2006 Cyprus MO/Lyceum/Problem 9
Revision as of 10:40, 27 April 2008 by I like pie (talk | contribs) (Standardized answer choices; eqnarray -> align; minor edits)
Problem
If ![$x=\sqrt[3]{4}$](http://latex.artofproblemsolving.com/c/c/2/cc27e42aa8ca74f047f8a090afeef7b57eb84829.png)
and ![$y=\sqrt[3]{6}-\sqrt[3]{3}$](http://latex.artofproblemsolving.com/6/1/8/61891fe4ee8185b4fd6dd052bcd60a6b5ef960e6.png)
, then which of the following is correct?
Solution
The question is asking us for an approximation of the ratio between 
. Thus, we are allowed to multiply both sides by a constant.
By difference of cubes,
![\begin{align*}\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\end{align*}](http://latex.artofproblemsolving.com/0/4/1/0419328786f0df4f4788702e695ca9ff07fae312.png)
We can approximate the terms on the LHS; ![$2\sqrt[3]{18} > 4$](http://latex.artofproblemsolving.com/3/e/b/3eb293723f1af999b2820ff9cd289aa04ef44dcb.png)
, ![$2\sqrt[3]{9} > 4$](http://latex.artofproblemsolving.com/a/1/4/a14b9e0a27fb90ea743567b0fbf75de14f398ccb.png)
, ![$\sqrt[3]{36} > 3$](http://latex.artofproblemsolving.com/0/c/0/0c0485765957d6082fd33d1c7d0cf3c6d432b85d.png)
, so the sum on the left side 
. Hence 
, and the answer is 
.
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 | |
| 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 | ||
