Difference between revisions of "2004 AMC 12A Problems/Problem 22"
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== Solution == | == Solution == | ||
+ | We draw the three spheres of radius <math>1</math>: | ||
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+ | [[File:2004AMC12A_22_1.png]] | ||
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+ | And then add the sphere of radius <math>2</math>: | ||
+ | |||
+ | [[File:2004AMC12A_22_2.png]] | ||
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+ | The height from the center of the bottom sphere to the plane is <math>1</math>, and from the center of the top sphere to the tip is <math>2</math>. | ||
+ | |||
[[Image:2004_AMC12A-22a.png]] | [[Image:2004_AMC12A-22a.png]] | ||
− | + | We now need the vertical height of the centers. If we connect the centers, we get a triangular [[pyramid]] with an [[equilateral triangle]] base. The distance from the vertex of the equilateral triangle to its [[centroid]] can be found by <math>30-60-90 \triangle</math>s to be <math>\frac{2}{\sqrt{3}}</math>. | |
[[Image:2004_AMC12A-22b.png]] | [[Image:2004_AMC12A-22b.png]] |
Revision as of 12:07, 28 November 2014
- The following problem is from both the 2004 AMC 12A #22 and 2004 AMC 10A #25, so both problems redirect to this page.
Problem
Three mutually tangent spheres of radius rest on a horizontal plane. A sphere of radius rests on them. What is the distance from the plane to the top of the larger sphere?
Solution
We draw the three spheres of radius :
And then add the sphere of radius :
The height from the center of the bottom sphere to the plane is , and from the center of the top sphere to the tip is .
We now need the vertical height of the centers. If we connect the centers, we get a triangular pyramid with an equilateral triangle base. The distance from the vertex of the equilateral triangle to its centroid can be found by s to be .
By the Pythagorean Theorem, we have . Adding the heights up, we get
See also
2004 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 12 Problems and Solutions |
2004 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Final Question | |
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.