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)

That's the way I've been thinking about this problem.  I assumed that every vertex would have the same arrangement of pentagons and hexagons.  But then there is the question of why there aren't two pentagons and a hexagon to each corner.

I thought I might just go ahead and see how that might work out, without all these math formulas (mainly because the math is a little beyond me).

So take a hexagon face in this theoretical solid.  There would be 6 pentagons surrounding it.  Next to the pentagons would be another row of alternating pentagons and hexagons.

I need to go no further, for I can already disprove such a solid.  After adding the six pentagons around the hexagon, there would be a jagged edge to add to.  The alternating hexagons and pentagons would go in the concave vertices of the jagged edge.  The problem here is that all of the concave vertices have the same angle (the pentagons all need edges with length equal to that of the hexagon).  You can't fit both pentagons and hexagons in the same size angles.  I believe the angle only fits the hexagons.

Now we have disproved the theoretical solid, and we only need to assume that there is at least one pentagon and one hexagons.  Therefore, each vertex has two hexagons and a pentagon.  As any soccerball will show, this pattern leads to and must come out with 12 pentagons and 20 hexagons.

Posted by Tristan on 2004-04-12 10:32:31