Arrange 8 points in 3-space so that for each of the 56 triples of points that they determine, at least two of the three distances between points in the triple are equal.
This seems to have certain similarities to "Four Points on a Plane" (pid 2532).
As a first approximation, I'd start with a regular octohedron (as the 3-D equivalent of a double-equilateral-triangle rhombus) This plots out 6 of the points.
If we were looking for seven points, the seventh would be the center of the octohedron, but placing the eighth becomes problematic.
I also considered the cube, but there are while there are 24 (1 - 1 - 2) triangles, and 8 (2 - 2 - 2) triangles, the other 24 triangles are (1 - 2 - 3)
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Posted by TomM
on 2005-02-13 17:22:25 |
First thoughts.
