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 "2023 AMC 12A Problems/Problem 16"

(→‎Solution 2)
m (→‎Solution 2)
Line 41: Line 41:
 
We are given that <math>1+z+z^2=c</math> where <math>c</math> is some complex number with magnitude <math>4</math>. Rearranging the quadratic to standard form and applying the quadratic formula, we have
 
We are given that <math>1+z+z^2=c</math> where <math>c</math> is some complex number with magnitude <math>4</math>. Rearranging the quadratic to standard form and applying the quadratic formula, we have
 
<cmath>z=\frac{-1\pm \sqrt{1^2-4(1)(1-c)}}{2}=\frac{-1\pm\sqrt{-3-4c}}{2}.</cmath>
 
<cmath>z=\frac{-1\pm \sqrt{1^2-4(1)(1-c)}}{2}=\frac{-1\pm\sqrt{-3-4c}}{2}.</cmath>
−
The imaginary part of <math>z</math> is maximized when c=4+0i, making it <math>i\sqrt{19}/2</math>. Thus the answer is <math>\boxed{21}</math>.
+
The imaginary part of <math>z</math> is maximized when <math>c=4</math>, making it <math>i\sqrt{19}/2</math>. Thus the answer is <math>\boxed{21}</math>.
  
 
~cantalon
 
~cantalon

Revision as of 00:47, 10 November 2023

Problem

Consider the set of complex numbers $z$ satisfying $|1+z+z^{2}|=4$. The maximum value of the imaginary part of $z$ can be written in the form $\tfrac{\sqrt{m}}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

$\textbf{(A)}~20\qquad\textbf{(B)}~21\qquad\textbf{(C)}~22\qquad\textbf{(D)}~23\qquad\textbf{(E)}~24$

Solution 1

First, substitute in $z=a+bi$.

\[|1+(a+bi)+(a+bi)^2|=4\] \[|(1+a+a^2-b^2)+ (b+2ab)i|=4\] \[(1+a+a^2-b^2)^2+ (b+2ab)^2=16\] \[(1+a+a^2-b^2)^2+ b^2(1+4a+4a^2)=16\]

Let $p=b^2$ and $q=1+a+a^2$

\[(q-p)^2+ p(4q-3)=16\] \[p^2-2pq+q^2 + 4pq -3p=16\]

We are trying to maximize $b=\sqrt p$, so we'll turn the equation into a quadratic to solve for $p$ in terms of $q$.

\[p^2+(2q-3)p+(q^2-16)=0\] \[p=\frac{(-2q+3)\pm \sqrt{-12q+73}}{2}\]

We want to maximize $p$, due to the fact that $q$ is always negatively contributing to $p$'s value, that means we want to minimize $q$.

Due to the trivial inequality: $q=1+a+a^2=(a+\frac 12)^2+\frac{3}4 \geq \frac{3}4$

If we plug $q$'s minimum value in, we get that $p$'s maximum value is \[p=\frac{(-2(\frac 34)+3)+ \sqrt{-12(\frac 34)+73}}{2}=\frac{\frac 32+ 8}{2}=\frac{19}{4}\]

Then \[b=\frac{\sqrt{19}}{2}\] and \[m+n=\boxed{21}\]

- CherryBerry

Solution 2

We are given that $1+z+z^2=c$ where $c$ is some complex number with magnitude $4$. Rearranging the quadratic to standard form and applying the quadratic formula, we have \[z=\frac{-1\pm \sqrt{1^2-4(1)(1-c)}}{2}=\frac{-1\pm\sqrt{-3-4c}}{2}.\] The imaginary part of $z$ is maximized when $c=4$, making it $i\sqrt{19}/2$. Thus the answer is $\boxed{21}$.

~cantalon

See Also

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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=2023_AMC_12A_Problems/Problem_16&oldid=201731"