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?
Define two adjacent vertices of the square
in the northern hemisphere,
Define a vertice of the square in the
southern hemisphere adjacent to A and B,
|AB|^2 = 2a^2
|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