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First we have a square inscribed in a circle with radius $1$. By connection vertexes of this square we have two diagonals, which divides square for $4$ rectangular triangles with (at least) one corner equals $\frac{\pi}{4}$. Then we draw bisector to each center right angle. By connection vertexes of square with two nearest intersections of bisector and circle, we have octagon. Ignoring divide by bisector first $4$ rectangular triangles (which form square), we have also new $8$ rectangular triangles with one corner equals $\frac{\pi}{8}$. Repeating this operation (animation) to $2^n$-gons we have $$C=\lim\limits_{n\to\infty}2^{n+1}\sin\left(\frac{\pi}{2^n}\right)=2\pi$$ $$S=\sum\limits_{k=2}^{n}2^k\sin\left(\frac{\pi}{2^{k-1}}\right)\sin^2\left(\frac{\pi}{2^k}\right)= 2^{n-1}\sin\left(\frac{\pi}{2^{n-1}}\right)$$ $$\lim\limits_{n\to\infty}2^{n-1}\sin\left(\frac{\pi}{2^{n-1}}\right)=\pi$$ where $C$ is a perimeter of $2^n$-gon and $S$ is area (as sum areas of $n$ new rectangular triangles from each iteration). By multiplying expression under the sum with $(-1)^{n}$ we have curve with same perimeter, but another area (animation). What is the type of curve is it? Looks like fractal with changing angle and alternate type of expansion (inside and outside).

user514787
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  • I don't understand your question, but it seems like you are trying to ask something very much like this one. – Xander Henderson Apr 26 '18 at 13:47
  • @XanderHenderson, no! Which part you don't understand? – user514787 Apr 26 '18 at 13:52
  • I do not understand the procedure by which you are attempting to approximate the circle, and I don't see what the series $C$ and $S$ have to do with that approximation. They say that a picture is worth a thousand words... – Xander Henderson Apr 26 '18 at 13:54
  • @XanderHenderson : The sequence of inscribed polygons is this one, Archimedes sequence of inscribed polygons for lower bounding $\pi$, not that this description makes that clear. The "one corner equals" phrases are describing the central angle subtending one edge of the polygon. – Eric Towers Apr 26 '18 at 13:58
  • $C$ and $S$ are undefined. There is only one sum in sight and multiplying its summand by $(-1)^n$ means that you have arbitrarily decided that some contributions to whatever $S$ is contribute negatively, whatever that means. However, multiplying an expression by a power of $-1$ doesn't magically produce a curve. What curve do you mean? How is it related to your modified sum? – Eric Towers Apr 26 '18 at 13:59
  • When you multiply your sum, you decide that some triangles have negative area. What does that mean? What curve do you intend when you declare that some triangles have negative area? – Eric Towers Apr 26 '18 at 15:10
  • @EricTowers Indeed, I meant to say the limit $C$ and the series $S$, but reversed them while typing and left out the word limit. – Xander Henderson Apr 26 '18 at 16:04
  • @EricTowers, I add animation. – user514787 Apr 26 '18 at 16:29
  • The triangles you are removing in your animation do not appear to have the same areas as the triangles you are negating in your sum. – Eric Towers Apr 26 '18 at 16:31
  • @EricTowers, I draw it in paint by hand, so, of course, they are not absolutely equal. – user514787 Apr 26 '18 at 16:37
  • @EricTowers, how completely are you sure in your statement? – user514787 Apr 26 '18 at 18:17

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