1973 IMO Shortlist Problems/Cuba 2: Difference between revisions
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* http://www.artofproblemsolving.com/Forum/viewtopic.php?p=603067 Discussion on AoPS/MathLinks] | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=603067 Discussion on AoPS/MathLinks] | ||
[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
Latest revision as of 18:53, 16 August 2006
Problem
Decompose the polynomial
into 
polynomials of second degree with real coefficients.
Solution
We use the quadratic formula to find that the polynomial is zero when
,
which clearly occurs when ![$\displaystyle x = |a| \left[ \cos \pm \left( \theta+\frac{ 2k\pi }{ m }\right)+i\sin \pm \left( \theta+\frac{ 2k\pi }{ m }\right) \right]$](http://latex.artofproblemsolving.com/5/5/2/552d8025e1762d33cc9cdec715c2d51c76f5c390.png)
, for all integral 
.
Hence
![$\begin{matrix} x^{2m}-2|a|^{m}x^{m}\cos ( m \theta )+a^{2m}& = & \displaystyle \prod_{k=0}^{m-1}\left[ x-|a| \mbox{ cis}\left( \theta+\frac{ 2k\pi }{ m }\right) \right] \left[ x-|a| \mbox{ cis} \left( -\left( \theta+\frac{ 2k\pi }{ m } \right) \right) \right] \\ & = & \displaystyle \prod_{k=0}^{m-1}\left( x^{2}-2 |a| \cos \left( \theta+\frac{2k\pi}{m}\right) x+|a|^{2}\right) \end{matrix}$](http://latex.artofproblemsolving.com/6/9/c/69cd62ca2eeae8769f5caf70a60b502dd50d9550.png)
Q.E.D.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
