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

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Angle <math>C</math> is split into three <math>30^{\circ}</math> angles. The shorter angle trisector will be the one closer <math>BC</math>. Let it intersect <math>AB</math> at point <math>P</math>. Let the perpendicular from point <math>P</math> intersect <math>BC</math> at point <math>R</math> and have length <math>x</math>. Thus <math>\triangle PRC</math> is a <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangle and <math>RC</math> has length <math>x\sqrt{3}</math>. Because <math>\triangle PBR</math> is similar to <math>\triangle ABC</math>, <math>RB</math> has length <math>\frac{3}{4}x</math>. <cmath>RC+RB=BC=x\sqrt{3}+\frac{3}{4}x=3</cmath>
 
Angle <math>C</math> is split into three <math>30^{\circ}</math> angles. The shorter angle trisector will be the one closer <math>BC</math>. Let it intersect <math>AB</math> at point <math>P</math>. Let the perpendicular from point <math>P</math> intersect <math>BC</math> at point <math>R</math> and have length <math>x</math>. Thus <math>\triangle PRC</math> is a <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangle and <math>RC</math> has length <math>x\sqrt{3}</math>. Because <math>\triangle PBR</math> is similar to <math>\triangle ABC</math>, <math>RB</math> has length <math>\frac{3}{4}x</math>. <cmath>RC+RB=BC=x\sqrt{3}+\frac{3}{4}x=3</cmath>
 
The problem asks for the length of <math>PC</math>, or <math>2x</math>. Solving for <math>x</math> and multiplying by two gives <math>\boxed{\textbf(A)}</math>.
 
The problem asks for the length of <math>PC</math>, or <math>2x</math>. Solving for <math>x</math> and multiplying by two gives <math>\boxed{\textbf(A)}</math>.
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==Video Solution==
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 +
https://youtu.be/ZdM2ou5Gsuw?t=374
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 +
~MathProblemSolvingSkills.com
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==See Also==
 
==See Also==
 
{{AHSME 40p box|year=1960 |before=[[Problem 39]]|after=[[1961 AHSME]]}}
 
{{AHSME 40p box|year=1960 |before=[[Problem 39]]|after=[[1961 AHSME]]}}

Latest revision as of 21:29, 28 December 2023

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


See Also

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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