## 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.