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): Solution by Charlie)

Interesting that Dej Mar saw a tetrahedron pair, while Bractal & brianjn jumping off point was the (nominally suggested) cube.

I too saw brianjn's "twisted cube", but I think it can be improved further with a further operation.

The twisting operation extends the minimal distance between points in different twist planes. The four points within a plane still lie on a square. That is to say, each point is closer to it's neighbour than to its diagonal counterpart (by a factor of sqr(2)). This suggests the next deformation:

Take a pair of diagonally opposed points within a square and move them (along the great circle connecting them) toward the square centre. Take the other two points and pull them away from the square centre. Repeat the process with the two pairs of points on the other square as well.

Thus we increase the distance between adjacent points within each square, doing so until the identity of closest pairs is about to change.

 

Posted by FrankM on 2008-02-02 07:11:34