Some Off Square

February 2024 : Solution

This month’s puzzle asked solvers to determine the likelihood that a circle with a randomly determined diameter — its endpoints chosen uniformly from within a square — would have some part of it protruding off the square.

One approach, sketched out in part above, is to instead consider the situation where the circle’s center and a point on the perimeter are chosen at random. One can show that there is a π/24 chance that the circle thus determined will be contained within the square.

Now, imagine choosing these circles by first selecting the point on the perimeter and then selecting the center. By doubling the vector from the perimeter point to the center, we get a vector which comprises the circle’s diameter. (That is, its new endpoint is the point one would need to select under the original problem statement in order to obtain the same circle.) However, with probability 3/4, that point is located outside the square! So the space of circles chosen by diameter is a uniform one-fourth of the circles chosen by radius. Crucially, all the circles that are entirely within the square can be selected by every diameter (by definition of them being entirely within the square), so the uniformity of the subspace is known for circles entirely contained within the square.

This means we have a π/6 chance of the circle being entirely within the square, so our answer is the complement1 of this value, namely 1−π/6.

Congrats to this month’s solvers!

  1. Which is not, as we got a handful of times, “5π/6” :-)