Number of triangles containing the center of regular 13-gon 
Show that the number of triangles whose vertices are vertices of a regular 13-gon and contain the center of 13-gon is 91.  Can you extend your solution to regular $2n+1-gon$?

If we draw a line from one vertex of 13-gon passing through the center,then the other two vertices must be selected from opposite sides of this line,now we should find and prove the condition that the formed triangle contains the center and count possible cases...
 A: Let's number the vertices of the tridecagon 1 to 13 clockwise.
If we pick vertex 1, then a triangle containing the center must be built with one and only one of the vertices 2,..,7.
With (1,2) there's only one third possible vertex: 8
With (1,3) there are two: 8,9
...
With(1,7) there are six: 8,9,10,11,12,13
Hence the number of solutions using vertex 1 is 1+...+6=21.
The total number of solutions is then $13*21/3=91$. (summing the numbers of triangles using each of the 13 vertices and dividing per 3 since each triangle is counted trice)
For generalization, for a $2n+1-gon$ there would be $\frac{n(n+1)}{2}*\frac{2n+1}{3}=\frac{n(n+1)(2n+1)}{6}$ possible triangles.
Note that $\frac{n(n+1)(2n+1)}{6}=\sum_{k=1}^n k^2$, so there may exist a more direct reasonning that would use the squares of integers. Maybe a recursive argument on n, but I don't really see how right now.
A further generalization would be to look for quadrilateres or convex $k-gons, k<2n+1$, built on vertices of the $2n+1-gon$ and containing the center.
A: For a given triangle, start at one vertex, go around the $13$-gon counting the number of edges to the next vertex of the triangle, then to the last vertex, and finally back to the first vertex again. This will give you three numbers that add to $13$ and that characterises the shape of the triangle. For convenience, say we choose to go around the $13$-gon in such a direction (and starting from such a vertex) that the three numbers come out in decreasing order. Note that this means that for any triangle, there is only one triple that represents it, although each triple represents several different triangles (more on that below).
Claim: The triangle contains the center iff the largest number of the triple is no greater than $6$.
Proof: If there are two vertices whose distance is $7$ or greater (meaning the number of edges between them along the arc of the $13$-gon that does not contain the third vertex), then that corresponds exactly to the whole triangle being on one side of the line mentioned in the question, using one of those two vertices as the origin of that line.
So we're looking for the number of triples of positive integers which


*

*is decreasing

*sums to $13$

*has no entries larger than $6$


Let us first assume that the largest number is $6$. That means the remaining two numbers must sum to $7$. There are three ways this can be done in $3$ ways: $(6, 4, 3), (6, 5, 2)$ and $(6, 6, 1)$. Next, we assume the largest number is $5$. Then the remaining two must sum to $8$, which can be done in two ways: $(5, 4, 4)$ and $(5, 5, 3)$. All in all we have two triples of distinct numbers, and three triples contining a pair.
Now we count how many triangles are represented by each triple. The triples of distinct numbers each represent $26$ different triangles, because we can choose starting vertex and direction around the $13$-gon. However, the triples containing a pair only represent $13$ triangles each, because for each such triangle, there is a second way of choosing starting point and direction that gives the same triangle (these triangles are isosceles and therefore bilaterally symmetric).
For instance, the triple $(6, 6, 1)$ gives the same triangle if we start from vertex $1$ and go clockwise as if we start at vertex $13$ and go counterclockwise (assuming the vertices of the $13$-gon are numbered in a clockwise fashion).
So the answer is that there are $2\cdot 26 + 3\cdot 13 = 91$ such triangles.
