Avor: Frankin B.P.

On a circle, $n$ points are marked. It turns out that exactly half of the triangles with vertices at these points are acute-angled.

Find all values of $n$ for which this is possible.
Obviously, $n>3$. Consider an arbitrary quadrilateral with vertices at the given points. If the center of the circle lies inside the quadrilateral and not on its diagonal (we will call such a quadrilateral good), then out of the four triangles formed by the vertices of the quadrilateral, exactly two are acute. In all other cases, there are fewer than two acute triangles. Therefore, the condition of the problem is satisfied only when all quadrilaterals formed by the given points are good. It is obvious that this is possible for $n=4$ and $n=5$ (for example, you can take the vertices of a regular pentagon).

Let $n>5$. Consider one of the given points $A$ and draw a diameter $A A^{\prime}$ through it. If point $A^{\prime}$ is marked, then the quadrilateral formed by $A, A^{\prime}$, and any two of the other points will not be good. Otherwise, there will be three marked points lying on one side of $A A^{\prime}$. The quadrilateral formed by these points and point $A$ is not good.

## Answer

$n=4$ or 5.

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