Math Gold Medalist

Lor

2023 AIME II 

Problem 3

Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$

Angle Chasing

Pythagorean Theorem

Sine Law

<APC=90

PC=20

AB=10sqrt(5)

.

   Solution