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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.

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