Max number of points that fit in/on a sphere of radius r, with minimum inter-point distance also r This just came up in a game, and I realized I don't know how to solve this. 
Given a sphere of radius r (say, 20'), what is the largest number of points that can be arranged within the sphere (the surface itself counts), with none of them coming closer than r from any other?
I looked for a prior solution with no luck so far. I think that the vertices of an icosahedron should work, because the radius of the circumscribed sphere is only ~95% or the edge length (so one could use an icosahedron of radius r+epsilon to be able to add a point at the center). Is it possible to do better?
Thanks!

Finding a collection of points with each point at distance at least $r$ from its nearest neighbor(s) is tantamount to finding a collection of disjoint balls of radius $\frac{1}{2}r$, which in turn relates to to sphere packing.
It's known that (in Euclidean space) no more than twelve balls of radius $\frac{1}{2}r$ can touch a ball of radius $\frac{1}{2}r$, so your guess is correct: Put one point at the center of the sphere, and arrange twelve points at the vertices of a regular icosahedron, or just sufficiently near the vertices of an icosahedron, as in the various dense lattice packings.
