Numbers with minimal sum at the vertices of a cube

Question

The eight vertices of a cube are marked with numbers from 1 to 8 such that the sum of any three numbers on any face is not less than 10.

What is the minimum sum of the four numbers on a face?

Answer 1

Old school proof of optimality:

Let n be the largest number on the face with the smallest sum. The remaining three numbers on that face must sum to at least 10 so cannot be all less than 5 (2+3+4=9) the largest must therefore be at least 6. + 10 = 16.

      2---3
     /|  /|
    5-+-6 |
    | 8-+-7
    |/  |/
    4---1
 

Answer 2

16:

 x y z
 0 0 0 5
 0 0 1 6
 0 1 0 2
 0 1 1 3
 1 0 0 4
 1 0 1 1
 1 1 0 7
 1 1 1 8
 

Answer 3

There is no unique way to label the vertices. So does the question mean “minimum” for a specific consistent cube? Or “minimum” over all consistent cubes? I choose the former because it’s more interesting, and it subsumes the other question

If the vertices 1&2 are diagonally opposite one another, then the labels 1...5 are fixed, up to rotation and reflection. But the other 3 vertices can be labelled 6...8 in any order, without risk of any triple being too light.

The vertex-sums of the faces are: 10+a, 26-a, 11+b, 25-b where a&b are distinct and drawn from 6...8. Some values appear twice. The minimum vertex-sum is 16 iff a is 6. The minimum is 18 (actually achieved by every face) iff a=8 and b=7. Otherwise the minimum is 17.

On the other hand if the 1&2 are on the same face (but not adjacent) then so are 7&8. The other 4 vertices can be labelled in 4 ways. Either all the sums are 18, or there is one 17/19 pair.

I didn’t know one could label the corners of a cube so that the sum of the 4 values on any face is the same: 18. Interesting