A Geometric Limit
Start with a regular triangle of perimeter $P.$ The midlines of the triangle cut it into four equal triangles, each being half as big as the original one. Therefore, the perimeter of each, the middle one in particular, is $P/2.$ Now apply the same procedure to the middle triangle (Pic 1). The small triangle at the center will have the perimeter $P/2^{2}.$ Continuation of that process generates a sequence of triangles inscribed into each other with diminishing periods of $P/2^{n},$ for the $(n+1)^{st}$ triangle.
The procedure can be modified without inflicting any change in the sequence of perimeters. One can readily observe (Pic 2) that the triangles in the sequence are not only inscribed into each other but also inscribed into the inscribed circles of their immediate predecessors. Inside those circles (Pic 3) the triangles may be rotated without changing perimeter.
So what we are doing is this. In a given triangle we inscribe a circle. Into that circle we inscribe a triangle, then again inscribe a circle, and again inscribe a triangle, and so on. Perimeters of the resulting triangles form a sequence $P,$ $P/2,$ $P/2^{2},$ $P/2^{3},\ldots$ Obviously the generic term $P/2^{\circ}$ of this sequence tends to $0$ as $N$ grows. The mathematical notation for that fact is
$\displaystyle\lim_{N\rightarrow\infty}\frac{p}{2^{N}}=0.$
It reads, "the limit of $P/2^{N}$ as $N$ tends to infinity is $0$."
When we keep inscribing circles into triangles and triangles into circles, circles shrink into a point. Now, let's modify the process just a little. Let's think of triangles as polygons. In the new construction, we shall be inscribing circles into polygons, and polygons into circles (Pic 4).
We'll have to answer two questions:
Find the perimeter $P$ of a regular $N$gon inscribed into a circle of circumference $C.$  Find the circumference $C$ of a circle inscribed into a regular $N$gon of perimeter $P.$  

In both cases the perimeter $P$ of the $N$gon relates to one half of its side a by $P = 2aN,$ while the circumference of a circle relates to its radius by  

 
We have the following basic relations  

 
So that  

 
which, after simplification, becomes  


Finally, we combine the two steps. Let there be a circle of circumference $C.$ Inscribe into the circle a regular $N$gon, and into the latter inscribe another circle. The circumference c of the latter satisfies
$c = C\cdot \cos(\pi /N).$
Next, we shall consider two problems. In the first, the number of sides of the inscribed polygons increases on every step by $1.$ In the second, it doubles with each step. If we start with inscribing a triangle, what would be the circumference of the last circle after $k$ steps?
From the above, we get
$c = C\cdot \cos(\pi /3)\cdot \cos(\pi /4)\cdot \cos(\pi /5)\cdot\ldots\cdot \cos(\pi /(k+2)),$
for the first problem, and
(2)
$c = C\cdot \cos(\pi /3)\cdot \cos(\pi /6)\cdot \cos(\pi /12)\cdot ...\cdot \cos(\pi /(3\cdot 2^{k1})),$
for the second.
If you remember the double angle formula for sine, $\sin(2\alpha ) = 2\sin(\alpha )\cos(\alpha ),$ which is an immediate consequence of the addition formula for \sine), you can readily simplify (2):
$\begin{align}\displaystyle c &= C\cdot \cos(\pi /3)\cdot \cos(\pi /6)\cdot \cos(\pi /12)\cdot\ldots\cdot \cos(\pi /(3\cdot 2^{k1}))\\ &= C\cdot \cos(\pi /3)\cdot \cos(\pi /6)\cdot\ldots\cdot \cos(\pi /(3\cdot 2^{k1}))\cdot \frac{\sin(\pi /(3\cdot 2^{k1}))}{\sin(\pi /(3\cdot 2^{k1}))} \\ &= C\cdot \cos(\pi /3)\cdot \cos(\pi /6)\cdot\ldots\cdot \cos(\pi /(3\cdot 2^{k2}))\cdot\frac{\sin(\pi /(3\cdot 2^{k2}))/2}{\sin(\pi /(3\cdot 2^{k1}))} \\ &= C\cdot \cos(\pi /3)\cdot \cos(\pi /6)\cdot\ldots\cdot \cos(\pi /(3\cdot 2^{k3})\cdot\frac{\sin(\pi /(3\cdot 2^{k3}))/4}{\sin(\pi /(3\cdot 2^{k1}))} \\ &= \ldots \\ &= C\cdot\frac{\sin(2\pi /3)/2^{k}}{\sin(\pi /(3\cdot 2^{k1}))} \\ &= C\cdot [\sin(2\pi /3)/(2\pi /3)]\cdot [\sin(\pi /(3\cdot 2^{k1}))/(\pi /(3\cdot 2^{k1}))] \\ \end{align}$
Now,
$\displaystyle\lim_{x\rightarrow 0}\frac{\sin x}{x}=1.$
is one of the most famous limits in all Calculus. Substituting $x = \pi /(3\cdot 2^{k1}),$ we see that, as k grows, the expression $\sin(\pi /(3\cdot 2^{k1}))/(\pi /(3\cdot 2^{k1}))$ tends to $1.$ We thus arrive at the conclusion that, if, in the second problem, the process is continued indefinitely, the circles do not shrink to a point but approach a circle with a nonzero circumference $\displaystyle C\cdot\frac{\sin(2\pi /3)}{(2\pi /3)}.$
What can be said about the first problem? Here the situation is more difficult. Termwise, with the exception of the first two factors, the terms in (1) are less than the terms in (2). The circles, however, as before, do not shrink to a point, but instead accumulate towards a circle of nonzero circumference. I do not know whether there exists a closedform expression for the radius (or the circumference) of that circle. It's possible to demonstrate its existence, nonetheless.
For the arguments $x,$ we encounter in the product (1), we have $\cos(x) \gt 1  x^{2}/2.$ Therefore, the product in (1) can't be less than the product
$\pi _{k} = C\cdot (1  (\pi /3)^{2}/2)\cdot (1  (\pi /4)^{2}/2)\cdot\ldots\cdot (1  (\pi /(k+2))^{2}/2).$
But the latter product has, as k grows, a nonzero limit. In general, it can be shown that a product
$(1 + a_{1})\cdot (1 + a_{2})\cdot (1 + a_{3})\cdot \dots$
converges to a nonzero limit iff the series
$a_{1} + a_{2} + a_{3}\ldots$
is convergent. With regard to $\pi _{k},$ this is obvious since the series $\sum k^{2}$ is indeed convergent.
This is also possible to circumscribe circles and polygons on top of each other. In this case the radii will grow with the reverse coefficient:
$\displaystyle \frac{1}{\cos(\pi /3)\cdot \cos(\pi /4)\cdot \cos(\pi /5)\cdot\ldots}.$
Cliff Pickover describes the problem and tells an amusing story [Pickover, 382383] . It appears that in the 1940s, the value of this coefficient was reported to be close to $12.$ That value has been published as late as 1964 in a German article. However, in 1965 Christoffel J. Bouwkamp published a paper with the value of $8.7000\ldots$ The correct value with 17 digits is $8.70003662520881945\ldots$ Half a century later, at the age when powerful computers became ubiquitous, it seems practically surrealistic to think that a value wrong by 50% could have survived over the space of two decades.
Reference
 C. Pickover, The Math Book, Sterling Publishing Co., 2009
Contact Front page Contents Geometry
Copyright © 19962018 Alexander Bogomolny