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(6): Further improvement by Charlie)

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.

Yes, indeed. Four nearest neighbours (not three, as I had erroneously written)!

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.

Right.

Perhaps I could have gotten the point across more directly by pointing out that this operation is enacted along lines of constant longitude..

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.

No. That is not right. After completing operation 1 (latitude perserving twist) and prior to performing operation 2 (longitude perserving shear) the points in the northern hemisphere no longer have any nearest neighbours in the south. That is to say, the closest points south of the equator are further removed than the near lying pair of northern neighbours. It is this fact that enables (limited) longitude perserving shear transforms without decreasing the minimum distance between pairs of points.

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BTW I feel like we are developing a new language with which blind men can discuss geometry. I miss being able to draw diagrams, but I have to admit its working better than I expected! 

Posted by FrankM on 2008-02-03 21:42:15