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Difference between revisions of "2019 IMO Problems/Problem 6"

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We define <math>A' = PG \cap AI.</math>
 
We define <math>A' = PG \cap AI.</math>
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<math>RD||AI, PRGD</math> cyclic <math>\implies \angle IAP = \angle DRP = \angle DGP.</math>
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−
<math>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</math> an inversion with respect <math>omega</math> swap <math>A</math> and <math>A' \implies A'</math> is the midpoint <math>EF.</math>
+
<math>RD||AI, PRGD</math> is cyclic <math>\implies \angle IAP = \angle DRP = \angle DGP.</math>
−
Let <math>DA'</math> meets <math>\omega</math> again at S (other than D). We define <math>T = PS \cap DI.</math>
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Opposite sides of any quadrilateral inscribed in the circle <math>omega</math> meet on the polar line of the intersection of the diagonals with respect to <math>\omega \implies DI</math> and <math>PS</math> meet on the line through <math>A</math> perpendicular to <math>AI.</math>
+
<math>RD||AI, RD \perp RG, RI=GI \implies \angle AIR = \angle AIG  \implies</math>
 +
<cmath>\triangle AIR \sim \triangle GIA' \implies  \frac {AI}{GI} = \frac {RI}{A'I}\implies A'I \cdot AI = R^2.</cmath>
 +
An inversion with respect <math>\omega</math> swap <math>A</math> and <math>A' \implies A'</math> is the midpoint <math>EF.</math>
 +
 
 +
Let <math>DA'</math> meets <math>\omega</math> again at <math>S.</math> We define <math>T = PS \cap DI.</math>
 +
 
 +
Opposite sides of any quadrilateral inscribed in the circle <math>\omega</math> meet on the polar line of the intersection of the diagonals with respect to <math>\omega \implies DI</math> and <math>PS</math> meet on the line through <math>A</math> perpendicular to <math>AI.</math>
 
The problem is reduced to proving that <math>Q \in PST.</math>
 
The problem is reduced to proving that <math>Q \in PST.</math>
 +
 +
<i><b>Step 2</b></i>

Revision as of 12:20, 29 August 2022

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

2019 6 s1.png

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

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