Here is a proof that for any convex polyhedron there exist 2 faces with equal number of edges. I was told that it doesn't work for non convex polyhedra but I can't find out why. Could someone show a counterexample or explain why it fails?
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what have you tried? – cineel Jan 25 '22 at 18:55
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I tried to come up with such non convex polyhedra but I couldn't find any that this proof wouldn't work for it. – Gh0st Jan 25 '22 at 19:07
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It is possible that two faces are connected by more than one edge. See for example the toroidal polyhedron on this page (about 3/4 of the way down).
Suppose $K$ is a face with the greatest number of edges, $n$. There may be fewer than $n$ distinct faces connected to $K$. That is where the proof fails.
However, any polyhedron like this is degenerate, in the sense that there are two faces in the same plane, or two edges on the same line.
Two distinct planes meet in at most one line, so two faces are connected by at most one edge, in a non-degenerate polyhedron. So the proof works in that case.
mr_e_man
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