  If we fix the angles at the vertices of a convex planar $n$-gon, the lengths
of its edges must satisfy two linear constraints in order for it to close up.
If we also require unit perimeter, our vectors of $n$ edge lengths form a
convex polytope of dimension $n-3$, each facet of which consists of those
$n$-gons in which the length of a particular edge has fallen to zero. Bavard
and Ghys require unit area instead, which gives them a hyperbolic polytope.
Those two polytopes are combinatorially equivalent, so either is fine for our
purposes.
  Such a fixed-angles polytope is combinatorially richer when the angles are
well balanced. We say that fixed external angles are "majority dominant" when
every consecutive string of more than half of them sums to more than $\pi$.
When $n$ is odd, we show that the fixed-angles polytope for any
majority-dominant angles is dual to the cyclic polytope $C_{n-3}(n)$. To extend
that result to even $n$, we require that the angles also have "dipole
tie-breaking": None of the $n$ strings of length $n/2$ sums to precisely $\pi$,
and the $n/2$ that sum to more than $\pi$ overlap as much as possible, all
containing a particular angle.
  Fixing the vertex angles is uncommon, however; people more often fix the edge
lengths. That is harder, in part because fixed-lengths $n$-gons may not be
convex, but mostly because fixing the lengths constrains the angles nonlinearly
-- so the resulting moduli spaces, called "polygon spaces", are curved. Using
Schwarz-Christoffel maps, Kapovich and Millson show that the subset of that
polygon space in which the $n$-gons are convex and traversed counterclockwise
is homeomorphic to the fixed-angles polytope above, for those same fixed
values. Each such subset is thus a topological polytope; and it is dual cyclic
whenever the fixed lengths are majority dominant and, for even $n$, have dipole
tie-breaking.
