solutions. Since the equation has more than one solution, we must have b 2 + c 2 4ad > 0, so that this is the equation of a circle centered at ( b, c 2a 2a and with radius b 2 +c 2 4ad 2a. Exercise 1.3. If x, y, z R 2 are distinct, there is a unique lircle through x, y, z. Solution. Consider the perpendicular bisectors L 1 and L 2 of xy and yz respectively, and note L 1 consists of the points that are equidistant from x, y and similarly for L 2 and y, z. Any circle that goes through x, y, z must have a center that is equidistant from x, y, z, so that it lies on both L 1 and L 2. If L 1 and L 2 are parallel, they do not intersect and there is no circle containing x, y, z. In this case the directions of xy and yz are equal, so that x, y, z are collinear, and they lie on a unique lircle. If L 1 and L 2 are not parallel, x, y, z are not collinear, and they do not lie on a line. In this case, L 1 and L 2 intersect at a unique point q, so that the circle centered at this point with radius given by the length of qx is the unique circle through x, y, z. Exercise 1.5. Show directly that the parallel postulate fails for H 2.

5 Solution. Let l be the hyperbolic line contained in the line x 0, and choose the point (1, 1. The hyperbolic line L 1 which is the arc of a circle centered at (1, 0 with radius 1 passes through (1, 1 but is disjoint from l, and the vertical hyperbolic line L 2 contained in x 1 passes through (1, 1 but is disjoint from l. Thus L 1 and L 2 are two distinct hyperbolic lines containing (1, 1 but parallel to l, so the parallel postulate fails for H 2. (In fact, it s not hard to see that the hyperbolic line through (c, 0 with radius c 2 2c + 2 contains (1, 1 but is disjoint from l, for any c > 1. Thus there are infinitely many lines through (1, 1 parallel to l. 5

We have set up our axioms to deal with the geometry of space but have not yet developed these ideas much. Let s redress that imbalance.

Solid geometry We have set up our axioms to deal with the geometry of space but have not yet developed these ideas much. Let s redress that imbalance. First, note that everything we have proven for the