Difference between revisions of "Talk:2021 Fall AMC 12B Problems/Problem 20"
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2. the sum of stablizer can be counted as <math>fix(g)</math> | 2. the sum of stablizer can be counted as <math>fix(g)</math> | ||
| − | 3. the sum of the <math>fix(g)/|G|</math> equals | + | 3. the sum of the <math>fix(g)/|G|</math> equals the # of orbit. |
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1. <math>\textbf{e (identity)}</math> | 1. <math>\textbf{e (identity)}</math> | ||
| − | For identity, there are <math>\frac{8!}{4!4!} = 70</math> | + | For identity, there are <math>\frac{8!}{4!4!} = 70</math> |
| − | 2. <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> | + | 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>\ | + | 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> | ||
| − | 4 | + | 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 | ~wwei.yu | ||
Latest revision as of 03:23, 29 November 2021
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 
3. the sum of the 
equals the # of orbit.
Let's start with defining the operator for a cube,
1. 
For identity, there are 
2. 
to be the rotation axis along three pair of opposite face,
each contains 
where 
therefore 
, and 
3. 
to the rotation axis along four cube diagnals.
each contains 
where 
therefore 
, and 
4. 
to be the rotation axis along 6 pairs of diagnally opposite sides
each contains 
where 
therefore 
5. The total number of operators are
Based on 1, 2, 3, 4 the total number of stablizer is
therefore the number of orbit 
~wwei.yu
