Assume a plane $P\subset\mathbb R^3$ has a hole $H$, and that the hole is topologically a compact disc. Being so, $P\setminus H$ does not separate the space. A regular tetrahedron $\sigma^3$ (of edge-length 1, say) wants to pass through the hole.

As far as I know, there are some papers about this.

  1. H. Maehara, N. Tokushige, A regular tetrahedron passes through a hole smaller than its face, preprint.

  2. J. Itoh, Y. Tanoue, T. Zamfirescu, Tetrahedra passing through a circular or square hole, Rendiconti del Circolo Matematico di Palermo, Suppl. 77 (2006), 349-354.

  3. J. Itoh, T. Zamfirescu, Simplicies passing through a hole, J. of Geometry, 83 (2005), 65-70.

The paper 1 shows that the minimum side-length of holes in the shape of a regular triangle is $\frac{1+\sqrt2}{\sqrt6}\approx0.985599$.

The paper 2 shows that the minimum diameter of circular holes is $\frac{t^2-t+1}{\sqrt{\frac{3}{4}t^2-t+1}}\approx0.8957$ $\left(3t=2+\sqrt[3]{\sqrt{43}-4}-\sqrt[3]{\sqrt{43}+4}\right)$ and that the minimum diagonal-length of holes in the shape of a square is 1.

The paper 3 shows that there exists a convex hole $H\subset P$ of diameter $\frac{\sqrt3}{2}$ and width $\frac{\sqrt2}{2}$ such that the regular tetrahedron $\sigma^3$ moving in $\mathbb R^3$ can pass through $H$. In the first paragraph of the proof, they say

"Take a square $Q\subset P$ of edge-length $\frac12$, with vertices $q_{\pm,+}=\left(\pm{\frac14}, \frac12\right)$ and $q_{\pm,-}=(\pm{\frac14}, 0)$. Denote the point $\left(0, \frac{\sqrt11}{4}\right)\in P$ by $v$. Take a disc $D$ of center $\left(0, \frac{\sqrt2}{4}\right)$ and radius $\frac{\sqrt2}{4}$. Define the hole $H$ as the convex hull of $D\cup T$, where $T$ is the triangle $vq_{+-}q_{--}$."

Then, here is my question.

Question: What is the shape of the holes which have the minimum area?

As far as I know, this question still remains unsolved. I have tried to solve this question, but I don't have any good idea. I suspect the paper 3 would be a key. I need your help.


1 Answer 1


Did you ever find any answer to this?
I find it intriguing that figuring out which shapes of holes a given solid object can pass through is widely considered to be a suitable puzzle for 2 year olds, yet we still don't know the smallest possible hole even for a regular simplex.

edit 29/04/21 — as pointed out by @mjqxxxx in the comments, there's a smaller area hole possible than what I described below by simply projecting onto a plane along the direction of one of the tetrahedron edges, giving a triangle with area $1/\sqrt{8}$ ≈ 0.3536.

I'll leave below my earlier answer, even though it looks rather silly in retrospect :)

Just playing I found a hole shape which I believe brings the upper bound down from anything I've seen described so far. It is not a convex shape and the tetrahedron does not pass through with a simple translation, but has to also perform multiple rotations.

Say instead of starting with the tetrahedron above the plane and figuring out how to get it below it, we start in the middle with the tetrahedron trapped in a square hole of side 0.5 — we can free one of its vertices by rotating it about one edge of this square, and the opposite edges sweep out a pair of hyperbolas, cutting away a slightly curved triangle. Once a vertex is free, the tetrahedron can slide out along the direction of one of the connected edges.

We can apply this sequence in reverse to get from above the plane to the middle, and then forwards about one of the other sides of the square to pass all the way through, cutting away a second curved triangle similar to the first.
Animation of the described shape passing through a hole

This gives a shape with a total area of ≈ 0.4044
For comparison, the minimal equilateral triangle has an area ≈ 0.4206,
and the minimal circle has area ≈ 0.6301.

I don't have any proof that this is the actual minimal-area hole, and indeed suspect that even smaller might be possible — for instance perhaps there is some way to twist the tetrahedron after it is half way through, so that the top half gets through using some of the same curved triangle we cut away for the lower half.

  • 5
    $\begingroup$ @AlexM. - could you clarify? Is there a problem with the comment? $\endgroup$
    – DPKR
    Feb 27, 2021 at 16:17
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    $\begingroup$ I voted this up since it represents interesting progress on a question that hadn't been answered in seven years. And I'm impressed by the animation; it's not so easy to make a clear illustration of the idea. I hope the author continues to contribute to MathOverflow. $\endgroup$ Feb 28, 2021 at 2:38
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    $\begingroup$ @DPiker: how did you compute $0.4044$? I think it's fine to post this as an answer, both because comments cannot contain images, and because in definitely ruling out certain other shapes as candidates for optimality, it is a sort of partial answer. $\endgroup$ Feb 28, 2021 at 13:56
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    $\begingroup$ The minimum-area shape that I've seen in the literature is just an isosceles triangle with base $1$ and height $\sqrt{2}/2$... for instance, "Problem 1" in this article (sciencedirect.com/science/article/pii/S0925772111000629). That area is just $\sqrt{2}/4 \approx 0.3536$, so, both smaller and simpler. $\endgroup$
    – mjqxxxx
    Apr 29, 2021 at 16:21
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    $\begingroup$ @mjqxxxx - that is indeed smaller and much simpler. It seems obvious in retrospect - not really sure how I forgot to check against that at the time. Ah well, live and learn! $\endgroup$
    – DPKR
    Apr 29, 2021 at 22:35

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