Inversions and Möbius transformations

 (This is Glen.)

At some point, Ker Yang wrote a post on Reim's theorem, which he used to solve two past year IMO problems. I remember commenting to him that I had solved neither of them with Reim (and no, I did not bash them). Later on, I tried reconstructing my solution to IMO 2017/4, and I noticed something interesting that made me find another (slightly weird) solution that (I think) isn't on AoPS. So that's what I'll be writing about today.

First, the problem:

(IMO 2017/4) Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.

Inversion

As the title of this post suggests, my solution involves inversion. I'm going to define it here, and then not say anything else about it; we'll derive the relevant properties later in the post. If you're not familiar with inversion, I suggest you skip the next section with my solution and come back to it later - verifying that the pictures I drew are indeed the inverted diagrams is a good exercise!

Anyway, given a circle with centre $O$ and radius $r$, an inversion is a geometric transformation which sends each point $P$ in the plane to the point $P'$ on the ray $\overrightarrow{OP}$ such that $OP\cdot OP' = r^2$.

This swaps certain pairs of points in the plain, except...you might have noticed that this definition doesn't quite work for the point $O$. However, notice that as $P$ gets closer to $O$, $P'$ goes further away, towards infinity. Hence, the usual way to think about how inversion acts on the centre of the circle is that it swaps it with a new point "at infinity".

The intuitive way to think about this is that inversion is "reflection across a circle" (instead of a line). But notice that inverting about different circles centred at the same point gives homothetic diagrams, and so we often don't really care about the radius of inversion, and just call it an "inversion about $O$".

Why is this useful? Well, inversion turns out to have a lot of nice geometric properties which we'll explore later. But for those who do have some basic knowledge of these properties, here's my original solution to the IMO problem:

A solution to IMO 2017/4

When I first tried this problem, my first instinct was to invert about $S$, because it looked like the centre of the universe. This looks like:

This had even more circles than the original question, so I abandoned it. I then tried to invert about $R$, which went better:

$\Omega$ and the line $AB$ now become parallel lines, and $T$ is the midpoint of $S$ and $\infty$ (recall that inversion swaps $R$ with the point at infinity). Then, $BSJA$ and $\infty KJA$ are isosceles trapeziums, so $\infty KSB$ is a parallelogram, and $T$ is the intersecction of its diagonals (even if it doesn't look like it in the picture). So, $(KT\infty)$ and $(STB)$ are rotations of each other about $T$ by $180^\circ$, so they're tangent to each other at $T$.

So we're done, and when I first did this back in 2017, I was happy and moved on. But when I got to this point again, something crossed my mind that will really only make sense if I talk about Möbius transformations.

Möbius transformations

Introduction

These aren't some fancy thing; a Möbius transformation is just a map of the form $$f:z\mapsto\frac{az+b}{cz+d}$$ where $a,b,c,d\in\mathbb{C}$ and $ad-bc\ne 0$. Now, you might complain and say that the denominator could be $0$, so really, we should define this as a map from $\mathbb{C}\cup\{\infty\}$ to itself, where we let $\frac{a\infty+b}{c\infty+d} = \frac{a}c$. Note that this is bijective; the inverse isn't hard to solve for. This $\mathbb{C}\cup\{\infty\}$ is a thing that people generally care about, and has a bunch of names such as the Riemann sphere or $\mathbb{CP}^1$. The intuition for "sphere" is the following:

Take a sphere; we can project points from north pole onto the plane containing the equator. This works for every point on the sphere except for the north pole itself; we think of this as the point at infinity. If you imagine points on the complex plane with $|z|\rightarrow\infty$, then the corresponding points on the sphere get closer to the north pole, so this does check out with our notion of infinity from inversion. In this setting, lines and circles on the plane correspond to circles on the sphere; lines correspond exactly to the circles which go through the north pole (i.e. the point at infinity). From this point of view, there is fundamentally no difference between lines and circles; a line is merely a special type of circle that happens to go through $\infty$.

But let's look at the plane because it's the setting we're (probably) more familiar with. What is a Möbius transformation, geometrically? Let's look at some special cases.

• When $b=c=0,d=1$, we have $f:z\mapsto az$, which is a spiral similarity about the origin: if $a=re^{i\theta}$, then we're scaling by $r$ and rotating by $\theta$. In particular, if $a\in\mathbb{R}$, then we have a homothety, and if $|a|=1$ we have a rotation.
• When $a=b=c=1,d=0$, we have $f:z\mapsto z+1$, which is a translation to the right by $1$.
• When $a=d=0,b=c=1$, we have $f:z\mapsto\frac1z$, which is...well, one is tempted to say that it's an inversion, but we have to be a little bit careful about this. Note that $\frac1z = \frac{\overline{z}}{|z|^2}$, so we're taking $z$, dividing it by the square of its norm, and then reflecting across the real axis. This is not an inversion, but an inverflection: the composition of inversion across a circle (in this case, the unit circle centred at the origin) and a reflection.

