For a $21$-sided regular polygon $A_ 1A _ 2A _ 3 ... A_{ 21}$ inscribed in a circle with centre O, how many triangles $A_iA_jA_k$ contain O? Let $A_1A _ 2A _ 3 ... A_{21}$ be a $21 -$ sided regular polygon inscribed in a circle with centre $O$ . How many triangles $A_iA_jA _ k, 1 \le i<j<k \le 21$, contain the point $O$ in their interior.
I got the answer $194.$ Just needed to confirm the answer and the solution.
PS: This is the sixth problem of $RMO$ $1995$.
 A: Starting at any point, $A$, we must choose another point, $B$ from the $10$ counter-clockwise from $A$. Once $B$ is chosen, the number of choices for $C$ from the $10$ clockwise from $A$ is the distance from $A$ to $B$. 
Thus, for each choice of $A$ there are $\sum\limits_{k=1}^{10}k=55$ choices for $B$ and $C$. There are $21$ choices for $A$, but for each triangle, we have counted $3$ vertices for $A$, so we must divide by $3$. This gives
$$
55\cdot\frac{21}3=385
$$
triangles.
A: To have the center within the triangle, the arc between any pair of points that does not include the third point must be less than half the circle.  We must select three point numbers $1\le a\lt b\lt  c\le 21$ so that $b-a \lt 11, c-b \lt 11, a-c \lt 21+11$  First there are seven equilateral triangles.  We will count the others in standard position, so that $A_a=A_1$ is one of the points, the longest side is from $A_1$ to $A_b$, if there is a tie for longest side the long ones are from $A_1$ to $A_b$ and $A_b$ to $A_c$.  In that case we can then multiply the count by $21$.  The next point has to be $A_9,A_{10},$ or $A_{11}$.  For $A_{11}$ we can choose any of the points $A_{13}$ to $A_{21}$.  We cannot choose $A_{12}$ because we would have two sides of $10$ and they would not be the first two.  This gives $9$ triangles.  For $A_{10}$ we can choose $A_{14}$ to $A_{19}$ for six choices.  For $A_9$ we can choose $A_{15}$ to $A_{17}$ for three choices.  This gives us $18$ non-equilateral triangles starting with $A_1$ but we could start anywhere, so we multiply by $21$.  Our final answer is $21 \cdot 18+7=385$
