Timeline for Sandwiching ellipses between planar convex bodies
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| S Oct 2 at 22:06 | history | bounty ended | CommunityBot | ||
| S Oct 2 at 22:06 | history | notice removed | CommunityBot | ||
| Sep 29 at 16:48 | comment | added | Guillaume Aubrun | If $K$ and $L$ are not strictly convex (as in @WillSawin's example), we can always arrange so that $K'$ and $L$ have a common segment in their boundary and therefore no ellipse can be sandwiched. | |
| Sep 29 at 12:58 | comment | added | Will Sawin | What about the opposite case: Let $L$ be the convex hull of a disc together with a point outside the disc but very close to it, and let $K$ be the intersection of a disc with a half-space which contains almost all but not all of the disc. Then can we squeeze $K$ ins such a way that it lies in $L$ but not the large disc inside $L$, and also does not lie in any other ellipse in $L$? | |
| Sep 25 at 7:58 | history | edited | Alex Ravsky |
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| Sep 24 at 20:54 | comment | added | Guillaume Aubrun | I agree that the idea works in some generality, but I'm missing arguments to cover remaining cases. | |
| S Sep 24 at 20:53 | history | bounty started | Guillaume Aubrun | ||
| S Sep 24 at 20:53 | history | notice added | Guillaume Aubrun | Draw attention | |
| Sep 23 at 12:41 | comment | added | Ilya Bogdanov | @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? | |
| Sep 23 at 6:18 | comment | added | Guillaume Aubrun | I'm not sure I understand your point. The argument is: if $K$ a triangle and if $L$ is a non-ellipse, it is possible to find affine images $K'$ and $L'$ such that any ellipse containing $K'$ has a larger area than any ellipse contained in $L'$. | |
| Sep 23 at 5:47 | comment | added | მამუკა ჯიბლაძე | Yes, but you seemingly need to turn a triangle inscribed in an ellipse into an equilateral one in such a way that this ellipse becomes a circle, no? | |
| Sep 22 at 21:06 | comment | added | Guillaume Aubrun | Yes. Up to affine bijection, there is only one triangle. | |
| Sep 22 at 20:55 | comment | added | მამუკა ჯიბლაძე | Does this also work for non-equilateral triangles? | |
| Sep 22 at 20:09 | comment | added | Guillaume Aubrun | 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. | |
| Sep 22 at 18:11 | comment | added | მამუკა ჯიბლაძე | Is this true for, say, a triangle and a near-circle? | |
| Sep 22 at 17:02 | history | asked | Guillaume Aubrun | CC BY-SA 4.0 |