## Archive for January 17, 2011

### Fair and Square Solution

A few days ago, I posted the following problem:

Three points are randomly chosen along the perimeter of a square. What is the probability that the center of the square will be contained within the triangle formed by these three points?

In that post, I mentioned that I had an elegant solution. If you want to think about the problem a little before seeing my solution, check out the Three Points page at the MJ4MF website. I created a really cool Excel file that allows you to explore 100 cases at a time. I think it’s worth a look.

But if you’re not interested in any of that, here’s my solution…

Without loss of generality, choose any two points along the perimeter. Call them A and B. Then, construct segments AD and BC that pass through point P, the center of the square. This gives four cases:

• If A and B are the two chosen points, then the third vertex must be chosen along segment CD to form a triangle that contains point P.
• If B and D are the two chosen points, then the third vertex must be chosen along segment CA.
• If D and C are the two chosen points, then the third vertex must be chosen along segment AB.
• If C and A are the two chosen points, then the third vertex must be chosen along segment BD.

What we’ve now done is selected two sets of points in four different ways, and the portions along the segments where the third point could be chosen collectively comprise the entire perimeter. Consequently, the probability that a point along the perimeter is 1, but this is divided among 4 cases, so the probability that a randomly selected triangle will contain the center is 1/4.

As a footnote, points A and B were chosen arbitrarily, so there is no loss of generality.