Difference between revisions of "1960 AHSME Problems/Problem 40"

Revision as of 15:32, 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 $\boxed{\textbf(A)}$ .

1960 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 39	Followed by 1961 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

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 ==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:
 <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>

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Difference between revisions of "1960 AHSME Problems/Problem 40"

Revision as of 15:32, 1 May 2016

Problem

Solution

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