Difference between revisions of "1972 USAMO Problems/Problem 4"

Revision as of 09:48, 29 September 2011

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 positive integers, then $a=e$ , $b=f$ , and $c=2d$ .

This problem needs a solution. If you have a solution for it, please help us out by adding it.

@@ Line 4: / Line 4: @@
 ==Solution==
+Note that when <math>R</math> approaches <math>\sqrt[3]{2}</math>, <math>\frac{aR^2+bR+c}{dR^2+eR+f}</math> must also approach <math>\sqrt[3]{2}</math> for the given inequality to hold. Therefore
+<cmath>\lim_{R\rightarrow \sqrt[3]{2}} \frac{aR^2+bR+c}{dR^2+eR+f}=\sqrt[3]{2}</cmath>
+which happens if and only if
+<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>.
 {{solution}}

1972 USAMO (Problems • Resources)
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Difference between revisions of "1972 USAMO Problems/Problem 4"

Revision as of 09:48, 29 September 2011

Problem

Solution

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