Timeline for Area of a lattice polygon in terms of its width
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Jul 14, 2013 at 9:57 | history | edited | Nikita Kalinin | CC BY-SA 3.0 |
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| Apr 5, 2013 at 12:22 | vote | accept | CommunityBot | ||
| Apr 5, 2013 at 12:22 | history | bounty ended | Nikita Kalinin | ||
| Apr 2, 2013 at 2:32 | answer | added | Fedor Petrov | timeline score: 3 | |
| Apr 2, 2013 at 0:33 | history | edited | Nikita Kalinin | CC BY-SA 3.0 |
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| Apr 1, 2013 at 23:56 | history | edited | Nikita Kalinin | CC BY-SA 3.0 |
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| Mar 31, 2013 at 7:11 | comment | added | Nikita Kalinin | @robot: I was looking for an estimation $area(M)\geq cd^2$ | |
| Mar 30, 2013 at 22:31 | comment | added | Vít Tuček | What is the question here? | |
| Mar 29, 2013 at 17:26 | answer | added | Nikita Kalinin | timeline score: 1 | |
| Mar 29, 2013 at 17:09 | history | edited | Nikita Kalinin | CC BY-SA 3.0 |
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| Mar 29, 2013 at 16:35 | history | edited | Nikita Kalinin | CC BY-SA 3.0 |
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| Mar 29, 2013 at 16:07 | history | edited | Nikita Kalinin | CC BY-SA 3.0 |
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| Mar 29, 2013 at 14:35 | history | edited | Nikita Kalinin | CC BY-SA 3.0 |
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| Mar 29, 2013 at 12:09 | history | bounty started | Nikita Kalinin | ||
| Mar 29, 2013 at 12:08 | comment | added | Nikita Kalinin | Thay gives us $area(M)\geq d^2/4$, it is worse than $g\geq d^2/{2\sqrt{3}}$ | |
| Mar 28, 2013 at 6:55 | comment | added | Ilya Bogdanov | Considering the quadrilateral $Q$ of maximal area inscribed into $M$. THen you may pass to the parallelogram with sides parallel to the diagonals of $Q$ --- its area is at most twice the area of $M$. Hence you may restrict yourself to the lattice parallelograms only. | |
| Mar 27, 2013 at 9:35 | history | asked | Nikita Kalinin | CC BY-SA 3.0 |