2004 AMC 10B Problem 16

Let $A, B, C$ be the centers of the three unit circles. Each pair of circles is tangent, so each pair of centers is $1 + 1 = 2$ apart: $ABC$ is an equilateral triangle of side $2$.

By symmetry the large circle's center $S$ is the center of this triangle. The altitude of an equilateral triangle of side $2$ is $\sqrt{3}$, and the center lies two-thirds of the way from a vertex along the median, so
$$SA = \frac{2}{3}\sqrt{3} = \frac{2}{\sqrt{3}}.$$

The large circle is internally tangent to the circle centered at $A$, so its radius $R$ satisfies $R - 1 = SA$ (the point of tangency lies on ray $SA$, one unit beyond $A$). Therefore
$$R = 1 + \frac{2}{\sqrt{3}} = 1 + \frac{2\sqrt{3}}{3} = \frac{3 + 2\sqrt{3}}{3}.$$

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

Tangency statements are distance statements between centers: external tangency means the centers are the sum of the radii apart, internal tangency means the difference. Drawing the centers replaces the circles by a triangle, and symmetric packings put the enclosing center at the triangle's centroid, whose distances to the vertices are standard.

Eliminate by size. The big circle must contain two small circles whose far points are $2 + 1 + 1 = 4$ apart, so $R \ge 2$, and $R = 2$ would force those two circles to touch the big circle at opposite ends of a diameter, leaving no room for the third; this rules out (A) and (B). Estimating $R \approx 1 + 1.155 = 2.15$ leaves (D), since (E) $\approx 2.37$ and (C) $\approx 3.12$.

• Using the full altitude sqrt(3) (or half of it) for the distance from the triangle's center to a vertex, instead of two-thirds of the altitude.
• Forgetting to add the unit radius at the end and answering $\frac{2\sqrt{3}}{3}$, or adding $2$ instead of $1$.