There are $n$ lines on the plane, any two of them intersect, but no three lines intersect at the same point. What is the maximum number of new lines we can get if we connect every pair of intersection points that are not connected by any of the original lines?
I. solution. The $n$ lines determine $\binom{n}{2}=p$ different intersection points, let one of them be the point $M$, and let the lines determining it be $a$ and $b$. By connecting $M$ with another intersection point $M'$, we always get a new line - relative to the given ones - if $M'$ is not on $a$ or $b$, that is, if $M'$ is among the intersection points of the other $(n-2)$ lines. The number of such points is $\binom{n-2}{2}=q$. The conceivable $pq$ connections in pairs always give the same line - $MM'$ and $M'M$ in roles -, so if no other coincidence occurs,

$$
\frac{1}{2} p q=\frac{1}{2}\binom{n}{2}\binom{n-2}{2}=\frac{1}{8} n(n-1)(n-2)(n-3)
$$

new lines are obtained, otherwise fewer.

János Pósfai (Szombathely, Nagy Lajos Gymn., IV. grade)

II. solution. It is enough to examine the question for $n \geq 4$, since it is easy to show that for $n=3$ there is no possibility for a new connection.

For $n=4$, let the lines be $a, b, c, d$. We get a new connecting line if we connect the intersection point of $a$ and another line with the intersection point of the remaining lines, e.g., the intersection point of $a$ and $b$ with the intersection point of $c$ and $d$. By replacing $b$ first with $c$ and then with $d$, we get 2 more connections, making a total of 3.

For cases where $n>4$, the solution can be reduced to the case of $n=4$. Taking another 4 lines from the given ones in the role of the previous $a, b, c, d$ lines, we get a different connecting line in place of each of the three previous new lines, meaning that at least one of the two intersection points determining the connection is different from the previous ones. (The line itself can be identical to one already obtained, i.e., in the case where a connecting line passes through another intersection point.) According to this, the number of connection possibilities is at most 3 times the number of quadruples that can be selected from the given $n$ lines:

$$
3\binom{n}{4}=\frac{n(n-1)(n-2)(n-3)}{1 \cdot 2 \cdot 4}
$$

József Angyal (Budapest, Berzsenyi D. Gymn., IV. grade)