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(3): Solution by Bractals)

Both the lines through the faces of the octahedron and the vertices of the double tetrahedron would be equivalent to the cube vertices.

Posted by Charlie on 2008-02-01 19:16:38