Difference between revisions of "2003 AIME II Problems/Problem 9"
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== Solution == | == Solution == | ||
| − | {{ | + | <math>{{Q(z_1)=0</math> therefore |
| + | <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> | ||
| + | Also <math>z_1^4-z_1^3-z_1^2=1 </math> | ||
| + | |||
| + | S0 <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> | ||
| + | <math>P(z_1)=z_1^6-z_1^5-z_1^4-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 <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 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> | ||
| + | |||
| + | <math>a_ns_2+a_n-1s_1+2a_n-1=0</math> | ||
| + | |||
| + | <math>(1)(s_2)+(-1)(1)+2(-1)=0</math> | ||
| + | |||
| + | <math>s_2-1-2=0</math> | ||
| + | |||
| + | <math>s_2=3</math> | ||
| + | |||
| + | So finally | ||
| + | <math>P(z_2)+P(z_1)+z_3)+P(z_4)=3+4-1=\box{6}</math> | ||
| + | |||
| + | |||
| + | }} | ||
== See also == | == See also == | ||
{{AIME box|year=2003|n=II|num-b=8|num-a=10}} | {{AIME box|year=2003|n=II|num-b=8|num-a=10}} | ||
Revision as of 23:42, 13 January 2008
Problem
Consider the polynomials 
and 
Given that 
and 
are the roots of 
find 
Solution
${{Q(z_1)=0$ (Error making remote request. No response to HTTP request) therefore
therefore 
Also 
S0 
So in 
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 $P(z_2)+P(z_1)+z_3)+P(z_4)=3+4-1=\box{6}$ (Error compiling LaTeX. Unknown error_msg)
}}
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 | ||
