2003 AMC 10A Problem 19

The lune is the small half-circle with a piece cut out, where the big half-circle overlaps it. So find the small half-circle's area and the overlap's area.

1. Small half-circle. Its diameter is $1$, so its radius is $\frac{1}{2}$ and its area is
$$\frac{1}{2} \pi \left(\frac{1}{2}\right)^2 = \frac{\pi}{8}.$$
2. Where they overlap. The small half-circle sits on a flat segment $PQ$ of length $1$ whose ends lie on the big circle. The overlap is the part of the big circle between that segment and the big arc.
3. Find the angle. Let $O$ be the center of the big circle, so $OP = OQ = 1$ (its radius). Also $PQ = 1$. All three sides of triangle $OPQ$ are $1$, so it is equilateral and the angle at $O$ is $60^\circ$.
4. Area of the overlap. The $60^\circ$ slice of the big circle (from $O$ out to the arc $PQ$) is made of the triangle $OPQ$ plus the overlap. So
$$\text{overlap} = \text{slice} - \text{triangle}.$$
- Slice: $\frac{60}{360} \times \pi \times 1^2 = \frac{\pi}{6}$.
- Equilateral triangle of side $1$: $\frac{\sqrt{3}}{4}$.
- Overlap: $\frac{\pi}{6} - \frac{\sqrt{3}}{4}$.
5. Lune.
$$\frac{\pi}{8} - \left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right) = \frac{\sqrt{3}}{4} + \frac{\pi}{8} - \frac{\pi}{6} = \frac{\sqrt{3}}{4} + \frac{3\pi}{24} - \frac{4\pi}{24} = \frac{\sqrt{3}}{4} - \frac{\pi}{24}.$$

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

Shapes bounded by two circle arcs break into slices, triangles, and the pieces between a straight segment and an arc (a slice minus a triangle). The helpful extra lines are radii to the ends of the segment, which reveal the angle. A segment equal in length to the radius always makes an equilateral triangle and a $60^\circ$ angle.

Check the choices by size. The lune is part of the small half-circle, so its area is positive and less than $\frac{\pi}{8} \approx 0.39$. Choice (A) is about $0.52 - 0.43 < 0$, and (D) and (E) are bigger than $0.43$, so they are out. That leaves (B), about $0.43 - 0.26 = 0.17$, and (C), about $0.43 - 0.13 = 0.30$. The overlap computed above settles it as (C).

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