Cutting into two equal parts, part two
With the first part can be familiarized here .
So, dear friends, in the previous part we talked about a parallel transfer, and today we will take a turn. It will be interesting. Now quickly recall the basic concepts - and forward.
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The rotation, as is known, is characterized by two parameters: the point around which everything rotates (the center of rotation) and the angle at which the rotation takes place. We will start by searching for a possible turning center. To begin with, we will creatively rework the terminology introduced in the first part.
Circumferencing circles with the center in the center of the turn (pardonte for a tautology) having at least one common point with figure A. Cross sections - the intersection of these circles with the figure. Boundaries are circles in relation to which the figure minus the section is either entirely inside or entirely outside. In the first case, it will be the outer boundary , in the second - the inner boundary . Note that the internal border exists if and only if the center of rotation is outside the figure (if we assume that the figure is closed, I would add a meticulous mathematician).
Under these conditions, Lemmas 1 and 2 from the first part remain valid, which we will use without prejudice.
Now I will draw a beautiful multi-colored picture. Each color denotes a locus of points for which, if they were the center of rotation, the far boundary section would consist of one particular vertex of A 0 (each vertex has its own color, this is shown in the picture). Since the boundary, and indeed any other section, cannot consist of a single point (see Lemma 1), the center of rotation can be located only on the black lines between the colored regions. If anyone is interested, where do we get the black lines - these are the median perpendiculars to the segments between the corresponding pairs of vertices.
Then I will draw another multi-colored picture, this time for the near borders. Note that the nearest point to the center of rotation, unlike the far, does not have to be a vertex - it can also lie on the side (the areas corresponding to the sides are colored with shades of gray). The boundaries between the areas corresponding to the side and the top at this side are not suitable for placing the center of rotation (the near boundary section would still consist of a single point), which is why I drew them in dotted lines. The thick black-yellow-black border is a piece of a parabola (the border between the area at the side and the area at the vertex that does not belong to this side has this form). I would like to say that the yellow stripe in the middle of a parabolic arc is a feature, but in fact it is an incomprehensible fatal artifact that arose at the junction of two curvilinear regions.
From this picture (all the similarities with the acid arrival are accidental and unintentional), it is clear that the restriction imposed on the near boundary leaves not so many places where the center of rotation could be located: the beam below and complex fucking (segment, turning into an arc of a parabola, which then goes to ray) above. Now it only remains to superimpose the pictures on each other and see which points are “valid” on both.
It is easy to see that there are only two such points (I designated them as O 1 and O 2 ). However, can they actually be centers of rotation that translate into each other hypothetical equal pieces of A 0 ? Obviously not. Then the arcs between pairs of points on the near and far border would have the same angular measure, since the angle of rotation does not depend on the distance to the center. At the same time, the angular measures of all four arcs (see figure) are different. Thus, the case of the center outside the figure can be considered closed.
If the figure A 0 and can be cut into two equal figures B and C, then the movement that translates B into C cannot be not only a parallel translation, but also a turn, the center of which is outside the figure. The case of turning with the center inside the figure, I decided to put in a separate article: due to technical difficulties that had not been noticed before, the volume of evidence for this case was fairly swollen. To be continued.
So, dear friends, in the previous part we talked about a parallel transfer, and today we will take a turn. It will be interesting. Now quickly recall the basic concepts - and forward.
')
Case 2: turn
The rotation, as is known, is characterized by two parameters: the point around which everything rotates (the center of rotation) and the angle at which the rotation takes place. We will start by searching for a possible turning center. To begin with, we will creatively rework the terminology introduced in the first part.
Circumferencing circles with the center in the center of the turn (pardonte for a tautology) having at least one common point with figure A. Cross sections - the intersection of these circles with the figure. Boundaries are circles in relation to which the figure minus the section is either entirely inside or entirely outside. In the first case, it will be the outer boundary , in the second - the inner boundary . Note that the internal border exists if and only if the center of rotation is outside the figure (if we assume that the figure is closed, I would add a meticulous mathematician).
Under these conditions, Lemmas 1 and 2 from the first part remain valid, which we will use without prejudice.
Case 2.1: the center of rotation outside the figure
Now I will draw a beautiful multi-colored picture. Each color denotes a locus of points for which, if they were the center of rotation, the far boundary section would consist of one particular vertex of A 0 (each vertex has its own color, this is shown in the picture). Since the boundary, and indeed any other section, cannot consist of a single point (see Lemma 1), the center of rotation can be located only on the black lines between the colored regions. If anyone is interested, where do we get the black lines - these are the median perpendiculars to the segments between the corresponding pairs of vertices.
Then I will draw another multi-colored picture, this time for the near borders. Note that the nearest point to the center of rotation, unlike the far, does not have to be a vertex - it can also lie on the side (the areas corresponding to the sides are colored with shades of gray). The boundaries between the areas corresponding to the side and the top at this side are not suitable for placing the center of rotation (the near boundary section would still consist of a single point), which is why I drew them in dotted lines. The thick black-yellow-black border is a piece of a parabola (the border between the area at the side and the area at the vertex that does not belong to this side has this form). I would like to say that the yellow stripe in the middle of a parabolic arc is a feature, but in fact it is an incomprehensible fatal artifact that arose at the junction of two curvilinear regions.
From this picture (all the similarities with the acid arrival are accidental and unintentional), it is clear that the restriction imposed on the near boundary leaves not so many places where the center of rotation could be located: the beam below and complex fucking (segment, turning into an arc of a parabola, which then goes to ray) above. Now it only remains to superimpose the pictures on each other and see which points are “valid” on both.
It is easy to see that there are only two such points (I designated them as O 1 and O 2 ). However, can they actually be centers of rotation that translate into each other hypothetical equal pieces of A 0 ? Obviously not. Then the arcs between pairs of points on the near and far border would have the same angular measure, since the angle of rotation does not depend on the distance to the center. At the same time, the angular measures of all four arcs (see figure) are different. Thus, the case of the center outside the figure can be considered closed.
Conclusion
If the figure A 0 and can be cut into two equal figures B and C, then the movement that translates B into C cannot be not only a parallel translation, but also a turn, the center of which is outside the figure. The case of turning with the center inside the figure, I decided to put in a separate article: due to technical difficulties that had not been noticed before, the volume of evidence for this case was fairly swollen. To be continued.
Source: https://habr.com/ru/post/178435/
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