# Mock Geometry AIME 2011 Problems/Problem 8

## Problem

Two circles $\omega_1,\omega_2$ have center $O_1,O_2$ and radius $25,39$ respectively. The smallest distance between a point on $\omega_1$ with a point on $\omega_2$ is $1$. Tangents from $O_2$ to $\omega_1$ meet $\omega_1$ at $S_1,T_1,$ and tangents from $O_1$ to $\omega_2$ meet $\omega_2$ at $S_2,T_2,$ such that $S_1,S_2$ are on the same side of line $O_1O_2.$ $O_1S_1$ meets $O_2S_2$ at $P$ and $O_1T_1$ meets $O_2T_2$ at Q. The length of $PQ$ can be expressed in the form $\frac{m} {n},$ where $m,n$ are relatively prime positive integers. Find $m+n.$

## Solution

First note that the diagram is symmetric about line $O_1O_2$; that is, when every point is reflected by $O_1O_2$, the diagram remains unchanged. This leads to two important observations: $\Delta PO_1O_2 \cong \Delta QO_1O_2$, and that $PQ \perp O_1O_2$. Then the distance from $P$ to $O_1O_2$ is half the length of $PQ$.

Second, note that the smallest distance between two points on two non-intersecting circles lies on the line connecting their centers. This can be demonstrated by letting $C$ be on $\omega_1$ and $D$ be on $\omega_2$, such that $CD$ is the smallest distance connecting both circles. Then $O_1C+CD+DO_2 \geq O_1O_2$; they are both paths from $O_1$ to $O_2$ and the smallest distance between two points is a straight line. This rearranges to $CD \geq O_1O_2-O_1C-DO_2$ with equality iff $C,D$ is on line $O_1O_2$. Hence $CD$ coincides with $O_1O_2$.

Then $O_1O_2=25+1+39=65$. By the Pythagorean Theorem on $\Delta S_1O_1O_2$, $65^2=(S_1O_2)^2+25^2$, and so $S_1O_2=60$. Also, from $\Delta S_2O_2O_1$, $65^2=(S_2O_1)^2+39^2$, and so $S_2O_1=52$.

Now let $H$ be the foot of the perpendicular from $P$ to $O_1O_2$. Let $h=PH, a=O_1H, b=O_2H$. Then $\tan{\angle PO_1O_2}=\frac{h}{a}=\frac{60}{25}$, where we have used right triangles $\Delta PO_1H, \Delta O_1S_1O_2$. Similarly, $\tan{\angle PO_2O_1}=\frac{h}{b}=\frac{52}{39}$ using right triangles $\Delta PO_2H, \Delta O_2S_2O_1$.

These equations rearrange into $h=\frac{12}{5}a=\frac{4}{3}b$. We also know $a+b=O_1O_2=65$. Solving this system of equations for $a$ yields $a=\frac{325}{14}$. Then $h=(\frac{12}{5})(\frac{325}{14})=\frac{390}{7}$. Then $2h=\frac{780}{7}=PQ$ and $m+n=780+7=\boxed{787}$.

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