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 ∠O1PO2=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, ∠O3PA=∠O1PA. Because AP>CP, triangle APD is acute and triangle CPD is obtuse with ∠CPD>90∘. It follows that ∠PO3D=2∠PAD=90∘ and ∠PO2D=2∠PCD=90∘. Note that O2P=O2D and O3P=O3D. Thus D2PO3 is a square. In particular, ∠O2PD=∠DPO3. Because ∠O3PA=∠O1PA and ∠O2PD=∠DPO3, it follows that ∠O1PO2=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.