Three points are chosen at random inside a square. Each point is chosen by choosing a random x-coordinate and a random y-coordinate.

A triangle is drawn with the three random points as the vertices. What is the probability that the center of the square is inside the triangle?

(In reply to Almost all the way by FrankM)

We can build on the ideas in my last comment without needing to look at multiple cases or worry about double integrals.

Consider the modified problem with a circle instead of a square. Here it is dead easy to compute the probability that the centre will lie inside the triangle. Following my earlier approach, the probability is the average angle between two radial segments divided by 2 pi. I.e., the probability is 1/4.

Now return to the case of the triangle embedded in a square. As before, let A and B be two vertices of the triangle. If we restrict A and B to lie in the circle inscribed to the square, then the probability we are looking for has been shown to be 1/4. But what if B = (x2,y2) lies in the portion of the square outside of the incircle? In this case we we match off this case with the case with B' =(-x2, -y2). Now BZB' is a straight line, so that AZB + AZB' = 180 degrees. It follows that, taken together, these two cases also contribute a probability of 1/4. Ergo, the probability that that the center of the square is inside the triangle is 1/4.

Posted by FrankM on 2008-03-17 19:56:21