Given a collection of Christmas ornaments in the shape of arbitrary polyhedra (such as, but not limited to cubes, tetrahedra, icosahedra, etc), suppose each face of each polyhedron is colored depending on the number of edges of the face. For example, every four-sided face (such as a face of a cube) has color number 4, and every triangular face has color number 3, and in general every n-sided face has color number n. Can there be a Christmas Ornament with all sides painted a different color? If we assume that two faces share no more than one common edge, and that there are a finite number of faces, then the answer is easily seen to be no. Simply list, in order, the numbers of neighbors for the n faces, so we have a list of n strictly increasing positive integers (because no two faces have the same number of neighbors). Thus at least one of the n faces must border on at least n other faces, which is impossible because there are only n-1 other faces. On the other hand, if we remove the restriction to finite-faced polyhedra, the answer is yes, because we can define a fractal structure that has no two surfaces with the same number of "sides". To illustrate this, consider a regular tetrahedron, each of whose four triangular faces has 3 sides. Now draw a triangle on one face, a square on another, a pentagon on another, and a 9-sided polygon on the remaining face. If we exclude these polygonal regions from their respective faces, then the original faces of the tetrahedron have 6, 7, 8, and 12 edges, and the excluded regions have 3, 4, 5, and 9 edges. Needless to say, these "excluded regions" aren't really distinct faces, because they reside entirely within the planes of the original faces. However, we can fix this by building little pyramids on each of those drawn polygons. This creates a large number of new triangular faces, each of which obviously has three sides. Now we can repeat the trick, drawing a 10-sided polygon on one of those little triangular faces, an 11-sided polygon on another, a 15-sided polygon on another, and so on. Then we build little pyramids on these polygonal bases and repeat the process. Thus, defining a solid as the limit of this construction process carried on indefinitely, we have a fractal ornament with infinitely many faces, no two of which has the same number of edges. Notice that, even though each face shares only one edge with each of its neighbors, this construction evades the earlier proof simply because it's not a finite construction, i.e., there is no finite integer n equal to the number of faces. So, in order for the answer to be no, we need to restrict our polyhedra to those with a finite number of faces. The question is whether there exists a finite polyhedron such that no two faces have the same number of edges - where "edge" is defined as a maximal *contiguuous* line segment contained in two specific faces. Thus, two faces can share more than one edge. (Note that a contiguous line segment counts as more than one edge for a given face if it intersects more than one other face.) I'll also assume the surface has genus 2 for the moment. Let's say a face is "normal" if it shares exactly one edge with each of its neighboring faces. We've seen that if all the faces of a solid are normal, it's impossible for all of them to have a different number of edges. Now let's consider a solid with one or more non-normal face(s), (as suggested by David Einstein) such as two edges adjoining the same face of neighbor ___ ____ \ \ / / \ \/ / \_____/ This implies the existence of two faces X and Y that are adjacent over (at least) two separate segments of their common line, as illustrated below: Face X e_________f c ___/ \ / d \ A_______B/ \C__________D \_______ j__/ l g\ / k \___/ h i Face Y Faces X and Y share the common edges AB and CD, but they do not connect to each other on the segment BC. Instead, within the region bounded by the cycle of vertices BcdefCkjihglB we have a set of other surfaces. Let's call a region like this an "anomaly" (borrowed from crystalography). Suppose all the surfaces enclosed by X and Y in this anomaly are normal. We can then apply our original argument (strengthened by the fact that no face can have fewer than 3 edges) to show that at least one of the n faces in this anomaly must have at least n+2 neighbors, which is impossible because even counting X and Y there are only n+1 available neighbors. Therefore, we conclude that not all the faces in this anomaly can be normal, i.e., there must be two faces that share more than one common edge. In other words, there must be a (strictly smaller) anomaly within the anomaly. We can now repeat the above argument to prove that THIS smaller anomaly must also contain an anomaly, and so on ad infinitum. This shows again that if we allow the polyhedron to have infinitely many faces the proof doesn't work - fortunately - because the theorem is false in that case. On the other hand, if we consider only polyhedra with a FINITE number of faces, then we must at some stage reach an anomaly that is filled with only normal faces, and so those faces cannot all have a different number of edges. What about the restriction to surfaces of genus 2? I imposed that because of the possibility that for a surface of higher genus an anomaly might "enclose itself", and thereby allow a finite solution, if the anomaly contains a "wormhole". There may be some way to close this loophole (literally!) for surfaces of higher genus, but I don't see it at the moment. "God hath given you one face, and you make yourself another."

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