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Difference between revisions of "1981 AHSME Problems/Problem 25"

(fixed formtting again)
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Also, by Stewart’s Theorem, <math>198+11d^2=2b^2+9c^2</math> and <math>330+11e^2=5b^2+6c^2</math>. Therefore, we have the following system of equations using our substitution from earlier:
 
Also, by Stewart’s Theorem, <math>198+11d^2=2b^2+9c^2</math> and <math>330+11e^2=5b^2+6c^2</math>. Therefore, we have the following system of equations using our substitution from earlier:
  
−
<math>\begin{cases}198=-\frac{3b^2}4+9c^2\\330=5b^2-\frac{75c^2}{4}\end{cases}</math>.
+
<cmath>\begin{cases}198=-\frac{3b^2}4+9c^2\\330=5b^2-\frac{75c^2}{4}\end{cases}.</cmath>
  
 
Thus, we have:
 
Thus, we have:
  
−
<math>\begin{cases}264=-b^2+12c^2\\264=4b^2-15c^2\end{cases}</math>.
+
<cmath>\begin{cases}264=-b^2+12c^2\\264=4b^2-15c^2\end{cases}.</cmath>
  
 
Therefore, <math>5b^2=27c^2</math>, so <math>b^2=\frac{27c^2}5</math>, thus our first equation from earlier gives <math>264=\frac{33c^2}{5}</math>, so <math>c^2=40</math>, thus <math>b^2=216</math>. So, <math>c<b</math> and the answer to the original problem is <math>c=\sqrt{40}=\boxed{2\sqrt{10}~\textbf{(A)}}</math>.
 
Therefore, <math>5b^2=27c^2</math>, so <math>b^2=\frac{27c^2}5</math>, thus our first equation from earlier gives <math>264=\frac{33c^2}{5}</math>, so <math>c^2=40</math>, thus <math>b^2=216</math>. So, <math>c<b</math> and the answer to the original problem is <math>c=\sqrt{40}=\boxed{2\sqrt{10}~\textbf{(A)}}</math>.

Revision as of 01:58, 9 August 2023

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)

1981 AHSME (Problems • Answer Key • Resources)
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Problem 24 		Followed by
Problem 26
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