These look terrifyingly different, but they actually are more similar than you think: recall that in our setting that lines and circles are basically the same thing, so reflections across lines are just special cases of inversions. Now, note that:

• The composition of the inversions about two circles of different radii with the same centre is a homothety.
• The composition of the reflections across two non-parallel lines is a rotation.
• The composition of the reflections across two parallel lines is a reflection.
• The composition of an inversion and a reflection is an inverflection (duh).

More generally, a Möbius transformation is the composition of (an even number of) arbitrary inversions (I believe four is enough to get you anything), though this isn't very helpful. Rather, I find it more useful to think of them as a general class of functions which unify all the usual geometric transformations we care about: spiral similarities, translations, inverflections.

Properties

I'm now going to list some properties of Möbius transformations; these should be fairly easy to verify if you do them in order.

1. Composition and inverses: The composition of $z\mapsto\frac{az+b}{cz+d}$ and $z\mapsto\frac{a'z+b'}{c'z+d}$ corresponds to multiplying the matrices $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $\begin{pmatrix} a' & b' \\ c' & d' \end{pmatrix}$. This should also tell you the formula for the inverse of a Möbius transformation. (Proof: Algebra bash.)
2. Generators: Every Möbius transformation can be written as a composition of the three examples $z\mapsto az$, $z\mapsto z+1$, $z\mapsto\frac1z$ given above. (Proof: Anything goes, really, but you might find it helpful to first show you can get all linear functions.)
3. Cross-ratios: Define the cross ratio of complex numbers $w,x,y,z$ to be $(w,x;y,z) = \frac{(w-y)(x-z)}{(w-z)(x-y)}.$ (If they all lie on a line or a circle, then this is exactly the usual notion of cross-ratio.) If, say, $w=\infty$, then we let infinities cancel and we're left with $\frac{x-z}{x-y}$. By this definition, Möbius transformations preserve the cross-ratio. (Proof: This looks scary to bash, but really, you just need to check it on the three generators.)
4. (Triple) transitivity: Let $x,y,z$ and $x',y',z'$ be two sets of three distinct points. Then, there's a unique Möbius transformation that sends $x\mapsto x',y\mapsto y',z\mapsto z'$. (Proof: WLOG $x'=1,y'=0,z'=\infty$. If $f$ works, then we must have $(w,x;y,z) = (f(w),1;0,\infty) = f(w)$. On the other hand, the LHS is a Möbius transformation in $w$.)
5. Circles and lines: Möbius transformations send circles/lines to circles/lines. (Proof: Notice that four points lie on a circle/line iff their cross-ratio is real. This is probably more familiar in the form $$\left. \frac{w-y}{w-z} \middle/ \frac{\overline{w-y}}{\overline{w-z}} \right. = \left. \frac{x-y}{x-z} \middle/ \frac{\overline{x-y}}{\overline{x-z}} \right.,$$ which checks that the (directed) angles at $w$ and $x$ are equal.)
   
6. Angles: Möbius transformations preserve angles between curves (there is a warning here: our usual notion of angles is of angles between lines, but lines aren't necessarily sent to lines!) Another way of saying this is that Möbius transformations are conformal. (Proof: Let $w$ be the intersection of the two curves, $y,z$ points on the curves, and $x$ some random point. If we look at $(w,x;y,z)/\overline{(w,x;y,z)}$, as $y,z$ move towards $w$, the $x$ terms tend to $1$, and the rest tends to twice the directed angle between the curves.
   
   
   Then, use the fact that Möbius transformations preserve the cross-ratio. If you want to be pedantic, you technically also need to use the fact that Möbius transformations are continuous. If you want to be even more pedantic, you also need to insist that the curves are continuously differentiable for angles to even make sense.)
7. Tangencies: Möbius transformations send tangent curves to tangent curves. This is particularly important if you notice that lines tangent to each other at infinity are the same thing as parallel lines (i.e. the circles they correspond to on the Riemann sphere are tangent at the north pole). (Proof: This is the previous point but with the angles being $0$.)

Revisiting inversions

How does all of this relate to inversions? The inversion about a circle with centre $o\in\mathbb{C}$ and radius $r$ is the map $z\mapsto\frac{r^2(z-o)}{|z-o|^2} + o = \overline{\frac{r^2}{z-o} + \overline{o}}$, which is the composition of a Möbius transformation and complex conjugation. So, the properties listed above hold, with some minor modifications:

• Inversions send cross-ratios to their complex conjugates. (On circles/lines, they are thus preserved).
• Inversions send circles/lines to circles/lines.
• Inversions preserve angles between curves - more accurately, they flip the signs of directed angles.
• Inversions preserve tangency. In particular, inversion about the tangency point of two circles/lines gives you two parallel lines.

This should be enough for you to follow along with my inverted diagrams for IMO 2017/4, if you couldn't before. Perhaps the most non-trivial part is showing that $T$ is the midpoint of $\infty S$ in the inversion about $R$: this comes from the fact that $(R,T;S,\infty) = -1$. Alternatively, it's a bit more obvious if you look at our original definition of an inversion.

