Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 1)
(→‎Solution 2)
 
(11 intermediate revisions by 6 users not shown)
Line 1: Line 1:
−
==Problem 15==
+
==Problem==
 
Recall that the conjugate of the complex number <math>w = a + bi</math>, where <math>a</math> and <math>b</math> are real numbers and <math>i = \sqrt{-1}</math>, is the complex number <math>\overline{w} = a - bi</math>. For any complex number <math>z</math>, let <math>f(z) = 4i\hspace{1pt}\overline{z}</math>. The polynomial <cmath>P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1</cmath> has four complex roots: <math>z_1</math>, <math>z_2</math>, <math>z_3</math>, and <math>z_4</math>. Let <cmath>Q(z) = z^4 + Az^3 + Bz^2 + Cz + D</cmath> be the polynomial whose roots are <math>f(z_1)</math>, <math>f(z_2)</math>, <math>f(z_3)</math>, and <math>f(z_4)</math>, where the coefficients <math>A,</math> <math>B,</math> <math>C,</math> and <math>D</math> are complex numbers. What is <math>B + D?</math>
 
Recall that the conjugate of the complex number <math>w = a + bi</math>, where <math>a</math> and <math>b</math> are real numbers and <math>i = \sqrt{-1}</math>, is the complex number <math>\overline{w} = a - bi</math>. For any complex number <math>z</math>, let <math>f(z) = 4i\hspace{1pt}\overline{z}</math>. The polynomial <cmath>P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1</cmath> has four complex roots: <math>z_1</math>, <math>z_2</math>, <math>z_3</math>, and <math>z_4</math>. Let <cmath>Q(z) = z^4 + Az^3 + Bz^2 + Cz + D</cmath> be the polynomial whose roots are <math>f(z_1)</math>, <math>f(z_2)</math>, <math>f(z_3)</math>, and <math>f(z_4)</math>, where the coefficients <math>A,</math> <math>B,</math> <math>C,</math> and <math>D</math> are complex numbers. What is <math>B + D?</math>
  
 
<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 11: Line 17:
 
<cmath>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).</cmath>
 
<cmath>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).</cmath>
 
Since <math>\overline{a}+\overline{b}=\overline{a+b},</math>
 
Since <math>\overline{a}+\overline{b}=\overline{a+b},</math>
−
<cmath>B=(4i)^2\overline{\left(z_1z_2+z_1z_3+\dots+z_3z_4\right)}=-16\overline{(3)}=-48</cmath>
+
<cmath>B=(4i)^2\left(\overline{z_1z_2+z_1z_3+\dots+z_3z_4}\right)=-16(\overline{3})=-48</cmath>
  
−
Also, <math>z_1z_2z_3z_4=1,</math> and <cmath>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.</cmath>
 
  
 
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 ==
 
−
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>.
+
==Solution 2==
  
−
For <math>P \left( z \right)</math>, following from Vieta's formula,
+
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
−
<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*}
 
\begin{align*}
−
B & = f \left( z_1 \right) f \left( z_2 \right)
+
P(z)=K*Q(4i\hspace{1pt}z)
−
+ 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*}
 
\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)
 
−
 
  
 +
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

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

Since $\overline{a}\cdot\overline{b}=\overline{ab},$ \[D=(4i)^4\overline{z_1z_2z_3z_4} = 256(\overline{1}) = 256\]

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 ~sl_hc

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)=K*Q(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 $\boxed{(\textbf{D}) \: 208}$.

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12A_Problems/Problem_15&oldid=231328"