CEMC Fermat, 2025

Solution 1:

We begin by placing $U$ and $V$ at the intersection of the gridlines
shown.

(You should confirm for yourself why $U$ and $V$ lie on $PQ$ and $RS$, respectively.)

Each of the line segments $PU$,
$UQ$, $QR$, $RV$, $VS$, $SP$, and $UV$ is a diagonal of a $1\times1$ square, and thus divides the
square into two triangles having equal areas.

Rectangle $PQRS$ contains 8 such
triangles, each with area $\frac12\times1\times1=\frac12$, and so the
area of $PQRS$ is $8\times\frac12=4$.

Solution 2:

We begin by constructing right-angled $\triangle STR$, as shown.

By the Pythagorean Theorem, $SR^2=ST^2+TR^2=2^2+2^2=8$, and so $SR=\sqrt8$ (since $\text{ST gt;0}$).

Similarly, $QR^2=1^2+1^2=2$, and so
$QR=\sqrt2$ (since $\text{QR gt;0}$).

The area of rectangle $PQRS$ is $SR\times QR=\sqrt8\times \sqrt2=\sqrt{16}=4$.