In [triangle](https://artofproblemsolving.com/wiki/index.php/Triangle) $ABC$ the [medians](https://artofproblemsolving.com/wiki/index.php/Median) $\overline{AD}$ and $\overline{CE}$ have lengths $18$ and $27$, respectively, and $AB=24$. Extend $\overline{CE}$ to intersect the [circumcircle](https://artofproblemsolving.com/wiki/index.php/Circumcircle) of $ABC$ at $F$. The area of triangle $AFB$ is $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$.
Applying [Stewart's Theorem](https://artofproblemsolving.com/wiki/index.php/Stewart%27s_Theorem) to medians $AD, CE$, we have:


\begin{align*} BC^2 + 4 \cdot 18^2 &= 2\left(24^2 + AC^2\right) \\ 24^2 + 4 \cdot 27^2 &= 2\left(AC^2 + BC^2\right)  \end{align*}


Substituting the first equation into the second and simplification yields $24^2 = 2\left(3AC^2 + 2 \cdot 24^2 - 4 \cdot 18^2\right)- 4 \cdot 27^2$ $\Longrightarrow AC = \sqrt{2^5 \cdot 3 + 2 \cdot 3^5 + 2^4 \cdot 3^3 - 2^7 \cdot 3} = 3\sqrt{70}$. 
By the [Power of a Point Theorem](https://artofproblemsolving.com/wiki/index.php/Power_of_a_Point_Theorem) on $E$, we get $EF = \frac{12^2}{27} = \frac{16}{3}$. The [Law of Cosines](https://artofproblemsolving.com/wiki/index.php/Law_of_Cosines) on $\triangle ACE$ gives 


\begin{align*} \cos \angle AEC = \left(\frac{12^2 + 27^2 - 9 \cdot 70}{2 \cdot 12 \cdot 27}\right) = \frac{3}{8} \end{align*}


Hence $\sin \angle AEC = \sqrt{1 - \cos^2 \angle AEC} = \frac{\sqrt{55}}{8}$. Because $\triangle AEF, BEF$ have the same height and equal bases, they have the same area, and $[ABF] = 2[AEF] = 2 \cdot \frac 12 \cdot AE \cdot EF \sin \angle AEF = 12 \cdot \frac{16}{3} \cdot \frac{\sqrt{55}}{8} = 8\sqrt{55}$, and the answer is $8 + 55 = \boxed{063}$.