Prove that it is impossible to make a football out of exactly 9 squares and $m$ octagons, where $m \ge 4$. (In this context, a “football” is a convex polyhedron in which at least 3 edges meet at each vertex.)
I think the number of faces, $F$, is $9+m$, but I don't know how to count the number of edges $E$, and the number of vertices, $V$.
How shall I count them? Is there a clever way of thinking?
First Euler's formula for convex polyhedrons is $$v-e+f=2$$ from the question we have $$f = 9+m,\quad e = 9\cdot 2 +m\cdot 4,\quad e \ge \frac{3}{2}v$$ For the inequality, for each vertex there are at least 3 edges, but an edge is shared between 2 vertices. Similar for the second equality 2 faces share one edge.
Now we plug everything together.
$$ v-e+f = 2 \:\Longleftrightarrow\: v - 18 - 4m + 9+m = 2 \:\Longleftrightarrow\: v = 11 + 3m $$
Hence we have $$ 11 + 3m = v \le \frac{2}{3} e = \frac{2}{3} 18 + \frac{2}{3} 4m \:\Longrightarrow\: \frac{1}{3}m \le 1 \:\Longleftrightarrow\: m \le 3. $$ which is not allowed because of the given constraint $m \ge 4$.$\Box$