Unbalance in cake-cutting

This is only a partial answer but too long for a comment

$n$ lines divide the plane into at most $m = \frac{n(n+1)}{2}+1$ pieces, which is known as the lazy caterer's sequence. This results in $\frac{n(n-1)}{2}$ intersections of two lines each, or equivalently $n-1$ intersections on every line. Notice that for odd $n$ the regular $\{n/\frac{n-1}{2}\}$ star polygons satisfy these criteria:

$n$ lines have $2n$ degrees of freedom which can be represented by the $x$ and $y$ coordinates of each point of the star. Now, depending on $r := \sqrt{x^2+y^2} < 1$ we can easily compute $A_i$ and $L_j$ because of the rotational symmetry and minimize the entropies.

I don't know if this results in the optimal arangement but it is a systematic approach.