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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>?
−
<cmath>\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</cmath>
 
  
−
==Solution==
+
<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:
+
==Solution 1==
 +
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>.
 +
 
 +
''Note'': A diagram of the six equilateral triangles is shown below.
 
<asy>
 
<asy>
 
size(200,200);
 
size(200,200);
Line 21: Line 21:
 
</asy>
 
</asy>
  
−
-programjames1
+
==Solution 1 but not exactly==
 +
WLOG <math>c</math> is the smallest of the <math>3</math>. Then the expression is equivalent to <math>a+b\omega</math>. To find the area of the region, we need only consider the extremities (<math>a=0, a=1, b=0, b=1</math>), as they will form a polygon which contains all points. So, when <math>a=0, b=0</math> we have the origin (diagram omitted). When <math>a=1, b=0</math> we have <math>1</math>:
 +
<asy>
 +
size(100,100);
 +
draw((-2,0)--(2,0), gray);
 +
draw((0,-2)--(0,2), gray);
 +
draw((0,0)--(1,0), red);
 +
</asy>
 +
 
 +
When <math>a=0, b=1</math> we have <math>-\frac{1}{2}, \frac{\sqrt{3}}{2}</math>:
 +
<asy>
 +
size(100,100);
 +
draw((-2,0)--(2,0), gray);
 +
draw((0,-2)--(0,2), gray);
 +
draw((0,0)--(-1/2,sqrt(3)/2), red);
 +
draw((0,0)--(1,0), red);
 +
</asy>
 +
 
 +
When <math>a=1, b=1</math> we have <math>\frac{1}{2}, \frac{\sqrt{3}}{2}</math>:
 +
<asy>
 +
size(100,100);
 +
draw((-2,0)--(2,0), gray);
 +
draw((0,-2)--(0,2), gray);
 +
draw((0,0)--(-1/2,sqrt(3)/2), red);
 +
draw((0,0)--(1,0), red);
 +
draw((-1/2,sqrt(3)/2)--(1/2,sqrt(3)/2), red);
 +
draw((1,0)--(1/2,sqrt(3)/2), red);
 +
</asy>
 +
 
 +
The area of this is <math>\frac{\sqrt{3}}{2}</math>. Multiply this by <math>3</math> to get <math>C</math>.
 +
 
 +
==Solution 2==
 +
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>
 +
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 <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>
 +
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 <math>a</math> 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 <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}}

Latest revision as of 20:23, 15 November 2022

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 1 but not exactly

WLOG $c$ is the smallest of the $3$. Then the expression is equivalent to $a+b\omega$. To find the area of the region, we need only consider the extremities ($a=0, a=1, b=0, b=1$), as they will form a polygon which contains all points. So, when $a=0, b=0$ we have the origin (diagram omitted). When $a=1, b=0$ we have $1$: [asy] size(100,100); draw((-2,0)--(2,0), gray); draw((0,-2)--(0,2), gray); draw((0,0)--(1,0), red); [/asy]

When $a=0, b=1$ we have $-\frac{1}{2}, \frac{\sqrt{3}}{2}$: [asy] size(100,100); draw((-2,0)--(2,0), gray); draw((0,-2)--(0,2), gray); draw((0,0)--(-1/2,sqrt(3)/2), red); draw((0,0)--(1,0), red); [/asy]

When $a=1, b=1$ we have $\frac{1}{2}, \frac{\sqrt{3}}{2}$: [asy] size(100,100); draw((-2,0)--(2,0), gray); draw((0,-2)--(0,2), gray); draw((0,0)--(-1/2,sqrt(3)/2), red); draw((0,0)--(1,0), red); draw((-1/2,sqrt(3)/2)--(1/2,sqrt(3)/2), red); draw((1,0)--(1/2,sqrt(3)/2), red); [/asy]

The area of this is $\frac{\sqrt{3}}{2}$. Multiply this by $3$ to get $C$.

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 (Problems • Answer Key • Resources)
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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