2019 AMC 12B Problems/Problem 24

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 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 $\boxed{\textbf{(C) } \frac{3}{2}\sqrt{3}}$.

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS