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

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==Problem==
 
==Problem==
 
Let <math>\omega=-\tfrac{1}{2}+\tfrac{1}{2}i\sqrt3.</math> Let <math>S</math> denote all points in the complex plane of the form <math>a+b\omega+c\omega^2,</math> where <math>0\leq a \leq 1,0\leq b\leq 1,</math> and <math>0\leq c\leq 1.</math> What is the area of <math>S</math>?
 
Let <math>\omega=-\tfrac{1}{2}+\tfrac{1}{2}i\sqrt3.</math> Let <math>S</math> denote all points in the complex plane of the form <math>a+b\omega+c\omega^2,</math> where <math>0\leq a \leq 1,0\leq b\leq 1,</math> and <math>0\leq c\leq 1.</math> What is the area of <math>S</math>?
 +
 
<math>\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</math>
 
<math>\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</math>
  
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-programjames1
 
-programjames1
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==Solution 2==
 +
We can add on each term one at a time. First off, the possible values of <math>\textstyle c\omega^2=c\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)</math> lie on the following graph:
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 +
<asy>
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size(100,100);
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draw((0,0)--(-1/2,-sqrt(3)/2), blue);
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draw((-2,0)--(2,0));
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draw((0,-2)--(0,2));
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</asy>
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 +
For each point on the line, we can add <math>\textstyle b\omega=b\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)</math>. This means that we can extend the area to
 +
 +
<asy>
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size(100,100);
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fill((0,0)--(1/2,-sqrt(3)/2)--(-1,0)--(-1/2,-sqrt(3)/2)--cycle, lightgray);
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draw((0,0)--(-1/2,-sqrt(3)/2), blue);
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draw((-2,0)--(2,0));
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draw((0,-2)--(0,2));
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</asy>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2019|ab=B|num-b=23|num-a=25}}
 
{{AMC12 box|year=2019|ab=B|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:11, 14 February 2019

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

Solution 2

We can add on each term one at a time. First off, the possible values of $\textstyle c\omega^2=c\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)$ lie on the following graph:

[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), blue); draw((-2,0)--(2,0)); draw((0,-2)--(0,2)); [/asy]

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

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