# Mock Geometry AIME 2011 Problems/Problem 11

## Problem

$C$ is on a semicircle with diameter $AB$ and center $O.$ Circle radius $r_1$ is tangent to $OA,OC,$ and arc $AC,$ and circle radius $r_2$ is tangent to $OB,OC,$ and arc $BC$. It is known that $\tan AOC=\frac{24}{7}$. The ratio $\frac{r_2} {r_1}$ can be expressed $\frac{m} {n},$ where $m,n$ are relatively prime positive integers. Find $m+n.$

## Solution

$[asy] unitsize(4mm); draw(Arc((0,0),12,0,180)); draw((12,0)--(-12,0)); draw(circle((-6,9/2),9/2)); draw(circle((4,16/3),16/3)); draw((0,0)--(-84/25,288/25)); draw((0,0)--(-48/5,36/5)); draw((0,0)--(36/5,48/5)); draw((-6,9/2)--(-6,0)); draw((4,16/3)--(4,0)); label("C",(-84/25,288/25),NNW); label("A",(-12,0),WSW); label("B",(12,0),ESE); label("O",(0,0),S); label("D",(-6,9/2),NNE); label("E",(4,16/3),NNW); label("F",(-6,0),S); label("G",(4,0),S); [/asy]$

Let the circle with radius $r_1$ have center $D$ and the circle with radius $r_2$ have center $E$. Let the projections of $D$ and $E$ onto $AB$ be $F$ and $G$, respectively.

$D$ is equidistant from $OA$ and $OC$, so it is on the angle bisector of $\angle AOC$.

We're given that $\tan AOC=\frac{24}{7}$, so $\cos AOC=\frac{7}{25}$. Now, we have $\tan FOD=\tan\frac{AOC}{2}=\sqrt{\frac{1-\cos AOC}{1+\cos AOC}}=\sqrt{\frac{1-\frac{7}{25}}{1+\frac{7}{25}}}=\frac{3}{4}$, which is positive because $\angle FOD<\frac{\pi}{2}$.

We therefore also have $\sin FOD=\frac{3}{5}$. Now, $DO=\frac{FD}{\sin FOD}=\frac{r_1}{\frac{3}{5}}=\frac{5r_1}{3}$, and we see that the radius of the large semicircle is $\frac{5r_1}{3}+r_1=\frac{8r_1}{3}$.

Similarly, $OE$ is the angle bisector of $\angle BOC$. Now, $\cos BOC=\cos(\pi-AOC)=-\cos AOC=-\frac{7}{25}$, and so $\tan BOE=\tan\frac{BOC}{2}=\sqrt{\frac{1-\cos BOC}{1+\cos BOC}}=\sqrt{\frac{1+\frac{7}{25}}{1-\frac{7}{25}}}=\frac{4}{3}$, again positive because $\angle BOE<\frac{\pi}{2}$, and $\sin BOE=\frac{4}{5}$.

Now, $OE=\frac{EG}{\sin BOE}=\frac{r_2}{\frac{4}{5}}=\frac{5r_2}{4}$. The radius of the large semicircle is thus $\frac{5r_2}{4}+r_2=\frac{9r_2}{4}$.

Since the radii of the large semicircle are equal, we have $\frac{8r_1}{3}=\frac{9r_2}{4}$, and so $\frac{r_2}{r_1}=\frac{32}{27}$, and $m+n=\boxed{059}$.