Difference between revisions of "2021 Fall AMC 12A Problems/Problem 15"
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<math>(\textbf{A})\: {-}304\qquad(\textbf{B}) \: {-}208\qquad(\textbf{C}) \: 12i\qquad(\textbf{D}) \: 208\qquad(\textbf{E}) \: 304</math> | <math>(\textbf{A})\: {-}304\qquad(\textbf{B}) \: {-}208\qquad(\textbf{C}) \: 12i\qquad(\textbf{D}) \: 208\qquad(\textbf{E}) \: 304</math> | ||
− | ==Solution== | + | ==Solution 1== |
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 | ||
+ | |||
+ | == Solution 2 == | ||
+ | Because all coefficients of <math>P \left( z \right)</math> are real, <math>\bar z_1</math>, <math>\bar z_2</math>, <math>\bar z_3</math>, and <math>\bar z_4</math> are four zeros of <math>P \left( z \right)</math>. | ||
+ | |||
+ | First, we compute <math>B</math>. | ||
+ | |||
+ | For <math>P \left( z \right)</math>, following from Vieta's formula, | ||
+ | <cmath> | ||
+ | \[ | ||
+ | 3 = \bar z_1 \bar z_2 + \bar z_1 \bar z_3 + \bar z_1 \bar z_4 + \bar z_2 \bar z_3 + \bar z_2 \bar z_4 + \bar z_3 \bar z_4 . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | For <math>Q \left( z \right)</math>, following from Vieta's formula, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | B & = f \left( z_1 \right) f \left( z_2 \right) | ||
+ | + f \left( z_1 \right) f \left( z_3 \right) | ||
+ | + f \left( z_1 \right) f \left( z_4 \right) | ||
+ | + f \left( z_2 \right) f \left( z_3 \right) | ||
+ | + f \left( z_2 \right) f \left( z_4 \right) | ||
+ | + f \left( z_3 \right) f \left( z_4 \right) \ | ||
+ | & = \left( 4 i \right)^2 \left( \bar z_1 \bar z_2 + \bar z_1 \bar z_3 + \bar z_1 \bar z_4 + \bar z_2 \bar z_3 + \bar z_2 \bar z_4 + \bar z_3 \bar z_4 \right) \ | ||
+ | & = - 16 \cdot 3 \ | ||
+ | & = - 48 . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Second, we compute <math>D</math>. | ||
+ | |||
+ | For <math>P \left( z \right)</math>, following from Vieta's formula, | ||
+ | <cmath> | ||
+ | \[ | ||
+ | 1 = \bar z_1 \bar z_2 \bar z_3 \bar z_4 . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | For <math>Q \left( z \right)</math>, following from Vieta's formula, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | D & = f \left( z_1 \right) f \left( z_2 \right) f \left( z_3 \right) f \left( z_4 \right) \ | ||
+ | & = \left( 4 i \right)^4 \bar z_1 \bar z_2 \bar z_3 \bar z_4 \ | ||
+ | & = 256 \cdot 1 \ | ||
+ | & = 256 . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Therefore, <math>B + D = - 48 + 256 = 208</math>. | ||
+ | |||
+ | Therefore, the answer is <math>\boxed{\textbf{(D) }208}</math>. | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
− | |||
Revision as of 20:48, 25 November 2021
Contents
[hide]Problem 15
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
Since
Also, and
Our answer is
~kingofpineapplz
Solution 2
Because all coefficients of are real, , , , and are four zeros of .
First, we compute .
For , following from Vieta's formula,
For , following from Vieta's formula,
Second, we compute .
For , following from Vieta's formula,
For , following from Vieta's formula,
Therefore, .
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
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.