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Jun 15, 2020 at 7:27 history edited CommunityBot
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Feb 16, 2017 at 18:40 comment added Wlodek Kuperberg @YaakovBaruch: I don't think it is obvious no matter which way it goes. So, If you find a proof that for some symmetric trapezoid the optimal ellipse is not aligned with the bases, please write it in as an answer. By the way, the idea of using symmetry for finding a counterexample was tried already, so far with no success.
Feb 16, 2017 at 9:00 comment added Yaakov Baruch @WlodekKuperberg: if an optimal ellipse is not aligned with the bases, then by symmetry there are at least 2 different optimal ellipses. I asked because, visually, it doesn't appear obvious to me that that is not the case (although I think it's unlikely).
Feb 15, 2017 at 23:39 comment added Wlodek Kuperberg @YaakovBaruch: what about them? Is it known that it does not? By the way, the problem is affine-invariant, so axial symmetry is irrelevant.
Feb 15, 2017 at 9:28 comment added Yaakov Baruch What about symmetric trapezoids? Is it known that the optimal ellipse has an axis parallel to the bases? Unless I missed some, I didn't see any comments mentioning trapezoids.
Feb 13, 2017 at 0:01 vote accept Wlodek Kuperberg
Feb 11, 2017 at 21:15 answer added Jairo Bochi timeline score: 5
Jan 4, 2017 at 5:24 answer added user44143 timeline score: 3
S Jan 4, 2017 at 5:06 history bounty ended Wlodek Kuperberg
S Jan 4, 2017 at 5:06 history notice removed Wlodek Kuperberg
Jan 1, 2017 at 23:42 comment added Wlodek Kuperberg In the above comment, I should have added that the metric based on the volume of the symmetric difference is topologically equivalent to the original Banach-Mazur metric. But if it happens that the ellipsoid in question is unique, then some nice geometric properties of the Banach-Mazur continuum would follow, such as its starlike shape, with the affine class of ellipsoid at the center, and straight-line segments connecting the center with all other points. (I wrote this in a separate comment because of the limit on the number of characters allowed in a comment.)
Jan 1, 2017 at 23:34 comment added Wlodek Kuperberg The question is equivalent to that of an ellipse $E$ (or ellipsoid) of area $1$ whose intersection with $K$ is of maximum area, and this version seems nicer for two reasons: (1) it sounds simpler and (2) the intersection is convex. Still, there is a reason why I asked the way I did. One can define the similar minimum value of the symmetric difference for two arbitrary convex bodies $K$ and $L$, each of volume $1$, using all bodies $L'$ affinely equivalent to $L$. The remark of Jairo Bochi on the affine invariance holds here, too. This leads to another metric on the Banach-Mazur compactum.
Dec 28, 2016 at 22:59 vote accept Wlodek Kuperberg
Dec 28, 2016 at 23:07
Dec 28, 2016 at 18:14 answer added Jairo Bochi timeline score: 3
S Dec 28, 2016 at 17:32 history bounty started Wlodek Kuperberg
S Dec 28, 2016 at 17:32 history notice added Wlodek Kuperberg Improve details
Dec 27, 2016 at 22:57 answer added Iosif Pinelis timeline score: 5
Dec 24, 2016 at 5:42 answer added Gerhard Paseman timeline score: 7
Dec 23, 2016 at 1:21 comment added Wlodek Kuperberg @JosephO'Rourke: The pair of "Banach-Mazur" homothetic ellipsoids is relevant, too, one contained in-, the other containing $K$, with the minimum homothety coefficient -- as in the definition of the Banach-Mazur metric. But uniqueness of the pair fails already in dimension 3, even for centrally symmetric $K$.
Dec 23, 2016 at 1:15 answer added Joseph O'Rourke timeline score: 11
Dec 22, 2016 at 23:19 comment added Joseph O'Rourke Besides the John-ellipsoid, I wonder if the Milman-ellipsoid is relevant? I can only find this PNAS paper on the M-ellipsoid: "An M-ellipsoid $E$ of a convex body $K$ has small covering numbers with respect to $K$."
Dec 22, 2016 at 23:00 comment added Wlodek Kuperberg @T.Amdeberhan: I believe so. But I could be wrong, of course.
Dec 22, 2016 at 22:40 comment added T. Amdeberhan Thanks. Replace $K$ by a square. Then, $E$ is a circle again?
Dec 22, 2016 at 22:31 comment added Wlodek Kuperberg @T.Amdeberhan: For a circle $K$, $E=K$. (A circle is an ellipse, too.)
Dec 22, 2016 at 22:28 comment added Wlodek Kuperberg @GerhardPaseman: I believe it is a circle, because for the (truncated) triangle each of the inscribed maximum-area ellipse and the circumscribed minimum-area ellipse is a circle. This is not a proof, but a reason to conjecture that it is so.
Dec 22, 2016 at 22:25 comment added T. Amdeberhan Suppose $K$ is the actual disk on the plane (normalized to have unit area). Assume $E$ is such an ellipse. Now, if you rotate the disk around wouldn't you get another $E'$?
Dec 22, 2016 at 22:12 comment added Gerhard Paseman I can imagine (but not readily prove) that there are three non circular ellipses which might serve for a truncated modification of an equilateral triangle (or perhaps no modification is needed). Can you prove that a circle would be minimal? Gerhard "Not Seeing The Numbers Yet" Paseman, 2016.12.22.
Dec 22, 2016 at 22:03 comment added domotorp The first sentence follows from a standard compactness argument. Disk is also used to denote any, typically convex set.
Dec 22, 2016 at 22:02 comment added Wlodek Kuperberg @T.Amdeberhan: A compact convex set of area $1$ is necessarily a convex disk (a convex $2$-dmensional body). It is known that each of, the maximum-area ellipsoid contained in a convex body $K$ in $\mathbb{R}^n$ and the minimum-area ellipsoid containing $K$ is unique.
Dec 22, 2016 at 22:01 comment added T. Amdeberhan Do you have a reference for the statement you made before the questions? Also, you title saying "a disk" but you statement in the first line says 'a compact set". Which one do you want?
Dec 22, 2016 at 21:59 history edited T. Amdeberhan CC BY-SA 3.0
added 38 characters in body
Dec 22, 2016 at 21:53 history asked Wlodek Kuperberg CC BY-SA 3.0