# 1958 AHSME Problems/Problem 18

## Problem

The area of a circle is doubled when its radius $r$ is increased by $n$. Then $r$ equals: $\textbf{(A)}\ n(\sqrt{2} + 1)\qquad \textbf{(B)}\ n(\sqrt{2} - 1)\qquad \textbf{(C)}\ n\qquad \textbf{(D)}\ n(2 - \sqrt{2})\qquad \textbf{(E)}\ \frac{n\pi}{\sqrt{2} + 1}$

## Solution

Since the new circle has twice the area of the original circle, its radius is $\sqrt{2}$ times the old radius. Thus, $$r + n = r\sqrt{2}$$ $$n = r\sqrt{2} - r$$ $$n = r(\sqrt{2} - 1)$$ $$r = \frac{n}{\sqrt{2} - 1}$$ Rationalizing the denominator yields $$r = \frac{n}{\sqrt{2} - 1} * \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = n(\sqrt{2} + 1)$$

Therefore, the answer is $\fbox{(A)}$

## See Also

 1958 AHSC (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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