How many triangles determined by three vertices of a regular $13$-gon contain the polygon's center? Can you help me solve this problem?

A regular 13-sided polygon is inscribed in a circle with centre $O$. Triangles can be formed by choosing three vertices of this polygon to be the vertices of the triangle.
How many of the triangles that can be formed in this way have the point $O$ inside?

I have no idea how to approach this question, so I tried to start off simple by  drawing the number of triangles that can be inscribed in a square and a pentagon. However, after finding the values 4 and 8, I have no idea where to go from here.
Any help is highly appreciated. Thank you!
 A: Choose two points. These are then a certain number of edges apart, up to $6$. A brief investigation will show that there are a number of points that can form a centre-including triangle with those points, and that number is equal to the separation of the points.
There are equal numbers of points with all valid separations, and each triangle will be counted three times, so the answer is
$$\binom{13}{2}\big / 6\cdot (1+2+3+4+5+6)/3 = 13\times 7 = 91$$
A: Let the paths between the three chosen vertices have lengths $a,b,c$. The number positive integer solutions to $a+b+c=13$ with $a,b,c\le6$, i.e. the number of nonnegative integer solutions to $a+b+c=10$ with $a,b,c\le5$. By stars and bars and inclusion and exclusion, this is
$$\binom{10+2}2-3\binom{4+2}2=21$$
These $21$ solutions reduce to five upon counting permutations as identical: three with two duplicated lengths ($166,355,544$) and two with all distinct lengths ($256,346$). Each of the first set counts $13$ times and each of the second $26$ times, for a final answer of $3×13+2×26=91$.
