Difference between revisions of "2017 AMC 10A Problems/Problem 11"
m (→Problem) |
m (→Solution) |
||
Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
− | 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>): | + | 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>): |
<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. | <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. |
Revision as of 20:20, 23 December 2019
Problem
The region consisting of all points in three-dimensional space within units of line segment
has volume
. What is the length
?
Solution
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
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.
Solving, we find that .
Solution 2
Because this is just a cylinder and "half spheres", and the radius is
, the volume of the
half spheres is
. Since we also know that the volume of this whole thing is
, we do
to get
as the area of the cylinder. Thus the height is
over the base, or
,
Diagram for Solution
http://i.imgur.com/cwNt293.png
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.