2004 AMC 10A Problem 25

Work with the centers. Each small sphere touches the plane, so its center is at height $1$. The three small spheres are mutually tangent, so their centers form an equilateral triangle of side $1 + 1 = 2$ lying in the plane at height $1$.

The large sphere touches each small sphere, so its center is at distance $2 + 1 = 3$ from each small center. Being equidistant from the three vertices, it lies on the vertical line through the triangle's center. The distance from the center of an equilateral triangle of side $2$ to a vertex (its circumradius) is $\frac{2}{\sqrt3}$, since the circumradius equals $\frac{s}{\sqrt3}$.

Let $h$ be the vertical rise from the small centers up to the large center. In the right triangle formed by a small center, the triangle's center, and the large center:
$$h^2 + \left(\frac{2}{\sqrt3}\right)^2 = 3^2 \quad\Longrightarrow\quad h^2 = 9 - \frac43 = \frac{23}{3} \quad\Longrightarrow\quad h = \sqrt{\frac{23}{3}} = \frac{\sqrt{69}}{3}.$$

Stack the heights: plane to small centers is $1$, small centers to large center is $h$, and large center to the top of the big sphere is its radius $2$. The total is
$$1 + \frac{\sqrt{69}}{3} + 2 = 3 + \frac{\sqrt{69}}{3}.$$

The answer is $\boxed{\textbf{(B)}\ 3 + \frac{\sqrt{69}}{3}}$.

Sphere-packing problems reduce to their centers: tangency fixes every center-to-center distance, and the surfaces contribute only the radii at the ends. The resulting skeleton here is a tetrahedron with an equilateral base and three equal lateral edges, whose height comes from one right triangle involving the base's circumradius. Splitting the vertical distance into "radius, center-to-center rise, radius" keeps the bookkeeping straight.

• Using the inradius 1/sqrt(3) (or the side 2) instead of the circumradius 2/sqrt(3) for the horizontal offset, or forgetting to add the radii 1 and 2.
• Placing the large center at distance $2$ (its own radius) from the small centers rather than $3$, the sum of the radii.