Difference between revisions of "2004 AMC 12A Problems/Problem 22"

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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?
 
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?
  
<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>
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<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>
  
 
== Solution ==
 
== Solution ==
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[[Image:2004_AMC12A-22a.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>. 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>.
 
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>. 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}}
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[[Image:2004_AMC12A-22b.png]]
  
 
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 = \frac{\sqrt{69} + 9}{3} \Rightarrow \mathrm{(B)}</math>.
 
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 = \frac{\sqrt{69} + 9}{3} \Rightarrow \mathrm{(B)}</math>.

Revision as of 17:20, 4 December 2007

Problem

Three mutually tangent spheres of radius $1$ rest on a horizontal plane. A sphere of radius $2$ rests on them. What is the distance from the plane to the top of the larger sphere?

$\text {(A)}\ 3 + \frac {\sqrt {30}}{2} \qquad \text {(B)}\ 3 + \frac {\sqrt {69}}{3} \qquad \text {(C)}\ 3 + \frac {\sqrt {123}}{4}\qquad \text {(D)}\ \frac {52}{9}\qquad \text {(E)}\ 3 + 2\sqrt2$

Solution

2004 AMC12A-22a.png

The height from the center of the bottom sphere to the plane is $1$, and from the center of the top sphere to the tip is $2$. 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 $30-60-90 \triangle$s to be $\frac{2}{\sqrt{3}}$.

2004 AMC12A-22b.png

By the Pythagorean Theorem, we have $\left(\frac{2}{\sqrt{3}}\right)^2 + h^2 = 3^2 \Longrightarrow h = \frac{\sqrt{69}}{3}$. Adding the heights up, we get $\frac{\sqrt{69}}{3} + 1 + 2 = \frac{\sqrt{69} + 9}{3} \Rightarrow \mathrm{(B)}$.

See also

2004 AMC 12A (ProblemsAnswer KeyResources)
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
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