1981 AHSME Problems/Problem 25
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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