4. In the plane, there are 1001 points. Three of them are vertices of a triangle in which the remaining 998 points are located. All points are arranged such that no three of them are collinear. How many non-overlapping triangles form the given triangle?
Solution. Let $A, B, C$ be the vertices of the triangle in which the remaining 998 points are located, and let point $M$ be one of the interior points. Point $M$ is a vertex of a certain number of non-overlapping triangles, so the sum of the interior angles of all these triangles is $360^{\circ}$. We have 998 points like point $M$, so the sum of the interior angles of the triangles whose vertices are these points is $998 \cdot 360^{\circ}$. If we add $180^{\circ}$ to this sum, we get the sum of the interior angles of all non-overlapping triangles whose vertices are the given 1001 points. Therefore, the number of all triangles is

$$
\frac{998 \cdot 360 + 180^{\circ}}{180^{\circ}} = 1997
$$