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
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