2011 AIME II Problems/Problem 13

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Problem:

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 $\angle O_{1}PO_{2} = 120^{\circ}$, then $AP = \sqrt{a} + \sqrt{b}$, where a and b are positive integers. Find a + b.

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