# 1969 AHSME Problems/Problem 13

## Problem

A circle with radius $r$ is contained within the region bounded by a circle with radius $R$. The area bounded by the larger circle is $\frac{a}{b}$ times the area of the region outside the smaller circle and inside the larger circle. Then $R:r$ equals: $\text{(A) }\sqrt{a}:\sqrt{b} \quad \text{(B) } \sqrt{a}:\sqrt{a-b}\quad \text{(C) } \sqrt{b}:\sqrt{a-b}\quad \text{(D) } a:\sqrt{a-b}\quad \text{(E) } b:\sqrt{a-b}$

## Solution

The area of the larger circle is $\pi R^2$, and the area of the region outside the smaller circle and inside the larger circle $\pi R^2 - \pi r^2$. Thus, $$\pi R^2 = \frac{a}{b} \cdot (\pi R^2 - \pi r^2)$$ $$R^2 = \frac{a}{b} \cdot R^2 - \frac{a}{b} \cdot r^2$$ $$\frac{a}{b} \cdot r^2 = (\frac{a}{b} -1)R^2$$ $$\frac{a}{b} \div \frac{a-b}{b} = \frac{R^2}{r^2}$$ $$\frac{a}{a-b} = \frac{R^2}{r^2}$$ $$\frac{\sqrt{a}}{\sqrt{a-b}} = \frac{R}{r}$$ The answer is $\boxed{\textbf{(B)}}$.

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