2019 IMO Problems/Problem 6

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 $A' = PG \cap AI.$ $RD||AI, PRGD$ cyclic $\implies \angle IAP = \angle DRP = \angle DGP.$ $RD||AI, RD \perp RG, RI=GI \implies \angle AIR = \angle AIG \implies \triangle AIR \sim \triangle GIA' \implies \frac {AI}{GI} = \frac {RI}{A'I}\implies A'I \cdot AI = R^2 \implies$ an inversion with respect $omega$ swap $A$ and $A' \implies A'$ is the midpoint $EF.$ Let $DA'$ meets $\omega$ again at S (other than D). 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 $Q \in PST.$

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2019 IMO Problems/Problem 6

Problem

Solution