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

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== Solution 5 ==
 
== Solution 5 ==
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<math>P(z_1)+P(z_2)+P(z_3)+p(z_4)=\sum_{i=1}^{4} z_{i}^{6}-\sum_{\substack{j=1 \\ j \neq 4}}^{5} \sum_{i=1}^{4} z_{i}^{j}</math>
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<cmath>P(z_1)+P(z_2)+P(z_3)+p(z_4)=\sum_{i=1}^{4} z_{i}^{6}-\sum_{\substack{j=1 \\ j \neq 4}}^{5} \sum_{i=1}^{4} z_{i}^{j}</cmath>
  
 
Let <math>S_j</math> = <math>\sum_{i=1}^{4} z_{i}^{j}</math>
 
Let <math>S_j</math> = <math>\sum_{i=1}^{4} z_{i}^{j}</math>

Revision as of 00:45, 11 September 2024

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

When we use long division to divide $P(x)$ by $Q(x)$, the remainder is $x^2-x+1$.

So, since $z_1$ is a root, $P(z_1)=(z_1)^2-z_1+1$.

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's 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 $P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\boxed{006}.$

Solution 2

Let $S_k=z_1^k+z_2^k+z_3^k+z_4^k$ then by Vieta's Formula we have \[S_{-1}=\frac{z_1z_2z_3+z_1z_3z_4+z_1z_2z_4+z_1z_2z_3}{z_1z_2z_3z_4}=0\] \[S_0=4\] \[S_1=1\] \[S_2=3\] By Newton's Sums we have \[a_4S_k+a_3S_{k-1}+a_2S_{k-2}+a_1S_{k-1}+a_0S_{k-4}=0\]

Applying the formula couples of times yields $P(z_1)+P(z_2)+P(z_3)+P(z_4)=S_6-S_5-S_3-S_2-S_1=\boxed{6}$.

~ Nafer

Solution 3

$P(x) = x^{2}Q(x)+x^{4}-x^{3}-x.$

So we just have to find: $\sum_{n=1}^{4} z^{4}_n - \sum_{n=1}^{4} z^{3}_n - \sum_{n=1}^{4} z_n$.

And by Newton's Sums this computes to: $11-4-1 = \boxed{006}$.

~ LuisFonseca123

Solution 4

If we scale $Q(x)$ by $x^2$, we get $x^6-x^5-x^4-x^2$. In order to get to $P(x)$, we add $x^4-x^3-x$. Therefore, our answer is $\sum_{n=1}^{4} z^4_n-z^3_n-z_n$. However, rearranging $Q(z_n) = 0$, makes our final answer $\sum_{n=1}^{4} z^2_n-z_n+1$. The sum of the squares of the roots is $1^2-2(-1) = 3$ and the sum of the roots is $1$. Adding 4 to our sum, we get $3-1+4 = \boxed{006}$.

~ Vedoral

Solution 5

\[P(z_1)+P(z_2)+P(z_3)+p(z_4)=\sum_{i=1}^{4} z_{i}^{6}-\sum_{\substack{j=1 \\ j \neq 4}}^{5} \sum_{i=1}^{4} z_{i}^{j}\]

Let $S_j$ = $\sum_{i=1}^{4} z_{i}^{j}$

By Newton's Sums,

$S_1-1=0$

$S_2-S_1-2=0$

$S_3-S_2-S_1=0$

$S_4-S_3-S_2-4=0$

$S_5-S_4-S_3-S_1=0$

$S_6-S_5-S_4-S_2=0$

Solving for $S_1,S_2,S_3,S_4,S_5,S_6$, we get $S_1=1, S_2=3, S_3=4, S_4=11, S_5=16, S_6=30$

$S_6-S_5-S_3-S_2-S_1=\boxed{006}$

Video Solution by Sal Khan

https://www.youtube.com/watch?v=ZSESJ8TeGSI&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=14 - AMBRIGGS


[rule]

Nice!-sleepypuppy

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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