Problem 1. (4 points) A billiard table has the shape of a $2 \times 1$ rectangle. Pockets are located at the corners and at the midpoints of the longer sides of the table. Balls are placed on the table such that each pocket lies on a line with at least two balls. What is the minimum number of balls placed? (The pockets and balls are considered as points.)
Solution. We will prove that at least 4 balls are placed. With the notation of the diagram, balls can be placed at the intersection points of $A D \cap C F, A D \cap C E, B D \cap C F$ and $B D \cap C E$.

![](https://cdn.mathpix.com/cropped/2024_06_03_436aec99131a0220e42fg-1.jpg?height=314&width=443&top_left_y=660&top_left_x=818)

We will prove that three balls are insufficient. A line through two balls intersects the boundary of the rectangle at two points. There are 6 pockets, so at least 3 lines are required. Three balls determine 3 lines only if these lines form a triangle. However, no three of the lines connecting two pockets $(A E, A D, B F, B E, B D, C F, C E)$ form a triangle with vertices in the interior of the rectangle.