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(1a^2))
B (0,a,sqrt(1a^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(1a^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 20080203 12:06:04 