## A Balanced Cube

Here’s another brainteaser from my friend, Alexander Karabegov.

Share:

Puzzle.Place the numbers from 1 to 8 at the vertices of a cube so that each face is balanced. On a balanced face, the sum of the numbers at the ends of one diagonal equals the sum of the numbers at the ends of the other diagonal.

## Ivan:

Let AB be the front side of the cube’s floor and EF be the front side of the cube’s ceiling. Then A=1, B=6, C=8, D=3, E=2, F=7, G=9, H=4.

2 November 2023, 12:59 pm## Guido:

Let’s name the vertices of the top face A, B, C, and D, that of the bottom face A’, B’, C’, and D’ correspondingly: we can write six balancing equations

(1) A + C = B + D

(2) A + B’= B + A’

(3) B + C’= C + B’

(4) C + D’= D + C’

(5) D + A’= A + D’

(6) A’+ C’= B’+ D’

With a bit of algebra, we can easily derive a seventh equation

(7) A + C’= B + D’= C + A’= D + B’

which states that in a Balanced Cube the sums along the diagonals of the cube are equal.

Let’s take equation (1) and 1 + 4 = 2 + 3 (which leaves 5 + 8 = 6 + 7)

A = 1 –> C = 4

B = 2 –> D = 3

Now we substitute A, B, C, and D in equation (7)

1 + C’= 2 + D’= 4 + A’= 3 + B’ –> A’= 5, B’= 6, C’=8, and D’=7

So 1-2-4-3/5-6-8-7 and the sum along the diagonals of the cube is 9

PS: checking the balancing equations

(1) 1 + 4 = 2 + 3

4 November 2023, 4:57 am(2) 1 + 6 = 2 + 5

(3) 2 + 8 = 4 + 6

(4) 4 + 7 = 3 + 8

(5) 3 + 5 = 1 + 7

(6) 5 + 8 = 6 + 7

## Lazar Ilic:

1 simple general solution as above is to consider instead isomorphically using the number from 0 to 7 at which point we may essentially take the coordinates of the vertices to binary valuations… so at (0,0,0) we put a 0, at (0,0,1) we put a 1, at (0,1,0) we put a 2, …, (1,1,1) we put a 7. Then by construction it would immediately follow that any 2 sets of points with the same gravicenter and same size would have the same sum of values.

10 November 2023, 5:23 pm## tanyakh:

Lazar, I was hoping to see your solution.

10 November 2023, 7:31 pm