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/2². Continuation of that process generates a sequence of triangles inscribed into each other with diminishing periods of P/2ⁿ, 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/2², P/2³, ... Obviously the generic term P/2^N of this sequence tends to 0 as N grows. The mathematical notation for that fact is

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
C = 2πR	c = 2πr
We have the following basic relations
a = R·sin(π/N)	r = a·cot(π/N)
So that
P = 2N·C/2π·sin(π/N)	c = 2π·P/2N·cot(π/N)
which, after simplification, becomes
P = C·N/π·sin(π/N)	c = P·π/N·cot(π/N)

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·cos(π/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

for the first problem, and

for the second.

If you remember the double angle formula for sine (sin(2a) = 2sin(a)cos(a), which is an immediate consequence of the addition formula for sine), you can readily simplify (2):

Now,

is one of the most famous limits in all Calculus. Substituting x = π/(3·2^k-1), we see that, as k grows, the expression sin(π/(3·2^k-1))/(π/(3·2^k-1)) 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 non-zero circumference cos(2π/3)/(2π/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 non-zero circumference. I do not know whether there exists a closed-form 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) > 1 - x²/2. Therefore, the product in (1) can't be less than the product

π_k = C·(1 - (π/3)²/2)·(1 - (π/4)²/2)·...·(1 - (π/(k+2))²/2)

But the latter product has, as k grows, a non-zero limit. In general, it can be shown that a product

(1 + a₁)·(1 + a₂)·(1 + a₂)· ...

converges to a non-zero limit iff the series

a₁ + a₂ + a₂ ...

is convergent. With regard to π_k, this is obvious since the series Sk^-2 is indeed convergent.

Limits in Geometry

• Two Circles and a Limit
  • Analytic Proof
  • Barbeau, Klamkin, Moser
  • Proof #2 by Stuart Anderson
  • Mariano Perez de la Cruz
  • Trigonometric Solution
• A Geometric Limit
• Iterations in Geometry, an example
  • Iterations in Geometry, a generalization
• Iterated Function Systems

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