Problem: Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be written as $\tfrac{m\sqrt{n}}p,$ where $m$ and $p$ are relatively prime positive integers, and $n$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$

Solution: [asy]         size(8cm); defaultpen(fontsize(10pt));          pair A,B,C,I,D,M,T,Y,Z,EE,F;         A=(0,3sqrt(7));         B=(-1,0);         C=(9,0);         I=incenter(A,B,C);         D=extension(A,I,B,C);         M=(A+D)/2;          draw(B--EE,gray+dashed);         draw(C--F,gray+dashed);         draw(A--B--C--A);         draw(A--D);         draw(B--(5,sqrt(28)));         draw(M--(5,sqrt(28)));         draw(C--(0,9sqrt(7)/7));         draw(M--(0,9sqrt(7)/7));         dot("$A$",A,NW);         dot("$B$",B,SW);         dot("$C$",C,SE);         dot("$D$",D,S);         dot("$E$",(5,sqrt(28)),N);          dot("$M$",M,dir(70));         dot("$F$",(0,9sqrt(7)/7),N);          label("$2$",B--D,S);         label("$3$",D--C,S);         label("$6$",A--C,N);         label("$4$",A--B,W);     [/asy]
As usual, we will use homogenized barycentric coordinates.
We have that $AD$ will have form $3z=2y$ . Similarly, $CF$ has form $5y=6x$ and $BE$ has form $5z=4x$ .
Since $A=(1,0,0)$ and $D=\left(0,\frac{3}{5},\frac{2}{5}\right)$ , we also have $M=\left(\frac{1}{2},\frac{3}{10},\frac{1}{5}\right)$ .
It remains to determine the equation of the line formed by the perpendicular bisector of $AD$
This can be found using EFFT. Let a point $T$ on $EF$ have coordinates $(x, y, z)$ .
We then have that the displacement vector $\overrightarrow{AD}=\left(-1, \frac{3}{5}, \frac{2}{5}\right)$ and that the displacement vector $\overrightarrow{TM}$ has form $\left(x-\frac{1}{2},y-\frac{3}{10},z-\frac{1}{5}\right)$ .
Now, by EFFT, we have $5^2\left(\frac{3}{5}\times\left(z-\frac{1}{5}\right)+\frac{2}{5}\times\left(y-\frac{3}{10}\right)\right)+6^2\left(-1\times\left(z-\frac{1}{5}\right)+\frac{2}{5}\times\left(x-\frac{1}{2}\right)\right)+4^2\left(-1\times\left(y-\frac{3}{10}\right)+\frac{3}{5}\times\left(x-\frac{1}{2}\right)\right)=0$ .
This equates to $8x-2y-7z=2$
Now, intersecting this with $BE$ , we have $5z=4x$ $8x-2y-7z=2$ , and $x+y+z=1$ .
This yields $x=\frac{2}{3}$ $y=-\frac{1}{5}$ , and $z=\frac{8}{15}$ , or $E=\left(\frac{2}{3},-\frac{1}{5},\frac{8}{15}\right)$
Similarly, intersecting this with $CF$ , we have $5y=6x$ $8x-2y-7z=2$ , and $x+y+z=1$ .
Solving this, we obtain $x=\frac{3}{7}$ $y=\frac{18}{35}$ , and $z=\frac{2}{35}$ , or $F=\left(\frac{3}{7},\frac{18}{35},\frac{2}{35}\right)$
We finish by invoking the Barycentric Distance Formula twice; our first displacement vector being $\overrightarrow{FE}=\left(\frac{5}{21},-\frac{5}{7},\frac{10}{21}\right)$ .
We then have $FE^2=-25\left(-\frac{5}{7}\cdot\frac{10}{21}\right)-36\left(\frac{5}{21}\cdot\frac{10}{21}\right)-16\left(\frac{5}{21}\cdot-\frac{5}{7}\right)=\frac{50}{7}$ , thus $FE=\frac{5\sqrt{14}}{7}$
Our second displacement vector is $\overrightarrow{AM}=\left(-\frac{1}{2},\frac{3}{10},\frac{1}{5}\right)$ .
As a result, $AM^2=-25\left(\frac{3}{10}\cdot\frac{1}{5}\right)-36\left(-\frac{1}{2}\cdot\frac{1}{5}\right)-16\left(-\frac{1}{2}\cdot\frac{3}{10}\right)=\frac{9}{2}$ , so $AM=\frac{3\sqrt{2}}{2}$
As ${AM}\perp{EF}$ , the desired area is $\frac{\frac{5\sqrt{14}}{7}\times\frac{3\sqrt{2}}{2}}{2}={\frac{15\sqrt{7}}{14}}\implies{m+n+p=\boxed{036}$ . ~Imayormaynotknowcalculus

Answer:  036