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 compiling LaTeX. Unknown error_msg) 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 | ||
