Problem 265

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

$r^8={\sqrt3}^8,$ so $r=\sqrt3.$
$\begin{cases} \begin{aligned} \cos(8\theta) &= 1 \\ \sin(8\theta) &= 0 \end{aligned}, \end{cases}$ 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

1. The given equation is $(z+6)^8 = 81$. To solve for $z$, we first take the eighth root of both sides:
$$z + 6 = \sqrt[8]{81} e^{\frac{2\pi i k}{8}}, \quad k \in \{0, 1, 2, \ldots, 7\}$$
Since $81 = 3^4$, we have:
$$\sqrt[8]{81} = \sqrt[8]{3^4} = 3^{\frac{4}{8}} = 3^{\frac{1}{2}} = \sqrt{3}$$
Therefore, the solutions are:
$$z + 6 = \sqrt{3} e^{\frac{2\pi i k}{8}}, \quad k \in \{0, 1, 2, \ldots, 7\}$$
Solving for $z$, we get:
$$z = \sqrt{3} e^{\frac{2\pi i k}{8}} - 6$$

2. These solutions form the vertices of a regular octagon centered at $-6$ with radius $\sqrt{3}$ in the complex plane.

3. To find the least possible area of $\triangle ABC$, we consider three consecutive vertices of the octagon. The vertices of the octagon are given by:
$$z_k = \sqrt{3} e^{\frac{2\pi i k}{8}} - 6, \quad k \in \{0, 1, 2, \ldots, 7\}$$

4. The area of $\triangle ABC$ is minimized when $A$, $B$, and $C$ are consecutive vertices. Let's denote these vertices as $z_0$, $z_1$, and $z_2$.

5. The angle between two consecutive vertices of the octagon is:
$$\theta = \frac{2\pi}{8} = \frac{\pi}{4}$$

6. The area of $\triangle ABC$ can be calculated using the formula for the area of a triangle formed by three points in the complex plane:
$$\text{Area} = \frac{1}{2} \left| \text{Im} \left( z_1 \overline{z_2} + z_2 \overline{z_0} + z_0 \overline{z_1} \right) \right|$$

7. Since the vertices are consecutive, we can use the fact that the area of a triangle formed by three points on a circle is given by:
$$\text{Area} = \frac{1}{2} R^2 \sin \theta$$
where $R$ is the radius of the circle and $\theta$ is the angle subtended by the arc between two points.

8. For our octagon, $R = \sqrt{3}$ and $\theta = \frac{\pi}{4}$:
$$\text{Area} = \frac{1}{2} (\sqrt{3})^2 \sin \left( \frac{\pi}{4} \right) = \frac{1}{2} \cdot 3 \cdot \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{4}$$

9. However, we need to consider the area of the triangle formed by three consecutive vertices, which involves subtracting the area of the overlapping triangles. The correct calculation involves:
$$[ABC] = [OAB] + [OBC] - [OAC]$$
where each of these areas is:
$$[OAB] = \frac{\sqrt{3} \cdot \sqrt{3} \cdot \sin \left( \frac{\pi}{4} \right)}{2} = \frac{3 \cdot \frac{\sqrt{2}}{2}}{2} = \frac{3\sqrt{2}}{4}$$
$$[OBC] = \frac{\sqrt{3} \cdot \sqrt{3} \cdot \sin \left( \frac{\pi}{4} \right)}{2} = \frac{3\sqrt{2}}{4}$$
$$[OAC] = \frac{\sqrt{3} \cdot \sqrt{3} \cdot \sin \left( \frac{\pi}{2} \right)}{2} = \frac{3 \cdot 1}{2} = \frac{3}{2}$$

10. Therefore, the area of $\triangle ABC$ is:
$$[ABC] = \frac{3\sqrt{2}}{4} + \frac{3\sqrt{2}}{4} - \frac{3}{2} = \frac{3\sqrt{2}}{2} - \frac{3}{2}$$

The final answer is $\boxed{\frac{3}{2}\sqrt{2} - \frac{3}{2}}$