The standard foot ball is made of 32 polygons, comprising of pentagons and hexagons.
Quick question: In a football How many pentagons and hexagons are in total?
There are 12 pentagons and 20 hexagons in football. Traditional way of coloring them is to have all pentagons in black and hexagon in white. This was done in 1970 to reportedly to enhance the visibility of the ball in television.
According to FIFA the definition of foot ball is something like this, it must be sphere with circumference between 68 to 70 centimeters with most 1.5% deviation from sphericity when inflated to a pressure of 0.8 atmosphere. Now let us see how mathematician defines a football. According to mathematicians the foot ball is one which satisfies these fundamental rules.
It is a polyhedron that consists only of pentagons and hexagons
The sides of each pentagon meet only hexagons
The sides of each hexagon alternately meet pentagons and hexagons
We have the famous Euler’s formula for graphs relating the (v) vertices, (e) edges and (f) faces.
Euler’s equation: v – e + f = 2 If we assume in a foot ball we have b black polygons and w white hexagons then the total number of faces is f = b + w. In all pentagons we have 5b edges and 6w edges in the case of hexagons. Since the edges are shared between pentagons and hexagons the number edges is (5b + 6w)/2. The total number of vertices (5b + 6w)/3 since each vertex is part of three polygons.
Substituting these values in the Euler’s formulae we get the number pentagons b to be 12. How ever there are no a priori limit for the number of hexagons w. Therefore there is no limit on the number of vertices. On imposing the additional condition (2) it can be shown that the number of hexagons needs to at least 20. As all edges of pentagon are also edges of hexagons and half of the edges of hexagons are also edges of pentagons we have (6w)/2 = 5b.
For b = 12 we have w = 20. There fore in a foot ball we have 12 pentagons and 20 hexagons.
Source : Dieter Kotschick beautiful article in American Scientist