Counting the pentagons and hexagons on the surface of a football (aka, soccer ball)

Question

For the Americans: I'm talking about soccer :-)

As you might know, a football consists of a collection of regular pentagons and hexagons, but how much of which ones?

The construction is simple: for every pentagon, there's a hexagon at each side and for every hexagon, the polygons at its sides are altering between a pentagon and a hexagon.

Let's say we have the following numbers:

• $p$ : the number of points
• $s$ : the number of sides
• $u$ : the number of surfaces.

There is the well-known formula: $$p - s + u = 2$$

Let's now say:

• $v$ : the number of pentagons ("vijfhoek" in Dutch)
• $h$ : the number of hexagons

Then we know that: $$v + h = u$$

Every point is related to three segments and three surfaces, and every segment is related to two surfaces. But how to enter this in a formula to calculate the number of pentagons and hexagons? Does anybody have an idea?

Thanks in advance

Answer 1

You have $v+h=u$. You also have $5v+6h=2s$ (each "side"/edge counts twice). You also have $5v+6h=3p$ (every "point"/vertex counts thrice). So:

$$\frac{1}{3}(5v+6h)-\frac{1}{2}(5v+6h)+(v+h)=2$$

i.e.

$$\frac{1}{6}v+0h=2$$

i.e. $v=12$. You must have $12$ pentagons. So, you can have plenty* of different "footballs", which only differ in the number of hexagons, the simplest one having no hexagons at all (i.e. a dodecahedron).

With your other conditions: if you say every hexagon is surrounded by $3$ pentagons and every pentagon is surrounded by $5$ hexagons you also have the relationship $3h=5v$ which on a football yields $h=20$.

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*Note: this is an overstatement: if you want the football to be a uniform polyhedron, then the dodecahedron and the F.A. football (truncated icosahedron) seem to be the only choices.

Answer 2

Adding a picture of the variant (in my opinion the most common), where the sides of the hexagons/pentagons are gotten by trisecting the sides of an icosahedron (aka BuckyBall). The edges of the circumscribed icosahedron show in this version. Look at the grey areas.