Let *S* be the common value for the twelve sums; we will confine ourselves to the search for the **minimum** of *S*: the maximum will then be obtained by changing any value *x* with *20-x* (thus *S* is replaced by *60-S*)

Let *C* be the central value, and let *V, I, E* be the 6-components arrays giving respectively the values on *Vertices, External* and *Internal* nodes; with *SumV, SumI, SumE* we will denote the sums of the elements of *V, I, E* respectively.

The most obvious relations are:

*V[j]+I[j]+C=S*, thus *SumV+SumI=6(S-C)*;
*2SumV+SumE=6S*;
*SumV+SumI+SumE+C = 1+...+19 = 190*.

Remark that the second relation implies that *SumE* must be even; then the first relation implies that also *SumI* must be even; then the third relation implies that also *C* must be even.

From *V[j]+I[j]=S-C (j=1,2,...6)* we see that the value *S-C* must be obtained 6 times; the minimum for *S*, because of *C* even, corresponds to *C=2, V[j]+I[j]=17* (say: 3+14, 4+13, ..., 8+9); thus we need *S > 18*.