# Koch's Snowflake

Is that true that if a sequence of curves converges (in some sense) to another curve then their lengths converge to the length of the latter? There is an example when this is so and another one when this is not true.

The applet below illustrates an additional possibility where the limit curve (which does exist) has no finite length at all, although each of the curves in the sequence does. Appropriately, the lengths of the curves grow without bound.

What if applet does not run? |

Start with a triangle - a collection of three line segments each of length, say, 1. To these we shall apply iterative modifications each producing another collection of line segments. After every iteration, the line segments so produced form a curve - a broken line that consists of four times more line segments, albeit each smaller than the ones on the previous iteration.

First we have a triangle (i.e. three line segments) of total length 3. Each of these is divided into three equal parts of which the two extreme are kept whilst the middle one is replaced with a "tent" formed by the sides of the equilateral triangle formed on that middle segment as the base. Obviously, each of the smaller segments have length 1/3; four of them that were created on one of the sides of the triangle have the total length of 4/3, and the whole curve that now consists of 12 segments has the length of 4.

On the next iteration we apply the same procedure to each of these 12 segments. There are going to be 48

The process continues. On iteration number n, the curve consists of 3×4^{n} line segments, each of length 1/3^{n}, to the total of _{n} = 3×(4/3)^{n}.*Koch's snowflake* does not have a finite length even though it is located in a bounded region around the original triangle.

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

71228914