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|450px|right]] | ||
+ | [[File:2019 6 s2.png|450px|right]] | ||
+ | [[File:2019 6 s3.png|390px|right]] | ||
+ | [[File:2019 6 s4.png|390px|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> is cyclic <math>\implies \angle IAP = \angle DRP = \angle DGP.</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> | ||
+ | |||
+ | <i><b>Step 2</b></i> | ||
+ | |||
+ | We find a simplified way to define the point <math>Q.</math> | ||
+ | |||
+ | We define <math>\angle BAC = 2 \alpha \implies \angle AFE = \angle AEF = 90^\circ – \alpha \implies</math> | ||
+ | <math>\angle BFE = \angle CEF = 180^\circ – (90^\circ – \alpha) = 90^\circ + \alpha = \angle BIC</math> | ||
+ | <math>(AI, BI,</math> and <math>CI</math> are bisectrices). | ||
+ | |||
+ | We use the Tangent-Chord Theorem and get <cmath>\angle EPF = \angle AEF = 90^\circ – \alpha.</cmath> | ||
+ | |||
+ | <math>\angle BQC = \angle BQP + \angle PQC = \angle BFP + \angle CEP =</math> | ||
+ | <math>=\angle BFE – \angle EFP + \angle CEF – \angle FEP =</math> | ||
+ | <math>= 90^\circ + \alpha + 90^\circ + \alpha – (90^\circ + \alpha) = </math> | ||
+ | <math>90^\circ + \alpha = \angle BIC \implies</math> | ||
+ | |||
+ | Points <math>Q, B, I,</math> and <math>C</math> are concyclic. | ||
+ | |||
+ | <i><b>Step 3</b></i> | ||
+ | |||
+ | We perform inversion around <math>\omega.</math> The straight line <math>PST</math> maps onto circle <math>PITS.</math> We denote this circle <math>\Omega.</math> We prove that the midpoint of <math>AD</math> lies on the circle <math>\Omega.</math> | ||
+ | |||
+ | In the diagram, the configuration under study is transformed using inversion with respect to <math>\omega.</math> The images of the points are labeled in the same way as the points themselves. Points <math>D,E,F,P,S,</math> and <math>G</math> have saved their position. Vertices <math>A, B,</math> and <math>C</math> have moved to the midpoints of the segments <math>EF, FD,</math> and <math>DE,</math> respectively. | ||
+ | |||
+ | Let <math>M</math> be the midpoint <math>AD.</math> | ||
+ | |||
+ | We define <math>\angle MID = \beta, \angle MDI = \gamma \implies</math> | ||
+ | <math>\angle IMA = \angle MID + \angle MDI = \beta + \gamma = \varphi.</math> | ||
+ | <math>DI = IS \implies \angle ISD = \gamma.</math> | ||
+ | |||
+ | <math>MI</math> is triangle <math>DAG</math> midline <math>\implies MI || AG \implies</math> | ||
+ | <cmath>MI || PG \implies \angle MAP = \angle AMI = \varphi.</cmath> | ||
+ | <cmath>\angle DPA = 90^\circ \implies PM = MA \implies</cmath> | ||
+ | <cmath>\angle PMA = \angle PMS = 180^\circ – 2 \varphi.</cmath> | ||
+ | <math>PI = IS \implies \angle PIS = 180^\circ – 2 \varphi =\angle DPA \implies</math> point <math>M</math> lies on <math>\Omega.</math> | ||
+ | <math>ABDC</math> is parallelogram <math>\implies M</math> is midpoint <math>BC.</math> | ||
+ | |||
+ | <i><b>Step 4</b></i> | ||
+ | |||
+ | We prove that image of <math>Q</math> lies on <math>\Omega.</math> | ||
+ | |||
+ | In the inversion plane the image of point <math>Q</math> lies on straight line <math>BC</math> (It is image of circle <math>BIC)</math> and on circle <math>PCE.</math> | ||
+ | |||
+ | <cmath>\angle PQM = \angle PQC = \angle PEC = \angle PED = \angle PSD = \angle PSM \implies</cmath> point <math>Q</math> lies on <math>\Omega</math>. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=2019|num-b=5|after=Last Problem}} |
Latest revision as of 01:52, 19 November 2023
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
is cyclic
An inversion with respect
swap
and
is the midpoint
Let meets
again at
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
Step 2
We find a simplified way to define the point
We define
and
are bisectrices).
We use the Tangent-Chord Theorem and get
Points and
are concyclic.
Step 3
We perform inversion around The straight line
maps onto circle
We denote this circle
We prove that the midpoint of
lies on the circle
In the diagram, the configuration under study is transformed using inversion with respect to The images of the points are labeled in the same way as the points themselves. Points
and
have saved their position. Vertices
and
have moved to the midpoints of the segments
and
respectively.
Let be the midpoint
We define
is triangle
midline
point
lies on
is parallelogram
is midpoint
Step 4
We prove that image of lies on
In the inversion plane the image of point lies on straight line
(It is image of circle
and on circle
point
lies on
.
vladimir.shelomovskii@gmail.com, vvsss
See Also
2019 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Problem |
All IMO Problems and Solutions |