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> | ||
− | 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> | + | 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 | ||
− | <cmath> (a+b+1)^2 - (a+b-1)^2 = 4a | + | <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> | ||
− | == | + | == 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] | ||
+ | {{MAA Notice}} | ||
[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] |
Latest revision as of 19:11, 18 July 2016
Problem
Let be a polynomial in the complex variable
, with real coefficients
. Suppose that
. Prove that there exist real numbers
and
such that
and
.
Solution
Let be the (not necessarily distinct) roots of
, so that
Since all the coefficients of
are real, it follows that if
is a root of
, then
, so
, the complex conjugate of
, is also a root of
.
Since
it follows that for some (not necessarily distinct) conjugates
and
,
Let
and
, for real
. We note that
Thus
Since
, these real numbers
satisfy the problem's conditions.
See Also
1989 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.