Difference between revisions of "1981 AHSME Problems/Problem 25"
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− | == Problem | + | __FORCETOC__ |
+ | |||
+ | == Problem == | ||
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("$A$",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> | ||
<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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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: | ||
− | < | + | <cmath> |
Thus, we have: | Thus, we have: | ||
− | < | + | <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>. | ||
− | [[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] | + | ~ [[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] |
− | {{AHSME box|year=1981| | + | {{AHSME box|year=1981|num-b=24|num-a=26}} |
Latest revision as of 20:01, 9 August 2023
Contents
[hide]Problem
In in the adjoining figure, and trisect . The lengths of , and are , , and , respectively. The length of the shortest side of is
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 .
1981 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |