4. What is the maximum number of intersection points of chords determined by eight points on a circle, if intersections on the circle are not counted?

## Solutions to the selection competition problems

Each problem is worth 7 points. When evaluating each problem, reasonably consider the provided scoring guide. A competitor should not receive more than 3 points for any problem if the partial solution does not clearly lead to the final solution of the problem.
IV/4. To get the maximum number of intersections, we must choose the points on the circle so that exactly two chords intersect at each intersection. Let's choose one of these eight points. If we connect this point to one of the other seven, there are $m$ points on one side of the chord, and $6-m$ points on the other side. It follows that the total number of intersections with chords starting from the chosen point is $5 \cdot 1 + 4 \cdot 2 + 3 \cdot 3 + 2 \cdot 4 + 1 \cdot 5 = 35$. There are 8 points in total, and since each intersection is determined by four points, the maximum number of intersections is 

![](https://cdn.mathpix.com/cropped/2024_06_07_7e12218b3c89a25adc33g-9.jpg?height=371&width=371&top_left_y=1845&top_left_x=1619)
equal to $\frac{8 \cdot 35}{4} = 70$. The image on the right shows that this number can indeed be achieved.

Combinatorial reasoning: 3 points. Calculation: 2 points. Example (or convincing justification) that the maximum number is actually achieved: 2 points.