Alternate proof of showing the below claim Prove that no matter how the vertices of a regular polygon with $2n+1$ , $n\geq 2$ vertices are coloured in two colours , there always exist three vertices of the same colour that forms an isosceles triangle .

Method:                          What i considered was for a odd cycle there will be surely two vertex adjacent to each other and having same color $(R)$ , from there we can say that as for isosceles we want the point (vertex) above the perpendicular bisector of that edge . Since two colours are already same $(R)$ the top most must be different lets say $(G)$ and observe that the two adjacent vertices to that edge are of different colours so its G both , but now observe that top most point and those vertices forms a isosceles triangle amd we are done  .


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*Though i understand this proof works but i was thinking if there was a alternate proof which somehow uses the total number of isosceles triangles and maximum non same coloured triangles ? And then showing the above claim by subtracting from total to get the minimum possible same coloured isosceles triangles ?

 A: It's a somewhat less direct argument, but we can do it.
First of all, there are $\binom{2n+1}{2} = n(2n+1)$ isosceles triangles total. One way to see this: each of the $\binom{2n+1}{2}$ chords is the base of a single isosceles triangle, whose third corner is at one end of the perpendicular bisector of that chord. (Here and further on, I will assume that each isosceles triangle has a specific edge which is the base, and the other two edges are equal - so an equilateral triangle corresponds to three different isosceles triangles, depending on which edge is chosen as the base. As long as we're consistent about this, it will not affect the argument.)
Now let's count the number of chords whose endpoints have the same color. Their quantity depends on the coloring, but it's going to be minimized when there are $n$ points of one color and $n+1$ points of the other color - in that case, the number of such chords is $\binom n2 + \binom{n+1}{2} = n^2$.
(We can see this intuitively, or prove it with algebra: if there are $x$ points of one color and $2n+1-x$ of the other, then there are $\binom x2 + \binom{2n+1-x}{2}$ chords whose endpoints have the same color, which can be written as $n^2 + (x - n - \frac12)^2 - \frac14$. Because $x$ is a integer, $|x - n - \frac12| \ge \frac12$, so $(x - n - \frac12)^2 - \frac14 \ge 0$.)
Each chord whose endpoints have the same color is part of three isosceles triangles: it is the base of one, and we can get the two others by choosing one endpoint to be the peak of the isosceles triangle, which uniquely determines the third corner. Since there are at least $n^2$ chords with endpoints of the same color, the $n(2n+1)$ isosceles triangles contain between them $3n^2$ pairs of corners of the same color.
For $n>1$, $n(2n+1) = 2n^2 + n < 3n^2$. Therefore at least one isosceles triangle gets more than one pair of corners of the same color. Then all three vertices of that triangle are the same color, and we are done.
