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2018 AMC 12B Problems/Problem 16

Revision as of 04:40, 29 September 2021 by MRENTHUSIASM (talk | contribs)

Problem

The solutions to the equation $(z+6)^8=81$ are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled $A,B,$ and $C$. What is the least possible area of $\triangle ABC?$

$\textbf{(A) } \frac{1}{6}\sqrt{6} \qquad \textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2} \qquad \textbf{(C) } 2\sqrt3-3\sqrt2 \qquad \textbf{(D) } \frac{1}{2}\sqrt{2} \qquad \textbf{(E) } \sqrt 3-1$

Solution 1 (Complex Numbers in Polar Form)

Recall that translations preserve the shapes and the sizes for all objects. We translate the solutions to the given equation $6$ units right, so they become the solutions to the equation $z^8=81.$

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

By De Moivre's Theorem, we have \[z^8=r^8\operatorname{cis}(8\theta)={\sqrt3}^8(1),\] from which

  1. $r^8={\sqrt3}^8,$ so $r=\sqrt3.$

  2. $\begin{cases} \begin{aligned} \cos(8\theta) &= 1 \\ \sin(8\theta) &= 0 \end{aligned}, \end{cases}$ so $\theta=0,\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\pi,\frac{5\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4}.$

In the complex plane, the solutions to the equation $z^8=81$ are the vertices of a regular octagon with center $0$ and radius $\sqrt3.$

The least possible area of $\triangle ABC$ occurs when $A,B,$ and $C$ are the consecutive vertices of the octagon. For simplicity purposes, let $A=\sqrt3\operatorname{cis}\frac{\pi}{4}=\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i, B=\sqrt3\operatorname{cis}\frac{\pi}{2}=\sqrt3i,$ and $C=\sqrt3\operatorname{cis}\frac{3\pi}{4}=-\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i,$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -2; int xMax = 2; int yMin = -2; int yMax = 2; 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*E); label("Im",(0,yMax),2*N); //The n such that we're taking the nth roots of unity multiplied by 2. int n = 8; pair A[]; for(int i = 0; i <= n-1; i+=1) { A[i] = rotate(360*i/n)*(sqrt(3),0); } label("$A$",A[1],1.5*NE,UnFill); label("$B$",A[2],1.5*NE,UnFill); label("$C$",A[3],1.5*NW,UnFill); fill(A[1]--A[2]--A[3]--cycle,green); draw(A[1]--A[3]^^A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,red); for(int i = 0; i< n; ++i) dot(A[i],red+linewidth(4.5)); [/asy] Note that $\triangle ABC$ has base $AC=\sqrt6$ and height $\sqrt3-\frac{\sqrt6}{2},$ so its area is \[\frac12\cdot\sqrt6\cdot\left(\sqrt3-\frac{\sqrt6}{2}\right)=\boxed{\textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2}}.\] ~MRENTHUSIASM

Solution 2 (Complex Numbers in Rectangular Form)

Recall that translations preserve the shapes and the sizes for all objects. We translate the solutions to the given equation $6$ units right, so they become the solutions to the equation $z^8=81.$

We have \begin{align*} z^8 &= 81 \\ z^4 &= 9, -9 \\ z^2 &= 3, -3, 3i, -3i. \\ \end{align*} Note that:

  1. For $z^2=3,$ we get $\boldsymbol{z=\sqrt3,-\sqrt3}.$

  2. For $z^2=-3,$ we get $\boldsymbol{z=\sqrt3i,-\sqrt3i}.$

  3. For $z^2=3i,$ let $z=a+bi$ for some real numbers $a$ and $b.$ We substitute and then expand: \begin{align*} (a+bi)^2 &= 3i \\ \left(a^2-b^2\right)+2abi &= 3i. \end{align*} We equate the real parts and the imaginary parts, respectively, then simplify: \begin{align*} a^2 &= b^2, &&(1) \\ ab &= \frac32. &&(2) \end{align*} Since $ab>0$ in $(2),$ we conclude that $a$ and $b$ must have the same sign. It follows that $(1)$ becomes $a=b.$ By substitution, we get $(a,b)=\left(\frac{\sqrt6}{2},\frac{\sqrt6}{2}\right),\left(-\frac{\sqrt6}{2},-\frac{\sqrt6}{2}\right),$ from which $\boldsymbol{z=\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i,-\frac{\sqrt6}{2}-\frac{\sqrt6}{2}i}.$

