[SPLIT] Koch snowflake: limiting value of area?
- Thread starter wind
- Start date
If the sides of the equilateral triangle are 1, then draw a line down the center from the point to bisect the base.
This makes 2 right triangles of which we can find the area.
Using Pythagoras we can find the height.
\(\displaystyle \L\\\sqrt{1^{2}-(\frac{1}{2})^{2}}=\frac{\sqrt{3}}{2}\)
\(\displaystyle A=\frac{bh}{2}\)
\(\displaystyle A=(\frac{1}{2})(\frac{\sqrt{3}}{2})=\frac{\sqrt{3}}{4}\)
The area of an equilateral triangle is given by \(\displaystyle \frac{\sqrt{3}}{4}s^{2}\)
Now, your next 'island' is the same as the first but with 3 smaller triangles added. There is a pattern.
The side length of the small triangles on the 2nd island is 1/3
This gives each an area of \(\displaystyle \frac{\sqrt{3}}{4}(\frac{1}{3})^{2}=\frac{\sqrt{3}}{36}\).
There are 3 with total area \(\displaystyle \frac{\sqrt{3}}{12}\)
Add this to the area of the original triangle: \(\displaystyle \frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{12}=\frac{\sqrt{3}}{3}\)
The side area of the next will have s=one third the previous. So, 1/3 of 1/3 is 1/9, and so on.
Here's the general case:
Actually, when we go from the nth island to the (n+1)st island, we have multiplied the number of sides by 4 and divided the length of each side by 3.
So, we have \(\displaystyle \L\\P_{n}=(\frac{4}{3})^{n}P_{0}\)
Where \(\displaystyle P_{0}\) is the perimeter of the original triangle.
If we let \(\displaystyle A_{n}\) be the combined areas of all the triangles up to, say, the nth island we have:
\(\displaystyle A_{n+1}=\frac{4}{9}A_{n}\), n=1,2,3,.......
If we add them all up, we have:
\(\displaystyle A_{1}+A_{2}+A_{3}+.......=A_{1}(1+\frac{4}{9}+\left(\frac{4}{9})^{2}+....\right)=\frac{9}{5}A_{1}\)
Let \(\displaystyle A_{0}\) be the area of the original triangle, then \(\displaystyle A_{1}+\frac{1}{3}A_{0}\). As we showed before.
So, \(\displaystyle \L\\\text{Area of the snowflake} =\sum_{n=0}^{\infty}A_{n}=\frac{8}{5}A_{0}\)
In your case the limit approaches:
\(\displaystyle \L\\\frac{2\sqrt{3}}{5}\)
The area of the (n+2)st island can be found by using \(\displaystyle \L\\\frac{\sqrt{3}}{12}(\frac{4}{9})^{n}\)
You have \(\displaystyle A_{0}\)(original) and \(\displaystyle A_{1}\)(6 pointed one)
The area of the third island would be \(\displaystyle \L\\\frac{\sqrt{3}}{12}(\frac{4}{9})^{1}=\frac{\sqrt{3}}{27}\)
Using the sum: \(\displaystyle \L\\\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{12}\sum_{0}^{\infty}(\frac{4}{9})^{n}=\frac{2\sqrt{3}}{5}\)
Which, as you can see, matches are above result of \(\displaystyle \L\\(\frac{\sqrt{3}}{4}){(\frac{8}{5})=\frac{2\sqrt{3}}{5}\)
Sorry to ramble on, but I always thought the Koch curve was very interesting. I just about spilled all I know about it.
A large part of this I derived myself.
Two other small things:
The perimeter is infinite, yet the area is finite.
It is continuous everywhere, but is not differentiable anywhere.
This makes 2 right triangles of which we can find the area.
Using Pythagoras we can find the height.
