Problem 746

Two lines lying in the same plane do not intersect if and only if they are parallel. Imagine we have placed 17 lines on the plane. Let's classify them, putting parallel lines in the same class. Suppose we have $s$ classes, with $k_{1}, \ldots$, and the last class has $k_{s}$ lines. Clearly, $k_{1}, \ldots, k_{s}$ are positive integers and

$$k_{1}+k_{2}+\ldots+k_{s}=17$$

No matter how we assign these numbers to satisfy the conditions, we can always find 17 lines on the plane such that no three lines intersect at the same point, and there are exactly $s$ classes, with $k_{1}, k_{2}, \ldots, k_{s}$ lines in each class.

We claim that the number of intersection points depends only on the numbers $k_{1}, \ldots, k_{s}$, not on the specific arrangement of the lines. Indeed, take a line from the i-th class. There are a total of $17-k_{i}$ intersection points on this line, because it is intersected by every line from other classes, and these intersection points are distinct. Thus, the number of intersection points is

$$\begin{aligned} & \left(17-k_{1}\right) k_{1}+\left(17-k_{2}\right) k_{2}+\ldots+\left(17-k_{s}\right) k_{s}= \\ = & 17\left(k_{1}+k_{2}+\ldots+k_{s}\right)-\left(k_{1}^{2}+\ldots+k_{s}^{2}\right)= \\ = & 17^{2}-\left(k_{1}^{2}+\ldots+k_{s}^{2}\right) \end{aligned}$$

half of this sum, because each intersection point is crossed by two lines, so we counted each intersection point twice. We need 101 intersection points, so the required lines exist if and only if there are positive integers $k_{1}, k_{2}, \ldots, k_{s}$ such that

$$\begin{aligned} & k_{1}+k_{2}+\ldots+k_{s}=17 \\ & k_{1}^{2}+k_{2}^{2}+\ldots+k_{s}^{2}=17^{2}-202=87 \end{aligned}$$

hold simultaneously.

After a brief trial, we find that $s=6, k_{1}=8, k_{2}=4, k_{3}=2, k_{4}=k_{5}=k_{6}=1$ is a solution. Thus, it is possible to give lines that meet the conditions.

Remark. It is not difficult to show that the above system of equations has only the following solutions satisfying $k_{1} \geq k_{2} \geq \ldots \geq k_{s}$:

$8,4,2,1,1,1 \quad(s=6)$

$8,3,3,2,1 \quad(s=5)$

$7,5,3,2 \quad(s=4)$

$6,5,5,1 \quad(s=4)$