Timeline for Triangle with largest perimeter in a convex region
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 2, 2014 at 9:57 | comment | added | Ilya Bogdanov | Oh, sorry; that was me who miscalculated... | |
| Mar 2, 2014 at 2:43 | comment | added | Yoav Kallus | So $r(0.821)=2.974$, as you calculated. And $r(a)$ is minimal at $a=1$. | |
| Mar 2, 2014 at 2:42 | comment | added | Yoav Kallus | I get the following expression for the perimeter $r$ as a function of the aspect ratio $a$: $\frac{2}{(1-a^2)\sqrt{a\pi}} (a^2\sqrt{2s-1-a^2}+(1-a^2)\sqrt{2s+2-a^2})$, where $s=\sqrt{1-a^2+a^4}$. | |
| Mar 2, 2014 at 1:20 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
= to approx
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| Mar 2, 2014 at 0:58 | history | answered | Joseph O'Rourke | CC BY-SA 3.0 |