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Difference between revisions of "2017 AMC 12B Problems/Problem 13"

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==Problem 13==
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==Problem==
 
In the figure below, <math>3</math> of the <math>6</math> disks are to be painted blue, <math>2</math> are to be painted red, and <math>1</math> is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
 
In the figure below, <math>3</math> of the <math>6</math> disks are to be painted blue, <math>2</math> are to be painted red, and <math>1</math> is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
  
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<math>\textbf{(A) } 6 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15</math>
 
<math>\textbf{(A) } 6 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15</math>
  
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==Solution==
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==Solution 1==
 
Looking at the answer choices, we see that the possibilities are indeed countable. Thus, we will utilize that approach in the form of two separate cases, as rotation and reflection take care of numerous possibilities. First, consider the case that the green disk is in a corner. This yields <math>6</math> possible arrangements for the <math>3</math> blue disks and <math>2</math> red disks in the remaining available slots. Now, consider the case that the green disk is on an edge. This yields <math>6</math> more possible arrangements for the <math>3</math> blue disks and <math>2</math> red disks in the remaining available slots. Thus, our answer is <math>6 + 6 = \boxed{\bold{(D)}\, 12}</math>
 
Looking at the answer choices, we see that the possibilities are indeed countable. Thus, we will utilize that approach in the form of two separate cases, as rotation and reflection take care of numerous possibilities. First, consider the case that the green disk is in a corner. This yields <math>6</math> possible arrangements for the <math>3</math> blue disks and <math>2</math> red disks in the remaining available slots. Now, consider the case that the green disk is on an edge. This yields <math>6</math> more possible arrangements for the <math>3</math> blue disks and <math>2</math> red disks in the remaining available slots. Thus, our answer is <math>6 + 6 = \boxed{\bold{(D)}\, 12}</math>
  
 
Solution by akaashp11
 
Solution by akaashp11
  
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Solution 2  
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==Solution 2 (Burnside's Lemma)==
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Since rotations that result in the same painting are considered equivalent this means that we only have two sets of paintings that can be produced one set would involve the green disk in a corner and the other set would involve the green disk on an edge. If the green disk is on a corner or edge it leaves open 5 spots for the 3 blue disks and 2 red disks. This means that there are 5 choose 3 ways to arrange the blue disks and there are 5 choose 2 ways to arrange the red disks (however these are equivalent since if you put down 3 disks in 5 spots that leaves two spots open for the remaining 2 disks). So the total number of ways to arrange the disks and take into consideration the rotation limit is 2(5 choose  3) (because you need to account for both the green being on an edge or in a corner) which equals 20. However you need to take into account that any painting that can be obtained by reflection is also equivalent. There are 4 such arrangements and for each 2 paintings can be obtained through reflection. Which means that our answer is 20-8 = 12
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Burnside's Lemma can be applied to the problem. Number the disks 1-6 from top to bottom and left to right. First, we count the number of group actions. There are <math>6</math> in total: <math>3</math> rotations, <math>0</math>°, <math>120</math>°, and <math>240</math>°. Additionally, there are <math>3</math> different reflections, which are over lines passing through the middle of disks <math>1</math>, <math>4</math>, and <math>6</math>. The <math>0</math>° rotation has the total number of diagrams fixed, which is <math>6!/3!\cdot2!=60</math>. The <math>120</math>° and <math>240</math>° rotations each have <math>0</math> fixed diagrams, as one would need the same color for disks  
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<math>1, 4, 6</math> and the same color for disks <math>2, 3, 5,</math> which isn't possible. Consider the reflection over the line passing through disk <math>1</math>.
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The green can be placed in <math>2</math> places, disk <math>1</math> or <math>5</math>, and then, the blue must go in either disks <math>2</math> and <math>3</math> or disks <math>4</math> and <math>6</math>, so the number of fixed diagrams is <math>2\cdot2=4</math>. The other two reflections also have <math>4</math> unique fixed diagrams. Therefore, applying Burnside's Lemma, we get <math>(60+4+4+4)/6 = \boxed{\bold{(D)}\, 12}</math>
  
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solution by brownboy
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Solution by Aadileo
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==Video Solution==
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https://youtu.be/T7eNQISBzsc?si=eOKSzNorVrHaV1px ~TheBeautyOfMath
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-aashrithm
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2017|ab=B|num-b=12|num-a=14}}
 
{{AMC12 box|year=2017|ab=B|num-b=12|num-a=14}}
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{{AMC10 box|year=2017|ab=B|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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 +
[[Category:Introductory Combinatorics Problems]]

Latest revision as of 09:00, 28 July 2024

Problem

In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?

[asy] size(100); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy]

$\textbf{(A) } 6 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15$

Solution 1

Looking at the answer choices, we see that the possibilities are indeed countable. Thus, we will utilize that approach in the form of two separate cases, as rotation and reflection take care of numerous possibilities. First, consider the case that the green disk is in a corner. This yields $6$ possible arrangements for the $3$ blue disks and $2$ red disks in the remaining available slots. Now, consider the case that the green disk is on an edge. This yields $6$ more possible arrangements for the $3$ blue disks and $2$ red disks in the remaining available slots. Thus, our answer is $6 + 6 = \boxed{\bold{(D)}\, 12}$

Solution by akaashp11

Solution 2 (Burnside's Lemma)

Burnside's Lemma can be applied to the problem. Number the disks 1-6 from top to bottom and left to right. First, we count the number of group actions. There are $6$ in total: $3$ rotations, $0$°, $120$°, and $240$°. Additionally, there are $3$ different reflections, which are over lines passing through the middle of disks $1$, $4$, and $6$. The $0$° rotation has the total number of diagrams fixed, which is $6!/3!\cdot2!=60$. The $120$° and $240$° rotations each have $0$ fixed diagrams, as one would need the same color for disks $1, 4, 6$ and the same color for disks $2, 3, 5,$ which isn't possible. Consider the reflection over the line passing through disk $1$. The green can be placed in $2$ places, disk $1$ or $5$, and then, the blue must go in either disks $2$ and $3$ or disks $4$ and $6$, so the number of fixed diagrams is $2\cdot2=4$. The other two reflections also have $4$ unique fixed diagrams. Therefore, applying Burnside's Lemma, we get $(60+4+4+4)/6 = \boxed{\bold{(D)}\, 12}$

Solution by Aadileo

Video Solution

https://youtu.be/T7eNQISBzsc?si=eOKSzNorVrHaV1px ~TheBeautyOfMath

-aashrithm

See Also

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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
2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 10 Problems and Solutions

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