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

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== Solution ==
 
== Solution ==
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<math>\fbox{}</math>
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Since the new circle has twice the area of the original circle, its radius is <math>\sqrt{2}</math> times the old radius.
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Thus,
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<cmath>r + n = r\sqrt{2}</cmath>
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<cmath>n = r\sqrt{2} - r</cmath>
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<cmath>n = r(\sqrt{2} - 1)</cmath>
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<cmath>r = \frac{n}{\sqrt{2} - 1}</cmath>
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Rationalizing the denominator yields
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<cmath>r = \frac{n}{\sqrt{2} - 1} *  \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = n(\sqrt{2} + 1)</cmath>
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Therefore, the answer is <math>\fbox{(A)}</math>
  
 
== See Also ==
 
== See Also ==

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 $\fbox{(A)}$

See Also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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All AHSME Problems and Solutions

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