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
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!