# 2019 IMO Problems/Problem 6

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## Problem

Let $I$ be the incenter of acute triangle $ABC$ with $AB \neq AC$. The incircle $\omega$ of $ABC$ is tangent to sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. The line through $D$ perpendicular to $EF$ meets $\omega$ again at $R$. Line $AR$ meets ω again at $P$. The circumcircles of triangles $PCE$ and $PBF$ meet again at $Q$. Prove that lines $DI$ and $PQ$ meet on the line through $A$ perpendicular to $AI$.

## Solution

Step 1

We find an auxiliary point $S.$

Let $G$ be the antipode of $D$ on $\omega, GD = 2R,$ where $R$ is radius $\omega.$

We define $A' = PG \cap AI.$

$RD||AI, PRGD$ is cyclic $\implies \angle IAP = \angle DRP = \angle DGP.$

$RD||AI, RD \perp RG, RI=GI \implies \angle AIR = \angle AIG \implies$ $$\triangle AIR \sim \triangle GIA' \implies \frac {AI}{GI} = \frac {RI}{A'I}\implies A'I \cdot AI = R^2.$$ An inversion with respect $\omega$ swap $A$ and $A' \implies A'$ is the midpoint $EF.$

Let $DA'$ meets $\omega$ again at $S.$ We define $T = PS \cap DI.$

Opposite sides of any quadrilateral inscribed in the circle $\omega$ meet on the polar line of the intersection of the diagonals with respect to $\omega \implies DI$ and $PS$ meet on the line through $A$ perpendicular to $AI.$ The problem is reduced to proving that $Q \in PST.$

Step 2

We find a simplified way to define the point $Q.$

We define $\angle BAC = 2 \alpha \implies \angle AFE = \angle AEF = 90^\circ – \alpha \implies$ $\angle BFE = \angle CEF = 180^\circ – (90^\circ – \alpha) = 90^\circ + \alpha = \angle BIC$ $(AI, BI,$ and $CI$ are bisectrices).

We use the Tangent-Chord Theorem and get $$\angle EPF = \angle AEF = 90^\circ – \alpha.$$

$\angle BQC = \angle BQP + \angle PQC = \angle BFP + \angle CEP =$ $=\angle BFE – \angle EFP + \angle CEF – \angle FEP =$ $= 90^\circ + \alpha + 90^\circ + \alpha – (90^\circ + \alpha) =$ $90^\circ + \alpha = \angle BIC \implies$

Points $Q, B, I,$ and $C$ are concyclic.

Step 3

We perform inversion around $\omega.$ The straight line $PST$ maps onto circle $PITS.$ We denote this circle $\Omega.$ We prove that the midpoint of $AD$ lies on the circle $\Omega.$

In the diagram, the configuration under study is transformed using inversion with respect to $\omega.$ The images of the points are labeled in the same way as the points themselves. Points $D,E,F,P,S,$ and $G$ have saved their position. Vertices $A, B,$ and $C$ have moved to the midpoints of the segments $EF, FD,$ and $DE,$ respectively.

Let $M$ be the midpoint $AD.$

We define $\angle MID = \beta, \angle MDI = \gamma \implies$ $\angle IMA = \angle MID + \angle MDI = \beta + \gamma = \varphi.$ $DI = IS \implies \angle ISD = \gamma.$

$MI$ is triangle $DAG$ midline $\implies MI || AG \implies$ $$MI || PG \implies \angle MAP = \angle AMI = \varphi.$$ $$\angle DPA = 90^\circ \implies PM = MA \implies$$ $$\angle PMA = \angle PMS = 180^\circ – 2 \varphi.$$ $PI = IS \implies \angle PIS = 180^\circ – 2 \varphi =\angle DPA \implies$ point $M$ lies on $\Omega.$ $ABDC$ is parallelogram $\implies M$ is midpoint $BC.$

Step 4

We prove that image of $Q$ lies on $\Omega.$

In the inversion plane the image of point $Q$ lies on straight line $BC$ (It is image of circle $BIC)$ and on circle $PCE.$

$$\angle PQM = \angle PQC = \angle PEC = \angle PED = \angle PSD = \angle PSM \implies$$ point $Q$ lies on $\Omega$.