2004 AMC 10A Problems/Problem 21

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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.)

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

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 $\theta$ :

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

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

2004 AMC 10A (Problems • 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

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2004 AMC 10A Problems/Problem 21

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