[Groovitz $\mathbf{B . M}$.

A regular polygon is drawn on the board. Vova wants to mark $k$ points on its perimeter so that there is no other regular polygon (not necessarily with the same number of sides) that also contains the marked points on its perimeter.

Find the smallest $k$ sufficient for any initial polygon.

#]
Example. Let's prove that five points are sufficient. Let $A, B, C, D$ be four consecutive vertices of the polygon (possibly $A=D$). Mark points $A, B$, an arbitrary point $X$ on side $AB$, point $Y$ on side $BC$, sufficiently close to $B$, and point $Z$ on side $CD$, sufficiently close to $C$.

Let $P$ be the original polygon, $Q$ be some regular polygon containing our five points on its perimeter, $\alpha$ and $\beta$ be the angles of these polygons. The line $AB$ must contain a side of polygon $Q$, as it contains three marked points. Let this side be $A'B'$. Further, let the side of $Q$ containing $Y$ lie on line $l$; then points $B$ and $Z$ must lie on the same side of it (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_0bf3a2f278e52693111ag-49.jpg?height=746&width=861&top_left_y=2005&top_left_x=610)

Since $Z$ is close to $C$, this means that the angle between lines $l$ and $BC$ is small. On the other hand, $\alpha = \angle ABY \geq \angle A'B'Y \geq \beta$, i.e., the number of sides of $Q$ is not greater than that of $P$. Therefore, the angle between $l$ and $BC$ can be sufficiently small only when these lines coincide. Thus, the line $BC$ also contains a side of $Q$, and point $B$ is then its vertex. Hence, $P$ and $Q$ are homothetic with center at point $B$, and the contour of $Q$ can contain $Z$ only when $P$ coincides with $Q$.

Estimate. Let's prove that for a sufficiently large number of sides $n$ of a regular polygon $P$, four points are not enough to determine it. Suppose that three of these points lie on one side $AB$ of the polygon $P$. The equilateral triangle constructed on $AB$ lies entirely inside $P$, so the fourth marked point lies outside it. Therefore, by applying a homothety to our triangle with center at the midpoint of $AB$ and a suitable coefficient greater than 1, we can ensure that the fourth marked point also lies on its contour. This means that in this case, the polygon is not uniquely determined. If our assumption is incorrect, then the marked points form a convex quadrilateral $ABCD$ "sufficiently close" to an inscribed one. Specifically, we will show that $\angle A + \angle C \geq 180^\circ - 360^\circ / n$. Let points $B$ and $C$ lie on sides $KL$ and $MN$ of the polygon $P$, and points $A$, $K$, and $N$ lie on the same side of $BD$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_0bf3a2f278e52693111ag-50.jpg?height=837&width=861&top_left_y=645&top_left_x=590)

Describe a circle $\omega$ with center $O$ around $P$. Then $\angle A + \angle C \geq \angle LAM + \angle KCN = 1/2(\angle LOM + \angle KON) = 180^\circ - 360^\circ / n$. Similarly,

$\angle B + \angle D \geq 180^\circ - 360^\circ / n$, so $\angle A + \angle C \leq 180^\circ + 360^\circ / n$.

We will show that $ABCD$ can be inscribed either in a square or in an equilateral triangle. Let angle $A$ be the largest angle of the quadrilateral $ABCD$, and $B \geq \angle D$. There are several cases.

1) $\angle B \geq 90^\circ$. Then it is clear that $ABCD$ can be inscribed in a square such that points $A, B$ fall on one of the sides - in one of the ways shown in the figure, depending on whether the projection of $CD$ on $AB$ or the projection of $ABCD$ on the line perpendicular to $AB$ is larger.

![](https://cdn.mathpix.com/cropped/2024_05_06_0bf3a2f278e52693111ag-50.jpg?height=532&width=1015&top_left_y=1972&top_left_x=499)
2) $\angle B \geq 240^\circ$ (since angle $A$ is the largest, $\angle B + \angle C < 240^\circ$). We will show that in this case, $ABCD$ can be inscribed in a square as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_06_0bf3a2f278e52693111ag-51.jpg?height=603&width=546&top_left_y=722&top_left_x=756)

Let $X$ and $Y$ be the projections of points $C$ and $D$ on $AB$; it is sufficient to check that $YB \leq CX$. But $\angle DCX = 270^\circ - (\angle A + \angle D) \leq 30^\circ$, so $XY \leq CX \tan 30^\circ \leq CX / \sqrt{3}$; on the other hand, $\angle B \geq 1/2(\angle B + \angle D) \geq 90^\circ - 180^\circ / n$, so for large $n$, $\tan \angle B \geq 10$; thus,

$XB = CX \cot \angle B \leq CX / 10$, and $YB \leq CX / \sqrt{3} + CX / 10 < CX$, as required.

## Answer

$k=5$.