2017 AMC 10A Problems/Problem 11

Revision as of 21:28, 30 November 2020 by Virjoy2001 (talk | contribs) (Solution 2)


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 $\textit{AB}$?

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

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

Because this is just a cylinder and $2$ "half spheres", and the radius is $3$, the volume of the $2$ half spheres is $\frac{4(3^3)\pi}{3} = 36 \pi$. Since we also know that the volume of this whole thing is $216 \pi$, we do $216-36$ to get $180 \pi$ as the area of the cylinder. Thus the height is $180 \pi$ over the base, or $180 \pi/9\pi=20$, so our answer is $\boxed{\textbf{(D)}\ 20}.$

~Minor edit by virjoy2001

Diagram for Solution


Video Solution


See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AMC 10 Problems and Solutions

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