That’s Unpossible! A Story In Three Parts

Since it’s Mathematics Awareness Month I thought I’d present to you something epic.  Epic I says.  So epic, I guarantee that it will blow your mind.  In fact, by the end of this you’ll probably believe that what I’m about to present to you is un-possible, but I assure you, that isn’t the case.

Let me introduce to you the very beautiful Koch Snowflake.  Just look at it; all full of snowflake-y nerdiness.  But what, pray tell, is the Koch Snowflake? It’s only a little slice of mathematical awesomeness.

Imagine starting with an equilateral triangle.  Simple.  Elegant.  Maybe slightly boring.  Clearly an equilateral triangle is not a snowflake.  So, let’s do something to it.  Specifically, as a first step, we’re going to remove the middle third of each side, replacing them with two lines that are 1/3 the length of the original side.  The result should look like the Star of David.  While this is more interesting than a basic equilateral triangle, it’s still rather simple.  We can do better.

Several iterations of the construction of the Koch Snowflake (http://www.search.com/reference/Koch_snowflake)

Let’s continue the process of removing bits and adding other bits.  So, remove the middle third from each of the remaining line segments, replacing them with two lines, each 1/3 the length of the original segment.  You should get something that looks like the image in the bottom left corner (to the right).

If you consider this process, it becomes clear that the perimeter of the snowflake has grown, as has the area.  Now, imagine continuing the process forever.  What becomes of the perimeter and the area?  Do they continue to grow forever, or do they approach some maximum value?

Let’s consider the perimeter.  With every update, each of the sides becomes \frac{4}{3} longer than the original length. That is, if one side originally has a length of 1, then after the first iteration it will have a length of \frac{4}{3}.  The second iteration will remove \frac{1}{9} from each of the 4 line segments making up the side, and replace them with \frac{2}{9}.  This gives us \frac{4}{3}-\frac{4}{9}+\frac{8}{9}=\frac{16}{9}=\left(\frac{4}{3}\right)^{2}.  Repeating the process would give us a length of \frac{64}{27}=\left(\frac{4}{3}\right)^{3}, etc. In general, after n iterations, we end up with a length for one side of \left(\frac{4}{3}\right)^{n}.   As n gets large, so does the perimeter.  In math speak, we say that the perimeter is unbounded.  That is

\displaystyle{\lim_{n\rightarrow\infty}}\left(\frac{4}{3}\right)^{n}=\infty.

In other words, the perimeter has infinite length.  With this in mind, you’d probably imagine that the area contained within must also be infinite.  Let’s check it out.

Imagine the area of the original triangle is A_{0}.   After the first iteration we have essentially added 3 new little triangle areas to the original triangle.  Specifically, the 3 additions are each \frac{1}{9} the area of the original.  Hence, we have added \frac{3}{9}\cdot A_{0} to the original area, or we can write A_{1}=A_{0}+\frac{3}{9}\cdot A_{0}=\frac{4}{3}\cdot A_{0}, where A_{1} is the area of the snowflake after the first iteration. On the second iteration, we add 12 tiny triangles each with area \frac{1}{81}\cdot A_{0}.  This gives an area of A_{2}=A_{1}+\frac{12}{81}\cdot A_{0}=\frac{40}{27}\cdot A_{0}.  A third iteration would give us A_{3}=A_{2}+\frac{48}{729}\cdot A_{0}=\frac{376}{243}\cdot A_{0}.

Do you see the pattern?  In general, after n iterations, the area A_{n} can be written as

A_{n}=A_{0}\left(1+\frac{1}{3}\displaystyle{\sum_{i=0}^{n}}\left(\frac{4}{9}\right)^{i}\right).

As n gets large (that is, as it heads off to infinity), we get the limiting area.  Specifically

\displaystyle{\lim_{n\rightarrow\infty}}A_{n}=A_{0}\left(1+\frac{1}{3}\displaystyle{\sum_{i=0}^{\infty}}\left(\frac{4}{9}\right)^{i}\right)=A_{0}\left(1+\frac{1}{3}\cdot\frac{9}{5}\right)=\frac{8}{5}\cdot A_{0}.

But how does \displaystyle{\sum_{i=0}^{\infty}}\left(\frac{4}{9}\right)^{i}=\frac{9}{5}?  I’ll explain that in another post.  Just trust me for now.

The point is, if we continue this process ad infinitum, the snowflake will end up with a finite area of \frac{8}{5} of the original area of the equilateral triangle, but it will be surrounded by an infinite border.  That means if I asked you to paint the area inside the snowflake, you’d be able to do it in no time flat.  But, if I asked you to walk around the snowflake, you’d not be able to.  Crazy right?

A Koch curve has an infinitely repeating self-...
Zooming in on the Koch Snowflake. Image via Wikipedia

Now the even weirder thing; would you believe me if I told you that this snowflake exists in a dimension that’s somewhere between the 1st and 2nd dimensions that we all know and love?  That is, most everyone understands that a line has 1 dimension (i.e., it has length, but not width or height).  A box on the other hand lies in 2 dimensions as it has both length and width.  The Koch Snowflake is all “screw you standard dimensions, I’m going to live in my own world and you aren’t invited”; choosing to live in some other dimension between that of a line and that of a box.  Crazy talk?  Quite possibly. But we’ll save that conversation for a later post.  For now, consider the image above that continually zooms in on the Koch Snowflake.  No matter how much we zoom, we keep seeing the same thing.

A few iterations of the snowflake are provided below.

7 first steps of the building of the von Koch ...
The Koch Snowflake. Image via Wikipedia

Until next time, Happy Mathematics Awareness Month all y’all.

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4 Comments Add yours

  1. Beth says:

    My mind just blew up into an infinite number of pieces!

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