# Difference between revisions of "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

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), 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);
D(cir1); D(cir2); D(cir3); D((14,0)--(-14,0)); D(A--B); D((-14,0)--D(MP("H",H,W))--A,linewidth(0.7) + linetype("4 4"));
D(MP("O_1",C1)); D(MP("O_2",C2)); D(MP("O_3",C3)); D(MP("T",T,N)); D(MP("A",A,NW)); D(MP("B",B,NE)); D(C1--MP("T_1",T1,N)); D(C2--MP("T_2",T2,N)); D(C3--T); D(rightanglemark(C_3,T,H));
(Error compiling LaTeX. D(MP("O_1",C1)); D(MP("O_2",C2)); D(MP("O_3",C3)); D(MP("T",T,N)); D(MP("A",A,NW)); D(MP("B",B,NE)); D(C1--MP("T_1",T1,N)); D(C2--MP("T_2",T2,N)); D(C3--T); D(rightanglemark(C_3,T,H));
^
b8976dddad76c795c2e7a7b702318c06de88d0f7.asy: 8.175: no matching variable 'C_3')

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 $O_3T = \frac{58}{7}$. 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 $m+n+p=\boxed{405}$.