Difference between revisions of "1958 AHSME Problems/Problem 18"

Latest revision as of 00:40, 22 December 2015

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 $\fbox{(A)}$

1958 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 18: / Line 18: @@
 Rationalizing the denominator yields
 <cmath>r = \frac{n}{\sqrt{2} - 1} *  \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = n(\sqrt{2} + 1)</cmath>
+Therefore, the answer is <math>\fbox{(A)}</math>
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Difference between revisions of "1958 AHSME Problems/Problem 18"

Latest revision as of 00:40, 22 December 2015

Problem

Solution

See Also