2004 AMC 10A Problem 25

Spheres that touch are easiest to handle through their centers, so find all the centers first.

1. Small centers. Each small sphere sits on the plane, so its center is $1$ above the plane. The three small spheres touch each other, so their centers are $1 + 1 = 2$ apart. They form an equilateral triangle with side $2$, lying flat at height $1$.
2. Big center. The big sphere touches each small one, so its center is $2 + 1 = 3$ from each small center. Being the same distance from all three, it sits directly above the middle point of the triangle.
3. Middle to corner. In an equilateral triangle with side $2$, the height is $\sqrt{2^2 - 1^2} = \sqrt3$. The middle point is $\tfrac23$ of the way down from a corner along that height, so it is $\tfrac23 \sqrt3 = \tfrac{2}{\sqrt3}$ from each corner.
4. The rise. Let $h$ be how far the big center is above the small centers. The big center, a small center and the middle point form a right triangle with long side $3$ and flat leg $\tfrac{2}{\sqrt3}$:
$$h^2 + \left(\frac{2}{\sqrt3}\right)^2 = 3^2 \quad\Longrightarrow\quad h^2 = 9 - \frac43 = \frac{23}{3}.$$
So $h = \sqrt{\dfrac{23}{3}} = \sqrt{\dfrac{69}{9}} = \dfrac{\sqrt{69}}{3}$.
5. Stack the heights. From the plane up to the top of the big sphere:
- plane to small centers: $1$,
- small centers to big center: $\dfrac{\sqrt{69}}{3}$,
- big center to its top: its radius, $2$.

The total is
$$1 + \frac{\sqrt{69}}{3} + 2 = 3 + \frac{\sqrt{69}}{3}.$$

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

With touching spheres, the centers carry all the information: every pair of touching spheres fixes a center-to-center distance, and the radii only matter at the very bottom and top. Here the centers form a pyramid with a triangle base, and one right triangle gives its height. Splitting the total height into "radius, rise, radius" keeps the bookkeeping clean.

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