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#853
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Drakkoska
Do you have a similar problem set?
No. Honestly!:D I just remember solving a similar problem a long time ago, specifically one involving the tilting of vessels. So when I saw that this was a "trick question," it immediately came to mind.

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#854
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Well done anyway. Go ahead and make a wish.
А, вот почему.

🏹🎪+⛓️♿=💕
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#855
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Does a 111-sided polygon exist where all its sides can be intersected by a single straight line? The answer, of course, needs to be justified. Hermit, this question is right up your alley.:D

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#856
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if I remember the definition of the intersection of two lines correctly, then there exists
..разым двазым трызым рызым пята лата сигерь мата локом боком крюк за крюк из бульмы бульма урюк..
Знание некоторых закономерностей освобождает от изучения многих фактов.
Мечты сбываются рано или поздно, так или иначе.
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#857
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All sides of this degenerate 111-gon, for example, have at least one point in common with line a, i.e., they intersect it: rolleyes:. I explained it.
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..разым двазым трызым рызым пята лата сигерь мата локом боком крюк за крюк из бульмы бульма урюк..
Знание некоторых закономерностей освобождает от изучения многих фактов.
Мечты сбываются рано или поздно, так или иначе.
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#858
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Sesame

all sides of this degenerate 111-sided polygon.
Such a shape doesn't count as a 111-sided polygon. The sides should not lie on the same line.

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#859
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what? A degenerate triangle, where all sides lie on the same line, exists, but a degenerate polygon does not?

Added after 46 seconds
Anyway, never mind. I have another idea. But I'll think about it later this evening.
..разым двазым трызым рызым пята лата сигерь мата локом боком крюк за крюк из бульмы бульма урюк..
Знание некоторых закономерностей освобождает от изучения многих фактов.
Мечты сбываются рано или поздно, так или иначе.
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#860
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Sesame
What are you talking about? Is a degenerate triangle, where all sides lie on the same line, considered to exist, but a degenerate polygon doesn't?
Yeah, and a regular triangle is actually a quadrilateral where 3 vertices lie on the same line. :D:D:D

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Specifically for people like you, I'll rephrase the question.
In city N (shape doesn't matter), there is a "ring road" consisting of 111 straight sections that do not intersect each other. A straight highway runs through the city. Is it possible for a car traveling on the highway through the city to cross all 111 sections of the "ring road"?

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In reply to Злобный Ящур
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#861
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Evil Lizard
Yeah, and a regular triangle is actually a 4-sided figure where 3 vertices lie on the same line. :D:D:D
I haven't encountered such cases, although anything is possible, but I have encountered a quadrilateral where ALL vertices lie on the same line.
But okay, let's say this is an incorrect option (something also confuses me in it, but not the vertices lying on the same line). I'll find another one.
Evil Lizard
Especially for people like you, I will rephrase the question.
In city N (the shape doesn't matter), there is a "ring" road consisting of 111 straight sections that do not intersect each other. A straight highway runs through the city. Is it possible for a car traveling along the highway through the city to cross all 111 sections of the "ring" road?
Hmm? Well, okay, I think I know how to cross them even in this formulation :p Although I haven't drawn it yet, I might be missing something.
..разым двазым трызым рызым пята лата сигерь мата локом боком крюк за крюк из бульмы бульма урюк..
Знание некоторых закономерностей освобождает от изучения многих фактов.
Мечты сбываются рано или поздно, так или иначе.
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#862
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It doesn't exist. I can't explain it yet, but I haven't been able to draw a line through a triangle or a pentagon, so I think it's impossible to do it for any polygon if the number of sides is odd.

Added 27 minutes later
The point is probably that a necessary condition for drawing a straight line through such a polygon that intersects all sides is to have a number of angles greater than 180o equal to the number of sides minus 2 divided by two, which is only possible with an even number of sides.
Всё не так плохо как Вы думаете. Всё намного хуже!
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#863
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If an external intersection is considered an intersection, then see line 1. If not, but a tangent at a boundary point is considered an intersection, then see line 2. If that's also not the case, it's impossible. Why? Because if we draw several connecting segments through a line and then try to connect the beginning of the first segment with the end of the last, regardless of the segments themselves, we have two options: 1. The connection of the segments lies on one side of the original line (and does not intersect it). 2. The connection of the segments intersects the line, but also intersects other segments, which does not create a polygon. Why does it intersect? Well, here the level of abstraction becomes even more complex. If necessary, I will try to finish the reasoning.
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#864
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Ah, here. The second option simply doesn't allow for an odd-sided polygon. With an odd-sided polygon, the beginning and end of the sequence of segments are guaranteed to be on the same side of the line. An odd number of angles equals an odd number of sides, which equals an even number of intersecting segments (not counting the closing segment). The beginning and end are on the same side.
If the problem is expanded to an even-sided polygon, then yes, the second option would need to be proven (but that shouldn't be too difficult either).
 

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#865
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I don't know if the figure can be considered a circular path or a polygon, but if so, then it intersects. The figure resembles a Ment figure (the number of teeth is such that there are 111 segments), only it is located diagonally; the only problem is whether it can be considered an intersection when a straight path passes through the point where the segments connect, marked with an X. There is also a version that this problem is solved not in a plane, but in a volume, that is, the polygon has an element protruding beyond the plane, or even more than one, which allows the straight line to intersect it. Or, in general, the polygon itself is located in a volume, like a staircase, that is, it does not have 3 segments that lie in the same plane.
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#866
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If an external intersection is considered an intersection
If not, but a tangent at a boundary point is considered an intersection.
No. It refers to an intersection within the segment, of course.
It's not possible. Why? Because if we draw several connecting segments through a line, and then try to connect the beginning of the first segment and the end of the last, we will have two options, regardless of the segments themselves:
1. The connection of the segments lies on one side of the original line (and does not intersect it)
2. The connection of the segments intersects the line, but also intersects other segments.
I'm sorry, you are wrong.
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#867
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Okay, so the second statement is also incorrect (it's understandable why it's difficult to grasp). But fortunately, it's not needed either, as it cannot be implemented for an odd-sided polygon.
 

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