Maximum number of acute triangles in a regular convex polygon triangulated into $n-2$ triangles by its diagonals. I have been reading about convex polygons, and I found the following:

We say that a simple polygon is convex if all its interior angles are less than $\pi$. If $P$, a regular convex polygon, is divided into $n-2$ triangles with diagonals, what is the maximum number of acute triangles one can have?

I don't understand what is meant by "$n-2$ triangles with diagonals".
Thank you.
Note: I have understood what you mean by "$ n-2 $ triangles with diagonals", but I have tried a lot to solve it using everything that has been written to me, and I have not succeeded, I do not know how to take the factor that the angles have to be sharp. I would appreciate it if you could help me solve it. Thank you.

When you have a convex polygon with $n$ vertices, numbering the vertices consecutively around the polygon, then a line segment between vertex $i$ and vertex $j$ is called a diagonal if $2 \le \lvert i - j \rvert \le n - 2$.  In other words, the diagonals discussed are line segments between vertices that do not already share an edge.
Because the polygon is convex, these "diagonals" are all contained within the polygon.  If you start from a pair of consecutive edges, form the first triangle by adding the diagonal between their endpoints, you can proceed through the polygon, dividing it into triangles, by picking the next edge (from either side), and connecting its endpoint to the next vertex in the other side.
This is at the core of polygon triangulation.  The different algorithms simply define rules on how to choose which edges to use for the next triangle to extract from the polygon, and work for different types of polygons from convex to monotone to concave.
If you count them, a convex polygon with $n$ vertices has $n - 1$ edges.  The first triangle takes two edges (and one diagonal), and each further triangle takes one edge (and an additional diagonal; the third side of the triangle being a previously added diagonal).  That is why you always end up with $n - 2$ triangles using (this kind of) polygon triangulation on a polygon with $n$ vertices.
