# Pack box inside “smaller” box

Now, there is a puzzle that is quite well known as far as I know, concerning the packing of rectangular boxes in 3-dimensional space. You also have a measurement of a box as the sum of the hight, width and length.

Now, I have read a proof that a bigger box, using this measurement, can be packed inside a smaller box. I do, however, remember that proof to be quite abstract. Is there a known example? Or is it just that I do not remember correctly, and it is actually impossible.

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The problem statement is somewhat unclear. I interpret it as follows: To every rectangular box, you assign the sum of its three dimensions as a measure of its "size". You then ask whether a box of greater "size" can be packed inside a box of lesser "size".

If you allow cutting up the box of greater "size", the answer is yes. For instance, a box with dimensions $3\times3\times3$ has "size" $9$ and volume $27$, whereas a box with dimensions $1\times4\times5$ has "size" $10$ and volume $20$, so you can cut the "bigger" one up, for instance into $20$ unit cubes, and pack it into the "smaller" one.

If cutting isn't allowed, this isn't possible. This is proved rather elegantly in Section 9 of this article. The basic idea is to consider the volume of space at a distance at most $\epsilon$ from the box. This has contributions in various powers of $\epsilon$, stemming from the various rounded bits surrounding the box, and the leading contribution that depends on the box dimensions is due to the quarter-cylinders parallel to each of the edges. This goes as $\epsilon^2$ times the sum of the box dimensions. Since for any given $\epsilon$ the inner box augmented by $\epsilon$ lies within the outer box augmented by $\epsilon$, this leading contribution can't be bigger for the inner box than for the outer box, and thus the "size" of the inner box can't exceed that of the outer box.

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It is impossible, even when tilting the interior box.

This is question 70 (Airline Luggage) of The art of mathematics: coffee time in Memphis by Béla Bollobás, where he proves twice over two pages the assertion

Let $l(B)=a+b+c$ be the linear measure of an $a\times b\times c$ box (rectangular parallelepiped). This linear measure is monotone in the sense that if a box $B$ is contained in a box $C$ then $l(B) \le l(C).$

A superficially similar question, with a different answer, is Prince Rupert's Cube, where a cube of side 1.06 can pass through a hole in a cube of side 1.

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