2. Determine the maximum number of points in a plane such that any three of them form the vertices of a right triangle.
2. The vertices of a rectangle satisfy the given condition. Let's prove that it is not possible to choose 5 points with the described property. Assume the opposite. Let $AB$ be the longest segment in the set of all segments whose endpoints are among the 5 observed points. Then the three remaining points must lie on the circle whose diameter is $AB$, and at least two of them must lie on the same arc $\overparen{AB}$. Let the arrangement of points on the arc be $A-C-D-B$. Then $\measuredangle ACD = \measuredangle ACB + \measuredangle BCD = 90^{\circ} + \measuredangle BCD > 90^{\circ}$, so $\triangle ACD$ is obtuse, which is a contradiction.

![](https://cdn.mathpix.com/cropped/2024_06_04_adcebcbe7d41887b6eefg-38.jpg?height=346&width=377&top_left_y=1442&top_left_x=1148)

From the above, it follows that the number of points sought is 4.