Divide the plane into 1000 parts using the fewest possible lines. (H)
First, let's examine how many parts \( n \) lines can divide the plane into. With the first two lines, \( 2+2 \) regions are created. It is easy to see that each additional line will create the maximum number of new regions if it is not parallel to any of the previously drawn lines and does not pass through the intersection point of any two previously drawn lines. Accordingly, for example, the fourth line intersects each of the first three lines at different points, and these three "new" intersection points divide this line into segments \( a, b, c, d \). Four "new" regions are created corresponding to these four intervals. Similarly, we can show that adding the \( n \)-th line can create at most \( n \) "new" regions. Therefore, \( n \) lines can divide the plane into at most \( N = 2 + 2 + 3 + \cdots + n \) parts.

It is known that \( 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2} \), so

\[
N = 1 + \frac{n(n+1)}{2}
\]

From formula (1), for \( n = 44 \), \( N = 991 \), so 44 lines are still insufficient.

We will show that 45 lines can divide the plane into 1000 parts. Since (1) gives a number greater than 1000 for \( n = 45 \), we try to draw more than two lines through a single point. Let's draw 10 lines through a single point; these lines divide the plane into 20 parts. The eleventh and each subsequent line should be drawn so that it does not pass through any previously created intersection point and is not parallel to any previously drawn line. Thus, the eleventh line creates 11, the twelfth 12, ..., and the forty-fifth 45 "new" regions. The number of regions is then

\[
20 + 11 + 12 + \cdots + 45 = 1000
\]

Lajos Deli (Hajdúszoboszló, Hógyes E. Gymn., 10th grade) Ta Vinh Thong (Fazekas M. Fóv. Gymn., 9th grade)

Remarks. 1. In the solution, we took it for granted, although it requires justification, that the \( n \)-th line can be drawn so that it intersects each previously drawn line and is not parallel to any of them.

2. Most solvers proved formula (1) by complete induction.

![](https://cdn.mathpix.com/cropped/2024_05_02_a0b7fbc0f33860502f02g-1.jpg?height=670&width=675&top_left_y=1237&top_left_x=730)