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): Further improvement by Charlie)

Hi Charlie,

I struggled to get my point across and perhaps was ambiguous. (Gee, I miss being able to show this on a blackboard, or at least a diagram).

Of course you are right that all points are constrained to the spherical surface: Points brought in means moving them along the great circle connecting the diagonal points and the face centre.

An elaboration may help: After the first operation (twisting one square plane), each point has two nearest neighbours. (Previously it had three). The new operation is to move  points along the great circle normal to the line connecting it's two nearest neigbours. Moving the right direction along this great circle increases the distance to the nearest neighbours. (Hope you are still following me!). The moving stops when the point again has three nearest neigbours.

 

Posted by FrankM on 2008-02-02 23:42:52