Let G be a polyhedron (or polyhedral graph), each of whose faces is bounded by a pentagon or a hexagon

(i) Use Euler's formula to show that G must have at least 12 pentagonal faces
(ii) Prove, in addition, that if there are exactly three faces meeting at each vertex, then G has exactly 12 pentagonal faces.


A corollary to Euler's Formula states that for a polyhedral graph with n vertices, m edges, and f faces, that $\displaystyle n - m + f = 2. $

I also know that since each face is bounded by at least 5 edges, that it follows that counting up the edges around each face that $\displaystyle 5f \leq 2m $ since each edge bounds two faces, but then I get

$\displaystyle 5(2-n+m) \leq 2m $
$\displaystyle 10 - 5n + 3m \leq 0$

but I can't see how this helps. Am I doing something wrong? Any help on (i) will be appreciated. After that, I'll try (ii) on my own.