6. There are 100 equally divided points on a circle. The number of obtuse triangles formed by these points as vertices is $\qquad$ .
$6.50 \times 49 \times 48$ (or 117600).
Let $A_{t}$ be the vertex of the obtuse angle. $\angle A_{\text{s}}$
intercepts an arc of $x$ equal parts, and the other two angles intercept arcs of $y, z$ equal parts, respectively, such that
$$
\left\{\begin{array}{l}
x+y+z=100, \\
x \geqslant 51, y \geqslant 1, z \geqslant 1 .
\end{array}\right.
$$

This can be rewritten as
$$
\left\{\begin{array}{l}
(x-50)+y+z=50 . \\ 
x-50 \geqslant 1, y \geqslant 1, z \geqslant 1 .
\end{array}\right.
$$
We can express (1) as the sum of 50 ones. There are 49 plus signs in between. By choosing any two of these 49 plus signs and merging the numbers they separate, we obtain a positive integer solution to equation (1); conversely, for any positive integer solution to equation (1), we can split it, corresponding to "choosing 2 out of 49 plus signs." This means the indeterminate equation (1) has $\mathrm{C}_{49}^{2}$ positive integer solutions.
Letting $i$ range from $1,2, \cdots, 100$ gives the number of obtuse triangles as $100 \mathrm{C}_{49}^{2}=50 \times 49 \times 48=117600$.