Timeline for Smallest dilation of a quadrilateral?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| May 19, 2017 at 12:28 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Image link broken; now fixed.
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Aug 16, 2010 at 13:41 | vote | accept | Joseph O'Rourke | ||
| Jul 21, 2010 at 0:41 | answer | added | TonyK | timeline score: 6 | |
| Jul 16, 2010 at 20:04 | comment | added | Joseph O'Rourke | @Ravi: Sorry for the confusion. I meant this. Fixing $P$, you take the max over all pairs of points on $P$. That is the worst dilation of $P$. Then one wants the min over all $P$ (the way I phrased it, for a fixed number $n$ of vertices). So if you fix $P$ to be an equilateral triangle, then its dilation is 2 as per the figure. It turns out that every triangle has at least this dilation. So the min dilation for triangles is 2. What is unknown is the min dilation for quadrilaterals. It is only known to be at least 1.66. | |
| Jul 16, 2010 at 19:34 | history | edited | Joseph O'Rourke | CC BY-SA 2.5 |
Addendum re open status.
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| Jul 12, 2010 at 11:20 | comment | added | Ravi Boppana | This problem looks interesting, but I'm a little confused by the definitions. In the definition of $\delta(x, y)$, what are we taking the maximum over? In the definition of $\delta(P)$, did you mean maximum, not minimum? | |
| Jul 10, 2010 at 14:58 | history | edited | Joseph O'Rourke | CC BY-SA 2.5 |
Improved syntax.
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| Jul 10, 2010 at 14:27 | history | edited | Joseph O'Rourke | CC BY-SA 2.5 |
Added that circle uniquely achieves pi/2.
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| Jul 10, 2010 at 12:52 | history | edited | Joseph O'Rourke | CC BY-SA 2.5 |
Reversed an inequality in Ex.2!
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| Jul 10, 2010 at 12:32 | history | edited | Joseph O'Rourke | CC BY-SA 2.5 |
Fixed broken link; revised last sentence.
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| Jul 10, 2010 at 12:15 | comment | added | Joseph O'Rourke | Convincing argument, Victor! | |
| Jul 10, 2010 at 9:14 | comment | added | Victor Protsak | A quadrilateral $P=ABCD$ in $\mathbb{R}^3$ may be viewed as two rigid triangles $ABC$ and $CDA$ hinged on the common edge $AC$ at a certain angle. After "unfolding" by increasing the angle to $\pi$, we get a flat quadrilateral, the Euclidean distance between any two points $x,y$ on the boundary $P$ weakly increases and the distance along the boundary stays the same, hence $\delta(x,y)$ weakly decreases. Conclusion: the smallest dilation is realized by a quadrilateral in $\mathbb{R}^2.$ | |
| Jul 10, 2010 at 2:02 | history | asked | Joseph O'Rourke | CC BY-SA 2.5 |