Difference between revisions of "1958 AHSME Problems/Problem 18"
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== Solution == | == Solution == | ||
− | <math>\fbox{}</math> | + | Since the new circle has twice the area of the original circle, its radius is <math>\sqrt{2}</math> times the old radius. |
+ | Thus, | ||
+ | <cmath>r + n = r\sqrt{2}</cmath> | ||
+ | <cmath>n = r\sqrt{2} - r</cmath> | ||
+ | <cmath>n = r(\sqrt{2} - 1)</cmath> | ||
+ | <cmath>r = \frac{n}{\sqrt{2} - 1}</cmath> | ||
+ | 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) n(\sqrt{2} + 1)}</math> | ||
== See Also == | == See Also == |
Revision as of 01:35, 22 December 2015
Problem
The area of a circle is doubled when its radius is increased by
. Then
equals:
Solution
Since the new circle has twice the area of the original circle, its radius is times the old radius.
Thus,
Rationalizing the denominator yields
Therefore, the answer is $\fbox{(A) n(\sqrt{2} + 1)}$ (Error compiling LaTeX. Unknown error_msg)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.