2007 AMC 10A Problem 24

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

Regions bounded by tangent segments and arcs are rarely computable directly; enclose them in a rectangle built from radii to the tangent points and subtract triangles and sectors. The tangent-radius right angle supplies the triangle's shape, and the triangle's acute angle tells you the sector's angle. Given $OA = 2\sqrt2$ with radius $2$, the $45^\circ$ angle is the designed simplification.

• Taking the sectors as $90$ degrees (quarter circles) instead of $45$ degrees, or forgetting that $OC = 2$ makes triangle $ACO$ isosceles right.
• Using $AB = 2\sqrt2$ (the half-length $OA$) for the rectangle's width, which halves the $8\sqrt2$ term.