Difference between revisions of "2020 AMC 12B Problems/Problem 20"
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| + | ==Solution 1.5== | ||
| + | (The above solution explains the solution very well; however the thought process is not explicitly addressed) | ||
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| + | We want P(two cubes can be rotated to match after they are painted); i.e. P(cubes are rotationally identical). By Polya's enumeration theorem (from Burnside's Lemma), there are <math>n^6 + 3n^4 +12n^3 +8n^2</math> distinct ways to color a cube using <math>n</math> colors. Plugging in <math>n = 2</math>, we find that there are 10 distinct ways to color a cube with 2 colors (meaning none of these 10 colorations can be rotated to match another). However, each coloration does not have the same probability of occuring (the cubes are painted randomly); for example, there is only 1 way to paint both cubes all black, as any rotations will not yield a visually different result. There are multiple ways to paint each cube with 1 white face and 5 black faces, as one cube can simply be rotated to match the other. Therefore, we have to do casework for 10 cases, where each case is a distinct way to color a cube. (The above solution sometimes addresses more than one case at once. | ||
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| + | Case 1: Both cubes are completely black <math>\right arrow</math> 1 way, P(Case 1) = <math>\frac{1^2}{4096}</math> (There are 2^12 or 4096 total colorations of the cubes) | ||
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| + | Case 2: Exactly one face is white, the rest are black <math>\right arrow</math> 6 ways, P(Case 2) = <math>\frac{6^2}{4096}</math> | ||
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| + | Case 3: 2 white faces share an edge, the rest are black <math>\right arrow</math> <math>\frac{6 choices for the first white face * 4 choices for the second adjacent white face}{2!(because the white faces are identical)}</math> = 12 ways, P(Case 3) = <math>\frac{12^2}{4096}</math> | ||
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| + | Case 4: 2 white faces opposite each other, the rest are black <math>\right arrow</math> <math>\frac{6 choices for the first white face * 1 choice for the second white face}{2!}</math> = 3 ways, P(Case 4) = <math>\frac{3^2}{4096}</math> | ||
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| + | Case 5: 3 white faces share a vertex, the rest are black <math>\right arrow</math> <math>\frac{6 choices for the first * 4 choices for the second * 3 choices for the third}{3!}</math> = 12 ways, P(Case 5) = <math>\frac{12^2}{4096}</math> | ||
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| + | Case 6: 3 white faces in a "row", the rest are black <math>\right arrow</math> #\frac{6 choices for the first * 4 choices for the second * 2 choice for the third}{3!}<math> = 8 ways, P(Case 5) = </math>\frac{8^2}{4096}<math> | ||
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| + | Cases 7, 8, 9, and 10: These cases are the same as cases 1, 2, 3, and 4, albeit with a color swap. Doubling the probabilities of cases 1-4 will cover these cases. | ||
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| + | Therefore we have P(cubes are rotationally identical) = </math>\frac{1^2 + 6^2 + 12^2 + 3^2 + 12^2 + 8^2 + 3^2 + 12^2 + 6^2 + 1^2}{4096}$ = <cmath>\frac{588}{4096} =\boxed{\textbf{(D) } \frac{147}{1024}}.</cmath> | ||
==See Also== | ==See Also== | ||
Revision as of 11:11, 23 June 2020
Contents
Problem
Two different cubes of the same size are to be painted, with the color of each face being chosen independently and at random to be either black or white. What is the probability that after they are painted, the cubes can be rotated to be identical in appearance?
Solution
Define two ways of painting to be in the same 
if one can be rotated to form the other.
We can count the number of ways of painting for each specific .
Case 1: Red-blue color distribution is 0-6 (out of 6 total faces)
Trivially 
way to paint the cubes.
Case 2: Red-blue color distribution is 1-5
Trivially all 
ways belong to the same , so 
ways to paint the cubes.
