Problem: 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 $m+n+p$
[asy] import cse5; size(12cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", (9,0), dir(-80)); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle, tpen); draw(M--A--N--cycle); fill(M--A--N--cycle, mediumgrey); pair shift = (-20.13, 0); pair B1 = MP("B", B+shift, dir(200)); pair A1 = MP("A", K+shift, dir(90)); pair C1 = MP("C", C+shift, dir(-20)); draw(A1--B1--C1--cycle, tpen);[/asy]

Solution: We let the original position of $A$ be $A$ , and the position of $A$ after folding be $D$ . Also, we put the triangle on the coordinate plane such that $A=(0,0)$ $B=(-6,-6\sqrt3)$ $C=(6,-6\sqrt3)$ , and $D=(3,-6\sqrt3)$
[asy] size(10cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = (9,0); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle); draw(M--A--N--cycle); label("$D$", A, S); pair X = (6,6*sqrt(3)); draw(B--X--C); label("$A$",X,dir(90)); draw(A--X); [/asy]
Note that since $A$ is reflected over the fold line to $D$ , the fold line is the perpendicular bisector of $AD$ . We know $A=(0,0)$ and $D=(3,-6\sqrt3)$ . The midpoint of $AD$ (which is a point on the fold line) is $(\tfrac32, -3\sqrt3)$ . Also, the slope of $AD$ is $\frac{-6\sqrt3}{3}=-2\sqrt3$ , so the slope of the fold line (which is perpendicular), is the negative of the reciprocal of the slope of $AD$ , or $\frac{1}{2\sqrt3}=\frac{\sqrt3}{6}$ . Then, using point slope form, the equation of the fold line is \[y+3\sqrt3=\frac{\sqrt3}{6}\left(x-\frac32\right)\] \[y=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}\] Note that the equations of lines $AB$ and $AC$ are $y=\sqrt3x$ and $y=-\sqrt3x$ , respectively. We will first find the intersection of $AB$ and the fold line by substituting for $y$ \[\sqrt3 x=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}\] \[\frac{5\sqrt3}{6}x=-\frac{13\sqrt3}{4} \implies x=-\frac{39}{10}\] Therefore, the point of intersection is $\left(-\tfrac{39}{10},-\tfrac{39\sqrt3}{10}\right)$ . Now, lets find the intersection with $AC$ . Substituting for $y$ yields \[-\sqrt3 x=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}\] \[\frac{-7\sqrt3}{6}x=-\frac{13\sqrt3}{4} \implies x=\frac{39}{14}\] Therefore, the point of intersection is $\left(\tfrac{39}{14},-\tfrac{39\sqrt3}{14}\right)$ . Now, we just need to use the distance formula to find the distance between $\left(-\tfrac{39}{10},-\tfrac{39\sqrt3}{10}\right)$ and $\left(\tfrac{39}{14},-\tfrac{39\sqrt3}{14}\right)$ \[\sqrt{\left(\frac{39}{14}+\frac{39}{10}\right)^2+\left(-\frac{39\sqrt3}{14}+\frac{39\sqrt3}{10}\right)^2}\] The number 39 is in all of the terms, so let's factor it out: \[39\sqrt{\left(\frac{1}{14}+\frac{1}{10}\right)^2+\left(-\frac{\sqrt3}{14}+\frac{\sqrt3}{10}\right)^2}=39\sqrt{\left(\frac{6}{35}\right)^2+\left(\frac{\sqrt3}{35}\right)^2}\] \[\frac{39}{35}\sqrt{6^2+\sqrt3^2}=\frac{39\sqrt{39}}{35}\] Therefore, our answer is $39+39+35=\boxed{113}$ , and we are done.

Answer:  113