# Inversion

Du¹an Ðukiæ

## Introduction

An inversion with respect to a given circle (sphere) is the map sending each point $$A$$ other than the center $$O$$ of the circle to the point $$A\prime$$ on ray $$OA$$ such that $$OA \prime=r^2/OA$$. What makes this map useful is the fact that it preserves angles and maps lines and circles onto lines or circles. Thus appropriate inversions can reduce the number of unpleasant circles (mapping them to lines) and often even turn a difficult problem into a quite simple one, as we show on a number of solved problems. Problems range from Ptolemy$$\prime$$s inequality, to Feuerbach$$\prime$$s theorem, and some of the hardest problems appearing on math competitions.

## General properties

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 $$A\prime =\Psi(A)$$ on the ray $$OA$$ such that $$OA\cdot OA\prime =r^2$$. From now on, unless noted otherwise, $$X\prime$$ 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". The next thing we observe is that $$\triangle P\prime OQ\prime \sim\triangle QOP$$ for all point $$P,Q\neq O$$ (for $$\angle P\prime OQ\prime =\angle QOP$$ and $$OP\prime /OQ\prime =(r^2/OP)/(r^2/OQ)= OQ/OP$$), with the ratio of similitude $$\frac{r^2}{OP\cdot OQ}$$. As a consequence, we have $\angle OQ\prime P\prime =\angle OPQ\quad\mbox{and} \quad P\prime Q\prime =\frac{r^2}{OP\cdot OQ}PQ.$

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 $$\angle OPQ=\angle OQ\prime P\prime$$ is right, so $$Q\prime$$ lies on circle $$k$$ with diameter $$OP\prime$$. 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 $$A\prime ,B\prime ,C\prime ,D\prime$$. The following angles are regarded as oriented. Let us show that $$\angle A\prime C\prime B\prime =\angle A\prime D\prime B\prime$$. We have $$\angle A\prime C\prime B\prime =\angle OC\prime B\prime -\angle OC\prime A\prime =\angle OBC-\angle OAC$$ and analogously $$\angle A\prime D\prime B\prime =\angle OBD-\angle OAD$$, which implies $$\angle A\prime D\prime B\prime -\angle A\prime C\prime B\prime =\angle CBD-\angle CAD=0$$, 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).
Remark. Based on what we have seen, it can be noted that inversion preserves angles between curves, in particular circles or lines. Maps having this property are called conformal.

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" circles and angles on a picture. Thus, some pictures cry" to be inverted:

• There are many circles and lines through the same point $$A$$. Invert through $$A$$.
Example (IMO 2003, Shortlist)

Let $$\Gamma_1,\Gamma_2,\Gamma_3,\Gamma_4$$ be distinct circles such that $$\Gamma_1,\Gamma_3$$ are externally tangent at $$P$$, and $$\Gamma_2,\Gamma_4$$ are externally tangent at the same point $$P$$. Suppose that $$\Gamma_1$$ and $$\Gamma_2$$; $$\Gamma_2$$ and $$\Gamma_3$$; $$\Gamma_3$$ and $$\Gamma_4$$; $$\Gamma_4$$ and $$\Gamma_1$$ meet at $$A,B,C,D$$, respectively, and that all these points are different from $$P$$. Prove that $\frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}.$

• There are many angles $$\angle AXB$$ with fixed $$A,B$$. Invert through $$A$$ or $$B$$.
Example (IMO 1996, problem 2)

Let $$P$$ be a point inside $$\triangle ABC$$ such that $$\angle APB-\angle C=\angle APC-\angle B$$. Let $$D,E$$ be the incenters of $$\triangle APB,\triangle APC$$ respectively. Show that $$AP$$, $$BD$$, and $$CE$$ meet in a point.

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

## Problems

Problem 1

Circles $$k_1,k_2,k_3,k_4$$ are such that $$k_2$$ and $$k_4$$ each touch $$k_1$$ and $$k_3$$. Show that the tangency points are collinear or concyclic.

Problem 2

Prove that for any points $$A,B,C,D$$, $$AB\cdot CD+BC\cdot DA \geq AC\cdot BD$$, and that equality holds if and only if $$A,B,C,D$$ are on a circle or a line in this order. (Ptolemy$$\prime$$s inequality)

Problem 3

Let $$\omega$$ be the semicircle with diameter $$PQ$$. A circle $$k$$ is tangent internally to $$\omega$$ and to segment $$PQ$$ at $$C$$. Let $$AB$$ be the tangent to $$k$$ perpendicular to $$PQ$$, with $$A$$ on $$\omega$$ and $$B$$ on segment $$CQ$$. Show that $$AC$$ bisects the angle $$\angle PAB$$.

