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Number of Pentagons and Hexagons on a Football

Given a standard football, regular hexagons and pentagons are drawn on it as shown in the picture. Find out the number of hexagons and pentagons.

Number of Pentagons and Hexagons on a Football 1

We can apply Euler Characteristics to find out number of Hexagons and Pentagons on a standard Football.
According to Euler Characteristics: For every surface S there exists an integer

    $$\\chi(S)$$

such that whenever a graph G with V vertices and E edges is embedded in S so that there are F faces(regions divided by the graph), we have:

     $$ V - E + F = \\chi(S)$$

For a sphere(shape of football)

    $$\\chi(S)$$

= 2.
Hence, the equation becomes V – E + F = 2.

Now, let number of pentagons P and number of hexagons H.

Number of vertices will be:
6 vertices for each of the hexagons, i.e. 6*H.
5 vertices for each of the pentagons, i.e. 5*P.
But we have counted each vertex thrice, once for each adjacent polygon, follow the picture
Number of Pentagons and Hexagons on a Football 2
Hence, number of vertices, V = (6*H + 5*P)/3.

Number of edges will be:
6 edges for each of the hexagons, i.e. 6*H.
5 edges for each of the pentagons, i.e. 5*P.
But we have counted each edge twice, once for each adjacent polygon, follow the picture
Number of Pentagons and Hexagons on a Football 3
Hence, number of edges, E = (6*H + 5*P)/2.

Number of faces will be:
There are H hexagons and P pentagons, each forming a face. Hence, total number of faces, F = (H + P).

So, we can write:

     $$ V - E + F = 2 $$ \\\\ $$ \\frac{6*H + 5*P}{3} - \\frac{6*H + 5*P}{2} + (H + P)  = 2$$

after solving this equation we will get P = 12. So, there are 12 Pentagons.
Now number of Hexagons:
We can see that each pentagon is surrounded by 5 Hexagons. So there should 5*P hexagons, but we have counted each hexagon thrice for each of its 3 adjacent pentagons. Hence, number of hexagons = 5*P/3 = 5*12/3 = 20.

Hence, there are 20 Hexagons and 12 Pentagons in a standard football.

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