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

Revision as of 19:45, 28 August 2015

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 $c=2d$ .

@@ Line 14: / Line 14: @@
 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}}
 ==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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Difference between revisions of "1972 USAMO Problems/Problem 4"

Revision as of 19:45, 28 August 2015

Problem

Solution

See Also