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

(Created page with "__TOC__ == Problem == Let <math>n</math> and <math>k</math> be positive integers with <math>k<n</math>. Let <math>P(x)</math> be a polynomial of degree <math>n</math> with re...")
 
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__TOC__
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== Problem == 
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Let <math>n</math> and <math>k</math> be positive integers with <math>k<n</math>. Let <math>P(x)</math> be a polynomial of degree <math>n</math> with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers <math>a_0, a_1, \dots, a_k</math> such that the polynomial <math>a_kx^k+\cdots+a_1x+a_0</math> divides <math>P(x)</math>, the product <math>a_0a_1\cdots a_k</math> is zero. Prove that <math>P(x)</math> has a nonreal root.
  
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== Problem ==
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== Solution ==
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Let <math>n</math> and <math>k</math> be positive integers with <math>k<n</math>. Let <math>P(x)</math> be a polynomial of degree <math>n</math> with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers <math>a_0,\,a_1,\,\ldots,\,a_k</math> such that the polynomial <math>a_kx^k+\cdots+a_1x+a_0</math> divides <math>P(x)</math>, the product <math>a_0a_1\cdots a_k</math> is zero. Prove that <math>P(x)</math> has a nonreal root.
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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>.
  
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== Solution ==
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\subsection*{Case <math>k=2</math>:}
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{{solution}}
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For any pair of roots <math>r_i, r_j</math>, consider:
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<cmath> Q(x) = (x-r_i)(x-r_j) = x^2 - (r_i+r_j)x + r_ir_j </cmath>
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The product of coefficients is:
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<cmath> -r_ir_j(r_i+r_j) = 0 </cmath>
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Since <math>r_i,r_j \neq 0</math>, we must have <math>r_i + r_j = 0</math> for all pairs.
  
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==See Also==
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But for three roots <math>r_1, r_2, r_3</math>, this gives:
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<cmath> r_1 + r_2 = 0 </cmath>
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<cmath> r_1 + r_3 = 0 </cmath>
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<cmath> r_2 + r_3 = 0 </cmath>
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which implies <math>r_1 = r_2 = r_3 = 0</math>, contradicting the nonzero constant term.
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\subsection*{General <math>k</math>:}
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For any <math>k</math> roots <math>r_{i_1}, \ldots, r_{i_k}</math>, the polynomial:
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<cmath> Q(x) = \prod_{m=1}^k (x-r_{i_m}) </cmath>
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must have some coefficient (other than constant term) equal to zero. For <math>k=3</math>, this requires:
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<cmath> r_i + r_j + r_m = 0 </cmath>
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for all triples, which is impossible for distinct non-zero reals when <math>n \geq 4</math>.
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Thus, <math>P(x)</math> must have at least one nonreal root. \hfill (by Jonathan Wang)
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== See Also ==
 
{{USAMO newbox|year=2025|num-b=1|num-a=3}}
 
{{USAMO newbox|year=2025|num-b=1|num-a=3}}
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{{MAA Notice}}
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{{WAA Notice}}

Revision as of 23:31, 27 March 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

Assume for contradiction that all roots of $P(x)$ are real. Let the distinct non-zero real roots be $r_1, r_2, \ldots, r_n$.

\subsection*{Case $k=2$:} For any pair of roots $r_i, r_j$, consider: \[Q(x) = (x-r_i)(x-r_j) = x^2 - (r_i+r_j)x + r_ir_j\] The product of coefficients is: \[-r_ir_j(r_i+r_j) = 0\] Since $r_i,r_j \neq 0$, we must have $r_i + r_j = 0$ for all pairs.

But for three roots $r_1, r_2, r_3$, this gives: \[r_1 + r_2 = 0\] \[r_1 + r_3 = 0\] \[r_2 + r_3 = 0\] which implies $r_1 = r_2 = r_3 = 0$, contradicting the nonzero constant term.

\subsection*{General $k$:} For any $k$ roots $r_{i_1}, \ldots, r_{i_k}$, the polynomial: \[Q(x) = \prod_{m=1}^k (x-r_{i_m})\] must have some coefficient (other than constant term) equal to zero. For $k=3$, this requires: \[r_i + r_j + r_m = 0\] for all triples, which is impossible for distinct non-zero reals when $n \geq 4$.

Thus, $P(x)$ 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

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