We have selected the vertices, the centroid, and the points that trisect the sides of an equilateral triangle. How many points can we retain from among them so that no three form an equilateral triangle?
Solution. Let's denote the points as $A, B, \ldots, J$ as shown in Figure 1. Since the triangle $ABC$ itself is equilateral, at least one of its vertices must be left out.

![](https://cdn.mathpix.com/cropped/2024_05_02_773e2fe68f2c5cbcc39cg-1.jpg?height=404&width=352&top_left_y=264&top_left_x=884)

Due to the symmetric role of the vertices, we can assume that the vertex $C$ is left out. If no other point were left out, the remaining points would form three equilateral triangles with no common vertices $(A D I, E B F$ and $H J G)$. Therefore, at least one vertex of each of these triangles must be left out to obtain a set of points that meets the conditions. Thus, at least $1+3=4$ points must be left out, meaning that at most $10-4=6$ points can be retained.

![](https://cdn.mathpix.com/cropped/2024_05_02_773e2fe68f2c5cbcc39cg-1.jpg?height=399&width=352&top_left_y=966&top_left_x=884)

We will show that this is indeed possible. Let's leave out the points $C, D, E$, and $J$. Then, among the remaining six points, three are collinear in pairs (Figure 2). If any three of them formed an equilateral triangle, we could assume, due to the symmetric role of the points, that two of its vertices belong to the set $\{A, I, H\}$. However, if one side of an equilateral triangle is $A I, A H$, or $I H$, then its third vertex could only be $D, E$, or $J$ respectively, but we have left out all these points.

Therefore, at most six points can be retained from the ten such that no three of them form an equilateral triangle.

Remark. It can be shown that if six points are retained, then no three of them can form an equilateral triangle only if the points are arranged as shown in Figure 2, i.e., three of them are on each side of the original equilateral triangle and differ from the intersection point of these two sides.