Timeline for Pairs not at maximal distance in compact set
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Nov 21, 2019 at 17:48 | comment | added | M. Winter | @TarasBanakh Your argument does not work if $x=y=p$ or if $K$ is empty. But I agree that the question should be modified to be interesting. | |
| Nov 21, 2019 at 16:50 | comment | added | Taras Banakh | Such a set $K$ does not exist: Let $x=y$ be a point in $K$ on the largest possible distance from $p$. Then after a rotation $x=y$ cannot lie in the interior of $K$. This means that in order to avoid such a trivial answer, one should add that the points $x,y$ are different. But in this case there appear another trivial answer: $|K|\le 2$ or $K$ is a 3-element set consisting of the vertices of an equilateral triangle. So, which conditions on $x,y$ and $K$ should be imposed in order to make this quastion non-trivial? | |
| Nov 21, 2019 at 14:10 | comment | added | M. Winter | I suppose you want to exclude the following sets for $K$: the empty set, a single point, two points. What about three points arranged in the shape of an equilateral triangle? | |
| Nov 21, 2019 at 13:22 | comment | added | Ramiro de la Vega | @TarasBanakh It seems that if $K$, as in your example, has a unique point at maximal distance from the center then $K$ won’t satisfy the required property (take $x$ as that unique point and $y$ some other point on $\partial K$ ). | |
| Nov 20, 2019 at 22:51 | comment | added | burtonpeterj | Can you elaborate on why? | |
| Nov 20, 2019 at 22:09 | comment | added | Taras Banakh | What about the set $K=\{(x,y):(x+1)^2+y^2\le 4\}\cap\{(x,y):(x-1)^2+y^2\le 4\}$ and point $p=(0,\sqrt{3})$? It seems that $K$ has the required property. | |
| Nov 20, 2019 at 21:51 | history | edited | burtonpeterj | CC BY-SA 4.0 |
deleted 67 characters in body
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| Nov 20, 2019 at 21:15 | history | asked | burtonpeterj | CC BY-SA 4.0 |