Area of the Limiting Polygon

We calculate explicitly that the area of the limiting polygon is $\dfrac{4}{7}A$ where $A$ is the area of the original triangle.

Note that on the $(n+1)$-th iteration we cut off twice as many triangles as we did in the $n$-th iteration. Consider the triangles we cut off in the $(n+1)$-th iteration. Each of these has $\dfrac{1}{3}$ the height and $\dfrac{1}{3}$ the base of a triangle cut off in the $n$-th iteration. Hence the ratio of the areas of these triangles is $\dfrac{1}{9}$. Since the total area of the first triangles we cut off is $\dfrac{1}{3}A$, we have that the area of the limiting polygon is

$$A-\sum\limits_{n=0}^\infty2^n\left(\frac{1}{3}A\right)\left(\frac{1}{9}\right)^n=A-\frac{1}{3}A\frac{1}{1-\frac{2}{9}}=\frac{4}{7}A$$

Whether this actually helps us determine the shape of the limiting figure I am not so sure.

This was motivated by Hagen von Eitzen's calculation and Michael's subsequent observation.

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EDIT: Explanation of $\dfrac{1}{3}$ base and $\dfrac{1}{3}$ height statement.

Consider the polygon at a particular vertex $V$ just before making the $n$-th iteration cuts. Choose a particular edge bordering $V$ and let it have length $9x$.

Now take the $n$-th and $(n+1)$-th iteration cuts. In the figure above, the green triangle is one that is cut off in the $n$-th iteration cuts, and the red triangle is one cut off in the $(n+1)$-th iteration cuts.

Because we trisect each time, one side of the green triangle has length $3x$ and one side of the red triangle has length $x$. Call these sides the base of each triangle. Consider now the length of the altitudes corresponding to these bases, letting the altitude of the green triangle be of length $3y$. By similar triangles we find that the altitude of the red triangle has length $y$ (the similar triangles are outlined in black).

Hence the ratio of area of the red triangle to the area of the green triangle is $$\frac{\frac{1}{2}xy}{\frac{1}{2}(3x)(3y)}=\frac{1}{9}$$ as claimed.

Fun note: we did not use the fact that the triangles in question are isosceles.

Numerically, after going all the way to the 12288-gon, the area of the shape (as proportion of the original triangle) is between $$ \frac{17932033916}{31381059609}= 0.5714285667\ldots$$and $$\frac{53796102772}{94143178827}= 0.5714285776\ldots$$

Going all the way to the 1572864-gon, the estimate improves to $$0.57142857142859\pm1.5\cdot10^{-13}.$$

I don't see any"recognizable" number in this, so I doubt that anything but numerics will help.