Proof that for any convex polyhedron there exist 2 faces with equal number of edges

Question

As the question states, I need to prove that for any convex polyhedron it is true that there exist two faces with same number of edges. My solution: Let face $K$ be the face with the greatest number of edges, $n$. Every adjacent face has $3,4,5,...,n-1,n$ edges. By Dirichlet's principle there must exists 2 faces with equal number of edges. Is my solution good? Are there any other solutions to this problem? Thanks!

Answer 1

Since I don't have 50 reputation, it is not a comment but an answer... But I am not sure it merits one on its own.

I was foiled by your mention of Dirichlet... not said like that in my country!

Basically, you take the face K with the higher number of edges (n in your case). There are exactly n other polygons linked to this first face K. You have maximum (n-1) numbers to distribute (in fact less so). Hence the absolute necessity to have at least two faces with the same number of edges.