Show that there is no regular planar graph (all vertices degree 3) so that all regions, including the unbounded region, are hexagonal.

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 $2$. Commented Mar 15, 2017 at 3:00
• (1) 3v = 2e (why?); (2) 2f = 6e (why?); (3) express v,f in terms of e; (4) v-e+f = 2 yields an easy contradiction. Commented Mar 15, 2017 at 3:02
• @quasi: I did it in terms of $f$ but it is the same thing. 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 $e=3f$. Commented Mar 15, 2017 at 3:07
• @Jack D'Aurizio: Your version is an ounce or two simpler, but I agree, in essence, they're the same. Commented Mar 15, 2017 at 3:12

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 $E=3F$.
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