Difference between revisions of "2017 AMC 10A Problems/Problem 11"

Revision as of 22:18, 31 July 2021

Problem

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 looks like. We know that, in a three dimensional space, 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 volume of the cylinder. Thus the height is $180 \pi$ divided by the area of the base, or $\frac{180 \pi}{9\pi}=20$ , so our answer is $\boxed{\textbf{(D)}\ 20}.$

~Minor edit by virjoy2001

Diagram for Solution

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

Video Solution

https://youtu.be/s4vnGlwwHHw

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 7: / Line 7: @@
 ==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 <math>3</math> units of a point would be a sphere with radius <math>3</math>.  However, we need to find the region containing all points within <math>3</math> 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 <math>216 \pi</math>):
+In order to solve this problem, we must first visualize what the region looks like. We know that, in a three dimensional space, the region consisting of all points within <math>3</math> units of a point would be a sphere with radius <math>3</math>.  However, we need to find the region containing all points within <math>3</math> 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 <math>216 \pi</math>):
 <math>\frac{4 \pi }{3} \cdot 3^3+9 \pi x=216 \pi</math>, where <math>x</math> is equal to the length of our line segment.

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