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.

  • $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ Commented Mar 21 at 13:24
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    $\begingroup$ Please make a title that describes a problem. Please use MathJax instead of Unicode whenever you type a formula on this site. It’s pretty easy. You just enclose your formulae into the $ signs, for the most part. $\endgroup$
    – Aig
    Commented Mar 21 at 13:24
  • $\begingroup$ What are “next two vertices”? One to the nearest to the left and one to the nearest to the right? Or one to nearest to the left and one to the left, next to the nearest? $\endgroup$
    – Aig
    Commented Mar 21 at 13:28
  • $\begingroup$ In the context of the given problem, "next two vertices" refers to the vertices adjacent to the vertex in consideration, one on the left and one on the right. So, when we say "connecting each vertex of the polygon to the next two vertices," we are referring to connecting each vertex to the two adjacent vertices in the polygon. This implies that for each vertex, we draw chords to the adjacent vertices on either side in the clockwise or counterclockwise direction around the circle. $\endgroup$
    – StudyME
    Commented Mar 21 at 13:32
  • $\begingroup$ To me it sounds like you need to find the perimeter of the polygon which will, obviously, converge to the circumference. $\endgroup$
    – Vasili
    Commented Mar 21 at 13:35

1 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. Pic1

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, $$ P = ln = 2\sin\left(\frac{\pi}{n}\right)rn $$


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