Can Cube = Cuboid?

[Shadow]

Hey all,

 

The title's a little misleading, what I'm really wondering about is, can a cube and a cuboid have the same areas AND volumes, without the trivial solution of a=b=c=s, where s is the side of the cube. I did some calculations, however I have absolutely no idea if they're right, and to be quite honest I sort of doubt it.

 

If a, b and c are the sides of a cuboid, and s is the side of a cube, this is what I came up with:

 

http://i44.tinypic.com/oi9e75.jpg

 

Cheers,

 

Gabe

[timo]

I think there should only be the trivial solution. Sketch for the proof I'd try:

- If (a,b,c,s) is a solution, then for any x>=0 (xa,xb,xc,xs) is a solution. Hence, it suffices to look for solutions with s=1. This fixes V=1, then.

- Let c be a fixed value between zero and one, you can show that the V=1 cuboid has its minimal surface when a=b

- Under the constraint that a=b, you can show that the V=1 cuboid has its minimal surface when a=b=c=1; i.e. when it is the cube.

=> A cuboid with V=1 will have at least the same surface as a V=1 cube. It will have the same surface when it is a cube.

=> There is only the trivial solution.

Edited March 20, 2009 by timo

[Shadow]

But what if [math]a \neq b \neq c \neq s[/math]?

[timo]

Hm, perhaps I should have explicitly stated what I mean by "solution": A solution is a set of numbers (a,b,c,s) for which the surface area A and the volume V of a cuboid with side-lengths a, b and c equal those of a cube with side-length s. So if for each solution a=b=c=s then there is no solution for which the cuboid is not a cube. So there is no non-cube cuboid that has the same surface area and volume as any cube.

[Shadow]

That last sentence was what I was after. Although I must ask, why won't the following equations yield the correct result?

 

[math]a\cdot b\cdot c = s^3[/math]

[math]2(ab+bc+ca) = 6s^2[/math]

 

The positive solutions of these equations are stated above...and while yes, logically the surface and volume of a non-cube cuboid should never be the same as those of a cube, these equations disagree. However, there is the not-so-slight possibility that my equations is just the wrong way to think about it.

 

Thanks for your answers,

 

Gabe

[timo]

I did not check your equations but a priori the fact that you can write the solutions in a complicated manner does not contradict the statement that there is only the trivial solution. But your equations are not a solution, only a way to find it (and the approach you presented in your previous post seems quite ok to me). Perhaps you should set s=1 (see first step in my first post to see why this is sufficient) and then try to find constraints on c (by looking at the possible c for which both square roots exist). I would not be too surprised if that straightforwardly leads to c=1 which then directly leads to a=b=1 (from plugging into your equations). And of course there can be something wrong with my sketch; I did not actually perform the steps too carefully.

[Shadow]

I tried out s=1, and c=1, 2, 3, and all the solutions are complex, so I wouldn't be at all surprised if a=b=c=s was the only solution. Unfortunately I have no clue how to put a restriction on a variable in Maple (any ideas by the way?), and I can't do it by hand for all possible values of c, so I guess I'll never be completely sure. But it's a reasonable estimate, and more than sufficient for my curiosity.

 

Thanks again,

 

Gabe

[timo]

Investigate the argument under the square root seperately, first. It only depends on c and must be >= 0.

For example you could plot the argument in gnuplot:

f(x) = 1. - 6.*x + 9.*x*x - 4.*x*x*x
set xrange[0:2]
set yrange[-5:5]
set grid
plot f(x) t "argument under the square root"

 

Sidenote: Mathematica, Maple and friends are great programs with nice capabilities. For some strange reason, gnuplot is better-suited for the task at hand in the majority of cases I encounter.