2013 AIME I Problems/Problem 10
Problem 10
There are nonzero integers , , , and such that the complex number is a zero of the polynomial . For each possible combination of and , let be the sum of the zeros of . Find the sum of the 's for all possible combinations of and .
Solution
Since is a root, by the Complex Conjugate Root Theorem, must be the other imaginary root. Using to represent the rational root, we have
Applying difference of squares, and regrouping, we have
So matching coefficients, we obtain
By Vieta's each so we just need to find the values of in each pair. We proceed by determining possible values for , , and and using these to determine and .
If , so (r, s) =
Similarly, for , so the pairs are
For , so the pairs are
Then, since only but not appears in the equations for and , we can ignore the plus minus sign for . The positive and negative values of r will cancel, so the sum of the for is times the number of distinct values (as each value of generates a pair ). Our answer is then .
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.