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(2): For Charlie's thoughtful consideration by FrankM)
"Could it be I missed a step in your transformation procedure? I understood that you twisted the northern hemisphere and that was all."
I assume you mean twisted from where it would be in a cube. But once the top plane was twisted (or the bottom planeit's relative), the height was adjusted so that the 8 triangles circumnavigating the globe would be equilateral; that was the whole point of the programto find where the squares sides would equal the other sides of the triangles. This enables the sides of the squares to be longer than they would be otherwise. But of course once the lengths match the lengths of the edges connecting the two squares, then the increase of the lengths of the sides of the squares would be pointless as the other sides of the triangles would become shorter, and it's the shortest edge in the whole thing that counts, and needs to be maximized.
"So perhaps after the rotation you move all 8 points isolongitudinally toward the equator. This latitudinal compression increases the distance between all pairs of points within the same hemisphere, including nearest neighbours."
Yes, as the distances when they were part of an inscribed cube were just not large enough. They had to be moved closer to the equator.
"This transformation can be sensibly continued until 'northsouth rapproachment' i.e. the recreation of nearest neighbour pairs across the equator."
Yes, forming two squares connected with 8 equilateral triangles. As I stated in my description: "But we'll also want to make the sides of the squares equal to the sides of the triangles, and so the triangles will also be equilateral, not just isosceles."
"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."
Any increase in the size of some distances will be negated and made worse by the decrease in the size of the others. That's counterproductive since in the optimum, they have the same length, so any decreased length is a worse solution.

Posted by Charlie
on 20080204 23:29:03 