🚀 go-pugleaf

In article <1993Jan26.235433.18005@cs.ucla.edu> byron elbows writes:
>I need help on a sphere packing question.  Suppose you have a white ping
>pong ball, and 13 red ones, all the same size.  Can you glue all of the red
>ping pong balls to the white one?  [...]
>
>It is a simple matter to do it with 12.  Simply place the spheres at the
>corners of an icosahedron centered at the center of the white ping pong ball.
>It turns out this leaves some extra room for spheres to be moved around [...]

This remark caused me to wonder what is the diameter of a sphere
circumscribed about an icosahedron of side 1.  (Presumably it's less
than 2, so that the above works.)  It would be nice if there were a
simple, painless answer to this, but I couldn't think of anything that
didn't involve lots of trigonometry.  Does anyone know a very simple
method to compute this?

By the way, you also get 12 spheres around a central one if you use
either hexagonal or cubic close packing, in which 6 of the spheres lie
in one plane (along with the center sphere) with centers at the
vertices of a regular hexagon.  In this configuration all the
surrounding spheres just barely fit in around the central one with no
"wiggling room," but it's easy enough to verify by diagram that the
configuration works.

-- David Karr (karr@cs.cornell.edu)