85. The diagonals of a convex 17-gon, drawn from one vertex, divide it into 15 triangles. Can a convex 17-gon be cut into 14 triangles?

What about a non-convex 17-gon? What is the smallest number of triangles into which a 17-gon can be cut?
85. Each of the angles of a convex polygon is less than $2 d$, so the vertices cannot lie on the sides of the triangles of the partition. When a convex polygon is divided into triangles (not necessarily by diagonals), all the angles of the polygon will be divided into parts, which will

![](https://cdn.mathpix.com/cropped/2024_05_21_d3653343768e91d6de9ag-184.jpg?height=545&width=726&top_left_y=919&top_left_x=239)

serve as angles of the triangles of the partition (Fig. 154). It follows that the sum of the angles of all the triangles of the partition must be no less than the sum of the angles of the polygon. But the sum of the angles of 14 triangles is $28 d$, while the sum of the angles of a 17-sided polygon is $30 d$. Therefore, a convex 17-sided polygon cannot be divided into 14 triangles.

It is easy to construct a non-convex 17-sided polygon that can be divided into six triangles (see, for example, Fig. 155, a, σ).

We will now show that no 17-sided polygon (convex or non-convex) can be cut into five triangles (if the 17-sided polygon does not have angles of $180^{\circ}$ ).
![](https://cdn.mathpix.com/cropped/2024_05_21_d3653343768e91d6de9ag-185.jpg?height=560&width=586&top_left_y=436&top_left_x=346)

Fig. 155.

Indeed, each vertex of the 17-sided polygon either coincides with the vertices of several (possibly one) triangles of the partition, or lies on the side of a triangle of the partition; in the latter case, the corresponding angle of the 17-sided polygon is obviously greater than $180^{\circ}$. Therefore, the remaining part of this angle must be filled by at least one more triangle of the partition; hence, in this case, the vertex also coincides with the vertex of at least one triangle. It follows that the total number of vertices of all the triangles of the partition is no less than the number of vertices of the 17-sided polygon, i.e., 17.

Therefore, if $T$ is the number of triangles in the partition, then

$$
3 T \geqslant 17
$$

from which

$$
T \geqslant 5 \frac{2}{3}
$$

Since $T$ is an integer, $T$ cannot be less than six.