Difference between revisions of "2003 AIME II Problems/Problem 9"
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<math>{{Q(z_1)=0</math> therefore | <math>{{Q(z_1)=0</math> therefore | ||
<math>z_1^4-z_1^3-z_1^2-1=0</math> | <math>z_1^4-z_1^3-z_1^2-1=0</math> | ||
+ | |||
therefore <math>-z_1^3-z^2=-z_1^4+1.</math> | therefore <math>-z_1^3-z^2=-z_1^4+1.</math> | ||
+ | |||
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> | ||
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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> | ||
− | <math>P(z_1)=z_1^6-z_1^5-z_1^ | + | |
+ | 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-z_2+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> | ||
− | Now <math>P(z_2)+P(z_1)+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> |
Now by Vieta's we know that <math>-z_4-z_3-z_2-z_1=-1</math> | Now by Vieta's we know that <math>-z_4-z_3-z_2-z_1=-1</math> | ||
So by Newton Sums we can find <math>z_1^2+z_2^2+z_3^2+z_4^2</math> | So by Newton Sums we can find <math>z_1^2+z_2^2+z_3^2+z_4^2</math> | ||
− | <math>a_ns_2+ | + | <math>a_ns_2+a_{n-1}s_1+2a_{n-2}=0</math> |
<math>(1)(s_2)+(-1)(1)+2(-1)=0</math> | <math>(1)(s_2)+(-1)(1)+2(-1)=0</math> | ||
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So finally | So finally | ||
− | <math>P(z_2)+P(z_1)+z_3)+P(z_4)=3+4-1=\boxed{6}</math> | + | <math>P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\boxed{6}</math> |
Revision as of 23:48, 13 January 2008
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
S0
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 |