7
$\begingroup$

Let $K$ and $L$ be planar convex bodies which are not ellipses. Does there exist an affine image $K'$ of $K$ such that

  1. $K' \subset L$
  2. No ellipse $E$ satisfies $K' \subset E \subset L$

I am also interested in the variant where $K$ and $L$ are centrally symmetric, and $K'$ is required to be a linear image of $K$.

Improve this question
$\endgroup$
10
  • $\begingroup$ Is this true for, say, a triangle and a near-circle? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 22 at 18:11
  • 2
    $\begingroup$ The answer is positive if $K$ is a triangle. Assume without loss of generality that the ellipse of maximal area inside $L$ is a disk $D$. If $K$ is any equilateral triangle inscribed in $D$, the only ellipse sandwichable between $K$ and $L$ is the disk $D$ (since the ellipse of minimal area containing $K$ is $D$). Since $L$ is not a disk, it is possible to choose the triangle $K$ such that one vertex lies in the interior of $L$, and now if we move away that vertex to enlarge the area of $K$, we obtain a configuration as required. $\endgroup$
    – Guillaume Aubrun
    Commented Sep 22 at 20:09
  • $\begingroup$ Does this also work for non-equilateral triangles? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 22 at 20:55
  • $\begingroup$ Yes. Up to affine bijection, there is only one triangle. $\endgroup$
    – Guillaume Aubrun
    Commented Sep 22 at 21:06
  • 3
    $\begingroup$ @GuillaumeAubrun It seems that the same idea works in larger generality. If we assume that $D$ is the largest area ellipse in $L$ and the smallest area ellipse containing $K$, the argument won’t work only if any rotation of $K$ contains points on the boundary of $D$ which lie on the boundary of $L$, and their convex hull contains the center of $D$. In this case, $K$ or $L$ should have circular segments on $\partial D$ of nonzero measure; maybe some different argument could finish this case? $\endgroup$
    – Ilya Bogdanov
    Commented Sep 23 at 12:41

0

Reset to default

You must log in to answer this question.