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

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Now we can make a formula for the area of the shaded region in terms of <math>\theta</math>:
 
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>
+
<math>\dfrac{2\theta}{2\pi} \cdot \pi +\dfrac{2(\pi-\theta)}{2\pi} \cdot (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>
  
 
Thus <math>3\theta=\dfrac{3\pi}{7}\Rightarrow \theta=\dfrac{\pi}{7}\Rightarrow\boxed{\mathrm{(B)}\ \frac{\pi}{7}}</math>
 
Thus <math>3\theta=\dfrac{3\pi}{7}\Rightarrow \theta=\dfrac{\pi}{7}\Rightarrow\boxed{\mathrm{(B)}\ \frac{\pi}{7}}</math>

Revision as of 11:03, 21 December 2015

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 $S=\dfrac{8}{13}\cdot \dfrac{39\pi}{7}=\dfrac{24\pi}{7}$.

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

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

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

See also

2004 AMC 10A (Problems • Answer Key • Resources)
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

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