Let the convex polygon of $n$ points by $C$ Then first noticing that a point $P$ is inside the circle with diameter $AB$ $\Leftrightarrow$ $\angle APB>\frac{\pi}{2}$ So convex polygon $C$ satisfies condition so defining the number of vertices of $C$ by $m$ $\pi \cdot (m-2)\le m\cdot \frac{\pi}{2}$ Solving this equation $m\le 4$ Now suppose there exists a point $Q$ inside the polygon. Then making polygon into $n-2$ triangle, at least one area $\triangle ABC$ posseses $Q$ Then $\angle APB+\angle BPC+\angle CPA=2\pi$ Contradiction So $n=3,4$ Equality holds for regular triangle and square

$n = 3, 4, 5$. Constructions are obvious: take equilateral triangle, square, and square plus center. Suppose $n \ge 6$. If convex hull has at least 5 vertices, then its average angle is $\ge 108^\circ$, so at least one angle is obtuse, clearly a contradiction. Thus convex hull has at most 4 vertices. If convex hull is triangle $ABC$, then there is a point $P$ inside it. At least one of $\angle APB, \angle BPC, \angle CPA$ is at least $120^\circ$, or obtuse, contradiction. If convex hull is quadrilateral $ABCD$, then it has all four right angles (since none are obtuse). Any point $P$ inside $ABCD$ satisfies $\angle APB = \angle BPC = \angle CPD = \angle DPA = 90^\circ$ for same reason. There is at most one such point, the intersection of circles of diameters $AB$ and $BC$ that is not $B$, so $n \le 5$. This completes the casework and the proof that $n = 3, 4, 5$.

suli wrote: $n = 3, 4, 5$. Constructions are obvious: take equilateral triangle, square, and square plus center. But isn't the square plus center not true? Take the diameter that is a diagonal of the square, then the center is in this circle.

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