Difference between revisions of "2025 USAMO Problems/Problem 2"
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Assume for contradiction that all roots of <math>P(x)</math> are real. Let the distinct non-zero real roots be <math>r_1, r_2, \ldots, r_n</math>. | Assume for contradiction that all roots of <math>P(x)</math> are real. Let the distinct non-zero real roots be <math>r_1, r_2, \ldots, r_n</math>. | ||
− | + | Case <math>k=2</math> | |
For any pair of roots <math>r_i, r_j</math>, consider: | For any pair of roots <math>r_i, r_j</math>, consider: | ||
<cmath> Q(x) = (x-r_i)(x-r_j) = x^2 - (r_i+r_j)x + r_ir_j </cmath> | <cmath> Q(x) = (x-r_i)(x-r_j) = x^2 - (r_i+r_j)x + r_ir_j </cmath> |
Revision as of 23:42, 27 March 2025
Problem
Let and
be positive integers with
. Let
be a polynomial of degree
with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers
such that the polynomial
divides
, the product
is zero. Prove that
has a nonreal root.
Solution
Assume for contradiction that all roots of are real. Let the distinct non-zero real roots be
.
Case
For any pair of roots
, consider:
The product of coefficients is:
Since
, we must have
for all pairs.
But for three roots , this gives:
which implies
, contradicting the nonzero constant term.
\subsection*{General :}
For any
roots
, the polynomial:
must have some coefficient (other than constant term) equal to zero. For
, this requires:
for all triples, which is impossible for distinct non-zero reals when
.
Thus, must have at least one nonreal root. \hfill (by Jonathan Wang)
See Also
2025 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |