Garfields's hexagon

First find which numbers will be common for all shaded triangles (center circle) so the remaining numbers can be ordered in three pairs with the same sum.

If I put $\boxed{1 \Rightarrow 2+7=3+6=4+5}$

If I put $2$ there's no a possible arrangament

If I put $3$ there's no a possible arrangament

If I put $\boxed{4 \Rightarrow 3+5=2+6=1+7}$

If I put $5$ there's no a possible arrangament

If I put $6$ there's no a possible arrangament

If I put $\boxed{7 \Rightarrow 1+6=2+5=3+4}$

So in the center I can put $3$ numbers, then I select one number for the first saded triangle $\binom{6}{1}$ , then by the second $\binom{4}{1}$ and last by the third $\binom{2}{1}$ . Result is $3\times\binom{6}{1}\binom{4}{1}\binom{2}{1}=\boxed{72}$ .

Note: I only select one number by triangle because the other one is definite by the first.