A corollary of Euler's formula tells us that the edge-vertex graph of every convex 3-polyhedron must have a face with either 3, 4, or 5 edges, using an argument about the average degree of vertices.


Does it imply anything further about the exact degree of the vertices of these faces?


In particular, is at least one of the following statements always true for such a polyhedron?


1. it has a face with three edges, OR
2. it has a face with four edges with at least one vertex of degree 3 on that face, OR
3. it has a face with five edges with at least two vertices of degree 3 on that face?


It's plainly true for simple 3-polytopes, as every vertex is degree 3. I see that it's also easily true for objects like the (rhombic triacontahedron)[http://en.wikipedia.org/wiki/File:Rhombictriacontahedron.svg] that are highly regular and composed of quadrilateral faces. 


Is there any 3-polytope where none of 1, 2, and 3 hold?



Does Euler's formula imply bounds on the degree of vertices in a 3-polytopal graph?
It does follow from Euler's theorem that every graph of a 3-polytope has either a triangle face or a vertex of degree 3. To see this note that if every face has 4 or more edges then $4F \le 2E$ (double count pairs (e,f) where e is an edge f is a face and f contains e.) and now look at Euler's theorem: $2V-2E+2F=4$ so $2V-E \ge 4$ and $E \le 2V-4$ and this imples that there is a vertex of degree 3. 


You may want to look at the question and answer on Eberhard's theorem [Characterizing faces of 3-dimensional polyhedra. (Related to Victor Eberhard's Theorem [1890]:)](https://mathoverflow.net/questions/10039/characterizing-faces-of-3-dimensional-polyhedra-related-to-victor-eberhards-th/24340#24340)


High dimensional analogues of the result I mentioned in the spirit of your question are very interesting. 


