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
Hint in the title? by Jer)
I was thinking about this yesterday, and decided that the answer is in the angles. Like Jer, I thought it too much work to figure out in detail, but I did hit on one point I found interesting. Perhaps someone can do something with this:
The interior angle of a hexagon is 120 degrees. For a pentagon it is 108 degrees. When two hexagons and a pentagon come together, the total of the interior angles is 348 degrees. Hence you have to bend them up to join the edges, and it is no longer flat.
The angle deficit at each vertex is 12 degrees. There are 60 vertices on the soccer ball. 60*12 = 720, which is exactly twice the number of degrees in a circle.
This seems too strange to be accidental.
In a cube, there are 8 corners and a 90 degree deficit at each corner. 8*90 = 720
In a tetrahedron there are 4 corners and 180 degree deficit at each corner. 4*180 = 720
In a icosohedron there are 20 corners and a deficit of 36 degrees at each corner. 20*36 = 720.
Looks like there is a therom here, which would indicate that 2 hexagons and a pentagon at each vertex REQUIRES 60 vertices, and hence the soccer ball pattern we have.
But I don't have the background or interest to try to prove it....
BTW -- what ever happened to the "preview" button this page used to have? With all my lines strung out to infinity I don't really know what this looks like before I post it.....
re: Hint in the title?
