Difference between revisions of "2003 AIME II Problems/Problem 9"
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Since <math> -z_1^3-z^2=-z_1^4+1.</math> and <math>z_1^6-z_1^5-z_1^4=z_1^2</math> | Since <math> -z_1^3-z^2=-z_1^4+1.</math> and <math>z_1^6-z_1^5-z_1^4=z_1^2</math> | ||
− | + | <cmath>P(z_1)=z_1^6-z_1^5-z_1^3-z_1^2-z_1</cmath> can now be | |
− | + | <cmath>P(z_1)=z_1^2-z_1+1</cmath> | |
+ | |||
Now this also follows for all roots of <math>Q(x)</math> | Now this also follows for all roots of <math>Q(x)</math> | ||
Now <math>P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1</math> | Now <math>P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1</math> |
Revision as of 23:33, 4 July 2015
Problem
Consider the polynomials and
Given that
and
are the roots of
find
Solution
therefore
therefore
Also
So
So in
Since and
can now be
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 |
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