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Difference between revisions of "1972 USAMO Problems/Problem 4"
(Added clarification regarding necessary/sufficient conditions)
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<cmath>\frac{a\sqrt[3]{4}+b\sqrt[3]{2}+c}{d\sqrt[3]{4}+e\sqrt[3]{2}+f}=\sqrt[3]{2}</cmath>
 
<cmath>\frac{a\sqrt[3]{4}+b\sqrt[3]{2}+c}{d\sqrt[3]{4}+e\sqrt[3]{2}+f}=\sqrt[3]{2}</cmath>
  
−
We cross multiply to get <math>a\sqrt[3]{4}+b\sqrt[3]{2}+c=2d+e\sqrt[3]{4}+f\sqrt[3]{2}</math>. It's not hard to show that, since <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math>, and <math>f</math> are positive integers, then <math>a=e</math>, <math>b=f</math>, and <math>c=2d</math>.
+
We cross multiply to get <math>a\sqrt[3]{4}+b\sqrt[3]{2}+c=2d+e\sqrt[3]{4}+f\sqrt[3]{2}</math>. It's not hard to show that, since <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math>, and <math>f</math> are integers, then <math>a=e</math>, <math>b=f</math>, and <math>c=2d</math>.
−
 
 
  
 +
Note, however, that this is a necessary but insufficient condition. For example, we must also have <math>a^2<2bc</math> to ensure the function does not have any vertical asymptotes (which would violate the desired property). A simple search shows that <math>a=0</math>, <math>b=1</math>, and <math>c=1</math> works.
  
 
==See Also==
 
==See Also==

Revision as of 14:09, 2 June 2018

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 $a=0$, $b=1$, and $c=1$ works.

See Also

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

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