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Two Circles Touching Internally and Externally

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Question

Circles $C_{1}$ and $C_{2}$ are externally tangent and they are both internally tangent to the circle $C_{3} .$ The radii of $C_{1}$ and $C_{2}$ are $4$ and $10,$ respectively and the centres of the three circles are collinear. A chord of $C_{3}$ is also a common internal 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 the value of $(m + n + p) .$

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Solution

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 $m + n + p = 405$

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