May be better to ask for help here. Let $v_{1}$, $v_{2}$, $\ldots$, $v_{m}$ be the vertices of a convex polygon in the plane and $v_{m+1}$ be a vertex in the interior of the convex polygon. Connect all the vertices by edges, and let $\alpha_{m}$ be the smallest angle among all the angles formed by two edges coming from the same vertex. Is it true that $m^{2}\alpha_{m}$ is bounded by an absolute constant (independent of $m$ and the $v$'s)? Any helpful answers would be greatly appreciated.

  • $\begingroup$ Consider an equilateral polygon. $alpha_m = (m - 2) \pi / m$, thus $m^2 alpha_m = m(m - 2) \pi$ which is unbounded $\endgroup$ – Squark Dec 6 '12 at 13:56
  • $\begingroup$ Squark, $v_{m+1}$ is also a vertex in the whole graph, sorry for unclearness. In an equilateral polygon, the smallest angle would be an interior angle of the triangle that has $v_{m+1}$ as a vertex, so it would be less than the one you expressed. $\endgroup$ – Palt Dec 6 '12 at 14:24

Let $m$ be odd, $v_1, \ldots, v_m$ be the vertices of a regular $m$-gon, and $v_{m+1}$ be its centre. The smallest angle only involving vertices of the $m$-gon is $\pi/m$, the angle over any edge when viewed from another vertex. The smallest angle obtained using the centre is half of this, $\pi/2m$. So $m^2\alpha_m$ is not bounded.

  • $\begingroup$ Ben, all the edges are included to form the angles, sorry for unclearness. The smallest angle obtained using the centre would be an interior angle of the triangle that contains the centre, a vertex on the boundary and its farthest vertex on the boundary, so it can be much less than $\pi/2m$. $\endgroup$ – Palt Dec 6 '12 at 14:35
  • 3
    $\begingroup$ Palt, I was considering the complete graph on $v_1, \ldots, v_{m+1}$, which should cover all possible interpretations of the question as by adding more edges the smallest angle cannot get larger. The angle you specify at the end of your comment is the one I'm claiming is $\pi/2m$; do you get a different value? $\endgroup$ – Ben Barber Dec 6 '12 at 14:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.