5.

Segments connecting an inner point of a convex non-equilateral $n$-gon with its vertices divide the $n$-gon into $n$ equal triangles.

For what smallest $n$ is this possible?
Let's prove that the specified situation is impossible for $n=3,4$.

The first method. For $n=3$, the angles of the triangles in the partition that meet at an internal point are equal, since the sum of any two different angles in them is less than $180^{\circ}$. But then the sides opposite to these angles, which are sides of the polygon, are also equal. This is a contradiction.

Suppose the quadrilateral $ABCD$ is divided into equal triangles by segments drawn from point $O$. Draw the broken line (or straight line) $AOC$. At point $O$, the sum of two angles on one side of $AOC$ is not less than $180^{\circ}$. Similarly to the case of $n=3$, we get that the corresponding sides are equal. Considering the broken line $BOD$, we get the equality of one of these sides and the third side. Let, for example, $AB = BC = CD = 1$. Suppose $AD \neq l$. In triangle $AOD$, there is a side of length $l$, say $AO$. Thus, in triangle $AOB$ (and all equal to it), there are two sides of length $l$. As a result, $AO = DO = l$. In triangle $BOC$, there are two sides of length $l$. One of them is $BC$, and the other, say, $BO$. But then in triangle $AOB$, all sides are equal. Therefore, this is also true for triangle $AOD$, i.e., $AD = l$. This is a contradiction.

The second method. Since the triangles into which the given triangle is divided are equal, the radii of the circumscribed circles around them and their areas are equal. The first implies that the point defining the division is the orthocenter of the triangle (see problem 56681a), and the second implies that it is the point of intersection of its medians (see problem 54992). However, the orthocenter and the point of intersection of the medians coincide only in an equilateral triangle (see problem 108458).

$\angle OAB = \angle OCB$, as angles opposite the same side of equal triangles. Similarly, $\angle OAD = \angle OCD$, $\angle OBC = \angle ODC$, $\angle OBA = \angle ODA$. Therefore, $\angle A = \angle C$, $\angle B = \angle D$, and $ABCD$ is a parallelogram. Since the segments from $O$ divide it into equal-area triangles, $O$ is the point of intersection of its diagonals, and then from the equality of the triangles, it follows that $ABCD$ is a rhombus.

For $n=5$, the specified situation is possible (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_c317f3f1f50d1d30304bg-16.jpg?height=717&width=920&top_left_y=202&top_left_x=569)

## Answer

$n=5$.