Timeline for Building a polyhedron from areas of its faces
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 8, 2017 at 10:45 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Image link broken; now fixed.
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| Dec 3, 2011 at 20:31 | vote | accept | Vladimir Reshetnikov | ||
| Dec 2, 2011 at 11:55 | comment | added | Pietro Majer | As a remark to the case of an n dimensional non-convex polytope $P\subset \mathbb{R}^n$, Gerhard's condition is still necessary. Indeed, for any face f , the orthogonal projection onto the hyperplane containing that face is 1-Lipschitz and maps the n−1 skeleton of P minus f surjectively onto f (the easy reason is: f has an inner and an outer side (essentially by hypothesis on P, so that the line orthogonal to f at x∈f also meet ∂P in another y∉f). The analogous argument seems to work also for k-faces (giving the constant 1/2, possibly non optimal) | |
| Dec 1, 2011 at 2:17 | comment | added | Gerhard Paseman | I should be more careful in my statements. Let a polytope live in R^d. For positive integral k less than d there is a constant c_(k,d) such that the sum of the measures of all k-faces of the polytope times that constant is greater than the k-measure of any single k-face. When k=d-1, I assert that the constant is 1/2. For smaller k, smaller constants are possible, and are likely to be c_(k,d)=1/(1+d-k). Gerhard "Ask Me About System Design" Paseman, 2011.11.30 | |
| Dec 1, 2011 at 2:17 | comment | added | Igor Pak | (cont'd) One curious extension of Zil'berberg's proof is that the polytope can be made simple (for even number of vertices). I bet your theorem extends to have all polytopes simpicial. | |
| Dec 1, 2011 at 2:16 | comment | added | Igor Pak | Let me mention that in a certain sense, the dual problem was resolved by Zil'berberg in 1962. This paper is not available in English I think, but is stated as Exercise 35.9 in my book: math.ucla.edu/~pak/book.htm There, the areas are replaced by curvatures, which satisfy the Gauss-Bonnet formula and the proof is via an easy reduction to Alexandrov's "ray theorem", which is dual to Minkowski's theorem (but neither easily implies another). | |
| Dec 1, 2011 at 1:53 | comment | added | Gerhard Paseman | Although I did not mention polygons, I suggested such in my comment above. There for n >=3 the condition is the same: no side must have length at least half the sum of all lengths. For arbitrary dimensions, I suspect the same holds true of any nontrivial polytope, convex or not: no k-face has k-measure at least half the sum of the k-measures of all k-faces. Triangular prisms should show that one cannot replace half by anything smaller. Gerhard "Ask Me About System Design" Paseman, 2011.11.30 | |
| Dec 1, 2011 at 1:36 | history | answered | Joseph O'Rourke | CC BY-SA 3.0 |