Difference between revisions of "2003 AIME II Problems/Problem 9"
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== Solution 2 == | == Solution 2 == | ||
− | Let <math>S_k=z_1^k+z_2^k+z_3^k+z_4^k</math> then by [[Vieta's Formula]] and [[Newton's | + | Let <math>S_k=z_1^k+z_2^k+z_3^k+z_4^k</math> then by [[Vieta's Formula]] and [[Newton's Sums]] we have |
+ | <cmath>S_0=4</cmath> | ||
+ | <cmath>S_1=1</cmath> | ||
+ | <cmath>S_2=</cmath> | ||
== See also == | == See also == |
Revision as of 12:55, 22 December 2019
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 Sums we can find
So finally
Solution 2
Let then by Vieta's Formula and Newton's Sums we have
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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