# 2005 AIME II Problems/Problem 8

## Problem

Circles $C_1$ and $C_2$ are externally tangent, and they are both internally tangent to circle $C_3.$ The radii of $C_1$ and $C_2$ are 4 and 10, respectively, and the centers of the three circles are all collinear. A chord of $C_3$ is also a common external tangent of $C_1$ and $C_2.$ Given that the length of the chord is $\frac{m\sqrt{n}}p$ where $m,n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m+n+p.$

## Solution $[asy] pointpen = black; pathpen = black + linewidth(0.7); size(200); pair C1 = (-10,0), C2 = (4,0), C3 = (0,0), H = (-10-28/3,0), T = 58/7*expi(pi-acos(3/7)); path cir1 = CR(C1,4.01), cir2 = CR(C2,10), cir3 = CR(C3,14), t = H--T+2*(T-H); pair A = OP(cir3, t), B = IP(cir3, t), T1 = IP(cir1, t), T2 = IP(cir2, t); draw(cir1); draw(cir2); draw(cir3); draw((14,0)--(-14,0)); draw(A--B); MP("H",H,W); draw((-14,0)--H--A, linewidth(0.7) + linetype("4 4")); draw(MP("O_1",C1)); draw(MP("O_2",C2)); draw(MP("O_3",C3)); draw(MP("T",T,N)); draw(MP("A",A,NW)); draw(MP("B",B,NE)); draw(C1--MP("T_1",T1,N)); draw(C2--MP("T_2",T2,N)); draw(C3--T); draw(rightanglemark(C3,T,H)); [/asy]$

Let $O_1, O_2, O_3$ be the centers and $r_1 = 4, r_2 = 10,r_3 = 14$ the radii of the circles $C_1, C_2, C_3$. Let $T_1, T_2$ be the points of tangency from the common external tangent of $C_1, C_2$, respectively, and let the extension of $\overline{T_1T_2}$ intersect the extension of $\overline{O_1O_2}$ at a point $H$. Let the endpoints of the chord/tangent be $A,B$, and the foot of the perpendicular from $O_3$ to $\overline{AB}$ be $T$. From the similar right triangles $\triangle HO_1T_1 \sim \triangle HO_2T_2 \sim \triangle HO_3T$, $$\frac{HO_1}{4} = \frac{HO_1+14}{10} = \frac{HO_1+10}{O_3T}.$$

It follows that $HO_1 = \frac{28}{3}$, and that $O_3T = \frac{58}{7}\dagger$. By the Pythagorean Theorem on $\triangle ATO_3$, we find that $$AB = 2AT = 2\left(\sqrt{r_3^2 - O_3T^2}\right) = 2\sqrt{14^2 - \frac{58^2}{7^2}} = \frac{8\sqrt{390}}{7}$$

and the answer is $m+n+p=\boxed{405}$. $\dagger$ Alternatively, drop an altitude from $O_1$ to $O_3T$ at $O_3'$, and to $O_2T_2$ at $O_2'$. Then, $O_2O_2'=10-4=6$, and $O_1O_2=14$. But $O_1O_3O_3'$ is similar to $O_1O_2O_2'$ so $O_3O_3'=\frac{6}{14} \cdot 10=\frac{30}{7}$. From rectangles, $O_3'T=O_1T_1=4$ so $O_3T=4+\frac{30}{7}=\frac{58}{7}$.

## Solution 2

Call our desired length $x$. Note for any $X$ on $\overline{AB}$ and $Y$ on $\overline{O_1O_2}$ such that $\overline{XY}\perp\overline{AB}$ that the function $f$ such that $f(\overline{O_1Y})=\overline{XY}$ is linear. Since $(0,4)$ and $(14,10)$, we can quickly interpolate that $f(10)=\overline{O_3T}=\frac{58}{7}$. Then, extend $\overline{O_3T}$ until it reaches the circle on both sides; call them $P,Q$. By Power of a Point, $\overline{PT}\cdot\overline{TQ}=\overline{AT}\cdot\overline{TB}$. Since $\overline{AT}=\overline{TB}=\frac{1}{2}x$, $$(\overline{PO_3}-\overline{O_3T})(\overline{QO_3}+\overline{O_3T})=\frac{1}{4}x^2$$ $$\left(14+\frac{58}{7}\right)\left(14-\frac{58}{7}\right)=\frac{1}{4}x^2$$ After solving for $x$, we get $x=\frac{8\sqrt{390}}{7}$, so our answer is $8+390+7=\boxed{405}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 