More triangles

Inscribe the 100-gon inside a circle. A triangle inscribed in this circle will be obtuse if all three of its vertices lie on one side of a diameter of the circle. (See this wiki for an explanation.)

With $100$ equidistant point on this circle, we can form $50$ diameters using the 100-gon vertices, where in each case $49$ vertices lie on one side of the diameter and $49$ on the other. Now in turn choose each of the $100$ vertices as the "starting" vertex moving clockwise, with this starting vertex as one end of a diameter. We can then choose $2$ of the $49$ vertices on the "clockwise" side of this diameter as the other two vertices of an obtuse triangle. In this way we account for all possible desired obtuse triangles, the number of which is

$100 \times \dbinom{49}{2} = 100 \times \dfrac{49*48}{2} = \boxed{117600}$ .