From Eli Luberoff (courtesy of Iñaki Huarte) comes a geometric curiosity:
Suppose you have a chain with infinitely many flat (i.e., one-dimensional) links. The first link has length 1, and the length of each successive link is a fraction ƒ of the previous link’s length. As you might expect, ƒ is less than 1. You place the chain flat on a table and some ink at the very end of the chain (i.e., the end with the infinitesimal links).
Initially, the chain forms a straight line segment, and the longest link is fixed in place. From there, the links are constrained to move in a very specific way: The angle between each chain and the next, smaller link is always the same throughout the chain. For example, if the Nth link and the N+1st link form a 40 degree clockwise angle, then so do the N+1st link and the N+2nd link.
After you move the chain around as much as you can, what shape is drawn by the ink that was at the tail end of the chain?
The length of all the paper clips as a straight line segment is:
$$1 + f + f^2 + ...$$
$$= \sum_{n=0}^{\infty} f^n$$
$$= \frac{1}{1-f}$$
If all the paper clips make 180° with their neighbors, the length of the clips is now
$$1 - f + f^2 - f^3 + f^4 - f^5 +...$$
$$ = (1-f) + f^2(1-f) + f^4(1-f)+ ...$$
$$= \sum_{n=0}^{\infty} f^{2n}(1-f)$$
$$ = \frac{1-f}{1-f^2} = \frac{1}{1+f}$$
Halfway between $\dfrac{1}{1-f}$ and $\dfrac{1}{1+f}$ is $\dfrac{1}{1-f^2}$.
The distance from the midpoint to either end is $\dfrac{f}{1-f^2}$.
If we lay all the paperclips down from the origin towards the positive x-axis, the end of the first paperclip will be at $(1, 0)$.
After moving the chain around, the end of the second paper clip can be represented as $1 + f \mathrm{cis}(\theta)$.
The end of the third paperclip can be represented as $1 + f \mathrm{cis}(\theta) + f^2 \mathrm{cis}(2 \theta).$
The end of the paper clip chain can be expressed as :
$$= \sum_{n=0}^{\infty} f^n \mathrm{cis}(n \theta)$$
$$= \sum_{n=0}^{\infty} \left(f\mathrm{cis}(\theta)\right)^n$$
$$= \dfrac{1}{1 - f \mathrm{cis}(\theta)}$$
To determine real and imaginary parts:
$$\dfrac{1}{1 - f \mathrm{cis}(\theta)}$$
$$= \dfrac{1}{1 - f\cos {\theta} - if\sin {\theta}} \cdot \dfrac{1 - f\cos {\theta} + if\sin {\theta}}{1 - f\cos {\theta} + if\sin {\theta}}$$
$$= \dfrac{1 - f\cos {\theta} + if\sin {\theta}}{\left(1 - f\cos {\theta}\right)^2 + \left(f\sin {\theta}\right)^2}$$
$$= \dfrac{1 - f\cos {\theta} + if\sin {\theta}}{1 - 2f\cos {\theta} + \left(f\cos {\theta}\right)^2+ \left(f\sin {\theta}\right)^2}$$
$$= \dfrac{1 - f\cos {\theta}}{1 - 2f\cos {\theta} + f^2} + \dfrac{if\sin {\theta}}{1 - 2f\cos {\theta} + f^2}$$
This point seems to lie on the circle with center $\left(\dfrac{1}{1-f^2}, 0 \right)$ and radius $\dfrac{f}{1-f^2}$.
Denoting the angle of the arc traced by the paperclips as $\alpha$, the endpoint can also be represented as $\dfrac{f}{1-f^2}\mathrm{cis}(\alpha) + \dfrac{1}{1-f^2}$.
Setting the $x$, or real, expressions equal to each other,
$$\dfrac{f}{1-f^2}\cos {\alpha} + \dfrac{1}{1-f^2} = \dfrac{1 - f\cos {\theta}}{1 - 2f\cos {\theta} + f^2}$$
$$f\cos {\alpha} + 1 = \dfrac{1 - f\cos {\theta}- f^2 + f^3\cos {\theta}}{1 - 2f\cos {\theta} + f^2}$$
$$f\cos {\alpha} + 1 = \dfrac{\left(1 - 2f\cos {\theta} + f^2 \right) + f\cos {\theta} -2f^2 + f^3\cos {\theta}}{1 - 2f\cos {\theta} + f^2}$$
$$f\cos {\alpha} = \dfrac{f\cos {\theta} -2f^2 + f^3\cos {\theta}}{1 - 2f\cos {\theta} + f^2}$$
The paper clip chain with fraction $f$ and chain angle $\theta$ will trace a circular arc of radius $\dfrac{f}{1-f^2}$ and angle $\alpha = \arccos {\left(\dfrac{\cos {\theta} -2f + f^2\cos {\theta}}{1 - 2f\cos {\theta} + f^2}\right)}$.