[Circles](https://artofproblemsolving.com/wiki/index.php/Circle) $C_1$ and $C_2$ are externally [tangent](https://artofproblemsolving.com/wiki/index.php/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](https://artofproblemsolving.com/wiki/index.php/Center) of the three circles are all [collinear](https://artofproblemsolving.com/wiki/index.php/Collinear). A [chord](https://artofproblemsolving.com/wiki/index.php/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](https://artofproblemsolving.com/wiki/index.php/Relatively_prime), and $n$ is not divisible by the square of any [prime](https://artofproblemsolving.com/wiki/index.php/Prime), find $m+n+p.$
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](https://artofproblemsolving.com/wiki/index.php/Right_triangle) $\triangle HO_1T_1 \sim \triangle HO_2T_2 \sim \triangle HO_3T$,
\[\frac{HO_1}{4} = \frac{HO_1+14}{10} = \frac{HO_1+10}{O_3T}.\]
It follows that $HO_1 = \frac{28}{3}$, and that $O_3T = \frac{58}{7}\dagger$. By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/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 $O_3T=4+\frac{30}{7}=\frac{58}{7}$.