2003 AMC 10A Problem 19

The small semicircle has radius $\tfrac12$, so its area is $\tfrac12 \pi \left(\tfrac12\right)^2 = \tfrac{\pi}{8}$.

The lune is what remains of the small semicircle after removing the part that overlaps the big semicircle. That overlap is the region of the big circle (radius $1$) lying above the chord that serves as the small semicircle's diameter: a circular segment.

Join the big circle's center $O$ to the two endpoints $P, Q$ of that chord. Then $OP = OQ = 1$ and $PQ = 1$, so triangle $OPQ$ is equilateral and $\angle POQ = 60^\circ$. Therefore
$$\text{segment} = \text{sector} - \text{triangle} = \frac{60}{360}\pi \cdot 1^2 - \frac{\sqrt{3}}{4} \cdot 1^2 = \frac{\pi}{6} - \frac{\sqrt{3}}{4}.$$

Lune $= \dfrac{\pi}{8} - \left(\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{4}\right) = \dfrac{\sqrt{3}}{4} - \dfrac{\pi}{24}$.

The answer is $\boxed{\textbf{(C)}\ \frac{\sqrt{3}}{4}-\frac{1}{24}\pi}$.

Regions bounded by arcs of two circles decompose into sectors, triangles and segments. The key auxiliary lines are radii to the chord's endpoints, which reveal the central angle; a chord equal to the radius always gives $60^\circ$. Then "segment = sector minus triangle" finishes it.

Sanity check: the answer must be positive and smaller than $\tfrac{\pi}{8} \approx 0.39$. Choice (C) is $0.433 - 0.131 = 0.30$, plausible; (A) is negative and (D), (E) exceed $\tfrac{\pi}{8}$, so only (B) and (C) survive, and the computation of the segment decides.

• Subtracting the whole 60-degree sector from the small semicircle without adding back the equilateral triangle, or using radius 1 for the small semicircle.
• Computing the big circle's area with radius $2$ (the diameter) so the sector becomes $\tfrac{2\pi}{3}$.