Let $G$ be a 3-regular connected planar graph with a planar embedding where each face has degree either 4 or 6 and each vertex is incident with exactly one face of degree 4. Determine the number of vertices, edges and faces of degree 4 and 6.
Using handshake lemmas and Euler Formula, I've come up with the following for $E$ edges and $n$ vertices:
$2E=3n$
$2E=4F_4+6F_6$
$n-E+F_4+F_6=2$
I'm missing an equation because I'm not sure how to use the restriction where each vertex is incident with one face of degree 4. Any help?
Best Answer
Consider the edges adjacent to any vertex $v\in V(G)$. We know the vertex is incident to only one face of degree 4, thus 2 of the three edges adjacent to it form part of the length of a face of degree 4. Thus, two thirds of all edges lie on the border of a face of length 4, and we have:
$f_4=\frac{1}{4}*\frac{2}{3} *e=\frac{1}{6}e$
Using this fact and the rest of your equations, you should be able to get all the values you need.