Three vertices of a regular 2n+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)/(4n-2).
Label the first vertex picked as 1 and the others as 2, 3, ... , 2n+1 (in order). There are 2n(2n-1)/2 ways of choosing the next two vertices. If the second vertex is 2 (or 2n+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 2n), then there are two (n+2, n+3) and so on. So the total number of favourable triangles is 2(1 + 2 + ... + n) = n(n+1). Thus the required probability is (n+1)/(4n-2).
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002