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If we fix the angles at the vertices of a convex planar $n$-gon, the lengthsof 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 aconvex 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. Bavardand Ghys require unit area instead, which gives them a hyperbolic polytope.Those two polytopes are combinatorially equivalent, so either is fine for ourpurposes.  Such a fixed-angles polytope is combinatorially richer when the angles arewell balanced. We say that fixed external angles are "majority dominant" whenevery 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 anymajority-dominant angles is dual to the cyclic polytope $C_{n-3}(n)$. To extendthat result to even $n$, we require that the angles also have "dipoletie-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, allcontaining a particular angle.  Fixing the vertex angles is uncommon, however; people more often fix the edgelengths. That is harder, in part because fixed-lengths $n$-gons may not beconvex, but mostly because fixing the lengths constrains the angles nonlinearly-- so the resulting moduli spaces, called "polygon spaces", are curved. UsingSchwarz-Christoffel maps, Kapovich and Millson show that the subset of thatpolygon space in which the $n$-gons are convex and traversed counterclockwiseis homeomorphic to the fixed-angles polytope above, for those same fixedvalues. Each such subset is thus a topological polytope; and it is dual cyclicwhenever the fixed lengths are majority dominant and, for even $n$, have dipoletie-breaking.

Dual-cyclic polytopes of convex planar polygons with fixed vertex angles