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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>
  
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==Solution==
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==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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~kingofpineapplz
 
~kingofpineapplz
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 +
==Solution 2==
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 +
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*}
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P(z)=KQ(4i\hspace{1pt}z)
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\end{align*}
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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>
  
 
==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}}

Revision as of 04:53, 3 November 2024

Problem

Recall that the conjugate of the complex number $w = a + bi$, where $a$ and $b$ are real numbers and $i = \sqrt{-1}$, is the complex number $\overline{w} = a - bi$. For any complex number $z$, let $f(z) = 4i\hspace{1pt}\overline{z}$. The polynomial \[P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1\] has four complex roots: $z_1$, $z_2$, $z_3$, and $z_4$. Let \[Q(z) = z^4 + Az^3 + Bz^2 + Cz + D\] be the polynomial whose roots are $f(z_1)$, $f(z_2)$, $f(z_3)$, and $f(z_4)$, where the coefficients $A,$ $B,$ $C,$ and $D$ are complex numbers. What is $B + D?$

$(\textbf{A})\: {-}304\qquad(\textbf{B}) \: {-}208\qquad(\textbf{C}) \: 12i\qquad(\textbf{D}) \: 208\qquad(\textbf{E}) \: 304$

Solution 1

By Vieta's formulas, $z_1z_2+z_1z_3+\dots+z_3z_4=3$, and $B=(4i)^2\left(\overline{z}_1\,\overline{z}_2+\overline{z}_1\,\overline{z}_3+\dots+\overline{z}_3\,\overline{z}_4\right).$

Since $\overline{a}\cdot\overline{b}=\overline{ab},$ \[B=(4i)^2\left(\overline{z_1z_2}+\overline{z_1z_3}+\overline{z_1z_4}+\overline{z_2z_3}+\overline{z_2z_4}+\overline{z_3z_4}\right).\] Since $\overline{a}+\overline{b}=\overline{a+b},$ \[B=(4i)^2\left(\overline{z_1z_2+z_1z_3+\dots+z_3z_4}\right)=-16(\overline{3})=-48\]


Our answer is $B+D=256-48=\boxed{(\textbf{D}) \: 208}.$

~kingofpineapplz

Solution 2

Since the coefficients of $P$ are real, the roots of $P$ can also be written as $\overline{z_1}, \overline{z_2}, \overline{z_3}, \overline{z_4}$. With this observation, it's easy to see that the polynomials $P(z)$ and $Q(4i\hspace{1pt}z)$ have the same roots. Hence, there exists some constant $K$ such that \begin{align*} P(z)=KQ(4i\hspace{1pt}z) \end{align*}

By comparing coefficients its easy to see that $K=\frac{1}{(4i)^4}$. Hence $\frac{B*(4i)^2}{(4i)^4}=3$ and $\frac{D}{(4i)^4}=1$. Hence $B=-48$, $D=256$. So $B+D=208$ and our answer is $\boxed{(\textbf{D}) \: 208}$

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

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