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: | ||
+ | |||
+ | <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), blue); | ||
+ | draw((-2,0)--(2,0)); | ||
+ | draw((0,-2)--(0,2)); | ||
+ | </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
Contents
[hide]Problem
Let Let denote all points in the complex plane of the form where and What is the area of ?
Solution
Let be the third root of unity. We wish to find the span of for reals . Note that if , then forms the same point as . Therefore, assume that at least one of them is equal to . If only one of them is equal to zero, we can form an equilateral triangle with the remaining two, of side length . Similarly for if two are equal to zero. So the area of the six equilateral triangles is
Here is a diagram:
-programjames1
Solution 2
We can add on each term one at a time. First off, the possible values of lie on the following graph:
For each point on the line, we can add . This means that we can extend the area to
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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