Timeline for Square lying on moving chord of a simple closed curve
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Aug 20, 2017 at 10:19 | history | edited | Taras Banakh | CC BY-SA 3.0 |
This is still another approach.
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| Aug 20, 2017 at 9:48 | history | undeleted | Taras Banakh | ||
| Aug 19, 2017 at 19:42 | history | deleted | Taras Banakh | via Vote | |
| Aug 19, 2017 at 18:28 | comment | added | Taras Banakh | @Luca Ghidelli Thanks for your comment. Indeed, your 1024 destroys the proof in the sense that one of the numbers $n_1$ or $n_2$ becomes zero. Indeed, this is a metric problem. Maybe some modification like minimizing $r(t)$ at $t=0$ and $t=1$ will save the proof. I will try to write the necessary correction. | |
| Aug 19, 2017 at 17:51 | comment | added | Luca Ghidelli | It's not true that $n_1$ and $n_2$ are both odd. On a more fundamental level, can you show where does your proof fail if you replace $c=z+i r \phi$ with $c=z+1024 i r\phi$? This is not a purely topological problem, it is also metric... | |
| Aug 19, 2017 at 17:07 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| Aug 19, 2017 at 17:01 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| Aug 19, 2017 at 10:50 | history | edited | Taras Banakh | CC BY-SA 3.0 |
Totally rewritten proof
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| Aug 19, 2017 at 9:49 | comment | added | Taras Banakh | @makkostya You are right there is a problem with this equivalence of intersections. Let me think a bit how to fix it. | |
| Aug 19, 2017 at 9:05 | comment | added | makkostya | But the first argument is not obvious for me. I mean why changing coordinate system in such a complex way preserves the fact of intersection? Is it some fundamental property? (I'm not at all specialist in this domain) | |
| Aug 19, 2017 at 9:05 | comment | added | makkostya | Correct me if I miss understood your arguments. You say: 1. $L$ intersects $C$ $\iff$ $C'$ intersects $C''$ 2. Then you show in details that $C'$ intersects $C''$. But I had problems with the first argument. For the second one I had an idea of proof : shapes bounded by $C'$ and $C''$ have the same area and at least one point in common (origin). I imagine that it is enough to say that $C'$ intersects $C''$. | |
| Aug 19, 2017 at 4:40 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| Aug 18, 2017 at 21:25 | history | edited | Taras Banakh | CC BY-SA 3.0 |
Added a more precise argument
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| Aug 18, 2017 at 20:30 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| Aug 18, 2017 at 20:18 | comment | added | makkostya | I thought about it, but the problem is that you don't take in acount the movement of the center of the chord. It's not trivial to show, that even with moving center, curves will intersect (at least I could not) | |
| Aug 18, 2017 at 19:23 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| Aug 18, 2017 at 19:17 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| Aug 18, 2017 at 18:52 | history | answered | Taras Banakh | CC BY-SA 3.0 |