Perimeter and Area of the Koch SnowflakeDate: 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/ |
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