bonanova Posted February 20, 2009 Report Share Posted February 20, 2009 Four identical cubical boxes, with interior dimensions of a cm on a side, and weighing b grams each, were packed with wooden spheres of density of d gram/cm3 Box 1 contained 1 single sphere; the largest one that would fit inside the box. Box 2 contained 8 identical spheres; the largest that would fit inside the box. Box 3 contained 27 identical spheres; again the largest that would fit. Box 4 contained 64 identical spheres; again the largest possible. Which loaded box [box plus contents] is heaviest, and what is its weight? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 21, 2009 Author Report Share Posted February 21, 2009 Long time reader but I think this is my first post. I noticed that most people are not doing the math, some at the end though.Most of you that did it mathematically are correct that all the boxes are the same, but the math is slightly off. "a" is the length of the sides of the cube, which would equal the diameter of the sphere in box 1, this makes r (which is the radius of the sphere) in the equation 4/3pi(r^3) equal to 1/2a. Thus turning the equation into: 4/3pi(1/2a)^3 or 1/6pi(a)^3 (by simplifying the fractions). The other boxes get slightly more confusing because we need to remember that "a" is split up more times and to multiply by the number of spheres, so for box 2: r=(1/4a) because we need the radius of one sphere, and there are two spheres on each side and 8 spheres total. The equation ends up being: 4/3pi(1/4a)^3(8) or 1/6pi(a)^3. This goes on for 3 and 4 with r equaling 1/6a and 1/8a respectively, and multiplied by 27 and 64 respectively. But by simplifying the fractions in each one we end up with: 1/6pi(a)^3(# spheres). And this will work for any perfect cube. Hi Freaky, and welcome [as a member now] to the Den! You're spot on with your comment. The correct result [they all weigh the same] was reached [the main question of the OP], even tho the weight in some calculations was off by a factor 23. Looking forward to seeing some of your favs posted here. Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted February 22, 2009 Report Share Posted February 22, 2009 I not sure about 64, but for 8 and 27 the cubic lattice packing is optimal. I assume the same would be true for 64 also. This has reminded me of a puzzle about octahedrons in a cube that I think I'll post. Thanks, bn. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 23, 2009 Author Report Share Posted February 23, 2009 A tesselation of 3-space perhaps? Quote Link to comment Share on other sites More sharing options...
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bonanova
Four identical cubical boxes, with interior dimensions of a cm on a side, and
weighing b grams each, were packed with wooden spheres of density of d gram/cm3
Box 1 contained 1 single sphere; the largest one that would fit inside the box.
Box 2 contained 8 identical spheres; the largest that would fit inside the box.
Box 3 contained 27 identical spheres; again the largest that would fit.
Box 4 contained 64 identical spheres; again the largest possible.
Which loaded box [box plus contents] is heaviest, and what is its weight?
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