\(\displaystyle \L\\\sqrt{1^{2}-(\frac{1}{2})^{2}}=\frac{\sqrt{3}}{2}\)
\(\displaystyle A=\frac{bh}{2}\)
\(\displaystyle A=(\frac{1}{2})(\frac{\sqrt{3}}{2})=\frac{\sqrt{3}}{4}\)
The area of an equilateral triangle is given by \(\displaystyle \frac{\sqrt{3}}{4}s^{2}\)
Now, your next 'island' is the same as the first but with 3 smaller triangles added. There is a pattern.
The side length of the small triangles on the 2nd island is 1/3
This gives each an area of \(\displaystyle \frac{\sqrt{3}}{4}(\frac{1}{3})^{2}=\frac{\sqrt{3}}{36}\).
There are 3 with total area \(\displaystyle \frac{\sqrt{3}}{12}\)
Add this to the area of the original triangle: \(\displaystyle \frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{12}=\frac{\sqrt{3}}{3}\)
The side area of the next will have s=one third the previous. So, 1/3 of 1/3 is 1/9, and so on.
Here's the general case:
Actually, when we go from the nth island to the (n+1)st island, we have multiplied the number of sides by 4 and divided the length of each side by 3.
So, we have \(\displaystyle \L\\P_{n}=(\frac{4}{3})^{n}P_{0}\)
Where \(\displaystyle P_{0}\) is the perimeter of the original triangle.
If we let \(\displaystyle A_{n}\) be the combined areas of all the triangles up to, say, the nth island we have:
\(\displaystyle A_{n+1}=\frac{4}{9}A_{n}\), n=1,2,3,.......
If we add them all up, we have:
\(\displaystyle A_{1}+A_{2}+A_{3}+.......=A_{1}(1+\frac{4}{9}+\left(\frac{4}{9})^{2}+....\right)=\frac{9}{5}A_{1}\)
Let \(\displaystyle A_{0}\) be the area of the original triangle, then \(\displaystyle A_{1}+\frac{1}{3}A_{0}\). As we showed before.
So, \(\displaystyle \L\\\text{Area of the snowflake} =\sum_{n=0}^{\infty}A_{n}=\frac{8}{5}A_{0}\)
In your case the limit approaches:
\(\displaystyle \L\\\frac{2\sqrt{3}}{5}\)
The area of the (n+2)st island can be found by using \(\displaystyle \L\\\frac{\sqrt{3}}{12}(\frac{4}{9})^{n}\)
You have \(\displaystyle A_{0}\)(original) and \(\displaystyle A_{1}\)(6 pointed one)
The area of the third island would be \(\displaystyle \L\\\frac{\sqrt{3}}{12}(\frac{4}{9})^{1}=\frac{\sqrt{3}}{27}\)
Using the sum: \(\displaystyle \L\\\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{12}\sum_{0}^{\infty}(\frac{4}{9})^{n}=\frac{2\sqrt{3}}{5}\)
Which, as you can see, matches are above result of \(\displaystyle \L\\(\frac{\sqrt{3}}{4}){(\frac{8}{5})=\frac{2\sqrt{3}}{5}\)
Sorry to ramble on, but I always thought the Koch curve was very interesting. I just about spilled all I know about it.
A large part of this I derived myself.
Two other small things:
The perimeter is infinite, yet the area is finite.
It is continuous everywhere, but is not differentiable anywhere.
See, the two right triangles formed by the equilateral triangle?.
By Pythagoras, the height is \(\displaystyle \L\\\sqrt{1^{2}-(\frac{1}{2})^{2}}=\frac{\sqrt{3}}{2}\)
\(\displaystyle \L\\A=\frac{(base)(height)}{2}\)
base = \(\displaystyle 1/2\) and height = \(\displaystyle \frac{\sqrt{3}}{2}\)
\(\displaystyle \L\\A=\underbrace{2}_{\text{there\\are\\2}}\cdot\frac{(\frac{\sqrt{3}}{2})(\frac{1}{2})}{2}\)
That funny wired E is the capital Greek letter Sigma. It represents summation; Adding a bunch of things up. For instance,
\(\displaystyle \L\\\sum_{n=1}^{10}n=55\)
This means we would add up all the numbers from 1 to 10. Which equals 55.