Difference between revisions of "2024 INMO"

(Solution 1)
(Solution)
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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.}
 
==Solution==
 
==Solution==
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To Prove: Points E,F,P,C are concyclic
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\newpage
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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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Notice that <cmath>\angle CBA = \angle FGA</cmath> because <math>CB \parallel EG</math>  <math>\Longrightarrow \angle FAG =\angle FGA</math>
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or <math>FA= FG</math>.
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\Here F is the circumcentre of \traingle EAG becuase  F lies on the Perpendicular bisector of AG.\\\\
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\implies <math>F</math> is the midpoint of <math>EG</math> \implies <math>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>.

Revision as of 13:05, 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

To Prove: Points E,F,P,C are concyclic \newpage

Observe: \[\angle CAB=\angle CBA=\angle EGA\] \[\angle ECB=\angle CEG=\angle EAB= 90^\circ\] Notice that \[\angle CBA = \angle FGA\] because $CB \parallel EG$ $\Longrightarrow \angle FAG =\angle FGA$

or $FA= FG$.

\Here F is the circumcentre of \traingle EAG becuase F lies on the Perpendicular bisector of AG.\\\\ \implies $F$ is the midpoint of $EG$ \implies $FP$ 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 $\triangle ECF$.