Maximum number of acute triangles in a regular convex polygon triangulated into $n-2$ triangles by its diagonals.

Question

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.

Answer 1

Rephrasing your question:

If a regular convex polygon is divided into $n-2$ triangles with diagonals, what is the maximum number of acute triangles you can have?

Solution:

If one wants to divide a regular convex polygon into $n-2$ triangles, one needs $n-3$ non-intersecting diagonals.

Proof:

Firstly, the $n-2$ triangles must be non-intersecting and their angles must contribute to the angles of the polygon only because the sum of interior angles of a triangle is $180^\circ$ and that of a regular convex polygon is $180^\circ(n-2)$, unlike

The triangulating diagonals must not intersect inside the polygon. If they do, the triangles don't follow the above statement.

Secondly, the $n-2$ triangles have $3(n-2)$ sides of which $n$ are the sides of the polygon, and $3(n-2)-n$ are due to diagonals, each doubled over two (not four as we have seen in the figure above) triangles. $$\Rightarrow 2d=3(n-2)-n\Rightarrow d=n-3$$ Think of it the opposite way- How many lines does one need to separate that polygon into its constituent $n-2$ triangles? The answer is $2d$. Above proves that dividing a regular convex polygon in any manner to get $n-2$ triangles requires non-intersecting $n-3$ diagonals or, as a converse, $n-3$ non-intersecting diagonals drawn in any way form $n-2$ triangles.

Howsoever one triangulates a regular convex polygon by non-intersecting diagonals, there will be at most one acute triangle.

Proof:

The acute triangle requires that at least $1$ vertex of the triangle lies on the other side of the diameter (of the circle circumscribing the polygon) as the other two vertices because having all vertices on one side of the diameter encloses arc lengths $\ge$ semicircle-arc length $\Rightarrow\ge90^\circ$ angles. Having drawn such a triangle, one can not draw another such triangle, not intersecting the first one. That is, the triangle must strictly contain the center to be acute.

Answer 2

From a particular point, rays can be sent to n-3 points (was written as n-2 previously). The rays to itself and the two adjacent points don't count because the lines formed are NOT diagonals. The diagonals with the n sides will form a total of n - 2 triangles.

Answer 3

The triangles don't have diagonals. The polygon has diagonals. You can take some of the diagonals of the polygon and with those diagonals (that is, by using those diagonals) divide the polygon into $n - 2$ triangles.

Answer 4

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.

Answer 5

To reword that sentence,

If $P$ is a convex polygon with $n$ sides then you can use several diagonals of $P$ to subdivide $P$ into $n-2$ triangles.

In fact we can be still more precise than that:

If $P$ is a convex polygon with $n$ sides then you can use $n-3$ diagonals of $P$ to subdivide $P$ into $n-2$ triangles.

For example, if $n=4$, and so $P$ is a quadrilateral, then you can use $1$ diagonal to subdivide $P$ into $2$ triangles.

Next, if $n=5$ and so $P$ is a pentagon, then you can use 2 diagonals of $P$ to subdivide $P$ into $3$ triangles.

And so on.