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 the regular icosahedron there are only 3 distances that points can be separated from one another: the edge length of the icosahedron; the distance from a vertex, past an edge to the next vertex encountered; and the distance diametrically across the icosahedron.

There are of course 20 triangles of edge length equal to the edge length of the icosahedron: the faces of the icosahedron itself.

There can't be any triangles with the side length equal to the diameter, as each point is involved in only one such distance.

From any one point there are 5 points that are the intermediate distance, and from each of those 5 points there are 2 points that lie the intermediate distance from each of these two points.  That's 10 triangles associated with each of the 12 vertices, but each triangle is associated with 3 vertices, so there are 40 triangles altogether of this size.

So there are 20 + 40 equilateral triangles that can be formed from the vertices of the regular icosahedron.

Posted by Charlie on 2005-08-26 18:13:54