What is the shortest distance between two opposite points (two vertices separated by 2R) in the surface of a regular icosahedron? The regular icosahedron's edges are 1ft in length.
(In reply to
Solution by Brian Wainscott)
OK, just wanted to get that in before someone else posted a solution. I'm pretty sure it it correct. Here is my method:
Since the whole thing is made up of equalateral triangles, you can take a portion of it and flatten it out in the plane. I can't draw a picture here, but I'll try to describe it.
Place one of the faces of the icosahedron on the plane, with a point at (0,0) the other corners at (+/- 0.5, sqrt(3)/2). Then unfold the icosahedron from there, so you get a triangular tiling of the plane. If the (0,0) point was the bottom of the icosahedron, the point at (0.5, 3*sqrt(3)/2) corresponds to the top point. The distance between them in the plane will be the same as the distance on the surface. This distance is sqrt(1/4+9*3/4)=sqrt(7).
re: Solution (details)
