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 (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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