# Packing a 3xMxN-box with copies of a single tetracube

There seems to be no solution for packing a 3xMxN-box with copies of this tetracube .

Does anybody know of a proof that this is impossible or is this still an open question?

Is it possible:

No. The reasoning is below:

First of all, $$M$$x$$N$$ must be divisible by 4 in a $$3$$x$$M$$x$$N$$ box because you're using shapes made up of 4 cubes. If we paint all the unit cubes like a 3D checkerboard, half of the "middle cubes" of the tetracubes must be of the opposite color of the rest. For this to be possible, $$M$$x$$N$$ must be divisible by 8.

Let's say there are $$t_{bottom}$$ tetracubes with three cubes at the bottom on the box's base, and $$o_{bottom}$$ with one. The base's area is $$3t_{bottom}+o_{bottom}$$. If $$3t_{bottom}+o_{bottom} = MN$$, then $$t_{bottom}>o_{bottom}$$ and $$t_{top}>o_{top}$$.

$$3t_{bottom}+o_{bottom} = MN$$
$$3t_{top}+o_{top} = MN$$
$$t_{bottom}+t_{top}+3o_{top}+3o_{bottom} = MN$$
$$3(MN-3t_{bottom}) + 3(MN-3t_{top}) + t_{bottom}+t_{top} = MN$$
$$6MN-8t_{bottom}-8t_{top} = MN$$
$$5MN = 8t_{bottom}+8t_{top}$$
$$t_{top} = 5MN/8 - t_{bottom}$$

Either the top or bottom layer has $$5MN/16$$ or more, and each has at least $$5MN/8 - MN/3 = 7MN/24$$. Both layers have a combined $$MN/8$$ lone legs of the tetracubes.

If two tetracubes combine into a cube, the latter can't be at a corner.

Let's look at the edges. There are $$4*(N+1)*M + 4*N*(M+1) + 3*(N+1)*(M+1) = 11MN + 7N + 7M + 3$$ in/on our prism.

Each tetracube has $$48$$ edges including the ones counted multiple times within the tetracube itself and there are $$3MN/4$$ tetracubes, so the product is $$36MN$$.

We also know that there are $$5MN/8$$ top- and bottom-centered ("justified") tetracubes in total and two edges at the top of each overlap with the surface.

Also, at least the 3 "deep" edges on every tetracube are counted only twice as well as the top/bottom 6 edges on each justified tetracube (the latter can overlap though). Assuming the rest are counted four times, the total count is:

4 times: $$11MN + 7N + 7M + 3 - (23MN/4) = 21MN/4 + 7N + 7M + 3$$
Twice: $$18MN/4$$
Once: $$5MN/4$$.

The total is at most $$41MN/4 + 21MN + 28N + 28M + 12$$.

$$41MN/4 + 21MN + 28N + 28M + 12 >= 36MN$$
$$MN/4 + 28N + 28M + 12 >= 5MN$$
$$28N + 28M + 12 >= 19MN/4$$
$$112N + 112M + 48 >= 19MN$$

Even assuming it could potentially work for sufficiently small $$N$$ and $$M$$ values, the fact that it's not applicable for any larger value (not even multiples) means it's impossible for any $$N$$ and $$M$$.

• "Each tetracube has 48 edges": Here, several edges are counted more than once, e.g., the "deep" edges are counted three times because they are shared by three cubes. Each tetracube has only 36 edges, so $3MN/4$ tetracubes contribute a total of only 27MN edges. Aug 31 at 4:52