Difference between revisions of "2003 AIME II Problems/Problem 9"
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== Solution 5 == | == Solution 5 == | ||
− | <math>P(z_1)+P(z_2)+P(z_3)+p(z_4)=\sum_{j=1}^{ | + | <math>P(z_1)+P(z_2)+P(z_3)+p(z_4)=\sum_{i=1}^{4} z_{i}^{6}</math>-\sum_{j=1}^{5} \sum_{i=1}^{4} z_{i}^{j}<math> |
− | Let <math>S_j< | + | Let </math>S_j<math> = </math>\sum_{i=1}^{4} z_{i}^{j}<math> |
By Newton's Sums, | By Newton's Sums, | ||
− | <math>S_1-1=0< | + | </math>S_1-1=0<math> |
− | <math>S_2-S_1-2=0< | + | </math>S_2-S_1-2=0<math> |
− | <math>S_3-S_2-S-1-3=0< | + | </math>S_3-S_2-S-1-3=0<math> |
− | <math>S_4-S_3-S_2-S_1-4=0 | + | </math>S_4-S_3-S_2-S_1-4=0$ |
== Video Solution by Sal Khan == | == Video Solution by Sal Khan == |
Revision as of 01:10, 11 September 2024
Contents
Problem
Consider the polynomials and
Given that
and
are the roots of
find
Solution
When we use long division to divide by
, the remainder is
.
So, since is a root,
.
Now this also follows for all roots of
Now
Now by Vieta's we know that ,
so by Newton's Sums we can find
So finally
Solution 2
Let then by Vieta's Formula we have
By Newton's Sums we have
Applying the formula couples of times yields .
~ Nafer
Solution 3
So we just have to find: .
And by Newton's Sums this computes to: .
~ LuisFonseca123
Solution 4
If we scale by
, we get
. In order to get to
, we add
. Therefore, our answer is
. However, rearranging
, makes our final answer
. The sum of the squares of the roots is
and the sum of the roots is
. Adding 4 to our sum, we get
.
~ Vedoral
Solution 5
-\sum_{j=1}^{5} \sum_{i=1}^{4} z_{i}^{j}
S_j
\sum_{i=1}^{4} z_{i}^{j}
S_1-1=0$$ (Error compiling LaTeX. Unknown error_msg)S_2-S_1-2=0$$ (Error compiling LaTeX. Unknown error_msg)S_3-S_2-S-1-3=0$$ (Error compiling LaTeX. Unknown error_msg)S_4-S_3-S_2-S_1-4=0$
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.