# Graph Theory: There is only one planar graph with all vertices of degree 4 and 10 regions

Let $G$ be a planar graph in which every vertex has degree 4 and there are 10 regions (faces). How many graphs up to isomorphism are there?

We can easily infer from the vertex sum and Euler's formula, that the number of vertices is 8.

I drew one but every other graph that I draw is isomorphic to it, so I suspect there is just one. Proving it is another matter. I tried using the fact that it must have an Eulerian cycle, defining the map from vertex to vertex by the edges in the path, but couldn't see an argument that the map must be well-defined and bijective. I tried using the fact that it must have a Hamiltonian path, and now the map is bijective but I could not see a way to prove that it satisfied the homomorphism property. I also thought about trying to construct an isomorphism by tracing out the graph, but there seemed to be too many choices that one could possibly make.

• Next you need two quadrilateral and six triangle faces or one pentagon and nine triangles. For the two quadrilaterals they can share an edge or not. – Ross Millikan Feb 5 '18 at 5:42