2004 AMC 10B Problem 25

From the figure, the two large circles are centered at $A$ and $B$, with $AB = 2$ (a diameter of the unit circle). The unit circle lies entirely inside the overlap of the two large disks, touching its boundary only at $A$ and $B$. So the shaded area is
$$(\text{lens where the two large disks overlap}) - (\text{area of the unit disk}).$$

Let $C$ and $D$ be the two points where the large circles cross. Since $CA = CB = 2 = AB$, triangle $ABC$ is equilateral, and likewise $ABD$. Thus at center $A$ the chord $CD$ spans an angle $\angle CAD = 60^\circ + 60^\circ = 120^\circ$.

The chord $CD$ splits the lens into two congruent circular segments. The segment on circle $A$'s side is the $120^\circ$ sector $ACD$ minus triangle $ACD$:

• sector: $\frac{120}{360} \cdot \pi \cdot 2^2 = \frac{4\pi}{3}$;
• triangle $ACD$: its base $CD$ has length $2\sqrt{3}$ (twice the altitude of an equilateral triangle of side $2$) and its height from $A$ is $1$ (half of $AB$), so its area is $\frac{1}{2} \cdot 2\sqrt{3} \cdot 1 = \sqrt{3}$.

Each segment is $\frac{4\pi}{3} - \sqrt{3}$, so the lens is $\frac{8\pi}{3} - 2\sqrt{3}$. Removing the unit disk of area $\pi$:
$$\frac{8\pi}{3} - 2\sqrt{3} - \pi = \frac{5\pi}{3} - 2\sqrt{3}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{5}{3}\pi - 2\sqrt{3}}$.

Overlaps of circles are built from circular segments, and a segment is always "sector minus triangle." Joining the intersection points to the centers exposes the key angles; here the equal radii and center distance make equilateral triangles, so every angle is $60^\circ$ or $120^\circ$ and every length involves $\sqrt{3}$. Decompose first, subtract the excluded piece last.

Answer-choice reasoning under time pressure: choices (C), (D), (E) each exceed a choice from (A), (B) by exactly $\pi$, the area of the unit disk, so they are the "forgot to subtract" traps. Angles of $60^\circ$ and $120^\circ$ generate $\sqrt{3}$, never $\sqrt{2}$, which leaves (B).

• Forgetting to remove the unit disk (giving 8pi/3 - 2sqrt(3), choice (E)), or double-counting the equilateral triangles when adding sectors.
• Using a $60^\circ$ sector instead of $120^\circ$ for each segment, which halves the $\pi$ term.