Case 3: Red-blue color distribution is 2-4
There are two 
for this case: the where the two red faces are touching and the other where the two red faces are on opposite faces. There are 
members of the latter since there are unordered pairs of 
opposite faces of a cube. Thus, there are 
members of the former . Thus, 
ways to paint the cubes for this case.
Case 4: Red-blue color distribution is 3-3
By simple intuition, there are also two for this case, the where the three red faces meet at a single vertex, and the other class where the three red faces are in a "straight line" along the edges of the cube. Note that since there are 
vertices in a cube, there are members of the former class and 
members of the latter class. Thus, 
ways to paint the cubes for this case.
Note that by symmetry (since we are only switching the colors), the number of ways to paint the cubes for Red-blue color distributions 4-2, 5-1, and 6-0 is 2-4, 1-5, and 0-6 (respectively).
Thus, our total answer is![\[\frac{2(6^2 + 1^2 + 12^2 + 3^2) + 12^2 + 8^2}{2^{12}} = \frac{588}{4096} = \boxed{\textbf{(D) } \frac{147}{1024}}.\]](http://latex.artofproblemsolving.com/4/0/9/4090b7116c16c14b3f455aea87e8b6520c02044a.png)
-fidgetboss_4000
Solution 1.5
(The above solution explains the solution very well; however the thought process is not explicitly addressed)
We want P(two cubes can be rotated to match after they are painted); i.e. P(cubes are rotationally identical). By Polya's enumeration theorem (from Burnside's Lemma), there are 
distinct ways to color a cube using 
colors. Plugging in 
, we find that there are 10 distinct ways to color a cube with 2 colors (meaning none of these 10 colorations can be rotated to match another). However, each coloration does not have the same probability of occuring (the cubes are painted randomly); for example, there is only 1 way to paint both cubes all black, as any rotations will not yield a visually different result. There are multiple ways to paint each cube with 1 white face and 5 black faces, as one cube can simply be rotated to match the other. Therefore, we have to do casework for 10 cases, where each case is a distinct way to color a cube. (The above solution sometimes addresses more than one case at once.
Case 1: Both cubes are completely black $\right arrow$ (Error compiling LaTeX. Unknown error_msg) 1 way, P(Case 1) = 
(There are 2^12 or 4096 total colorations of the cubes)
Case 2: Exactly one face is white, the rest are black $\right arrow$ (Error compiling LaTeX. Unknown error_msg) 6 ways, P(Case 2) = 
Case 3: 2 white faces share an edge, the rest are black $\right arrow$ (Error compiling LaTeX. Unknown error_msg) 
= 12 ways, P(Case 3) = 
Case 4: 2 white faces opposite each other, the rest are black $\right arrow$ (Error compiling LaTeX. Unknown error_msg) 
= 3 ways, P(Case 4) = 
Case 5: 3 white faces share a vertex, the rest are black $\right arrow$ (Error compiling LaTeX. Unknown error_msg) 
= 12 ways, P(Case 5) =
Case 6: 3 white faces in a "row", the rest are black $\right arrow$ (Error compiling LaTeX. Unknown error_msg) #\frac{6 choices for the first * 4 choices for the second * 2 choice for the third}{3!}
\frac{8^2}{4096}$Cases 7, 8, 9, and 10: These cases are the same as cases 1, 2, 3, and 4, albeit with a color swap. Doubling the probabilities of cases 1-4 will cover these cases.
Therefore we have P(cubes are rotationally identical) =$ (Error compiling LaTeX. Unknown error_msg)\frac{1^2 + 6^2 + 12^2 + 3^2 + 12^2 + 8^2 + 3^2 + 12^2 + 6^2 + 1^2}{4096}$ = ![\[\frac{588}{4096} =\boxed{\textbf{(D) } \frac{147}{1024}}.\]](http://latex.artofproblemsolving.com/7/8/b/78bc6374e814cfc26e84952d2923a03157a569a2.png)
See Also
| 2020 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 19 |
Followed by Problem 21 |
| 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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