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Difference between revisions of "2020 AMC 12B Problems/Problem 20"

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==Solution==
 
==Solution==
 
Define two ways of painting to be in the same <math>class</math> if one can be rotated to form the other.
 
Define two ways of painting to be in the same <math>class</math> if one can be rotated to form the other.
 +
 
We can count the number of ways of painting for each specific <math>class</math>.
 
We can count the number of ways of painting for each specific <math>class</math>.
 +
 
Case 1: Red-blue color distribution is 0-6 (out of 6 total faces)
 
Case 1: Red-blue color distribution is 0-6 (out of 6 total faces)
 +
 
Trivially <math>1^2 = 1</math> way to paint the cubes.
 
Trivially <math>1^2 = 1</math> way to paint the cubes.
 +
 
Case 2: Red-blue color distribution is 1-5
 
Case 2: Red-blue color distribution is 1-5
 +
 
Trivially all <math>\dbinom{6}{5} = 6</math> ways belong to the same <math>class</math>, so <math>6^2</math> ways to paint the cubes.
 
Trivially all <math>\dbinom{6}{5} = 6</math> ways belong to the same <math>class</math>, so <math>6^2</math> ways to paint the cubes.
 +
 
Case 3: Red-blue color distribution is 2-4
 
Case 3: Red-blue color distribution is 2-4
 +
 
There are two <math>classes</math> for this case: the <math>class</math> where the two red faces are touching and the other <math>class</math> where the two red faces are on opposite faces. There are <math>3</math> members of the latter <math>class</math> since there are <math>3</math> unordered pairs of <math>2</math> opposite faces of a cube. Thus, there are <math>\dbinom{6}{4} - 3 = 12</math> members of the former <math>class</math>. Thus, <math>12^2 + 3^2</math> ways to paint the cubes for this case.
 
There are two <math>classes</math> for this case: the <math>class</math> where the two red faces are touching and the other <math>class</math> where the two red faces are on opposite faces. There are <math>3</math> members of the latter <math>class</math> since there are <math>3</math> unordered pairs of <math>2</math> opposite faces of a cube. Thus, there are <math>\dbinom{6}{4} - 3 = 12</math> members of the former <math>class</math>. Thus, <math>12^2 + 3^2</math> ways to paint the cubes for this case.
 +
 
Case 4: Red-blue color distribution is 3-3
 
Case 4: Red-blue color distribution is 3-3
 +
 
By simple intuition, there are also two <math>classes</math> for this case, the <math>class</math> 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 <math>8</math> vertices in a cube, there are <math>8</math> members of the former class and <math>\dbinom{6}{3} - 8 = 12</math> members of the latter class. Thus, <math>12^2 + 8^2</math> ways to paint the cubes for this case.
 
By simple intuition, there are also two <math>classes</math> for this case, the <math>class</math> 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 <math>8</math> vertices in a cube, there are <math>8</math> members of the former class and <math>\dbinom{6}{3} - 8 = 12</math> members of the latter class. Thus, <math>12^2 + 8^2</math> ways to paint the cubes for this case.
 +
 
Note that by symmetry (since we are literally 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).
 
Note that by symmetry (since we are literally 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<cmath>\frac{2(6^2 + 1^2 + 12^2 + 3^2) + 12^2 + 8^2}{2^{12}} = \frac{588}{4096} = \boxed{\frac{147}{1024}}.</cmath>
 
Thus, our total answer is<cmath>\frac{2(6^2 + 1^2 + 12^2 + 3^2) + 12^2 + 8^2}{2^{12}} = \frac{588}{4096} = \boxed{\frac{147}{1024}}.</cmath>
  
  
 
-fidgetboss_4000
 
-fidgetboss_4000

Revision as of 16:03, 7 February 2020

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?

$\textbf{(A)}\ \frac{9}{64} \qquad\textbf{(B)}\ \frac{289}{2048} \qquad\textbf{(C)}\ \frac{73}{512} \qquad\textbf{(D)}\ \frac{147}{1024} \qquad\textbf{(E)}\ \frac{589}{4096}$


Solution

Define two ways of painting to be in the same $class$ if one can be rotated to form the other.

We can count the number of ways of painting for each specific $class$.

Case 1: Red-blue color distribution is 0-6 (out of 6 total faces)

Trivially $1^2 = 1$ way to paint the cubes.

Case 2: Red-blue color distribution is 1-5

Trivially all $\dbinom{6}{5} = 6$ ways belong to the same $class$, so $6^2$ ways to paint the cubes.

Case 3: Red-blue color distribution is 2-4

There are two $classes$ for this case: the $class$ where the two red faces are touching and the other $class$ where the two red faces are on opposite faces. There are $3$ members of the latter $class$ since there are $3$ unordered pairs of $2$ opposite faces of a cube. Thus, there are $\dbinom{6}{4} - 3 = 12$ members of the former $class$. Thus, $12^2 + 3^2$ ways to paint the cubes for this case.

Case 4: Red-blue color distribution is 3-3

By simple intuition, there are also two $classes$ for this case, the $class$ 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 $8$ vertices in a cube, there are $8$ members of the former class and $\dbinom{6}{3} - 8 = 12$ members of the latter class. Thus, $12^2 + 8^2$ ways to paint the cubes for this case.

Note that by symmetry (since we are literally 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\[\frac{2(6^2 + 1^2 + 12^2 + 3^2) + 12^2 + 8^2}{2^{12}} = \frac{588}{4096} = \boxed{\frac{147}{1024}}.\]


-fidgetboss_4000

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