In the Koch snowflake, the zeroth iteration is an equilateral triangle, and the n-th iteration is made by adding an equilateral triangle directly in the middle of each side of the previous iteration. The area of the Koch snowflake is $8/5$ the area of the starting triangle.

If I wanted to generalize this to other regular polygons, such as squares, pentagons, etc, the area, counting overlap, is $\frac{8}{8-n}$ times the area of the starting polygon, where $n$ is the number of sides, and $n < 8$. The area does not converge for $n \ge 8$.

The problem with this area is that it counts overlap multiple times (in the case of $n > 4$; $n = 3$ and $n = 4$ have no overlap). If a section of the "generalized snowflake" is covered multiple times, it counts all of those, not just one. How can I find the area of the generalized snowflake, counting areas covered multiples times only once?

Edit: In these snowflakes, the side of a polygon added at the $n$th iteration has $1/3$ the length of a side of a polygon added at the $(n-1)$th iteration.

Edit 2: I think the area of the hexagon ($n = 6$) case is $12/5$ the area of the original hexagon. I'm not sure about this, though.

  • $\begingroup$ In the square snowflake, are the second-generation squares $1/3$ the size of the original or $1/4$ the size? $\endgroup$
    – saulspatz
    May 12, 2019 at 13:37
  • $\begingroup$ Each side is $1/3$ the length of the previous generation (so the square is directly in the middle of the previous line). $\endgroup$ May 12, 2019 at 13:42
  • 1
    $\begingroup$ Like this? There is no overlap in the "square snowflake." $\endgroup$
    – kccu
    May 12, 2019 at 13:44
  • $\begingroup$ Yes that's what it looks like for the case of a square. $\endgroup$ May 12, 2019 at 13:44
  • 2
    $\begingroup$ I just notice after suitable scaling, the picture for hexagon ($n = 6$) looks exactly like the one for the vanilla koch curve ($n = 3$) If that is the case, the area ratio will be $\frac32 \times \frac85 = \frac{12}{5}$, the number you believe. $\endgroup$ May 15, 2019 at 6:10

1 Answer 1


As first conjectured by OP, for the hexagon ($n = 6$) case, the area of the generalized Koch snowflake indeed equals to $\frac{12}{5}$ of that of the seed hexagon.

This comes down to following observation. When one scale the seed hexagon to make its area two-third of that of a seed triangle, the generalized Koch snowflake generated from the seed hexagon "looks" the same as the Koch snowflake generated from the seed triangle.

Following is an illustration for what happens at iteration level $\ell = 0,1,2,3$
(upper-left, upper-right, lower-left, lower-right).

Generalized Koch curve at different level

The seed hexagon and the shapes generated from it are colored in red. The seed triangle and the shapes generated from it are colored in yellow.

As you can see, there is nothing in red in above figure. Instead, we see a bunch of orange regions. This is because we have rendered the seed triangles and its descendants in $50\%$ opacity and superpose them on top of the seed hexagon and its descendants.

There is nothing in red because in each iteration, the descendant from the triangle completely cover the descendant from the hexagon.

At each iteration level $\ell$, their "difference" is a bunch of yellow triangles of side $\frac{1}{3^{\ell+1}}$ of that of seed triangle. If the seed triangle has unit area, the area of each yellow triangle is $\frac{1}{9^{\ell+1}}$. Let $n_\ell$ be the number of these triangles. If one compare the yellow triangles in iteration level $\ell$ to that in iteration $\ell-1$. We find we can group the triangles in level $\ell$ in units of three. An unit may come from a triangle in level $\ell-1$ or newly spawned at an edge of descendant of the triangle at level $\ell-1$. This leads to following recursive relation for $n_\ell$.

$$n_\ell = \begin{cases} 3, & \ell = 0\\ 3(n_{\ell-1} + 3\cdot 4^{\ell-1}), &\ell > 0\end{cases}$$

Solving this give us $n_\ell = 9\cdot 4^\ell - 6\cdot3^\ell$. The total area of the yellow triangles is $\frac{n_\ell}{9^{\ell+1}} = \left(\frac49\right)^\ell - \frac{2}{3^{\ell+1}}$. Since this converges to $0$ as $\ell \to \infty$ and we know the area of Koch snowflakes converge to $\frac85$. The area of descendants of hexagon also converge to $\frac85$.

As a result, the area of the generalized Koch snowflake is $\frac85\left/\frac23\right. = \frac{12}{5}$ of that of the seed hexagon.

  • $\begingroup$ Thanks for showing my guess was right. I don't think it can, but can this same method of comparing the generalized snowflake with the Koch snowflake be used for other values of $n$? $\endgroup$ May 18, 2019 at 18:05
  • $\begingroup$ @automaticallyGenerated I don't think so. This work because the vertices generated by hexagon and triangle share the same hexagonal lattice (after suitable scaling). It seems a lattice is possible only for $n = 3, 4, 6$. $\endgroup$ May 18, 2019 at 18:12

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