[Math] Finding the interior angles of an irregular polygon inscribed on a circle

geometrytrigonometry

Is there any way to calculate the interior angles of an irregular N-sided polygon inscribed on a circle?

I only have a list of edge lengths (in order). I don't know any of the interior angles nor the radius of the circle the polygon is inscribed upon.

Here is an example of what I'm trying to figure out:

Irregular polygon points inscribed on a circle

The polygon can have any number of sides, but I'll always know the lengths of each side (for example, in the picture above I know what the lengths are for AB, BC, CD, DE, EF, and FA) and the polygon is always guaranteed to be inscribed on a circle.

I have found numerous solutions for solving triangles (which is easy, that's trig 101) and quadrangles (or rather, "cyclic quadrilaterals"). I'm wondering if a similar solution exists for N-sided irregular polygons, but so far I've not been able to find anything.

Best Answer

[Improved answer, based on excellent comment from David]

We have two different cases: one with the center inside the polygon and one with the center outside of it.

enter image description here

WLOG, we can assume that $a_1$ is the length of the longest edge. If the length of the $i^{th}$ edge of the polygon is $a_i$ and the radius of the circumscribed circle is $R$, the corresponding central angle would be (in both cases):

$$\alpha_i=2\arcsin \frac{a_i}{2R}$$

Case on the left: Sum of all central angles is $2\pi$:

$$2\pi=\sum_{i=1}^{N}\alpha_i=2\arcsin \frac{a_1}{2R}+\sum_{i=2}^{N}2\arcsin \frac{a_i}{2R}$$

$$\pi=\arcsin \frac{a_1}{2R}+\sum_{i=2}^{N}\arcsin \frac{a_i}{2R}$$

$$\pi-\arcsin \frac{a_1}{2R}=\sum_{i=2}^{N}\arcsin \frac{a_i}{2R}$$

$$\frac{a_1}{2R}=\sin(\sum_{i=2}^{N}\arcsin \frac{a_i}{2R})\tag{1}$$

Case on the rigth: $$2\pi=(2\pi-2\arcsin \frac{a_1}{2R})+\sum_{i=2}^{N}2\arcsin \frac{a_i}{2R}$$

$$\arcsin \frac{a_1}{2R}=\sum_{i=2}^{N}\arcsin \frac{a_i}{2R}$$

$$\frac{a_1}{2R}=\sin(\sum_{i=2}^{N}\arcsin \frac{a_i}{2R})\tag{2}$$

Actually, both equations (1),(2) are the same. The equation (1,2) has only one unknown ($R$) but I doubt that it is possible to solve it in a closed form for an arbitrary $N$. So numerical approach seems to be your only option. Once you have $R$ you can calculate pretty much anything, all internal angles included.

Example for the case on the left with $a_1=a_2=a_3=6$, $a_4=3$:

a1 = 6
a2 = 6
a3 = 6
a4 = 3
Solve[a1/(2 R) == Sin[ArcSin[a2/(2 R)] + ArcSin[a3/(2 R)] + ArcSin[a4/(2 R)]], R]

The result is $R=6\sqrt\frac25$.

Example for the case on the right with $a1=4$, $a2=3$, $a3=2$:

a1 = 4
a2 = 3
a3 = 2
Solve[a1/(2 R) == Sin[ArcSin[a2/(2 R)] + ArcSin[a3/(2 R)]], R]

This gives one solution: $R={8\over\sqrt{15}}$

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