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
Contents
[hide]Problem
Given right with legs . Find the length of the shorter angle trisector from to the hypotenuse:
Solution
Angle is split into three angles. The shorter angle trisector will be the one closer . Let it intersect at point . Let the perpendicular from point intersect at point and have length . Thus is a triangle and has length . Because is similar to , has length . The problem asks for the length of , or . Solving for and multiplying by two gives .
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 |