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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?

Sinai Robins
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1 Answers1

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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.

Will Sawin
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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
  • Not as easy as it could be, because I don't think there's a global function of the $4$ angles that is center-invariant. – Will Sawin Nov 17 '12 at 21:31
  • Ummm... it seems to me like the manifold of all $k$-sided polygons is $(2\cdot k)$-dimensional, and the manifold of polygons with those angles from a point is $(k+3)$-dimensional. $:$ In any case, Will neglected the important point that the space of convex $k$-sided polygons is an open subset of the manifold of all $k$-sided polygons. $;;$ –  Nov 17 '12 at 22:39
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    I was modding out by the symmetry group, which is $3$-dimensional. – Will Sawin Nov 17 '12 at 22:52
  • Isn't that group 4-dimensional? $:$ (translate, rotate, scale) $;;$ –  Nov 17 '12 at 22:56
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    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
  • As per Gerry's query, I think we need an explicit example to be totally convincing... – Joseph O'Rourke Nov 18 '12 at 00:42
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    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 01:26
  • Yes, sounds right. Is transcendence theory strong enough that we can explicitly find such sets of $8$ numbers? Actually if just one coordinate is independent of the other $7$, we're probably safe. – Will Sawin Nov 18 '12 at 02:55
  • @O'Rourke and @Gerry: Thank you, and I agree with you and Gerry that some explicit examples of quadrilaterals that don't allow any rational viewing points would be very nice. – Sinai Robins Nov 18 '12 at 08:23
  • @Will: I think your dimension count argument is generally valid, though requires more details I think, as far as showing that most polygons don't have a rational viewing point. On the other hand, it leaves out infinitely many interesting polygons that DO have a rational viewing angle, when we restrict attention to algebraic vertices. But perhaps I should post this as a separate MO question, with more examples, as it seems to be separated from your arguments above. – Sinai Robins Nov 18 '12 at 08:27
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    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