[math-fun] "Stable polyhedron" (?) problem
I'm not sure of the correct name of this old problem, but it's been around since at least the mid 1800's. (There's a legend about George Boole's discussing this with a teacher.): Does there exist a convex polyhedron P in 3-space such that, if it has constant positive 3D density, then when its center of gravity is projected orthogonally onto the planes of its faces, at least one such projection lies inside the face. (I.e., there is a face F such that if P is placed on F on a horizontal plane under gravity, it won't fall over.) An old heuristic nonexistence proof is that if P existed, then placing P on an infinite plane under gravity would lead to a perpetual motion machine. (I don't find this argument persuasive even in the realm of physics.) But in any case, does anyone know whether this problem has been settled? Reference(s) ? (And what is it called?) Thanks, Dan
You've garbled the wording of your question, but I'll answer it as it seems to be intended (unless I'm missing something?). It doesn't matter whether or not the density is constant. Your remark about perpetual motion machines can be readily mathematized: Consider any point x inside P. If its projection to the plane of a face F is outside the face, then the length of the perpendicular from x to this plane is less than the distance from x to F. The perpendicular intersects the boundary of P in some other closed face, which is therefore closer to x than F. (A special case: when the foot of the perpendicular lies on the boundary of F --- if you think about it, some other face is still strictly closer). Therefore, any face with least distance to x will do (i.e. any local minimum of potential energy gives a face that it orthogonally projects inside) It's interesting to look at the smooth version of this: suppose you have a smooth, strictly convex shape, so that its dual, using the point x as the origin (i.e. the dual is the set Q = {Y| for all X inside P, Y . X < 1 ). When P is smooth and strictly convex, then there's a homeomorphism from the boundary of P to the boundary of Q, taking X to the unique Y that maximizes the inner product. Every orientation preserving map of S^2 to itself has a fixed point, QED. This of course doesn't work in odd dimensions, and the potential energy proof is more elementary anyway. This is related to another question that Peter Doyle was fond of, about, something like Penelope's boulder. Penelope has a big boulder (some kind of polyhedron) that she was browbeaten by her suitors to roll one "notch" every evening until it sits over a spot on the floor, at which point she has to choose one of the suitors. The rule is that she must roll the boulder across an edge whose extension separates the face on which it rests from the goal. Is there any shape of polyhedron that allows Penelope to delay reaching the goal indefinitely? I won't try to spoil it by trying to remember what I once knew about it. Bill
I'm not sure of the correct name of this old problem, but it's been around since at least the mid 1800's. (There's a legend about George Boole's discussing this with a teacher.):
Does there exist a convex polyhedron P in 3-space such that, if it has constant positive 3D density, then when its center of gravity is projected orthogonally onto the planes of its faces, at least one such projection lies inside the face. (I.e., there is a face F such that if P is placed on F on a horizontal plane under gravity, it won't fall over.)
An old heuristic nonexistence proof is that if P existed, then placing P on an infinite plane under gravity would lead to a perpetual motion machine.
(I don't find this argument persuasive even in the realm of physics.)
But in any case, does anyone know whether this problem has been settled? Reference(s) ? (And what is it called?)
Thanks,
Dan
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If "It doesn't matter whether or not the density is constant.", then it doesn't matter if the object is convex. Any non-convex object can be padded out to its convex hull by adding zero-density fill. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com on behalf of Bill Thurston Sent: Sun 4/30/2006 9:26 PM To: dasimov@earthlink.net; math-fun Subject: Re: [math-fun] "Stable polyhedron" (?) problem You've garbled the wording of your question, but I'll answer it as it seems to be intended (unless I'm missing something?). It doesn't matter whether or not the density is constant. Your remark about perpetual motion machines can be readily mathematized: Consider any point x inside P. If its projection to the plane of a face F is outside the face, then the length of the perpendicular from x to this plane is less than the distance from x to F. The perpendicular intersects the boundary of P in some other closed face, which is therefore closer to x than F. (A special case: when the foot of the perpendicular lies on the boundary of F --- if you think about it, some other face is still strictly closer). Therefore, any face with least distance to x will do (i.e. any local minimum of potential energy gives a face that it orthogonally projects inside) It's interesting to look at the smooth version of this: suppose you have a smooth, strictly convex shape, so that its dual, using the point x as the origin (i.e. the dual is the set Q = {Y| for all X inside P, Y . X < 1 ). When P is smooth and strictly convex, then there's a homeomorphism from the boundary of P to the boundary of Q, taking X to the unique Y that maximizes the inner product. Every orientation preserving map of S^2 to itself has a fixed point, QED. This of course doesn't work in odd dimensions, and the potential energy proof is more elementary anyway. This is related to another question that Peter Doyle was fond of, about, something like Penelope's boulder. Penelope has a big boulder (some kind of polyhedron) that she was browbeaten by her suitors to roll one "notch" every evening until it sits over a spot on the floor, at which point she has to choose one of the suitors. The rule is that she must roll the boulder across an edge whose extension separates the face on which it rests from the goal. Is there any shape of polyhedron that allows Penelope to delay reaching the goal indefinitely? I won't try to spoil it by trying to remember what I once knew about it. Bill
I'm not sure of the correct name of this old problem, but it's been around since at least the mid 1800's. (There's a legend about George Boole's discussing this with a teacher.):
Does there exist a convex polyhedron P in 3-space such that, if it has constant positive 3D density, then when its center of gravity is projected orthogonally onto the planes of its faces, at least one such projection lies inside the face. (I.e., there is a face F such that if P is placed on F on a horizontal plane under gravity, it won't fall over.)
An old heuristic nonexistence proof is that if P existed, then placing P on an infinite plane under gravity would lead to a perpetual motion machine.
(I don't find this argument persuasive even in the realm of physics.)
But in any case, does anyone know whether this problem has been settled? Reference(s) ? (And what is it called?)
Thanks,
Dan
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Don't you mean that "Is it true that all convex polyhedra have at least one face such that the polyhedron centroid, projected onto the plane containing this face, lies within this face?" And isn't it necessary to assume that all polyhedra with a nonconvex face are nonconvex? (Maybe this is obvious, but it might need proof.) Steve Gray ----- Original Message ----- From: <dasimov@earthlink.net> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Sunday, April 30, 2006 5:53 PM Subject: [math-fun] "Stable polyhedron" (?) problem
I'm not sure of the correct name of this old problem, but it's been around since at least the mid 1800's. (There's a legend about George Boole's discussing this with a teacher.):
Does there exist a convex polyhedron P in 3-space such that, if it has constant positive 3D density, then when its center of gravity is projected orthogonally onto the planes of its faces, at least one such projection lies inside the face. (I.e., there is a face F such that if P is placed on F on a horizontal plane under gravity, it won't fall over.)
An old heuristic nonexistence proof is that if P existed, then placing P on an infinite plane under gravity would lead to a perpetual motion machine.
(I don't find this argument persuasive even in the realm of physics.)
But in any case, does anyone know whether this problem has been settled? Reference(s) ? (And what is it called?)
Thanks,
Dan
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participants (4)
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Bill Thurston -
dasimov＠earthlink.net -
Schroeppel, Richard -
Steve Gray