# 1972 USAMO Problems/Problem 4

## Problem

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 $R$,

$\left|\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt[3]{2}\right|<|R-\sqrt[3]{2}|$

## Solution

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

$$\frac{a\sqrt[3]{4}+b\sqrt[3]{2}+c}{d\sqrt[3]{4}+e\sqrt[3]{2}+f}=\sqrt[3]{2}$$

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.