Sierpinski GasketDate: 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 Hi, 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 |
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