Back to IMO 2017/4

Let's look back at the final line of my original proof:

$(KT\infty)$ and $(STB)$ are rotations of each other about $T$ by $180^\circ$, so they're tangent to each other at $T$.

Wait a second. We've performed an inversion, and then this nice Möbius transformation (rotation by $180^\circ$) happens to send a bunch of points to each other. But what if we perform our inversion, then the rotation, and then invert back? This is the composition of three Möbius transformations (we're complex conjugating twice, once for each inversion), and so we should be getting another Möbius transformation. What is it?

We know that Möbius transformations are triply transitive, so we just need to see where three of the points are sent. The images of $T$ (and $R$, which is now at infinity) under the inversion are fixed by the rotation, so the overall composition must fix $T$ and $R$. On the other hand, the rotation swaps the images of $S$ and $\infty$, so the composition must swap $S$ and $\infty$. It's not hard to guess what this Möbius transformation is: it's the inversion about $S$ with radius $SR=ST$, followed by a reflection across $RT$.

That's pretty nice. Can we solve the question directly from this? At this point, you might be thinking, "didn't we previously try inverting about $S$ and fail?" But that's not exactly right. We're not merely trying to transform the diagram to get a nicer one (we already know that we won't); we're trying to send points to each other to see what new information we get.

For reference, here's the diagram again:

Now $\Omega = (SJR)$ is sent to a line through $R$. But we know more than that: Möbius transformations preserve directed angles, so it has to be sent to the tangent line at $R$, i.e. $RA$. Likewise, $\Gamma = (STA)$ is sent to the tangent line at $T$. $A,B$, i.e. the intersections of $RA$ and $\Gamma$, are thus sent to the intersections of $\Omega$ and the tangent at $T$. But wait! One of these is supposed to be $K$. From the configuration, the Möbius transformation must swap $B$ and $K$, but why?

Before we proceed, note that if we show this, we're done: if the Möbius transformation swaps $B,K$, then it swaps $KT = (KT\infty)$ with $(STB) = \Gamma$, which, as mentioned before, means that $\Gamma$ and $KT$ are tangent at $T$.

Now, note that $B$ lies on $RA$ and $K$ lies on $\Omega$ and the Möbius transformation swaps $RA$ and $\Omega$. So, we just have to show that $\angle KST = \angle TSB$. But we have $\angle KST = 180^\circ - \angle RSK = 180^\circ - \angle RJK = \angle AJR = \angle KRA$ while $\angle TSB = \angle TAB$, so we're just left with showing that $KR$ and $TA$ are parallel, which is...oh, Reim. Sorry, Ker Yang.

I would argue that this is still more motivatable than constructing a parallelogram.

Conjugating symmetries

What I've just illustrated above is an example of a more general phenomenon. I first learnt this when David did a guest session on inversion during the 2016 IMO intensive training, which really changed how I thought of inversion. The naive usage of inversion generally goes along the lines of: invert about a nice point in the diagram, and then this new diagram happens to look a lot nicer and you can solve the problem directly. Sometimes, this is because the circles all turn into lines, but often, it's because you end up with some nice symmetric diagram which gives you a lot of information for free. (For an example, take the diagram on page 3 of David's notes and invert it about $A$.)

However, in such cases, there is often another inversion that does the same trick: instead of inverting to get a nice diagram, you invert to send points to each other. This gives you information; for instance, in David's example, you get some collinearity for free. This version of inversion is a lot more useful in general: you can't always make the diagram nicer just by inverting about points, but sending points to each other is far more achievable.

Why exactly does this work? What we want to show is that doing an inversion $f$, performing a reflection about a line $\ell$, and then undoing $f$ gives you some other inversion. In fact, it is the inversion about $f^{-1}(\ell)$. To see this, note that the composition fixes $f^{-1}(\ell)$. On the other hand, note that inversion and reflection both satisfy the following property: suppose the inversion/reflection swaps $P$ and $P'$. Then any circle through $P$ that's orthogonal to the circle/line we're inverting/reflecting across also passes through $P'$ (and vice versa).

Since inversions (or more generally, Möbius transformations) send circles/lines to circles/lines and preserve the property of being orthogonal, the inversion about $f^{-1}(\ell)$ is exactly the map we wanted.

(By the way, this argument is really why reflection is a special case of inversion.)

In the previous example, we performed an inversion, a rotation by $180^\circ$, and then undid the inversion. But a rotation is just the composition of two reflections across lines which meet each other at right angles. It's thus not much of a surprise that it ends up corresponding to an inverflection, which is an inversion about a circle followed by a reflection across a line that's orthogonal to the circle.

In both of these examples, we compared functions $g$ with $f^{-1}gf$. The latter is known as the conjugate of $g$ by $f$ and is something that people care about in group theory. In general, the conjugate $f^{-1}gf$ does the "same thing" as $g$, but in a reparametrised version of the universe (where the reparametrisation is given by $f$). Conjugates do the same thing as each other, but from a different perspective.

I hope this post, too, has let you view things from a different perspective.