There are finitely many points on the plane such that no three of them are on the same line. For at most how many points can the segments connecting the points be colored with three colors so that the following conditions are satisfied:

a) no point can have three segments of different colors starting from it;

b) among the points, it is not possible to find three such that the three segments they determine have the same color;

c) all three colors are used.
For four points, a coloring that meets the conditions is visible in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_bce463c860d11ae2907fg-1.jpg?height=404&width=629&top_left_y=210&top_left_x=737)

We claim that for more points, it is no longer possible to color the segments appropriately.

First, we show that no point can have three segments of the same color. Indeed, select any four points and label them $A, B, C,$ and $D$. Contrary to our claim, connect $A$ to $B, C,$ and $D$ with segments of the same color, say red. Then, neither $BC$, nor $BD$, nor $CD$ can be red, as this would violate condition (b). Let $BC$ be green; then, by condition (a), $BD$ and $CD$ must also be green. However, this means that the sides of triangle $BCD$ are all the same color, contradicting condition (b).

Therefore, from any point, at most two segments of the same color can originate, and since only two colors can be used for the segments originating from a single point, at most four segments can originate from any point. This already implies that the number of points can be at most 5. For 5 points, if no three points are collinear, they determine a total of 10 segments. If we select one point and draw two red segments from it, then from the points to which these segments lead, one more red segment must originate. By condition (b), these are different segments, so there are at least 4 red segments. Since this is true for the other two colors as well, and all three colors must be used, at least 12 segments are required. However, only 10 segments can be drawn between 5 points, so it is not possible to color the segments in such a way that all conditions are met for 5 points. This proves our claim.

Based on the work of Borsodi Brigitta (Tata, Eötvös J. Gymnasium, 2nd year)

Note. Many incorrect solutions were submitted due to a misunderstanding of the problem. Some examined the fulfillment of the three conditions separately, whereas they must all be satisfied simultaneously.

Some did not find the correct coloring for 4 points because they tried to add a point to a solution for 3 points. Indeed, in any selection of 3 points from 4, there are two segments of the same color among the segments they determine.

Those who showed a single coloring and claimed that it is impossible for 5 points did not receive points. This does not mean that there is no solution that satisfies the problem's conditions. We saw this with 4 points, where both good and bad colorings can be found.

Most incomplete solutions proved the impossibility of coloring for 5 points but did not show that it is also impossible for more than 5 points. The question, however, was "what is the maximum number of points" that can be colored.