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Difference between revisions of "2023 AMC 12A Problems/Problem 16"

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==Solution==
 
==Solution==
 +
First, substitute in <math>z=a+bi</math>.
 +
 +
<cmath>|1+(a+bi)+(a+bi)^2|=4</cmath>
 +
<cmath>|(1+a+a^2-b^2)+ (b+2ab)i|=4</cmath>
 +
<cmath>(1+a+a^2-b^2)^2+ (b+2ab)^2=16</cmath>
 +
<cmath>(1+a+a^2-b^2)^2+ b^2(1+4a+4a^2)=16</cmath>
 +
 +
Let <math>p=b^2</math> and <math>q=1+a+a^2</math>
 +
 +
<cmath>(q-p)^2+ p(4q-3)=16</cmath>
 +
<cmath>p^2-2pq+q^2  + 4pq -3p=16</cmath>
 +
 +
We are trying to maximize <math>b=\sqrt p</math>, so we'll turn the equation into a quadratic to solve for <math>p</math> in terms of <math>q</math>.
 +
 +
<cmath>p^2+(2q-3)p+(q^2-16)=0</cmath>
 +
<cmath>p=\frac{(-2q+3)\pm \sqrt{-12q+73}}{2}</cmath>
 +
 +
We want to maximize <math>p</math>, due to the fact that <math>q</math> is always negatively contributing to <math>p</math>'s value, that means we want to minimize <math>q</math>.
 +
 +
Due to the trivial inequality:
 +
<math>q=1+a+a^2=(a+\frac 12)^2+\frac{3}4 \geq \frac{3}4</math>
 +
 +
If we plug <math>q</math>'s minimum value in, we get that <math>p</math>'s maximum value is
 +
<cmath>p=\frac{(-2(\frac 34)+3)+ \sqrt{-12(\frac 34)+73}}{2}</cmath>
 +
<cmath>p=\frac{\frac 32+ 8}{2}</cmath>
 +
<cmath>p=\frac{19}{4}</cmath>
 +
 +
Then
 +
<cmath>b=\frac{\sqrt{19}}{2}</cmath>
 +
and
 +
<cmath>m+n=\boxed{21}</cmath>
 +
 +
- CherryBerry
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2023|ab=A|num-b=15|num-a=17}}
 
{{AMC12 box|year=2023|ab=A|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:34, 9 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

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 \[p=\frac{(-2(\frac 34)+3)+ \sqrt{-12(\frac 34)+73}}{2}\] \[p=\frac{\frac 32+ 8}{2}\] \[p=\frac{19}{4}\]

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

- CherryBerry

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

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