2
$\begingroup$

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. 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"?

I know the polygon must be continuous but isn't differenciable, and that it has symmetry $S_3$. I also know the area ($4/7$). It's not a circle, constant diameter shape, or a finite collection of beizier curves. It is convex.

$\endgroup$
5
  • 1
    $\begingroup$ See also mathoverflow.net/questions/51008/… and mathoverflow.net/questions/438022/… $\endgroup$
    – lhf
    Feb 4 at 11:24
  • $\begingroup$ Gregory and Qu proved that the limit is not smooth in this case. Smoothness of the limit requires cutting at strictly less than $1/3$. See their paper Nonuniform corner cutting (p 771). $\endgroup$
    – lhf
    Feb 4 at 11:28
  • $\begingroup$ but the limit has a continuous derivative? $\endgroup$
    – Sny
    Feb 4 at 11:46
  • $\begingroup$ It seems that the centers of the sides of the triangle are remaining untouched and such each side center that is result of the cuts. It seems that you could get close to a circle as much as you want. $\endgroup$
    – Moti
    Feb 5 at 5:20
  • $\begingroup$ No, the area doesn't match one of a circle's. $\endgroup$
    – Sny
    Feb 5 at 5:42

0

You must log in to answer this question.

Browse other questions tagged .