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

(Solution)
(Solution)
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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
 
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>
 
<cmath>\boxed{\text{(C) }\frac{3\sqrt{3}}{2}}</cmath>
 +
 +
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);
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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);
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draw((-2,0)--(2,0));
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draw((0,-2)--(0,2));
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</asy>
 +
 
-programjames1
 
-programjames1
  
 
==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}}

Revision as of 12:09, 14 February 2019

Problem

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

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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