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"

m (→‎Solution 3 (Geometry + Logic))
(→‎Solution 2)
 
(12 intermediate revisions by 6 users not shown)
Line 22: Line 22:
 
<cmath>p=\frac{(-2q+3)\pm \sqrt{-12q+73}}{2}</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>.
+
We want to maximize <math>p</math>. Since <math>q</math> is always negatively contributing to <math>p</math>'s value, we want to minimize <math>q</math>.
  
 
Due to the trivial inequality:
 
Due to the trivial inequality:
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{4c-3}}{2}.</cmath>
 
<cmath>z=\frac{-1\pm \sqrt{1^2-4(1)(1-c)}}{2}=\frac{-1\pm\sqrt{4c-3}}{2}.</cmath>
−
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
+
The imaginary part of <math>z</math> is maximized when <math>c=-4</math>. (Why? See note below.)
 +
 
 +
Thus <math>z=-\tfrac 12 \pm i \tfrac{\sqrt{19}}{2}</math>, and so the answer is <math>\boxed{\textbf{(B)}~21}</math>.
 +
 
 +
~cantalon, centslordm
 +
 
 +
Note: You can cheese/fakesolve <math>c</math> by assuming <math>c</math> is real. But there is a good reason for it, within the parameters of this problem:
 +
 
 +
<math>c</math> lies on a circle of radius 4, centered at the origin. Thus, <math>{4c-3}</math> is also a circle, centered at <math>-3</math>. Luckily, the center of this adjusted circle is a negative real number, so the most negative point on that circle (<math>w = -19</math>) has the largest magnitude (and so its square root also has the largest magnitude of any square root), and its positive square root is the most purely imaginary square root ("vertical"). [We ignore the alternative negative square root, since the problem asks for the most positive value.]
 +
 
 +
Thus, <math>\text{Im}(\sqrt w) = r \sin(\theta)</math> has both the largest <math>r</math> and largest <math>\sin(\theta)</math> of any square root of <math>4c-3</math>, and so the largest product <math>r \sin(\theta)</math>. (See note2 for more details)
 +
 
 +
[This argument and overall solution would not work if <math>-3</math> were instead a positive or non-real number, which would happen if the original problem had <math>|C+Bz+Az^2|=4</math> with <math>B^2-4AC > 0 </math>.]
 +
~oinava
 +
 
 +
Note 2:
 +
<asy>
 +
size(250);
 +
import TrigMacros;
 +
rr_cartesian_axes(-22,19,-19,19,complexplane=true, usegrid = false);
 +
Label f;
 +
f.p=fontsize(4);
 +
dot((-3, 0), red);
 +
label("$(-3, 0)$", (-3, 0), NW);
 +
dot((0,0));
 +
draw(circle((-3, 0), 16), red);
 +
dot((-19, 0), blue);
 +
label("$w(-19, 0)$", (-19, 0), NW);
 +
</asy>
 +
 
 +
The circle in the graph is all the possible values of <math>4c - 3</math>.
 +
Let <math>{4c - 3} = re ^ {i\theta}</math>.
 +
 
 +
<math>{4c - 3} = re ^ {i\theta} \implies \sqrt{4c - 3} = \sqrt{re ^ {i\theta}} = \sqrt{r} e ^ {i\frac{\theta}{2}}</math>.
 +
Therefore, <math>Im(\sqrt{4c - 3}) = \sqrt{r}\sin(\frac{\theta}{2})</math>.
 +
 
 +
At <math>w(-19, 0)</math> as in the graph, <math>\theta = \pi</math> and <math>r = 19</math>.
 +
So <math>\sqrt{r}</math> and <math>\sin(\frac{\theta}{2})</math> are all at their maximum, which means that <math>Im(\sqrt{4c - 3})</math> is at its maximum.
 +
 
 +
~JiYang
  
 
==Solution 3 (Geometry + Logic)==
 
==Solution 3 (Geometry + Logic)==
Line 71: Line 109:
  
  
−
This can non-rigorously be seen by sketching the ellipse which is the locus of <math>v</math>.  
+
This can non-rigorously be seen by sketching the oval which is the locus of <math>v</math>.  
  
  
−
This gives <math>19 + 2 \implies \boxed{\textbf{(B)~21}}</math>
+
This gives <math>19 + 2 \implies \boxed{\textbf{(B)}~21}</math>
  
