2002 USAMO Problems/Problem 3
Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree with real coefficients is the average of two monic polynomials of degree with real roots.
Lemma. For any set of ordered pairs of reals with for all , there exists a unique monic real polynomial of degree such that for all integers .
Proof 1. By the Lagrange Interpolation Formula, there exists a unique real polynomial of degree less than such that . Then it is necessary and sufficient that .
Proof 2. Again by the Lagrange Interpolation Formula, there exists a unique real polynomial of degree less than such that . Then it is necessary and sufficient that .
Let be the monic polynomial of degree given in the problem. Let be a sequence of strictly increasing reals, and choose such that for odd , , but for even , . Let be the unique monic polynomial such that . We shall prove that and satisfy the problem's conditions.
Note that for , and have different signs, so has a real root between and . It follows that has at least real roots. But a polynomial with real coefficients must have an even number of non-real roots, so must have real roots. Similarly, must have real roots. Q.E.D.
Let be our polynomial. If , then we may let , which is the average of the polynomials and , each of which has a real root. Otherwise, let
We will prove that for sufficiently large , and satisfy the problem's conditions.
We note that for the values of , alternates in sign, and always has magnitude at least 1 (since it is the product of several factors of magnitude at least one. On the other hand, there exists some such that for all , and . Let be greater than . Then both and alternate in sign for . It follows that each has at least real roots. But since a polynomial with real coefficients can only have an even number of complex roots, this means that both and must have roots. Q.E.D.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
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