Perimeter and Area of the Koch Snowflake

Date: 12/02/98 at 12:25:16
From: Anne Clayton
Subject: Snowflake problem

Have you heard of the snowflake problem? To set it up, start with an 
equilateral triangle. Then divide each side of the triangle into 
thirds. Where the side is divided (the cut portions) make another 
equilateral triangle, so the above side looks like this:

         ^
        / \      
       /   \                
   ---  ---  ---  (imagine this is equilateral)  

Then you do the same steps as above repeatedly, so the original 
triangle looks like a snowflake. Now the question: prove that the 
perimter is infinite and the area is finite.

Date: 12/02/98 at 13:35:33
From: Doctor Anthony
Subject: Re: Snowflake problem

You are probably thinking of a figure like the Koch Snowflake. To draw 
this you start with an equilateral triangle of side a. Now divide each 
side into three equal parts and on the middle third of each side 
construct an equilateral triangles pointing outwards from the original 
triangle. The total perimeter is now (4/3)(3a) = 4a. We now further 
subdivide each straight edge into 3 parts and construct equilateral 
triangles on the middle third of each side - again pointing outwards 
from the original figure. This process will enlarge the perimeter by a 
further factor of 4/3. There is no overlapping of the extra sides with 
those already present. The above process is repeated indefinitely, at 
each stage the perimeter being increased by a factor of 4/3. So we 
have:

   perimeter = (3a)(4/3)(4/3)(4/3) ...  to infinity.

Clearly the perimeter will increase without bound and become infinite, 
but the area of the figure will be less than the area of the 
circumcircle of the original equilateral triangle. So this figure has 
an infinite perimeter but a finite area. It is the defining property 
of a fractal shape that has self-similarity to an infinite depth. That 
is, you can enlarge a portion of the boundary to any magnitude and find 
shapes similar to the original figure.

If A = area of original triangle then the area follows the pattern:
 
   A[1 + 1/3 + 4/27 + 16/243 + 64/2187 + ... 
        where after the first term we have a GP with common ratio 4/9
    = A + A[1/3 + 4/27 + .....]
    = A + A(1/3)/(1-4/9)
    = A + A(1/3)/(5/9)
    = A + 3A/5
    = 8A/5 

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/