0
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 $$\boxed{\frac{n}{\binom{2n}{2}}}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.