Eight points are placed on the surface of a sphere with a radius of 1. The shortest distance between any two points is greater than 1.2. How can the points be arranged?
Hint: They are not arranged as a cube. The cube would have an edge length of only 2/sqrt(3) = 1.1547.
(In reply to
re(4): Further improvement by Charlie)
BTW, when the square's opposite corners are pulled in, and the squares bent along that diagonal to make new triangles, you'd probably come up with a snub disphenoid, described in Mathworld and Wikipedia. I don't know if that's inscribable within a sphere without making the triangles nonequilateral.

Posted by Charlie
on 20080202 12:28:56 