1972 USAMO Problems/Problem 4


Let $R$ denote a non-negative rational number. Determine a fixed set of integers $a,b,c,d,e,f$, such that for every choice of $R$,



Note that when $R$ approaches $\sqrt[3]{2}$, $\frac{aR^2+bR+c}{dR^2+eR+f}$ must also approach $\sqrt[3]{2}$ for the given inequality to hold. Therefore

\[\lim_{R\rightarrow \sqrt[3]{2}} \frac{aR^2+bR+c}{dR^2+eR+f}=\sqrt[3]{2}\]

which happens if and only if


We cross multiply to get $a\sqrt[3]{4}+b\sqrt[3]{2}+c=2d+e\sqrt[3]{4}+f\sqrt[3]{2}$. It's not hard to show that, since $a$, $b$, $c$, $d$, $e$, and $f$ are integers, then $a=e$, $b=f$, and $c=2d$.

Note, however, that this is a necessary but insufficient condition. For example, we must also have $a^2<2bc$ to ensure the function does not have any vertical asymptotes (which would violate the desired property). A simple search shows that $a=0$, $b=2$, and $c=2$ works.

See Also

1972 USAMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5
All USAMO Problems and Solutions

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