2006 Cyprus MO/Lyceum/Problem 9

Revision as of 12:38, 24 October 2007 by Inscrutableroot (talk | contribs) (Problem)

Problem

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

A. $x=y$

B. $x<y$

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[3]{4}(\sqrt[3]{36}+\sqrt[3]{18}+\sqrt[3]{9}) &:& (\sqrt[3]{6}-\sqrt[3]{3})(\sqrt[3]{36}+\sqrt[3]{18}+\sqrt[3]{9})\\
2\sqrt[3]{18} + 2\sqrt[3]{9} + \sqrt[3]{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; $2\sqrt[3]{18} > 4$, $2\sqrt[3]{9} > 4$, $\sqrt[3]{36} > 3$, 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.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 8
Followed by
Problem 10
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