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Difference between revisions of "2025 USAMO Problems/Problem 2"

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Consider any subset of <math>k</math> roots <math>\{r_{i_1}, r_{i_2}, \ldots, r_{i_k}\}</math> and form the polynomial:
 
Consider any subset of <math>k</math> roots <math>\{r_{i_1}, r_{i_2}, \ldots, r_{i_k}\}</math> and form the polynomial:
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<cmath> Q(x) = \prod_{j=1}^k (x - r_{i_j}) = x^k + a_{k-1}x^{k-1} + \cdots + a_0 </cmath>
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<cmath> Q(x) = \prod_{j=1}^k (x - r_{i_j}) =a_{k} x^k + a_{k-1}x^{k-1} + \cdots + a_0 </cmath>
  
 
By Vieta's formulas:
 
By Vieta's formulas:
  
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<math>a_0 = (-1)^k \prod_{j=1}^k r_{i_j} \neq 0</math>
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<math>a_0 = (-1)^k \prod_{j=1}^k r_{i_j} \neq 0</math>
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<math>a_{k-1} = -\sum_{j=1}^k r_{i_j}</math>
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<math>a_{k-1} = -\sum_{j=1}^k r_{i_j}</math>
 
<math>a_{k-2} = \sum_{1\leq m<n\leq k} r_{i_m}r_{i_n}</math>
 
<math>a_{k-2} = \sum_{1\leq m<n\leq k} r_{i_m}r_{i_n}</math>
  
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For any <math>k</math> roots, some symmetric sum must be zero. For <math>k=3</math>, this would require:
 
For any <math>k</math> roots, some symmetric sum must be zero. For <math>k=3</math>, this would require:
 
<cmath> r_i + r_j + r_m = 0 \quad \text{for all triples} </cmath>
 
<cmath> r_i + r_j + r_m = 0 \quad \text{for all triples} </cmath>
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which leads to contradictions when <math>n \geq 4</math> as it would force roots to be equal.
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which leads to contradictions for any <math>n \geq 4</math> as it would force roots to be equal.
  
 
Thus, our initial assumption is false, and <math>P(x)</math> must have at least one nonreal root.~Jonathan
 
Thus, our initial assumption is false, and <math>P(x)</math> must have at least one nonreal root.~Jonathan

Latest revision as of 14:43, 5 June 2025

Problem

Let $n$ and $k$ be positive integers with $k<n$. Let $P(x)$ be a polynomial of degree $n$ with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers $a_0, a_1, \dots, a_k$ such that the polynomial $a_kx^k+\cdots+a_1x+a_0$ divides $P(x)$, the product $a_0a_1\cdots a_k$ is zero. Prove that $P(x)$ has a nonreal root.

Solution

We proceed by contradiction. Assume that all roots of $P(x)$ are real. Let the distinct roots be $r_1, r_2, \ldots, r_n$, all nonzero since the constant term is nonzero.

Consider any subset of $k$ roots $\{r_{i_1}, r_{i_2}, \ldots, r_{i_k}\}$ and form the polynomial: \[Q(x) = \prod_{j=1}^k (x - r_{i_j}) =a_{k} x^k + a_{k-1}x^{k-1} + \cdots + a_0\]

By Vieta's formulas:

$a_0 = (-1)^k \prod_{j=1}^k r_{i_j} \neq 0$ $a_{k-1} = -\sum_{j=1}^k r_{i_j}$ $a_{k-2} = \sum_{1\leq m<n\leq k} r_{i_m}r_{i_n}$


The given condition requires that $a_0a_1\cdots a_k = 0$. Since $a_0 \neq 0$, at least one other coefficient must be zero.

Case $k=2$: For any pair of roots $(r_i, r_j)$, we have: \[Q(x) = x^2 - (r_i+r_j)x + r_ir_j\] The condition implies $-r_ir_j(r_i+r_j) = 0$, so $r_i + r_j = 0$ for all pairs. But with $n \geq 3$, considering three roots $r_1, r_2, r_3$ gives: \[r_1 + r_2 = 0 \quad \text{and} \quad r_1 + r_3 = 0 \implies r_2 = r_3\] contradicting distinct roots. In General $k$: For any $k$ roots, some symmetric sum must be zero. For $k=3$, this would require: \[r_i + r_j + r_m = 0 \quad \text{for all triples}\] which leads to contradictions for any $n \geq 4$ as it would force roots to be equal.

Thus, our initial assumption is false, and $P(x)$ must have at least one nonreal root.~Jonathan

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

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