In the plane, there are 6 given points, no three of which are on the same line. How many segments must be drawn between the point pairs so that there is certainly a triangle whose vertices are among the given points?
As can be seen in Figure 1, it is not sufficient to draw 9 connecting segments. The 9 segments do not yet form a triangle whose vertices are among the given points.

It is easy to see that a total of $\frac{6 \cdot 5}{2}=15$ segments can be drawn. Let's draw 10 of these. The 10 segments have a total of 20 endpoints. If at most 3 segments were to start from each of the 6 given points, then there could be at most $6 \cdot 3=18$ endpoints in total. Given that there are 20 endpoints, there must be at least one point among the 6 points from which at least 4 segments start. In Figure 2, this point is $A$, and the other endpoints of the segments starting from $A$ are $B, C, D, E$. If we want to avoid forming a triangle, then we cannot draw the segments $B C, B D, B E, C D, C E, D E$. Therefore, since there are a total of 15 segments, only 9 would be drawable without forming a triangle whose vertices are among the 6 given points. Thus, 10 segments will already form at least one triangle.

Ágnes Tóth (Hajdúszoboszló, Hőgyes E. Gymnasium, 8th grade)
![](https://cdn.mathpix.com/cropped/2024_05_02_0b7f15d7c4ce2d97bc3eg-1.jpg?height=876&width=506&top_left_y=641&top_left_x=798)