As I wrote in my previous post, we draw the Koch snowflake by starting with an equilateral triangle, and then following some simple rules ad infinitum. At each iteration we will have a snowflake, a regular polygon of sides, each of length . As we have seen before, at each iteration, the number of sides of the snowflake is
and the length of each side is given by
The perimeter of the snowflake is simply
where is the perimeter of the initial snowflake (i.e., the equilateral triangle whose side is ). Note that the perimeter of the snowflake is given by a geometric progression of common ratio (which is ), and therefore the progression will diverge: the perimeter of the snowflake will tend to infinity as . The Koch snowflake has a finite area, but a boundary of infinite length! Wow! Ain’t that cool!?