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 "2020 AMC 12A Problems/Problem 15"

m (→‎Solution)
m (→‎Solution 1)
 
(14 intermediate revisions by the same user not shown)
Line 4: Line 4:
 
<math>\textbf{(A) } 2\sqrt{3} \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 2\sqrt{21} \qquad \textbf{(E) } 9+\sqrt{3}</math>
 
<math>\textbf{(A) } 2\sqrt{3} \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 2\sqrt{21} \qquad \textbf{(E) } 9+\sqrt{3}</math>
  
−
== Solution ==
+
== Solution 1 ==
−
Realize that <math>z^{3}-8=0</math> will create an equilateral triangle on the complex plane with the first point at <math>2+0i</math> and two other points with equal magnitude at <math>-1{\pm}i\sqrt{3}</math>.  
+
We solve each equation separately:
 +
<ol style="margin-left: 1.5em;">
 +
  <li>We solve <math>z^{3}-8=0</math> by De Moivre's Theorem.<p>
 +
Let <math>z=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta,</math> where <math>r</math> is the magnitude of <math>z</math> such that <math>r\geq0,</math> and <math>\theta</math> is the argument of <math>z</math> such that <math>0\leq\theta<2\pi.</math> <p>
 +
We have <cmath>z^3=r^3\operatorname{cis}(3\theta)=8(1),</cmath> from which
 +
<ul style="list-style-type:square;">
 +
  <li><math>r^3=8,</math> so <math>r=2.</math></li><p>
 +
  <li><math>\begin{cases}
 +
\begin{aligned}
 +
\cos(3\theta) &= 1 \\
 +
\sin(3\theta) &= 0
 +
\end{aligned},
 +
\end{cases}</math> so <math>3\theta=0,2\pi,4\pi,</math> or <math>\theta=0,\frac{2\pi}{3},\frac{4\pi}{3}.</math> </li><p>
 +
</ul>
 +
The set of solutions to <math>z^{3}-8=0</math> is <math>\boldsymbol{A=\left\{2,-1+\sqrt{3}i,-1-\sqrt{3}i\right\}}.</math> In the complex plane, the solutions form the vertices of an equilateral triangle whose circumcircle has center <math>0</math> and radius <math>2.</math></li><p>
 +
  <li>We solve <math>z^{3}-8z^{2}-8z+64=0</math> by factoring by grouping.</li><p>
 +
We have
 +
<cmath>\begin{align*}
 +
z^2(z-8)-8(z-8)&=0 \\
 +
\bigl(z^2-8\bigr)(z-8)&=0.
 +
\end{align*}</cmath>
 +
The set of solutions to <math>z^{3}-8z^{2}-8z+64=0</math> is <math>\boldsymbol{B=\left\{2\sqrt{2},-2\sqrt{2},8\right\}}.</math>
 +
</ol>
 +
In the graph below, the points in set <math>A</math> are shown in red, and the points in set <math>B</math> are shown in blue. The greatest distance between a point of <math>A</math> and a point of <math>B</math> is the distance between <math>-1\pm\sqrt{3}i</math> to <math>8,</math> as shown in the dashed line segments.
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(200);
  
−
Also, realize that <math>z^{3}-8z^{2}-8z+64</math> can be factored through grouping: <math>z^{3}-8z^{2}-8z+64=(z-8)(z^{2}-8).</math> <math>(z-8)(z^{2}-8)</math> will create points at <math>8+0i</math> and <math>\pm2\sqrt{2}+0i.</math>
+
int xMin = -10;
 +
int xMax = 10;
 +
int yMin = -10;
 +
int yMax = 10;
 +
int numRays = 24;
  
