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?
Current License: CC BY-SA 4.0
22 events
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| Mar 29, 2022 at 12:52 | vote | accept | M. Winter | ||
| Mar 28, 2022 at 12:56 | answer | added | Joseph Doolittle | timeline score: 3 | |
| Dec 14, 2021 at 15:17 | history | edited | M. Winter | CC BY-SA 4.0 |
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| Dec 14, 2021 at 14:59 | history | edited | M. Winter |
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| Dec 14, 2021 at 14:16 | history | edited | M. Winter | CC BY-SA 4.0 |
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| Dec 14, 2021 at 4:33 | answer | added | Anton Petrunin | timeline score: 0 | |
| Dec 14, 2021 at 1:31 | comment | added | Joseph O'Rourke | It is no doubt a matter of taste, but to me the uniqueness for the 2D case is simple, as illustrated in the figure I just posted. | |
| Dec 14, 2021 at 1:30 | answer | added | Joseph O'Rourke | timeline score: 0 | |
| Dec 14, 2021 at 0:25 | history | edited | M. Winter | CC BY-SA 4.0 |
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| Dec 14, 2021 at 0:24 | comment | added | M. Winter | @JosephO'Rourke Thank you for your comments. To your first comment: do you think this fact gives a shorter proof for $n=m=2$? I would be happy to see this as an asnwer. To your second comment: I explicitly want this to hold as general as possible. In general, I would say "simple" is too much of a restriction, but of course, seeing a proof for simple 3-polytopes might be a start. I skimmed throught the paper and couldn't find anything specific that seems to help here. I might be wrong of course. | |
| Dec 13, 2021 at 23:33 | comment | added | Joseph O'Rourke | Perhaps you could exploit the special conditions that must hold for a polytope to be inscribable? Arnau Padrol, Günter M. Ziegler: Six Topics on Inscribable Polytopes. E.g., maybe restrict attention to simple 3-polytopes, where a graph theoretic characterization is available. | |
| Dec 13, 2021 at 23:25 | comment | added | Joseph O'Rourke | For $n=m=2$, the lengths of the edges of $P$ uniquely determine the circumscribed circle. See Cyclic polygons generalized to higher dimensions. | |
| Dec 13, 2021 at 21:42 | comment | added | M. Winter | Let us continue this discussion in chat. | |
| Dec 13, 2021 at 21:40 | comment | added | M. Winter | @Matt If there is a way to make the question clearer, I would love to know. But I do not understand your objection. What is wrong/misleading about my phrasing? In the example of my last comment, $P$ is the pentagon (a convex polytope), and the pentagram is formed by the $p_i$ (in the sense that if you draw the lines between $p_i$ and $p_j$ for all $ij\in E(G)$, then you will see a pentagram). | |
| Dec 13, 2021 at 21:17 | comment | added | M. Winter | @Matt How about a pentagon and a pentagram, the faces are defined by different sets of vertices? But in general, just let the $p_i$ be the vertices of an arbitrary inscribed polytope (in any dimension if you want) that you scaled down sufficiently, so that the edge length constraints are satisfied. | |
| Dec 13, 2021 at 19:07 | history | edited | M. Winter | CC BY-SA 4.0 |
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| Dec 13, 2021 at 16:35 | comment | added | M. Winter | @Matt Surely the convex hull of the $p_i$ has some faces (if it is not of a lower dimension, but I believe we can ignore this). But why should these faces be in one-to-one correspondence with the faces of the polytope? Or are we not talking about the polytop's faces? I agree that summing over the fractions of the faces of the convex hull of the $p_i$ will give 1. | |
| Dec 13, 2021 at 0:36 | comment | added | M. Winter | @Ravenclaw Since I do not require the circumcenter to be in the interior of the convex hull, I agree with your argument if $m<n$, even without the perturbation. However, if $m>n$, then, as far as I understand, you try to perturb the vertices of the polytope. This might create a polytope with new edges, the lengths of which might have not been respected by the initial higher-dimensional point arrangement. Have I misunderstood something? | |
| Dec 13, 2021 at 0:10 | comment | added | RavenclawPrefect | The final question at least can be answered: it does not change if $n=m$, and you can reduce to that case without loss of generality. This is because you can embed the smaller-dimension point set into a sphere of the larger dimension, and perturb the points by epsilon until they strictly contain the center - then one of the point sets, viewed in a large-dimension sphere, will have strictly greater edge lengths than the other. (This also shows that if the conjecture holds in dimension $n$, it does so in all lower dimensions as well.) | |
| Dec 12, 2021 at 23:49 | comment | added | M. Winter | @Matt I am not sure that I can follow your argument. What follows from the larger fractions? I can imagine an argument going the other way: if the radius of the sphere were larger, then the fractions would be smaller and can no longer sum up to 1. But it seems hard to show that "summing over faces yields at least 1" is a necessary criterion, as the subsets of points that form faces have special meaning only in the polytope, but not in the point arrangement $p_1,...,p_s$ (at least not obviously). | |
| Dec 12, 2021 at 21:15 | history | edited | LSpice | CC BY-SA 4.0 |
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| Dec 12, 2021 at 20:14 | history | asked | M. Winter | CC BY-SA 4.0 |