Area of the Limiting Polygon

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.

What is the area of the limiting "polygon"?

My first thought was that it would be the circle tangent to the midpoint of each side. However, by making some pictures, it becomes clear quite quickly that the limiting polygon is not a circle and, in fact, might perhaps be an ellipse. The problem with that is that while there is a unique circle passing through any three points in general position, I need five generic points to specify a conic.

Can anyone suggest any hints as to how to find the area? By "hint" I don't mean "what you think might work", I mean "what you have tried and you know will work". Thanks in advance.

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The result is definitely not an ellipse (unless it is a circle), just think of the $120^\circ$ symmetry. –  Hagen von Eitzen Jul 7 at 20:36
@HagenvonEitzen I've drawn the shape up to the fourth iteration, and it looks suspiciously ellipse-like. –  Fly by Night Jul 7 at 20:50
–  lhf Jul 7 at 22:02
I doubt it could be an ellipse, since the construction preserves the symmetry of the triangle. –  Leonhard Jul 8 at 7:33

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.

$\\$

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.

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Thanks for including the motivation - so often the end result can seem obscure, and what led us to an answer is often more useful - particularly for novices - than the actual result. –  Glen_b Jul 8 at 1:11
Could you explain exactly what you mean by base and height? Every cut-away triangle, after the first iteration, is isosceles, and the sides of the same length do indeed go down by a factor of $\frac{1}{3}$ each time. However, the included angle gets bigger with every iteration and so the cut-aways are not similar. Moreover, the perpendicular heights do not scale by a factor of $\frac{1}{3}$. Starting with an equilateral triangle with sides $3$, the first cut-aways have perpendicular height of about $0.87$. The second cut-aways have a p-height of about $0.17$. –  Fly by Night Jul 8 at 15:48
@FlybyNight I apologize for being rather informal in my original answer. I have added an edit to my post which hopefully will clear up confusion. –  Peter Woolfitt Jul 8 at 17:43

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.

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Is that just 4/7? –  Michael Jul 7 at 21:11
@Michael Given that $\frac{4}{7}\approx 0.5714285714286$, it does seem like a good candidate! –  Fly by Night Jul 7 at 21:35
How did you arrive at these estimates? –  Fly by Night Jul 7 at 21:35

Snipping off corners as you've described you get an infinite-sided regular polygon: a circle.

It will be tangent to the original triangle at the midpoints of the sides.

If the length of the side of your triangle is $1$, the radius of the inscribed circle is $\tfrac{1}{2}$ and the area is $\tfrac{\pi}{4}$.

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The result is definitely not a circle. The edge midpoints of each intermediate polynomial will be on the limit figure, and already the next step, the 12gon, is irregular (has two different side lengths). –  Hagen von Eitzen Jul 7 at 20:40
You obviously didn't read all of my post... –  Fly by Night Jul 7 at 20:48