Timeline for Given the skeleton of an inscribed polytope. If I move the vertices so that no edge increases in length, can the circumradius still get larger?
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| Dec 18, 2021 at 17:24 | comment | added | Anton Petrunin | @M.Winter I did not get (3), for (2) --- I do not see a nice argument at the moment, but it must be true. | |
| Dec 18, 2021 at 17:22 | comment | added | Anton Petrunin | @M.Winter (1) if a polygon does not contain its circumcenter then decreasing its longest side we increase its area. If we take two of such polygons that share the longest side then their total area increases, BUT this picture cannot come from the central projection of a convex polyhedron --- so you have to use the property discusses in the previous sentence. | |
| Dec 18, 2021 at 14:10 | comment | added | M. Winter | But in general, the structure of the argument seems plausible to me: (1) the projections of the 2-faces of $Q$ cover the full sphere. (2) the projections of the $P$-equivalents of the 2-faces of $Q$ cover the same sphere. (3) but these $P$-equivalents have smaller volume. Contradiction. | |
| Dec 18, 2021 at 14:05 | comment | added | M. Winter | (3) "Now, for each edge of $Q$ draw a line segment between the corresponding vertices of $P$". Where is this part of the argument used? | |
| Dec 18, 2021 at 14:04 | comment | added | M. Winter | (2) "Since the center of sphere is inside of $P$ the spherical polygons that correspond to facets of $Q$ will cover the whole sphere". This is not completely obvious to me. Consider the case $m=n=2$ in my post: it is possible that an inscribed polygonal chain whose convex hull contains the circumcenter still has winding number zero and does not cover the full circle. Of course, the edge length contractions should prevent this for $m=n=2$, but elaborating this for $m=n=3$ might need some work. Still, I believe this should be possible. | |
| Dec 18, 2021 at 13:59 | comment | added | M. Winter | Thank you very much for the update :). I have some questions: (1) "if one decreases the edge $e$ while keeping $F$ and $G$ cyclic, then the total area of $F$ and $G$ decreases", this seems plausible (I believe there should be many ways to see this), but I do not understand how this follows from the discussion prior to this sentence. By the way, I already know that this fails in higher dimensions: there is a polytope, if I shorten all of its edges, the volume can still increase. | |
| Dec 18, 2021 at 9:42 | comment | added | Anton Petrunin | @M.Winter Now it is a complete solution for $m=n=3$. | |
| Dec 18, 2021 at 9:41 | history | edited | Anton Petrunin | CC BY-SA 4.0 |
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| Dec 17, 2021 at 15:34 | history | undeleted | Anton Petrunin | ||
| Dec 17, 2021 at 15:34 | history | edited | Anton Petrunin | CC BY-SA 4.0 |
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| Dec 14, 2021 at 20:33 | history | deleted | Anton Petrunin | via Vote | |
| Dec 14, 2021 at 13:49 | comment | added | M. Winter | This is definitely a very promising approach! I was not aware of the Kirszbraun Theorem. But I cannot yet see (and I put some thought into this) why $q_i\mapsto p_i$ is non-expanding. This is clear if we consider distances between $p_i,p_j$ with $ij\in E(G)$, but there are other pairs that do not correspond to edges in $G$. I believe this is true, but I do not yet know how to use the polytope structure to conclude this (but I will think a lot more in this direction). | |
| Dec 14, 2021 at 4:33 | history | answered | Anton Petrunin | CC BY-SA 4.0 |