# Difference between revisions of "2019 AMC 12B Problems/Problem 24"

## Solution

Let $\omega=e^{\frac{2i\pi}{3}}$ be the third root of unity. We wish to find the span of $a+b\omega+c\omega^2)$ for reals $0\le a,b,c\le 1$. Note that if $a,b,c>0$, then $a-\min(a,b,c), b-\min(a,b,c), c-\min(a,b,c)$ forms the same point as $a,b,c$. Therefore, assume that at least one of them is equal to $0$. If only one of them is equal to zero, we can form an equilateral triangle with the remaining two, of side length $1$. Similarly for if two are equal to zero. So the area of the six equilateral triangles is $$\boxed{\text{(C) }\frac{3\sqrt{3}}{2}}$$

Here is a diagram: $[asy] size(200,200); draw((0,0)--(1,0)--(1/2,sqrt(3)/2)--cycle); draw((0,0)--(1/2,sqrt(3)/2)--(-1/2,sqrt(3)/2)--cycle); draw((0,0)--(-1/2,sqrt(3)/2)--(-1,0)--cycle); draw((0,0)--(-1,0)--(-1/2,-sqrt(3)/2)--cycle); draw((0,0)--(-1/2,-sqrt(3)/2)--(1/2,-sqrt(3)/2)--cycle); draw((0,0)--(1/2,-sqrt(3)/2)--(1,0)--cycle); draw((-2,0)--(2,0)); draw((0,-2)--(0,2)); [/asy]$

-programjames1