Difference between revisions of "2021 Fall AMC 12A Problems/Problem 15"
(→Solution) |
(→Solution 2) |
||
(5 intermediate revisions by 3 users not shown) | |||
Line 4: | Line 4: | ||
<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_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> | ||
Line 16: | Line 22: | ||
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== | ||
+ | |||
+ | Since the coefficients of <math>P</math> are real, the roots of <math>P</math> can also be written as <math>\overline{z_1}, \overline{z_2}, \overline{z_3}, \overline{z_4}</math>. With this observation, it's easy to see that the polynomials <math>P(z)</math> and <math>Q(4i\hspace{1pt}z)</math> have the same roots. Hence, there exists some constant <math>K</math> such that | ||
+ | \begin{align*} | ||
+ | P(z)=K*Q(4i\hspace{1pt}z) | ||
+ | \end{align*} | ||
+ | |||
+ | By comparing coefficients, its easy to see that <math>K=\frac{1}{(4i)^4}</math>. Hence <math>\frac{B*(4i)^2}{(4i)^4}=3</math> and <math>\frac{D}{(4i)^4}=1</math>. Hence <math>B=-48</math>, <math>D=256</math>, so <math>B+D=208</math> and our answer is <math>\boxed{(\textbf{D}) \: 208}</math>. | ||
+ | |||
+ | ~tsun26, inspired by mAth_SUN | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021 Fall|ab=A|num-b=14|num-a=16}} | {{AMC12 box|year=2021 Fall|ab=A|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 05:36, 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, inspired by mAth_SUN
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.