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
re: Solution (details) by Brian Wainscott)
The way I figure your interpretation (measure along the surface of the icosahedron), I would get a strip of triangles laid out up/down/up/down. Ten of them make up the strip (5 up and 5 down, to match up with the top and bottom caps of 5 triangles each). If a point is at (0,0), and another point on this triangle is at (.5,sqrt(3)/2), then I'd go out horizontally 2 more units (along the tops of two downward pointing triangles) to arrive at (2.5,sqrt(3)/2) giving √(2.5²+(√(3)/2)²) = √7. The same answer, but I understand it better this way.

Posted by Charlie
on 20031210 16:10:10 