Difference between revisions of "2004 AMC 12A Problems/Problem 22"
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| − | By the [[Pythagorean Theorem]], we have <math>\left(\frac{2}{\sqrt{3}}\right)^2 + h^2 = 3^2 \Longrightarrow h = \frac{\sqrt{69}}{3}</math>. Adding the heights up, we get <math>\frac{\sqrt{69}}{3} + 1 + 2 | + | By the [[Pythagorean Theorem]], we have <math>\left(\frac{2}{\sqrt{3}}\right)^2 + h^2 = 3^2 \Longrightarrow h = \frac{\sqrt{69}}{3}</math>. Adding the heights up, we get <math>\frac{\sqrt{69}}{3} + 1 + 2 \Rightarrow\boxed{\mathrm{(B)}\ 3+\frac{\sqrt{69}}{3}}</math> |
== See also == | == See also == | ||
Revision as of 01:05, 6 December 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. 
