Difference between revisions of "2017 AMC 10A Problems/Problem 11"
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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> | ||
| − | == | + | ==Solution 2== |
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. | 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. | ||
Revision as of 17:44, 8 February 2017
Contents
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 
Solution 2
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
