Find number of arrangements of a cube if sum of numbers on each face must be same

Question

Each vertex of a cube is to be labelled with an integer 1 through 8, without repetition, such that sum of numbers of the four vertices of a face is the same for each face. Arrangements that can be obtained through rotations of the cube are considered to be the same. How many different arrangements are possible?

My attempt: Fix 8 on E.

Then, D+C=F+G (Since H is common to both faces)

Similarly, C+H=A+F and D+A=H+G

We can extend this analogy to other sides, and we see that D+E=B+G, E+H=A+B, E+F=B+C.

This means that {1,2}, {1,3}, {7,8}, {8,6} cannot be on one line segment since there are no other distinct numbers which add up to give 3,4,15 or 14 respectively.

How can I get the other conditions?

Edit: As @John and @Alexander mentioned in the comments, the sum of numbers on each face should be 18. The possible sets of numbers on opposite faces will be {8,7,1,2}, {3,4,5,6}; {8,6,1,3}, {4,5,7,2}; {8,5,2,3}, {1,4,7,6} and {8,5,1,4}, {2,3,6,7}.

Case 2 will not have any possibilities since there are no 2 numbers except for 8,3 which add up to 11. In case 4, 4 must be opposite to 8 and in case 1, 7 must be opposite to 8.

Now we can try to count it on a case by case basis, but I can't think of a foolproof method without repetitions. Can someone help :)

Answer 1

Note that the sum of the numbers on each face must be $18$ because $\frac{1+2+\cdots+8}{2}=18$.

So now consider the opposite edges (two edges that are parallel but not on the same face of the cube); they must have the same sum value too. Now think about the points $1$ and $8$. If they are not on the same edge, they must be endpoints of opposite edges, and we should have $1+X=8+Y$. However, this scenario would yield no solution for $[2,7]$, which is a contradiction. The points $1$ and $8$ are therefore on the same side and all edges parallel must also sum to $9 .$ Now we have $4$ parallel sides $1-8,2-7,3-6,4-5$. Thinking about $4$ endpoints, we realize they need to sum to 18. It is easy to notice only $1-7-6-4$ and $8-2-3-5$ would work. So if we fix one direction $1-8($ or $8-1)$ all other 3 parallel sides must lay in one particular direction. $(1-8,7-2,6-3,4-5)$ or $(8-1,2-7,3-6,5-4)$ Now, the problem is the same as arranging 4 points in a two-dimensional square, which is $\frac{4 !}{4}=6$.

It is also available here. There is also a video solution here.

Answer 2

The answer is $6$.

Identify the set of vertices with $\{(a,b,c)\mid a,b,c\in \{0,1\}\}$. I write $C(a,b,c)$ for the color of the vertex $(a,b,c)$. Suppose that $C(0,0,0)=1$ and $C(1,1,1)=1+x$. Then $$\sum_{i,j}C(i,j,0)=\sum_{k,l}C(k,1,l)$$ yields $$C(1,0,0)=C(0,1,1)+x.$$ Similarly, we get $$C(0,1,0)=C(1,0,1)+x$$ and $$C(0,0,1)=C(1,1,0)+x.$$ In summary, $C(a,b,c)=C(1-a,1-b,1-c)+x$ holds for all $(a,b,c)$ with $a+b+c$ odd. Conversely, any labeling with this property satisfies the required condition. There are three cases:

• $x=1$ and $\{C(0,1,1), C(1,0,1), C(1,1,0)\}=\{3,5,7\}$.
• $x=2$ and $\{C(0,1,1), C(1,0,1), C(1,1,0)\}=\{2,5,6\}$.
• $x=4$ and $\{C(0,1,1), C(1,0,1), C(1,1,0)\}=\{2,3,4\}$.

For each cases we have two distinct labelings, so there are exactly $3×2=6$ labelings.