Area of the Koch Snowflake « Reasonable Deviations

Area of the Koch Snowflake

By Rod Carvalho

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

step 1: divide each line segment into three segments of equal length.

step 2: draw an equilateral triangle that has the middle segment from step 1 as its base and points outward.

step 3: remove the line segment that is the base of the triangle from step 2.

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

[ image source ]

What is the area of the Koch snowflake at each iteration $k \in \mathbb{Z}_0^{+}$ ? 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.

__________

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 $k \in \mathbb{Z}_0^{+}$ . 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.

__________

Some links:

Koch Snowflake

Area of the Koch Snowflake

Koch curve

Recursion and making a Koch Snowflake with Maple

Koch snowflake

Tags: Fractals, Koch Snowflake

This entry was posted on December 26, 2007 at 10:52 pm and is filed under Brain Teasers, Mathematics, Recreational Math. 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”

Perimeter of the Koch Snowflake « Reasonable Deviations Says:
December 28, 2007 at 4:17 am | Reply
[...] Reasonable Deviations :: alea jacta est :: « Area of the Koch Snowflake [...]
jkwiens.com: Answer: Area of a Koch Snowflake Says:
January 4, 2008 at 5:58 am | Reply
[...] routine. As for finding the area of the Koch Snowflake, Reasonable Deviations has already posted a solution to this problem. Also, Rod has posted a variation of the problem which finds the perimeter of a [...]

Reasonable Deviations

Area of the Koch Snowflake

2 Responses to “Area of the Koch Snowflake”

Leave a Reply