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): Further improvement by FrankM)

If the opposite points on the squares are brought in, to make a diamond, the resulting planar shape will no longer have its vertices on the spherical surface. The diamond face would have to change into two triangles by bending along the short diagonal. The result would now be a 12-sided dodecahedron, but it of necessity couldn't be regular, as a regular dodecahedron has pentagonal faces.

The Wikipedia article referenced by brianjn mentions:

When eight points are distributed on the surface of a sphere with the aim of maximising the distance between them in some sense, then the resulting shape corresponds to a square anti-prism rather than a cube. Different examples include maximising the distance to the nearest point, or using electrons to maximise the sum of all reciprocals of squares of distances.

Posted by Charlie on 2008-02-02 10:45:59