@sand15 

......using a well-known "classic" method. Aside from the technique involving a ray originating from the point under investigation, the only other method I recall involves the use of angle sums. In this approach, line segments are drawn from the selected point to the vertices of the polygon. The angles lying between adjacent segments are then calculated. If the sum of these angles is less than Pi, the point lies outside the polygon. Interior points yield an angle sum of 2*Pi, while points situated on the boundary yield an angle sum of exactly Pi.

As I have just read in one of your sources, the angle method is described there.

p.s.

Long-established methods from complex analysis (such as Cauchy’s theorems and Laurent series) used to present quite a challenge in practical application. It was only after some pondering over your solution that I realized - just now, in fact - how effectively these methods can yield solutions through programming... provided, of course, that one knows how to do it (which I don't be able :-( ). Back in my student days, the winding number was jokingly referred to in academic circles as a "zero-counter." In Maple, however, this is now a quick, "dry" process. In this respect, your solution has opened up a glimpse into a "new world" for me.

@vv @Rouben Rostamian  

...Your solutions - much like the others published here - is certainly worth keeping. From a programming standpoint, however, I am still far from capable of writing such command sequences myself. In the meantime, though, I have learned to read programs with understanding. Consequently, I can utilize your solutions as building blocks in my own small projects. And regarding the topic of "lattice polygons," there is a multitude of problems available - not least among them those from the IMO.
Thank you very much!

@sand15 

Your reference to literature is welcome. Familiar topics are presented in a methodologically sound and easily understandable way. Apparently, French mathematicians have a special knack for geometry and number theory (Desargue, Fermat, ...) ;-) .

@janhardo 

... is not my native language. Therefore, my translation may have been misunderstood - my apologies. Hence, a new attempt: It is surprising that a program as powerful as Maple lacks such a simple and convenient approach to the subject of lattice points and lattice polygons - unlike, for instance, the program Cabri. As a relatively inexperienced Maple user, I would find this very helpful.

@Rouben Rostamian  

These are exactly the kinds of polygons I want to generate in a straightforward way. Apparently, Maple only offers the option to do this using lists of coordinates. But regardless: Thanks—that’s exactly what I needed. It's a shame that it's not as comfortable as in Cabri.

@nm 

...and how is a grid polygon drawn with pinpoint precision?

@sand15 

After copying your algorithm, the following error message appears:

Error, illegal use of an object as a name

What am I doing wrong?

@Jean-Michel 

...Mr. Peano is to blame ;-) . His final formulation of the axioms named after him defines zero as a natural number. And for high school students and undergraduates, this is entirely sufficient and practically applicable. Naturally, in so-called "higher mathematics" a class-based definition grounded in set theory is employed. However in the otherwise excellent book *Exposé moderne des mathématiques élémentaires* by Lucienne Félix (page 553) zero is omitted during the introduction of the natural numbers, and the sequence begins with 1. It is not until the introduction of the integers that zero necessarily makes its appearance. In this respect, the book reverts to the original form of Peano's axioms.

@vv 

It works. Which type of "input" did you use? I would like to learn more about the different input methods.

@vv 

...I have made an inquiry. Unfortunately, after copying your algorithm, I encountered an error message, as shown in the attached file "test1." I am unable to locate the source of the error and would therefore like to request permission to copy a working version. Naturally, I fully respect copyright laws.
Your program performs a numerical sorting operation. It is a fascinating challenge that simply begs to be tested out.
I came across this problem in a delightful biography of Euler (Rüdiger Thiele, "Leonhard Euler", Teubner-Verlag, 1982, p. 158). The book offers a detailed description of Euler's teaching methods, his modes of thought, and his working style. Regarding this specific problem, the biography notes that Euler nearly despaired while attempting to solve it.
And since my inquiry has once again branched off into a broader discussion, I will take the liberty of adding the keyword "numeri idonei" here as well. ;-)

test1.mw

@vv 

...I wasn't familiar with that yet. I only knew that limitations on chord- length exist. But on page 408, this is now clarified. Many thanks! :-)

@dharr 

...the true intent of the puzzle has been sidetracked into a discussion regarding the rigor of Holditch's original proof. Naturally, with the aid of methods from convex analysis, it is now possible to provide a "rock-solid" proof—even for a generalization—of this theorem. The actual aim of the puzzle was, rather, to serve as a reminder of this elegant theorem and—perhaps with the help of Maple—not only to perform a calculation of the area (and thus of Pi) but also to generate an animation. It would be interesting to conduct numerical-graphical experiments starting from various initial curves, all of which invariably yield Pi. This approach could be utilized in the classroom to introduce differential and integral calculus, or—in more advanced semesters—within the context of functional analysis/Theory of Functions. Regrettably, however, I have long since ceased to be involved in teaching.

@vv 

To solve the puzzle, it is not necessary to invoke the theory of convex bodies or support functions. Since You have already "given away" the result, allow me to point out a simpler theoretical background—the almost forgotten theorem of Holditch.

@dharr 

...the solution path of the ODE simplifies if the initial equation is differentiated once. Assuming y'(x) is not equal to zero, the result is "test".

Download test.mw

@sand15 

"The largest octahedron (up to some symmetries) is necessarily the octahedron whose the base is the largest square S  contained in the cube."

... but not sufficient, it remains to be shown that the two octahedron vertices/tips land where they have to belong. However, this can be easily verified by calculation. Prince Rupert's cube is an interesting connection. That was news to me. It is also interesting that the cube in which the maximal octahedron is situated in a dual position has an edge length of 3/2.