7.1. Divide a regular $n$-gon $(n \geqslant 5)$ into the smallest number of:

a) acute triangles,

b) obtuse triangles.

Each side of any triangle in the partition must simultaneously be a side of another triangle in the partition or a side of the polygon.

It is important that no vertex of one triangle divides the side of another triangle. Such a partition is called a usual triangulation.

## DISSECTION OF A RECTANGLE
7.1. a) The segments connecting the center of the given $n$-gon with its vertices divide it into $n$ acute triangles. It is impossible to divide the given polygon into a smaller number of acute triangles, since two different sides of the $n$-gon cannot belong to one acute triangle (for $n \geqslant 5$, all angles of a regular convex $n$-gon are obtuse).

b) It is easy to divide an acute triangle into 3 obtuse triangles - for example, by connecting the point of intersection of the angle bisectors of this triangle with its vertices (prove this).

Furthermore, any regular $n$-gon can be divided by its diagonals into $(n-3)$ obtuse and one

![](https://cdn.mathpix.com/cropped/2024_05_21_20eb9f6ef2c568605a63g-54.jpg?height=279&width=295&top_left_y=595&top_left_x=135)
a)

![](https://cdn.mathpix.com/cropped/2024_05_21_20eb9f6ef2c568605a63g-54.jpg?height=289&width=289&top_left_y=588&top_left_x=469)

b)

![](https://cdn.mathpix.com/cropped/2024_05_21_20eb9f6ef2c568605a63g-54.jpg?height=340&width=297&top_left_y=566&top_left_x=787)

b)

Fig. 7.1.

acute triangle (see figures a and b), corresponding to odd and even $n$; the acute triangle in the figure is shaded.) By dividing this acute triangle into 3 obtuse triangles, we obtain a division of the given polygon into $n$ obtuse triangles.

Now we will show that a division into obtuse triangles cannot contain fewer than $n$ triangles. First, we will prove that a regular $n$-gon cannot be divided into obtuse triangles by drawing only diagonals. Assume the contrary. Let $AC$ be the largest of the diagonals participating in the division. It is a common side of two triangles in the division - $ABC$ and $ADC$. Describe a circle around the given polygon (see figure b). Since the quadrilateral $ABCD$ is inscribed in a circle, $\angle ABC + \angle ADC = 180^{\circ}$, and at least one of these angles, say $\angle ABC$, does not exceed $90^{\circ}$. In triangle $ABC$, this angle is the largest - it is opposite the largest side $AC$, so triangle $ABC$ is not obtuse. Contradiction.

Thus, among the vertices of the triangles in the division, there is at least one point $M$ lying inside

![](https://cdn.mathpix.com/cropped/2024_05_21_20eb9f6ef2c568605a63g-55.jpg?height=318&width=323&top_left_y=356&top_left_x=143)

Fig. 7.2. the given polygon. Consider the sum of the angles of all triangles in the division. The angles of the triangles with vertex at point $M$ sum up to $360^{\circ}$, and the angles with vertices coinciding with the vertices of the polygon sum up to $180^{\circ}(n-2)$. Therefore, the number of triangles participating in the division cannot be less than $\left[180^{\circ}(n-2) + 360^{\circ}\right] : 180^{\circ} = n$.