Question 62
Point $P$ lies inside equilateral triangle $ABC$ so that $\angle BPC = 120^{\circ}$ and $AP \sqrt{2} = BP + CP$. $\frac{AP}{AB}$ can be written as $\frac{a \sqrt{b}}{c}$, where $a, b, c$ are integers, $c$ is positive, $b$ is square-free, and $\operatorname{gcd}(a, c) = 1$. Find $100a + 10b + c$.