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 20040412 09:47:25 