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Difference between revisions of "2003 AIME II Problems/Problem 9"

(→‎Solution)
(→‎Solution)
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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>
  
Line 11: Line 13:
  
 
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_1^4-z_1+1</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-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+a_n-1s_1+2a_n-1=0</math>
+
<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>
Line 28: Line 33:
  
 
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 $P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x$ and $Q(x) = x^{4} - x^{3} - x^{2} - 1.$ Given that $z_{1},z_{2},z_{3},$ and $z_{4}$ are the roots of $Q(x) = 0,$ find $P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).$

Solution

${{Q(z_1)=0$ (Error compiling LaTeX. Unknown error_msg) therefore $z_1^4-z_1^3-z_1^2-1=0$

therefore $-z_1^3-z^2=-z_1^4+1.$

Also $z_1^4-z_1^3-z_1^2=1$

S0 $z_1^6-z_1^5-z_1^4=z_1^2$

So in $P(z_1)=z_1^6-z_1^5-z_1^3-z_1^2-z_1$

Since $-z_1^3-z^2=-z_1^4+1.$ and $z_1^6-z_1^5-z_1^4=z_1^2$ 

     $P(z_1)=z_1^6-z_1^5-z_1^3-z_2+1$ can now be 
     $P(z_1)=z_1^2-z_1+1$

Now this also follows for all roots of $Q(x)$ Now $P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1$

Now by Vieta's we know that $-z_4-z_3-z_2-z_1=-1$ So by Newton Sums we can find $z_1^2+z_2^2+z_3^2+z_4^2$

$a_ns_2+a_{n-1}s_1+2a_{n-2}=0$

$(1)(s_2)+(-1)(1)+2(-1)=0$

$s_2-1-2=0$

$s_2=3$

So finally $P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\boxed{6}$


}}

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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