2019 AMC 10A Problem 16

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

Tangent-circle figures are solved by connecting centers: tangency converts to fixed distances (sum of radii), and the centers form triangles with known sides. In a hexagonal packing the centers lie on equilateral triangles, so $30$-$60$-$90$ ratios give every distance. The final subtraction $13$ small circles from one big one is bookkeeping once $R$ is known.

Coordinates: put $O$ at the origin. Neighbors of the center circle are at $(\pm1, \sqrt3)$, and the top circle, tangent to both, has center $(0, y)$ with $1 + (y - \sqrt3)^2 = 4$, so $y = 2\sqrt3$. Hence $R = 2\sqrt3 + 1$ and the area is $4\sqrt3\pi$.

• Assuming the big radius is 5 (from the vertical column of five circles as if they were stacked), giving 25 pi - 13 pi = 12 pi.
• Taking $R = 2\sqrt3$ (the distance to the top center) and forgetting to add the top circle's radius, which gives $12\pi - 13\pi < 0$.