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 $A' = PG \cap AI.$

$RD||AI, PRGD$ 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.

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2019 IMO Problems/Problem 6

Problem

Solution