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)

1. That's a concise piece of programming, I like it.  I realised that I needed to come to terms with a dihedral but all I had was for three faces at a vertex.  And looking at the time between my post and yours, about 1.25 hrs!!  How long to write the program??

2.  Why I ruled out equilateral triangles I don't know; I mean, if two basal point are 'x' units distant then the third can also be!

3.  By virtue of my statement re "iscoceles" triangles there is nothing in the problem text that excludes the idea, we are just told that "The shortest distance between any two points is greater than 1.2". 
Charlie's tabulation validates that comment, but barely.

Posted by brianjn on 2008-02-02 03:46:26