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(4): More for Charlie's thoughtful consideration
"Call your eight points position P0 and the corresponding nearest
neighbour distance d0. Derive P1 from P0 by taking two diagonally
opposite points in the southern square and moving them
isolongitudinally toward the south pole, while, similarly, moving two
diagonally opposite points in the northern square and moving them
toward the north pole. Of course, we ensure that the points remain on
the spherical surface."
If we move two diagonally opposite (separated by 180-degrees of longitude) points in the southern hemisphere without moving the other two southern points north toward the equator, they will be closer to those other two southern points, as they are on a lesser circle around the south pole, not the great circle that is the equator. So the separation is less than d0 even within that southern square.
Posted by Charlie
on 2008-02-06 00:36:23