Results 1 to 2 of 2

Thread: Sierpinski's triangle area formula(s).

  1. #1
    Member integral's Avatar
    Dec 2009

    Sierpinski's triangle area formula(s).

    I was playing around with Sierpinski's triangle and derived this equation for the total area for all the down triangles with an infinite number of iterations. (the area of the triangle as a whole =1)
    A\downarrow=\lim_{n\rightarrow \infty}\sum_{n=1}^{\infty}(\frac{1}{4})^n\cdot 3=1$
    (Edit:I realized the sum was kind of useless... But I will keep it there considering it looks cooler and the equation means the same thing with or without )

    So the opposite must also be true.
    A\uparrow=\lim_{n\rightarrow \infty}1-(\frac{3}{4})^n=0$

    I am not sure if this is right, and I would like someone to check it please
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Apr 2005
    Exactly what triangle are you starting with? My understanding of the "Sierpinski triangles" is that you start with an equilateral triangle, divide it into four triangles by connecting the bisectors of the sides, then remove the middle triangle. Then repeat that for the remaining three equilateral triangles. If the original triangle was 1, then the area of the three sierpinski triangles is 3/4. Since that is already less than 1 and you continue by always removing triangles, the total area of all remaining triangles, as you repeat infinitely, cannot be 1.

    In fact, after n repetitions, the area of the remaining triangles is $\displaystyle (3/4)^n$ which goes to 0. That's why the "Sierpinski triangles" is a fractal set. The measure of a set of dimension "d" is proportional to $\displaystyle x^d$ for some "typical" length x. At each step we are dividing all lengths by 1/3 while multiplying the number of the triangles by 4: the total area would be proportional to $\displaystyle (4(1/3)^d)^n$. If $\displaystyle 4/3^d> 1$ that will go to infinity. If $\displaystyle 4/3^d< 1$ that will go to 0. We can only get a non-zero finite measure if $\displaystyle 4/3^d= 1$ or $\displaystyle 3^d= 4$. Taking logarithms of both sides, d log(3)= log(4) so d= log(4)/log(3) which is about 1.26. The fractal dimension of the Sierpinski triangles is about 1.26.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Find area of triangle [Heron's Formula]
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Jan 11th 2009, 10:11 PM
  2. Replies: 1
    Last Post: Oct 28th 2008, 07:02 PM
  3. Replies: 7
    Last Post: Jul 19th 2008, 06:53 AM
  4. Replies: 27
    Last Post: Apr 27th 2008, 10:36 AM
  5. need a formula for triangle
    Posted in the Geometry Forum
    Replies: 3
    Last Post: Dec 6th 2007, 10:13 AM

Search tags for this page

Click on a term to search for related topics.

Search Tags

/mathhelpforum @mathhelpforum