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

1981 AHSME (Problems • Answer Key • Resources)

Preceded byProblem 24

Followed byProblem 26

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