  4. For $z^2=-3i,$ we get $\boldsymbol{z=\frac{\sqrt6}{2}-\frac{\sqrt6}{2}i,-\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i}$ by a similar reasoning.

We continue with the last two paragraphs of Solution 1 to obtain the answer $\boxed{\textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2}}.$

~MRENTHUSIASM

Solution 3 (Regular Octagon)

The polygon formed will be a regular octagon since there are $8$ roots of $z^8=81$. By normal math computation, we can figure out that two roots of $z^8=81$ are $\sqrt{3}$ and $-\sqrt{3}$. These will lie on the real axis of the plane. Since it's a regular polygon, there has to be points on the vertical plane also which will be $\sqrt{3}i$ and $-\sqrt{3}i$.

Clearly, the rest of the points will lie in each quadrant. The next thing is to get their coordinates (note that to answer this question, we do not need all the coordinates, only 3 consecutive ones are needed).

The circumcircle of the octagon will have the equation $i^2+r^2=3$. The coordinates of the point in the first quadrant will be equal in magnitude and both positive, so $i=r$. Solving gives $i=r=\frac{\sqrt{3}}{\sqrt{2}}$ (meaning that the root represented is $\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i$).

This way we can deduce the values of the $8$ roots of the equation to be \[\sqrt{3},-\sqrt{3},-\sqrt{3}i,\sqrt{3}i,\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i,-\frac{\sqrt{3}}{\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}}i,\frac{\sqrt{3}}{\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}}i,-\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i.\]

To get the area, $3$ consecutive points such as $\sqrt{3},$ $\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i,$ and $\sqrt{3}i$ can be used. The area can be computed using different methods like using the Shoelace Theorem, or subtracting areas to find the area. The answer you get is $\boxed{\textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2}}$.

(This method is not actually as long as it seems if you understand what you're doing while doing it. Also calculations can be made a little easier by solving using $x^8=1$ and multiplying your answer by $\sqrt{3}$).

~OlutosinNGA

Solution 4 (Roots of Unity)

Now, we need to solve the equation $(z+6)^8 = 81$ where $z = a+bi$. We can substitute this as \[(a+6+bi)^8 = 81\] Now, let $a+6 = q$ for some $q \in \mathbb{Z}$. Thus, the equation becomes $((q+bi)^2)^4 = 81$. Taking it to the other side, we get the equation to be $(q+bi)^2 = 3$. Rearranging variables, we get $(q+bi) = \sqrt{3}$. Plotting this in the complex place, this is a circle centered at the origin and of radius $\sqrt{3}$.

The graph of the original equation $(a+6+bi) = \sqrt{3}$ is merely a transformation which doesn't change affect the area. Thus, we can find the minimum area of the transformed equation $(q+bi)^2 = 3$. Using Roots of Unity, we know that the roots of the equation lie at $0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \ldots, 2\pi$ radians from the origin.

We can quickly notice that the area of the roots will be smallest with points at $\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}$. Using trigonometry, we get the respective roots to be $(\operatorname{Re}(z), \operatorname{Im}(z)) \in \left\{\left(\sqrt{3},0\right), \left(\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}\right), \left(0,\sqrt{3}\right)\right\}$. Using the Shoelace Theorem, the area quickly comes out to be $\frac{3\sqrt{2}-3}{2} = \boxed{\textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2}}.$

See Also

2018 AMC 12B (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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