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Difference between revisions of "1990 AHSME Problems/Problem 14"

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</asy>
 
</asy>
  
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An acute isosceles triangle, <math>ABC</math>, is inscribed in a circle. Through <math>B</math> and <math>C</math>, tangents to the circle are drawn, meeting at point <math>D</math>. If <math>\angle{ABC=\angle{ACB}=2\angle{D}</math> and <math>x</math> is the radian measure of <math>\angle{A}</math>, then <math>x=</math>
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An acute isosceles triangle, <math>ABC</math>, is inscribed in a circle. Through <math>B</math> and <math>C</math>, tangents to the circle are drawn, meeting at point <math>D</math>. If <math>\angle{ABC}=\angle{ACB}=2\angle{D}</math> and <math>x</math> is the radian measure of <math>\angle{A}</math>, then <math>x=</math>
  
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<math>\text{(A) } \frac{2\pi}{7}\quad
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<math>\text{(A) } \frac{3\pi}{7}\quad
 
\text{(B) } \frac{4\pi}{9}\quad
 
\text{(B) } \frac{4\pi}{9}\quad
 
\text{(C) } \frac{5\pi}{11}\quad
 
\text{(C) } \frac{5\pi}{11}\quad
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== Solution ==
 
== Solution ==
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<math>\fbox{A}</math>
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We can make two equations (assume angle D is y): <math>y+2x=180</math> and <math>4y+x=180</math>. We find that <math>x=\dfrac{540}{7}</math>. Now we have to convert this to radians. 360 degrees is <math>2\pi</math> radians, so since we have <math>\dfrac{540}{7}</math> degrees, the answer is <math>\dfrac{3\pi}{7}</math> which is <math>\fbox{A}.</math>
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We can make two equations (assume angle D is y): <math>y+2x=180</math> and <math>4y+x=180</math>. We find that <math>x=\dfrac{360}{7}</math>. Now we have to convert this to radians. 360 degrees is <math>2\pi</math> radians, so since we have <math>\dfrac{360}{7}</math> degrees, the answer is <math>\dfrac{2\pi}{7}</math>.
 
  
 
== See also ==
 
== See also ==

Latest revision as of 13:09, 2 April 2023

Problem

[asy] draw(circle((0,0),1),black); draw((0,1)--(cos(pi/14),-sin(pi/14))--(-cos(pi/14),-sin(pi/14))--cycle,dot); draw((-cos(pi/14),-sin(pi/14))--(0,-1/cos(3pi/7))--(cos(pi/14),-sin(pi/14)),dot); draw(arc((0,1),.25,230,310)); MP("A",(0,1),N);MP("B",(cos(pi/14),-sin(pi/14)),E);MP("C",(-cos(pi/14),-sin(pi/14)),W);MP("D",(0,-1/cos(3pi/7)),S); MP("x",(0,.8),S); [/asy]

An acute isosceles triangle, $ABC$, is inscribed in a circle. Through $B$ and $C$, tangents to the circle are drawn, meeting at point $D$. If $\angle{ABC}=\angle{ACB}=2\angle{D}$ and $x$ is the radian measure of $\angle{A}$, then $x=$

$\text{(A) } \frac{3\pi}{7}\quad \text{(B) } \frac{4\pi}{9}\quad \text{(C) } \frac{5\pi}{11}\quad \text{(D) } \frac{6\pi}{13}\quad \text{(E) } \frac{7\pi}{15}$

Solution

We can make two equations (assume angle D is y): $y+2x=180$ and $4y+x=180$. We find that $x=\dfrac{540}{7}$. Now we have to convert this to radians. 360 degrees is $2\pi$ radians, so since we have $\dfrac{540}{7}$ degrees, the answer is $\dfrac{3\pi}{7}$ which is $\fbox{A}.$

See also

1990 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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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