Given 2006 points on the plane. What is the maximum number of points among them that can form an acute-angled triangle with any other two points?
I. solution. There can only be one point that satisfies the condition. We show such a point system: on a unit circle centered at $O$, mark a $\frac{\pi}{2}$ length arc, and place 2005 points inside this arc. Any two of these points, together with $O$, form an isosceles triangle where the angle at $O$ is less than $90^{\circ}$.

However, there cannot be two such points, because by choosing another two points with them, all triangles determined by the four points would have to be acute-angled. This is not possible: If four points form a convex quadrilateral, then one of its angles is obtuse or right. If one of the four points lies inside the triangle formed by the other three, then from this point, at most one side is visible at an acute angle.

Remark. Show a point system where there is no point that forms an acute-angled triangle with any other two points.

For the part of the problem that states there can be at most two such points, we provide two other proofs.

II. solution. Suppose $P$ is a suitable point. If $A$ is a different point, then the perpendicular to $PA$ at $P$ must contain all 2006 points on the side that contains $A$, because a point $B$ on the other side would form an obtuse angle $\angle APB$. Therefore, $P$ lies on the perimeter of the convex hull of the 2006 points.

The convex hull must have acute angles at $P$ and at the two vertices adjacent to $P$. However, a convex polygon can have at most three acute angles. Thus, if the convex hull has at least four vertices, there cannot be another suitable point besides $P$.

If, however, the convex hull is a triangle, then none of its vertices can be suitable, because from any interior point of a triangle, at most one side is visible at an acute angle.

Therefore, there can be at most one suitable point among the 2006.

III. solution. Suppose that among the given points, $A$ and $B$ both have the desired property. Draw perpendiculars to the segment $AB$ at both endpoints. The other points must lie between these two lines. Take the points farthest from $AB$ on both sides of $AB$ (if there are multiple, choose one). Let these points be $C$ and $D$. Then all points are in one of the four rectangles shown in the diagram. However, each rectangle is contained within the Thales circle of its diagonal, so any point in it forms an obtuse (or right) triangle with the endpoints of the diagonal. This contradicts our assumption, so there can be at most one suitable point among the 2006.

![](https://cdn.mathpix.com/cropped/2024_05_02_0df33d3dcb04364ff5b2g-1.jpg?height=507&width=339&top_left_y=1324&top_left_x=882)