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2020 AMC 12A Problems/Problem 15

Revision as of 11:38, 13 September 2021 by MRENTHUSIASM (talk | contribs) (→‎Solution 1)
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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

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