Difference between revisions of "2019 AMC 12B Problems/Problem 24"
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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); | ||
| + | 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 | -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 
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
![\[\boxed{\text{(C) }\frac{3\sqrt{3}}{2}}\]](http://latex.artofproblemsolving.com/5/a/1/5a163e3683b65af361716346b4c015a1ea7950fc.png)
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]](http://latex.artofproblemsolving.com/4/2/d/42dee9da6ce367a03187683e00b3061794ee81a5.png)
-programjames1
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 | |
