Difference between revisions of "1981 AHSME Problems/Problem 25"
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== Problem == | == 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 |
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 | |
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All AHSME Problems and Solutions |