Difference between revisions of "2020 AMC 12A Problems/Problem 15"
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<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 == |
− | + | 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); | ||
− | + | 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 <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
Contents
Problem
In the complex plane, let be the set of solutions to and let be the set of solutions to What is the greatest distance between a point of and a point of
Solution 1
We solve each equation separately:
- We solve by De Moivre's Theorem.
Let where is the magnitude of such that and is the argument of such that
We have from which
- so
- so or
- We solve by factoring by grouping.
We have The set of solutions to is
In the graph below, the points in set are shown in red, and the points in set are shown in blue. The greatest distance between a point of and a point of is the distance between to as shown in the dashed line segments. By the Distance Formula, the answer is ~lopkiloinm ~MRENTHUSIASM
Solution 2
Alternatively, we can solve by the difference of cubes:
- If then
- If then by either completing the square or the quadratic formula.
The set of solutions to is
Following the rest of Solution 1 gives the answer
~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.