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== |
− | 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 | + | 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 real root, we have |
<math>(x-q)(x-r-si)(x-r+si) = x^3 -ax^2 + bx -65</math> | <math>(x-q)(x-r-si)(x-r+si) = x^3 -ax^2 + bx -65</math> | ||
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Applying difference of squares, and regrouping, we have | Applying difference of squares, and regrouping, we have | ||
− | <math>(x-q)(x^2 - 2rx + (r^2 + s^2) = x^3 -ax^2 + bx -65</math> | + | <math>(x-q)(x^2 - 2rx + (r^2 + s^2)) = x^3 -ax^2 + bx -65</math> |
So matching coefficients, we obtain | So matching coefficients, we obtain | ||
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<math>(\pm2, \pm 3), (\pm3, \pm 2)</math> | <math>(\pm2, \pm 3), (\pm3, \pm 2)</math> | ||
− | For <math>q = 13</math> ,<math> r^2 + s^2 = 5</math> so the pairs <math> (r,s)</math> are | + | For <math>q = 13</math>, <math>r^2 + s^2 = 5</math> so the pairs <math> (r,s)</math> are |
<math>(\pm2, \pm 1), (\pm1, \pm 2)</math> | <math>(\pm2, \pm 1), (\pm1, \pm 2)</math> | ||
− | + | Now we can disregard the plus minus signs for s because those cases are included as complex conjugates of the counted cases. | |
The positive and negative values of r will cancel, so the sum of the <math> {p}_{a,b} = a </math> for <math>q = 1</math> is <math>q</math> times the number of distinct <math>r</math> values (as each value of <math>r</math> generates a pair <math>(a,b)</math>). | The positive and negative values of r will cancel, so the sum of the <math> {p}_{a,b} = a </math> for <math>q = 1</math> is <math>q</math> times the number of distinct <math>r</math> values (as each value of <math>r</math> generates a pair <math>(a,b)</math>). | ||
− | Our answer is then <math>(1)(8) + (5)(4) + (13)(4) = \boxed{ | + | Our answer is then <math>(1)(8) + (5)(4) + (13)(4) = \boxed{080}</math>. |
== See also == | == See also == | ||
{{AIME box|year=2013|n=I|num-b=9|num-a=11}} | {{AIME box|year=2013|n=I|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} |
Latest revision as of 11:05, 10 August 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
Since is a root, by the Complex Conjugate Root Theorem, must be the other imaginary root. Using to represent the real 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
Now we can disregard the plus minus signs for s because those cases are included as complex conjugates of the counted cases. 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.