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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 $m\neq n$) 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.

Community
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Quiet_waters
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1 Answers1

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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 $m\ge4$, what values of $n\ge3$ yield polyhedra? All the football polyhedra are listed below.

  • $m=4,n\ge3$ 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°.

Parcly Taxel
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  • This is a wonderful answer ... Thank you. Some doubts: 1) Do you know the truncated polyhedras 'by mind' (just a curiosity)? 2) At this question, it's not explicit "regular" ... Do you think there's some problem? 3) Do not you use Euler's polyhedron formula in any local, right? Thanks for all the help. – Quiet_waters Nov 12 '18 at 14:13
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    @Na'omi All of these can be called uniform polyhedra (all vertices the same, but not all faces the same), which is one step below regular. These, too, have been completely classified, and I know them by heart. Euler's formula can be used fo find the number of faces for each of these uniform polyhedra, but the more usual construction is by modifying the Platonic solids to yield new polyhedra, such as by truncation. – Parcly Taxel Nov 12 '18 at 14:16
  • Thank you once again. – Quiet_waters Nov 12 '18 at 14:29