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: Summary? by Charlie)

Define two adjacent vertices of the square 
in the northern hemisphere,

   A (a,0,sqrt(1-a^2))

   B (0,a,sqrt(1-a^2))

Define a vertice of the square in the
southern hemisphere adjacent to A and B,

   C (a*sqrt(2)/2,a*sqrt(2)/2,-sqrt(1-a^2))

Then

   |AB|^2 = 2a^2

       and

   |AC|^2 = 4 - a^2*(2+sqrt(2))

Setting |AB|^2 = |AC|^2 gives

   a^2 = 4/(4+sqrt(2))

Therefore, for the twisted squares, 
the maximum minimum is

   |AB| = |AC| = sqrt(2a^2)

        = sqrt(8/(4+sqrt(2))) ~= 1.215562524

Edited on February 3, 2008, 12:08 pm

Posted by Bractals on 2008-02-03 12:06:04