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$\{v_0, v_n : n\in \Bbb{N}\}$ are points on the circumference of a unit circle, having the following three properties:

  • The (straight line) distances between neighboring points in the sequence form a convergent geometric series. That is, for all $n>0$, $d(v_n,v_{n+1}) = r\cdot d(v_{n-1},v_n) $ with $r$ a constant ratio for the entire set of verticies, and $0<r<1$.

  • $\lim_{n_to\infty} d(v_n,v_0) = 0$. That is, the "polygon" formed by these vertices forms a closed shape (in the limit) without having to add a "last" side going back to $v_0$.

  • For all $n>0$, the shorter of the two arcs from $v_n$ to $v_{n+1}$ does not contain $v_0$. That is, the sides go around the circle only once.

I want to find the area of this "polygon" in terms of the side ratio $r$. (Although the number of sides needed to close the shape is infinite, one can easily define the area as the limit, as $k$ goes to infinity, of the area formed by the first $k$ sides and an added last side connecting $v_{k-1}$ to $v_0$.)

Several things are immediately clear:

  • This area is not defined for $r<\frac12$ because the set of lines cannot from a closed (in the limit) "polygon" unless the sum of the lengths of all the sides other than the first is longer than the first side. So the function $A(r)$ need only be found on the open interval $(\frac12,1)$.

  • For any $r \in( \frac12,1)$, there exists a value of $L_0 \equiv d(v_0,v_1)$ such that the figure exactly closes without overshooting. $L(r)$ has the property that $$ \sum_{n=0}^\infty \sin^{-1}\left(\frac{L}{2}r^n\right) = \pi$$


EDIT For $r$ below about $0.6$, the first arc covers more than half the circle, and the area of the polygon does not include that first "super-wedge." In that circumstance, $L(r)$ would have the property that

$$ \sum_{n=1}^\infty \sin^{-1}\left(\frac{L}{2}r^n\right) = \sin^{-1}\left(\frac{L}{2}\right) $$

However, my numerical fiddling around indicates that you can't satisfy this, and that makes sense since arcsin is concave upwards.


  • $\lim_{r\to\frac12} A(r) =0$ and $\lim_{r\to 1} A(r) =\pi$. The former represents almost perfectly retracing the first side along a very short arc that is almost a straight line; the latter represents going around the circle in what is almost a regular inscribed polygon.

EDIT The nearly straight line case does not happen; see the edit above.


I'm also almost sure that for small positive $\epsilon$, $A(r)$ has a positive second derivative at $r=\frac12 + \epsilon$, and a negative second derivative at $r = 1-\epsilon$, and that $A(r)$ has just one point of inflection.

My question, then, is to find $A(r)$, or failing that, to describe some non-trivial properties of $A(r)$.

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  • $\begingroup$ NB Irregular polygon... $\endgroup$ – user301988 Nov 8 '16 at 3:33
  • $\begingroup$ Taking into account that $L_0\le2$ I think that it should be $r>0.624$ (about). $\endgroup$ – Intelligenti pauca Nov 8 '16 at 12:41
  • $\begingroup$ My question as psed was flawed in that when $r$ is close to $\frac12$ the first of the arcs is greater than $\pi$ so that the first term in the sum should be $\pi - \sin^{-1}(L/2)$ instead of $\sin^{-1}(L/2)$. That is the reason that you can have such a polygon with $r$ just over $\frac12$. I will edit a change. $\endgroup$ – Mark Fischler Nov 8 '16 at 21:08
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EDIT.

Function $L(r)$ is implicitly defined by the equation: $$ \sum_{n=0}^\infty\arcsin\left({1\over2}L(r)\ r^n\right)=\pi $$ if $r\ge r_0\approx0.602527$, and by the equation $$ \sum_{n=1}^\infty\arcsin\left({1\over2}L(r)\ r^n\right)=\arcsin\left({1\over2}L(r)\right) $$ if ${1/2}<r<r_0$. Here's a plot of $L(r)$ obtained by a numerical computation:

enter image description here

If $r\ge r_0$ the polygon is the sum of isosceles triangles $T_n$, having vertex at the center of the circle and base $L_n=L(r)r^n$, so its area is given by $$ A(r)=\sum_{n=0}^\infty{1\over2}L(r)\ r^n\sqrt{1-{1\over4}L(r)^2r^{2n}}. $$ If $r<r_0$ the first triangle must be subtracted by the sum of the others, leading to $$ A(r)=-{1\over2}L(r)\sqrt{1-{1\over4}L(r)^2}+ \sum_{n=1}^\infty{1\over2}L(r)\ r^n\sqrt{1-{1\over4}L(r)^2r^{2n}}. $$ Here's a plot of $A(r)$:

enter image description here

For a better visualization of the behaviour near $r=1/2$ here's a close-up of the same plot:

enter image description here

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