Show that there is no regular planar graph (all vertices degree 3) so that all regions, including the unbounded region, are hexagonal.
I am sure this has something to do with the fact that for planar graphs the sum of degrees of regions is twice the number of edges. Looking for a hint to push me in the right direction.
The Euler characteristic of a planar graph is $2$. However, if every face is hexagonal and every vertex has degree $3$, $V+F-E$ cannot be $2$, because it is $ 2F+F-3F = 0$.
Explanation: there are $6$ vertices on every face and every vertex belongs to $3$ faces, hence $V=2F$. Every face has six edges and every edge belongs to two faces, hence $E=3F$.
However, we have such kind of graph embedded in a torus (courtesy of R. M.):
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