2011 AIME II Problem 13

Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP>CP$. Let $O_1$ and $O_2$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB=12$ and $\angle O_1 P O_2 = 120^\circ$, then $AP=\sqrt{a}+\sqrt{b}$ where $a$ and $b$ are positive integers. Find $a+b$.

Leading zeroes must be inputted, so if your answer is 34, then input 034. Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.