Comments by Michael Reid:

interesting! i can show that no perfect hexagon is possible,

i assume that's what you're asking.

however, it is possible to make hexagonal annuli with arbitrarily

small inner hexagon and arbitrarily large outer hexagon.

now that i think about it some more, i'm less excited about it.

4 fins make a 60 degree parallelogram with edges 3 + sqrt(3) and

2 + 2 sqrt(3). and these can be packed into a parallelogram

with edges m(3 + sqrt(3)) and n(2 + 2 sqrt(3)) ,
for any

positive integers m and n . note that 12 copies
of an a x b

60 degree parallelogram make a hexagonal annulus where the outer

hexagon has edge a + b and the inner hexagon has edge
|a - b| .

so the arbitrarily good packing comes from making the difference

m(3 + sqrt(3)) - n(2 + 2 sqrt(3)) small.

to show that a perfect hexagon is impossible, suppose its edge is
s .

note that s = a + b sqrt(3) for some integers a
and b .

then its area is 3/2 sqrt(3) s^2 . the fin itself has area

((1 + sqrt(3))^2 + 1 (2 + sqrt(3)))/2 = (6 + 3 sqrt(3))/2.

now suppose that there are n of them in the perfect packing.

equating areas, we get (6 + 3 sqrt(3)) n / 2 = 3/2 sqrt(3)
s^2,

which gives (2 + sqrt(3)) n = sqrt(3) s^2 . multiply by
sqrt(3)

to get (3 + 2 sqrt(3)) n = 3 s^2 .

but now take the conjugate of this equation. if s = a +
b sqrt(3) ,

then it says (3 - 2 sqrt(3)) n = 3 (a - b sqrt(3))^2 , which
is

impossible, since the left hand side is negative!