Three points have been chosen randomly from the vertices of a tesseract (4-cube).

What is the probability that they form (a) an acute triangle; (b) a right triangle?

(In reply to computer solution by Charlie)

Of course, the first simplification could have been to choose one point always to include in the triangle without loss of generality. That would bring the number of cases down to C(15,2)=105, but still a lot to consider manually. The cases could probably be categorized into groups of cases to get an analytic solution, but that involves some 4-dimensional thinking.

Posted by Charlie on 2010-12-21 13:51:41