# Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 9"

## Problem

If $x=\sqrt{4}$ and $y=\sqrt{6}-\sqrt{3}$, then which of the following is correct?

A. $x=y$

B. $x

C. $x=2y$

D. $x>2y$

E. None of these

## Solution

The question is asking us for an approximation of the ratio between $x : y$. Thus we are allowed to multiply both sides by a constant. So (by difference of cubes)

\begin{eqnarray*}\sqrt{4}(\sqrt{36}+\sqrt{18}+\sqrt{9}) &:& (\sqrt{6}-\sqrt{3})(\sqrt{36}+\sqrt{18}+\sqrt{9})\\
2\sqrt{18} + 2\sqrt{9} + \sqrt{36} &:& 3 (Error compiling LaTeX. ! LaTeX Error: \begin{eqnarray*} on input line 20 ended by \end{document}.)

We can approximate the terms on the LHS; $2\sqrt{18} > 4$, $2\sqrt{9} > 4$, $\sqrt{36} > 3$, so the sum on the left side $> 11$. Hence $x > 2y$, and the answer is $\mathrm{(D)}$.

Remark: There doesn't seem to be any direct way to calculate a simple ratio between the two terms, but various variations can involve approximating terms by multiplying by certain quantities.