2019 AMC 12B Problems/Problem 24

Revision as of 12:10, 14 February 2019 by Programjames1 (talk | contribs) (Problem)

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

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

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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All AMC 12 Problems and Solutions
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