# 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 $\boxed{\textbf(A)}$.

## See Also

 1960 AHSC (Problems • Answer Key • Resources) Preceded byProblem 39 Followed by1961 AHSME 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 All AHSME Problems and Solutions
Invalid username
Login to AoPS