Rupert Polyhedra: Cube

Last week, I wrote about Rupert polyhedra, and how a tetrahedron has the Rupert property. The idea dates back to the 1600s, when Prince Rupert of the Rhine won a bet that it was possible to make a hole in a cube that was large enough for an identical cube to pass through, so let’s look at how the Rupert property works for a cube.

As with the tetrahedron, we start by looking at the shadows that can be cast by a cube. We’re looking for either a shadow that will fit entirely within one face of the cube, or a shadow that can entirely enclose one face.

Looking at the shadows, we see squares, rectangles and hexagons.

Cube with square shadow
Cube rotated by 45 degrees about x-axis, producing rectangular shadow.
Cube rotated by 45 degrees about x-axis and arccos(root(2/3)), producing regular hexagon shadow.

Rotating a cube with side length $1$ by an angle of $45^\circ$ around the $x$-axis and $\cos^{-1}\left(\sqrt{\frac{2}{3}}\right)$ around the $y$-axis produces a shadow that is a regular hexagon, with side length $\sqrt{\frac{2}{3}}$. Will a square of side length $1$ fit inside a hexagon of side length $\sqrt{\frac{2}{3}}$? To work this out, consider a hexagon of side length 1, and work out the maximum size square that will fit inside.

Let $x$ be the length of the side of the square.

The largest square that can fit in a regular hexagon. Square has side length x. Perpendicular distance from side of square to corner of hexagon is d.

Using trigonometry, we know that $\dfrac{d}{\frac{x}{2}} = \tan\left(30^\circ\right)= \dfrac{1}{\sqrt{3}}$. Since the distance from the centre of a regular hexagon to a corner is the same as the side length, we also know that $d+\frac{x}{2}=1\Rightarrow d=1-\frac{x}{2}$. Substitute this value for $d$ into the first equation and rearrange:

\frac{2-x}{x} &=\frac{1}{\sqrt{3}}\\\\
x&=\frac{2\sqrt{3}}{ 1+\sqrt{3}}\approx 1.267\dots

So for a hexagon of side length $h$, the largest square that will fit inside has side length $\frac{2\sqrt{3}}{ 1+\sqrt{3}}h$.

The side length of the shadow hexagon is $\sqrt{\frac{2}{3}}$, so the maximum square that will fit inside it has side length

\left(\frac{2\sqrt{3}}{ 1+\sqrt{3}}\right)\left(\sqrt{\frac{2}{3}}\right)&=\frac{2\sqrt{2}}{1+\sqrt{3}}\\\\
&\approx 1.035…

So we can fit a square of side length $1$ inside the shadow and still have a little space left over.

Square of side length 1 inside regular hexagon of side length root(2/3).

Now imagine extruding that square upwards to make a tunnel through the rotated cube. Because the square fits fully inside the shadow, the tunnel will fit fully inside the cube, producing a cube with the desired size of hole.

In fact, a slightly different rotation will make a shadow than can contain an even larger square. This was discovered by Pieter Nieuwland, who calculated that for a cube of edge length $1$, the maximum size cube that can fit through a hole has edge length $\frac{3}{4}\sqrt{2}=1.0606\dots$. As I mentioned when talking about the tetrahedron, this number is called the Nieuwland constant – the maximum ratio by which a polyhedron can be increased in size and still fit through a hole in the original-sized polyhedron.

In practice, this slightly larger ratio allows enough wriggle room to make a 3D-printed Rupert cube. You can download the OpenSCAD file for this, as well as the Geogebra file for a rotating cube and its shadow, from GitHub.

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





3 responses to “Rupert Polyhedra: Cube”

  1. Gerard Westendorp avatar
    Gerard Westendorp

    Nice weblog!
    I used your OpenScad file to also print 2 Rupert cubes, I posted them on Twitter:

    1. Sam Hartburn avatar
      Sam Hartburn

      Thanks Gerard! Your cubes look great – I didn’t think of printing two with the hole in and passing them through each other.

  2. Dave avatar

    Can I ask what settings you used for supports for this? The removal of supports generally leaves some material behind which would make the small margins potentially not work. Just curious if you had some printing tips. Thanks!

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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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