<The von Koch Snowflake Investigation> properties of fractals is self-similarity. It means that we can magnify them many times and after every

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1 Jiwon MYP 5 Math Ewa Puzanowska 18th of Oct 2012 <The von Koch Snowflake Investigation> About Fractal... In geometry, a fractal is a shape made up of parts that are the same shape as itself and are of smaller and smaller sizes. It is a never ending pattern that repeats itself at different scales. The most important properties of fractals is self-similarity. It means that we can magnify them many times and after every step we will see the same shape. Fractals may be exactly the same at every scale, or, they may be nearly the same at different scales. We can easily see fractals all over the nature in a different scales. We find the same patterns again and again, from the tiny branching of our blood vessels and neurons to the branching of trees, lightning bolts, and river networks. Regardless of scale, these patterns are all formed by repeating a simple branching process. One of the most famous fractal, The Sierpinski Triangle, is made by repeatedly removing the middle triangle from the prior generation. The number of colored triangles increases by a factor of 3 each step,

2 1,3,9,27,81,243,729, etc. -The Sierpinski Triangle- Another famous fractal, The Koch Curve, which is made by repeatedly replacing each segment of a generator shape with a smaller copy of the generator. At each step, or iteration, the total length of the curve gets longer, eventually approaching infinity. Much like a coastline, the length of the curve increases the more closely you measure it. -The Koch Curve- Below are fractals that I have made. At first I had 7 points and 6 lines. Then I pulled some points to make the shape of clown. As I increased the iterations I could see that a fractal is made by repeating a simple process again and again.

3 It was clear that at each iteration, the total length of the curve got longer eventually approaching infinity. I could also see that the length of the curve increases the more closely as I measure it. Below are more fractals that I ve made, 1 iteration > 2nd iteration 1 iteration > 4th iteration 1 iteration > 4th iteration

4 What is von Koch Snowflake? Swedish mathematician Helge von Koch introduced the "Koch curve" in Starting with a line segment, recursively replace the line segment. von Koch Snowflake is also one of the fractals built by starting with an equilateral triangle instead of line, using the same rule as other fractals. Removing the inner third of each side and building another equilateral triangle at the location where the side was removed. The steps in creating the Koch Curve are then repeatedly applied to each side of the equilateral triangle, creating a "snowflake" shape. In this mathematical task, I am going to investigate how the area and perimeter of a shape/curve changes and find out whether they increase by the same number every time, as the following process is repeated. iteration 1st 2nd Perimeter 3L 4L 3rd 16/3L Looking at the table, I ve realized that no matter which step it is, if I multiply 4/3 to it, I ll get the next step. For example, If I multiply 4/3 to the perimeter of the 1st iteration, which is 3L, I ll get 4L, which is the perimeter of the 2nd iteration. And the same thing applies to the 2nd iteration, 3rd iteration and so on. How does this work? Let s first look at the 1st and the 2nd iteration.

5 If supposed that the length of one side of the 1st iteration triangle is L, As we replaced 1/3 of the L with two 1/3L, the total perimeter of the 2nd iteration triangle will be, 3L - 3(1/3L) +3(2/3L) = 3L + L = 4L Here we can simplify the formula as 3L(1+1/3) and can see how we got the perimeter of the 2nd iteration triangle by multiplied 4/3 to the original triangle. Now, let s look at the 2nd and the third iteration, Just looking in to one small triangle, which each one side is 1/3L, We once again replace 1/3 of 1/3L, which is 1/9L, with two 2/9L. So the total perimeter of the 3rd iteration triangle will be, 4L(Which is the 2nd triangle's perimeter) - (2*1/9L*6) + (2*1/9L*12) = 4L - 4/3L+8/3L = 4L + 4/3L = 16/3L Here we can simplify the formula as 4L(1+1/3) and can clearly see how we got the perimeter of the 3rd iteration triangle by multiplied 4/3 to the 2nd iteration triangle. Seeing from above formula I can even predict what the perimeter for iteration 4 to 6 would look like. Iteration Perimeter 4th 16/3L(4/3) = 64/9L 5th 64/9L(4/3) = 256/27L 6th 256/27L(4/3) = 1024/81L While demonstrating, I found that if I want to find the perimeter of the certain iteration, I just need to

