Difference between revisions of "2024 INMO"
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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== | ||
+ | To Prove: Points E,F,P,C are concyclic | ||
+ | \newpage | ||
+ | |||
+ | Observe: <cmath>\angle CAB=\angle CBA=\angle EGA</cmath> <cmath>\angle ECB=\angle CEG=\angle EAB= 90^\circ</cmath> | ||
+ | Notice that <cmath>\angle CBA = \angle FGA</cmath> because <math>CB \parallel EG</math> <math>\Longrightarrow \angle FAG =\angle FGA</math> | ||
+ | or <math>FA= FG</math>. | ||
+ | \Here F is the circumcentre of \traingle EAG becuase F lies on the Perpendicular bisector of AG.\\\\ | ||
+ | \implies <math>F</math> is the midpoint of <math>EG</math> \implies <math>FP</math> is the perpendicular bisector of <math>EG</math>.\\ | ||
+ | This gives<cmath>\angle EFP =90^\circ</cmath>.\\ | ||
+ | And because <cmath>\angle EFP+\angle ECP=180^\circ</cmath>. Points E,F,P,C are concyclic.\\ | ||
+ | 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 , \text{point E lies on the circumcircle of} \text{triangle ABC such that} . \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: Notice that because
or .
\Here F is the circumcentre of \traingle EAG becuase F lies on the Perpendicular bisector of AG.\\\\ \implies is the midpoint of \implies is the perpendicular bisector of .\\ This gives.\\ And because . Points E,F,P,C are concyclic.\\ Hence proven that the centre of the circumcircle of lies on the circumcircle of .