Replace the letters in the diagram with a different number from 0 to 9, such that the sum of the four numbers on all edges are the same, and the sum of the three numbers on all three corners are the same.
D E F
G H I J
In other words, A+C+F+J = A+B+D+G = G+H+I+J and
A+B+C = D+G+H = F+I+J.
I seem to remember that the three number sums were originally different when this problem was waiting in the queue - possibly when it was re-formatted the intent was altered? I could easily be wrong of course, but the three number equivalent sums were originally;
D+E+F = B+E+I = C+E+H
I also don't know if this would reduce the redundancies, as I never worked through this one.