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]
Let $B=(0,0)$ and $BC$ be the line $y=0$ .
We compute that $\cos{\angle{ABC}}=\frac{1}{8}$ , so $\tan{\angle{ABC}}=3\sqrt{7}$ .
Thus, $A$ lies on the line $y=3x\sqrt{7}$ . The length of $AB$ at a point $x$ is $8x$ , so $x=\frac{1}{2}$
We now have the coordinates $A=\left(\frac{1}{2},\frac{3\sqrt{7}}{2}\right)$ $B=(0,0)$ and $C=(5,0)$ .
We also have $D=(2,0)$ by the angle-bisector theorem and $M=\left(\frac{5}{4},\frac{3\sqrt{7}}{4}\right)$ by taking the midpoint.
We have that because $\cos{\angle{ABC}}=\frac{1}{8}$ $\cos{\frac{\angle{ABC}}{2}}=\frac{3}{4}$ by half angle formula.
We also compute $\cos{\angle{ACB}}=\frac{3}{4}$ , so $\cos{\frac{\angle{ACB}}{2}}=\frac{\sqrt{14}}{4}$
Now, $AD$ has slope $-\frac{\frac{3\sqrt{7}}{2}}{2-\frac{1}{2}}=-\sqrt{7}$ , so it's perpendicular bisector has slope $\frac{\sqrt{7}}{7}$ and goes through $\left(\frac{5}{4},\frac{3\sqrt{7}}{4}\right)$
We find that this line has equation $y=\frac{\sqrt{7}}{7}x+\frac{4\sqrt{7}}{7}$
As $\cos{\angle{CBI}}=\frac{3}{4}$ , we have that line $BI$ has form $y=\frac{\sqrt{7}}{3}x$ .
Solving for the intersection point of these two lines, we get $x=3$ and thus $E=\left(3, \sqrt{7}\right)$
We also have that because $\cos{\angle{ICB}}=\frac{\sqrt{14}}{4}$ $CI$ has form $y=-\frac{x\sqrt{7}}{7}+\frac{5\sqrt{7}}{7}$
Intersecting the line $CI$ and the perpendicular bisector of $AD$ yields $-\frac{x\sqrt{7}}{7}+\frac{5\sqrt{7}}{7}=\frac{x\sqrt{7}}{7}+\frac{4\sqrt{7}}{7}$
Solving this, we get $x=\frac{1}{2}$ and so $F=\left(\frac{1}{2},\frac{9\sqrt{7}}{14}\right)$
We now compute $EF=\sqrt{\left(\frac{5}{2}\right)^2+\left(\frac{5\sqrt{7}}{14}\right)^2}=\frac{5\sqrt{14}}{7}$ .
We also have $MA=\sqrt{\left(\frac{3}{4}\right)^2+\left(\frac{3\sqrt{7}}{4}\right)^2}=\frac{3\sqrt{2}}{2}$
As ${MA}\perp{EF}$ , we have $[\triangle{AEF}]=\frac{1}{2}\left(\frac{3\sqrt{2}}{2}\times\frac{5\sqrt{14}}{7}\right)=\frac{15\sqrt{7}}{14}$
The desired answer is $15+7+14=\boxed{036}$ ~Imayormaynotknowcalculus

Answer:  036