- 27 Apr '05 15:33

What do you mean? Do you mean the largest cube that you could pass through the unit cube no matter what its orientation? Or just the larget cube that could pass through with a specific orientation (which I think would just be a unit cube)?*Originally posted by THUDandBLUNDER***What is the size of the largest cube that can pass through a cube of unit length?** - 27 Apr '05 16:11 / 1 edit

I think you're going to have to elaborate on what you mean by 'pass through'. I suspect your question is equivalent to "what is the largest cube such that, for some projections of this cube and the unit cube onto the plane, the projection of the larger cube fits inside the projection of the smaller cube?"*Originally posted by THUDandBLUNDER***Yes (and No).** - 27 Apr '05 16:45 / 1 edit

That's correct, Acolyte.*Originally posted by Acolyte***I think you're going to have to elaborate on what you mean by 'pass through'. I suspect your question is equivalent to "what is the largest cube such that, for some projections of this cube and the unit cube onto the plane, the project ...[text shortened]... the larger cube fits inside the projection of the smaller cube?"**

While being more precise, I'm not sure if mentioning 'projections onto the plane' would actually makes things clearer to most people.

- 27 Apr '05 17:08

OK. In that case, what we want to do is maximize the projection area of the unit cube, and minimize the projection area of the bigger cube.*Originally posted by THUDandBLUNDER***That's correct, Acolyte.**

While being more precise, I'm not sure if mentioning 'projections onto the plane' would actually makes things clearer to most people.

To maximize the projection area of the unit cube, we have to look for the longest diagonal, which is the one going from one corner of the cube to the opposite corner. This has length 3^(1/2). By symmetry, we know that a pair of these exist. So the biggest projection you could have would be a square having diagonal length 3^(1/2), resulting in a side length (3/2)^(1/2).

To minimize the projection area of the bigger cube, we have to look for the smallest diagonal. This is the diagonal going from one corner to the opposite corner on the same face. We set this equal to the diagonal length of the unit cube, or 3^(1/2), resulting in a side length of (3/2)^(1/2), which is approximately equal to 1.225.

So the biggest cube has side length (3/2)^(1/2).