(In reply to
10 balls by rixar)
Given that the centers of the corner balls are really the vertices of a regular tetrahedron, which would have to be of edge length 2d, I completely agree with your answer. My calculator gives 2.632993162d as the height, which is very plausibly close to, but less than, 3d.
It isn't clear at all to me, though, how to show that the centers of the corner balls will be the vertices of a regular tetrahedron. Some will no doubt say that this was a "given" of the problem and doesn't need to be proved. Perhaps it is "obvious," but if so, why it is obvious needs to be pointed out, I think.
Edited on August 3, 2004, 9:44 pm
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Posted by Richard
on 2004-08-03 21:42:11 |
re: 10 balls
