I was studying for my graph theory test and I got stucked with the following question:
Let G be a plane 3-regular connected graph formed with hexagonal and petagonal regions, where every vertex of G is exactly in only one of the pentagons. Determine the order, size and number of faces of the graph.
I started by naming $p$ to the number of pentagonal faces of the graph, and $f$ to the number of faces of the graph. Therefore, $|V|=5p$. Moreover, as G is planar and the smallest cicle is 5, we have that $5f \leq 2|A| = |V| \delta (G) = 15p$. Thus, $f \leq 3p$. We could write also $f=p+h+1$ where $h$ denotes the number of hexagonal faces of the graph and use Euler's Identity, getting the expression $6p+h-|E|=1$ . However, I didn't have any further ideas, so I would thank any hint or idea to continue with the activity.
Actually, I had also the bound $|E| <= \frac{5}{3} (|V|-2)= \frac{5}{3}(5p-2)$ as G is a planar graph with cicles of order higher than $5$, though I can't see any further conclusion from this inequality. Moreover, as G is 3-regular, $|E|=\frac{3}{2} |V| = \frac{15p}{2}$, and the number of vertices is, therefore, an even multiple of 5.