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

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(Solution)
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<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24</math>
 
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24</math>
  
==Solution==
+
==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 3 units of a segment.  Therefore, 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 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 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 <math>216 \pi</math>):
  
 
<math>\frac{4 \pi }{3}3^3+9 \pi x=216</math>
 
<math>\frac{4 \pi }{3}3^3+9 \pi x=216</math>
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We isolate <math>x</math>.  This comes out to be <math>\boxed{\textbf{(D)}\ 20}</math>
 
We isolate <math>x</math>.  This comes out to be <math>\boxed{\textbf{(D)}\ 20}</math>
 +
 +
==Visualizing the Region==
 +
To envision what the region must look like, we simplify the problem to finding all points within <math>3</math> 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.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=A|num-b=10|num-a=12}}
 
{{AMC10 box|year=2017|ab=A|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:54, 8 February 2017

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 $216 \pi$):

$\frac{4 \pi }{3}3^3+9 \pi x=216$

Where $x$ is equal to the length of our line segment.

We isolate $x$. This comes out to be $\boxed{\textbf{(D)}\ 20}$

Visualizing the Region

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.

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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