# That’s Unpossible! One Giant Nerdgasm

Whilst perusing the Twitterverse, I stumbled on something most excellent.  Something that goes by the name of Hexaflake, which is a type of N-flake.  I won’t get into N-flakes, but trust me, they are cool.  Instead, I’m going to chat a little bit about this Hexaflake thing-y.  What it is, some of its very cool properties, and how it’s related to the Koch Snowflake that we talked about here, here, and here.

Trust me, you are going to want to keep reading.

Basically, I follow a super giant math nerd named John D Cook1.  He has several twitter accounts (most of which I follow) that are devoted to tips and interesting tidbits of nerdery.  These include the accounts outlined in the table below (which I copied directly from his blog – which can be found by clicking here).

Seriously – if you are into computer science, probability, $\LaTeX$, statistics or any number of other things, and if you are on Twitter, I recommend you follow his posts.  You shan’t be disappointed.

Anyway, while I was checking out his Twitter account, I noticed that he followed someone known as CarnivalOfMath.  The name spoke to me, so I checked out his blog and his Twitter feed and was immediately glad I did.  I now follow CarnivalOfMath as well2.  One particular tweet jumped out at me:

So I clicked to find out more.  And lo and behold, I was staring at the Hexaflake.

What is the Hexaflake?  Start with a regular hexagon, and remove some triangles from the edges.  But also remove some triangles from the centre so that after the first iteration you have 7 copies of the original hexagon, just scaled so that they fit within the boundaries of the original.  Do this again with the 7 copies.  Then again with the resulting 49 copies.  Then again.  And again.  And again, ad nauseam.  The resulting image is a very cool looking snowflake.

Now, if you remember back to the Koch Snowflake, we were adding to the edges.  As we did this an infinite number of times, we learned that the length of one edge of the Koch Snowflake was infinite.  Thus, the perimeter of the Koch Snowflake was also infinite.  Further, no matter how much we zoomed into the edge, we’d see the same thing over and over because the Koch Snowflake has the property known as self-similarity.  And this is why when we determined that the dimension was between 1 and 2, it seemed to make some sort of sense.

First, if you look at one edge of the Hexaflake as it’s constructed, you’ll notice that it looks like an edge of the Koch Snowflake (pay attention to the white area, not the black area).  And you’d be correct.  In fact, if you look at the triangles that are removed from the centre of the hexagon, you’ll also see more edges to the Koch Snowflake here.  In truth, as the triangles are removed ad infinitum, you can see that we are actually removing an infinite number of Koch Snowflakes (of varying size) from the original hexagon.  Clearly, since the edges of the original hexagon become edges of a Koch Snowflake, we can see that the perimeter of this object will also be infinite.

But what about its area?  This is where it gets uber cool.

The original hexagon is composed of 54 triangles.  On the first iteration, we end up removing 12 of them.  Hence the area is $1-\frac{12}{54}=1-\frac{2}{9}=\frac{7}{9}$.  At this point, we have 7 smaller copies of the original hexagon.  If we scaled them up by a factor of 3, each would be the size of the original hexagon.  If we imagine that each of the 7 hexagons are composed of 54 little triangles, 12 of which are removed, and each being $\frac{1}{9}$ the size of the previously removed triangle, our area becomes $1-\frac{2}{9}-7\frac{2}{9}\frac{1}{9}=1-\frac{2}{9}-\frac{14}{81}=\frac{49}{81}$.  In general, after $n$ iterations, the area of the Hexaflake would be

$1-\frac{2}{9}\displaystyle{\sum_{i=0}^{n}}\left(\frac{7}{9}\right)^{i}$

If we allow the iterations to go off to infinity, we get

$1-\frac{2}{9}\displaystyle{\sum_{i=0}^{\infty}}\left(\frac{7}{9}\right)^{i}=1-\frac{2}{9}\left(\frac{1}{1-\frac{2}{9}}\right)=1-\frac{2}{9}\left(\frac{9}{2}\right)=1-1=0$

WHAT THE WHAT?!?

This shape has no area?  So freaking cool I can barely stand it.  But there you have it.  The area is 0.  What’s even more amazing is that the perimeter is infinite, and yet it contains no area.  If that doesn’t mess with your mind, I don’t know what will.  AWESOME.  Total freaking nerdgasm.

Before I end this little trip through mathematical nerdery, let’s figure out the dimension of this bad boy.  Recall that our formula was

$m=\frac{\log{(s)}}{\log{(n)}},$

where $m$ represents dimension, $s$ is the number of copies of the original object that need to be scaled up by a factor of $n$ to be of the same size as the original object.  Here $s=7$, and $n=3$.  Hence our dimension is

$m=\frac{\log{(7)}}{\log{(3)}}\approx 1.7712,$

which seems entirely satisfying since it seems weird to have a 2-dimensional object that is without area.

Happy Mathematics Awareness Month all y’all.

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1 I follow several nerdy Tweeters, but John’s tweets are especially full of a lot of math-y awesomeness. Some of it is simply interesting from a mathematical/statistical point of view. The rest of it includes useful tips and tricks, and references to other nerdery. As such, his Twitter accounts are a great resource for someone like me. Clearly all Twitter nerds should follow him.

2 And this dear readers, is why I believe Twitter is an AWESOME resource. I have had several nay-sayers nay-say the use of Twitter as something for teenage girls and those with no lives. I beg to differ, it’s also for NERDS! w00t!