MathDB
Anna Kareninested internally tangent circles

Source: 2017 AIME II #12

March 23, 2017
AMCAIMEAIME IIAIME I

Problem Statement

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,... and a sequence of points on the circles A1,A2,A3,...A_1,A_2,A_3,... 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 of C0C_0 to BB is mn\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+nm+n.
[asy] size(6cm); real r = 0.8;
pair nthCircCent(int n){ pair ans = (0, 0); for(int i = 1; i <= n; ++i) ans += rotate(90 * i - 90) * (r^(i - 1) - r^i, 0); return ans; }
void dNthCirc(int n){ draw(circle(nthCircCent(n), r^n)); }
dNthCirc(0); dNthCirc(1); dNthCirc(2); dNthCirc(3);
dot("A0A_0", (1, 0), dir(0)); dot("A1A_1", nthCircCent(1) + (0, r), dir(135)); dot("A2A_2", nthCircCent(2) + (-r^2, 0), dir(0)); [/asy]
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