, 2013

Solution:

Answer: $8 r^{2}$

Let $U$ denote the union of the four circles, so we seek
$$U-([O]-U)=2U-[O]=2\left[(2r)^{2}+4 \cdot \frac{1}{2} \pi r^{2}\right]-\pi(2r)^{2}=8 r^{2}$$
(Here we decompose $U$ into the square $S$ with vertices at $(\pm r, \pm r)$ and the four semicircular regions of radius $r$ bordering the four sides of $U$.)

Solution:

There are three different kinds of regions: let $x$ be an area of a small circle that does not contain the two intersections with the other two small circles, $y$ be an area of intersection of two small circles, and $z$ be one of those four areas that is inside the big circle but outside all of the small circles.

Then the key observation is $y = z$. Indeed, adopting the union $U$ notation from the previous solution, we have $4z = [O] - U = \pi(2r)^{2} - U$, and by the inclusion-exclusion principle, $4y = [A] + ** + [C] + [D] - U = 4\pi r^{2} - U$, so $y = z$. Now $U = (4x + 4y) - (4z) = 4x$. But the area of each $x$ is simply $2r^{2}$ by moving the curved outward parts to fit into the curved inward parts to get a $r \times 2r$ rectangle. So the answer is $8 r^{2}$.