From: propp@math.mit.edu (Jim Propp) Newsgroups: sci.math.research Subject: antipodes Date: 23 Jan 1996 22:10:35 -0500 Given a centrally symmetric convex compact body K in 3-space and a point p on its surface, the point on the surface farthest from p in the surface metric need not be the antipode of p. (Cf. the Knuth-Kotani puzzle, which asks one to find the point on the surface of a 1-by-1-by-2 box that is farthest from one of the corners; it is not the opposite corner.) However, one might still conjecture that the maximum distance between two points in the surface metric, as BOTH of them vary over the surface, is achieved by a pair of antipodal points. (This is believed to be true in the case of an a-by-b-by-c box for all a, b, and c, though to my knowledge no proof is known even for this very special case.) Can anyone prove or disprove the conjecture? (Is the convexity condition necessary?) Jim Propp Department of Mathematics M.I.T. ============================================================================== "Knuth-Kotani example" has this property: position the box with a 1x1 square at the base and p at the front, left, lower corner (0,0,0) (say). Its antipode q is at the back, right, upper corner, (1,1,2). The distance from p to q is sqrt(8), as seen by the path across the front and right faces. But r = (3/4, 3/4, 2) is of distance sqrt(8 1/4) from p no matter which of several routes is used. -- djr. ============================================================================== From: George Russell Newsgroups: sci.math,rec.puzzles,alt.math.recreational Subject: Re: Box & Ribbon Date: Mon, 21 Dec 1998 19:57:41 +0000 David M Einstein wrote > . . . which is why I am > curious as to whether or not the most distant pair of points on a box > are opposite each other, I believe that they are but to show it for a > general box will be messy, and will put it off till I have more time > for grinchliness. > If it is true for the box is it true for all centrally symettric > solids? how would one even start to prove this? It can't be true for all centrally symmetric solids. Imagine a polyhedron in the shape of a torus with very ridged faces making travel from the outer circle to the inner circle very expensive; then the most distant pair of points would have to be on the outer and inner circle. Perhaps it might be true for centrally symmetric solids which are also convex.