, 2019

Solution:

Extending three sides of the equiangular hexagon gives an equilateral triangle. Thus, if the sides are $a, b, c, d, e, f$, in order, then $a+b+c = a+f+e \Rightarrow b+c = e+f \Rightarrow f-c = b-e$. By a symmetric argument, we see that $d-a = f-c = b-e$ holds, which means that they must be separated into three groups of two with equal differences.

If the grouping is $(1,2),(3,4),(5,6)$, then we have $1,4,5,2,3,6$ around the hexagon. If the grouping is $(1,4),(2,5),(3,6)$, then we get $1,5,3,4,2,6$ as the other possibility. Finally, we can use our equilateral triangle trick to find the areas.

For the first, we get a big triangle of side $1+4+5=10$, and must subtract smaller triangles of sides $1,5,3$. This gives $(100-1-25-9) \sqrt{3} / 4 = 65 \sqrt{3} / 4$.

For the other, we get $(81-1-9-4) \sqrt{3} / 4 = 67 \sqrt{3} / 4$.

The positive difference between these is $\sqrt{3} / 2$.