Problem:
Point P lies on the diagonal AC of square ABCD with AP>CP. Let O1​ and O2​ be the circumcenters of triangles ABP and CDP, respectively. Given that AB=12 and ∠O1​PO2​=120∘, then AP=a​+b​, where a and b are positive integers. Find a+b.
Solution:
Let O3​ be the circumcenter of triangle ADP. By symmetry, ∠O3​PA=∠O1​PA. Because AP>CP, triangle APD is acute and triangle CPD is obtuse with ∠CPD>90∘. It follows that ∠PO3​D=2∠PAD=90∘ and ∠PO2​D=2∠PCD=90∘. Note that O2​P=O2​D and O3​P=O3​D. Thus D2​PO3​ is a square. In particular, ∠O2​PD=∠DPO3​. Because ∠O3​PA=∠O1​PA and ∠O2​PD=∠DPO3​, it follows that ∠O1​PO2​=2∠DPA, implying that ∠DPA=60∘. Thus triangle ADP has ∠A=45∘,∠APD=60∘, and AD=12. Let E be the foot of the perpendicular from D to AP. Then △ADE is an isosceles right triangle with DE=EA=62​=72​ and PDE is a 30∘−60∘−90∘ triangle with PE=DE/3​=26​=24​, from which it follows that AP=72​+24​ and a+b=96​.