Mongolian Mathematical Olympiad

Answer: 5. The four vertices and the center of a square satisfy the conditions stated in the problem. Let's now prove that this is the only possibility. Consequently, it is not possible to find six points that satisfy the conditions.

Consider a triangle $ABC$ formed by three non-collinear points $A$, $B$, and $C$, with an additional point $D$. If point $A$ lies inside triangle $ABC$, then there exists an obtuse triangle since $2\pi /3$ is greater than $\pi /2$. If point $D$ lies on one of the sides of triangle $ABC$, then $D$ must be the foot of the perpendicular from the opposite vertex. Therefore, there can be at most three points on a line. If point $D$ lies outside triangle $ABC$, then $ABCD$ must be a rectangle since $2\pi /4$ equals $\pi /2$.

Now let's consider the fifth point. If three points form a triangle, and two of them are the feet of the heights of that triangle, then these two feet and the opposite vertex form an obtuse triangle. Thus, we can assume that they form a rectangle and an additional point. If this point lies outside the rectangle, there will be an obtuse triangle. Moreover, this point cannot lie on any side of the rectangle. Therefore, the only possibility is that this point lies inside the rectangle. In this case, this point and any two vertices of the rectangle form a right-angled triangle. Hence, the rectangle must be a square, and the point must be its center.

In conclusion, the only configuration that satisfies the conditions is a square with its four vertices and center.