In $\triangle ABC$ in the adjoining figure, $AD$ and $AE$ trisect $\angle BAC$. The lengths of $BD$, $DE$ and $EC$ are $2$, $3$, and $6$, respectively. The length of the shortest side of $\triangle ABC$ is

$\textbf{(A)}\ 2\sqrt{10}\qquad \textbf{(B)}\ 11\qquad \textbf{(C)}\ 6\sqrt{6}\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ \text{not uniquely determined by the given information}$
Let $AC=b$, $AB=c$, $AD=d$, and $AE=e$. Then, by the Angle Bisector Theorem, $\frac{c}{e}=\frac{2}{3}$ and $\frac{d}{b}=\frac12$, thus $e=\frac{3c}2$ and $d=\frac b2$.
Also, by Stewart’s Theorem, $198+11d^2=2b^2+9c^2$ and $330+11e^2=5b^2+6c^2$. Therefore, we have the following system of equations using our substitution from earlier:
\[\begin{cases}198=-\frac{3b^2}4+9c^2\\330=5b^2-\frac{75c^2}{4}\end{cases}.\]
Thus, we have:
\[\begin{cases}264=-b^2+12c^2\\264=4b^2-15c^2\end{cases}.\]
Therefore, $5b^2=27c^2$, so $b^2=\frac{27c^2}5$, thus our first equation from earlier gives $264=\frac{33c^2}{5}$, so $c^2=40$, thus $b^2=216$. So, $c<b$ and the answer to the original problem is $c=\sqrt{40}=\boxed{2\sqrt{10}~\textbf{(A)}}$.
~ [Aops-g5-gethsemanea2](https://artofproblemsolving.com/wiki/index.php/User:Aops-g5-gethsemanea2)



[1981 AHSME](https://artofproblemsolving.com/wiki/index.php/1981_AHSME) ([Problems](https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems) • [Answer Key](https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Answer_Key) • [Resources](https://artofproblemsolving.comhttp://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44&year=1981))


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