2004 AMC 10B Problem 16

Circles are hard to work with directly, so work with their centers.

1. Call the centers of the small circles $A$, $B$, $C$. Two small circles that touch have centers $1 + 1 = 2$ apart. So $ABC$ is an equilateral triangle with all sides $2$.
2. By symmetry, the big circle's center $S$ is the center of this triangle.
3. Find the height of the triangle. Cut it in half down the middle: you get a right triangle with longest side $2$ and short side $1$, so the height is $\sqrt{2^2 - 1^2} = \sqrt{3}$.
4. The center of an equilateral triangle sits two thirds of the way down the height from a corner. So
$$SA = \frac{2}{3}\sqrt{3}.$$
5. Now go from $S$ through $A$ to the point where the small circle touches the big one. That point is one more unit past $A$ (the small radius). So the big radius is
$$R = \frac{2}{3}\sqrt{3} + 1 = \frac{2\sqrt{3}}{3} + \frac{3}{3} = \frac{3 + 2\sqrt{3}}{3}.$$
6. Rough check: $\sqrt{3} \approx 1.73$, so $R \approx 1 + 1.15 = 2.15$. That is a bit more than $2$, which makes sense.

The answer is $\boxed{\textbf{(D)}\ \frac{3 + 2\sqrt{3}}{3}}$.

"Tangent" is really a statement about the distance between centers. Two circles touching from outside have centers (radius + radius) apart; a small circle touching a big one from inside has its center (big radius - small radius) from the big center. Replacing circles by their centers turns the picture into a triangle, and a symmetric picture puts the big center at the center of that triangle.

Estimate and use the choices. Two small circles side by side already stretch $4$ units from end to end, so the big circle's diameter is more than $4$ once the third circle has to fit too; the radius is a bit more than $2$. The choices are about (A) $1.48$, (B) $2$, (C) $3.12$, (D) $2.15$, (E) $2.37$. The estimate $2.15$ from the triangle picture matches (D).