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USAMO 1972 #4 *July 26, 2009*

*Posted by lumixedia in Problem-solving.*

Tags: algebra, contest math, olympiad math, USAMO, USAMO 1972

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Tags: algebra, contest math, olympiad math, USAMO, USAMO 1972

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**USAMO 1972 #4.** Let denote a non-negative rational number. Determine a fixed set of integers , , , , , such that, for *every* choice of ,

**Solution.** From the desired inequality, we can conclude that

So

Since is not the root of any quadratic polynomial with integer coefficients, is continuous at and therefore

For this equation to be satisfied, we must have , , . So the LHS of the inequality becomes the absolute value of

Dividing both sides of the inequality by and multiplying by , we find that it is equivalent to

This is certainly satisfied if we choose so both and are always positive for nonnegative . Any with ,,, , and will work. One example is .

It might possibly be interesting to find conditions on which are necessary and sufficient for the inequality to hold. It might also not be very interesting at all.

I suggested in the AoPS thread on the problem here that an interesting follow-up would be to figure out which choices take convergents of the continued fraction of to convergents. It’s probable no choice with this property exists since the continued fraction of non-square roots are aperiodic.

Thanks for the link and the suggestion. Do you have any ideas as to how to tackle a problem like that? I know very little about continued fractions.

No, not really; it’s not easy to identify the convergents of an aperiodic continued fraction. Maybe a more reasonable question would be to identify choices that take best rational approximations to best rational approximations.