What Is Inversion?

Inversion

If this is the first time you're learning inversion, your reaction might be : "What? Why in the world do we have to invert?". Yet, as we shall see, inversion is really powerful.

(as shown in the diagram, $X_1$ is the image of $X$ under inversion with respect to $\omega$.)

and vice versa.

This is why this transformation is called inversion!

Why? You might ask. The proof is below

By the definition of inversion, $\frac{IX'}{IY} = \frac{r^2}{IY \cdot IX} = \frac{IY'}{IX}$. Thus, we can conclude that the two triangles are indeed similar.

This provides us with a formula to calculate lengths after inversion, namely $X'Y' = XY \cdot \frac{r^2}{IX \cdot IY}$ and $XY = X'Y' \cdot \frac{r^2}{IX' \cdot IY'}$.

Now, let's see what happens to lines and circles after inversion. Note that I'll give the following facts without proof. If you're interested, you can try to prove them yourself. (they're not really hard)

From now, inverting about a point means inverting about a circle with radius $1$ and center that point.

Firstly, let $l'$ be the image of line $l$ under inversion about $I$.

Next, we look at circles.

Let $\omega'$ be the image of a circle $\omega$ with center $O$ under inversion about $I$.

Hopefully inversion will be helpful to you in Olympiad problems!

$\sqrt{bc}$-inversion

$\sqrt{bc}$-inversion is a combination of inversion and reflection. Let's see what I mean :

Given a triangle $ABC$ we consider the transformation that first reflects point $X$ over the $A$-angle bisector into $X'$ and then inverts $X'$ about $A$ with radius $\sqrt{bc}$ into $X''$. We call $X''$ the image of $X$ in $\sqrt{bc}$-inversion.

Note that $AB = c, BC = a, AC = b$ for convention.

Now, what does this whole complicated thing even do? Well, this might not seem clear right now, but first we note a few properties :

(a) $B$ maps to $C$ and $C$ maps to $B$.

This is obvious and doesn't require further explanation.

(b) $BC$ maps to the circumcircle of $ABC$ and vice versa.

This is also obvious.

(c) $AX$ and $AX'$ are isogonal for $X \neq A$.

This is clearly true as they are the reflections across the $A$-angle bisector.

I'll show the power of $\sqrt{bc}$-inversion with the following example :

(Serbia $2008$) Let $ABC$ be a triangle. $D$ and $E$ lie on $AB$ such that $AD = AC, BE = BC$ and the points $D, A, B, E$ are collinear in that order. The bisectors of angle $A$ and $B$ intersect $BC, AC$ at $P$ and $Q$ respectively, and the circumcircle of $ABC$ at $M$ and $N$ respectively. Let $O_1$ be the circumcenter of $BME$ and $O_2$ be the circumcenter of $AND$. $AO_1$ and $BO_2$ intersect at $X$. Prove that $CX \perp PQ$.

Try to solve this without looking at the solution first :

Let's try $\sqrt{bc}$-inversion again.

and vice versa.

and vice versa.

However, this clearly implies that $CX$ is isogonal with $CO$ in $\triangle CPQ$. Since $H$ and $O$ are isogonal conjugates (a well-known fact which is not hard to prove), $CX$ is an altitude in $\triangle CPQ$! Thus, $CX \perp PQ$!

(Iran $1995$) Let $M, N, P$ be points where the incircle of scalene triangle $ABC$ touches its sides $BC, CA, AB$ respectively. Prove that the orthocenter of $MNP$, the incenter of $\triangle ABC$ and the circumcenter of $\triangle ABC$ are collinear.