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Difference between revisions of "1989 USAMO Problems/Problem 3"

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Let <math>z_1, \dotsc, z_n</math> be the (not necessarily distinct) roots of <math>P</math>, so that
 
Let <math>z_1, \dotsc, z_n</math> be the (not necessarily distinct) roots of <math>P</math>, so that
 
<cmath> P(z) = \prod_{j=1}^n (z- z_j) . </cmath>
 
<cmath> P(z) = \prod_{j=1}^n (z- z_j) . </cmath>
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Since all the coefficients of <math>P</math> are real, it follows that if <math>w</math> is a root of <math>P</math>, then <math>P( \overline{w}) = \overline{ P(w)} = 0</math>, so <math>\overline{w}</math>, the [[complex conjugate]] of <math>\overline{w}</math>, is also a root of <math>P</math>.
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Since all the coefficients of <math>P</math> are real, it follows that if <math>w</math> is a root of <math>P</math>, then <math>P( \overline{w}) = \overline{ P(w)} = 0</math>, so <math>\overline{w}</math>, the [[complex conjugate]] of <math>w</math>, is also a root of <math>P</math>.
  
 
Since
 
Since
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<cmath> \lvert z_i-i \rvert \cdot \lvert z_j-i \rvert < 1. </cmath>
 
<cmath> \lvert z_i-i \rvert \cdot \lvert z_j-i \rvert < 1. </cmath>
 
Let <math>z_i = a+bi</math> and <math>z_j = a-bi</math>, for real <math>a,b</math>.  We note that
 
Let <math>z_i = a+bi</math> and <math>z_j = a-bi</math>, for real <math>a,b</math>.  We note that
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<cmath> (a+b+1)^2 - (a+b-1)^2 = 4a^2 + 4b^2 . </cmath>
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<cmath> (a+b+1)^2 - (a+b-1)^2 = 4a + 4b . </cmath>
 
Thus
 
Thus
 
<cmath> \begin{align*}
 
<cmath> \begin{align*}
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Since <math>P(a+bi) = P(z_i) = 0</math>, these real numbers <math>a,b</math> satisfy the problem's conditions.  <math>\blacksquare</math>
 
Since <math>P(a+bi) = P(z_i) = 0</math>, these real numbers <math>a,b</math> satisfy the problem's conditions.  <math>\blacksquare</math>
  
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== Resources ==
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== See Also ==
  
 
{{USAMO box|year=1989|num-b=2|num-a=4}}
 
{{USAMO box|year=1989|num-b=2|num-a=4}}
  
 
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356637#356637 Discussion on AoPS/MathLinks]
 
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356637#356637 Discussion on AoPS/MathLinks]
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{{MAA Notice}}
  
  
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]

Latest revision as of 18:11, 18 July 2016

Problem

Let $P(z)= z^n + c_1 z^{n-1} + c_2 z^{n-2} + \cdots + c_n$ be a polynomial in the complex variable $z$, with real coefficients $c_k$. Suppose that $|P(i)| < 1$. Prove that there exist real numbers $a$ and $b$ such that $P(a + bi) = 0$ and $(a^2 + b^2 + 1)^2 < 4 b^2 + 1$.

Solution

Let $z_1, \dotsc, z_n$ be the (not necessarily distinct) roots of $P$, so that \[P(z) = \prod_{j=1}^n (z- z_j) .\] Since all the coefficients of $P$ are real, it follows that if $w$ is a root of $P$, then $P( \overline{w}) = \overline{ P(w)} = 0$, so $\overline{w}$, the complex conjugate of $w$, is also a root of $P$.

Since \[\lvert i- z_1 \rvert \cdot \lvert i - z_2 \rvert \dotsm \lvert i - z_n \rvert = \lvert P(i) \rvert < 1,\] it follows that for some (not necessarily distinct) conjugates $z_i$ and $z_j$, \[\lvert z_i-i \rvert \cdot \lvert z_j-i \rvert < 1.\] Let $z_i = a+bi$ and $z_j = a-bi$, for real $a,b$. We note that \[(a+b+1)^2 - (a+b-1)^2 = 4a + 4b .\] Thus \begin{align*} (a^2+b^2+1)^2 &= (a^2+b^2-1)^2 + 4a^2 + 4b^2 = \lvert a^2 + b^2 - 1 - 2ai \rvert ^2 + 4b^2 \\ &= \lvert (a-i)^2 - (bi)^2 \rvert^2 + 4b^2 \\ &= \bigl( \lvert a+bi - i \rvert \cdot \lvert a-bi -i \rvert \bigr)^2 + 4b^2 \\ &= \bigl( \lvert z_i - i \rvert \cdot \lvert z_j - i \rvert \bigr)^2 + 4b^2 < 1+4b^2. \end{align*} Since $P(a+bi) = P(z_i) = 0$, these real numbers $a,b$ satisfy the problem's conditions. $\blacksquare$

See Also

1989 USAMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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