Difference between revisions of "2003 AIME II Problems/Problem 9"
Ilikepie333 (talk | contribs) m (→Solution) |
|||
Line 16: | Line 16: | ||
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> | ||
− | <math>P(z_1)=z_1^6-z_1^5-z_1^3- | + | <math>P(z_1)=z_1^6-z_1^5-z_1^3-z_1^2-z_1</math> can now be |
<math>P(z_1)=z_1^2-z_1+1</math> | <math>P(z_1)=z_1^2-z_1+1</math> | ||
Now this also follows for all roots of <math>Q(x)</math> | Now this also follows for all roots of <math>Q(x)</math> |
Revision as of 17:47, 6 October 2013
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
![]()
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.