# 3-regular planar graph

Yet another question I was going over and struggled.

Given a 3-regular connected planar graph, so that every vertex lies on the edge of a face of length 4, of a face of length 6 and of a face of length 8. Find the number of vertices, edges and faces of each length.

Obviously, in order to solve it one will have to use:

$n-e+f=2$

$2e=3n$

$2e=4f_4+6f_6+8f_8$

$f=f_4+f_6+f_8$

But using just these three equations doesn't lead to any useful result. The best I managed to get out of them is:

$f_4=6+f_8$

And some identities connection the edges to number of vertices to number of faces - none enough to solve the question. It seems as if I miss some identity that lies with the fact that "each vertex lies on the edge of a face of length 4, of a face of length 6 and of a face of length 8" - it can't be that the number of each of these faces is equal, they seem to be certainly different actually. Yet $f_6$ disappears along the way and I don't know how to go on. Any help/direction/ideas will be extremely helpful!

Consider the edges incident to some vertex $v$. Exactly two of these three edges border a face of length $4$. Therefore two thirds of all edges lie on the border of a face of length $4$. Hence $f_4=\frac{1}{4}\left(\frac{2}{3}\right)e=\frac{1}{6}e$. Using the same justification, we have $f_6=\frac{1}{6}\left(\frac{2}{3}\right)e=\frac{1}{9}e$, and $f_8=\frac{1}{8}\left(\frac{2}{3}\right)e=\frac{1}{12}e$.
Hence we have $$f=f_4+f_6+f_8=e\left(\frac{1}{6}+\frac{1}{9}+\frac{1}{12}\right)=\frac{13}{36}e.$$
Plugging this into our first equations, we have $$2=n-e+f=\frac{2}{3}e-e+\frac{13}{36}e=\frac{e}{36}.$$
Hence $e=72$, and plugging things back in, we get $n=\frac{2}{3}e=48$, $f=\frac{13}{36}e=26$, $f_4=\frac{1}{6}e=12$, $f_6=\frac{1}{9}e=8$, and $f_8=\frac{1}{12}e=6$.
• Even simpler(?): Count vertex-face incidences to find $n=4f_4=6f_6=8f_8$ – Hagen von Eitzen Jun 29 '14 at 9:01