Prove that for any polyhedron there are two faces with the s

Prove that for any polyhedron there are two faces with the same number of vertices.

Solution

Let us assume the opposite and the face with the greatest number of edges (it is only one!) has k edges. This face has k neighbor faces, but since all of them have different number of edges, these faces are at most k-3, (being eventually with 3, 4, 5, ..., k-1 edges) - contradiction.

Euler\'s formula : F + V - E = 2

The Duality Principle for polyhedra (vertices faces) allows to conclude similarly that at least 2 vertices have same number of faces.