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(5): Further improvement by FrankM)

Some clarification, and let's see if I understand:

"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."

In my final outcome, each chord (and each arc it subtends) is equal in length, so each point lies at a vertex of one square and of three triangles and therefore has four nearest neighbors--two on the same square and two that are on the opposite square, in the other hemisphere.

Does this describe what you are saying?:

On the northern square, increase the latitude (closer to the north pole) of diametrically opposite points and decrease the latitude (closer to the equator) of the other pair of opposite points.

In the southern square do the same thing.

If this is the case, then:

However, take one of the northern points that is moved closer to the equator. It has two nearest neighbors in the southern square, one of which will also move closer to the equator, and the two points in question will then be closer than before.

Posted by Charlie on 2008-02-03 11:50:52