Difference between revisions of "2013 AIME I Problems/Problem 10"
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− | ==Problem | + | ==Problem == |
There are nonzero integers <math>a</math>, <math>b</math>, <math>r</math>, and <math>s</math> such that the complex number <math>r+si</math> is a zero of the polynomial <math>P(x)={x}^{3}-a{x}^{2}+bx-65</math>. For each possible combination of <math>a</math> and <math>b</math>, let <math>{p}_{a,b}</math> be the sum of the zeros of <math>P(x)</math>. Find the sum of the <math>{p}_{a,b}</math>'s for all possible combinations of <math>a</math> and <math>b</math>. | There are nonzero integers <math>a</math>, <math>b</math>, <math>r</math>, and <math>s</math> such that the complex number <math>r+si</math> is a zero of the polynomial <math>P(x)={x}^{3}-a{x}^{2}+bx-65</math>. For each possible combination of <math>a</math> and <math>b</math>, let <math>{p}_{a,b}</math> be the sum of the zeros of <math>P(x)</math>. Find the sum of the <math>{p}_{a,b}</math>'s for all possible combinations of <math>a</math> and <math>b</math>. | ||
− | == Solution == | + | == Solution 1== |
Since <math>r+si</math> is a root, by the Complex Conjugate Root Theorem, <math>r-si</math> must be the other imaginary root. Using <math>q</math> to represent the rational root, we have | Since <math>r+si</math> is a root, by the Complex Conjugate Root Theorem, <math>r-si</math> must be the other imaginary root. Using <math>q</math> to represent the rational root, we have |
Latest revision as of 19:09, 24 January 2021
Problem
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 1
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 |
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