Putnam 2021, Problem B-1

Suppose that the plane is tiled with an infinite checkerboard of unit squares. If another unit square is dropped on the plane at random with position and orientation independent of the checkerboard tiling, what is the probability that it does not cover any of the corners of the squares of the checkerboard?

Solution:
The probability is $2 - \frac{6}{\pi}$.

Set coordinates so that the original tiling includes the (filled) square $S = \{(x,y) : 0 \leq x,y \leq 1\}$. It is then equivalent to choose the second square by first choosing a point uniformly at random in $S$ to be the center of the square, then choosing an angle of rotation uniformly at random from the interval $[0, \pi/2]$.

For each $\theta \in [0, \pi/2]$, circumscribe a square $S_\theta$ around $S$ with angle of rotation $\theta$ relative to $S$; this square has side length $\sin\theta + \cos\theta$. Inside $S_\theta$, draw the smaller square $S_\theta'$ consisting of points at distance greater than $1/2$ from each side of $S_\theta$; this square has side length $\sin\theta + \cos\theta - 1$.

We now verify that a unit square with angle of rotation $\theta$ fails to cover any corners of $S$ if and only if its center lies in the interior of $S_\theta'$. In one direction, if one of the corners of $S$ is covered, then that corner lies on a side of $S_\theta$ which meets the dropped square, so the center of the dropped square is at distance less than $1/2$ from that side of $S_\theta$. To check the converse, note that there are two ways to dissect the square $S_\theta$ into the square $S_\theta'$ plus four $\sin\theta \times \cos\theta$ rectangles. If $\theta \neq 0, \pi/4$, then one of these dissections has the property that each corner $P$ of $S$ appears as an interior point of a side (not a corner) of one of the rectangles $R$. It will suffice to check that if the center of the dropped square is in $R$, then the dropped square covers $P$; this follows from the fact that $\sin\theta$ and $\cos\theta$ are both at most 1.

It follows that the conditional probability, given that the angle of rotation is chosen to be $\theta$, that the dropped square does not cover any corners of $S$ is $(\sin\theta + \cos\theta - 1)^2$. We then compute the original probability as the integral
$$\frac{2}{\pi} \int_0^{\pi/2} (\sin\theta + \cos\theta - 1)^2 d\theta$$
$$= \frac{2}{\pi} \int_0^{\pi/2} (2 + \sin 2\theta - 2\sin\theta - 2\cos\theta) d\theta$$
$$= \frac{2}{\pi} \left(2\theta - \frac{1}{2}\cos 2\theta + 2\cos\theta - 2\sin\theta\right)_0^{\pi/2}$$
$$= \frac{2}{\pi} (\pi + 1 - 2 - 2) = 2 - \frac{6}{\pi}.$$

**Remark:** Noam Elkies has some pictures illustrating this problem: image 1, image 2.