AMC12 2019 B

Answer (C): Note that $\omega^2=\bar{\omega}$ and $\omega^3=1$. Then $1+\omega=\frac{1}{2}+\frac{1}{2}i\sqrt{3}\in S$, $\omega+\bar{\omega}=-1\in S$, and $1+\bar{\omega}=\frac{1}{2}-\frac{1}{2}i\sqrt{3}\in S$. These three points, together with the original points $1$, $\omega$, and $\omega^2$, form the vertices of a regular hexagon inscribed in the unit circle $|z|=1$. The segment from $0$ to $1$ is a convex set in the complex plane. The segment from $0$ to $\omega$ is also a convex set, as is the segment from $0$ to $\bar{\omega}$. The sum of convex sets (that is, the set of all numbers of the form $s+t$, where $s$ is in one set and $t$ is in the other) is convex. Thus the set in question contains not just the hexagon whose vertices are the sixth roots of unity, but also its interior. Now consider the largest imaginary part of any point $a+b\omega+c\omega^2$ in $S$. Because \[ \operatorname{Im}(a)=0,\qquad \operatorname{Im}(b\omega)\le \frac{1}{2}\sqrt{3},\qquad \text{and}\qquad \operatorname{Im}(c\bar{\omega})\le 0, \] all points in $S$ have imaginary part less than or equal to $\frac{1}{2}\sqrt{3}$. Similarly, the imaginary parts of points in $S$ are all greater than or equal to $-\frac{1}{2}\sqrt{3}$. Thus $S$ lies in a strip of width $\sqrt{3}$. Furthermore, $\omega S=\omega^2 S=S$. Hence $S$ lies in the intersection of three strips, and this intersection is the hexagon already obtained. The area of the hexagon is $6$ times the area of an equilateral triangle with side length $1$, namely $\frac{3}{2}\sqrt{3}$.