Circle Inversion and the Pappus Chain

There is a pledge of the big and of the small in the infinite.

-Dejan Stojanović

In the next two posts, we are going to look at two interesting geometric ideas of the 19th century involving circles. Next time, we will consider Poincaré’s disk model for hyperbolic geometry. Today, though, we immerse ourselves in the universe of inversive geometry.

Consider a circle in the infinite 2-dimensional plane:


This circle divides the plane into two regions: the bounded region inside the circle and the unbounded region outside the circle (let’s say that the points on the circle belong to both regions). A natural thing to want to do, now, especially in the context of this blog, would be to try to exchange these two regions, to map the infinite space outside the circle into the bounded space of the circle, and vice versa, in a “natural” way.

I could be bounded in a nutshell, and count myself a king of infinite space.

-William Shakespeare, Hamlet

Upon first reflection, one might be tempted to say that we want to “reflect” points across the circle. And this is sort of right, but reflection already carries a meaning in geometry. Truly reflecting points across the circle would preserve their distance from the circle, so the inside of the circle could only be mapped onto a finite ring whose outer radius is twice that of the circle two. Moreover, it would not be clear how to reflect points from outside this ring into the circle.

Instead, we want to consider a process known as “inversion.” Briefly speaking, we want to arrange so that points arbitrarily close to the center of the circle get sent to points arbitrarily far away from the center of the circle, and vice versa. For simplicity, let us suppose that the circle is centered at the origin of the plane and has a radius of 1. The most natural way to achieve our aim is to send a point P to a point P' that lies in the same direction from the origin as P and whose distance from the origin is the reciprocal of the distance from P to the origin. Here’s an example:

P and P’ get swapped by inversion.

One can check that, algebraically, this inversion sends a point P with coordinates (x,y) to a point P' with coordinates (\frac{x}{x^2+y^2}, \frac{y}{x^2+y^2}). Points inside the circle are sent to points outside the circle, points outside the circle are sent to points inside the circle, and points on the circle are sent to themselves. Moreover, as one might expect from the name, the inversion map is its own inverse: applying it twice, we end up where we started. Perfect!

Wait a second, though. We’re being a little too hasty. What about the origin? Where is it sent? Our procedure doesn’t seem to tell us, and if we try to use our algebraic expression, we end up dividing by zero. Since the origin is inside the circle, it should certainly be sent to a point outside the circle, but all of those points are already taken. Also, since points arbitrarily close to the origin get mapped to points arbitrarily far from the origin, we want to send the origin to a point as far away from itself as possible. At first glance, we might seem to be in a quandary here, but longtime readers of this blog will see an obvious solution: the origin gets mapped to a point at infinity! (And the point at infinity, in turn, gets mapped to the origin.)

(Technical note: Since we’ve added a point at infinity, the inversion map should be seen not as a map on the plane \mathbb{R}^2, but on its one-point compactification (or Alexandroff compactification), \hat{\mathbb{R}}^2. In fact, the inversion map is a topological homeomorphism of \hat{\mathbb{R}}^2 with itself.)

Let’s examine what the inversion map does to simple geometric objects. We have already seen what happens to points. It should also be obvious that straight lines through the origin get mapped to themselves. For example, in the image above, the line connecting P and P' gets mapped to itself. (Here we are specifying, of course, that every line contains the point at infinity.)

A bit of thought and calculation will convince you that lines not passing through the origin get sent to circles that do pass through the origin.

The red line, when inverted across the black circle, gets sent to the red circle.

Since the inversion map is its own inverse, circles passing through the origin get mapped to lines that don’t pass through the origin. Circles that don’t pass through the origin, on the other hand, get mapped to other circles that don’t pass through the origin.

The red circle on the left is sent to the red circle on the right through inversion.

There’s an important special case of this phenomenon: a circle that is met perpendicularly by the circle through which we are inverting gets mapped to itself.

The red circle is perpendicular to the circle of inversion and is thus sent to itself.

We thus have a sort of duality between lines and circles that has been revealed through the process of circle inversion. Lines, when seen in the right light, are simply circles with an infinite radius. We’re going to move on to some applications of circle inversion in just a sec, but, first, a pretty picture of an inverted checkerboard.

