Cfrgtkky

I am dealing with the test of the OBM (Brasilian Math Olympiad), University level, 2016, phase 2.

As I've said at this topic (question 1), this other (question 2) and this (question 3), I hope someone can help me to discuss this test. Thanks for any help.

The question 5 says:

A soccer ball is usually obtained from a polyhedral figura that has two kinds of faces, hexagons and pentagons, and in each vertex focus three faces, which are two hexagons and one pentagon.

We say that a polyhedra is "soccer" if, as the soccer ball, has faces that are $m$-agons and $n$-agons (with $mneq n$) and in each vertex focus three faces, which are two $m$-agons and one $n$-agons.

(i) Show that $m$ is even.

(ii) Find all the soccer polyhedrals.

I'm trying to use $V+F=A+2$. It's trivial that $A=frac{3}{2}V$, so $F=frac{1}{2}V+2$ (particularly, $V$ is even).

I have $frac{2V}{m}$ $m$-agons and $frac{V}{n}$ $n$-agons, so $F=V(frac{2}{m}+frac{1}{n})$.

Then, $V(frac{2}{m}+frac{1}{n}-frac{1}{2})=2$...

Thank you for a help.

The first part does not require Euler's polyhedron formula. It is trivial to show that $m$- and $n$-gons alternate around any $m$-gon, which immediately implies that $m$ is even (otherwise there would be a vertex figure not of the form given).

The second part is just casework: for each even $mge4$, what values of $nge3$ yield polyhedra? All the football polyhedra are listed below.

• 
  $m=4,nge3$ is the set of prisms


• 
  $m=6,n=3$ is the truncated tetrahedron


• 
  $m=6,n=4$ is the truncated octahedron


• 
  $m=6,n=5$ is the truncated icosahedron, or a normal football


• 
  $m=8,n=3$ is the truncated cube


• 
  $m=10,n=3$ is the truncated dodecahedron

I stopped at $(m,n)=(6,6),(8,4),(10,4)$ and $m=12$ because at those points the finite footballs turn into tilings (i.e. there are an infinite number of faces), either in the Euclidean or hyperbolic plane – the sum of angles around each point becomes greater than 360°.