Difference between revisions of "1972 USAMO Problems/Problem 4"
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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>. | 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= | + | 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=2</math>, and <math>c=2</math> works. |
==See Also== | ==See Also== |
Latest revision as of 14:09, 2 June 2018
Problem
Let denote a non-negative rational number. Determine a fixed set of integers , such that for every choice of ,
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 integers, then , , and .
Note, however, that this is a necessary but insufficient condition. For example, we must also have to ensure the function does not have any vertical asymptotes (which would violate the desired property). A simple search shows that , , and 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.