From: David G Radcliffe Subject: Re: Prove or disprove. Date: 29 Aug 1999 05:07:28 GMT Newsgroups: sci.math Keywords: recover volume from areas of projections Andres Koropecki wrote: : Prove or disprove: : If S is a convex solid, and S_x, S_y, S_z are the projections : of S over the planes yz, xz, xy, respectively, then : Volume(S) <= (area(S_x)*area(S_y)*area(S_z))^.5 I first prove this for the case that S is a polycube whose faces are parallel to the coordinate axes. A polycube is a solid that is formed by gluing congruent cubes together along common faces -- see http://treasure-troves.wri.com/math/Polycube.html . Every convex solid is the limit of a sequence of polycubes, so that the volumes of the polycubes converge to the volume of the convex solid, and the areas of the projections of the polycubes converge to the areas of the projections of the convex solid. Therefore, it is sufficient to prove the inequality when S is a polycube. The proof is by induction on the number of layers of S. (Each layer is parallel to the xy-plane.) It is easy to see that the inequality holds if S has only one layer. Now assume that S has more than one layer. We cut S into two smaller polycubes A and B by a horizontal plane. Then A and B have fewer layers than S, so by the inductive hypothesis, Volume(A) <= (area(A_x)*area(A_y)*area(A_z))^.5 and Volume(B) <= (area(B_x)*area(B_y)*area(B_z))^.5. We use the following observations: . Volume(S) = Volume(A) + Volume(B) . area(S_x) = area(A_x) + area(B_x) . area(S_y) = area(A_y) + area(B_y) . area(A_z) <= area(S_z) . area(B_z) <= area(B_z) Let p=area(A_x)/area(S_x) and q=area(A_y)/area(S_y). Combining all of this data yields Volume(S) <= (area(S_x)*area(S_y)*area(S_z))^.5 * [(pq)^.5 + ((1-p)(1-q))^.5] I claim that (pq)^.5 + ((1-p)(1-q))^.5 <= 1. This is because (pq)^.5 + ((1-p)(1-q))^.5 is the inner product of the unit vectors (p^.5,(1-p)^.5) and (q^.5,(1-q)^.5). Therefore Volume(S) <= [area(A_x)*area(A_y)*area(A_z)]^.5 as claimed.