Probability of a random triangle containing the center of a polygon Consider a regular polygon of $2n+1$ sides. Let three random vertices be chosen at random to get a triangle. The probability that the chosen triangle contains the center of the polygon is 5/14.
What is n?
I'm confused as to how to proceed on the cases that we will get. They do not form any definite pattern.

The probability seems to work out for a regular $9$-gon (so $n=4$).
Consider a circle with $9$ equally spaced points marked.  Let's label the points $0, 1, 2, \dots, 8$ going clockwise around the circle.  These points form our regular $9$-gon.  When we choose $3$ of these vertices and form a triangle, the triangle contains the center if and only if the triangle is acute. This in turn occurs if and only if as you go from one chosen vertex to the next you never skip more than $3$ vertices.  
By symmetry, we can freely choose vertex $0$.  And then we choose two other vertices $a$ and $b$ with $a<b$.  We need the spacing between these consecutive vertices to be at most $4$.  This means that $a$ must be $1$, $2$, $3$, $4$.
For $a=1$ we must have $b=5$ to contain the center.
For $a=2$ we must have $b=5$ or $6$ to contain the center.
For $a=3$ we must have $b=5$ or $6$ or $7$.
For $a=4$ we must have $b=5$ or $6$ or $7$ or $8$.
This gives us a total of $10$ good triangles out of ${8\choose 2}=28$ triangles that use vertex $0$.
So the probability of containing the center is $\frac{10}{28}=\frac{5}{14}$. 
