A paper equilateral triangle $ABC$ has side length $12$. The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$. The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$.
Let $M$ and $N$ be the points on $\overline{AB}$ and $\overline{AC}$, respectively, where the paper is folded. Let $D$ be the point on $\overline{BC}$ where the folded $A$ touches it.

We have $AF=6\sqrt{3}$ and $FD=3$, so $AD=3\sqrt{13}$. Denote $\angle DAF = \theta$; we get $\cos\theta = 2\sqrt{3}/\sqrt{13}$. 
In triangle $AXY$, $AY=\tfrac 12 AD = \tfrac 32 \sqrt{13}$, and $AX=AY\sec\theta =\tfrac{13}{4}\sqrt{3}$. 
In triangle $AMX$, we get $\angle AMX=60^\circ-\theta$ and then use sine-law to get $MX=\tfrac 12 AX\csc(60^\circ-\theta)$; similarly, from triangle $ANX$ we get $NX=\tfrac 12 AX\csc(60^\circ+\theta)$. Thus \[MN=\tfrac 12 AX(\csc(60^\circ-\theta) +\csc(60^\circ+\theta)).\] Since $\sin(60^\circ\pm \theta) = \tfrac 12 (\sqrt{3}\cos\theta \pm \sin\theta)$, we get
\begin{align*}     \csc(60^\circ-\theta) +\csc(60^\circ+\theta) &= \frac{\sqrt{3}\cos\theta}{\cos^2\theta - \tfrac 14} = \frac{24 \cdot \sqrt{13}}{35} \end{align*}
Then \[MN = \frac 12 AX \cdot \frac{24 \cdot \sqrt{13}}{35} = \frac{39\sqrt{39}}{35}\]
The answer is $39 + 39 + 35 = \boxed{113}$.