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Difference between revisions of "2006 AMC 10B Problems/Problem 4"

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== Problem ==
 
== Problem ==
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Circles of [[diameter]] 1 inch and 3 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?  
+
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> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 9 </math>
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<math> \textbf{(A) } 2\qquad \textbf{(B) } 3\qquad \textbf{(C) } 6\qquad \textbf{(D) } 8\qquad \textbf{(E) } 9 </math>
  
 
== Solution ==
 
== 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.  
+
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:
  
−
<math> A_{red}=\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4} </math>
+
<cmath>\begin{align*}
−
 
+
A_{red}&=\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4}\\
−
<math> A_{blue}=\pi\left(\frac{3}{2}\right)^2-\pi\left(\frac{1}{2}\right)^2=2\pi </math>
+
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:  
  
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<math> \frac{2\pi}{\frac{\pi}{4}}=8\Rightarrow D </math>
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<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 ==

Latest revision as of 09:47, 19 December 2021

Problem

Circles of diameter $1$ inch and $3$ 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?

2006amc10b04.gif

$\textbf{(A) } 2\qquad \textbf{(B) } 3\qquad \textbf{(C) } 6\qquad \textbf{(D) } 8\qquad \textbf{(E) } 9$

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:

\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*}

So the desired ratio is:

$\frac{A_{blue}}{A_{red}}=\frac{2\pi}{\frac{\pi}{4}}=2\cancel{\pi}\cdot \frac{4}{\cancel{\pi}}=\boxed{\textbf{(D) }8}.$

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. AMC logo.png

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