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2007 Cyprus MO/Lyceum/Problem 11

Revision as of 12:25, 6 May 2007 by Leonariso (talk | contribs) (→‎Solution: You cannot use calculators in this competition (This was not a nescessary operation))

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

$X=\frac{1}{2007 \sqrt{2006}+2006 \sqrt{2007}}$ $=\frac{1}{2007 \sqrt{2006}+2006 \sqrt{2007}}\cdot\frac{2007 \sqrt{2006}-2006 \sqrt{2007}}{2007 \sqrt{2006}-2006 \sqrt{2007}}$ $=\frac{2007\sqrt{2006}-2006\sqrt{2007}}{2006 \times 2007}$ $Y=\frac{1}{\sqrt{2006}}-\frac{1}{\sqrt{2007}}=\frac{\sqrt{2006}}{2006}-\frac{\sqrt{2007}}{2007}=\frac{2007\sqrt{2006}-2006\sqrt{2007}}{2006 \times 2007}$

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

See also

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