@vv 

I am still fairly familiar with the theorems of analysis relevant to this interesting task. I was confused by the discrepancy with numerical experiments for large n using Mathcad15 (forget it) and derive. I was interested in the numerically comprehensible convergence behavior. I would also have liked to understand the details of the Maple calculation of the limit. But now everything is clear.

This is not a proof of the convergence of the sequence defined according to (2). A calculation in long number arithmetic (limit replaced by large n, derive) yields the number 2.4567 for n=10^12...
Unfortunately, Maple hides the calculation steps that lead to the result 0.0004937... Please explain ;-)

@Kitonum 

Your reference to literature was a pleasure. It also reminded me of my remaining linguistic knowledge, when books by Michlin, Pontrjagin and Shilov/Gurevich were in great demand in the original, for example.

As I am more interested in solution paths and ways of thinking in mathematics than the solution itself, I have tried to create a solution path that is clear and easy to understand for students. It should work with the help of typical school "tools". That is why it looks complicated and unprofessional at first glance - so please don't laugh. As I still make mistakes with Maple, I had to use good old derive and create the attached files from it.

Seite_1_und_3.pdf

Seite_2_und_4.pdf

@dharr ... I thank you for Your solution. It helped me to understand the geometry package a little. This means I am now able to solve my own special private puzzles. In the future I will offer much more difficult problems from my old collection here. These will be suitable for Maple and come from the areas of analysis, geometry and number theory. I hope you like them.

@dharr 

I hadn't expected such a difficult solution. Nevertheless, it helped me to understand the Maple command sequence and to recognize the logic behind the package selection.
The following solution is simpler:
The n-gon has a side length of 1. The interior angle between adjacent polygon sides is known to be (n-2)*pi/n. This angle is opposite the side length we are looking for in the triangle. The side lengths on the sides of this angle (n-2)*pi/n are 1/2. The cosine theorem then gives us the side length we are looking for of the inscribed polygon as 1/2-1/2*cos((n-2)*pi/n) and the limit n--->oo is obvious.

@vv 

Please excuse me. I will try to improve. Thanks for the reference.

@Rouben Rostamian  

...I solved this problem in exactly the same way in 1977 using pen and paper.

@vv 

I would have liked to have had these options 60 years ago. I would have been spared many full wastepaper baskets. ;-)

At the moment, it is very important to me that theoretical backgrounds are also understood in principle. That is why I like to refer to literature in the usual way.

There used to be a humorous saying:
If Kamke doesn't say anything, it can't be solved.

@Alfred_F 

I forgot to refer to "Niven I., Zuckerman H.S., Montgomery H.L. An introduction to the theory of numbers", page 183.

This problem is quite old. If you follow the sources in "Dickson", Volume 1, Chapter IX and the older sources that follow from it, then this and similar questions were already being dealt with around 1880. My attempt to use school-typical methods (factorization) to provide an understandable solution for amateurs was doomed to fail. Otherwise, mathematicians would have come up with it back then. The solution therefore requires more sophisticated number theory methods (Legendre). And one solution is shown in Carmichael's book "The Theory of Numbers", Mathematical Monographs No. 13, on pages 26 to 28, in principle using a slightly weaker, similar problem. The current problem here is given by Carmichael as exercise No. 9 with an asterisk (for a special level of difficulty).
Overall, your question is very interesting. It was a pleasure to study old sources. And in my view it is a gain to now see a solution formulated with the help of Maple's CAS.

@Ronan 

The task does not require that a solution be found in the Euclidean plane. There are surfaces in R^3 that are created by straight lines. And a quadrilateral with straight sides can lie on them. But that is then a very difficult task.

@Kitonum 

It would therefore be worthwhile to delve into the world of polygons. In a lattice plane (point lattice) Pick's theorem could then be of help.

@Kitonum 

Your program is instructive for me. I have now learned not only to be able to read command structures, but also to understand them.

The theoretical/topological background is well known. The quoted sentence only proves the existence of area division. You have gone a step further and constructed a solution. This raises the question: Are there further divisions for the same Jordan curve? At what point in your program would something have to be changed?

It is worth trying to substitute: ln(y(x))=z(x). The left-hand side of the differential equation becomes z´(x) and for x>1 the right-hand side is continuous. Solvable problems of a similar structure can be found, for example, in the terrifying collections:
- EqWorld
- Kamke differential equations. Mathematica 12.3 and Maple 2021, Nasser M. Abbasi
- https://dlmf.nist.gov

@Ronan,  @vv

I found this task while cleaning up one of my hard drives. A long time ago, with some help, a solution was found in Mathcad 14 code. This solution is very long, but it makes the way of thinking on the way to the solution clear even for interested amateurs. That was my goal at the time. Unfortunately, I can only do simple tasks in Maple so far. But the solution presented here helps again. The brevity in dealing with ODE and graphics is impressive.