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(3): For Charlie's thoughtful consideration by Charlie)

Charlie,

Your latest response was certainly helpful. I'm satisfied that we understand each other now. I will make one final verbal effort to bring my point home. If that falls flat, I may resort to writing a small program to identify the points and enhanced minimum distances. (Call me old fashioned if you wish: program generated solutions are inelegant to me, and I'm getting a lot of good practice communicating geometric notions from our exchange - two reasons why I've hesitated until now to grab for the compiler!)

Let's take your eight points as a starting position. This ought to make the comparison easy. I propose a two step transformation to increase the minimum distance between pairs of points.

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.

How does P1 differ from P0? In P1 the distances between nearest neighbours within the same hemisphere have increased. (We ensure that the shift was small enough so that the pole pushed points don't encroach on d0). Each of the four pole pushed points have also been repelled from their cross hemisphere nearest neighbours. Only the four static points still have neighbours within d0: Each static point is a distance d0 away from the two static points on the other hemisphere (i.e. it has two nearest neighbours).

We are now ready to effect the second and final transformation. Note that the cross hemisphere distance between P1-static points can be increased by moving these points away from the equator. We will want to move them only a small amount (so that the intra-hermispherical minimum separation stays above d0). The minimal separation in P2 then exceeds d0.

 

 

Edited on February 5, 2008, 9:38 pm

Edited on February 5, 2008, 9:39 pm

Posted by FrankM on 2008-02-05 21:37:04