Van Koch Snowflake Perimeter and Area

Date: 10/31/98 at 02:12:02
From: Brad Dooley
Subject: Van Koch's snowflake

I am required to find the answer to this problem for homework. It's 
more of a research thing as we haven't done fractals yet, and I have 
no idea whatsoever how to do it.

Question: If the Van Koch's snowflake process is repeated indefinitely, 
find the perimeter and the area, given an initial area 'A'.

Thank you in advance for any help you can give.
Brad Dooley

Date: 10/31/98 at 15:27:44
From: Doctor Anthony
Subject: Re: Van Koch's 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 triangle pointing outward 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 
outward 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, the perimeter being increased at each stage 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.  

The area follows the pattern:
 
   A[1 + 1/3 + 4/27 + 16/243 + 64/2187 + ....  ]

where after the first term we have a geometric progression with common 
ratio 4/9. Thus, the above is:

   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/