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Difference between revisions of "1983 AHSME Problems/Problem 10"

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==Solution==
 
==Solution==
  
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Note that since <math>AB</math> is a diameter, <math>\angle AEB = 90^{\circ}</math>, which means <math>AB</math> is an altitude of equilateral triangle <math>ABC</math>. It follows that <math>\triangle ABE</math> is a <math>30^{\circ},60^{\circ},90^{\circ}</math> triangle, and so <math>AE = AB \cdot \frac{\sqrt{3}}{2} = (2 \cdot 1) (\frac{\sqrt{3}}{2}) = \boxed{\textbf{(D)}\ \sqrt{3}}</math>.
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Note that since <math>AB</math> is a diameter, <math>\angle AEB = 90^{\circ}</math>, which means <math>AB</math> is an altitude of equilateral triangle <math>ABC</math>. It follows that <math>\triangle ABE</math> is a <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangle, and so <math>AE = AB \cdot \frac{\sqrt{3}}{2} = (2 \cdot 1) (\frac{\sqrt{3}}{2}) = \boxed{\textbf{(D)}\ \sqrt{3}}</math>.
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==See Also==
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{{AHSME box|year=1983|num-b=9|num-a=11}}
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{{MAA Notice}}

Revision as of 23:46, 19 February 2019

Problem

Segment $AB$ is both a diameter of a circle of radius $1$ and a side of an equilateral triangle $ABC$. The circle also intersects $AC$ and $BD$ at points $D$ and $E$, respectively. The length of $AE$ is

$\textbf{(A)} \ \frac{3}{2} \qquad \textbf{(B)} \ \frac{5}{3} \qquad \textbf{(C)} \ \frac{\sqrt 3}{2} \qquad \textbf{(D)}\ \sqrt{3}\qquad \textbf{(E)}\ \frac{2+\sqrt 3}{2}$

Solution

Note that since $AB$ is a diameter, $\angle AEB = 90^{\circ}$, which means $AB$ is an altitude of equilateral triangle $ABC$. It follows that $\triangle ABE$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, and so $AE = AB \cdot \frac{\sqrt{3}}{2} = (2 \cdot 1) (\frac{\sqrt{3}}{2}) = \boxed{\textbf{(D)}\ \sqrt{3}}$.

See Also

1983 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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
All AHSME Problems and Solutions


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