Apollonian Snowflake
From Math Images
Apollonian Snowflake |
---|
Apollonian Snowflake
- This is a combination of the Apollonian Gasket and the Koch Snowflake, both of which are fractals. The result will be an endless fractal made from two existing fractals. Its really a half-Apollonian Gasket because I'm only iterating the largest inscribed circle for each triangle. To start, a construction of the Koch Snowflake must be made because this will be the layout for the circles. On Geometers Sketchpad (GSP, a very useful program that I highly recommend to those who read this page) I am able to make a Koch Curve tool. This tool will let me apply the Koch Curve to any line segment. (Look at the bottom for specific instructions). Simultaneously, I add inscribed circles to my Curve tool. These circles will be my Apollonian aspect of my image. When I increase the iterations of the snowflake, the circles will iterate as well. As cool as this sounds already, adding colors will really make the snowflake ten times more awesome.
Contents |
Basic Description
The goal is to have an Apollonian Gasket in a Koch Snowflake since both fractals are endless. There will be circles within triangles, which make up the Koch Curve. The Koch Curve Tool that I made in Geometers sketchpad only takes care of one part. Where are the circles? By incorporating circles into my tool, the circles and triangles will make more copies of itself as I increase the iteration. The basis of this Koch Curve tool is shown in Iteration 0 (picture below). The more circles and the more detail I add to my tool (size, color, etc.) the more detailed the Snowflake will be. Below are the Iterations that I used for my image.
CREATING THE KOCH CURVE TOOL
Draw a horizontal line segment with points A and B. Mark point A as the center and click on B. Click Dilate under the Transform menu. Set the ratio to 1/3 and click Dilate. Mark point B as the center and click point A. Go under the Transform menu and click Dilate. (The ratio should still be 1/3. Click dilate). Now points A’ and B’ are at one third increments of the whole line segment. Mark A’ as the center and rotate point B’ 60 degrees. Select point B’ again but this time Dilate it to the ratio ½. Mark that as the center. Click point B’’ (the point rotated 60 degrees) and dilate it to ratio 2/3. Hide the original line segment, segment AB. Connect the points. The two points in the middle will form the circle (center and radius). Select points A and B and click Iterate. Iterate for all the line segments. |
This is how I organized my areas, by using different sizes of triangle in the Koch Snowflake to help find the area. Type 1 is the largest triangle and Type 5 is the smallest triangle.
A More Mathematical Explanation
- Note: understanding of this explanation requires: *Fractals
FINDING THE AREA OF THE KOCH SNOWFLAK [...] |
FINDING THE AREA OF THE KOCH SNOWFLAKE AND ITS FUNCTION |
Oohhh! Now the mathematical part. My first step was to find the area of the actual Koch Snowflake and see how the area changes. These tables show the area of the snowflake at each iteration, up to iteration 4. (In my tables, the variable A represents the side length of the triangle in Iteration 0. This is true for my whole page.) I made the last chart to see how the area at each iteration changes as A increases.
Okay so now it is simple to see how the area increases with each iteration. But what I'm looking for is a function that when I sub in the Iteration number, I will be able to find the area. This calls for two variables, the iteration number and a constant. The constant is the length of the side of the triangle in Iteration 0. By looking at the charts, I see that the areas are expressed through a long string of math. The string is made up of segments and each segment corresponds to an iteration number. With each additional iteration, another segment of arithmetic is added onto the previous string. So the area at a specific iteration is dependent on the area of the previous iteration. Likewise, my function will have a similar pattern where the knowledge of the previous area will allow me to find the area of the current iteration. Confused??? Let me explain further.
Basically, I am trying to find out how the area changes with each iteration. The change is from where subscript n is the iteration number. If you look at the table above where I represent the areas in terms of A, there is an apparent pattern starting from Iteration 1. The pattern is represented by the expression where N is the iteration number and A is the length of the side of the triangle in iteration 0. is what's being added onto my string of math as I increase the iterations. So where N is the iteration number and A is the constant. However, the problem is that this function is unable to find the area at Iteration 0. THE AREA AT ITERATION 0 WILL ALWAYS BE GIVEN. This is the only way to find the area of the Koch snowflake using this function.
But this is how the area changes. What do I do if I want to find the area and all I have is the number of iterations? The function will help. For this function, the knowledge of the area of the previous iteration will help you find the area of the current iteration. The tedious part of this is that you would need to know all the areas of the iterations before. So if you wanted to find the area of Iteration 3, you would need to find the area at Iteration 2 and you can only find the area at Iteration 2 if you know the area of Iteration 1. The more iterations, the more tedious work must be done.
FINDING THE AREA OF THE CIRCLES AND ITS FUNCTION |
First off, I was able to inscribe two different circles in every equilateral triangle within the Koch Snowflake, large and small circles. So I had to find a function for finding the area of the large circles, a function to find the small circles, and a function to find the total area. Below are the circle types that I mentioned before.
Like with the area of the Koch Snowflake, I made a table to show the area of the circles. The first and second columns show the areas of the Large and Small circles for each iteration. The bolded numbers are the simplified version of each expression. The variable A is still the constant, or the length of the side of the triangle in Iteration 0.
Does this seem familiar? The area of the circles are represented in a long string of math, just like the area of the Koch Snowflake. The first term in the area of the large circles, , is constant and repeats in every iteration. The only thing that changes is the later terms. The function for the large circle area is . However, there is a catch. The iteration number tells you how many times you use the expression . So if you want to find the area of the large circles in iteration three, you use for n=3, n=2, and n=1 and then add . This goes the same for the area of the small circles. The function for finding the area of these circles is . You have to use the expression multiple times, depending on the iteration number and then add to it. So now to find the total area of the circles. I used the regression function on my calculator. I got the function . By just plugging in the iteration number, the area is calculated. Simple yes?
WHAT TO TAKE AWAY FROM THIS:
Area of Large circles ---> (IMPORTANT! Make sure to use the correct number of times) Area of Small Circles ---> (IMPORTANT! Make sure to use the correct number of times) |
</div>
Why It's Interesting
Honestly, there's a whole lot of patterns in these fractals. I've only talked about two, the areas of the Koch Snowflake and the circles. And plus, when you add colors to it, it just looks so freaking awesome. I mean look at that! When you add in color into the circles, the possibilities are endless. So yes there's math to this, lots of it. But the result is also beautiful to look at.
Teaching Materials
- There are currently no teaching materials for this page. Add teaching materials.
About the Creator of this Image
Victor has worked countless hours with this image, mastering Geometers Sketchpad along the way. The sweat and effort that he has put into this is what makes it amazing.
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.