Problem:
Let AB be a chord of a circle ω, and let P be a point on the chord AB. Circle ω1​ passes through A and P and is internally tangent to ω. Circle ω2​ passes through B and P and is internally tangent to ω. Circles ω1​ and ω2​ intersect at points P and Q. Line PQ intersects ω at X and Y. Assume that AP=5,PB=3,XY=11, and PQ2=nm​, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let O,O1​, and O2​ denote the centers of ω,ω1​, and ω2​, respectively. Points O1​ and O2​ lie on AO and BO, respectively, as shown in the figure below. It is clear that △AOB,△AO1​P, and △BO2​P are isosceles and similar to each other, and PO2​​∥AO and PO1​​∥BO, and therefore PO1​OO2​ is a parallelogram. In particular, O and P lie on opposite sides of line O1​O2​. Also note that P and Q lie on opposite sides of line O1​O2​.
Because PO1​OO2​ is a parallelogram, OO2​=O1​P=O1​Q and OO1​= O2​P=O2​Q. It follows from the last two equations that △OO1​O2​ is congruent to △QO2​O1​ by SSS. Then O1​OQO2​ is a trapezoid with OQ​∥O1​O2​​. Because PQ​ is the common chord of ω1​ and ω2​,O1​O2​​⊥PQ​. Thus OQ​⊥PQ​, and therefore Q is the midpoint of XY and QX=QY=211​. By the Power of a Point Theorem,
15=AP⋅PB=PX⋅PY=(QX−PQ)(PQ+QY)=4121​−PQ2
so PQ2=4121​−15=461​. The requested sum is 61+4=65.
The problems on this page are the property of the MAA's American Mathematics Competitions