4
$\begingroup$

Does there exist a compact subset $K$ of $\mathbb{R}^2$ and a point $p$ in $\mathbb{R}^2$ with the following property?

Let $x$ and $y$ be a pair of points in $K$ such that \begin{equation} |x-y| < \sup_{z,w \in K}|z-w| \end{equation} Then there exists a rotation $R$ fixing $p$ such that $R(K)$ still contains both $x$ and $y$ and such that at least one if $x$ and $y$ lies in the interior of $R(K)$.

Intuitively one would think that $p$ should in some sense be the ‘center’ of $K$, but that’s not required for the application I have in mind.

$\endgroup$
6
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Nov 20, 2019 at 22:09
  • 1
    $\begingroup$ Can you elaborate on why? $\endgroup$ Commented Nov 20, 2019 at 22:51
  • 2
    $\begingroup$ @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 $\partial K$ ). $\endgroup$ Commented Nov 21, 2019 at 13:22
  • $\begingroup$ 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? $\endgroup$
    – M. Winter
    Commented Nov 21, 2019 at 14:10
  • $\begingroup$ 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? $\endgroup$ Commented Nov 21, 2019 at 16:50

0

You must log in to answer this question.

Browse other questions tagged .