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 The 720 degree deficit by Larry)

If we continue with the assumption that this is right:

We want a soccer ball to be nice and round(ish). Thats why an Archimedian solid is used. In an Archimedian solid every vertex has the same arrangement of verticies.

A regular pentagon has 108 degree corners where they are 120 for a regular hexagon.

There are only two possibilities.
1. Each vertex is two pentagons and one hexagon.
2. Each vertex is one pentagon and two hexagons.

In case 1 each vertex is 336 degrees or 24 short of 360. There would need to be 30 of these verticies for a 720 degree deficit.

In case 2 each vertex is 348 degrees or 12 short of 360. There would need to be 60 of these verticies for a 720 degree deficit.

From there I'm stuck. I've been trying Euler's formula F+V=E+2 using the ratios between F and E required. I keep showing that both cases are impossible. That ain't good.

-Jer

Posted by Jer on 2004-04-12 09:47:25