Difference between revisions of "2003 AIME II Problems/Problem 9"
(→Solution 2) |
(→Solution 2) |
||
Line 32: | Line 32: | ||
<cmath>S_1=1</cmath> | <cmath>S_1=1</cmath> | ||
<cmath>S_2=3</cmath> | <cmath>S_2=3</cmath> | ||
− | + | By [[Newton's Sums]] we have | |
+ | <cmath>a_4S_k+a_3S_{k-1}+a_2S_{k-2}+a_1S_{k-1}+a_0S_{k-4}=0</cmath> | ||
− | + | Applying the formula couples of times yields <math>P(z_1)+P(z_2)+P(z_3)+P(z_4)=S_6-S_5-S_3-S_2-S_1=\boxed{6}</math>. | |
== See also == | == See also == |
Revision as of 14:24, 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 we have
By Newton's Sums we have
Applying the formula couples of times yields .
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.