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:

von_koch_curve.gif

[ 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.

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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{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.

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2 Responses to “Area of the Koch Snowflake”

  1. Perimeter of the Koch Snowflake « Reasonable Deviations Says:

    [...] Reasonable Deviations :: alea jacta est :: « Area of the Koch Snowflake [...]

  2. jkwiens.com: Answer: Area of a Koch Snowflake Says:

    [...] 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 [...]

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