Snowflake By Haese Mathematics Site

Since ( A_0 = \frac{\sqrt{3}}{4} ), the final area is: [ A_{\infty} = \frac{8}{5} \cdot \frac{\sqrt{3}}{4} = \frac{2\sqrt{3}}{5} ]

Introduction One of the most captivating paradoxes in mathematics is the Koch Snowflake, a fractal curve first described by Swedish mathematician Helge von Koch in 1904. In the context of Haese Mathematics (particularly for IB Diploma Analysis & Approaches SL/HL), the snowflake serves as a perfect case study for geometric sequences , limits to infinity , and the surprising behaviour of infinite series . Construction: The Iterative Process We begin with an equilateral triangle of side length ( a_0 ). This is Stage 0. snowflake by haese mathematics

Using the sum of a geometric series with ratio ( r = \frac{4}{9} < 1 ), as ( n \to \infty ): [ A_{\infty} = A_0 \left[ 1 + \frac{1}{3} \times \frac{1}{1 - \frac{4}{9}} \right] = A_0 \left[ 1 + \frac{1}{3} \times \frac{9}{5} \right] = A_0 \left[ 1 + \frac{3}{5} \right] = \frac{8}{5} A_0 ] Since ( A_0 = \frac{\sqrt{3}}{4} ), the final

This simplifies to: [ A_n = A_0 \left[ 1 + \sum_{k=1}^n \frac{3 \times 4^{k-1}}{9^k} \right] = A_0 \left[ 1 + \frac{1}{3} \sum_{k=1}^n \left(\frac{4}{9}\right)^{k-1} \right] ] This is Stage 0

Repeat the process: take every straight line segment, divide it into three equal parts, and replace the middle third with two segments of that length.

For each side, remove the middle third and replace it with two segments of the same length (forming an equilateral "bump"). The number of sides increases.