I started by thinking about the total number of triangles: 8 choose 3, or 56. That seemed too hard.
So I focused on what types of triangles were possible:
a) all 3 vertices on the same face, ie sides formed of 2 edges and a face diagonal, ie side lengths of 1 1 and sqrt(2) right triangle.
b) 2 of the vertices are opposite corners of the cube, ie sides formed of an edge, a face diagonal, and a "thru the air" diagonal, ie side lengths of 1 sqrt(2) and sqrt(3), again right triangle
c) all 3 edges are face diagonals, ie sides are all sqrt(2) in length; so type "c" triangles are equilateral, ie acute.
OK, so statistically, say one vertex is already chosen. To get an equilateral triangle, the other 2 vertices have to be along a face diagonal from my first point. There are 3 such face diagonals going away from my first point. So there are 3 vertices that are acceptable. I have a total of 7 vertices to choose from. To get an equilateral, the probablity of success (if I were picking at random) would be: 3/7 * 2/6 = 1/7.
Posted by Larry
on 2004-12-06 03:23:49