Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

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

(→‎See also)
(Created higher-quality images)
Line 9: Line 9:
 
We draw the three spheres of radius <math>1</math>:
 
We draw the three spheres of radius <math>1</math>:
  
−
[[File:2004AMC12A_22_1.png]]
+
[[File:2004AMC12A_22_1_rerender.png]]
  
 
And then add the sphere of radius <math>2</math>:
 
And then add the sphere of radius <math>2</math>:
  
−
[[File:2004AMC12A_22_2.png]]
+
[[File:2004AMC12A_22_2_rerender.png]]
  
 
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>.
 
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>.

Revision as of 13:12, 17 January 2016

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 $1$ 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

We draw the three spheres of radius $1$:

2004AMC12A 22 1 rerender.png

And then add the sphere of radius $2$:

2004AMC12A 22 2 rerender.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 $2$.

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 $30-60-90 \triangle$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 $\frac{\sqrt{69}}{3} + 1 + 2 \Rightarrow\boxed{\mathrm{(B)}\ 3+\frac{\sqrt{69}}{3}}$

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
Last 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12A_Problems/Problem_22&oldid=74614"