2006 AMC 10B Problem 19

Place $O$ at the origin with $A = (1, 0)$, $C = (0, 1)$, $B = (1, 1)$. Then $D$ is on the line $x = 1$ and on the circle, so $D = (1, \sqrt3)$; similarly $E = (\sqrt3, 1)$.

In right triangle $OAD$, the legs are $1$ and $\sqrt3$ with hypotenuse $2$, so $\angle AOD = 60^\circ$. By the same token $\angle COE = 60^\circ$, so $\angle EOA = 30^\circ$ and $\angle DOE = 60^\circ - 30^\circ = 30^\circ$.

Draw $OD$ and $OE$. The sector $ODE$ is made of the shaded region plus triangles $OBD$ and $OBE$.

• Sector: $\frac{30}{360} \cdot \pi \cdot 2^2 = \frac{\pi}{3}$.
• Triangle $OBD$: base $BD = \sqrt3 - 1$ on the line $x = 1$, and $O$ is at distance $1$ from that line, so area $\frac{1}{2}(\sqrt3 - 1)(1)$. Triangle $OBE$ is its mirror image with the same area.

Shaded area $= \frac{\pi}{3} - 2 \cdot \frac{\sqrt3 - 1}{2} = \frac{\pi}{3} - \sqrt3 + 1$.

The answer is $\boxed{\textbf{(A)}\ \frac{\pi}{3} + 1 - \sqrt3}$.

Regions bounded by an arc and segments are handled by connecting the center to the arc's endpoints: the sector is easy, and what remains are triangles. The lengths $1$ and $2$ produce a $30$-$60$-$90$ triangle, which fixes both the sector angle and the segment $\sqrt3 - 1$. Choosing triangles $OBD$ and $OBE$ (rather than $ODE$ and $DBE$) keeps the arithmetic minimal.

Both (A) and (E) contain the sector's $\frac{\pi}{3}$, but the shaded region is a proper part of the sector, so its area must be less than $\frac{\pi}{3}$. Only (A) qualifies.

• Subtracting triangle ODE (area 1) from the sector and stopping, forgetting to add back the small triangle DBE, which gives pi/3 - 1.
• Taking $\angle DOE$ to be $60^\circ$ or $45^\circ$ instead of $30^\circ$, which changes the sector to $\frac{2\pi}{3}$ or $\frac{\pi}{2}$.