USAMO 1972 #4 July 26, 2009Posted by lumixedia in Problem-solving.
Tags: algebra, contest math, olympiad math, USAMO, USAMO 1972
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
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.