Post Number:#2 by Soumitra Das » Mon Sep 25, 2017 3:48 am
If $ABC$ is an arbitrary triangle and the value of $n$ satisfies the given condition,obviously it will be also true for certain $ABC$.Consider a triangle $ABC$ with angle $$A=120$$, $$AB=AC$$, circumcenter $O$, midpoint of $$AB=M$$, midpoint of $$AC=N$$, center of the regular polygon constructed outside of $AB$ is $Z$ and $AC$ is $Y$ and $BC$ is $A$.
for all $n$, $YZ$ is parallel to $MN$ and so triangle $ZYO$ is similar to $MNO$, which is equilateral. Since $X$ belongs to the perpendicular bisector of $BC$, $$X=O$$ and which is possible only then, when $$n=3$$.
So, $$n=3$$ can be one solution and it is($XYZ$ is a Nepoleon triangle).So, there exist exactly one value of $n$ which is $3$.