Let us now take a closer look at the **hex-fractal** we sliced last week. Chopping a level 0, 1, 2, and 3 Menger sponge through our slanted plane gives the following:

This suggests an iterative recipe to generate the hex-fractal. Any time we see a hexagon, chop it into six smaller hexagons and six triangles as illustrated below. Similarly, any time we see a triangle, chop it into a hexagon and three triangles like this:

In the limit, each triangle and hexagon in the above image becomes a hex-fractal or a **tri-fractal**, respectively. The final hex-fractal looks something like this (click for larger image):

Now we are in a position to answer last week’s question: how can we compute the **Hausdorff dimension** of the hex-fractal? Let *d* be its dimension. Like last week, our computation will proceed by trying to compute the “*d*-dimensional volume” of our shape. So, start with a “large” hex-fractal and tri-fractal, each of side-length 1, and let their *d*-dimensional volumes be *h* and *t* respectively.^{[1]}

Break these into “small” hex-fractals and tri-fractals of side-length 1/3, so these have volumes \(h/3^d\) and \(t/3^d\) respectively (this is how “*d*-dimenional stuff” scales). Since $$\begin{gather*}(\text{large hex}) = 6(\text{small hex})+6(\text{small tri}) \quad \text{and}\\ (\text{large tri}) = (\text{small hex})+3(\text{small tri}),\end{gather*}$$ we find that \(h=6h/3^d + 6t/3^d\) and \(t=h/3^d+3t/3^d\). Surprisingly, this is enough information to solve for the value of \(3^d\).^{[2]} We find \(3^d = \frac{1}{2}(9+\sqrt{33})\), so $$d=\log_3\left(\frac{9+\sqrt{33}}{2}\right) = 1.8184\ldots,$$ as claimed last week.

As a final thought, why did we choose to slice the Menger sponge on *this* plane? Why not any of the (infinitely many) others? Even if we only look at planes parallel to our chosen plane, a mesmerizing pattern emerges:

### More Information

It takes a bit more work to turn the above computation of the hex-fractal’s dimension into a full proof, but there are a few ways to do it. Possible methods include **mass distributions**^{[3]} or **similarity graphs**^{[4]}.

This diagonal slice through the Menger sponge has been proposed as an exhibit at the Museum of Math. Sebastien Perez Duarte seems to have been the first to slice a Menger sponge in this way (see his rendering), and his animated cross section inspired my animation above.

Thanks for reading!

### Notes

- We’re assuming that the hex-fractal and tri-fractal have the same Hausdorff dimension. This is true, and it follows from the fact that a scaled version of each lives inside the other. [↩]
- There are actually two solutions, but the fact that
*h*and*t*are both positive rules one out. [↩] - Proposition 4.9 in: Kenneth Falconer.
*Fractal Geometry: Mathematical Foundations and Applications.*John Wiley & Sons: New York, 1990. [↩] - Section 6.6 in: Gerald Edgar.
*Measure, Topology, and Fractal Geometry*(Second Edition). Springer: 2008. [↩]