 
~AtharvNaphade
 
~AtharvNaphade
 +
 +
==Solution 4==
 +
To start, we factor <math>1+z+z^2</math> to get:
 +
 +
<cmath>|(z-\frac{-1+\sqrt{3}i}{2})(z-\frac{-1-\sqrt{3}i}{2})|=4</cmath>
 +
 +
Note that since the magnitude of a product of complex numbers is equal to the product of the magnitudes:
 +
 +
<cmath>|(z+\frac{1-\sqrt{3}i}{2})||(z+\frac{1+\sqrt{3}i}{2})|=4</cmath>
 +
 +
Now, we substitute <math>z=a+bi</math> (Note that we are trying to maximize b):
 +
 +
<cmath>|a+\frac{1}{2}+(b+\frac{\sqrt{3}}{2})i||a+\frac{1}{2}+(b-\frac{\sqrt{3}}{2})i|=4</cmath>
 +
 +
Since we are trying to maximize <math>b</math>, we want the real parts of the components to be as small as possible, which we can do by setting <math>a=-\frac{1}{2}</math>. This leaves us with:
 +
 +
<cmath>|(b+\frac{\sqrt{3}}{2})i||(b-\frac{\sqrt{3}}{2})i|=4</cmath>
 +
<cmath>(b+\frac{\sqrt{3}}{2})(b-\frac{\sqrt{3}}{2})=4</cmath>
 +
<cmath>b^2-\frac{3}{4}=4</cmath>
 +
<cmath>b^2=\frac{19}{4}</cmath>
 +
<cmath>b=\frac{\sqrt{19}}{2}</cmath>
 +
 +
This gives us <math>19 + 2 \implies \boxed{\textbf{(B)}~21}</math>.
 +
 +
~LTHMath
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 05:44, 11 November 2024

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 $p$. Since $q$ is always negatively contributing to $p$'s value, 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 \[m+n=\boxed{\textbf{(B)}~21}\]

- 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 \[z=\frac{-1\pm \sqrt{1^2-4(1)(1-c)}}{2}=\frac{-1\pm\sqrt{4c-3}}{2}.\]

The imaginary part of $z$ is maximized when $c=-4$. (Why? See note below.)

Thus $z=-\tfrac 12 \pm i \tfrac{\sqrt{19}}{2}$, and so the answer is $\boxed{\textbf{(B)}~21}$.

~cantalon, centslordm

Note: You can cheese/fakesolve $c$ by assuming $c$ is real. But there is a good reason for it, within the parameters of this problem:

$c$ lies on a circle of radius 4, centered at the origin. Thus, ${4c-3}$ is also a circle, centered at $-3$. Luckily, the center of this adjusted circle is a negative real number, so the most negative point on that circle ($w = -19$) has the largest magnitude (and so its square root also has the largest magnitude of any square root), and its positive square root is the most purely imaginary square root ("vertical"). [We ignore the alternative negative square root, since the problem asks for the most positive value.]

Thus, $\text{Im}(\sqrt w) = r \sin(\theta)$ has both the largest $r$ and largest $\sin(\theta)$ of any square root of $4c-3$, and so the largest product $r \sin(\theta)$. (See note2 for more details)

[This argument and overall solution would not work if $-3$ were instead a positive or non-real number, which would happen if the original problem had $|C+Bz+Az^2|=4$ with $B^2-4AC > 0$.] ~oinava

Note 2: [asy] size(250); import TrigMacros; rr_cartesian_axes(-22,19,-19,19,complexplane=true, usegrid = false); Label f; f.p=fontsize(4); dot((-3, 0), red); label("$(-3, 0)$", (-3, 0), NW); dot((0,0)); draw(circle((-3, 0), 16), red); dot((-19, 0), blue); label("$w(-19, 0)$", (-19, 0), NW); [/asy]

The circle in the graph is all the possible values of $4c - 3$. Let ${4c - 3} = re ^ {i\theta}$.

${4c - 3} = re ^ {i\theta} \implies \sqrt{4c - 3} = \sqrt{re ^ {i\theta}} = \sqrt{r} e ^ {i\frac{\theta}{2}}$. Therefore, $Im(\sqrt{4c - 3}) = \sqrt{r}\sin(\frac{\theta}{2})$.

At $w(-19, 0)$ as in the graph, $\theta = \pi$ and $r = 19$. So $\sqrt{r}$ and $\sin(\frac{\theta}{2})$ are all at their maximum, which means that $Im(\sqrt{4c - 3})$ is at its maximum.

~JiYang

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 oval which is the locus of $v$.


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

~AtharvNaphade

Solution 4

To start, we factor $1+z+z^2$ to get:

\[|(z-\frac{-1+\sqrt{3}i}{2})(z-\frac{-1-\sqrt{3}i}{2})|=4\]

Note that since the magnitude of a product of complex numbers is equal to the product of the magnitudes:

\[|(z+\frac{1-\sqrt{3}i}{2})||(z+\frac{1+\sqrt{3}i}{2})|=4\]

Now, we substitute $z=a+bi$ (Note that we are trying to maximize b):

\[|a+\frac{1}{2}+(b+\frac{\sqrt{3}}{2})i||a+\frac{1}{2}+(b-\frac{\sqrt{3}}{2})i|=4\]

Since we are trying to maximize $b$, we want the real parts of the components to be as small as possible, which we can do by setting $a=-\frac{1}{2}$. This leaves us with:

\[|(b+\frac{\sqrt{3}}{2})i||(b-\frac{\sqrt{3}}{2})i|=4\] \[(b+\frac{\sqrt{3}}{2})(b-\frac{\sqrt{3}}{2})=4\] \[b^2-\frac{3}{4}=4\] \[b^2=\frac{19}{4}\] \[b=\frac{\sqrt{19}}{2}\]

This gives us $19 + 2 \implies \boxed{\textbf{(B)}~21}$.

~LTHMath

Video Solution

https://youtu.be/Aq0iIp0ipyM

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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=233231"