2013 AMC 10A Problem 22

For touching spheres, only the distances between centers matter:
- two spheres touching from outside: centers are (radius $+$ radius) apart;
- a small sphere touching a big one from inside: centers are (big radius $-$ small radius) apart.

1. The small spheres' centers. A regular hexagon with side $2$ is made of $6$ equilateral triangles with side $2$. So each corner is $2$ away from the center $O$.
2. The big sphere. Each small sphere (radius $1$) touches the big sphere from inside, and its center is $2$ from $O$. So the big radius is $2 + 1 = 3$.
3. Where the eighth sphere goes. It touches all six small spheres the same way, so its center $P$ sits on the line through $O$ that stands straight up from the hexagon's flat plane. Call its radius $r$ and the height $OP = h$.
4. Touching the big sphere. From inside: $h = 3 - r$.
5. Touching a small sphere. Let $S$ be one small center. Then $PS = r + 1$. The triangle $O\,P\,S$ has a right angle at $O$ (one leg lies flat, the other goes straight up), with legs $OS = 2$ and $OP = 3 - r$.
6. Pythagoras.
$$(3 - r)^2 + 2^2 = (r + 1)^2.$$
Multiply out: $9 - 6r + r^2 + 4 = r^2 + 2r + 1$. The $r^2$ cancels: $13 - 6r = 2r + 1$, so $12 = 8r$ and $r = \tfrac32$.
7. Check. $h = 3 - \tfrac32 = \tfrac32$, and $PS = \sqrt{2^2 + (\tfrac32)^2} = \sqrt{\tfrac{25}{4}} = \tfrac52 = r + 1$. It fits.

The answer is $\boxed{\textbf{(B)}\ \frac{3}{2}}$.

Touching spheres (or circles) are all about the distance between centers: add the radii when they touch from outside, subtract when one is inside the other. Slice the picture with a flat cut through $O$, $P$ and one small center, and the 3D problem becomes a right triangle in a plane.