1981 AHSME Problems/Problem 25

Problem 25

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

[asy] defaultpen(linewidth(.8pt)); pair A = (0,11); pair B = (2,0); pair D = (4,0); pair E = (7,0); pair C = (13,0); label(" $A$ ",A,N); label(" $B$ ",B,SW); label(" $C$ ",C,SE); label(" $D$ ",D,S); label(" $E$ ",E,S); label(" $2$ ",midpoint(B--D),N); label(" $3$ ",midpoint(D--E),NW); label(" $6$ ",midpoint(E--C),NW); draw(A--B--C--cycle); draw(A--D); draw(A--E); [/asy]

$\textbf{(A)}\ 2\sqrt{10}\qquad \textbf{(B)}\ 11\qquad \textbf{(C)}\ 6\sqrt{6}\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ \text{not uniquely determined by the given information}$

Solution

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 (talk) 01:47, 9 August 2023 (EDT)

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1981 AHSME Problems/Problem 25

Problem 25

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