Htykuut

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.

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.

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.

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

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 $ein Bbb Z^{+}$ denotes the total of number of edges of $G$. Hence $$f_{1}+f_{2}+...+f_{F}=2e.$$

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.