Dividing each edge of a polygon into one-thirds

Question

Begin with an equilateral triangle. Divide each edge into $3$ equal parts and cut off the corners. Continue this procedure indefinitely. Does this create a regular polygon at each stage?

I don't think that it creates that.

Answer 1

No, only the first one (an hexagon) is a regular polygon. Once you reach the second step (a dodecagon), it is already a non-regular polygon.

I have programmed it (see below).

Have a look

• at the general shape (Fig. 1) where all the intermediate constructions are displayed and

• at a magnification of one side (Fig. 2).

Edit : (an answer to the area issue) : Taking $n=10$ successive cropping operations for this equilateral triangle $T$ with sidelength $2$ , I find 0.989743318... for the approximate value of the area of the limit shape, and I guess that the exact value is $4 \sqrt{3}/7$ (coincidence on nearly 15 digits for this value $n=10$ !). As the area of $T$ is $\sqrt{3}$, it means that the ratio of the area of the inside of the limit shape to the area of initial triangle should be the very simple result $4/7$.

How can we obtain a proof of this result ? I there a trick of the kind Archimedes was using for computing areas, or maybe with more sophisticated tools ?

Matlab program :

 clear all;close all;hold on;axis equal;axis off;
 i=complex(0,1);
 T=[-1,1,sqrt(3)*i]; % equilateral triangle with sidelength 2
 plot([T,T(1)],'color',[0,0,1]);
 n=5; % depth
 Q=zeros(1,3*2^n);
 for k=1:n
    R=[T,T(1:3)]; % T is "augmented" for cycling purposes
    for p=0:length(T)
         Q(2*p+1)=(R(p+1)+2*R(p+2))/3;
         Q(2*p+2)=(2*R(p+2)+R(p+3))/3;
    end;
    T=Q(1:3*2^k);
    plot([T,T(1)],'color',[k/n,0,1-k/n]);
 end;
 [~,A]=convhull(real(T),imag(T));% A = area