# Problem 7.5 (7 points)

On a plane, 6 lines are drawn and several points are marked. It turned out that on each line exactly 3 points are marked. What is the minimum number of points that could have been marked?
# Solution:

The vertices of the triangle, the midpoints of its sides, and the point of intersection of the medians - 7 points lying in threes on 6 lines (3 sides and 3 medians).

P.S. It doesn't have to be medians specifically.

Proof of the estimate: If we have a point through which at least 4 lines pass, then we will have no fewer than $1+4 \cdot 2=9$ marked points (on these 4 lines, there are 2 other marked points each). If we have

![](https://cdn.mathpix.com/cropped/2024_05_06_893b058e592ad8944ddeg-2.jpg?height=557&width=616&top_left_y=1195&top_left_x=1231)
a point through which exactly 3 lines pass, then we will have no fewer than $1+3 \cdot 2=7$ marked points. If, however, through each marked point, no more than two lines pass, then the number of marked points will be no less than $6 \cdot 3 / 2=9$ (6 lines with 3 points each, each point is counted no more than twice). Therefore, we will have no fewer than 7 marked points.

| Criteria | Points |
| :--- | :---: |
| Complete solution | 7 |
| Proof of the estimate without an example | 4 |
| Example without the estimate | 3 |
| Incorrect answer | 0 |

## Answer: 7 points