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 $\boxed{\textbf{(B)}}$ .

1969 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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1969 AHSME Problems/Problem 13

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