2013 AMC 10A Problem 22

Let $O$ be the center of the hexagon. In a regular hexagon of side $2$, each vertex is at distance $2$ from the center, so each small sphere's center is $2$ from $O$. The small spheres are internally tangent to the big sphere, so its radius is $R = 2 + 1 = 3$.

By symmetry the eighth sphere's center $P$ lies on the axis through $O$ perpendicular to the hexagon's plane. Let its radius be $r$ and let $OP = h$.

Internal tangency to the big sphere: $h + r = 3$, so $h = 3 - r$.

External tangency to a small sphere with center $S$: $PS = r + 1$. Triangle $OPS$ has a right angle at $O$ with legs $OS = 2$ and $OP = h$, so
$$(3 - r)^2 + 2^2 = (r + 1)^2.$$
Expanding: $9 - 6r + r^2 + 4 = r^2 + 2r + 1$, hence $12 = 8r$ and $r = \frac{3}{2}$.

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

Sphere tangency is entirely about distances between centers: sum of radii when externally tangent, difference when internally tangent. Connecting the centers produces a right triangle whose legs are the hexagon's circumradius and the height of the new sphere above the plane, and a single Pythagorean equation finishes the job. Draw the vertical cross-section through $O$ and one small center; the problem is then planar.

• Putting the eighth sphere in the plane of the hexagon (at the center, radius 1) and forgetting it can rise above the plane to grow larger.
• Using $h - r = 3$ or $PS = r - 1$, mixing up internal and external tangency.