# 2019 AMC 12B Problems/Problem 24

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## Problem

Let $\omega=-\tfrac{1}{2}+\tfrac{1}{2}i\sqrt3.$ Let $S$ denote all points in the complex plane of the form $a+b\omega+c\omega^2,$ where $0\leq a \leq 1,0\leq b\leq 1,$ and $0\leq c\leq 1.$ What is the area of $S$? $\textbf{(A) } \frac{1}{2}\sqrt3 \qquad\textbf{(B) } \frac{3}{4}\sqrt3 \qquad\textbf{(C) } \frac{3}{2}\sqrt3\qquad\textbf{(D) } \frac{1}{2}\pi\sqrt3 \qquad\textbf{(E) } \pi$

## Solution 1

Notice that $\omega=e^{\frac{2i\pi}{3}}$, which is one of the cube roots of unity. We wish to find the span of $(a+b\omega+c\omega^2)$ for reals $0\le a,b,c\le 1$. Observe also that if $a,b,c>0$, then replacing $a$, $b$, and $c$ by $a-\min(a,b,c), b-\min(a,b,c),$ and $c-\min(a,b,c)$ leaves the value of $a+b\omega+c\omega^2$ unchanged. Therefore, assume that at least one of $a,b,c$ is equal to $0$. If exactly one of them is $0$, we can form an equilateral triangle of side length $1$ using the remaining terms. A similar argument works if exactly two of them are $0$. In total, we get $3+{3 \choose 2} = 6$ equilateral triangles, whose total area is $6 \cdot \frac{\sqrt{3}}{4} = \boxed{\textbf{(C) } \frac{3}{2}\sqrt3}$.

Note: A diagram of the six equilateral triangles is shown below. $[asy] size(200,200); draw((0,0)--(1,0)--(1/2,sqrt(3)/2)--cycle); draw((0,0)--(1/2,sqrt(3)/2)--(-1/2,sqrt(3)/2)--cycle); draw((0,0)--(-1/2,sqrt(3)/2)--(-1,0)--cycle); draw((0,0)--(-1,0)--(-1/2,-sqrt(3)/2)--cycle); draw((0,0)--(-1/2,-sqrt(3)/2)--(1/2,-sqrt(3)/2)--cycle); draw((0,0)--(1/2,-sqrt(3)/2)--(1,0)--cycle); draw((-2,0)--(2,0)); draw((0,-2)--(0,2)); [/asy]$

## Solution 1 but not exactly

WLOG $c$ is the smallest of the $3$. Then the expression is equivalent to $a+b\omega$. To find the area of the region, we need only consider the extremities ( $a=0, a=1, b=0, b=1$), as they will form a polygon which contains all points. So, when $a=0, b=0$ we have the origin (diagram omitted). When $a=1, b=0$ we have $1$: $[asy] size(100,100); draw((-2,0)--(2,0), gray); draw((0,-2)--(0,2), gray); draw((0,0)--(1,0), red); [/asy]$

When $a=0, b=1$ we have $-\frac{1}{2}, \frac{\sqrt{3}}{2}$: $[asy] size(100,100); draw((-2,0)--(2,0), gray); draw((0,-2)--(0,2), gray); draw((0,0)--(-1/2,sqrt(3)/2), red); draw((0,0)--(1,0), red); [/asy]$

When $a=1, b=1$ we have $\frac{1}{2}, \frac{\sqrt{3}}{2}$: $[asy] size(100,100); draw((-2,0)--(2,0), gray); draw((0,-2)--(0,2), gray); draw((0,0)--(-1/2,sqrt(3)/2), red); draw((0,0)--(1,0), red); draw((-1/2,sqrt(3)/2)--(1/2,sqrt(3)/2), red); draw((1,0)--(1/2,sqrt(3)/2), red); [/asy]$

The area of this is $\frac{\sqrt{3}}{2}$. Multiply this by $3$ to get $C$.

## Solution 2

We can add on each term one at a time. Firstly, the possible values of $\textstyle c\omega^2=c\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)$ lie on the following line: $[asy] size(100,100); draw((0,0)--(-1/2,-sqrt(3)/2), blue); draw((-2,0)--(2,0)); draw((0,-2)--(0,2)); [/asy]$

For each point on the line, we can add $\textstyle b\omega=b\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)$. This means that we can extend the area to $[asy] size(100,100); fill((0,0)--(-1/2,sqrt(3)/2)--(-1,0)--(-1/2,-sqrt(3)/2)--cycle, lightgray); draw((0,0)--(-1/2,sqrt(3)/2)--(-1,0)--(-1/2,-sqrt(3)/2)--cycle); draw((0,0)--(-1/2,sqrt(3)/2), red); draw((0,0)--(-1/2,-sqrt(3)/2), blue); draw((-2,0)--(2,0)); draw((0,-2)--(0,2)); [/asy]$

by "moving" the blue line along the red line. Finally, we can add $a$ to every point, giving $[asy] size(100,100); fill((-1/2,sqrt(3)/2)--(-1,0)--(-1/2,-sqrt(3)/2)--(1/2,-sqrt(3)/2)--(1,0)--(1/2,sqrt(3)/2)--cycle, lightgray); draw((-1/2,sqrt(3)/2)--(-1,0)--(-1/2,-sqrt(3)/2)--(1/2,-sqrt(3)/2)--(1,0)--(1/2,sqrt(3)/2)--cycle); draw((0,0)--(-1/2,sqrt(3)/2), red); draw((0,0)--(-1/2,-sqrt(3)/2), blue); draw((-2,0)--(2,0)); draw((0,-2)--(0,2)); draw((0,0)--(1,0), heavygreen); [/asy]$

by "moving" the previous area along the green line. This leaves us with a regular hexagon with side length $1$, so, as in Solution 1, the total area is $\boxed{\textbf{(C) } \frac{3}{2}\sqrt{3}}$.

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