Question

Circle $C_0$ has radius $1,$ and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r \lt 1$ and is internally tangent to $C_0$ at point $A_0.$ Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^\circ$ counterclockwise from $A_0$ on $C_1.$ Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1.$ In this way a sequence of circles $C_1, C_2, C_3, \ldots$ and a sequence of points on the circles $A_1, A_2, A_3, \ldots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1},$ and point $A_n$ lies on $C_n$ $90^\circ$ counterclockwise from point $A_{n-1},$ as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \frac{11}{60},$ the distance from the center of $C_0$ to $B$ is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Solution:

Work in the complex plane with $C_0$ centered at $O_0 = 0$ and $A_0 = 1,$ and let $O_n$ be the center of $C_n.$ Inductively, $A_n = O_n + r^n i^n:$ this holds for $n = 0,$ and since $C_{n+1}$ is internally tangent to $C_n$ at $A_n,$ its center is $O_{n+1} = A_n - r^{n+1} i^n;$ then $A_n$ sits in direction $i^n$ from $O_{n+1},$ so rotating $90^\circ$ counterclockwise gives $A_{n+1} = O_{n+1} + r^{n+1} i^{n+1}.$

Therefore $O_{n+1} - O_n = (r^n - r^{n+1})\, i^n = (1 - r)(ir)^n.$ The circles are nested, and their radii shrink to $0,$ so the common point $B$ is the limit of the centers: $B = (1 - r)\sum_{n=0}^{\infty} (ir)^n = \frac{1 - r}{1 - ir},$ at distance $\frac{1 - r}{|1 - ir|} = \frac{1 - r}{\sqrt{1 + r^2}}$ from the origin.

For $r = \frac{11}{60}$ this equals $\frac{49/60}{\sqrt{3721}/60} = \frac{49}{61},$ so $m + n = 49 + 61 = 110.$

2017 AIME II Problem 12

Solution:

Problem 12 in Other Years