2004 AMC 10A Problem 21

Let $\theta$ be the acute angle between the lines. The lines cut the plane into two opposite wedges of angle $\theta$ and two of angle $\pi - \theta$. A pair of opposite wedges of angle $\alpha$ takes up the fraction $\frac{2\alpha}{2\pi} = \frac{\alpha}{\pi}$ of any ring centered at $O$.

The three rings have areas $\pi$ (inner disk), $4\pi - \pi = 3\pi$ (middle ring) and $9\pi - 4\pi = 5\pi$ (outer ring). In the diagram, the inner disk and the outer ring are shaded in the acute wedges, while the middle ring is shaded in the obtuse wedges. So
$$S = \frac{\theta}{\pi}\cdot\pi + \frac{\pi - \theta}{\pi}\cdot 3\pi + \frac{\theta}{\pi}\cdot 5\pi = \theta + 3\pi - 3\theta + 5\theta = 3\pi + 3\theta.$$

The whole figure has area $9\pi$. Shaded is $\tfrac{8}{13}$ of unshaded, so shaded is $\frac{8}{8 + 13} = \frac{8}{21}$ of the total:
$$3\pi + 3\theta = \frac{8}{21}\cdot 9\pi = \frac{24\pi}{7} \quad\Longrightarrow\quad 3\theta = \frac{24\pi}{7} - \frac{21\pi}{7} = \frac{3\pi}{7}.$$

Thus $\theta = \dfrac{\pi}{7}$, and the answer is $\boxed{\textbf{(B)}\ \frac{\pi}{7}}$.

Sector areas are proportional to their angles, so the shaded area is linear in $\theta$ once the figure is decomposed into rings. Converting the part-to-part ratio $8 : 13$ into the part-to-whole fraction $\tfrac{8}{21}$ is the other essential step. Sanity check: at $\theta = 0$ only the full middle ring ($3\pi$) would be shaded, matching the constant term.

• Taking the shaded area to be 8/13 of the total area instead of 8/13 of the unshaded area.
• Misreading which wedges are shaded in the middle ring, which produces $S = 9\theta$ and no valid answer.