HMMT February Guts Round, 2024

Solution:

Let $x = AP$ and $y = BP$. Rotate $\triangle BAP$ by $90^\circ$ around $B$ to get $\triangle BCQ$. Then, $\triangle BPQ$ is right isosceles, and from $\angle BQC = 135^\circ$, we get $\angle PQC = 90^\circ$. Therefore, by Pythagorean's theorem, $PC^2 = x^2 + 2y^2$. Similarly, $PD^2 = y^2 + 2x^2$.

Thus, $y^2 = \frac{2PC^2 - PD^2}{3} = 21$, and similarly $x^2 = 102 \Longrightarrow xy = 3\sqrt{238}$.

Thus, by the Law of Cosines, the area of the square is
$$$$
$$\begin{aligned} AB^2 & = AP^2 + BP^2 - 2 \cos(135^\circ)(AP)(BP) \\ & = x^2 + y^2 + \sqrt{2}xy \\ & = 123 + 6\sqrt{119} \end{aligned}$$