2021 Fall AMC 10A Problems/Problem 24
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
Since we want the sum of the edges of each face to be , we need there to be s and s on each face. Through experimentation, we find that either four of them 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 1
Since we want the sum of the edges of each face to be , we need there to be s and 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 2
We see that each face has to have 2 1's and 2 0's. We can try all the cases, starting with edges connecting to A.
Case 1
In this case, we can completely fill in the rest of the cube.
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:
In this case, we have different ways to fill out and
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 2 ways to go from here. More casework!
Case 2.2.1
This goes to:
We can see that this is the inverse of case 1. 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 it all together
We see that if the 3 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 3 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.
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.