2004 AMC 10A Problem 21

1. Call the acute angle between the lines $\theta$. The two lines cut every circle into four wedges: two opposite wedges with angle $\theta$ and two with angle $\pi - \theta$.
2. A full turn is $2\pi$, so the two opposite wedges with angle $\theta$ together take up $\dfrac{2\theta}{2\pi} = \dfrac{\theta}{\pi}$ of any ring. The other two take up $\dfrac{\pi - \theta}{\pi}$.
3. Find the area of each ring, from the inside out:
- inner disk (radius $1$): $\pi$,
- middle ring (between radius $1$ and $2$): $4\pi - \pi = 3\pi$,
- outer ring (between radius $2$ and $3$): $9\pi - 4\pi = 5\pi$.
4. From the diagram, the inner disk and the outer ring are shaded in the two narrow wedges, and the middle ring is shaded in the two wide wedges. So the shaded area is
$$\frac{\theta}{\pi}\cdot\pi + \frac{\pi - \theta}{\pi}\cdot 3\pi + \frac{\theta}{\pi}\cdot 5\pi = \theta + (3\pi - 3\theta) + 5\theta = 3\pi + 3\theta.$$
5. The whole figure has area $9\pi$. Shaded is to unshaded as $8$ is to $13$, so the shaded part is $8$ parts out of $8 + 13 = 21$. That means
$$3\pi + 3\theta = \frac{8}{21} \times 9\pi = \frac{24\pi}{7}.$$
6. Solve: $3\theta = \dfrac{24\pi}{7} - \dfrac{21\pi}{7} = \dfrac{3\pi}{7}$, so $\theta = \dfrac{\pi}{7}$.

The answer is $\boxed{\textbf{(B)}\ \frac{\pi}{7}}$.

A wedge of a circle takes up the same fraction of the area as its angle takes of the full turn. Breaking the figure into rings makes the shaded area a simple expression in $\theta$. The other key step is changing "shaded is $\tfrac{8}{13}$ of unshaded" into "shaded is $\tfrac{8}{21}$ of everything." Quick check: if $\theta$ were $0$, only the middle ring would be shaded, and the formula gives $3\pi$.