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Investigation – Von Koch’s Snowflake Curve

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Adrian Zwierzchowski
2 IB
Investigation – Von Koch’s snowflake curve

In this investigation I am going to consider a limit curve named after the Swedish mathematician Niels Fabian Helge von Koch. I will try to investigate the perimeter and area of Von Koch’s curve.

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The Koch’s curve has an infinite length because each time the steps above are performed on each line segment of the figure there are four times as many line segments, the length of each being one-third the length of the segments in the previous stage.

First of all I am going to suppose c1 has a perimeter of 3 units. I will try to find the perimeter of c2, c3, c4 and c5. c1 s1 = 1 (s – side length) c2 s2 = 1/3 c3 s3 = 1/9 …show more content…

The surface of these rectangles is one ninth of the unit surface of the triangle drawn at the preceeding iteration.

The area of the curve on the next iteration continues to increase but to a smalller extent: 16 small triangles are added but their area are now 81 times smaller than the very first triangle of the Von Koch’s curve.

The following figures show the area increase:

|iteration |triangle |triangles |curve |
|number |number |area |area |
|1 |0 |0 |0.00 |
|2 |1 |1 |1.00 |
|3 |4 |0.44444 |1.44444 |
|4 |16 |0.19753 |1.64197 |
|5 |64 |0.08779 |1.72976 |
|... |... |... |... |
|10 |65536 |0.00152 |1.79878 |

Still, this very interesting property of the Von Koch curve: its area converges rapidly to a finite limit while the total length of the segments that composed that curve have no limit.

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becomes
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so 3 parts become 4 parts. cn has a perimeter of 3 · (1 ⅓) n-1. The total length increases by one third.

Where it comes to an area of the figure, it is equal to An = 2/5 · (√3) · a2 where a is the length of the one side of the equilateral triangle.

Now I am going to investigate the sequence of curves obtained by ‘pushing out’ squares on successive curves from

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