Convex n-sided polygons whose exterior angles expressed in degrees are in arithmetic progression

Solution 1:

Edit: I just realized that I hadn't answered in a convenient way to the question which looks for values of $n$. Here is a re-writing of my solution.

First remark: we do not exclude the case of regular polygons ("regular" with the meaning "equal external angles").

The well known expression of the terms of an Arithmetic Progression is $a+bk$ for given constants $a,b$, variable $k$ taking values $0,1, \cdots$ (here parameters $a\ge 0,b \ge 0$ are integers; had we wanted to exclude regular polygons, we should have taken $b > 0$).

Therefore, we must have, for a $n$-gon,

$$\sum_{k=1}^n (a+bk)=360$$

Otherwise said:

$$na+b \frac12 n(n+1)=360 \iff n(2a+b(n+1))=720 \ \text{with} \ 720=2^4.3^2.5\tag{1}$$

We have to solve this equation understood as the existence of integers $a \ge 0,b \ge 0,n>0$ fulfilling it.

The values of $n$ have to be chosen among the $(4+1) \times (2+1)\times (1+1)=30$ divisors of $720$:

$$n \in \{1, \ 2,\ 3,\ 4,\ 5,\ 6,\ 8,\ 9,\ 10,\ 12,\ 15,\ 16,\ 18,\ 20,\ 24,\ 30,\ 36,\ 40,\ 45 ,\ 48,\ 60,\ 72,\ 80,\ 90,\ 120,\ 144,\ 180,\ 240, \ 360, \ 720 \}$$

with pairs of complementary divisors: $(1,720), (2,360), (3,240)....$ that I will call "twins" in the following.

(in fact only $27$ of them can be used: because polygons with $n=1, \ 2$ aren't polygons and polygon with 720 vertices has 720 sides, and would generate non integer external angles).

Let us browse all possible values of $n$ and attempt to solve (1) :

• $n=3:$ twined with $240$ gives $(2a+4b)=240$ which has (many) solutions: therefore $n=3$ is eligible.

• $n=4:$ twined with $180$ is eligible because (1) gives $2a+5b=180$ which has solutions.

• $n=5:$ is eligible because (1) gives $2a+6b=144$ which has solutions ($a=0,b=24$ for example).

• etc. etc.

• an impossibility occurs for $n=48, n=80, n=144, n=240$ because none of the corresponding equations $2a+49b=15, \ 2a+81b=9, \ 2a+145b=5, 2a+241 b=3$ resp. have nonnegative solutions.

This rules out four values for $n$.

Therefore the answer is that there are $27-4=23$ possible values of $n$.

You have suggested that the answer could/should be $22$, i.e., one less. I must have a closer look and see if I haven't "lost" a case...

Fig. 1: An example: A graphical representation of the solution $n=8, a=0, b=10$ with external angles $10°, 20°, 30°, \cdots 80°$ in Arithmetic Progression.

Remark 1: A discussion should take place about the sidelengths of the polygon, for which there is a large "degree of freedom".

Remark 2: about the adjective "external" (vs. exterior) angles see here.