Difference between revisions of "2017 AMC 10A Problems/Problem 11"
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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. | + | 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 has volume 216. What is the length ?
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 units of a point would be a sphere with radius . 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 ):
Where is equal to the length of our line segment.
We isolate . This comes out to be
Visualizing the Region
To envision what the region must look like, we simplify the problem to finding all points within 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 (Problems • Answer Key • Resources) | ||
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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