Difference between revisions of "1972 USAMO Problems/Problem 4"
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{{USAMO box|year=1972|num-b=3|num-a=5}} | {{USAMO box|year=1972|num-b=3|num-a=5}} | ||
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[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
[[Category:Olympiad Inequality Problems]] | [[Category:Olympiad Inequality Problems]] | ||
Revision as of 17:52, 3 July 2013
Problem
Let 
denote a non-negative rational number. Determine a fixed set of integers 
, such that for every choice of ,
![$\left|\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt[3]{2}\right|<|R-\sqrt[3]{2}|$](http://latex.artofproblemsolving.com/e/4/8/e48c707907f83429e12a32d297e18dcbe4621168.png)
Solution
Note that when approaches ![$\sqrt[3]{2}$](http://latex.artofproblemsolving.com/9/f/d/9fdfe572d1673f00f286d5ed753942d18f401cbb.png)
, 
must also approach for the given inequality to hold. Therefore
![\[\lim_{R\rightarrow \sqrt[3]{2}} \frac{aR^2+bR+c}{dR^2+eR+f}=\sqrt[3]{2}\]](http://latex.artofproblemsolving.com/b/5/6/b5698b2d22cba7c2484b0687423a6dc646d225d0.png)
which happens if and only if
![\[\frac{a\sqrt[3]{4}+b\sqrt[3]{2}+c}{d\sqrt[3]{4}+e\sqrt[3]{2}+f}=\sqrt[3]{2}\]](http://latex.artofproblemsolving.com/8/a/c/8ac236958b2b7da10530f59d2a29fe7e268838b1.png)
We cross multiply to get ![$a\sqrt[3]{4}+b\sqrt[3]{2}+c=2d+e\sqrt[3]{4}+f\sqrt[3]{2}$](http://latex.artofproblemsolving.com/5/4/0/5408a32deb83531265c1e9de11f508f374699260.png)
. It's not hard to show that, since 
, 
, 
, 
, 
, and 
are positive integers, then 
, 
, and 
.
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See Also
| 1972 USAMO (Problems • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
