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 non-equilateral.

Posted by Charlie on 2008-02-02 12:28:56