Difference between revisions of "2003 AIME II Problems/Problem 9"
Baijiangchen (talk | contribs) m (→Solution: Typo) |
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Also <math>z_1^4-z_1^3-z_1^2=1 </math> | Also <math>z_1^4-z_1^3-z_1^2=1 </math> | ||
− | + | So <math>z_1^6-z_1^5-z_1^4=z_1^2</math> | |
So in <math>P(z_1)=z_1^6-z_1^5-z_1^3-z_1^2-z_1</math> | So in <math>P(z_1)=z_1^6-z_1^5-z_1^3-z_1^2-z_1</math> |
Revision as of 22:08, 28 February 2012
Problem
Consider the polynomials and
Given that
and
are the roots of
find
Solution
${{Q(z_1)=0$ (Error compiling LaTeX. Unknown error_msg) therefore
therefore
Also
So
So in
Since and
can now be
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Now this also follows for all roots of
Now
Now by Vieta's we know that
So by Newton Sums we can find
So finally
}}
See also
2003 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |