Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 11"

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==Solution==
 
==Solution==
<math>X=\frac{1}{2007 \sqrt{2006}+2006 \sqrt{2007}}</math>
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Define <math>a = 2007,\ b = 2006</math>.
<math>=\frac{1}{2007 \sqrt{2006}+2006 \sqrt{2007}}\cdot\frac{2007 \sqrt{2006}-2006 \sqrt{2007}}{2007 \sqrt{2006}-2006 \sqrt{2007}}</math>
 
<math>=\frac{2007\sqrt{2006}-2006\sqrt{2007}}{2006 \cdot 2007}</math>
 
<math>Y=\frac{1}{\sqrt{2006}}-\frac{1}{\sqrt{2007}}=\frac{\sqrt{2006}}{2006}-\frac{\sqrt{2007}}{2007}=\frac{2007\sqrt{2006}-2006\sqrt{2007}}{2006 \cdot 2007}</math>
 
  
<math>X=Y\Rightarrow\mathrm{ C}</math>
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<math>X = \frac{1}{a\sqrt{b} + b\sqrt{a}} \cdot \left(\frac{a\sqrt{b} - b\sqrt{a}}{a\sqrt{b} - b\sqrt{a}}\right) = \frac{a\sqrt{b} - b\sqrt{a}}{ab(a-b)} = \frac{a\sqrt{b} - b\sqrt{a}}{ab} = \frac{\sqrt{b}}{b} - \frac{\sqrt{a}}{a} = \frac{1}{\sqrt{b}} - \frac{1}{\sqrt{a}} = Y</math>
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<math>X=Y\Longrightarrow\mathrm{ C}</math>
  
 
==See also==
 
==See also==
*[[2007 Cyprus MO/Lyceum/Problems]]
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{{CYMO box|year=2007|l=Lyceum|num-b=10|num-a=12}}
 
 
*[[2007 Cyprus MO/Lyceum/Problem 10|Previous Problem]]
 
  
*[[2007 Cyprus MO/Lyceum/Problem 12|Next Problem]]
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[[Category:Introductory Algebra Problems]]

Latest revision as of 16:56, 6 May 2007

Problem

If $X=\frac{1}{2007 \sqrt{2006}+2006 \sqrt{2007}}$ and $Y=\frac{1}{\sqrt{2006}}-\frac{1}{\sqrt{2007}}$, which of the following is correct?

$\mathrm{(A) \ } X=2Y\qquad \mathrm{(B) \ } Y=2X\qquad \mathrm{(C) \ } X=Y\qquad \mathrm{(D) \ } X=Y^2\qquad \mathrm{(E) \ } Y=X^2$

Solution

Define $a = 2007,\ b = 2006$.

$X = \frac{1}{a\sqrt{b} + b\sqrt{a}} \cdot \left(\frac{a\sqrt{b} - b\sqrt{a}}{a\sqrt{b} - b\sqrt{a}}\right) = \frac{a\sqrt{b} - b\sqrt{a}}{ab(a-b)} = \frac{a\sqrt{b} - b\sqrt{a}}{ab} = \frac{\sqrt{b}}{b} - \frac{\sqrt{a}}{a} = \frac{1}{\sqrt{b}} - \frac{1}{\sqrt{a}} = Y$

$X=Y\Longrightarrow\mathrm{ C}$

See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 10
Followed by
Problem 12
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