No! Not a firing squad nor the need for a continuous line to cross all line segments just once!
To each vertex labeled A to L apply a different value from 1 to 12. Let V, W, X, Y and Z be the sums of their respective surrounding vertices.
Provide at least one example where V=W=X=Y=Z, or offer a reason why this, like the continuous line, is impossible.
I had addressed the corners question earlier. If each sum is of the four vertices there are 2816 solutions, varying from 23 to 29 but none for sum=26 (as Charlie noted) -- this was the "interesting" feature I suggested; we agree there are 980 solutions, with every sum from 25 to 32 possible, if all points on a perimeter are added. I discovered all those by "brute force" and still wonder what analytic approaches might have done to "prove" something along these lines. With all the options, it would not be very difficult to find a counter example, which of course would disprove the generalization that such was an impossibility.
four-square solutions
