View source for Talk:2021 Fall AMC 12B Problems/Problem 20

==Solution 3 (Burnside Lemma)==

Burnside lemma is used to counting number of orbit where the element on the same orbit can be achieved by the defined operator, naming rotation, reflection and etc. 

The fact for Burnside lemma are 

1. the sum of stablizer on the same orbit equals to the # of operators; 

2. the sum of stablizer can be counted as <math>fix(g)</math> 

3. the sum of the <math>fix(g)/|G|</math> equals the # of orbit. 


Let's start with defining the operator for a cube, 

1. <math>\textbf{e (identity)}</math>

For identity, there are <math>\frac{8!}{4!4!} = 70</math> 


2. <math>{\bf r^{1}, r^{2}, r^{3}}</math> to be the rotation axis along three pair of opposite face, 

each contains <math>r^{i}_{90}, r^{i}_{180}, r^{i}_{270}</math> where <math>i= 1, 2, 3</math>

<math>fix(r^{i}_{90}) = fix(r^{i}_{270}) = 2\cdot1 = 2 </math>

<math>fix(r^{i}_{180}) = \frac{4!}{2!\cdot2!} = 6 </math>

therefore <math>fix(\bf r^{i}) = 2+2+6 = 10 </math>, and <math>fix(\bf r^{1})+fix(\bf r^{2})+fix(\bf r^{3}) = 30 </math>
 

3. <math>{\bf r^{4}, r^{5}, r^{6}, r^{7}}</math> to the rotation axis along four cube diagnals. 

each contains <math>r^{i}_{120}, r^{i}_{240}</math>   where <math>i= 4, 5, 6, 7</math>

<math>fix(r^{i}_{120}) = fix(r^{i}_{240}) = 2\cdot1\cdot2\cdot1 = 4 </math>

therefore <math>fix(\bf r^{i}) = 4+4 = 8 </math>, and <math>fix(\bf r^{4})+fix(\bf r^{5})+fix(\bf r^{6})+fix(\bf r^{7}) = 32 </math>



4. <math>{\bf r^{8}, r^{9}, r^{10}, r^{11}, r^{12}, r^{13}}</math> to be the rotation axis along 6 pairs of diagnally opposite sides

each contains <math>r^{i}_{180}</math>   where <math>i= 8, 9, 10, 11, 12, 13</math>

<math>fix(r^{i}_{180}) = \frac{4!}{2!\cdot2!} = 6 </math>

therefore <math>fix(\bf r^{8})+fix(\bf r^{9})+fix(\bf r^{10})+fix(\bf r^{11})+fix(\bf r^{12})+fix(\bf r^{13}) = 36 </math>


5. The total number of operators are 

<math>|G| = 1 + 3\cdot3 + 4\cdot2 + 6\cdot1 = 24 </math>

Based on 1, 2, 3, 4 the total number of stablizer is 
<math>70 + 30 + 32 + 36 = 168</math>

therefore the number of orbit <math>= \frac{168}{G=24} = \boxed{7} </math>


~wwei.yu