Timeline for A non-self-intersecting unit side length polygon in a unit square has odd number of sides unless it is the square itself
Current License: CC BY-SA 4.0
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| Nov 11 at 7:10 | comment | added | 1001 | If, at some point, there are many consecutive left turns, we end up with a spiral. It could be possible to prove (by considering the angles) that the spiral must be traced back using the same number of right turns. | |
| Nov 11 at 7:03 | comment | added | 1001 | Probably the convex hull of P would be better to consider. | |
| Nov 11 at 3:51 | comment | added | 1001 | If you follow the edges of the polygon, you always make an acute turn left or right. One idea is to distinguish cases based on the number of consecutive turns in the same direction (e.g. the example has 4, and perhaps this is the only possibility). The simplest case would be alternating left and right turns. Considering alternating points on the polygon, we obtain two new polygons: an outer polygon P and an inner polygon Q. Consider the area of the triangles based on the inside of the edges of polygon P. Fixing the edge lengths of P, its area is maximized when its vertices are concyclic. | |
| Nov 9 at 2:15 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Oct 29 at 22:51 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Oct 28 at 18:59 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Oct 28 at 12:06 | comment | added | JetfiRex | @TimothyChow Thank you, but I have thought about this before. But generally If we place the segments in a good way, there can be two segments such that one segment has end points in regions 1, 4, And the other can have the segment has endpoint in regions 2, 3. For example, the segment connects $(0,0)$ and $(\sqrt 3/2-\epsilon_1,1/2+\epsilon_2)$; and the segment connects $(1-\sqrt 3/2,1)$ to $(1-\epsilon_3, 1/2-\epsilon_4)$ for some small $\epsilon$'s (There does not need to be the same $\epsilon$...) | |
| Oct 28 at 11:46 | comment | added | Timothy Chow | @JetfiRex I think you may be able to prove that any edge from 1 to 4 must cross any edge from 2 to 3 by considering where the endpoints of a diagonal edge must be. There is a forbidden region in the center of the original unit square (any point in this forbidden region is less than 1 unit away from any other point in the square). | |
| Oct 27 at 21:50 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Oct 27 at 12:49 | comment | added | JetfiRex | @GerryMyerson You can try it yourself here... In the diagram all the lengrhs are $1$, And if you consider the extreme case that $F, H$, etc Is on the lower side, The height for each isosceles triangle does not exceed $1$. Now move $F$ and $H$ little up, then you can get the polygon. Sorry for the picture because it's not really easy to be distinguishable by human eyes because the distance to the edges is $O(1/k^2)$ So it looks like $E,F,G,H,I$ are on the edge, but it's not... I am sorry for the confusion. I will try to update the picture with smaller points and thinner lines. | |
| Oct 27 at 8:43 | comment | added | Gerry Myerson | I don't understand the first diagram. It looks like all the points are on a side of the square, so most of the lengths exceed one. | |
| Oct 26 at 18:04 | history | asked | JetfiRex | CC BY-SA 4.0 |