2007 AMC 10A Problem 24

The shaded region has curved edges, so build a simple shape around it and subtract the extra pieces.

1. Make a rectangle. Draw radii $AE$ and $BF$. A radius to a point where a line touches the circle is perpendicular to that line, so $AE$ and $BF$ both stand straight up from $AB$, with length $2$. So $ABFE$ is a rectangle.
2. Rectangle area. $AB = 2 \times OA = 4\sqrt2$ and the height is $2$, so the area is $2 \times 4\sqrt2 = 8\sqrt2$.
3. What is extra? Near $A$, the rectangle also contains triangle $ACO$ and the slice of circle $A$ between radii $AC$ and $AE$. The same two pieces appear near $B$.
4. Triangle $ACO$. $OC$ touches circle $A$ at $C$, so $\angle ACO = 90^\circ$. With $AC = 2$ and $AO = 2\sqrt2$, the Pythagorean theorem gives
$$OC^2 = (2\sqrt2)^2 - 2^2 = 8 - 4 = 4, \qquad OC = 2.$$
Both legs are $2$, so the area is $\frac12 \times 2 \times 2 = 2$, and the angle at $A$ is $45^\circ$.
5. Slice $CAE$. $AE$ is at $90^\circ$ to $AB$, and $AC$ is at $45^\circ$ to $AB$, so the slice between them is $90^\circ - 45^\circ = 45^\circ$. That is $\frac{45}{360} = \frac18$ of the circle:
$$\frac18 \times \pi \times 2^2 = \frac{\pi}{2}.$$
6. Subtract both pairs. By symmetry, the pieces near $B$ are the same size:
$$8\sqrt2 - 2 \times 2 - 2 \times \frac{\pi}{2} = 8\sqrt2 - 4 - \pi.$$

The answer is $\boxed{\textbf{(B)}\ 8\sqrt {2} - 4 - \pi}$.

Areas bounded by arcs and tangent lines are rarely found directly. Draw radii to the touching points to make a rectangle, then take away triangles and circle slices. The right angle between a radius and a tangent line tells you the triangle's shape, and the triangle's angle tells you how big the slice is.

Draw radii $AC$, $AE$, $BD$, $BF$. Since $EF$ is tangent to both circles at their topmost points, $AE$ and $BF$ are perpendicular to $EF$ and to $AB$, so $ABFE$ is a rectangle with $AB = 2 \cdot 2\sqrt2 = 4\sqrt2$ and height $2$: area $8\sqrt2$.

The shaded region is this rectangle with four pieces cut away from the bottom corners and the middle: triangle $ACO$, sector $CAE$, triangle $BDO$, and sector $DBF$.

Triangle $ACO$: $OC$ is tangent at $C$, so $\angle ACO = 90^\circ$. With $AC = 2$ and $AO = 2\sqrt2$, the Pythagorean theorem gives $OC = 2$. So $ACO$ is an isosceles right triangle with area $\frac12 \cdot 2 \cdot 2 = 2$, and $\angle CAO = 45^\circ$.

Sector $CAE$: $\angle EAO = 90^\circ$ and $\angle CAO = 45^\circ$, so the sector spans $45^\circ$: area $\frac{45}{360} \cdot \pi \cdot 2^2 = \frac{\pi}{2}$.

By symmetry the pieces at $B$ are identical. Subtract everything:
$$8\sqrt2 - 2 \cdot 2 - 2 \cdot \frac{\pi}{2} = 8\sqrt2 - 4 - \pi.$$

The answer is $\boxed{\textbf{(B)}\ 8\sqrt {2} - 4 - \pi}$.