The Koch Snowflake

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IB HL Math Year 1

Felipe F Fagundes

        The purpose of doing this investigation is so that we can learn about the different patterns that the Koch Snowflake presents. The simple fact of the Koch Snowflake is a fractal already says a lot about it. A fractal is a figure which repeats itself in smaller scales. For example if stage 12 was looked closer stage two would be found and so would be stage 3 and so on.

        In this investigation we explored the patterns that emerge from one stage to the other. We tried to find any patterns for when n= 0,1,2,3 that could be applied to find any other n term. We specifically looked for patterns in N= the number of sides, L= the length of one side, P= the total perimeter, and A= the total area. And in order to come up with these patterns we used many resources like drawing in order to prove that our results were accurate.

N= I was able to come up with the number of sides (N) by counting the number of sides of the triangles in the drawing of the fractals. When I had to count the number of sides in stage 3 I was able to see that a pattern had already been established. I was able to notice that the answer was increasing constantly by four (4). Using the pattern I was able to find I multiplied 48 which was my answer for n=2 by 4.  I got the answer that n=3 is equal to 192. In order to prove that my answer was accurate I had to actually count the number of sides in n=3. In order to do that I got in the internet a high resolution drawing of stage 3 this allowed me to precisely count the number of sides. The fact that I got 192 proved that a pattern had already been established for N.

L= I was able to come up with the length of each sides by using the property of fractals. In stage 0 I knew that the side length of each triangle was 1. Since I knew that in order to add a new side I had to make it ,1-3.the size of the previous triangle I know the base of the triangle would be ,1-3. of the previous triangle and so would be the other sides since it is a equilateral triangle. And so I was able to realize that there was a pattern that each time the side length would get smaller by,1-3. . So for the next stages I was able to simply multiply my previous answer by ,1-3. . For example the side length in stage 0 is 1. If I multiply that by ,1-3. I will get the side length of stage 1 which is,1-3.. And if I multiply that by ,1-3. once again I will be able to get the side length for stage 2 which is,1-9..

P= In order to come up with the perimeter of each side I simply had to multiply my answers obtained for N and L. Since when finding the perimeter the length of all sides are added together, I was able to realize that if I multiplied the answers I would get my answer for perimeter. In order to prove my idea I estimated my answer for n=2 which was,16-3.. I than counted added up the length of every side with my calculator using the drawing of stage 2. And I was able to get the same answers proving my way of quickly getting the perimeter of a fractal.

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A= In order to calculate the area of each stage I had to use the formula A=,1-2.×𝑏× which is the formula to find the area of a triangle. In order to find the height of each triangle I had to use the Pythagorean theorem (a2+b2=c2) (Note that in the drawings of triangles and when I solved I used b instead of h for the height of the triangle.)  For every stage I would find the area of one small triangle than multiply that by the number of new triangles I added in this stage and then I would add the answer ...

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