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== | ||
+ | [img]https://i.imgur.com/ivcAShL.png[/img] | ||
To Prove: Points E, F, P, C are concyclic | To Prove: Points E, F, P, C are concyclic | ||
Revision as of 13:16, 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
[img]https://i.imgur.com/ivcAShL.png[/img] To Prove: Points E, F, P, C are concyclic
Observe: Notice that because . Here F is the circumcentre of because lies on the Perpendicular bisector of AG is the midpoint of 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 .