Difference between revisions of "1972 USAMO Problems/Problem 4"
(New page: ==Problem== Let <math>R</math> denote a non-negative rational number. Determine a fixed set of integers <math>a,b,c,d,e,f</math>, such that for ''every'' choice of <math>R</math>, <center...) |
(made some progress) |
||
Line 4: | Line 4: | ||
==Solution== | ==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}} | {{solution}} | ||
Revision as of 10:48, 29 September 2011
Problem
Let denote a non-negative rational number. Determine a fixed set of integers
, such that for every choice of
,
![$\left|\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt[3]{2}\right|<|R-\sqrt[3]{2}|$](http://latex.artofproblemsolving.com/e/4/8/e48c707907f83429e12a32d297e18dcbe4621168.png)
Solution
Note that when approaches
,
must also approach
for the given inequality to hold. Therefore
which happens if and only if
We cross multiply to get . It's not hard to show that, since
,
,
,
,
, and
are positive integers, then
,
, and
.
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See also
1972 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |