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Difference between revisions of "2019 AMC 12B Problems/Problem 24"
(soln 2)
Line 38: Line 38:
 
<asy>
 
<asy>
 
size(100,100);
 
size(100,100);
−
fill((0,0)--(1/2,-sqrt(3)/2)--(-1,0)--(-1/2,-sqrt(3)/2)--cycle, lightgray);
+
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((0,0)--(-1/2,-sqrt(3)/2), blue);
 
draw((-2,0)--(2,0));
 
draw((-2,0)--(2,0));
 
draw((0,-2)--(0,2));
 
draw((0,-2)--(0,2));
 
</asy>
 
</asy>
 +
 +
by "moving" the blue line along the red line. Finally, we can add <math>a</math> to every point
 +
 +
<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 the total area is <math>\boxed{\textbf{(C) } \frac{3}{2}\sqrt{3}}</math>.
 +
 +
~~IYN~~
  
 
==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 18:19, 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)--(-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

[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 the total area is $\boxed{\textbf{(C) } \frac{3}{2}\sqrt{3}}$.

~~IYN~~

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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