Problem 4

Points $$A,B,C$$ are given on a line in this order. Semicircles $$\omega,\omega_1,\omega_2$$ are drawn on $$AC,AB$$, $$BC$$ respectively as diameters on the same side of the line. A sequence of circles $$(k_n)$$ is constructed as follows: $$k_0$$ is the circle determined by $$\omega_2$$ and $$k_n$$ is tangent to $$\omega,\omega_1, k_{n-1}$$ for $$n\geq1$$. Prove that the distance from the center of $$k_n$$ to $$AB$$ is $$2n$$ times the radius of $$k_n$$.

Problem 5

A circle with center $$O$$ passes through points $$A$$ and $$C$$ and intersects the sides $$AB$$ and $$BC$$ of the triangle $$ABC$$ at points $$K$$ and $$N$$, respectively. The circumscribed circles of the triangles $$ABC$$ and $$KBN$$ intersect at two distinct points $$B$$ and $$M$$. Prove that $$\measuredangle OMB=90^{\circ}.$$ (IMO 1985-5.)

Problem 6

Let $$p$$ be the semiperimeter of a triangle $$ABC$$. Points $$E$$ and $$F$$ are taken on line $$AB$$ such that $$CE=CF=p$$. Prove that the circumcircle of $$\triangle EFC$$ is tangent to the excircle of $$\triangle ABC$$ corresponding to $$AB$$.

Problem 7

Prove that the nine-point circle of triangle $$ABC$$ is tangent to the incircle and all three excircles. (Feuerbach$$\prime$$s theorem)

Problem 8

The incircle of a triangle $$ABC$$ is tangent to $$BC,CA,AB$$ at $$M,N$$ and $$P$$, respectively. Show that the circumcenter and incenter of $$\triangle ABC$$ and the orthocenter of $$\triangle MNP$$ are collinear.

Problem 9

Points $$A,B,C$$ are given in this order on a line. Semicircles $$k$$ and $$l$$ are drawn on diameters $$AB$$ and $$BC$$ respectively, on the same side of the line. A circle $$t$$ is tangent to $$k$$, to $$l$$ at point $$T\neq C$$, and to the perpendicular $$n$$ to $$AB$$ through $$C$$. Prove that $$AT$$ is tangent to $$l$$.

Problem 10

Let $$A_1A_2A_3$$ be a nonisosceles triangle with incenter $$I$$. Let $$C_i$$, $$i=1,2,3$$, be the smaller circle through $$I$$ tangent to $$A_iA_{i+1}$$ and $$A_iA_{i+2}$$ (the addition of indices being mod 3). Let $$B_i$$, $$i=1,2,3$$, be the second point of intersection of $$C_{i+1}$$ and $$C_{i+2}$$. Prove that the circumcenters of the triangles $$A_1B_1I,A_2B_2I,A_3B_3I$$ are collinear. (IMO 1997 Shortlist)

Problem 11

If seven vertices of a hexahedron lie on a sphere, then so does the eighth vertex.

Problem 12

A sphere with center on the plane of the face $$ABC$$ of a tetrahedron $$SABC$$ passes through $$A,B$$ and $$C$$, and meets the edges $$SA,SB,SC$$ again at $$A_1,B_1,C_1$$, respectively. The planes through $$A_1,B_1,C_1$$ tangent to the sphere meet at a point $$O$$. Prove that $$O$$ is the circumcenter of the tetrahedron $$SA_1B_1C_1$$.

Problem 13

Let $$KL$$ and $$KN$$ be the tangents from a point $$K$$ to a circle $$k$$. Point $$M$$ is arbitrarily taken on the extension of $$KN$$ past $$N$$, and $$P$$ is the second intersection point of $$k$$ with the circumcircle of triangle $$KLM$$. The point $$Q$$ is the foot of the perpendicular from $$N$$ to $$ML$$. Prove that $$\angle MPQ=2\angle KML$$.

Problem 14

The incircle $$\Omega$$ of the acute-angled triangle $$ABC$$ is tangent to $$BC$$ at $$K$$. Let $$AD$$ be an altitude of triangle $$ABC$$ and let $$M$$ be the midpoint of $$AD$$. If $$N$$ is the other common point of $$\Omega$$ and $$KM$$, prove that $$\Omega$$ and the circumcircle of triangle $$BCN$$ are tangent at $$N$$. (IMO 2002 Shortlist)

2005-2018 IMOmath.com | imomath"at"gmail.com | Math rendered by MathJax