2017 AMC 10A Problems/Problem 11

Problem

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

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$

Solution 1

In order to solve this problem, we must first visualize what the region contained looks like. We know that, in a three dimensional plane, the region consisting of all points within $3$ units of a point would be a sphere with radius $3$ . However, we need to find the region containing all points within 3 units of a segment. It can be seen that our region is a cylinder with two hemispheres on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal $216 \pi$ ):

$\frac{4 \pi }{3} \cdot 3^3+9 \pi x=216 \pi$ , where $x$ is equal to the length of our line segment.

Solving, we find that $x = \boxed{\textbf{(D)}\ 20}$ .

Solution 2

To envision what the region must look like, we simplify the problem to finding all points within $3$ units from a point. This is a sphere. To account for the line, we drag the sphere's center across the line, sweeping out the desired volume. As stated above, this is a cylinder with two hemispheres on both ends.

Diagram

http://i.imgur.com/cwNt293.png

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2017 AMC 10A Problems/Problem 11

Contents

Problem

Solution 1

Solution 2

Diagram

See Also