Difference between revisions of "2019 IMO Problems/Problem 6"
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Let <math>I</math> be the incenter of acute triangle <math>ABC</math> with <math>AB \neq AC</math>. The incircle <math>\omega</math> of <math>ABC</math> is tangent to sides <math>BC</math>, <math>CA</math>, and <math>AB</math> at <math>D</math>, <math>E</math>, and <math>F</math>, respectively. The line through <math>D</math> perpendicular to <math>EF</math> meets <math>\omega</math> again at <math>R</math>. Line <math>AR</math> meets ω again at <math>P</math>. The circumcircles of triangles <math>PCE</math> and <math>PBF</math> meet again at <math>Q</math>. | Let <math>I</math> be the incenter of acute triangle <math>ABC</math> with <math>AB \neq AC</math>. The incircle <math>\omega</math> of <math>ABC</math> is tangent to sides <math>BC</math>, <math>CA</math>, and <math>AB</math> at <math>D</math>, <math>E</math>, and <math>F</math>, respectively. The line through <math>D</math> perpendicular to <math>EF</math> meets <math>\omega</math> again at <math>R</math>. Line <math>AR</math> meets ω again at <math>P</math>. The circumcircles of triangles <math>PCE</math> and <math>PBF</math> meet again at <math>Q</math>. | ||
Prove that lines <math>DI</math> and <math>PQ</math> meet on the line through <math>A</math> perpendicular to <math>AI</math>. | Prove that lines <math>DI</math> and <math>PQ</math> meet on the line through <math>A</math> perpendicular to <math>AI</math>. | ||
+ | |||
+ | ==Solution== | ||
+ | [[File:2019 6 s1.png|500px|right]] | ||
+ | <i><b>Step 1</b></i> | ||
+ | |||
+ | We find an auxiliary point <math>S.</math> | ||
+ | |||
+ | Let <math>G</math> be the antipode of <math>D</math> on <math>\omega, GD = 2R,</math> where <math>R</math> is radius <math>\omega.</math> | ||
+ | |||
+ | We define <math>A' = PG \cap AI.</math> | ||
+ | <math>RD||AI, PRGD</math> cyclic <math>\implies \angle IAP = \angle DRP = \angle DGP.</math> | ||
+ | <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> | ||
+ | Let <math>DA'</math> meets <math>\omega</math> again at S (other than D). 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> |
Revision as of 12:07, 29 August 2022
Problem
Let be the incenter of acute triangle
with
. The incircle
of
is tangent to sides
,
, and
at
,
, and
, respectively. The line through
perpendicular to
meets
again at
. Line
meets ω again at
. The circumcircles of triangles
and
meet again at
.
Prove that lines
and
meet on the line through
perpendicular to
.
Solution
Step 1
We find an auxiliary point
Let be the antipode of
on
where
is radius
We define
cyclic
an inversion with respect
swap
and
is the midpoint
Let
meets
again at S (other than D). We define
Opposite sides of any quadrilateral inscribed in the circle
meet on the polar line of the intersection of the diagonals with respect to
and
meet on the line through
perpendicular to
The problem is reduced to proving that