Consider a $(2n+1)$ sided regular polygon. Find the probability that three vertices chosen at random form the vertices of an isosceles triangle.

My Attempt

If I choose $3$ vertices containing two sets of $r$ consecutive sides then the triangle so formed by the $3$ chosen vertices(i.e. the $2r$ sides are all consecutive) is clearly isosceles. So number of ways to do so will be $$\sum_{r=1}^{n}r=\frac{n(n+1)}{2}$$So required probability$$=\frac{\sum_{r=1}^{n}r}{\binom{2n+1}{3}}$$.

Is it correct.


Choose the first vertex at random(it is indistinguishable from the other vertices).

Next, we drop an axis of symmetry at the point. For each of the remaining $2n$ points, it suffices to pick a single vertex() from one side of the axis of symmetry(the other vertex will be determined my reflecting across the axis of symmetry).

The number of ways to choose 2 points at random frorm the remaininig points is $\binom{2n}{2},$ which gives an overall probability of $$\boxed{\frac{n}{\binom{2n}{2}}}.$$


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