Hello, and welcome to another edition of Tinkerer! Today, I will be sharing my writeup of why there are only $8$ of those beloved tetracubes that we all cherish and play with.
A bit of a note though: There's going to be some personal stories of how I discovered them in the first two sections. Feel free to follow this link to visit the writeup directly.
A Small Setback
Due to me lacking motivation to study just a few months before Texas MATHCOUNTS State in 2026, I really, really did MATHCOUNTS State badly. For whatever miracle I personally was still in top 30, and our school did pretty well! For the privacy of everyone involved, this is all I'm going to say [0]. I also nearly solved Sprint #30 (the last problem on the speed, no-calculator round), but messed up on a lower bound for an infinite sum and missed my bragging rights [1].
But after Texas MATHCOUNTS State, we still occasionally did MATHCOUNTS practice at my school. Now, I learnt to cherish every single bit of it, as my last MATHCOUNTS was that state competition, and I knew that middle school competition problems will soon become a past, distant, and heartbreaking memory of mine.
So whenever I saw an interesting problem that I missed in the practice we did after State (just for fun in our larger team, including people we did not send for Chapter), I'd make sure to take the time to properly study it and make all my knowledge about it not only as rigorous, but as intuitive as possible.
The problem went (roughly) like this:
How many different solids can be formed by gluing 4 unit cubes together? Solids are considered the same if they can be repositioned or rotated to match one another.
The answer given was $8$. After the practice test we did, I took a lot of time to work out the correct $8$ configurations as I couldn't do so in practice. And, I did!
Tetracubes!
Turns out, this class of shapes is known as tetracubes, a term I was previously unfamiliar with. We all know that arranging $1$, $2$, and $3$ cubes together under rotational equivalence is trivial ($1$, $1$, and $2$ respectively), but $4$ is when the game gets complex: there are $8$.
It should also be noted that after we know the possible tetracubes, it's interesting to try to stack them in weird ways to make interesting shapes. [2] and [3] in the Footnotes and References sections are some great resources to start looking at for that, as these things are a bit beyond me. It's also notable that if you want to play with those shapes, you need to make them yourself, which can be a bit of a challenge if you're not that of a practical person in terms of wood and glue (I'm not.)
But I couldn't find any post on the internet explaining how one can derive the fact that there are $8$ tetracubes in an ad-hoc but organized way; and as with any inquisitive person would do, I wrote it up!
The Writeup
So... here we go!
The following content (until the next section) has previously been sent in private to my math teacher, and is being reposted here on this blog. It has been edited, and this should serve as the canonical version from here on out.
Firstly, it should be rare for the cases where a cube has $3$ adjacent cubes since there are only $4$ cubes. In these cases, there is a central cube holding the other $3$ cubes together, and none of the other cubes can be considered as another central cube. We consider this case first.
Without loss of generality, the central cube has a cube extending to the top, and not a cube extending to the bottom. This is always possible, as the number of cubes to extend ($3$) is odd, so one must always have $2$ opposite directions with one with a cube and one without. Now, we have to choose two other lateral faces to extend cubes out of; this is either two faces opposite to each other, or two faces adjacent to each other (as more specific variations are all rotationally equivalent). There are $2$ possibilities for this case:
Now, we consider the other case of a solid that does not have a cube with $3$ connections.
In this case, the solid can be traced out by a path in a single stroke. Consider the longest completely straight run (a maximal sequence of cubes in a line) in the tracing; if this consists of $4$ cubes, then the figure is just a straight line, like this—for $1$ case:
Finally, the case remains where there’s no $3$ cubes in a line. You can still imagine “drawing” the cube out in 3D from one end to another, but in this case the cubes can form a “loop”. Since every step needs to turn with only $4$ cubes in total, the $4$-th cube must turn back and connect to the first cube, like this:
This is because we can see that any $3$ cubes that form an L shape like this can be rotated into this configuration, so we do it for the first $3$ cubes from $1$ endpoint. To “turn” after the second cube on the second row, there are $3$ valid directions as seen below.
We should also check that tracing each figure from the other endpoint does not result in a duplicate configuration with another figure; in this case, none of them duplicates, so we’re in the clear!
There are a total of $8$ different solids.
5 or more?
You might be wondering: How many pentacubes, hexacubes... are there? Is there a formula for the number of possible $n$-cubes in general?
Sadly, no. The sequence of the number of possible $n$-cubed is tracked as OEIS A000162 [4], but there is no general closed-form solution.
The possible ways to compute this sequence are all a bit beyond me right now; but if you have any good explanations, leave it in the comments below!
Bonus chatter: All diagrams drawn in MS Paint and Word.
Footnotes and References
[0] As the results are public, it's trivial (but not left as exercise) to reverse engineer my school, coach, and everything if I had given more information. I will say, our school did really well.
[1] Programmers here, $1$-index or $0$-index?
[2] https://www.mathematische-basteleien.de/tetracube.htm
[3] https://polyominoes.blogspot.com/2019/06/tetracubes-very-short-introduction.html
[4] https://oeis.org/A000162
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