Difference between revisions of "2006 AMC 10B Problems/Problem 4"
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== Problem == | == Problem == | ||
| − | Circles of | + | Circles of diameter <math>1</math> inch and <math>3</math> inches have the same center. The smaller circle is painted red, and the portion outside the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area? |
[[Image:2006amc10b04.gif]] | [[Image:2006amc10b04.gif]] | ||
| − | <math> \ | + | <math> \textbf{(A) } 2\qquad \textbf{(B) } 3\qquad \textbf{(C) } 6\qquad \textbf{(D) } 8\qquad \textbf{(E) } 9 </math> |
== Solution == | == Solution == | ||
| − | The | + | The area painted red is equal to the area of the smaller circle and the area painted blue is equal to the area of the larger circle minus the area of the smaller circle. |
| − | So: | + | So we have: |
| − | < | + | <cmath>\begin{align*} |
| − | + | A_{red}&=\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4}\\ | |
| − | + | A_{blue}&=\pi\left(\frac{3}{2}\right)^2-\pi\left(\frac{1}{2}\right)^2=2\pi\\ | |
| + | \end{align*}</cmath> | ||
So the desired ratio is: | So the desired ratio is: | ||
| − | <math> \frac{2\pi}{\frac{\pi}{4}}= | + | <math> \frac{A_{blue}}{A_{red}}=\frac{2\pi}{\frac{\pi}{4}}=2\cancel{\pi}\cdot \frac{4}{\cancel{\pi}}=\boxed{\textbf{(D) }8}.</math> |
== See Also == | == See Also == | ||
| Line 23: | Line 24: | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 09:47, 19 December 2021
Problem
Circles of diameter 
inch and 
inches have the same center. The smaller circle is painted red, and the portion outside the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area?
Solution
The area painted red is equal to the area of the smaller circle and the area painted blue is equal to the area of the larger circle minus the area of the smaller circle.
So we have:
So the desired ratio is:
See Also
| 2006 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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. 
