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In the case of an odd-sided polygon, the beginning and end of the sequence of segments are guaranteed to lie on the same side of the line.

That's correct. Now we need to figure out why.

There is also a theory that this problem is solved not in a plane, but in a volume, meaning that the polygon has an element that protrudes beyond the plane, or even more than one, which allows a straight line to intersect it.

Heh, but in principle, nothing is said about the plane...

That's right. Now we need to figure out why.

Because when constructing a series of intersecting segments, they can be arranged in completely different ways, but before the closing segment is constructed, there will be an even number of them. The closing segment is an odd number. Each segment is either "there" or "back" relative to the line. An even number means that "there" = "back," sorry for being unserious.)

Heh, but basically, nothing is said about the plane...

If the polygon is not flat, then the intersection of all its sides is impossible, because a straight line intersecting two of its sides lies entirely in the plane of these sides and cannot intersect sides lying in other planes.
So, of course, it refers to a flat 111-sided polygon.

If the polygon is not flat, then the intersection of all its sides is impossible, because a straight line intersecting two of its sides lies entirely in the plane of those sides and cannot intersect sides lying in other planes.

Well, it just won't work. I would say it like this: if you build two connected segments and intersect them with a straight line, then the next segment, connected to the previous one and intersecting the same straight line, lies in the same plane. Repeat n times.

Each segment is either "there" or "back" relative to the straight line. An even number means that "there" = "back", sorry for being unserious.

Well, in principle, it's correct. The straight line divides the plane into 2 parts (let's call them "white" and "black"). Let's number the vertices from 1 to 111. If 1 is "white"/"black", then 2 is "black"/"white", 3 is "white"/"black", and so on. 111 turns out to be "white"/"black", which is impossible, because 2 adjacent vertices (1 and 111) turn out to be the same color.

Go ahead, make your guess.

Let's try this:

There is a plane randomly colored with two colors: green and magenta. Is it always possible to find two points of the same color on this plane with a given distance x between them? Prove your answer.

If 2 adjacent areas have different colors, what color is the border between these areas considered to be?

Damn, I was writing a response, but the message didn't send. Anyway, the main thing we need is that the points are not of a single specific color, but simply identical (otherwise, the answer is no, as we also have a case where the entire plane is green, and it is impossible to find 2 magenta points on it). So, we are looking for the boundary between two differently colored areas and placing a point (1) at a distance of x/2 along the perpendicular to the tangent to this boundary (assuming it is uneven). Next, we try to find another point of the same color as (1) along the entire circumference with a radius of x (taking (1) as the center of the circle). If we don't find such a point, the entire circumference coincides with (1) in color, then we place point (2) on that line, which is perpendicular to the tangent (and on which point (1) is already located), at a distance of x from (1), and then we move the segment along this perpendicular. Within the diameter of the circle 2x, there must be an even number of color changes in the plane, so the desired distance x must be found. Yes, if it suddenly turns out that it cannot be found because our plane is divided by color like concentric rings from a diffraction grating, and we have chosen (1) as the center of these rings, then we can always find 2 identical points on one of the chords of the rings. This essentially simplifies the answer. If there are no color matches with (1) along the circumference with a radius of x from (1), then there are 2 points of the same color on the circumference itself, and a chord can be constructed between them, the length of which is equal to the radius.

we are looking for the boundary between two differently colored areas

Considering that the coloring can be anything, there might not be any boundaries at all. :D For example, how about this coloring: all points with both coordinates being rational are green, and the rest are magenta. :D

Added 2 minutes later

If there are no color matches on a circle of radius x from (1) with (1), then there are 2 points of the same color on the circle itself, and a chord can be constructed between them, equal in length to the radius.

And this seems to be what we need!

If a polygon is not flat, then the intersection of all its sides is impossible, because a straight line intersecting two of its sides lies entirely in the plane of those sides and cannot intersect sides lying in other planes.

A plane is a set of points. Each point can have its own color. If the points are combined into solid areas (which can only happen by chance), then the boundary of the areas can be... Any color, and not necessarily the same.
I will try to process Uranus's answer now.

Yes, Uran, the last paragraph is correct (although it's best to forget about the others; Yashur explained the reason).
Actually, the solution can be made even a little simpler. Namely, take an equilateral triangle with sides of length x. Since the triangle has three vertices and there are only two colors, at least two vertices will be the same color. But in principle, it's almost the same as a circle.
Go ahead and guess.

Let's keep it simple. In the Lorwyn set of the Magic: The Gathering collectible card game, there were cards featuring elves, fairies, goblins, and others. However, cards featuring THEM, strangely enough, were not released, even though THEY appeared in every set before and after. Name THEM.

I want to present my solution to the riddle of who stole the teacher's wallet once again. Yes, Yashchur solved it correctly before me; I just want to write down my correct solution, as I intended to do that time, since the previous one was incorrect.

Let the statements that they didn't do it be:
Lilian - a,
Judy - b,
David - c,
Theo - d,
Margaret - e.

Lilian states that she has never stolen anything in her life - f
Judy's father is rich - g,
Margaret knows who stole it - h,
David is not acquainted with Margaret - k,

Then:
Lilian: a+f+(1-d)=2, a>=f
Judy: b+g+h=2
David: c+k+(1-d)=2
Theo: d+(1-e)+d=2
Margaret: e+(1-b)+(1-k)=2
a+b+c+d+e=4

Then, from the fourth equation, we get that d=1 and e=1
From the third equation, we get that c=1 and k=1
And from the fifth equation, we get that b=0

The trick is that Lilian's statement should be written as a+f+(1-d)=2, a>=f, because the situation where a=1, f=1 is possible, a=1, f=0 is possible, a=0, f=0 is possible, if we don't take into account that two of her statements must be true, but a=0, f=1 is not possible... that is, we need to add a>=f

In the Lorwyn cycle of the Magic: The Gathering collectible card game, there were cards featuring elves, fairies, goblins, and others. However, cards featuring THEM were not released, even though THEY appeared in every cycle before and after.
Name THEM.

I think the most difficult part here is choosing just one answer. Elves, fairies, goblins... This set definitely includes humans, orcs, and probably gnomes/dwarves.
Lorwyn is clearly about elven forests. In principle, there could be humans there, but I'm more inclined to think that "others" are some kind of centaurs. Or halflings. Or gnomes (but not dwarves). However, remembering the lore of Might & Magic, I'm not sure that it isn't humans who have settled in the elven forests. They're like that; they can easily do it. They go everywhere; that's just our kind.
And yet... If "THEY" are not humans, then it's too difficult to choose between orcs/dwarves and others. So my answer is humans.