Difference between revisions of "1958 AHSME Problems/Problem 18"
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The area of a circle is doubled when its radius <math> r</math> is increased by <math> n</math>. Then <math> r</math> equals: | The area of a circle is doubled when its radius <math> r</math> is increased by <math> n</math>. Then <math> r</math> equals: | ||
| − | <math> \textbf{(A)}\ n(\sqrt{2} | + | <math> \textbf{(A)}\ n(\sqrt{2} + 1)\qquad |
| − | \textbf{(B)}\ n(\sqrt{2} | + | \textbf{(B)}\ n(\sqrt{2} - 1)\qquad |
\textbf{(C)}\ n\qquad | \textbf{(C)}\ n\qquad | ||
| − | \textbf{(D)}\ n(2 | + | \textbf{(D)}\ n(2 - \sqrt{2})\qquad |
| − | \textbf{(E)}\ \frac{n\pi}{\sqrt{2} | + | \textbf{(E)}\ \frac{n\pi}{\sqrt{2} + 1}</math> |
== 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)}</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 
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,
![\[r + n = r\sqrt{2}\]](http://latex.artofproblemsolving.com/3/8/a/38a95ad1b0b19e8d52644cd0ff4a5306096f1cbc.png)
![\[n = r\sqrt{2} - r\]](http://latex.artofproblemsolving.com/3/5/3/35345cf545001faecf14fed200491db39d34c19e.png)
![\[n = r(\sqrt{2} - 1)\]](http://latex.artofproblemsolving.com/4/6/e/46e0cbdbb3f4c9c42cb06a9e7af2ce9761aad6e5.png)
![\[r = \frac{n}{\sqrt{2} - 1}\]](http://latex.artofproblemsolving.com/2/f/2/2f20b3a5424b38a52436ad02c48b10c65d5ca6af.png)
Rationalizing the denominator yields
![\[r = \frac{n}{\sqrt{2} - 1} * \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = n(\sqrt{2} + 1)\]](http://latex.artofproblemsolving.com/8/5/8/85862773da08a3512a5251f9f4bdbd3b9356de09.png)
Therefore, the answer is 
See Also
| 1958 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
