Difference between revisions of "2021 Fall AMC 12A Problems/Problem 15"
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==Solution 1== | ==Solution 1== | ||
+ | |||
+ | By Vieta's formulas, <math>z_1z_2z_3z_4=1</math>, and <math>D= | ||
+ | (4i)^4\overline{z}_1\,\overline{z}_2\,\overline{z}_3\,\overline{z}_4.</math> | ||
+ | |||
+ | Since <math>\overline{a}\cdot\overline{b}=\overline{ab},</math> | ||
+ | <cmath>D=(4i)^4\overline{z_1z_2z_3z_4} = 256(\overline{1}) = 256</cmath> | ||
By Vieta's formulas, <math>z_1z_2+z_1z_3+\dots+z_3z_4=3</math>, and <math>B=(4i)^2\left(\overline{z}_1\,\overline{z}_2+\overline{z}_1\,\overline{z}_3+\dots+\overline{z}_3\,\overline{z}_4\right).</math> | By Vieta's formulas, <math>z_1z_2+z_1z_3+\dots+z_3z_4=3</math>, and <math>B=(4i)^2\left(\overline{z}_1\,\overline{z}_2+\overline{z}_1\,\overline{z}_3+\dots+\overline{z}_3\,\overline{z}_4\right).</math> | ||
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Our answer is <math>B+D=256-48=\boxed{(\textbf{D}) \: 208}.</math> | Our answer is <math>B+D=256-48=\boxed{(\textbf{D}) \: 208}.</math> | ||
− | ~kingofpineapplz | + | ~kingofpineapplz ~sl_hc |
==Solution 2== | ==Solution 2== |
Revision as of 05:00, 3 November 2024
Contents
[hide]Problem
Recall that the conjugate of the complex number , where and are real numbers and , is the complex number . For any complex number , let . The polynomial has four complex roots: , , , and . Let be the polynomial whose roots are , , , and , where the coefficients and are complex numbers. What is
Solution 1
By Vieta's formulas, , and
Since
By Vieta's formulas, , and
Since Since
Our answer is
~kingofpineapplz ~sl_hc
Solution 2
Since the coefficients of are real, the roots of can also be written as . With this observation, it's easy to see that the polynomials and have the same roots. Hence, there exists some constant such that
By comparing coefficients, its easy to see that . Hence and . Hence , , so and our answer is .
~tsun26
See Also
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.