# 1979 AHSME Problems/Problem 16

## Problem 16

A circle with area $A_1$ is contained in the interior of a larger circle with area $A_1+A_2$. If the radius of the larger circle is $3$, and if $A_1 , A_2, A_1 + A_2$ is an arithmetic progression, then the radius of the smaller circle is

$\textbf{(A) }\frac{\sqrt{3}}{2}\qquad \textbf{(B) }1\qquad \textbf{(C) }\frac{2}{\sqrt{3}}\qquad \textbf{(D) }\frac{3}{2}\qquad \textbf{(E) }\sqrt{3}$

## Solution

Solution by e_power_pi_times_i

The area of the larger circle is $A_1 + A_2 = 9\pi$. Then $A_1 , 9\pi-A_1 , 9\pi$ are in an arithmetic progression. Thus $9\pi-(9\pi-A_1) = 9\pi-A_1-A_1$. This simplifies to $3A_1 = 9\pi$, or $A_1 = 3\pi$. The radius of the smaller circle is $\boxed{\textbf{(E) } \sqrt{3}}$.

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