2021 Fall AMC 10A Problems/Problem 24
Contents
Problem
Each of the edges of a cube is labeled or . Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the faces of the cube equal to ?
Solution 1
For simplicity, we will name this cube by vertices, as shown below. Note that for each face of this cube, two edges are labeled and two edges are labeled For all twelve edges of this cube, we conclude that six edges are labeled and six edges are labeled
We apply casework to face Recall that there are ways to label its edges:
- Opposite edges have the same label.
- Opposite edges have different labels.
There are ways to label the edges of We will consider one of the ways, then multiply the count by Without loss of generality, we assume that are labeled respectively:
We apply casework to the label of as shown below. We have such labelings for this case.
There are ways to label the edges of We will consider one of the ways, then multiply the count by Without loss of generality, we assume that are labeled respectively:
We apply casework to the labels of and as shown below. We have such labelings for this case.
Therefore, we have such labelings in total.
~MRENTHUSIASM
Solution 2
Since we want the sum of the edges of each face to be , we need there to be two s and two s on each face. Through experimentation, we find that either or all of them have s adjacent to s and s adjacent to on each face. WLOG, let the first face (counterclockwise) be . In this case we are trying to have all of them be adjacent to each other. First face: . Second face: choices: or . After that, it is basically forced and everything will fall in to place. Since we assumed WLOG, we need to multiply by to get a total of different arrangements.
Secondly, of the faces have all of them adjacent and of the faces do not: WLOG counting counterclockwise, we have . Then, we choose the other face next to it. There are two cases, which are and . Therefore, this subcase has different arrangements. Then, we can choose the face at front to be . This has cases. The sides can either be or . Therefore, we have another cases.
Summing these up, we have . Therefore, our answer is .
Remark
It is very easy to get disorganized when counting, resulting in incorrect calculations, so when doing this problem, make sure to draw a diagram of the cube. Labeling is a bit harder, since we often confuse one side with another. Try doing the problem by labeling sides on the lines (literally letting the lines pass through your s and s.) I found that to be very helpful when solving this problem.
~Arcticturn
Solution 3
We see that each face has to have 2 1's and 2 0's. We can start with edges connecting to A.
Case 1
This goes to:
We can see that we choose diametrically opposite vertices to put 's on the connecting edges. As a result, this case has orientations.
Case 2
Filling out a bit more, we have:
Let's try filling out and first.
Case 2.1
This goes to:
We can see that it consists of chains of three 's, with the middle of each chain being opposite edges. As a result, this case has orientations.
Case 2.2
Oh no... We have different ways of filling out and . More casework!
Case 2.2.1
This goes to:
We can see that this is the inverse of case 1 (Define inverse to mean swapping 's for 's and 's for 's). Therefore, this should also have orientations.
Case 2.2.2
This goes to:
This is the inverse of case 2.1, so this will also have orientations.
Putting Them All Together
We see that if the edges connecting to has two 's, and one , it would have the same solutions as if it had two 's, and one . The solutions would just be inverted. As case 2.1 and case 2.2.2 are inverses, and case 2.2.1 has case 1 as an inverse, there would not be any additional solutions.
Similarly, if the edges connecting to has three 's, it would be the same as the inverse of case 1, or case 2.2.1, resulting in no new solutions.
Putting all the cases together, we have solutions.
~ConcaveTriangle
Solution 4
The problem states the sum of the labels on the edges of each of the faces of the cube equal to . That is, the sum of the labels on the edges of a face is equal to . The labels can only be or , meaning edges are labeled , the other are labeled .
This problem can be approached by Graph Coloring of Graph Theory. Note that each face of the cube connects to other faces, each with a shared edge. We use the following graph to represent the problem. Each vertex represents a face, each edge represent the cube's edge. Each vertex has edges connecting to other vertices. The edges can be colored red or blue, with red as label , and blue as label . Each vertex must have red edges and blue edges.
: adjacent red edges from vertex A. There are ways to choose red edges adjacent to each other and connect to vertices with an edge between them as shown below.
: adjacent red edges from vertex form a closed loop with a third red edge. There is only case as shown below.
: adjacent red edges from vertex does not form a closed loop with a third red edge. There are cases as shown below.
In case , there are total ways.
: red edges from vertex with blue edge in between. There are ways to choose red edges with blue edge in between.
There are only cases as shown below.
In case , there are total ways.
From both case and case , there are ways in total.
See Also
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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