jmc

Applying Stewart's Theorem to medians $AD,CE$, we have: $$\begin{array}{rl}B{C}^2+4\cdot 18^2& =2\left(24^2+A{C}^2\right)\\ 24^2+4\cdot 27^2& =2\left(A{C}^2+B{C}^2\right)\end{array}$$ Substituting the first equation into the second and simplification yields $24^2=2\left(3A{C}^2+2\cdot 24^2-4\cdot 18^2\right)-4\cdot 27^2$ $⟹AC=\sqrt{2^5\cdot 3+2\cdot 3^5+2^4\cdot 3^3-2^7\cdot 3}=3\sqrt{70}$. By the Power of a Point Theorem on $E$, we get $EF=\frac{12^2}{27}=\frac{16}{3}$. The Law of Cosines on $△ACE$ gives $$\begin{array}{r}\cos \angle AEC=\left(\frac{12^2+27^2-9\cdot 70}{2\cdot 12\cdot 27}\right)=\frac{3}{8}\end{array}$$ Hence $\sin \angle AEC=\sqrt{1-{\cos }^2\angle AEC}=\frac{\sqrt{55}}{8}$. Because $△AEF,BEF$ have the same height and equal bases, they have the same area, and $[ABF]=2[AEF]=2\cdot \frac{1}{2}\cdot AE\cdot EF\sin \angle AEF=12\cdot \frac{16}{3}\cdot \frac{\sqrt{55}}{8}=8\sqrt{55}$, and the answer is $8+55=\boxed{63}$.