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>?
  
==Solution==
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<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>
Let <math>\omega=e^{\frac{2i\pi}{3}}</math> be the third root of unity. We wish to find the span of <math>a+b\omega+c\omega^2)</math> for reals <math>0\le a,b,c\le 1</math>.
 
Note that if <math>a,b,c>0</math>, then <math>a-\min(a,b,c), b-\min(a,b,c), c-\min(a,b,c)</math> forms the same point as <math>a,b,c</math>. Therefore, assume that at least one of them is equal to <math>0</math>. If only one of them is equal to zero, we can form an equilateral triangle with the remaining two, of side length <math>1</math>. Similarly for if two are equal to zero. So the area of the six equilateral triangles is
 
<cmath>\boxed{\text{(C) }\frac{3\sqrt{3}}{2}}</cmath>
 
  
Here is a diagram:
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==Solution 1==
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Notice that <math>\omega=e^{\frac{2i\pi}{3}}</math>, which is one of the cube roots of unity. We wish to find the span of <math>(a+b\omega+c\omega^2)</math> for reals <math>0\le a,b,c\le 1</math>.
 +
Observe also that if <math>a,b,c>0</math>, then replacing <math>a</math>, <math>b</math>, and <math>c</math> by <math>a-\min(a,b,c), b-\min(a,b,c),</math> and <math>c-\min(a,b,c)</math> leaves the value of <math>a+b\omega+c\omega^2</math> unchanged. Therefore, assume that at least one of <math>a,b,c</math> is equal to <math>0</math>. If exactly one of them is <math>0</math>, we can form an equilateral triangle of side length <math>1</math> using the remaining terms. A similar argument works if exactly two of them are <math>0</math>. In total, we get <math>3+{3 \choose 2} = 6</math> equilateral triangles, whose total area is <math>6 \cdot \frac{\sqrt{3}}{4} = \boxed{\textbf{(C) } \frac{3}{2}\sqrt3}</math>.
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''Note'': A diagram of the six equilateral triangles is shown below.
 
<asy>
 
<asy>
 
size(200,200);
 
size(200,200);
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</asy>
 
</asy>
  
-programjames1
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==Solution 2==
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We can add on each term one at a time. Firstly, 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 line:
 +
 
 +
<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>
 +
 
 +
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)--(-1,0)--(-1/2,-sqrt(3)/2)--cycle);
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draw((0,0)--(-1/2,sqrt(3)/2), red);
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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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 +
by "moving" the blue line along the red line. Finally, we can add <math>a</math> to every point, giving
 +
 
 +
<asy>
 +
size(100,100);
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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);
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draw((0,0)--(-1/2,sqrt(3)/2), red);
 +
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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draw((0,0)--(1,0), heavygreen);
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</asy>
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 +
by "moving" the previous area along the green line. This leaves us with a regular hexagon with side length <math>1</math>, so, as in Solution 1, the total area is <math>\boxed{\textbf{(C) } \frac{3}{2}\sqrt{3}}</math>.
  
 
==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}}

Revision as of 00:26, 19 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 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

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