Area of the Koch Snowflake

A Koch Snowflake is a fractal curve that can be constructed quite easily. Start with an equilateral triangle, and then iterate as follows:

Divide each line segment into three segments of equal length.
Draw an equilateral triangle that has the middle segment from step 1 as its base and points outward.
Remove the line segment that is the base of the triangle from step 2.

The following animated image illustrates the first seven iterations of this algorithm:

[ source ]

What is the area of the Koch snowflake at each iteration $k \in \mathbb{N}$ ? I have solved this problem years ago, when I was in high school. Nonetheless, a post on the Koch snowflake at Jeffrey Wiens‘ blog made me want to solve the problem yet once again.

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My solution:

I will use the following variables:

$A_k$ : area of the snowflake.

$n_k$ : number of sides of the snowflake.

$l_k$ : length of the sides of the snowflake.

where the subscript denotes iteration $k \in \mathbb{N}$ . Let us also define function $a(s) = \frac{\sqrt{3}}{4} s^2$ , which gives us the area of the equilateral triangle whose side is $s > 0$ . The variables above are initialized as follows:

$A_0 = a(x), \quad n_0 = 3, \quad{} l_0 = x$

since that the initial snowflake is an equilateral triangle with side $x$ . It can easily be seen that the length of each side is reduced by a factor of $3$ with each iteration, and therefore

$l_{k+1}= 3^{-1} l_k$ .

Since $l_0 = x$ , we have $l_k = 3^{-k} x$ . It can also be seen that the number of sides is increased by a factor of $4$ with each iteration, and therefore

$n_{k+1} = 4 n_k$ .

Since $n_0 = 3$ , we have $n_k = 4^k 3$ . The area of the snowflake at each iteration is given by the following recursion

$A_{k+1} = A_k + n_k a(l_{k+1})$ .

We know that $A_0 = a(x)$ , and we can now compute the area at each iteration:

$A_1 = A_0 + n_0 a(l_1) = a(x) + 3 a(x/3)$

$A_2 = A_1 + n_1 a(l_2) = a(x) + 3 a(x/3) + 4.3 a(x/3^2)$

$A_3 = A_2 + n_2 a(l_3) = a(x) + 3 a(x/3) + 4.3 a(x/3^2) + 4^2 3 a(x/3^3)$

and, finally, we obtain

$A_k = a(x) + \displaystyle\frac{3}{4} \displaystyle\sum_{m=1}^k 4^m a(x/3^m)$ .

Note that

$a(x/3^m) = a(3^{-m}x) = 3^{-2 m} a(x) = 9^{-m} a(x)$

and therefore we can compute the area of the snowflake at each iteration

$A_k = a(x) \left[1 + \displaystyle\frac{3}{4} \displaystyle\sum_{m=1}^k (4/9)^m \right]$ .

We have a sum of $k$ terms of a geometric progression with common ratio $r = \frac{4}{9}$

$\displaystyle\sum_{m=1}^k r^m = r \displaystyle\sum_{m=0}^{k-1} r^m = r \left[\frac{1-r^k}{1-r}\right] = \left(\frac{r}{1-r}\right)(1 - r^k)$ ,

and therefore

$\displaystyle\sum_{m=1}^k (4/9)^m = \left(\frac{4/9}{1-4/9}\right)(1 - (4/9)^k) = \frac{4}{5} (1 - (4/9)^k)$

and finally we have the area of the snowflake at iteration $k$

$A_k = a(x) \left[\displaystyle\frac{8}{5} - \displaystyle\frac{3}{5} \left(\frac{4}{9}\right)^k\right]$ .

Note that the Koch snowflake is a fractal curve, not a polygon. Therefore, the Koch snowflake is the curve to which the snowflake “converges” when we apply the recursion above ad eternum. Note also that for finite $k$ we have a polygon of $n_k$ sides, not a fractal curve. Having said that, the area of the Koch snowflake is thus given by

$\displaystyle\lim_{k \rightarrow +\infty} A_k = \displaystyle\frac{8}{5} a(x)$ .

Since $a(x) = \frac{\sqrt{3}}{4} x^2$ , we have

$\displaystyle\lim_{k \rightarrow +\infty} A_k = \displaystyle\frac{8}{5} \frac{\sqrt{3}}{4} x^2 = \displaystyle\frac{2 \sqrt{3}}{5}x^2$ .

The area of a Koch snowflake is thus finite if $x$ is finite.

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Some links:

Koch Snowflake

Area of the Koch Snowflake

Koch curve

Recursion and making a Koch Snowflake with Maple

Koch snowflake

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Tags: Fractals, Koch Snowflake

This entry was posted on December 26, 2007 at 22:52 and is filed under Mathematics, Puzzles. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “Area of the Koch Snowflake”

Sreema Says:
October 29, 2012 at 03:50 | Reply
Hi, I am just wondering, how did you come up with the formula for the are of the snowflake at iteration k? More specifically, how did you get the “8/5 – 3/5″ section in the formula. I am quite confused about that, A reply would be much apprieciated. :)
- Rod Carvalho Says:
  November 2, 2012 at 09:32 | Reply
  $\begin{array}{rl} A_k &= a(x) \left[1 + \displaystyle\frac{3}{4} \displaystyle\sum_{m=1}^k (4/9)^m \right]\\\\ &= a(x) \left[1 + \displaystyle\frac{3}{4} \left(\frac{4}{5} (1 - (4/9)^k)\right) \right]\\\\ &= a(x) \left[1 + \displaystyle\frac{3}{5} (1 - (4/9)^k) \right] \\\\ &= a(x) \left[\displaystyle\frac{8}{5} - \displaystyle\frac{3}{5} \left(\frac{4}{9}\right)^k\right]\end{array}$

Rod Carvalho

Area of the Koch Snowflake

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