1960 AHSME Problems/Problem 40

Problem

Given right $\triangle ABC$ with legs $BC=3, AC=4$ . Find the length of the shorter angle trisector from $C$ to the hypotenuse:

$\textbf{(A)}\ \frac{32\sqrt{3}-24}{13}\qquad\textbf{(B)}\ \frac{12\sqrt{3}-9}{13}\qquad\textbf{(C)}\ 6\sqrt{3}-8\qquad\textbf{(D)}\ \frac{5\sqrt{10}}{6}\qquad\textbf{(E)}\ \frac{25}{12}\qquad$

Solution

Angle $C$ is split into three $30^{\circ}$ angles. The shorter angle trisector will be the one closer $BC$ . Let it intersect $AB$ at point $P$ . Let the perpendicular from point $P$ intersect $BC$ at point $R$ and have length $x$ . Thus $\triangle PRC$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle and $RC$ has length $x\sqrt{3}$ . Because $\triangle PBR$ is similar to $\triangle ABC$ , $RB$ has length $\frac{3}{4}x$ . $RC+RB=BC=x\sqrt{3}+\frac{3}{4}x=3$ The problem asks for the length of $PC$ , or $2x$ . Solving for $x$ and multiplying by two gives $\boxed{\textbf(A)}$ .

Video Solution

https://youtu.be/ZdM2ou5Gsuw?t=374

~MathProblemSolvingSkills.com

1960 AHSC (Problems • Answer Key • Resources)
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1960 AHSME Problems/Problem 40

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