Timeline for What is the sequence that maximizes this distance?
Current License: CC BY-SA 3.0
10 events
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| Mar 14, 2017 at 12:41 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Image links broken; now fixed.
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| Dec 28, 2012 at 1:52 | comment | added | Joseph O'Rourke | @Gerhard: I like your problem (which is different from the OP's)! | |
| Dec 28, 2012 at 1:25 | comment | added | Gerhard Paseman | In fact, with all the alphas equal and using my intepretation, it seems clear a length increasing ordering achieves the longest endpoints. From this I think one can also show that a monotonic angle measure ordering is required for optimality when the exterior angles alpha are different. Gerhard "Thinking Outside All The Boxes" Paseman, 2012.12.27 | |
| Dec 28, 2012 at 1:16 | comment | added | Gerhard Paseman | I think the problem is more tractable and still interesting if the alpha_i are supplementary (or complementary? Anyway, the angles exterior to a convex polygon which add up to 2pi.) angles. Gerhard "Geometry Was So Long Ago" Paseman, 2012.12.27 | |
| Dec 28, 2012 at 1:05 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
added 410 characters in body
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| Dec 28, 2012 at 0:48 | comment | added | Joseph O'Rourke | So the above illustration fails your criteria, because the $\pi/8$ angles are not each measured clockwise? Between $l_2$ and $l_1$, the angle is clockwise. But between $l_3$ and $l_2$, it is counterclockwise. Am I correct that my drawing fails your criteria? | |
| Dec 28, 2012 at 0:47 | comment | added | Tomás | Yes you are right. | |
| Dec 28, 2012 at 0:39 | comment | added | Tomás | making an angle $\alpha_1$ with respect to $l_1$. All angles are assumed to be "clockwise". | |
| Dec 28, 2012 at 0:24 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
deleted 36 characters in body; deleted 18 characters in body; added 2 characters in body
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| Dec 28, 2012 at 0:18 | history | answered | Joseph O'Rourke | CC BY-SA 3.0 |