Prove that in any convex polyhedron, the number of faces that have an odd number of edges is even.

Question

Prove that in any convex polyhedron, the number of faces that have an odd number of edges is even.

I attempted to prove this by contradiction but didn't make any progress.

Answer 1

Let the faces be $F_1,\ldots,F_k$. Consider the sum $S=n_1+\cdots+n_k$ where face $F_i$ has $n_i$ edges. This sum counts each edge twice: $S$ is even. So the number of $j$ for which $n_j$ is odd must be even.

Answer 2

I'll elaborate a bit more, which'll helpfully help anybody interested get off on the right foot.

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Let $G$ be a graph depicting the planar representation of a convex polyhedron. Let $F \in \Bbb Z^{+}$ denote the total number of faces in $G$ and $f_{1},f_{2},...,f_{F}\in \Bbb Z^{+}$ denote the respective number of edges of each of the $F$ faces of $G$. Recall $$\sum_{i=1}^F f_{i} =2e,$$ where $e\in \Bbb Z^{+}$ denotes the total of number of edges of $G$. Hence $$f_{1}+f_{2}+...+f_{F}=2e.$$

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There we have it. You just use some basic properties of number theory/ arithmetic/ algebra to deduce that there must be an even number of odd $f$'s so that ultimately their sum is an even number. For example, note that $2+3+5=10$ is an even number, but $2+3+5+7=17$ is not.

Answer 3

Create a graph in the following way:

• Vertices are faces of the polyhedron
• Two vertices are connected if their corresponding faces share an edge

Observe that the odd-edged faces are exactly the vertices of odd degree in our graph. We know that the number of vertices of odd degree in any graph $G$ is even.