The problem comes with two different of scissor(s) congruence. Previously I cited this paper: < http://link.springer.com/article/10.1007/BF02759727#page-1 >, which defines scissors congruence as congruence by a dissection into topological disks; two shapes are scissor congruent when each disk of one's dissection is isometric to a (one-to-one) corresponding disk of a dissection of the other. That's how I was using the word, but another definition for "scissors congruence is a synonym for dissectable into the same set of polygons (-hedra, -topes), up to isometry of each. The paper excludes the topological disk definition from applying to the square and round disk of the same area. --Dan On Apr 28, 2014, at 12:03 PM, Bill Gosper <billgosper@gmail.com> wrote:
DanA>
Is there a proof that a 4-dimensional cube and 4-dimensional ball (sphere + interior) of the same volume are not scissors congruent? <DanAsimov
I missed something. Is there a proof that a TWO-dimensional "cube" and TWO-dimensional "ball" (disk) of the same area are not interdissectible w/o A/C?
It seems to me there's a lot of middle ground between "piece-wise smooth" (if that is indeed what scissors require) and "non-measurable".
Also, I need to be educated as to how the following two conditions can be inequivalent:
WDS> However, Dehn showed in one of the first Hilbert problem solves, that a regular tetrahedron and cube (same volume) are NOT scissors congruent. -----
DanA>I thought Dehn proved only that they could not each be dissected into the same finite set of polyhedra (up to isometries), analogous to the Bolyai dissections. -----
--rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun