HMMT February, 2025

Solution:

Let $h_{1}$, $h_{2}$, $h_{3}$, and $h_{4}$ be the lengths of the altitudes from $P$ to sides $AB$, $BC$, $CD$, and $DA$, respectively. Then, the problem statement implies that $\{h_{1}, h_{2}, h_{3}, h_{4}\} = \{x, 2x, 3x, 4x\}$ for some $x$. Furthermore, the area of the square is $1 + 2 + 3 + 4 = 10$, so we have
$$h_{1} + h_{3} = \sqrt{10} = h_{2} + h_{4}.$$
Hence either $(\{h_{1}, h_{3}\} = \{x, 4x\}$ and $\{h_{2}, h_{4}\} = \{2x, 3x\}$, or $(\{h_{2}, h_{4}\} = \{x, 4x\}$ and $\{h_{1}, h_{3}\} = \{2x, 3x\}$. In any case, we get $5x = \sqrt{10}$ from above, and finish via the Pythagorean theorem:
$$PA \cdot PB \cdot PC \cdot PD = \sqrt{x^{2} + (2x)^{2}} \cdot \sqrt{(2x)^{2} + (4x)^{2}} \cdot \sqrt{(4x)^{2} + (3x)^{2}} \cdot \sqrt{(3x)^{2} + x^{2}}$$
$$\qquad = 50\sqrt{10}x^{4}$$
$$\qquad = (50\sqrt{10}) \cdot \left(\frac{2}{5}\right)^{2}$$
$$\qquad = \boxed{8\sqrt{10}}.$$