Seeing the vertices of a polygon with rational angles

Given any convex polygon in the plane, is it always possible to find a point $p$ in its interior such that when we draw the line segments from $p$ to each of its vertices, the angles formed at $p$ are all (not necessarily equal) rational multiples of $\pi$?

For a triangle $T$, it's easy to construct such a point, namely the Steiner point $p$ will do, enjoying three angles of measure $2\pi/3$ each, between $p$ and any two adjacent vertices of $T$. But is this known in general?

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What a beautiful question! –  Joseph O'Rourke Nov 17 '12 at 21:50

Consider the manifold of all $k$-sided polygons. This is $2k-3$ dimensional. Given a set of angles $\theta_1,...,\theta_k$, the manifold of polygons with those angles from a point is $k$-dimensional, since it's determined by the distances of the vertices from the point.
You're not going to cover a $2k-3$ - dimensional manifold with countably many $k$-dimensional manifolds if $k\geq 4$. In particular, these maps are real algebraic maps, and thus aren't anything like space-filling curves.
I was modding out by the symmetry group, which is $3$-dimensional. –  Will Sawin Nov 17 '12 at 22:52
I think $\{(0,0),(1,0),(0,1),(1,\pi)\}$ is an explicit example, because for each list of four angles, the coordinates would have to satisfy some equation defined over the field generated by $\cos$ and $\sin$ of the angles, which would certainly be nontrivial in each coordinate. –  Will Sawin Nov 18 '12 at 16:40