Difference between revisions of "2021 Fall AMC 12B Problems/Problem 20"
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==Solution 1 (Direct Counting)== | ==Solution 1 (Direct Counting)== | ||
| − | Divide the <math>2 \times 2 \times 2</math> cube into two layers. | + | Divide the <math>2 \times 2 \times 2</math> cube into two layers, say, front and back. Rotate the cube such that the front layer has the same or greater number of white cubes than black cubes. |
| − | # '''Case 1: Each layer contains 2 cubes of each color.''' Note that we only need to consider the | + | # '''Case 1: Each layer contains 2 cubes of each color.''' Note that we only need to consider the configuration of the white cubes because all the other cubes will be black cubes. There are 2 ways that the 2 white cubes in each layer can be arranged: adjacent or diagonal to each other. |
| − | ## '''Case 1.1: Both layers have | + | ## '''Case 1.1: Both layers have 2 white cubes adjacent to each other.''' Rotate the cube such that there are white cubes along the top edge of the front layer. Now, the white cubes in the back layer can be along the top, bottom, right, or left edges.<sup>See note 1.</sup> So, case 1.1 results in <math>4</math> constructions. |
| − | ## '''Case 1.2: One layer has | + | ## '''Case 1.2: One layer has 2 white cubes adjacent to each other, and the other has 2 white cubes diagonal from each other.''' Rotate the cube such that there are white cubes along the top of the front layer. The white cubes in the back layer can be at the top-left and bottom-right or at the top-right and bottom-left. If we rotate the latter case by 90 degrees clockwise, it becomes the same as the former case. So, case 1.2 results in <math>1</math> additional construction. |
| − | ## '''Case 1.3: Both layers have white cubes diagonal from each other.''' Rotate the cube such that there is a white cube at the top-left | + | ## '''Case 1.3: Both layers have white cubes diagonal from each other.''' Rotate the cube such that there is a white cube at the top-left and bottom-right of the front layer. The back layer could also have white cubes at the top-left and bottom-right, but this is the same as case 1.1 with the white cubes in the back layer along the bottom edge. Alternatively, the back layer could have white cubes at the top-right and the bottom-left. This is a distinct case. So, case 1.3 results in <math>1</math> additional construction. |
| − | ## So case 1 results in <math>4+1+1=6</math> | + | ## So, case 1 results in <math>4+1+1=6</math> distinct constructions. |
| − | # '''Case 2: | + | # '''Case 2: The front layer contains 3 white cubes.''' In this case, unless the sole black and white cubes in the front and back layers are on opposite corners of the <math>2 \times 2 \times 2</math> cube, then the <math>2 \times 2 \times 2</math> cube can be split into left and right layers with 2 cubes of each color in each (these constructions were counted in case 1). So, case 2 results in <math>1</math> additional construction. |
| − | # '''Case 3: | + | # '''Case 3: The front layer contains 4 white cubes.''' Only 1 construction can result from this case. However, if we split this contsruction into its left and right layers, then each layer will have 2 cubes of each color. So, this construction is covered in case 1, and case 3 results in <math>0</math> additional constructions. |
| − | Therefore, our answer is <math>6+1=\boxed{\textbf{(A)}\ 7}</math>. | + | Therefore, our answer is <math>6+1+0=\boxed{\textbf{(A)}\ 7}</math>. |
===Notes=== | ===Notes=== | ||
Revision as of 11:12, 26 November 2021
- The following problem is from both the 2021 Fall AMC 12B #20 and 2021 Fall AMC 12B #24, so both problems redirect to this page.
Problem
A cube is constructed from 
white unit cubes and black unit cubes. How many different ways are there to construct the 
cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)
Solution 1 (Direct Counting)
Divide the cube into two layers, say, front and back. Rotate the cube such that the front layer has the same or greater number of white cubes than black cubes.
- Case 1: Each layer contains 2 cubes of each color. Note that we only need to consider the configuration of the white cubes because all the other cubes will be black cubes. There are 2 ways that the 2 white cubes in each layer can be arranged: adjacent or diagonal to each other.
- Case 1.1: Both layers have 2 white cubes adjacent to each other. Rotate the cube such that there are white cubes along the top edge of the front layer. Now, the white cubes in the back layer can be along the top, bottom, right, or left edges.See note 1. So, case 1.1 results in
constructions. - Case 1.2: One layer has 2 white cubes adjacent to each other, and the other has 2 white cubes diagonal from each other. Rotate the cube such that there are white cubes along the top of the front layer. The white cubes in the back layer can be at the top-left and bottom-right or at the top-right and bottom-left. If we rotate the latter case by 90 degrees clockwise, it becomes the same as the former case. So, case 1.2 results in
additional construction. - Case 1.3: Both layers have white cubes diagonal from each other. Rotate the cube such that there is a white cube at the top-left and bottom-right of the front layer. The back layer could also have white cubes at the top-left and bottom-right, but this is the same as case 1.1 with the white cubes in the back layer along the bottom edge. Alternatively, the back layer could have white cubes at the top-right and the bottom-left. This is a distinct case. So, case 1.3 results in additional construction.
- So, case 1 results in
distinct constructions.
- Case 1.1: Both layers have 2 white cubes adjacent to each other. Rotate the cube such that there are white cubes along the top edge of the front layer. Now, the white cubes in the back layer can be along the top, bottom, right, or left edges.See note 1. So, case 1.1 results in
- Case 2: The front layer contains 3 white cubes. In this case, unless the sole black and white cubes in the front and back layers are on opposite corners of the
cube, then the cube can be split into left and right layers with 2 cubes of each color in each (these constructions were counted in case 1). So, case 2 results in additional construction. - Case 3: The front layer contains 4 white cubes. Only 1 construction can result from this case. However, if we split this contsruction into its left and right layers, then each layer will have 2 cubes of each color. So, this construction is covered in case 1, and case 3 results in
additional constructions.
additional construction.
distinct constructions.
additional constructions.Therefore, our answer is 
.
Notes
1: To prove the 3rd and 4th cases distinct, we can model them with our hands. Extend our thumbs and pointer fingers into an L. These fingers represent the three white cubes on the top layer. Our left and right hands represent the 3rd and 4th cases respectively. The 4th white cube in each case extends down from the tip of each pointer finger towards the rest of each hand. If we overlap our thumbs and pointer fingers, then the 4th cube in each situation will extend outwards in opposite directions, so these cases are distinct.
See Also
| 2021 Fall 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 | |
| 2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
