Difference between revisions of "2004 AMC 10A Problems/Problem 21"

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==Solution==
 
==Solution==
{{solution}}
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Let the area of the shaded region be <math>S</math>, the area of the unshaded region be <math>U</math>, and the acute angle that is formed by the two lines be <math>\theta</math>. We can set up two equations between <math>S</math> and <math>U</math>:
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<math>S+U=9\pi</math>
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<math>S=\dfrac{8}{13}U</math>
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Thus <math>\dfrac{21}{13}U=9\pi</math>, and <math>U=\dfrac{39\pi}{7}</math>, and thus <math>S=\dfracc{8}{13}\cdot \dfrac{39\pi}{7}=\dfrac{24\pi}{7}</math>.
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Now we can make a formula for the area of the shaded region in terms of <math>\theta</math>:
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<math>\dfrac{2\theta}{2\pi}*\pi +\dfrac{2(\pi-\theta)}{2\pi}*(4\pi-\pi)+\dfrac{2\theta}{2\pi}(9\pi-4\pi)=\theta +3\pi-3\theta+5\theta=3\theta+3\pi=\dfrac{24\pi}{7}</math>
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Thus <math>3\theta=\dfrac{3\pi}{7}\Rightarrow \theta=\dfrac{\pi}{7}\Rightarrow \mathrm{(B) \ }</math>
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== See also ==
 
== See also ==
 
{{AMC10 box|year=2004|ab=A|num-b=20|num-a=22}}
 
{{AMC10 box|year=2004|ab=A|num-b=20|num-a=22}}
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[[Category:Introductory Geometry Problems]]

Revision as of 08:53, 8 August 2008

Problem

Two distinct lines pass through the center of three concentric circles of radii 3, 2, and 1. The area of the shaded region in the diagram is $\frac{8}{13}$ of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: $\pi$ radians is $180$ degrees.)

AMC10 2004A 21.png

$\mathrm{(A) \ } \frac{\pi}{8} \qquad \mathrm{(B) \ } \frac{\pi}{7} \qquad \mathrm{(C) \ } \frac{\pi}{6} \qquad \mathrm{(D) \ } \frac{\pi}{5} \qquad \mathrm{(E) \ } \frac{\pi}{4}$

Solution

Let the area of the shaded region be $S$, the area of the unshaded region be $U$, and the acute angle that is formed by the two lines be $\theta$. We can set up two equations between $S$ and $U$:

$S+U=9\pi$

$S=\dfrac{8}{13}U$

Thus $\dfrac{21}{13}U=9\pi$, and $U=\dfrac{39\pi}{7}$, and thus $S=\dfracc{8}{13}\cdot \dfrac{39\pi}{7}=\dfrac{24\pi}{7}$ (Error compiling LaTeX. Unknown error_msg).

Now we can make a formula for the area of the shaded region in terms of $\theta$:

$\dfrac{2\theta}{2\pi}*\pi +\dfrac{2(\pi-\theta)}{2\pi}*(4\pi-\pi)+\dfrac{2\theta}{2\pi}(9\pi-4\pi)=\theta +3\pi-3\theta+5\theta=3\theta+3\pi=\dfrac{24\pi}{7}$

Thus $3\theta=\dfrac{3\pi}{7}\Rightarrow \theta=\dfrac{\pi}{7}\Rightarrow \mathrm{(B) \ }$

See also

2004 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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
All AMC 10 Problems and Solutions