2017 AMC 10A Problem 11

First picture the shape, then split it into pieces whose volumes you know.

1. Next to the segment. Stand anywhere along the segment. The points within $3$ units of it, measured straight out sideways, fill a circle of radius $3$. Stack those circles along the whole length and you get a solid cylinder of radius $3$ and length $AB$.
2. Past the ends. Beyond endpoint $A$, the closest point of the segment is $A$ itself. So the points there within $3$ units form half of a ball of radius $3$ centered at $A$. The same happens at $B$.
3. Put the ends together. Two half-balls of radius $3$ make one full ball of radius $3$. So the region is a cylinder plus one ball, like a capsule or pill.
4. Volumes. Call the length $L$.
- Cylinder: $\pi r^2 \times \text{length} = \pi \cdot 3^2 \cdot L = 9\pi L$.
- Ball: $\tfrac43 \pi r^3 = \tfrac43 \pi \cdot 27 = 36\pi$.
5. Set the total equal to $216\pi$:
$$9\pi L + 36\pi = 216\pi.$$
Divide everything by $\pi$: $9L + 36 = 216$, so $9L = 180$ and $L = 20$.

The answer is $\boxed{\textbf{(D)}\ 20}$.

"All points within a distance of a shape" means the shape puffed out by that distance. Around a segment, that is a cylinder with rounded caps. Break any such solid into familiar pieces (cylinder, half-balls) and add their volumes. The caps are what people forget, and together they are worth one full ball.

Picture the set of points within $3$ units of the segment. Alongside the segment it is a solid cylinder of radius $3$ whose length equals $AB$. Beyond each endpoint it is a half-ball of radius $3$ (all points within $3$ of that endpoint on the far side). The two half-balls together make one full sphere of radius $3$.

Let $L = AB$. The volume is
$$\pi \cdot 3^2 \cdot L + \frac{4}{3}\pi \cdot 3^3 = 9\pi L + 36\pi.$$
Setting this equal to $216\pi$:
$$9\pi L = 180\pi \quad\Longrightarrow\quad L = 20.$$

The answer is $\boxed{\textbf{(D)}\ 20}$.