Koch Snowflake - Part 2

This question is hard if you do not know where to start.

The statement about subdivision into 7 smaller snowflakes of the same size is wrong.
Proof by contradiction, assume it is correct. By considering the areas, the smaller snowflakes must each have side length $\frac{ 1} { \sqrt{7} } s$ . This clearly cannot fit into the snowflake.

What is true though, is that it can be subdivided into 7 smaller snowflakes. 6 of size $\frac{1}{3}$ in each of the 'corners' and 1 of size $\frac{1}{ \sqrt{3} }$ in the center.

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Here is a tessellation of the snowflake in the plane. It surprised me when I first saw it.

image

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Let's tackle an easy one, namely the area. Starting with an equilateral triangle of side length $s$ , it has area $X = \frac{ \sqrt{3} } { 4} s^2$ .
In the first step, there are 3 line segments.
In the second step, there are 12 line segments.
At the nth step, there are $3 \times 4^{n-1}$ line segments.

In the first step, the area of the small triangles added is $\frac {X}{9}$ .
In the second step, the area of the small triangles added is $\frac{X}{9^2}$ .
In the nth step, the area of the small triangles added is $\frac{X}{9^n}$ .

Hence, the total area is $X + \frac{3}{9} X + \frac{ 3 \times 4 } {9^2} X + \ldots + \frac{ 3 \times 4^{n-1} } { 9^n} X$ .
Ignoring the first two terms, we get a geometric progression with common ratio $\frac{4}{9}$ , and hence the sum is $\frac{ 3 \times 4 } { 9^2 } X \times \frac{ 1} { 1 - \frac{ 4}{9}} = \frac{ 4}{15} X$ . Hence, the total area is $\frac{ 8}{5} X = \frac{ 2 \sqrt{3} } { 5} s^2$ .
Thus, the statement on area is correct!

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I leave the fact about a non-differentiable perimeter to you. It will require some work.