6 multiple 4/3 to the perimeter of the previous stage triangle. So it could be expressed as, Perimeter = Number of sides * Length of a slide P = 3*(4/3)^n Considering this, I could generalize this theory. The total perimeter for the nth stage would be : Pn = 4/3 Pn-1 Conclusion : As the perimeter grows by 4/3 (common ration) in each iteration, which is greater than 1, when n gets larger, the perimeter of each new curve will continue to increase with no bounds. The value of perimeter after each iteration will get close to infinity Now, let s look at the area if the Von Koch Snowflake. First of all, let s suppose that the area of the first iteration triangle is A. Below is the table that I have made to show the area of a triangle. Stage Area of a small triangle Number of triangle Area of the added triangle 1 A 1 A 2 1/9A 3 1/3A 3 1/9 of 1/9A = (1/3^4)A 12 4/3^3A 4 1/9 of 1/81A = (1/3^6)A 48 4^2/3^5A 5 1/9 of 1/729A = (1/3^8)A 192 4^3/3^7A 6 1/9 of 1/2187A = (1/3^10)A 768 4^4/3^9A Stage Total Area of the triangle 1 A 2 A + 1/3A = 4/3A

7 3 4/3A + 4/3^3A = 40/27A 4 40/27A + 4^2/3^5A = 376/243A 5 376/243A + 4^3/3^7A =1.5766A A + 4^4/3^9A = A Looking at the table, I ve realized that no matter which step it is, the area can be get from adding the total sum of the previous area and the area of the extra added equilateral triangles after each iterative process. I noticed the pattern that if I multiplied For example, If I multiply 4/3 to the perimeter of the 1st iteration, which is 3L, I ll get 4L, which is the perimeter of the 2nd iteration. And the same thing applies to the 2nd iteration, 3rd iteration and so on. How does this work? Let s first look at the 1st and the 2nd iteration.

8 Looking at the second iteration triangle, we can notice that 3 more little triangle were added. As the ratio of the length between two triangle is 1:3, the ratio of the area between two triangle would be 1^2 : 3^2 = 1 : 9. Thus, the area of the small triangle would be 1/9 of A. So the area of the 2nd iteration would be A + 3*(1/9A) = 4/3A. Now, let s look at the 2nd and the third iteration, Here, we can see that : to the 2nd iteration triangle, 12 more small little triangle were added. We can also notice that the ratio of the length between this triangle and this small triangle is 1:3. Thus, the ratio of the area between two triangle would be 1^2 : 3^2 = 1 : 9. However as the area of this triangle is already 1/9A, the area of the small triangle would be 1/9 of 1/9A = 1/81A. So the area of the 3rd iteration would be 4/3A + 12*(1/81A) = 40/27A. Lastly, let s look at the 3rd and the 4th iteration, Similarly, in 4th iteration triangle, we can once again see that to the 3rd iteration triangle, 48 more small little triangle were added. We can also notice that the ratio of the length between this triangle and the

9 smaller triangle is 1:3. Thus, the ratio of the area between two triangle would be 1^2 : 3^2 = 1 : 9. However as the area of this triangle is already 1/81A, the area of the small triangle would be 1/9 of 1/81A = 1/729A. So the area of the 3rd iteration would be 40/ *(1/729A) = 376/243A. From above examples, I could find some repeated pattern while finding the area of the each triangle. The number of triangle increased by 4(except for the first iteration). The area for the nth iteration of Von Koch curve, can be expressed as : An-1 + (4^n-2/3^2n-3)A. A n A nà A nà nà The area is the total sum of the previous area and extra equilateral triangles(which continue to be reduced in size) being added after each iterative process. Here, I ve noticed that while the perimeter increases with no limitation when N is infinitely large, the area reaches to a finite value. I can clearly prove this with one circle. Though the N is infinitely large, the area of this figure will not exceed the area over the circle around it. It clearly proves that the area of the Koch curve reaches to a finite value. In conclusion : As the Koch curve endlessly increases its number of sides, its perimeter increases infinitely, while its area reaches to a finite value.

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