7.8. A) On a plane, three lines and $n$ points are drawn such that on either side of each line there are exactly two points (points lying on the line itself do not belong to either side). For which values of $n$ is this possible?
Answer: for $n$ equal to $4, 5, 6$ or 7.

Solution. Let's count how many points lie on the specified lines. Since for each line, four points do not lie on it, there should be $n-4$ points on the line. In this case, either all three lines intersect at one point, and if this is one of the marked points, we have counted it three times, or the lines intersect in no more than three points, and if any of them are marked, such points have been counted twice. Therefore, there are at least $3(n-4)-3=3n-15$ points on the lines. On the other hand, there are no more than $n$ points, so $3n-15 \leq n$, from which it follows that $n \leq 7$. It is also clear that there are no fewer than four points. An example for each value of $n$ from 4 to 7 is shown in the figure.
![](https://cdn.mathpix.com/cropped/2024_05_06_402f7649344ac77c021fg-6.jpg?height=360&width=1468&top_left_y=361&top_left_x=401)

b) On a plane, three lines intersect at one point, and several points are marked such that there are exactly two points on each side of each line (points lying on the line itself do not belong to either side). Additionally, a total of $n$ points lie on the lines. For which values of $n$ is this possible?

Answer: for $n$ equal to $0, 1, 3, 4, 6$ or 7.

Solution. Let $m$ points be marked, of which $n$ lie on the lines. For each line, 4 points do not lie on it, and $m-4$ points lie on the line. If the intersection point of the lines is not marked, then $n=3(m-4)$, and if it is marked, we have counted it three times, so $n=3(m-4)-2$. Thus, the number $n$ is either divisible by 3 or gives a remainder of 1 when divided by 3. Also, $n=3(m-4)-2 \leq m$, so $m \leq 7$. Therefore, $n \leq 7$, meaning that a total of $0, 1, 3, 4, 6$ or 7 points can lie on the lines. An example for each of these values is shown in the figure.

Note. The fact that the total number of points is no more than seven can be proven differently. If there are at least 8 marked points, then on each line, there are at least 4 of them. Consider any of the three lines. Outside of it, on the second line, there are at least 3 points and on the third line, there are at least 3 points, totaling at least 6 points. This is impossible.
![](https://cdn.mathpix.com/cropped/2024_05_06_402f7649344ac77c021fg-6.jpg?height=730&width=1024&top_left_y=1636&top_left_x=640)

## 8th grade