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?

First point is chosen at random – does not affect the triangle yet.

For the second point: We have 7 corners to choose from (since the first point takes a corner). There are 4 corners that automatically force the triangle to be right, no matter where the third point ends up. Those are the 3 corners that are adjacent to the first point, PLUS the corner that is on the opposite side of the cube. So we have 4/7 chance of definitely having a right triangle, and 3/7 chance of possibly having an acute triangle just based on the first two points.

Well, let’s say we chose the second point to be one of the 3 that gave us a possibility of having an acute triangle (these are the 3 corners that are diagonally across a face from the first point). What are the chances that the third point will make the triangle acute/right?

Well, there are 6 corners left to choose from (since the first and second point took two corners). 4 of them will result in the triangle being right (the same 4 vertices mentioned above when discussing the second point), and 2 of them will result in the triangle definitely being acute.

By symmetry, this applies to each of the three vertices that gave us the possibility of having an acute triangle.

So the total chances for the two types of triangles are:

(a) = 3/7 * 2/6 = 1/7
(b) = 4/7 + 3/7 * 4/6 = 4/7 + 2/7 = 6/7

Fyew! They add up to 1.

Posted by nikki on 2004-12-06 14:16:01