Using only the vertices of a regular icosahedron as the corners, how many equilateral triangles can you make?

What if you could only use the vertices of a regular dodecahedron?

(In reply to re(3): part 2 -- spoiler by Tristan)

OK.  Looking back at part 1--the icosahedron: it looks as if in the counting of the larger triangles, I actually counted each triangle twice, as the "2 points" from which the second point can connect are one CW and one CCW, but the same set is counted by reversing the order of choosing the points, that is, when the third point chosen becomes the second point, the CCW becomes the CW, to make the original second point be the third point of the twice-counted triangle.  So the large triangles are really only 5 * 12 / 3 = 20 rather than 40, making the total triangles of any size 20 + 20 = 40.  I hope it's correct this time.

Posted by Charlie on 2005-08-28 02:16:37