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/
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