Left: A checkerboard. Right: A checkerboard inverted across a circle centered at the middle of the board with radius equal to the side length of one checkerboard square. (from Mathographics by R. Dixon)

The introduction of the method of circle inversion is widely attributed to the Swiss mathematician Jakob Steiner, who wrote a treatise on the matter in 1824. When combined with the more familiar rigid transformations of rotation, translation, and reflection, the decidedly non-rigid transformation of inversion gives rise to inversive geometry, which became a major topic of study in nineteenth geometry. It was perhaps most notably applied by William Thomson (later to become 1st Baron Kelvin, immortalized in the name of a certain temperature scale), at the age of 21, to solve problems in electrostatics. Circle inversion also allows for extremely elegant proofs of classical geometric facts. We end today’s post with an example.

Consider three half-circles, all tangent to one another and centered on the same horizontal line, with two placed inside the third, as follows:

An arbelos. (Original by Julio Reis, new version by Rubber Duck, CC BY-SA 3.0)

This figure (or, more precisely, the grey region enclosed by the semicircles) is known as an arbelos, and its first known appearance dates back to The Book of Lemmas by Archimedes. A remarkable fact about the arbelos is that, starting with the smallest of the semicircles in the figure, one can nestle into it an infinite sequence of increasingly small circles, each tangent to the two larger semicircles and the circle appearing before it, thus creating the striking Pappus chain, named for Pappus of Alexandria, who investigated the figure in the 3rd century AD:

A Pappus chain. (By Pbroks13, CC BY 3.0)

Let us label the circles in the Pappus chain (starting with the smallest semicircle in the arbelos) \mathcal{C}_0, \mathcal{C}_1, \mathcal{C}_2, etc. (So, in the picture above, P_1 is the center of \mathcal{C}_1, P_2 is the center of \mathcal{C}_2, and so on.) Clearly, the size of \mathcal{C}_n decreases as n increases, but it is natural to ask how quickly it decreases. It is also natural to ask how the position of the point P_n changes as n increases. In particular, what is the height of P_n above the base of the figure? It turns out that the answers to these two questions are closely related, a fact discovered by Pappus through a long and elaborate derivation in Euclidean geometry, and which we will derive quickly and elegantly through circle inversion.

Let d_n denote the diameter of the circle \mathcal{C}_n, and let h_n denote the height of the point P_n above the base of the Pappus chain (i.e., the line segment AB). We will prove the remarkable formula:

For all n \in \mathbb{N}h_n = n \cdot d_n.

For concreteness, let us demonstrate the formula for \mathcal{C}_3. The same argument will work for each of the circles in the Pappus chain. As promised, we are going to use circle inversion. Our first task is to find a suitable circle across which to invert our figure. And that circle, it turns out, will be the circle centered at A and perpendicular to \mathcal{C}_3:

We will invert our figure across the red circle.

Now, what happens when we invert our figure? First, consider the two larger semicircles in the arbelos, with diameters AC and AB. The circles of which these form the upper half pass through the center of our circle of inversion and thus, as discussed above, are mapped to straight lines by our inversion. Moreover, since the centers of these circles lie directly to the right of A, a moment’s thought should convince you that they are mapped to vertical lines.

Now, what happens to the circles in the Pappus chain? Well, none of them pass through A, so they will all get mapped to circles. \mathcal{C}_3 is perpendicular to the circle of inversion, so it gets mapped to itself. But, in the original diagram, \mathcal{C}_3 is tangent to the larger semicircles in the arbelos. Since circle inversion preserves tangency, in the inverted diagram, \mathcal{C}_3 is tangent to the two vertical lines that these semicircles are mapped to. And, of course, the same is true of all of the other circles in the Pappus chain. Finally, note that, since the center of \mathcal{C}_0 lies on the base of the figure, which passes through the center of our inversion circle, it also gets mapped to a point on the base of the figure. Putting this all together, we end up with the following striking figure:

Pappus chain inversion: before and after. (from “Reflections on the Arbelos” by Harold P. Boas)

The circle with diameter AB gets mapped to the vertical line through B', and the circle with diameter AC gets mapped to the vertical line through C'. Our Pappus chain, meanwhile, is transformed by inversion into an infinite tower of circles, all of the same size, bounded by these vertical lines. Moreover, the circle \mathcal{C}_3 and the point P_3 are left in place by the inversion. It is now straightforward to use this tower to calculate the height h_3 of P_3 in terms of the diameter d_3 of \mathcal{C}_3. To get from P_3 down to the base, we must first pass through half of \mathcal{C}_3, which has a height of \frac{d_3}{2}. We then must pass through the image of \mathcal{C}_2 under the inversion, which has a height of d_3. Then the image of \mathcal{C}_1, which also has a height of d_3. And, finally, the image of the smallest semicircle of the arbelos, which has a height of \frac{d_3}{2}. All together, we get:

h_3 = \frac{d_3}{2} + d_3 + d_3 + \frac{d_3}{2} = 3d_3.

