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Difference between revisions of "2023 AMC 12A Problems/Problem 16"
(→‎Solution 3 (Geometry))
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~cantalon
 
~cantalon
  
−
==Solution 3 (Geometry)==
+
==Solution 3 (Geometry + Logic)==
−
 
+
<asy>
 +
size(250);
 +
import TrigMacros;
 +
rr_cartesian_axes(-6,5,-5,5,complexplane=true, usegrid = true);
 +
Label f;
 +
f.p=fontsize(6);
 +
xaxis(-6,5,Ticks(f, 1.0));
 +
yaxis(-5,5,Ticks(f, 1.0));
 +
dot((0,0));
 +
draw(circle((-3/4, 0), 4), red + dashed);
 +
dot((-19/4, 0), blue);
 +
label("$\phi$", (-19/4, 0), NW);
 +
dot((0, 2.18), blue);
 +
label("$v'$", (0, 2.18), NE);
 +
draw(ellipse((0,0),1.8,2.18), green);
 +
</asy>
 
We can write the given condition as <cmath>\left|\left(z+\frac{1}{2}\right)^2 + \frac{3}{4}\right| = 4.</cmath>
 
We can write the given condition as <cmath>\left|\left(z+\frac{1}{2}\right)^2 + \frac{3}{4}\right| = 4.</cmath>
 
Letting <math>u = \left(z+\frac{1}{2}\right)^2</math>, the equation <math>\left|u + \frac{3}{4}\right| = 4</math> equates to the circle centered at <math>-\frac{3}{4}</math> with radius <math>4</math> in the complex plane, call it <math>\omega</math>. Thus the locus of <math>\left(z+\frac{1}{2}\right)^2</math> is <math>\omega</math>. Let <math>v = z+\frac{1}{2}</math>, and since the <math>+\frac{1}{2}</math> does not change <math>z</math>'s imaginary part, we now need to find <math>v</math> with the largest imaginary part such that <math>v^2</math> lies on <math>\omega</math>.  
 
Letting <math>u = \left(z+\frac{1}{2}\right)^2</math>, the equation <math>\left|u + \frac{3}{4}\right| = 4</math> equates to the circle centered at <math>-\frac{3}{4}</math> with radius <math>4</math> in the complex plane, call it <math>\omega</math>. Thus the locus of <math>\left(z+\frac{1}{2}\right)^2</math> is <math>\omega</math>. Let <math>v = z+\frac{1}{2}</math>, and since the <math>+\frac{1}{2}</math> does not change <math>z</math>'s imaginary part, we now need to find <math>v</math> with the largest imaginary part such that <math>v^2</math> lies on <math>\omega</math>.  
  
−
Note that the point on <math>\omega</math> with largest magnitude is <math>19/4</math> and argument <math>\pi</math> (The leftmost point on <math>\omega</math>). The value <math>v</math> with positive imaginary part which correspond to this point has an argument of <math>\frac{\pi}{2}</math> and a magnitude of <math>\frac{\sqrt{19}}{2}</math>.  
+
Note that the point on <math>\omega</math> with largest magnitude is <math>19/4</math> and has argument <math>\pi</math>, call it <math>\phi</math> (The leftmost point on <math>\omega</math>). The value <math>v'</math> with positive imaginary part such that <math>(v')^2 = \phi</math> has an argument of <math>\frac{\pi}{2}</math> and a magnitude of <math>\frac{\sqrt{19}}{2}</math>.  
  
  
−
Since across all values of <math>v</math> the imaginary part is given by <math>r\sin{\theta}</math> and this value has the largest possible <math>r</math> and the largest possible value of <math>\sin{\theta},</math> it must be the largest value.  
+
Since across all values of <math>v</math> the imaginary part is given by <math>r\sin{\theta}</math> and <math>v'</math> has the largest possible <math>r</math> and the largest possible value of <math>\sin{\theta},</math> it must have the largest imaginary part.  
  
  
−
This can non-rigorously be seen by sketching the conic figure which is the locus of <math>v</math>.  
+
This can non-rigorously be seen by sketching the ellipse which is the locus of <math>v</math>.  
  
  

Revision as of 13:14, 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

Solution 3 (Geometry + Logic)

[asy] size(250); import TrigMacros; rr_cartesian_axes(-6,5,-5,5,complexplane=true, usegrid = true); Label f; f.p=fontsize(6); xaxis(-6,5,Ticks(f, 1.0)); yaxis(-5,5,Ticks(f, 1.0)); dot((0,0)); draw(circle((-3/4, 0), 4), red + dashed); dot((-19/4, 0), blue); label("$\phi$", (-19/4, 0), NW); dot((0, 2.18), blue); label("$v'$", (0, 2.18), NE); draw(ellipse((0,0),1.8,2.18), green); [/asy] We can write the given condition as \[\left|\left(z+\frac{1}{2}\right)^2 + \frac{3}{4}\right| = 4.\] Letting $u = \left(z+\frac{1}{2}\right)^2$, the equation $\left|u + \frac{3}{4}\right| = 4$ equates to the circle centered at $-\frac{3}{4}$ with radius $4$ in the complex plane, call it $\omega$. Thus the locus of $\left(z+\frac{1}{2}\right)^2$ is $\omega$. Let $v = z+\frac{1}{2}$, and since the $+\frac{1}{2}$ does not change $z$'s imaginary part, we now need to find $v$ with the largest imaginary part such that $v^2$ lies on $\omega$.

Note that the point on $\omega$ with largest magnitude is $19/4$ and has argument $\pi$, call it $\phi$ (The leftmost point on $\omega$). The value $v'$ with positive imaginary part such that $(v')^2 = \phi$ has an argument of $\frac{\pi}{2}$ and a magnitude of $\frac{\sqrt{19}}{2}$.


Since across all values of $v$ the imaginary part is given by $r\sin{\theta}$ and $v'$ has the largest possible $r$ and the largest possible value of $\sin{\theta},$ it must have the largest imaginary part.


This can non-rigorously be seen by sketching the ellipse which is the locus of $v$.


This gives $19 + 2 \implies \textbf{(B) 21}$

~AtharvNaphade

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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