Timeline for A polytope with congruent facets and an insphere that is not facet-transitive?
Current License: CC BY-SA 4.0
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| Mar 29, 2022 at 11:53 | comment | added | RavenclawPrefect | I don't see why every such example gives a polytope in $\mathbb{R}^3$ - this tiling of the sphere, for example, does not yield a monohedral polyhedron if one takes the convex hull of its vertices, because points that are colinear on the sphere don't remain so in Euclidean space (in particular, the problem arises when we have vertices which lie on the edges of some of the tiles). | |
| Jun 4, 2020 at 16:12 | comment | added | M. Winter | Thank you. This direction now seems plausible to me for faces of all dimensions, as long as they are congruent. | |
| Jun 4, 2020 at 15:28 | comment | added | Wlodek Kuperberg | @M.Winter, one more remark, in a similar vein and with a similar proof, that may interest you: If a convex polytope with edges of the same length is inscribed in a sphere, then it has a sphere tangent to all edges as well, and the two spheres are concentric. | |
| Jun 4, 2020 at 13:11 | vote | accept | M. Winter | ||
| Jun 4, 2020 at 12:53 | history | edited | Wlodek Kuperberg | CC BY-SA 4.0 |
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| Jun 4, 2020 at 12:49 | comment | added | Wlodek Kuperberg | Proof. Let P be a convex polytope with congruent facets and inscribed in a sphere S centered at the origin. Obviously, the origin is an interior point of P. Now, let S_0 be the largest sphere centered at the origin and contained in P. Then S_0 touches the boundary of P at at least one point, and the point must lie on some facet F of P. Since P is inscribed in a sphere, the facet is inscribed in a circle lying on S, hence S_0 touches F at the center of the circle. Since all facets are congruent, the distance from the origin to the center of each circle circumscribed about any facet is the same. | |
| Jun 4, 2020 at 8:52 | comment | added | M. Winter | I know it is not relevant for the answer (which I am going to accept soon), but how can I see that your modified first paragraph indeed holds? | |
| Jun 3, 2020 at 23:28 | history | edited | Wlodek Kuperberg | CC BY-SA 4.0 |
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| Jun 3, 2020 at 23:27 | comment | added | Wlodek Kuperberg | You are right, an additional assumption is needed, namely the facets should be inscribed in a sphere of one dimension lower. Thank you, I am correcting my mistake. | |
| Jun 3, 2020 at 22:10 | comment | added | M. Winter | Thank you for your answer! Very interesting linked question of yours. However, I object to the claim in the first paragraph: the rhombic dodecahedron has congruent faces, an insphere that touches all faces, but no circumsphere that contains all vertices. | |
| Jun 3, 2020 at 21:58 | history | edited | Wlodek Kuperberg | CC BY-SA 4.0 |
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| Jun 3, 2020 at 21:31 | history | edited | Wlodek Kuperberg | CC BY-SA 4.0 |
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| Jun 3, 2020 at 21:17 | history | undeleted | Wlodek Kuperberg | ||
| Jun 3, 2020 at 21:17 | history | edited | Wlodek Kuperberg | CC BY-SA 4.0 |
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| Jun 3, 2020 at 21:03 | history | deleted | Wlodek Kuperberg | via Vote | |
| Jun 3, 2020 at 20:56 | history | answered | Wlodek Kuperberg | CC BY-SA 4.0 |