Answer: Area of a Koch Snowflake

January 3rd, 2008

I am a bit late posting a solution to this problem because I was on holidays. However, I am now back to my regular 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 Koch Snowflake. Since the problem has already been solved, I will only do a quick overview of the solution.

Answer: The easiest way to solve this problem is to calculate the area added to the Koch Snowflake after each iteration.

Area of the first iteration can easily be calculated by using Pythagoras.

x^2 = (\frac{x}{2})^2 + h^2
h^2 = \frac{3}{4} x^2
h = \frac{\sqrt{3}}{2} x

\therefore area = (\frac{\sqrt{3}}{2} x)(\frac{x}{2}) = \frac{\sqrt{3}}{4} x^2

Using the solution for the first iteration, we can easily calculate the area of the second iteration. The area will be the area of the original equilateral triangle plus the area of 3 smaller equilateral triangles.

area = \frac{\sqrt{3}}{4} x^2 + 3 \frac{\sqrt{3}}{4} (\frac{x}{3})^2

The area of the third iteration would, therefore, be:

area = \frac{\sqrt{3}}{4} x^2 + 3 \frac{\sqrt{3}}{4} (\frac{x}{3})^2 + 3 \cdot 4 \frac{\sqrt{3}}{4} (\frac{x}{3^2})^2

If we continue doing this, we will get the following summation.

area = \frac{\sqrt{3}}{4} x^2 + 3 \frac{\sqrt{3}}{4} (\frac{x}{3})^2 + 3 \cdot 4 \frac{\sqrt{3}}{4} (\frac{x}{3^2})^2 + 3 \cdot 4 \cdot 4 \frac{\sqrt{3}}{4} (\frac{x}{3^3})^2 \ldots
area = \frac{\sqrt{3}}{4} x^2 + \sum_{i=0}^{\infty}3 \cdot 4^i \frac{\sqrt{3}}{4} \frac{x^2}{3^{2(i+1)}}
area = \frac{\sqrt{3}}{4} x^2 + \sum_{i=0}^{\infty} \frac{3^{\frac{3}{2}}}{4} x^2 \frac{4^i}{9^{i+1}}
area = \frac{\sqrt{3}}{4} x^2 + \sum_{i=0}^{\infty} \frac{1}{4 \sqrt{3}} x^2 (\frac{4}{9})^i

Since the summation is a geometric series, we know that \sum_{i=0}^{\infty} a r^i = \frac{a}{1-r} when r < 1.

Therefore,

area = \frac{\sqrt{3}}{4} x^2 + \frac{1}{4 \sqrt{3}} x^2 \cdot \frac{1}{1 - \frac{4}{9}}
area = \frac{\sqrt{3}}{4} x^2 + \frac{1}{4 \sqrt{3}} x^2 \cdot \frac{9}{5}
area = \frac{2 \sqrt{3}}{5} x^2

Comment by rod

Made Thursday, 3 of January , 2008 at 10:09 pm

This is a neat solution. It helps to dislay the snowflake at each iteration.

By the way, did you notice the generalization I wrote about in one of the comments in my blog? To cut a long story short, let’s say that the triangles “radiate” inwards, not outwards. That would mean that we would remove area from the initial triangle, instead of adding area. Do you think that the inwards-radiating triangles will overlap? If they do overlap, then there’s a big problem because we will be removing the same area twice, and in that case, the simple recursion formula does not work.

Comment by eldila

Made Thursday, 3 of January , 2008 at 11:29 pm

I have been skimmming the comments about the “radiate” inward snowflake. I have a feeling that the inwards-radiating triangles would overlap, but I haven’t proven it yet. I will look into it more.

Comment by Kellen Myers

Made Friday, 26 of December , 2008 at 1:00 pm

The “inward-radiating” snowflake is called an anti-snowflake. It does NOT have double-counting (ie no overlap). The same formula works.

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