2024 INMO

Revision as of 13:05, 25 April 2024 by L13832 (talk | contribs) (Solution)

==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 $\triangle ECF$.