3650.23 – Big Rectangular Solid


Imagine a rectangular solid composed of small unit cubes. The rectangle measures xx cubes by yy cubes by zz cubes. Now imagine building a shell of unit cubes around the entire solid. In other words, the shell is 1 cube thick in all dimensions.

First: what is an equation for the number of unit cubes in the shell in terms of xx, yy, and zz? Second: for what xx, yy, and zz is the number of cubes in the rectangular solid equal to the number of cubes in the shell?


Solution

The original solid has xyzxyz unit cubes. The new one (see the figure), cube plus shell, has

(x+2)(y+2)(z+2)=(xy+2x+2y+4)(z+2)=xyz+2xy+2xz+4x+2yz+4y+4z+8.(x+2)(y+2)(z+2) = (xy + 2x + 2y + 4)(z + 2) = xyz + 2xy + 2xz + 4x + 2yz + 4y + 4z + 8.

unit cubes. So the shell alone has 2(xy+xz+yz)+4(x+y+z)+82(xy + xz + yz) + 4(x + y + z) + 8 cubes. That answers the first question.

As for the second: when does the shell alone equals the original cube in volume, this requires that we solve the equation:

xyz=2(xy+xz+yz)+4(x+y+z)+8, xyz = 2(xy + xz + yz) + 4(x + y + z) + 8,

in three variables, which is, umum, not easily solved and we're not going to try.

It would be an interesting exercise to program a spreadsheet and fool around with different dimensions to see if anything turns up. Also, try having the original solid be a cube itself, i.e. x=y=zx = y = z. Then

x3=6x2+12x+8x^3 = 6x^2 + 12x + 8. HmmmHmmm. Programming a spreadsheet gives us x between 7 and 8, with no integer solution. But there is an explicit solution. Mathematica gives

x=2(1+23+43).x = 2(1 + \sqrt[3]{2} + \sqrt[3]{4}).

3650_23_solution_7f48380e06.png