Difference between revisions of "2020 AMC 12B Problems/Problem 20"
Argonauts16 (talk | contribs) (→Solution) |
|||
| Line 28: | Line 28: | ||
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{\textbf{(D) } \frac{147}{1024}}.</cmath> |
Revision as of 20:42, 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?
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 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![\[\frac{2(6^2 + 1^2 + 12^2 + 3^2) + 12^2 + 8^2}{2^{12}} = \frac{588}{4096} = \boxed{\textbf{(D) } \frac{147}{1024}}.\]](http://latex.artofproblemsolving.com/4/0/9/4090b7116c16c14b3f455aea87e8b6520c02044a.png)
-fidgetboss_4000
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
