Three points have been chosen randomly from the vertices of a cube. What is the probability that they form (a) an acute triangle; (b) a right triangle?

I come up with a 1/7 chance of an equilateral triangle, and a 6/7 chance of a right triangle.

There are (8 x 7 x 6) /(3 x 2 X 1) = 56 distinct sets of points that can be selected.

If any two of the points are on the same edge, then the resulting triangle is a right triangle.

The only sets which do not have a common edge are the 8 sets where each member of the set shares a common edge with one corner of the cube, and these all form equilateral triangles.

8/56 = 1/7 chance of an equilateral triangle.
6/7 chance of a right triangle.

Posted by Steve Herman on 2004-12-04 17:25:13