# 2006 AMC 10B Problems/Problem 4

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 