Pretty nice!

For further reading on circle inversion, see Harold P. Boas’ excellent article, “Reflections on the Arbelos.”

Cover image: René Magritte, The false mirror


Infinite Acceleration: Risset Rhythms

In our most recent post, we took a look at and a listen to Shepard tones and their cousins, Shepard-Risset glissandos, which are tones or sequences of tones that create the illusion of perpetually rising (or falling) pitch. The illusion is created by overlaying a number of tones, separated by octaves, rising in unison. The volumes gradually increase from low pitch to middle pitch and gradually decrease from middle pitch to high pitch, leading to a fairly seamless continuous tone.

The same idea can be applied, mutatis mutandis, to percussive loops instead of tones, and to speed instead of pitch, thus creating the illusion of a rhythmic track that is perpetually speeding up (or slowing down). (The mechanism is exactly the same as that of the Shepard tone, so rather than provide an explanation here, I will simply refer the reader to the previous post.) Such a rhythm is known as a Risset rhythm.

I coded up some very basic examples on Supercollider. Here’s an accelerating Risset rhythm:

And a decelerating Risset rhythm:

Here’s a more complex Risset rhythm:

And, finally, a piece of electronic music employing Risset rhythms: “Calculus,” by Stretta.


Infinite Ascent: Shepard Tones

Have you ever been watching a movie and noticed that the musical score was seeming, impossibly, to be perpetually rising, ratcheting up the intensity of the film more and more? Or perhaps it seemed to be perpetually falling, creating a deeper and deeper sense of doom onscreen? If so, it is likely that this effect was achieved using a Shepard tone, a way of simulating an unbounded auditory ascent (or descent) in a bounded range.

To understand how Shepard tones work, let’s look at a simplified implementation of one. We will have three musical voices (middle, low, and high), with an octave between successive voices. The voices then start to move, in unison, and always an octave apart, up through a single octave, over, say, five seconds. As they go, though, they also change their volumes: the middle voice stays at full volume the whole time, the low voice gradually increases from zero volume to full volume, and the high voice gradually decreases from full volume to zero volume. The result will simply sound like a tone rising through an octave, and it can be represented visually as follows.


This by itself is nothing special, though. The trick of the Shepard tone is that this pattern is then repeated over, and over, and over again. Each repetition of the pattern sounds like a tone ascending an octave, but, because of the volume modulation, successive patterns are aurally glued together: the low voice from one cycle leads seamlessly to the middle voice of the next, the middle voice from one cycle leads seamlessly to the high voice of the next, and the high voice simply fades away. The result sounds like a perpetually increasing tone.


Note the similarity to the visual barber pole illusion, in which a rotating pole causes stripes to appear to be perpetually rising. Also, this whole story can be turned upside down, which will lead to a perpetually falling tone.

Let’s hear some Shepard tones in action! Now, in practice, using only three voices does not create a particularly convincing illusion, so, to make these sounds, I used nine voices, spread across nine octaves. Also, linearly varying the volume, as in the above visualization, seems to make it more noticeable when voices enter or fade away, so I used something more like a bell curve.

(Technical notes: These Shepard tones were created in Supercollider, using modified code written by Eli Fieldsteel, from whose YouTube tutorials I have learned a great deal of what I know about Supercollider. Also, I used a formant oscillator instead of the more traditional sine oscillator.)

First, a simple ascending Shepard tone:

The effect becomes more convincing, and the tone more interesting, if multiple Shepard tones are played simultaneously at a fixed interval. Here, we have two ascending Shepard tones separated by a tritone, a.k.a. the devil’s interval, a.k.a. half an octave:

Next, three descending Shepard tones, arranged in a minor triad:

Finally, two Shepard tones, with one ascending and the other descending:

The origins of the Shepard tone lie with Roger Shepard, a 20th-century American cognitive scientist, as a sequence of discrete notes. The continuous Shepard scale, or Shepard-Risset glissando, which our code approximates, was introduced by French composer Jean-Claude Risset, who perhaps most notably used it in his Computer Suite from Little Boy from 1968.

More recently, it has prominently been deployed by Christopher Nolan and Hans Zimmer, as the basis for the Batpod sound in The Dark Knight and in the Dunkirk soundtrack.

Cover image: M.C. Escher, Waterfall