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

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==Solution==
 
==Solution==
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[[File:2023 IMO 2o0.png|400px|right]]
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Denote the point diametrically opposite to a point <math>S</math> through <math>S' \implies AS'</math> is the internal angle bisector of <math>\angle BAC</math>.
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Denote the crosspoint of <math>BS</math> and <math>AS'</math> through <math>H, \angle ABS = \varphi.</math>
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<cmath>AE \perp BC, SS' \perp BC \implies \overset{\Large\frown} {AS} = \overset{\Large\frown} {ES'} = 2\varphi \implies</cmath>
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<cmath>\angle EAS' = \varphi = \angle ABS \implies \angle DAH = \angle ABH \implies</cmath>
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<cmath>\triangle AHD \sim \triangle BAH \implies \frac {AH}{BH} = \frac {DH}{AH} \implies AH^2 = BH \cdot DH.</cmath>
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To finishing the solution we need only to prove that <math>PH = AH.</math>
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Denote <math>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} =</math>
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<math>=\angle FCB + \varphi \implies \angle FBS = \angle ABC \implies H</math> is incenter of <math>\triangle ABF.</math>
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Denote <math>T = S'B \cap SA \implies SB \perp TS', S'A \perp TS \implies H</math> is the orthocenter of <math>\triangle TSS'.</math>
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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.
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<math>\angle EBS' = \varphi, \angle LBG = \angle LDG = 90^\circ = \angle DBS' \implies \angle DBG = \varphi = \angle SBF \implies</math>
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points <math>B, G,</math> and <math>F</math> are colinear <math>\implies GF</math> is symmetric to <math>AF</math> with respect <math>TF.</math>
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We use the lemma and complete the proof.
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==Solutions==
 
https://www.youtube.com/watch?v=JhThDz0H7cI [Video contains solutions to all day 1 problems]
 
https://www.youtube.com/watch?v=JhThDz0H7cI [Video contains solutions to all day 1 problems]

Revision as of 08:53, 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.

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

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

We use the lemma and complete the proof.

Solutions

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

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