Recently I stumbled on the problem quoted here about a geometric proof of the AM-GM inequality $$(a_1+\cdots+a_n)^n\ge n^n a_1\cdots a_n$$ by packing $n^n$ rectangular $ n$-dimensional boxes of sides $a_1,\dots,a_n$ into a hypercube of side $a_1+\dots+a_n$.

For $n=2$ this is almost trivial $-$ just leave a square of side $|a_2-a_1|$ in the middle and pack four $a_1\times a_2$ rectangles around it. But already for $n=3$ it is highly non trivial, and the quoted article shows in several ways graphically a packing of 27 $a\times b\times c$ blocks into a cube of side $a+b+c$. Presumably this packing was found by a program using trial and error, and it has hardly any symmetries. There should be many non isomorphic packings like this, and I am wondering:

Is there a packing of 27 $a\times b\times c$ blocks into a cube of side $a+b+c$ with a higher degree of symmetry? E.g.

opposite sides of the cube have the same pattern, or

a certain $S_3$ isomorphism like $(a,b,c)\mapsto (c,b,a)$ results in some rotation of the packed cube? Or if not,

- is there at least a packing where in each direction (or to start with: in just one direction), each one of the layers has three separated holes?
- or, why not, each layer has just
tworectangular holes?

As far as the $S_3$ isomorphisms, I'd guess that since the order of the three sides is crucial, if we require wlog $a<b<c$, the isomorphism $(a,b,c)\mapsto (c,b,a)$ should be, if at all, the only one where it might be possible ... Note that because of the special position of the middle length $b$, it also seems hopeless to find a kind of symmetry where $a,b,c$ correspond to the three directions in $ \mathbb R^3$.

The idea behind all this is of course that if some high symmetry patterns exist, there might be a tiny chance to generalize them to higher dimensions. Well, tiny... even from $ \mathbb R^2$ to $ \mathbb R^3$ that is not the case.

notoccur to me during packing my suitcase. :) $\endgroup$ – Wolfgang Aug 9 '18 at 19:14The Mathematical Gardner, books.google.com/… . There are photos. There is also an assertion that there are 21 solutions. $\endgroup$ – Zach Teitler Aug 9 '18 at 19:41