Tough geometry problem

Question

Problem: A regular polygon of $n$ sides is inscribed in a circle of radius $r$. Chords are drawn connecting each vertex of the polygon to the next two vertices. find the sum of the lengths of these chords.

My attempt

In a regular $n$-gon inscribed in a circle, there are two types of chords to consider:

Diagonals: These connect vertices that are not next to each other. However, not all regular polygons have diagonals that satisfy the condition of connecting every other vertex to the next two. Diagonals are only possible for $n ≥ 4$.

Long Chords: These connect vertices that are next to each other but skip one vertex in between. These are present in all regular polygons regardless of the number of sides ($n$). I analyzed each type of chord and then found out the total sum for the entire figure.

Long Chords:

Then I imagined the polygon divided into smaller triangles by connecting each vertex to the center of the circle. The long chord will be the hypotenuse of one of these right triangles.

The central angle formed by two sides of the polygon is $360/n$ (degrees). Half of this angle ($\alpha/2$) is the angle between a radius and the long chord (since the long chord is the hypotenuse of a right triangle formed by a radius and a side of the polygon). We are given the radius ($r$). Using trigonometry (sine function), now i can find the length of the long chord ($h$) as:

$$h = 2r\sin(\alpha/2) = 2r\sin(180/n)$$

There are $n$ long chords in the polygon (one for each side).

Diagonals $(n ≥ 4)$:

For diagonals to exist, $n$ must be greater than or equal to $4$. Diagonals also form right triangles with the radius, but they connect vertices that are further apart.

The central angle formed by the two sides that form the diagonal is $2\cdot(360/n) = 720/n$ degrees. Similar to long chords, half of this angle ($\alpha/2$) is the angle between a radius and the diagonal. Using sine function again: Diagonal length: $$ d = 2r\sin(\alpha/2) = 2r\sin(360/n)$$

From there I infered that There are $n(n-3)/2$ diagonals in an $n$-sided polygon (a formula to count combinations).

At this point I'm not able to connect the dots. I am working on this problem for 3 days what should be the continuation.

Answer 1

I will assume this is what you meant, just in your case not pentagon, but a $n$-sided regular polygon, and you are trying to find the perimeter.

We can just dissect the regular polygon into $n$ equal isosceles triangles, with the vertex angle $\theta = \frac{2\pi}{n}$, and to find the base: $l = 2\sin(\frac{\theta}{2})r$ Then the perimeter would equal too, $$ P = ln = 2\sin\left(\frac{\pi}{n}\right)rn $$