In this answer https://mathoverflow.net/a/19823/5239, it is indicated that it is impossible to make a polyhedron (with 3 faces meeting at each vertex) out of 12 pentagons and 1 hexagon.
There is certainly an easy reason why one hexagon is not possible but I have to admit that I am missing it. (In the reference given, it is stated as 'obvious'). Any help?
We can build up such a polyhedron starting from the unqique hexagon: Each of its edges must be an edge of a pentagon. By the three-faces-per-vertex condition, pentagons touching neighbouring edges must be distinct and share an edge. Even if we consider only the planar graph, not a 3d polyhedron, this makes it impossible that any two of theses six pentagons coincide. Using the three-faces-per-vertex condition again, there must be a second ring of six pentagons around this. These leave us with six edges than cannot be "closed".