Determine the minimum number of lines that can be drawn on the plane so that they intersect in exactly $200$ distinct points.
(Note that for $3$ distinct points, the minimum number of lines is $3$ and for $4$ distinct points, the minimum is $4$)
The answer is $21$. The configuration can be achieved with $10$ pairs of parallel lines, and $1$ line which is not parallel to any of the other lines. The line that is not parallel to any of the other lines contributes $20$ points, while each of the other lines contributes $19$. So we have \[19 \cdot 20 + 20= 200\]lines. (Note that each of the pairs of parallel lines are not parallel to any of the other lines. )
Now we prove that this is the smallest . Clearly for any $n$, the maximum possible number of intersections with $n$ lines is $\frac{n(n-1)}{2}$. This number has to be greater than or equal to $200$. It is easy to check that $21$ is the smallest n which satisfies this requirement.