Difference between revisions of "2021 Fall AMC 10A Problems/Problem 24"
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Revision as of 02:49, 8 December 2021
Contents
[hide]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 purposes, we 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 the 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 the 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
~Arcticturn
Remark: It is very easy to get disorganized when counting, 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.
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
See Also
2021 Fall AMC 10A (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.