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Sierpinski Gasket by Trema RemovalOne of the ways to construct the Sierpinski gasket (also known as the Sierpinski triangle) is by starting with a triangle. Then splitting the triangle into four and removing the (open) middle one. The same procedure then applies to the remaining three triangles. The term trema refers to the removed pieces. Professor McWorter drew my attention to the fact that this is not the only way to obtain the Sierpinski gasket by removing tremas. The applet on the right illustrates his approach. The original shape is a square which is split into four smaller squares with the south-eastern one subsequently removed. The same procedure then applies to the remaining three squares. Click on the applet to perform one additional step. When you reach the last step allowed with the accuracy of a square the size of a pixel, it becomes absolutely obvious that the two ways (and the two trema shapes) lead to exactly same result. Let's ask how much area we remove in this process. The original square had side of length 1. Skip the first step and start with the inverted L-shape of area 3/4. On the next step, me remove 3 smaller squares, then 9, then 27, and so on. On every step, the side of the removed squares is divided by 2. The first 3 squares had side 1/4 and area 1/16. The 9 smaller ones had area 1/64=(1/8)2 each. Afterwards, me removed 27 squares of the total area 27/162. And so on. We get a series: 3/42 + 9/82 + 27/162 + ... = 3/16(1 + 3/4 + 9/16 + 27/64 + ...) = 3/16·1/(1-3/4) = 3/4. This is exactly the area of the inverted L-shape. Therefore, Sierpinski gasket's area is 0. Can we obtain the same result when removing triangular tremas? Assume the whole triangle's area is 1. The number of triangles removed at each step is multiplied by 3 whereas the area of any one triangle removed is divided by 4. Check that the process leads to a geometric series that sums up to 1. Another derivation does not directly appeal to the notion of limit. It applies in exactly the same way to both trema removal procedures. In both cases, assume the starting shape(a square or a triangle) has area 1. Let's denote the total removed area as S. We want to show that S = 1. On the first step, we remove the area of 1/4 and leave 3 shapes similar to the starting one with size half as big. Each of the remaining triangles has area 1/4, i.e., one fourth of the big triangle. The removed portion of the smaller triangles must be S/4 (again, one fourth of the removed total.) This leads to the equation: S = 1/4 + 3·S/4, solving which we get the desired S = 1. 
Copyright © 1996-2008 Alexander Bogomolny |
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