# Fractal Geometry

## Von Koch Snowflake Algorithm

One of the simplest examples of a classic fractal is the von Koch "snowflake curve". Created in 1904 by the Swedish mathematician Helge von Koch, the snowflake curve has a truly remarkable property, as we will see shortly. But, let's begin by looking at how the snowflake curve is constructed. The initiator of this curve is an equilateral triangle with side $s=1.$ Let ${P}_{1}$ be the perimeter of curve 1, then ${P}_{1}=3.$

Trisect each side of the triangle. On the middle third of each of the three sides, construct an equilateral triangle with sides of length 1/3. Remove the base of each of the three new triangles. The result is a six-pointed "Star of David". Curve 2 is the result of the first iteration of the Koch snowflake algorithm.

Repeat the above steps on each of the twelve sides of curve 2 to get curve 3, the second iteration of the Koch snowflake.

Repeat the process on the 48-sided curve 3. You should be able to see the similarity between this third iteration of the Koch curve and an actual snowflake.

Repeat the process ad infinitum to complete the Koch snowflake curve.

### Perimeter of the Koch Snowflake

Recall that the initiator of the Koch snowflake curve is an equilateral triangle with side $s=1.$ Let ${P}_{1}$ be the perimeter of the triangle, then ${P}_{1}=3.$ At the conclusion of the first iteration, each side of the triangle has been trisected and reconstructed to become four sides of the second figure. The four sides have a length of 4/3. It follows that the perimeter of curve 2 is

$\begin{array}{l}{P}_{2}={P}_{1}\left(\frac{4}{3}\right)\\ {P}_{2}=3\left(\frac{4}{3}\right)\end{array}$

Repeat the process to get curve 3 with perimeter

$\begin{array}{l}{P}_{3}={P}_{2}\left(\frac{4}{3}\right)\\ {P}_{3}=3\left(\frac{4}{3}\right)\left(\frac{4}{3}\right)\\ {P}_{3}=3{\left(\frac{4}{3}\right)}^{2}\end{array}$

Repeat the process again to get curve 4 with perimeter

$\begin{array}{l}{P}_{4}={P}_{3}\left(\frac{4}{3}\right)\\ {P}_{4}=3{\left(\frac{4}{3}\right)}^{2}\left(\frac{4}{3}\right)\\ {P}_{4}=3{\left(\frac{4}{3}\right)}^{3}\end{array}$

Continue the process to derive the general formula for the perimeter of the Koch snowflake.

${P}_{n}=3{\left(\frac{4}{3}\right)}^{n-1}$
Perimeter of the Koch Snowflake
n ${P}_{n}$
1 3
2 4
3 5.3
10 39.95
20 709.51
40 223,734
100 7,020,000,000,000

The table at the right lists the perimeter of the Koch snowflake at various stages of construction. It appears that as

Also notice that the sequence of perimeters listed below form a geometric sequence with ratio $r=\frac{4}{3}.$

Since $r>1,$ the sequence of diverges. Thus $\underset{n\to \infty }{\mathrm{lim}}{P}_{n}=\infty .$ In other words, as n increases without bound, the perimeter of the snowflake curve becomes infinitely long. When we consider the area of the Koch snowflake in comparison to its perimeter, we stumble upon a surprising, paradoxical property.

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