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Let's make a construction involving regular polygons:

► First, we begin with a equilateral triangle, with side $\ell_3 = 1;$

► After, we draw a square on the middle point each side of the initial triangle, with side $\ell_4 = \frac{1}{2} = \frac{\ell}{2}.$

Now, the construction continues, taking one of these steps:

► If the regular polygon have an even number of sides $n$ with length $\ell_n$, then we draw two regular polygons with $n + 1$ sides of length $\ell_{n+1} = \frac{\ell_n}{2},$ from the middle point of the extreme segments.

► If the regular polygon have an odd number of sides $n$ with length $\ell_n$, then we draw one regular polygons with $n + 1$ sides of length $\ell_{n+1} = \frac{\ell_n}{2},$ from the middle point of the unique extreme segment in this case.

To clarify the explanation, we will obtain a figure like the one below:

enter image description here

I have two questions about this:

Q1. This figure is inside a circumference with center in the incenter of the initial equilateral triangle? In affirmative case, what is the radius $R$ of the circumference?

Q2. The sequence of the lengths I adopted in the construction is $$\ell_n = \frac{1}{2^{n-3}}, \quad \forall n \ge 3 $$ If I consider other sequence $\ell_n$, when exists a circumference with center in the incenter of the initial equilateral triangle and radius $R$ in which the figure is inside?

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    $\begingroup$ Consider a path that starts at a corner of the triangle and meets the center of the opposite square, then the center of a connected pentagon, then a connected hexagon, etc, etc, etc. This path bends, so its length does not measure distance from the triangle's center; however, if the "limit" of the path's length is finite, then so is the "limit" of the distance from the center. Well, the path consists of a circumradius and an inradius of each $n$-gon; the formulas to calculate these from side-length are straightforward. The infinite sum, less so, but it's a place to start. $\endgroup$ – Blue Nov 20 '18 at 10:38

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