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Are perpendicular bisectors of an arbitrary $n$-gon concurrent ? If so, how to prove it ?

Dr. user44690
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  • Surely you mean a regular $n$-gon, right? – Sarvesh Ravichandran Iyer Apr 05 '19 at 04:23
  • @астонвіллаолофмэллбэрг No, I mean arbitrary convex polygon. It is okay, I just realized they won't be. – Dr. user44690 Apr 05 '19 at 04:24
  • @астонвіллаолофмэллбэрг Sorry, they maybe. I am not sure. – Dr. user44690 Apr 05 '19 at 04:24
  • Exactly, I think even for $n=4$ it won't be the case. In fact, thinking of it now, can you find a (convex) quadrilateral where two of the perpendicular bisectors are in fact parallel? (So that these two, and therefore all of them cannot meet). Think of something trapezium like – Sarvesh Ravichandran Iyer Apr 05 '19 at 04:24
  • @астонвіллаолофмэллбэрг Ah! Thanks. – Dr. user44690 Apr 05 '19 at 04:30
  • @астонвіллаолофмэллбэрг Your name reminds me of a friend of mine. He is doing a PhD in mathematics now. His name was also Ravichandran Iyer and he also liked to write stuff in Russian. And he liked football as well. – Dr. user44690 Apr 05 '19 at 04:33
  • Interesting how so many details seem to match. I think, with high probability, I am your friend. My first name is сарвеш (now how do you say that?) – Sarvesh Ravichandran Iyer Apr 05 '19 at 05:34
  • @астонвіллаолофмэллбэрг Wonderful. You are Sarvesh from IISc. – Dr. user44690 Apr 06 '19 at 05:10
  • I was in IISc, now I am somewhere else, but you are correct!To be frank I cannot guess who you are, and for some reason I'd actually like to continue guessing! Also, +1. – Sarvesh Ravichandran Iyer Apr 06 '19 at 13:44
  • I actually thought it might be you! How are you doing? I am now fully into Stochastic PDE, and it also turns out my main focus at first is on the KPZ equation, which apparently has some applications in conformal field theory. I am wondering if you have come across it – Sarvesh Ravichandran Iyer Apr 08 '19 at 05:05

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Hint What does it mean for all perpendicular bisectors of an $n-$gon $A_1A_2...A_n$ to be concurrent at $P$?

Well, it means, that all segments $PA_1, PA_2, ..., PA_n$ are equal in length. Why?

enter image description here

Observe that the triangles $\triangle A_1A_2P, \triangle A_2, A_3P...$ are all isosceles since $P$ lies on their perpendicular bisector. All of them being isosceles means that $$A_1P=A_2P\quad A_2P=A_3P\quad A_3P=A_4P...\implies A_1P=A_2P=A_3P=A_4P=...$$

$P$ is thus the circumcentre of the $n-$ gon.

But this is only possible if you polygon is $\color{blue}{cyclic}$

Dr. Mathva
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