−
Plotting the points and looking at the graph will make you realize that <math>-1{\pm}i\sqrt{3}</math> and <math>8+0i</math> are the farthest apart and through Pythagorean Theorem, the answer is revealed to be <math>\sqrt{\sqrt{3}^{2}+(8-(-1))^{2}}=\sqrt{84}=\boxed{\textbf{(D) } 2\sqrt{21}.}</math>
+
//Draws a polar grid that goes out to a number of circles
 +
//equal to big, with numRays specifying the number of rays:
 +
void polarGrid(int big, int numRays)
 +
{
 +
  for (int i = 1; i < big+1; ++i)
 +
  {
 +
    draw(Circle((0,0),i), gray+linewidth(0.4));
 +
  }
 +
  for(int i=0;i<numRays;++i)
 +
  draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4));
 +
}
  
−
~lopkiloinm
+
//Draws the horizontal gridlines
 +
void horizontalLines()
 +
{
 +
  for (int i = yMin+1; i < yMax; ++i)
 +
  {
 +
    draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
 +
  }
 +
}
  
−
==Remark==
+
//Draws the vertical gridlines
−
In the graph below, the solutions to <math>z^{3}-8=0</math> are shown in red, and the solutions to <math>z^{3}-8z^{2}-8z+64=0</math> are shown in blue. The greatest distance between one red point and one blue point is shown in a black dashed line segment.
+
void verticalLines()
 +
{
 +
  for (int i = xMin+1; i < xMax; ++i)
 +
  {
 +
    draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));
 +
  }
 +
}
  
−
[[File:2020 AMC 12A Problem 15.png|center]]
+
horizontalLines();
 +
verticalLines();
 +
polarGrid(xMax,numRays);
 +
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
 +
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
 +
label("Re",(xMax,0),(2,0));
 +
label("Im",(0,yMax),(0,2));
  
−
Graph in Desmos: https://www.desmos.com/calculator/uylcxkffak
+
//The n such that we're taking the nth roots of unity multiplied by 2.
 +
int n = 3;
 +
 
 +
pair A[];
 +
for(int i = 0; i <= n-1; i+=1) {
 +
  A[i] = rotate(360*i/n)*(2,0);
 +
}
 +
 
 +
draw(Circle((0,0),2),red);
 +
draw(A[1]--(8,0),dashed);
 +
draw(A[2]--(8,0),dashed);
 +
 
 +
for(int i = 0; i< n; ++i) dot(A[i],red+linewidth(4.5));
 +
 
 +
dot((2*sqrt(2),0),blue+linewidth(4.5));
 +
dot((-2*sqrt(2),0),blue+linewidth(4.5));
 +
dot((8,0),blue+linewidth(4.5));
 +
</asy>
 +
By the Distance Formula, the answer is <cmath>\sqrt{(-1-8)^2+\left(\pm\sqrt{3}-0\right)^2}=\sqrt{84}=\boxed{\textbf{(D) } 2\sqrt{21}}.</cmath>
 +
~lopkiloinm ~MRENTHUSIASM
 +
 
 +
==Solution 2==
 +
Alternatively, we can solve <math>z^{3}-8=0</math> by the difference of cubes: <cmath>(z-2)\left(z^2+2z+4\right)=0.</cmath>
 +
* If <math>z-2=0,</math> then <math>z=2.</math>
 +
 
 +
* If <math>z^2+2z+4=0,</math> then <math>z=-1\pm\sqrt{3}i</math> by either completing the square or the quadratic formula.
 +
 
 +
The set of solutions to <math>z^{3}-8=0</math> is <math>A=\left\{2,-1+\sqrt{3}i,-1-\sqrt{3}i\right\}.</math>
 +
 
 +
Following the rest of Solution 1 gives the answer <math>\boxed{\textbf{(D) } 2\sqrt{21}}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Latest revision as of 11:38, 13 September 2021

Problem

In the complex plane, let $A$ be the set of solutions to $z^{3}-8=0$ and let $B$ be the set of solutions to $z^{3}-8z^{2}-8z+64=0.$ What is the greatest distance between a point of $A$ and a point of $B?$

$\textbf{(A) } 2\sqrt{3} \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 2\sqrt{21} \qquad \textbf{(E) } 9+\sqrt{3}$

