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Sierpinski Gasket

Date: Tue, 29 Aug 1995 08:05:54 -0500 (CDT)
From: Anonymous
Subject: sierpinski's gasket

We are 8th grade algebra students from Oshkosh, Wisconsin.  We are stumped 
by this problem: A Sierpinski Gasket is formed by removing successfully 
smaller equilateral triangles.  What is the area of a Sierpinski gasket?  
We've done research and our resources are exhausted! 

					Thank You!
					Stephanie Koch

Date: 8/29/95 at 10:10:15
From: Doctor Heather
Subject: Re: sierpinski's gasket


I think you may be digging yourselves into a hole.  Let me explain why.  
Well, first, did you find a question that asks this?  

First, the definition of a Sierpinski gasket, which you may already know, is:

Start with an equilateral triangle plane.  Split it into four equilateral
triangles and remove the central triangle.  Now, repeat the process on the 
remaining triangles.  Do this process infinitely many times.

Now, let's try to calculate the area.  We need to do this by seeing how the 
area decreases every time we repeat the process.  We will assume that our
initial sidelength is 1.  So, the first area is 

1/2 x 1 x (sqr root 3)/2   (I got (sqr root 3)/2 by using trigonometric 
equations) = (sqr root 3)/4

The second area is three times one of the remaining triangles, which has a 
baselength of 1/2, and a height of (sqr root 3)/4 (again, you need to use
Pythagorean theorem).

So, let's just make a list of the areas so far, and I'll continue with a few
more and see if we can see a pattern.

1/2 x 1 x (sqr root 3)/2 = (sqr root 3)/4
1/2 x 1/2 x (sqr root 3)/4 x 3 = 3(sqr root 3)/16
1/2 x 1/4 x (sqr root 3)/8 x 9 = 9(sqr root 3)/64
1/2 x 1/8 x (sqr root 3)/16 x 27 = 27(sqr root 3)/256
So, we can see a pattern for the area.  Here is a formula:

1/4 x (sqr root 3) x 3^n x (1/4)^n = 1/4 x (sqr root 3) x (3/4)^n

Now, the easiest way to solve this is to use calculus.  I'm assuming since
this is an eighth grade class you probably will not understand the terminology.
But let's just think what happens with the quantity (3/4)^n as n gets larger.

3/4, 9/16, 27/64, 81/256, 243/1024, ....

The fractions are decreasing.  If we keep doing this process infintely, how
large is (3/4)^n going to be?  It keeps getting smaller and smaller, and it
will approach zero.  

So, let's regroup.  We have the formula:

(sqr root 3)/4 x (3/4)^n = area

What happens to the value of the area as (3/4)^n gets really really small?
The area gets really, really small.  As (3/4)^n gets close to zero, the area
of the Sierpinski gasket gets close to zero.  So, I would say that the area
of a Sierpinski gasket is zero.  Please tell me if an answer book has it 
listed differently, but I am pretty certain of this answer.

Another interesting aspect of the Sierpinski gasket is that its dimension
is not 1 or 2.  It's somewhere in between.  A dimension measure called the
capacity dimension is used to measure the dimension of fractals.  If you would
like to know more about how to find it's dimension, write back.

Fractals can be a lot of fun!

-Doctor Heather,  The Geometry Forum
Associated Topics:
Middle School Algebra

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