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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 20"

(→‎Solution 1 (Direct Counting))
(→‎Solution 1 (Direct Counting): something is still wrong lol)
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Case 1.1: Both layers have two white cubes adjacent to each other.  
 
Case 1.1: Both layers have two white cubes adjacent to each other.  
  
−
Rotate the cube such that there is a white cube along the top of the first layer.  
+
Rotate the cube such that there are white cubes along the top of one layer. Now, consider the placement of the white cubes in the other layer:  
−
 
 
−
Now, consider the placement of the white cubes in the other layer:  
 
  
 
Case 1.1.1: The white cubes are along the top.
 
Case 1.1.1: The white cubes are along the top.
  
 
Case 1.1.2: The white cubes are along the bottom.  
 
Case 1.1.2: The white cubes are along the bottom.  
−
 
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This is hopefully obviously distinct from case 1.1.1.
 
  
 
Case 1.1.3: The white cubes are along the left.  
 
Case 1.1.3: The white cubes are along the left.  
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Case 1.1.4: The white cubes are along the right.  
 
Case 1.1.4: The white cubes are along the right.  
  
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There can be both adjacent and on top of each other, both adjacent but the two layers diagonal to each other, or both diagonal and  
+
All four of these cases are distinct (1.1.3 and 1.1.4 can be verified so by sketching the other faces of each).
 +
 
 +
Case 1.2: One layer has two white cubes adjacent to each other, while the other has two white cubes diagonal from each other.
 +
 
 +
Rotate the cube such that there are white cubes along the top of one layer. Now, consider the placement of the white cubes in the other layer:
 +
 
 +
Case 1.2.1: The white cubes are at the top-left and bottom-right.
 +
 
 +
Case 1.2.3: The white cubes are at the top-right and bottom-left.
 +
 
 +
These two cases can be verified to be distinct by sketching the other faces.
 +
 
 +
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 of one layer. Therefore, there must be a white cube at the bottom-right of that layer.
 +
 
 +
Case 1.3.1: There other layer also has white cubes at the top-left and the bottom-right. This is the same as case 1.1.2 approached from the perpendicular direction.
  
−
both diagonal but the two layers on top of each other
+
Case 1.3.2: The other layer has white cubes at the top-right and the bottom-left. This is a distinct case.
  
−
one adjacent and one diagonal (it doesn't matter diagonal which way), both diagonal and the two layers across from each other,  In the last case, this is the same as the second case. So we have four arrangements here.  
+
So now we have seven cases.  
  
 
Case 2: Each layer contains 3 cubes of one color and 1 cube of the other color.
 
Case 2: Each layer contains 3 cubes of one color and 1 cube of the other color.
Line 47: Line 59:
 
Only 1 possible <math>2 \times 2 \times 2</math> cube can result from this case. However, if we divide up this <math>2 \times 2 \times 2</math> cube in the direction perpendicular to the way we first did, then we see this is a repeat of the case where we have 2 white cubes in the first layer and the second layer is arranged such that whites and blacks are straight with each other.
 
Only 1 possible <math>2 \times 2 \times 2</math> cube can result from this case. However, if we divide up this <math>2 \times 2 \times 2</math> cube in the direction perpendicular to the way we first did, then we see this is a repeat of the case where we have 2 white cubes in the first layer and the second layer is arranged such that whites and blacks are straight with each other.
  
−
Therefore, our answer is <math>6 + 1 + 0 = \boxed{\textbf{(A)}\ 7}</math>.
+
Therefore, our answer is <math>\boxed{\textbf{(A)}\ 7}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 20:09, 25 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 $4$ white unit cubes and $4$ black unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)

$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$

Solution 1 (Direct Counting)

Divide the $2 \times 2 \times 2$ cube into two layers.

Case 1: Each layer contains 2 cubes of each color.

Note that we only need to consider the layout of the white cubes because all the other cubes will be black cubes.

There are 2 ways that the two white cubes can be arranged within each layer: adjacent or diagonal to each other.

Case 1.1: Both layers have two white cubes adjacent to each other.

Rotate the cube such that there are white cubes along the top of one layer. Now, consider the placement of the white cubes in the other layer:

Case 1.1.1: The white cubes are along the top.

Case 1.1.2: The white cubes are along the bottom.

Case 1.1.3: The white cubes are along the left.

Case 1.1.4: The white cubes are along the right.

All four of these cases are distinct (1.1.3 and 1.1.4 can be verified so by sketching the other faces of each).

Case 1.2: One layer has two white cubes adjacent to each other, while the other has two white cubes diagonal from each other.

Rotate the cube such that there are white cubes along the top of one layer. Now, consider the placement of the white cubes in the other layer:

Case 1.2.1: The white cubes are at the top-left and bottom-right.

Case 1.2.3: The white cubes are at the top-right and bottom-left.

These two cases can be verified to be distinct by sketching the other faces.

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 of one layer. Therefore, there must be a white cube at the bottom-right of that layer.

Case 1.3.1: There other layer also has white cubes at the top-left and the bottom-right. This is the same as case 1.1.2 approached from the perpendicular direction.

Case 1.3.2: The other layer has white cubes at the top-right and the bottom-left. This is a distinct case.

So now we have seven cases.

Case 2: Each layer contains 3 cubes of one color and 1 cube of the other color.

Split this cube into two layers; the sole white cube on one layer must be on the opposite corner of the sole black cube on the other layer, otherwise there will be some way to spit the cube into two layers such that there are 2 cubes of each color on each layer.

Case 3: One layer contains 0 white cubes and the other layer contains 4 white cubes.

Only 1 possible $2 \times 2 \times 2$ cube can result from this case. However, if we divide up this $2 \times 2 \times 2$ cube in the direction perpendicular to the way we first did, then we see this is a repeat of the case where we have 2 white cubes in the first layer and the second layer is arranged such that whites and blacks are straight with each other.

Therefore, our answer is $\boxed{\textbf{(A)}\ 7}$.

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

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