Please look at the problem here. What is the shape of the polygon at the end?

Start with an equilateral triangle with unit area. Trisect each of the sides and then cut-off the corners. In this case, we get a regular hexagon - see the picture below. Next, trisect each of the sides of the hexagon and cut-off the corners. This will give a dodecagon, but not a regular one.

Continue this process ad infinitum.

What is the SHAPE of the limiting "polygon"?

Not sure but I don't think it's going to be a circle

  • $\begingroup$ Hmmmm.. I'm pretty sure that one thing we can be certain of is that the midpoint of each generated segment remains forever. That might help. $\endgroup$ – Dan Uznanski Mar 31 '18 at 10:08
  • $\begingroup$ Yes, that does remain forever $\endgroup$ – mathwizard1 Mar 31 '18 at 10:12
  • $\begingroup$ Any idea to find out? $\endgroup$ – mathwizard1 Mar 31 '18 at 10:42

The limiting polygon appears to be rather nondescript, though it's clearly not a circle.

I wrote a program using Sage. It starts with an ordered list of the vertices of the equilateral triangle. On each iteration it creates a new list, replacing the vertices in the list with new vertices at the points $1/3$ and $2/3$ of the way along each edge.

The plot below is after $12$ iterations. At this point the polygon has $3\times 2^{12}=12288$ sides. The initial equilateral triangle is shown for reference.

The number of sides far exceeds the image resolution so this should be a good approximation to the limiting polygon. Note that I manually scaled the original plot to get the aspect ratio close to $1:1$ in the final image.

enter image description here


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