Difference between revisions of "2004 AMC 12A Problems/Problem 22"
NumberNinja (talk | contribs) m (→Solution) |
(→Problem) |
||
| Line 2: | Line 2: | ||
== Problem == | == Problem == | ||
| − | Three mutually [[tangent]] [[sphere]]s of [[radius]] <math>1</math> rest on a horizontal [[plane]]. A sphere of radius <math>2</math> rests on them. What is the [[distance]] from the plane to the top of the larger sphere? | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Three mutually [[tangent]] [[sphere]]s of [[radius]] <math>1</math> rest on a horizontal [[plane]]. A sphere of radius <math>2</math> rests on them. What is the [[distance]] from the plane to the top of the larger sphere?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
<math>\text {(A)}\ 3 + \frac {\sqrt {30}}{2} \qquad \text {(B)}\ 3 + \frac {\sqrt {69}}{3} \qquad \text {(C)}\ 3 + \frac {\sqrt {123}}{4}\qquad \text {(D)}\ \frac {52}{9}\qquad \text {(E)}\ 3 + 2\sqrt2</math> | <math>\text {(A)}\ 3 + \frac {\sqrt {30}}{2} \qquad \text {(B)}\ 3 + \frac {\sqrt {69}}{3} \qquad \text {(C)}\ 3 + \frac {\sqrt {123}}{4}\qquad \text {(D)}\ \frac {52}{9}\qquad \text {(E)}\ 3 + 2\sqrt2</math> | ||
Revision as of 17:53, 27 March 2015
- 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. 
