Difference between revisions of "2021 Fall AMC 12A Problems/Problem 15"

Revision as of 01:01, 26 November 2021

Problem 15

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

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 $\overline{a}+\overline{b}=\overline{a+b},$ $B=(4i)^2\overline{\left(z_1z_2+z_1z_3+\dots+z_3z_4\right)}=-16\overline{(3)}=-48$

Also, $z_1z_2z_3z_4=1,$ and $D=(4i)^4\left(\overline{z}_1\,\overline{z}_2\,\overline{z}_3\,\overline{z}_4\right)=256\overline{\left(z_1z_2z_3z_4\right)}=256\overline{(1)}=256.$

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

~kingofpineapplz

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.

@@ Line 18: / Line 18: @@
 ~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>
-\[
-= \bar z_1 \bar z_2 + \bar z_1 \bar z_3 + \bar z_1 \bar z_4 + \bar z_2 \bar z_3 + \bar z_2 \bar z_4 + \bar z_3 \bar z_4 .
-\]
-</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>
-\[
-= \bar z_1 \bar z_2 \bar z_3 \bar z_4  .
-\]
-</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)
 ==See Also==
 {{AMC12 box|year=2021 Fall|ab=A|num-b=14|num-a=16}}
 {{MAA Notice}}

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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 15"

Revision as of 01:01, 26 November 2021

Problem 15

Solution

See Also