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> | ||
− | ==Solution== | + | ==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> | ||
− | - | + | ==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 21:23, 15 November 2022
Problem
Let Let
denote all points in the complex plane of the form
where
and
What is the area of
?
Solution 1
Notice that , which is one of the cube roots of unity. We wish to find the span of
for reals
.
Observe also that if
, then replacing
,
, and
by
and
leaves the value of
unchanged. Therefore, assume that at least one of
is equal to
. If exactly one of them is
, we can form an equilateral triangle of side length
using the remaining terms. A similar argument works if exactly two of them are
. In total, we get
equilateral triangles, whose total area is
.
Note: A diagram of the six equilateral triangles is shown below.
Solution 1 but not exactly
WLOG is the smallest of the
. Then the expression is equivalent to
. To find the area of the region, we need only consider the extremities (
), as they will form a polygon which contains all points. So, when
we have the origin (diagram omitted). When
we have
:
When we have
:
When we have
:
The area of this is . Multiply this by
to get
.
Solution 2
We can add on each term one at a time. Firstly, the possible values of lie on the following line:
For each point on the line, we can add . This means that we can extend the area to
by "moving" the blue line along the red line. Finally, we can add to every point, giving
by "moving" the previous area along the green line. This leaves us with a regular hexagon with side length , so, as in Solution 1, the total area is
.
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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