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: For Charlie's thoughtful consideration by Charlie)

Hi Charlie,

I respect your hard work and carefully precise calculation. I haven't done the calculation - and you've shown restraint not to challenge me on that score. I respect you for that as well. Please continue with me a while longer.

Could it be I missed a step in your transformation procedure? I understood that you twisted the northern hemisphere and that was all. Such a transformation alienates each northern point from its southern nearest neighbour, leaving each point with just two nearest neighbours. But you talk about four nearest neighbours. So perhaps after the rotation you move all 8 points iso-longitudinally toward the equator. This latitudinal compression increases the distance between all pairs of points within the same hemisphere, including nearest neighbours. This transformation can be sensibly continued until "north-south rapproachment"- i.e. the re-creation of nearest neighbour pairs across the equator.

That's not bad. But consider: with twist + compression all points remain constrained within two planes (parallel with and equidistant from the equator). The twist + shear transformation breaks this symmetry. Intuitively I expect the abandonment of a constraint to yield a higher optimum.

Of course, the only way to tell for sure would be to do the work.

Edited on February 4, 2008, 7:40 pm

Edited on February 4, 2008, 7:41 pm

Posted by FrankM on 2008-02-04 19:37:38