2006 AMC 10B Problem 19

1. Put the square on a grid: $O = (0,0)$, $A = (1,0)$, $B = (1,1)$, $C = (0,1)$. The circle has radius $2$.
2. Find $D$. Side $AB$ lies on the line $x = 1$. Going up that line until the distance from $O$ is $2$: in right triangle $OAD$, $OA = 1$ and $OD = 2$, so $AD = \sqrt{4 - 1} = \sqrt3$. So $D = (1, \sqrt3)$. In the same way, $E = (\sqrt3, 1)$.
3. Find the angle $DOE$. Triangle $OAD$ has sides $1$, $\sqrt3$, $2$, so it is a $30$-$60$-$90$ triangle with the $60^\circ$ angle at $O$: $\angle AOD = 60^\circ$. By the same reasoning $\angle COE = 60^\circ$, so $\angle AOE = 90^\circ - 60^\circ = 30^\circ$. Then
$$\angle DOE = \angle AOD - \angle AOE = 60^\circ - 30^\circ = 30^\circ.$$
4. Draw $OD$ and $OE$. They cut out a slice (sector) of the circle. The slice is made of three parts: the shaded region, triangle $OBD$, and triangle $OBE$.
5. Slice area. A $30^\circ$ slice is $\frac{30}{360} = \frac1{12}$ of the circle: $\frac{1}{12} \cdot \pi \cdot 2^2 = \frac{\pi}{3}$.
6. Triangle $OBD$. Its side $BD$ runs along $x = 1$ from height $1$ to height $\sqrt3$, so $BD = \sqrt3 - 1$. The point $O$ is $1$ unit sideways from that line, so the height is $1$. Area $= \frac12(\sqrt3 - 1)(1) = \frac{\sqrt3 - 1}{2}$. Triangle $OBE$ is its mirror image, with the same area.
7. Shaded area = slice minus the two triangles:
$$\frac{\pi}{3} - 2\cdot\frac{\sqrt3 - 1}{2} = \frac{\pi}{3} - \sqrt3 + 1.$$

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

For a region with a curved edge, connect the circle's center to the ends of the arc. That makes a slice whose area is easy, and the rest of the slice is made of triangles. The lengths $1$ and $2$ give a $30$-$60$-$90$ triangle, which tells you both the slice angle and the length $\sqrt3 - 1$.

Quick check with the choices. The shaded region is only part of the $\frac{\pi}{3}$ slice, so its area is less than $\frac{\pi}{3}$. In (A), $1 - \sqrt3$ is negative, so (A) is less than $\frac{\pi}{3}$. In (E), $-1 + \sqrt3$ is positive, so (E) is too big. This rules out (E), which is what you get by adding the two thin triangles instead of taking them away.

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