Solution 1

We solve each equation separately:

  1. We solve $z^{3}-8=0$ by De Moivre's Theorem.

    Let $z=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta,$ where $r$ is the magnitude of $z$ such that $r\geq0,$ and $\theta$ is the argument of $z$ such that $0\leq\theta<2\pi.$

    We have \[z^3=r^3\operatorname{cis}(3\theta)=8(1),\] from which

    • $r^3=8,$ so $r=2.$

    • $\begin{cases} \begin{aligned} \cos(3\theta) &= 1 \\ \sin(3\theta) &= 0 \end{aligned}, \end{cases}$ so $3\theta=0,2\pi,4\pi,$ or $\theta=0,\frac{2\pi}{3},\frac{4\pi}{3}.$

    The set of solutions to $z^{3}-8=0$ is $\boldsymbol{A=\left\{2,-1+\sqrt{3}i,-1-\sqrt{3}i\right\}}.$ In the complex plane, the solutions form the vertices of an equilateral triangle whose circumcircle has center $0$ and radius $2.$

  2. We solve $z^{3}-8z^{2}-8z+64=0$ by factoring by grouping.

    3. We have \begin{align*} z^2(z-8)-8(z-8)&=0 \\ \bigl(z^2-8\bigr)(z-8)&=0. \end{align*} The set of solutions to $z^{3}-8z^{2}-8z+64=0$ is $\boldsymbol{B=\left\{2\sqrt{2},-2\sqrt{2},8\right\}}.$

In the graph below, the points in set $A$ are shown in red, and the points in set $B$ are shown in blue. The greatest distance between a point of $A$ and a point of $B$ is the distance between $-1\pm\sqrt{3}i$ to $8,$ as shown in the dashed line segments. [asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -10; int xMax = 10; int yMin = -10; int yMax = 10; int numRays = 24; //Draws a polar grid that goes out to a number of circles //equal to big, with numRays specifying the number of rays: void polarGrid(int big, int numRays) { for (int i = 1; i < big+1; ++i) { draw(Circle((0,0),i), gray+linewidth(0.4)); } for(int i=0;i<numRays;++i) draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4)); } //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } horizontalLines(); verticalLines(); polarGrid(xMax,numRays); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("Re",(xMax,0),(2,0)); label("Im",(0,yMax),(0,2)); //The n such that we're taking the nth roots of unity multiplied by 2. int n = 3; pair A[]; for(int i = 0; i <= n-1; i+=1) { A[i] = rotate(360*i/n)*(2,0); } draw(Circle((0,0),2),red); draw(A[1]--(8,0),dashed); draw(A[2]--(8,0),dashed); for(int i = 0; i< n; ++i) dot(A[i],red+linewidth(4.5)); dot((2*sqrt(2),0),blue+linewidth(4.5)); dot((-2*sqrt(2),0),blue+linewidth(4.5)); dot((8,0),blue+linewidth(4.5)); [/asy] By the Distance Formula, the answer is \[\sqrt{(-1-8)^2+\left(\pm\sqrt{3}-0\right)^2}=\sqrt{84}=\boxed{\textbf{(D) } 2\sqrt{21}}.\] ~lopkiloinm ~MRENTHUSIASM

Solution 2

Alternatively, we can solve $z^{3}-8=0$ by the difference of cubes: \[(z-2)\left(z^2+2z+4\right)=0.\]

  • If $z-2=0,$ then $z=2.$
  • If $z^2+2z+4=0,$ then $z=-1\pm\sqrt{3}i$ by either completing the square or the quadratic formula.

The set of solutions to $z^{3}-8=0$ is $A=\left\{2,-1+\sqrt{3}i,-1-\sqrt{3}i\right\}.$

Following the rest of Solution 1 gives the answer $\boxed{\textbf{(D) } 2\sqrt{21}}.$

~MRENTHUSIASM

See Also

2020 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=2020_AMC_12A_Problems/Problem_15&oldid=162187"