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 Solution by Dej Mar)

I believe that this goes very close to the required arrangement but the rotation has been excessive.

Consider two squares each on a parallel plane which intersects the sphere.  Let the squares be rotated 45º in relation to each other. The solid which would be described by the eight vertices would form a square antiprism.

The net would be two squares with a band of 8 congruent triangles separating them; I haven't determined any measurements but the triangles would probably be iscoseles with the base being the shorter side.

One location describing this object (and there are certainly more) is at http://en.wikipedia.org/wiki/Square_antiprism

Edited on February 2, 2008, 1:09 am

Posted by brianjn on 2008-02-02 00:38:36