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The Koch Snowflake

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Introduction

The Koch Snowflake:

Description:

In 1904 the swedish mathematician Helge Von Koch created the Koch snowflake. The Koch snowflake is a fractal curve, a geometric shape which can be subdivided into various parts. This fractal curve is built starting with a normal equilateral triangle (stage 0) and then to get to the following stage the inner third of each side is removed and another equilateral triangle is built where the side was removed. This process can be repeated various times. The first 4 stages are shown below:

image00.png

      Stage 0                            Stage 1                             Stage 2                          Stage 3

The aim of this investigation is to find an expression linking the area of all stages.

Method:

Part 1:

We’ll start by creating a table that shows various values at each different stage

N n = the number of sides of each stage

L n = the length of a single side of each stage

P n = The length of the perimeter of each stage

A n = The area of the snowflake of each stage

Stage 0

Stage 1

Stage 2

Stage 3

N n

3

12

48

192

L n

1

1/3

1/9

1/27

P n

3

4

5 1/3

7 1/9

A n

image01.png

image09.png

image16.png

image23.png

...read more.

Middle

image17.png

Total area of Stage 2 =  image11.png  +  image18.png

= image19.png

=image16.png

image20.png

Area of Stage 3 = Area of Stage 2 + Area of 48 small shaded triangles

image12.png

Height= cos -1 ((image21.png) / (image22.png)) = 60º = ½

= tan 60 (image05.png) x image21.png = image24.png

Area of shaded small triangle

= image22.png x image24.png

          2

= image25.png

Total area of Stage 3 = image16.png + image26.png

= image27.png

=image23.png

N n ) The number of sides increases from stage to stage by a factor of 4. This is because we remove the inner third of each stage and we then build an equilateral triangle where it was removed, therefore the number of sides of each stage is equal to the number of sides of the previous stage, plus 3 (number of sides of the equilateral triangle) times the number of sides of the previous stage. This is equal to 4 times the number of sides of the previous stage.

L n ) The length of a single side decreases by a factor of 3 because the inner third of each side is removed therefore if the original length is 1 and you decrease it by a factor of 3 (multiply by 1/3) you’ll get the following length to be 1/3

P n )

...read more.

Conclusion

image54.png, becomes smaller and smaller and will be converging to zero therefore the Area for the nth stage must be converging to a certain number so it must be finite.                                                                                                                                    

In order to find the number to which it converges we can simply use the spreadsheet and from looking at a part of the spreadsheet below we can see that as n is getting larger the value for An is remaining the same therefore it must be converging to that number.

We can see that as n becomes larger than 25 (as n image51.png) the area remains to be 0.692820323 therefore it’s converging to that value. To find an exact value for this I can simply use my trigonometry knowledge…image56.png= 0.7071067812, therefore by trial and improvement I discovered that 0.692820323 = image57.png which in turn is equal toimage58.png.image55.png

In conclusion we can state that the Koch Snowflake has an infinite perimeter and a finite area as n image51.png. We can also see that as n image51.png, the length of each side (Ln ) decreases and the number of sides Nn increases.

...read more.

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