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.
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
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. Repeat the process: take every straight line segment,
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} ] Construction: The Iterative Process We begin with an
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 ]
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] ]
