2019 AMC 10A Problem 16

The shaded area is the big circle minus $13$ small circles of area $\pi \cdot 1^2 = \pi$ each:
$$\text{shaded} = \pi R^2 - 13\pi,$$
where $R$ is the big radius. So everything comes down to finding $R$.

1. Touching circles. Two circles of radius $1$ that touch have centers exactly $1 + 1 = 2$ apart. So draw segments between the centers of touching circles.
2. Set up coordinates. Put the center of the middle circle at $O = (0,0)$. The middle row has circles at $(-2,0)$, $(0,0)$ and $(2,0)$.
3. The row just above. Each circle in this row sits in the dip between two circles of the middle row. Its center, with the two centers below it, makes an equilateral triangle of side $2$. The height of that triangle is $\sqrt{2^2 - 1^2} = \sqrt3$. So the two inner circles of this row have centers $(-1, \sqrt3)$ and $(1, \sqrt3)$.
4. The top circle sits in the dip between those two, so by the same reasoning its center is another $\sqrt3$ higher: $T = (0, 2\sqrt3)$.
5. The big radius. The top circle touches the big circle, and the big circle is centered at $O$. So $R$ is the distance from $O$ to $T$ plus the top circle's radius:
$$R = 2\sqrt3 + 1.$$
(The outer circles of the upper row, at $(\pm3, \sqrt3)$, are also $\sqrt{9 + 3} = 2\sqrt3$ from $O$, so they touch the big circle too, matching the picture.)
6. Area.
$$R^2 = (2\sqrt3 + 1)^2 = 12 + 4\sqrt3 + 1 = 13 + 4\sqrt3.$$
$$\text{shaded} = \pi(13 + 4\sqrt3) - 13\pi = 4\sqrt3\,\pi.$$

The answer is $\boxed{\textbf{(A)}\ 4\pi\sqrt3}$.

With touching circles, always connect the centers: touching turns into a known distance (the sum of the radii). Circles packed like this make equilateral triangles of side $2$, whose height $\sqrt3$ gives every other distance. Notice how neatly it ends: the $13$ in $R^2$ cancels the $13$ small circles.

Work in the triangle $O$, $(1, \sqrt3)$, $T$ without coordinates. Both sides from the point $(1,\sqrt3)$ have length $2$, and the angle there is $120^\circ$ (two $60^\circ$ angles from equilateral triangles). Cut this triangle down the middle into two $30$-$60$-$90$ triangles with hypotenuse $2$; the long leg of each is $\sqrt3$. So $OT = 2\sqrt3$, the same as before.

The shaded area is $\pi R^2 - 13\pi$ where $R$ is the radius of the large circle, so we only need $R$.

Let $O$ be the common center of the large circle and the middle small circle. Tangent unit circles have centers $2$ apart. In the picture the circles are arranged in a hexagonal packing: six circles surround the middle one at distance $2$ from $O$, and the topmost circle touches two of those (the ones at the upper left and upper right).

Take the top circle's center $T$ and the two neighbors' centers $L$ and $R'$. Then $OL = LT = 2$, and $L$ sits at $60^\circ$ above the horizontal, while $T$ is straight above $O$. Triangle $OLT$ has $OL = LT = 2$ with $\angle OLT = 120^\circ$, so it splits into two $30$-$60$-$90$ triangles and
$$OT = 2 \cdot 2\cos 30^\circ = 2\sqrt3.$$
The top circle is tangent to the big circle, so $R = OT + 1 = 2\sqrt3 + 1$.

Shaded area:
$$\pi(2\sqrt3 + 1)^2 - 13\pi = \pi(12 + 4\sqrt3 + 1) - 13\pi = 4\sqrt3\,\pi.$$

The answer is $\boxed{\textbf{(A)}\ 4\pi\sqrt3}$.