the regular polygon has $ 2n + 1$ vertices, $ v_1,v_2,...,v_{2n + 1}$, and a center $ O$. drawing a line from one vertice $ v_a$ through the center $ O$ will hit the the midpoint of an oposite segment at $ m_a$. Also, since the polygon is symmetrical we can assume that one of the vertices we pick is $ v_1$. Now we will count the number of triangles, with vertex $ v_1$, that contain $ O$. Suppose the second vertex picked is $ v_{1 + i}$. if the triangle is to contain $ O$ then the thrid vertex must lie between $ m_1$ and $ m_{1 + i}$ (the smaller arc), and there are $ i$ vertices between $ m_1$ and $ m_{1 + i}$. To avoid overcounting we only need to consider the vertices $ v_{1 + i}$ with $ 1\le i \le n$ This gives $ \sum_{i = 1}^n i = \frac {n(n + 1)}{2}$ triangles with contain $ O\qquad(1)$. The number of traingles that include vertex $ v_1$ is given by $ \binom{2n}{2} = n(2n - 1)\qquad(2)$ From $ (1)$ and $ (2)$ it follows that the desired probability is $ \frac {n + 1}{2(2n - 1)}$

I believe I have a slightly different solution.

Anyone looking for a similar challenge might want to see the 10th question of the 2006 Euclid 12 Contest from Canada. http://www.cemc.uwaterloo.ca/contests/past_contests/2006/2006EuclidContest.pdf

There are $\binom{2n+1}{3}$ ways how to pick the three vertices. We will now count the ways where the interior does NOT contain the center. These are obviously exactly the ways where all three picked vertices lie among some $n+1$ consecutive vertices of the polygon. We will count these as follows: We will go clockwise around the polygon. We can pick the first vertex arbitrarily ($2n+1$ possibilities). Once we pick it, we have to pick $2$ out of the next $n$ vertices ($\binom{n}{2}$ possibilities). Then the probability that our triangle does NOT contain the center is \[ p = \frac{ (2n+1){\binom{n}{2}} }{ {\binom{2n+1}{3} } } = \frac{ (1/2)(2n+1)(n)(n-1) }{ (1/6)(2n+1)(2n)(2n-1) } = \frac{ 3(n)(n-1) }{ (2n)(2n-1) } \] And then the probability we seek is \[ 1-p = \frac{ (2n)(2n-1) - 3(n)(n-1) }{ (2n)(2n-1) } = \frac{ n^2+n }{ 4n^2 - 2n } = \boxed{\frac{n+1}{4n-2}} \]

This problem has appeared several times before, and notably in a well known problem on the Putnam contest, I think, where it gave a sphere, and told you to take four points making a tetrahedron and compute the probability it contains the center. First, we can circumscribe a circle around it. Label this polygon's vertices, starting with $A_1$, and going clockwise, $A_2, A_3, ... , A_{2n}, A_{2n+1}$. We will proceed by complementary counting. Specifically, we will count the number of triangles that do not have center $O$ as its vertex and has $A_1$ as its "most counterclockwise vertex". In order for those two conditions to be met, we can choose our next two vertices among only the vertices $A_2, A_3, ...A_n, A_{n+1}$. Since these are $n+1-2+1 = n$ possible selections, the number of ways to form such triangles is $\displaystyle {n} \choose {2}$. By symmetry, this property applies for all $2n+1$ vertices. Thus, the total number of ways to form a triangle from 3 vertices of this regular polygon such that it does not contain center $O$ in its interior is $\displaystyle {n} \choose {2}$$(2n+1)$. Finally, note that there are $\displaystyle {2n+1} \choose {3}$ total triangles. Since we are counting the complement, we must subtract from 1; thus, our desired probability is: $1- \frac{\frac{(n)(n-1)(2n+1)}{2}}{\frac{(2n+1)(2n)(2n-1)}{6}} = 1-\frac{3n-3}{4n-2}=\boxed{\frac{n+1}{2(2n-1)}} $. $\blacksquare$ got it, thanks no more need

huashiliao2020 wrote: This problem has appeared several times before, and notably in a well known problem on the Putnam contest, I think, where it gave a sphere, and told you to take four points making a tetrahedron and compute the probability it contains the center. First, we can circumscribe a circle around it. Label this polygon's vertices, starting with $A_1$, and going clockwise, $A_2, A_3, ... , A_{2n}, A_{2n+1}$. We will proceed by complementary counting. Specifically, we will count the number of triangles that do not have center $O$ as its vertex and has $A_1$ as its "most counterclockwise vertex". In order for those two conditions to be met, we can choose our next two vertices among only the vertices $A_2, A_3, ...A_n, A_{n+1}$. Since these are $n+1-2+1 = n$ possible selections, the number of ways to form such triangles is $\displaystyle {n} \choose {2}$. $\square$ \square By symmetry, this property applies for all $2n+1$ vertices. Thus, the total number of ways to form a triangle from 3 vertices of this regular polygon such that it does not contain center $O$ in its interior is $\displaystyle {n} \choose {2}$$(2n+1)$. Finally, note that there are $\displaystyle {2n+1} \choose {3}$ total triangles. Since we are counting the complement, we must subtract from 1; thus, our desired probability is: $1- \frac{\frac{(n)(n-1)(2n+1)}{2}}{\frac{(2n+1)(2n)(2n-1)}{6}} = 1-\frac{3n-3}{4n-2}=\boxed{\frac{n+1}{2(2n-1)}} $. $\blacksquare$ \blacksquare Also, could someone pm/comment on this thread how to latex add a black square/white square at the end of my proof?