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)

I don't think the intent was finding the distance along the surface. I interpreted the "...in the surface" phrase to refer just to the "opposite points in the surface", and the distance as being the distance in space (the 2R, representing twice the radius of the circumscribing sphere).

As such, I got:

Project the icosahedron outward to its circumscribing sphere, of radius R. Consider one spherical triangle face. The great circle arc of one side of this spherical triangle subtends a chord which is one edge of the icosahedron.

Each corner angle of this spherical triangle is 72 degrees, as five fit together to completely surround each vertex of the icosahedron.

Then we use the law of cosines for spherical triangle, for angles:

cos 72 = -cos²72 + sin²72 cos S,
where S is the side length sought.

This solves to S=63.424948... degrees.

The chord for the arc has length 2Rsin(S/2), or R*1.051462224238267... Since we are given that the length of the chord (the edge of the icosahedron) is 1 foot, 1 = R * 1.051462224238267, so R=1/1.051462224238267 and 2R=1.902113032590307... feet.

As SilverKnight has said, it's quite similar to the soccer ball problem.

Posted by Charlie on 2003-12-10 15:57:25