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Geometry Difficulty 4.9 AIME Prove it United States

Problem:

Point $P$ is inside a square $ABCD$ such that $\angle APB = 135^\circ$ , $PC = 12$ , and $PD = 15$ . Compute the area of this square.

Solution

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

$\begin{aligned} AB^2 & = AP^2 + BP^2 - 2 \cos(135^\circ)(AP)(BP) \\ & = x^2 + y^2 + \sqrt{2}xy \\ & = 123 + 6\sqrt{119} \end{aligned}$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.