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2017 AMC 10A Problem 11

2017 AMC 10A Problem 11AMC 10 Step by Step

Name: Date: September 13, 2026

Solid Geometry · difficulty 2 of 5 · about 3 min · position P11-15

The region consisting of all points in three-dimensional space within $3$ units of line segment $\overline{AB}$ has volume $216\pi$ . What is the length $\textit{AB}$ ?

A)
$6$
B)
$12$
C)
$18$
D)
$20$
E)
$24$

Answer

Key insight

Points within 3 of a segment form a cylinder of radius 3 capped by two hemispheres, i.e. a cylinder plus one full sphere of radius 3.

Solution

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 $\boxed{\textbf{(D)}\ 20}$ .

Why this works

"All points within distance $r$ of a shape" is the shape thickened by $r$ ; for a segment this is a capsule. Decompose it into pieces with known formulas: a cylinder for the interior and two hemispherical caps for the ends. The caps are the part students forget, and they contribute a full sphere's volume.

The trap

Forgetting the rounded ends and solving 9 pi L = 216 pi to get L = 24, or adding two full spheres instead of two hemispheres.

Common mistakes

Forgetting the rounded ends and solving 9 pi L = 216 pi to get L = 24, or adding two full spheres instead of two hemispheres.
Using diameter $3$ instead of radius $3$ in the formulas.

Techniques

Set up the equation/formula and compute; no special trick needed · Cut the figure into known shapes (triangles, rectangles, sectors)