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Difference between revisions of "2024 INMO"

(Created page with "==Problem 1 \text {In} triangle ABC with <math>CA=CB</math>, \text{point E lies on the circumcircle of} \text{triangle ABC such that} <math>\angle ECB=90^\circ</math>. \text{...")
 
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\text {In} triangle ABC with <math>CA=CB</math>, \text{point E lies on the circumcircle of} \text{triangle ABC such that} <math>\angle ECB=90^\circ</math>. \text{The line through E parallel to CB intersect CA in F} \text{and AB in G}.\text{Prove that}\\ \text{the centre of the circumcircle of} triangle EGB \text{lies on the circumcircle of triangle ECF.}
 
\text {In} triangle ABC with <math>CA=CB</math>, \text{point E lies on the circumcircle of} \text{triangle ABC such that} <math>\angle ECB=90^\circ</math>. \text{The line through E parallel to CB intersect CA in F} \text{and AB in G}.\text{Prove that}\\ \text{the centre of the circumcircle of} triangle EGB \text{lies on the circumcircle of triangle ECF.}
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==Solution 1==
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==Solution==
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[[file:///Users/lakshyapamecha/Downloads/INMO%20.png]]
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To Prove: Points E, F, P, C are concyclic
  
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\includegraphics[width=1.25\textwidth]{INMO 2024 P1.png}
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Observe: <cmath>\angle CAB=\angle CBA=\angle EGA</cmath> <cmath>\angle ECB=\angle CEG=\angle EAB= 90^\circ</cmath>
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To Prove: Points E,F,P,C are concyclic
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Notice that <cmath>\angle CBA = \angle FGA</cmath> because <math>CB \parallel EG</math>  <math>\Longrightarrow \angle FAG =\angle FGA \Longrightarrow FA= FG</math>.
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\newpage
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Here F is the circumcentre of <math>\triangle EAG</math> because  <math>F</math> lies on the Perpendicular bisector of AG
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<math>\Longrightarrow F</math> is the midpoint of <math>EG \Longrightarrow FP</math> is the perpendicular bisector of <math>EG</math>.
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This gives <cmath>\angle EFP =90^\circ</cmath>
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And because <cmath>\angle EFP+\angle ECP=180^\circ</cmath> Points E, F, P, C are concyclic.
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Hence proven that the centre of the circumcircle of <math>\triangle EGB</math> lies on the circumcircle of <math>\triangle ECF</math>.
  
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Observe: <cmath>\angle CAB=\angle CBA=\angle EGA</cmath> <cmath>\angle ECB=\angle CEG=\angle EAB= 90^\circ</cmath>
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∼Lakshya Pamecha
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\text{Notice that} <cmath>\angle CBA = \angle FGA</cmath> because <math>CB \parallel EG</math>} \implies <math>\angle FAG =\angle FGA</math>
 
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\:\text{or} \:<math>FA= FG</math>.\\
 
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\text{Here F is the circumcentre of \traingle EAG becuase  F lies on the Perpendicular bisector of AG.}\\\\
 
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\implies \text{<math>F</math> is the midpoint of <math>EG</math>} \implies \text{<math>FP</math> is the perpendicular bisector of <math>EG</math>.}\\
 
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\text{This gives} \:<cmath>\angle EFP =90^\circ</cmath>.\\
 
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\text{And because}<cmath>\angle EFP+\angle ECP=180^\circ</cmath>. \:\text{Points E,F,P,C are concyclic.}\\
 
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\text{Hence proven that the centre of the circumcircle of <math>\triangle EGB</math> lies on the circumcircle of <math>\triangle ECF</math>.}
 

Revision as of 14:20, 25 April 2024

==Problem 1

\text {In} triangle ABC with $CA=CB$, \text{point E lies on the circumcircle of} \text{triangle ABC such that} $\angle ECB=90^\circ$. \text{The line through E parallel to CB intersect CA in F} \text{and AB in G}.\text{Prove that}\\ \text{the centre of the circumcircle of} triangle EGB \text{lies on the circumcircle of triangle ECF.}

Solution

File:///Users/lakshyapamecha/Downloads/INMO .png To Prove: Points E, F, P, C are concyclic

Observe: \[\angle CAB=\angle CBA=\angle EGA\] \[\angle ECB=\angle CEG=\angle EAB= 90^\circ\] Notice that \[\angle CBA = \angle FGA\] because $CB \parallel EG$ $\Longrightarrow \angle FAG =\angle FGA \Longrightarrow FA= FG$. Here F is the circumcentre of $\triangle EAG$ because $F$ lies on the Perpendicular bisector of AG $\Longrightarrow F$ is the midpoint of $EG \Longrightarrow FP$ is the perpendicular bisector of $EG$. This gives \[\angle EFP =90^\circ\] And because \[\angle EFP+\angle ECP=180^\circ\] Points E, F, P, C are concyclic. Hence proven that the centre of the circumcircle of $\triangle EGB$ lies on the circumcircle of $\triangle ECF$.

∼Lakshya Pamecha

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