## Problem 3

Three vertices of a regular $2 n+1$ sided polygon are chosen at random. Find the probability that the center of the polygon lies inside the resulting triangle.


## Solution

Answer: $(n+1) /(4 n-2)$.

Label the first vertex picked as 1 and the others as $2,3, \ldots, 2 n+1$ (in order). There are $2 n(2 n-1) / 2$ ways of choosing the next two vertices. If the second vertex is 2 (or $2 n+1$ ), then there is just one way of picking the third vertex so that the center lies in the triangle (vertex $n+2$ ). If the second vertex is 3 (or $2 n$ ), then there are two $(n+2, n+3)$ and so on. So the total number of favourable triangles is $2(1+2+\ldots+n)=n(n+1)$. Thus the required probability is $(n+1) /(4 n-2)$.

