## Cutting a triangle into infinitely many Koch snowflakes

I want to explain the construction behind a piece that I posted recently without explanation.  Starting from a single triangle, I’ll show you how to cut it up into pieces, rearrange those pieces, and get this:

## The construction

Mark off each edge into thirds, and connect those points with lines using this pattern:

This divides up the triangle into into seven smaller triangles, three equilateral and three isosceles.  Remove the three isosceles triangles to get this:

But now, notice that the three triangles you removed can each be cut in half and then taped back together along an edge, so that you get three equilateral triangles. Do that, and place the new equilateral triangles so that they stick out from the sides of the original triangle, and you will get this:

That’s it for step 1!

Now, for step 2, repeat this whole process for each of the 7 equilateral triangles obtained in step 1.  Step 3: Do the same for each of the 49 triangles obtained in step 2.  And so on. My original picture, at the top of this post, is what you get after step 5.

Notice that each step is area-preserving, so in particular, the total area of all of the black triangles in my original picture is the same as the are of the triangle I started with.

Here’s an animation showing the first five steps in the sequence, and then those same steps backwards, going back to the original triangle:

The reason the picture seems to get darker in the latter steps is that the triangles are drawn with black edges, and eventually there are a lot of edges.  Since there’s a limit to how thin the edges can be drawn, eventually, the picture is practically all edges.

## How many snowflakes do you see?

The outline of the entire picture is clearly a Koch curve, so we have generated a Koch curve from a triangle.  But, what I really love about this construction is that every triangle that occurs at any step in the recursive process also spawns a Koch curve!  That’s a lot of Koch curves.

To make this precise, we can assume that triangles at each step are closed subsets of the plane.  Admittedly, the “cutting” analogy falls apart slightly here, since two pieces resulting from a “cut” each contain a copy of the edge the cut was made along, but that’s OK.   With this closure assumption, each of the Koch curves, one for each triangle formed at any stage in the process, is a subset of the intersection over all steps.