Math 168 Weekly Assignment #1 Solutions 1. Write

Math 168 Weekly assignment #1 Solutions

1. Write a definition of what a polyhedron is. First note that any plane S in space (we want to save the letter P for polyhedra) creates two closed half spaces S≤ and S≥ which each consist of the plane itself and all of the points on one side of the plane. (It doesn’t matter which you call which, as long as you’re consistent for any given plane.) Now given a plane arrangement, which is just a finite list of distinct planes S1,...,Sn, a closed cell of the plane arrangement r1 rn is any intersection of the form C = S1 ∩ · · · ∩ Sn where each ri is either ≤ or ≥, such that this intersection contains some sphere (no matter how small). A closed cell is called bounded if it lies entirely inside some sphere (however large). Two closed cells are called adjacent if exactly one of their ri differs. A collection of objects is called connected if for any pair, there is a sequence starting with the first and ending with the last such that each element of the sequence is adjacent to the next one. Finally, a polyhedron is the union C1 ∪· · ·∪Cm of a finite connected set of bounded closed cells of a plane arrangement with the final property that if the intersection of two cells is nonempty, then the collection of cells that contain their intersection is connected. Explain why it is a good definition. This definition captures that fact that polyhedra result from chopping up space by planes, and that they are finite, but it allows for “bowtie-shaped” polyhedra. The connected condition prevents things with more than one piece from being considered a single polyhedron, and the final condition avoids objects with (for example) an “arm” that extends out and then has a spike that just touches another face in a single point. The definition is also very concrete and unambiguous, we should be able to refer precisely to the pieces that make up the polyhedron when we are working with one. 2. Now define the faces, edges, and vertices of a polyhedron. Given a polyhedron P , let C be any closed cell of its plane arrangement (not necessarily bounded) which is adjacent to some cell of P but is not in P . Then C ∩ P is called a facet of P . Two facets are called adjacent if their intersection contains more than one point. A face of P is the union of a connected collection of facets in the same plane, such that no other facet of P in that plane is adjacent to any facet in the collection. (We have to go to this trouble because it is possible for a polyhedron to have two faces far away from each other that happen to be coplanar, so we can’t just say something like the intersection of the polyhedron and a bounding plane.) Any line segment contained in two distinct faces, but not contained in any larger line segment which is contained in both faces, is called an edge. (Again, we have to be careful because a face might have non- consecutive edges that happen to be collinear.) Finally, an endpoint of an edge is called a vertex. 3. Must any polyhedron P have all of these elements? Yes. Any plane arrangement has unbounded cells, so there is some cell which is not in the polyhedron. By proceeding from there to all of its adjacent cells, and so on, we must eventually come to a cell that is not in the polyhedron but is adjacent to one that is. The intersection of these two is a facet, and the largest connected collection of coplanar facets containing this one is a face. Any face F of P lying in a plane Si of the arrangement must be bounded (since the polyhedron is), so choose a facet F0 on its boundary. Consider all of the cells of the arrangement which intersect F0 in more than one point. We know that one of them is in P and one of them is not in order to create facet F0, but there must be another pair of such cells, one in P and one not, whose intersection does not lie in the same plane Si, or else F0 could not lie on the boundary of F . Therefore, we have another facet G0which intersects F0 in more than one point (and hence in a line segment) but which lies in a distinct plane. Extend G0 to a face G of the polyhedron. G intersects F in at least some line segment, and so their intersection contains at least one edge of P . (Yes, it could actually contain two or more.) Finally, any edge is bounded since the polyhedron is, so it must have at least one (actually two) endpoints, so P must have a vertex. 4. What is the minimum number of each of these elements a polyhedron can have? 4 faces, 6 edges, and 4 vertices. 5. What is the minimum number of faces that can meet at a given vertex? From the definition, two faces meet at any edge. A vertex is an endpoint of that edge, so at least the two faces meeting at the edge will meet at the vertex. However, because the vertex is the endpoint of the edge,

1 we can argue similarly to problem (3) in ﬁnding an edge: if we consider all of the cells intersecting the vertex, there must be an adjacent pair in/not in P which intersect in a third plane, distinct from the plane of the ﬁrst two faces, or else the edge would not end. This pair will give us a facet in that third plane intersecting the vertex, and hence a third face meeting that vertex.

6. There must be at least one convex vertex in any polyhedron. Choose a plane T not parallel to any plane or edge of the polyhedron, and not intersecting it. This is possible because the arrangement has finitely many planes and the polyhedron has finitely many edges and the polyhedron is bounded. Now imagine sliding the plane T (in direction perpendicular to itself) toward the polyhedron. At some instant, it must first touch the polyhedron. At this point, it cannot touch along a face or an edge, because T was not parallel to any face or edge. So it much touch at a single vertex. And at this instant of first touch, the entire polyhedron lies on one side of T . Therefore, the vertex of first touch must be convex. 7. What can you say about the sum of the vertex angles of the faces that meet at a convex vertex? Nothing. The vertex angles can add up to any amount: less than, equal to, or more than a full circle. I will show samples in class. [Note full credit will be awarded for any answer to this question, as the question was flawed.] 8. Can a regular polyhedron have any non-convex vertices? No. We showed in 6 that every polyhedron has a convex vertex (in the sense defined in this problem set, which is not mathematically standard). Clearly if two vertices are congruent and one is convex, the other is as well. So all vertices of a regular polyhedron are convex.

9. In light of your answers to questions 5, 7, and 8, which regular polygons can occur as the faces of a regular polyhedron? Since we could not reach any conclusions in 7, there’s no way to rule out any regular polygons here. [Again, full credit will be awarded for any answer to this question.]

10. For each of the polygons in question 9, what are the possible numbers of that kind of polygon that might meet at a vertex of the polyhedron? Same situation: we can’t make any conclusions. [Full credit will be awarded.] 11. What is the maximum possible number of different regular polyhedra? Again, there is nothing we can say on the basis of these problems. [Full credit will be awarded.] Of course, it is the case that there really are only five regular polyhedra. There are various fixes to the flawed step 7 in this proof outline. One is to assume that the polyhedron satisfies the usual definition of convex: for any two points in the polyhedron, the entire line segment between the two points lies in the polyhedron. Then you really can say in step 7 that the sum of the angles must be less than a full circle. But you don’t really need that assumption: you could instead just assume that the polyhedron has no “holes” or “loops” in it. In that case, you can use something called Euler’s formula to show that there must be an overall angle deficit, and hence the same angle deficit (sum less than a full circle) at each vertex, and the proof again goes through. Laura may show this version in review session. But in fact, you shouldn’t even really need this assumption on the shape of the polyhedron; there really aren’t any polyhedra with holes satisfying the other assumptions about a regular polyhedron. However, this appears to be quite difficult to prove, and I don’t currently know a convincing proof outline. I will provide an update if the situation clarifies. However, we should all take this as an iconic tale: even simple questions can lead to deep and interesting mathemat- ics, which is one of the wonderful things about the subject.