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:  
 +
 
<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 $\triangle ABC$ 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 39
Followed by
1961 AHSME
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