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

(Fixed grammar and formatting)
 
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draw((0,-2)--(0,2));
 
draw((0,-2)--(0,2));
 
</asy>
 
</asy>
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 +
==Solution 1 but not exactly==
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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>:
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<asy>
 +
size(100,100);
 +
draw((-2,0)--(2,0), gray);
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draw((0,-2)--(0,2), gray);
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draw((0,0)--(1,0), red);
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</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);
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draw((1,0)--(1/2,sqrt(3)/2), red);
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</asy>
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 +
The area of this is <math>\frac{\sqrt{3}}{2}</math>. Multiply this by <math>3</math> to get <math>C</math>.
  
 
==Solution 2==
 
==Solution 2==

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 (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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