Difference between revisions of "1981 AHSME Problems/Problem 25"
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− | Problem 25 | + | ## Problem 25 |
In <math>\triangle ABC</math> in the adjoining figure, <math>AD</math> and <math>AE</math> trisect <math>\angle BAC</math>. The lengths of <math>BD</math>, <math>DE</math> and <math>EC</math> are <math>2</math>, <math>3</math>, and <math>6</math>, respectively. The length of the shortest side of <math>\triangle ABC</math> is | In <math>\triangle ABC</math> in the adjoining figure, <math>AD</math> and <math>AE</math> trisect <math>\angle BAC</math>. The lengths of <math>BD</math>, <math>DE</math> and <math>EC</math> are <math>2</math>, <math>3</math>, and <math>6</math>, respectively. The length of the shortest side of <math>\triangle ABC</math> 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("<math>A</math>",A,N); label("<math>B</math>",B,SW); label("<math>C</math>",C,SE); label("<math>D</math>",D,S); label("<math>E</math>",E,S); label("<math>2</math>",midpoint(B--D),N); label("<math>3</math>",midpoint(D--E),NW); label("<math>6</math>",midpoint(E--C),NW); draw(A--B--C--cycle); draw(A--D); draw(A--E); [/asy] | [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("<math>A</math>",A,N); label("<math>B</math>",B,SW); label("<math>C</math>",C,SE); label("<math>D</math>",D,S); label("<math>E</math>",E,S); label("<math>2</math>",midpoint(B--D),N); label("<math>3</math>",midpoint(D--E),NW); label("<math>6</math>",midpoint(E--C),NW); draw(A--B--C--cycle); draw(A--D); draw(A--E); [/asy] | ||
+ | |||
<math>\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}</math> | <math>\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}</math> | ||
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+ | ## Solution | ||
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+ | Let <math>AC=b</math>, <math>AB=c</math>, <math>AD=d</math>, and <math>AE=e</math>. Then, by the Angle Bisector Theorem, <math>\frac{c}{e}=\frac{2}{3}</math> and <math>\frac{d}{b}=\frac12</math>, thus <math>e=\frac{3c}2</math> and <math>d=\frac b2</math>. | ||
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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: | ||
+ | |||
+ | <math> | ||
+ | |||
+ | Thus, we have: | ||
+ | |||
+ | <math> | ||
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+ | 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>. | ||
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+ | [[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] ([[User talk:Aops-g5-gethsemanea2|talk]]) 01:47, 9 August 2023 (EDT) |
Revision as of 00:48, 9 August 2023
- Problem 25
In in the adjoining figure, and trisect . The lengths of , and are , , and , respectively. The length of the shortest side of 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,N); label("",B,SW); label("",C,SE); label("",D,S); label("",E,S); label("",midpoint(B--D),N); label("",midpoint(D--E),NW); label("",midpoint(E--C),NW); draw(A--B--C--cycle); draw(A--D); draw(A--E); [/asy]
- Solution
Let , , , and . Then, by the Angle Bisector Theorem, and , thus and .
Also, by Stewart’s Theorem, and . Therefore, we have the following system of equations using our substitution from earlier:
.
Thus, we have:
.
Therefore, , so , thus our first equation from earlier gives , so , thus . So, and the answer to the original problem is .
Aops-g5-gethsemanea2 (talk) 01:47, 9 August 2023 (EDT)