Soccer balls are usually covered with a design based on regular pentagons and hexagons.

How many pentagons/hexagons MUST there be, and why?

(In reply to re: The 720 degree deficit by Jer)

Thanks for the idea.  I completely forgot about Euler's formula.

Let H = number of hexagons
Let P = number of  pentagons

F = H+P
E = (6H+5P)/2
V = (6H+5P)/3

(H+P) + (5P+6H)/3 = (5P+6H)/2 + 2
3H + 8P/3 = 5P/2 + 3H + 2
16P = 15P + 12
P = 12

There can be any number of hexagons (including zero) but there will always be 12 pentagons.

Posted by Brian Smith on 2004-04-12 10:26:10