I'm curious about if there is any guarantee about the amount of angles that can be concave in a given polygon. I'm wondering if there's a relation between the number of convex/concave angles, and specifically if it is possible to have a larger or equal number of concave angles than convex ones.


1 Answer 1


Yes, it's possible. The simple example with the equal number of "convex" and "concave" angles could be the following: hexagon with the vertices (0,0), (7,0), (4,1), (2,2), (1,4), (0,7). And you can add an arbitrary number of "concave" angles, so an equality isn't a requirement.

  • $\begingroup$ +1. To see the picture think of the figure formed by two tangents to a circle. Then approximate the circular arc between the points of tangency by as many short chords as you like. $\endgroup$ Commented Jan 9, 2017 at 13:40
  • $\begingroup$ You are completely right. As a followup question, could there be then a limit based on the total angle covered by the angles, rather than the number of angles itself? $\endgroup$
    – kace91
    Commented Jan 9, 2017 at 14:20
  • $\begingroup$ @kace91: Well, the total sum for "convex" angles can span from zero to infinity (if you don't bound the total number of corners). $\endgroup$ Commented Jan 9, 2017 at 15:53
  • $\begingroup$ @SergeiGolovan my bad, I didn't express the idea well. I was thinking of something like total angle of convex angles minus total angle of concave angles, and expressing number in relation to the number of vertices... $\endgroup$
    – kace91
    Commented Jan 9, 2017 at 15:57
  • $\begingroup$ @kace91: As far as I can see, the difference can be anything from minus to plus infinity. Imagine a moon-like shape. Then fix some vertices at the outer curve and choose arbitrary many points at the inner one - this way the difference can go to minus infinity. The opposite: fix a few points at the inner curve and select many-many ones at the outer one - the difference will go to plus infinity. $\endgroup$ Commented Jan 9, 2017 at 16:05

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