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AIME 2017b Problem 12

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Circle C0C_0 has radius 11, and the point A0A_0 is a point on the circle. Circle C1C_1 has radius r<1r<1 and is internally tangent to C0C_0 at point A0A_0. Point A1A_1 lies on circle C1C_1 so that A1A_1 is located 90∘90^{\circ} counterclockwise from A0A_0 on C1C_1. Circle C2C_2 has radius r2r^2 and is internally tangent to C1C_1 at point A1A_1. In this way a sequence of circles C1,C2,C3,…C_1,C_2,C_3,\ldots and a sequence of points on the circles A1,A2,A3,…A_1,A_2,A_3,\ldots are constructed, where circle CnC_n has radius rnr^n and is internally tangent to circle Cn−1C_{n-1} at point An−1A_{n-1}, and point AnA_n lies on CnC_n 90∘90^{\circ} counterclockwise from point An−1A_{n-1}, as shown in the figure below. There is one point BB inside all of these circles. When r=1160r = \frac{11}{60}, the distance from the center C0C_0 to BB is mn\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+nm+n. [asy] draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label("A0A_0",(125,0),E); dot((25,100)); label("A1A_1",(25,100),SE); dot((-55,20)); label("A2A_2",(-55,20),E); [/asy]

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