# 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.

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So, it shouldn't be hard to give an explicit example of, say, a quadrilateral with no rational viewing point? –  Gerry Myerson Nov 17 '12 at 21:11
I was modding out by the symmetry group, which is $3$-dimensional. –  Will Sawin Nov 17 '12 at 22:52
Oh I wasn't counting scaling. You could also mod out by scaling, the result would be the same. But I think unmodded is actually the best way to go. –  Will Sawin Nov 17 '12 at 23:34
Isn't it the case then that almost all quadrilaterals have no rational viewing points? In particular, choose any one where all coordinates of the vertices are algebraically independent and you should be safe. –  j.c. Nov 18 '12 at 1:26
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