Difference between revisions of "1960 AHSME Problems/Problem 40"
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==Problem== | ==Problem== | ||
Given right <math>\triangle ABC</math> with legs <math>BC=3, AC=4</math>. Find the length of the shorter angle trisector from <math>C</math> to the hypotenuse: | Given right <math>\triangle ABC</math> with legs <math>BC=3, AC=4</math>. Find the length of the shorter angle trisector from <math>C</math> to the hypotenuse: | ||
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<math> \textbf{(A)}\ \frac{32\sqrt{3}-24}{13}\qquad\textbf{(B)}\ \frac{12\sqrt{3}-9}{13}\qquad\textbf{(C)}\ 6\sqrt{3}-8\qquad\textbf{(D)}\ \frac{5\sqrt{10}}{6}\qquad\textbf{(E)}\ \frac{25}{12}\qquad</math> | <math> \textbf{(A)}\ \frac{32\sqrt{3}-24}{13}\qquad\textbf{(B)}\ \frac{12\sqrt{3}-9}{13}\qquad\textbf{(C)}\ 6\sqrt{3}-8\qquad\textbf{(D)}\ \frac{5\sqrt{10}}{6}\qquad\textbf{(E)}\ \frac{25}{12}\qquad</math> | ||
Latest revision as of 16:32, 1 May 2016
Problem
Given right with legs . Find the length of the shorter angle trisector from to the hypotenuse:
Solution
Angle is split into three angles. The shorter angle trisector will be the one closer . Let it intersect at point . Let the perpendicular from point intersect at point and have length . Thus is a triangle and has length . Because is similar to , has length . The problem asks for the length of , or . Solving for and multiplying by two gives .
See Also
1960 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 39 |
Followed by 1961 AHSME | |
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All AHSME Problems and Solutions |