Olympiad training materials (main page)

Inversion \(\Psi\) is a map of a plane or space without a fixed point \(O\) onto itself, determined by a circle \(k\) with center \(O\) and radius \(r\), which takes point \(A\neq O\) to the point \(A' =\Psi(A)\) on the ray \(OA\) such that \(OA\cdot OA' =r^2\). From now on, unless noted otherwise, \(X' \) always denotes the image of object \(X\) under a considered inversion.

Clearly, map \(\Psi\) is continuous and inverse to itself, and maps the interior and exterior of \(k\) to each other, which is why it is called ``inversion_quot_. The next thing we observe is that \(\triangle P' OQ' \sim\triangle QOP\) for all point \(P,Q\neq O\) (for \(\angle P' OQ' =\angle QOP\) and \(OP' /OQ' =(r^2/OP)/(r^2/OQ)= OQ/OP\)), with the ratio of similitude \(\frac{r^2}{OP\cdot OQ}\). As a consequence, we have \[\angle OQ' P' =\angle OPQ\quad\mbox{and} \quad P' Q' =\frac{r^2}{OP\cdot OQ}PQ.\]

What makes inversion attractive is the fact that it maps lines and circles into lines and circles. A line through \(O\) (\(O\) excluded) obviously maps to itself. What if a line \(p\) does not contain \(O\)? Let \(P\) be the projection of \(O\) on \(p\) and \(Q\in p\) an arbitrary point of \(p\). Angle \(\angle OPQ=\angle OQ' P' \) is right, so \(Q' \) lies on circle \(k\) with diameter \(OP' \). Therefore \(\Psi(p)=k\) and consequently \(\Psi(k)=p\). Finally, what is the image of a circle \(k\) not passing through \(O\)? We claim that it is also a circle; to show this, we shall prove that inversion takes any four concyclic points \(A,B,C,D\) to four concyclic points \(A' ,B' ,C' ,D' \). The following angles are regarded as oriented. Let us show that \(\angle A' C' B' =\angle A' D' B' \). We have \(\angle A' C' B' =\angle OC' B' -\angle OC' A' =\angle OBC-\angle OAC\) and analogously \(\angle A' D' B' =\angle OBD-\angle OAD\), which implies \(\angle A' D' B' -\angle A' C' B' =\angle CBD-\angle CAD=0\), as we claimed. To sum up:

- A line through \(O\) maps to itself.
- A circle through \(O\) maps to a line not containing \(O\) and vice-versa.
- A circle not passing through \(O\) maps to a circle not passing through \(O\) (not necessarily the same).

When should inversion be used? As always, the answer comes with experience and cannot be put on a paper. Roughly speaking, inversion is useful in destroying ``inconvenient_quot_ circles and angles on a picture. Thus, some pictures ``cry_quot_ to be inverted:

- There are many circles and lines through the same point \(A\). Invert through \(A\).

- There are many angles \(\angle AXB\) with fixed \(A,B\). Invert through \(A\) or \(B\).

*Caution:* Inversion may also bring new inconvenient circles and
angles.