Rupert Polyhedra: Octahedron

A polyhedron is Rupert if you can cut a hole in it that’s large enough for an identical polyhedron to fit through. I’ve already talked about the Rupert tetrahedron and the Rupert cube. Now it’s the turn of the octahedron.

In the case of the tetrahedron and the cube, I said that we needed either a shadow that would fit entirely within one face of the solid, or a shadow that could entirely enclose one face. This wasn’t entirely accurate. What we actually need are two shadows, one of which will fit entirely inside the other. So let’s look at the shadows made by an octahedron.

Similarly to the cube, we see squares, rectangles and hexagons, and also some parallelograms. When the octahedron is positioned with every vertex on an axis, its shadow is square. From this position, if it is rotated by $45^\circ$ around the $x$-axis and $\theta=\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)-\cos^{-1}\left(\sqrt\frac{2}{3}\right)$ around the $y$-axis it makes a hexagonal shadow that fits entirely inside the square, as shown below.

Octahedron rotated to show two shadows, one inside the other.

We can calculate the factor by which the octahedron can be scaled, and still fit through the hole (the Nieuwland constant) by finding the ratio of $AB$ to $A’B’$. Assuming that the octahedron has an edge length of $1$, the two points of the original octahedron that sit on the $x$-axis are projected to $A\left(\frac{-1}{\sqrt{2}},0,0\right)$ and $B \left(\frac{1}{\sqrt{2}},0,0\right) $. $A’$ and $B’$ are the projections of these points after the rotations described above.

Digram of the shadows.

We can find the coordinates of the corresponding points on the rotated octahedron by applying the rotation matrices

1 & 0 & 0\\
0 & \cos(45) &-\sin(45)\\
0 & \sin(45) &\cos(45)\\


\cos(\theta) & 0 & \sin(\theta) \\
0 & 1 & 0\\
\sin(\theta) & 0 &\cos(\theta)\\

By the magic of cancelling trigonometry (which I invite you to work through yourself) this gives $\left(-\frac{2}{3},0,\frac{1}{3\sqrt{2}}\right)$ and $\left(\frac{2}{3},0,-\frac{1}{3\sqrt{2}}\right)$. Their projections onto the $x$-$y$ plane are $A’=\left(-\frac{2}{3},0,0\right)$ and $B’=\left(\frac{2}{3},0,0\right)$.

So the ratio between the lengths $AB$ and $A’B’$ is

&\approx 1.06066

This is the current best estimate for the Nieuwland constant for the octahedron, but it has not been proved to be the biggest. As with the cube, this ratio gives just enough wriggle room to make a 3D-printed version, although there’s not much of the original octahedron left once the tunnel has been cut. You can download the OpenSCAD file from GitHub.

Reference: Most of the information required to write this came from the article Platonic Passages (Jerrard, Wetzel and Yuan, 2017).





2 responses to “Rupert Polyhedra: Octahedron”

  1. Juan Daniel Castanier avatar
    Juan Daniel Castanier

    Hey, I love your work, and I have a question. How would you approximate the problem of mathematically finding the position that gives you the largest and smallest shadows of a polyhedron? This is a step you didn’t mention and I find fascinating. I’m thinking of a function of A dependent on the rotation around each axis, it’s derivative would be the an easy answer.

    Thanks for your time.

    1. Juan Daniel Castanier avatar
      Juan Daniel Castanier

      Btw, I came here from Matt’s video.

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Hello! I’m Sam Hartburn, a freelance maths author, editor and animator. I also dabble in music and write mathematical songs. Get in touch by emailing or using any of the social media buttons above.

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