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Difference between revisions of "2023 IMO Problems/Problem 2"

(→‎Solution)
(→‎Solution)
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Denote <math>G = PS' \cap AE \implies \angle BPG = \angle BPS' = \angle BSS' = \angle BDG \implies B, L, P, D,</math> and <math>G</math> are concyclic.
 
Denote <math>G = PS' \cap AE \implies \angle BPG = \angle BPS' = \angle BSS' = \angle BDG \implies B, L, P, D,</math> and <math>G</math> are concyclic.
−
+
[[File:2023 IMO 2 lemma.png|300px|right]]
 
<math>\angle EBS' = \varphi, \angle LBG = \angle LDG = 90^\circ = \angle DBS' \implies \angle DBG = \varphi = \angle SBF \implies</math>
 
<math>\angle EBS' = \varphi, \angle LBG = \angle LDG = 90^\circ = \angle DBS' \implies \angle DBG = \varphi = \angle SBF \implies</math>
  
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<i><b>Lemma 1</b></i>
 
<i><b>Lemma 1</b></i>
 +
 
[[File:2023 IMO 2b Lemma.png|300px|right]]
 
[[File:2023 IMO 2b Lemma.png|300px|right]]
 
Let acute triangle <math>\triangle ABC, AB > AC</math> be given.
 
Let acute triangle <math>\triangle ABC, AB > AC</math> be given.
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The point <math>E</math> is symmetric to <math>D</math> with respect to <math>AH.</math>
 
The point <math>E</math> is symmetric to <math>D</math> with respect to <math>AH.</math>
 +
 
The line <math>BE</math> meets <math>\Omega</math> again at <math>F \neq B</math>.
 
The line <math>BE</math> meets <math>\Omega</math> again at <math>F \neq B</math>.
 
   
 
   
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Let <math>\omega</math> be the circle centered at <math>H</math> with radius <math>HD.</math>
 
Let <math>\omega</math> be the circle centered at <math>H</math> with radius <math>HD.</math>
 +
 
The <math>\omega</math> meets <math>\Omega</math> again at <math>F' \neq D, HD = HF'.</math>  
 
The <math>\omega</math> meets <math>\Omega</math> again at <math>F' \neq D, HD = HF'.</math>  
 +
 
Let <math>\omega</math> meets <math>BF'</math> again at <math>E' \neq F'</math>.
 
Let <math>\omega</math> meets <math>BF'</math> again at <math>E' \neq F'</math>.
−
We use Reim’s theorem for <math>\omega, \Omega</math> and lines <math>CDD</math> and <math>BE'F'</math> and get <math>E'D || BC</math> (the idea was recommended by Leonid Shatunov).
+
 
 +
We use Reim’s theorem for <math>\omega, \Omega</math> and lines <math>CDD</math> and <math>BE'F'</math> and get <math>E'D || BC</math>
 +
 
 +
(this idea was recommended by <i><b>Leonid Shatunov</b></i>).
 +
 
 
<math>AH \perp BC \implies AH \perp E'D \implies</math>
 
<math>AH \perp BC \implies AH \perp E'D \implies</math>
 +
 
The point <math>E'</math> is symmetric to <math>D</math> with respect to <math>AH \implies E' = E \implies F' = F \implies HF = HD.</math>
 
The point <math>E'</math> is symmetric to <math>D</math> with respect to <math>AH \implies E' = E \implies F' = F \implies HF = HD.</math>
  

Revision as of 09:09, 23 July 2023

Problem

2023 IMO 2o.png

Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$.

Solution

2023 IMO 2o0.png

Denote the point diametrically opposite to a point $S$ through $S' \implies AS'$ is the internal angle bisector of $\angle BAC$.

Denote the crosspoint of $BS$ and $AS'$ through $H, \angle ABS = \varphi.$

\[AE \perp BC, SS' \perp BC \implies \overset{\Large\frown} {AS} = \overset{\Large\frown} {ES'} = 2\varphi \implies\]

\[\angle EAS' = \varphi = \angle ABS \implies \angle DAH = \angle ABH \implies\] \[\triangle AHD \sim \triangle BAH \implies \frac {AH}{BH} = \frac {DH}{AH} \implies AH^2 = BH \cdot DH.\] To finishing the solution we need only to prove that $PH = AH.$

Denote $F = SS' \cap AC \implies \angle CBS = \frac {\overset{\Large\frown} {CS}}{2} = \frac {\overset{\Large\frown} {BS}}{2} = \frac {\overset{\Large\frown} {AB}}{2} + \frac {\overset{\Large\frown} {AS}}{2} =$ $=\angle FCB + \varphi \implies \angle FBS = \angle ABC \implies H$ is incenter of $\triangle ABF.$

Denote $T = S'B \cap SA \implies SB \perp TS', S'A \perp TS \implies H$ is the orthocenter of $\triangle TSS'.$

Denote $G = PS' \cap AE \implies \angle BPG = \angle BPS' = \angle BSS' = \angle BDG \implies B, L, P, D,$ and $G$ are concyclic.

2023 IMO 2 lemma.png

$\angle EBS' = \varphi, \angle LBG = \angle LDG = 90^\circ = \angle DBS' \implies \angle DBG = \varphi = \angle SBF \implies$

points $B, G,$ and $F$ are collinear $\implies GF$ is symmetric to $AF$ with respect $TF.$

We use the lemma and complete the proof.

Lemma 1

2023 IMO 2b Lemma.png

Let acute triangle $\triangle ABC, AB > AC$ be given.

Let $H$ be the orthocenter of $\triangle ABC, BHD$ be the height.

Let $\Omega$ be the circle $BCD. BC$ is the diameter of $\Omega.$

The point $E$ is symmetric to $D$ with respect to $AH.$

The line $BE$ meets $\Omega$ again at $F \neq B$.

Prove that $HF = HD.$

Proof

Let $\omega$ be the circle centered at $H$ with radius $HD.$

The $\omega$ meets $\Omega$ again at $F' \neq D, HD = HF'.$

Let $\omega$ meets $BF'$ again at $E' \neq F'$.

We use Reim’s theorem for $\omega, \Omega$ and lines $CDD$ and $BE'F'$ and get $E'D || BC$

(this idea was recommended by Leonid Shatunov).

$AH \perp BC \implies AH \perp E'D \implies$

The point $E'$ is symmetric to $D$ with respect to $AH \implies E' = E \implies F' = F \implies HF = HD.$

vladimir.shelomovskii@gmail.com, vvsss

Solutions

https://www.youtube.com/watch?v=JhThDz0H7cI [Video contains solutions to all day 1 problems]

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