Timeline for Surface area of convex vs. non-convex polyhedra with same volume
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Jun 4, 2013 at 17:03 | comment | added | Yoav Kallus | @Dima: no, consider the catenoid (approximated by a polyhedron). | |
| Jun 4, 2013 at 2:01 | vote | accept | SSL | ||
| Jun 3, 2013 at 21:52 | answer | added | Gerardo Arizmendi | timeline score: 1 | |
| Nov 14, 2012 at 15:31 | comment | added | SSL | "An orthogonal polyhedron is one all of whose faces meet at right angles, and all of whose edges are parallel to axes of a Cartesian coordinate system." [wikipedia.org] | |
| Nov 14, 2012 at 15:23 | comment | added | Gerald Edgar | What is an "orthogonal" polyhedron? One whose faces are all parallel to a coordinate plane? | |
| Nov 14, 2012 at 14:46 | comment | added | SSL | I have edited the question based on the comments. Can anyone suggest an appropriate tag? | |
| Nov 14, 2012 at 14:46 | history | edited | SSL | CC BY-SA 3.0 |
added 204 characters in body
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| Nov 14, 2012 at 13:03 | comment | added | Dima Pasechnik | Difficult reduction? Won't taking the convex closure of the non-convex polyhedron suffice? | |
| Nov 14, 2012 at 11:43 | comment | added | Douglas Zare | If you pick a point inside the polyhedron from which it does not look star-like, and you replace the polyhedron with a star-like figure with that center with the same mass in each direction, does that reduce the surface area? Unfortunately, the result is not necessarily convex, and not necessarily a polyhedron, but perhaps if you start with a polyhedron some finite sequence of these would produce a convex figure. | |
| Nov 14, 2012 at 9:11 | comment | added | Pietro Majer | Of course, but the difficult reduction is from nonconvex to convex (because it requires non local variations). | |
| Nov 14, 2012 at 7:55 | comment | added | Igor Pak | Um, for every polyhedron (i.e. even convex one), there exists a convex polyhedron with the same volume and smaller surface area. | |
| Nov 14, 2012 at 7:03 | comment | added | Pietro Majer | What is true is: for any nonconvex polyhedron, there exists a convex polyhedron with the same volume and smaller surface area. | |
| Nov 14, 2012 at 5:11 | comment | added | Steven Gubkin | You can't. Counterexample: Cube vs. Sphere with a small dent in it. | |
| Nov 14, 2012 at 5:00 | comment | added | Yemon Choi | Why does this have the algebraic-geometry tag? | |
| Nov 14, 2012 at 4:46 | history | asked | SSL | CC BY-SA 3.0 |