In $\triangle{ABC}$ medians $\overline{AD}$ and $\overline{BE}$ intersect at $G$ and $\triangle{AGE}$ is equilateral. Then $\cos(C)$ can be written as $\frac{m\sqrt p}n$, where $m$ and $n$ are relatively prime positive integers and $p$ is a positive integer not divisible by the square of any prime. What is $m+n+p?$ $\textbf{(A) }44 \qquad$ $ \textbf{(B) }48 \qquad$ $ \textbf{(C) }52 \qquad$ $ \textbf{(D) }56 \qquad$ $ \textbf{(E) }60$
Solution
Assume $AE=EG=AG=2$, so $DG=1, CE=2, BG=4$.
$BD^2=1^2+4^2-1\cdot 4=13$, $BD=\sqrt{13}$, so $CD=\sqrt{13}$.
Using law of cosines with triangle $ACD$, $\cos C=\frac{4^2+13-9}{2\cdot 4\cdot\sqrt{13}}=\frac{5\sqrt{13}}{26}$.
