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] | + | [img]file:///Users/lakshyapamecha/Downloads/INMO%20.png[/img] |
To Prove: Points E, F, P, C are concyclic | To Prove: Points E, F, P, C are concyclic | ||
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And because <cmath>\angle EFP+\angle ECP=180^\circ</cmath> Points E, F, P, C are concyclic. | 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>. | 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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| + | ∼Lakshya Pamecha | ||
Revision as of 13:19, 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]file:///Users/lakshyapamecha/Downloads/INMO%20.png[/img] To Prove: Points E, F, P, C are concyclic
Observe: ![\[\angle CAB=\angle CBA=\angle EGA\]](http://latex.artofproblemsolving.com/8/0/4/804eea72b35117cbeb648bd29eb74cd965fc3103.png)
![\[\angle ECB=\angle CEG=\angle EAB= 90^\circ\]](http://latex.artofproblemsolving.com/7/2/2/722712b65d8613d3797ea63c4f56a91c07bc74bb.png)
Notice that ![\[\angle CBA = \angle FGA\]](http://latex.artofproblemsolving.com/d/2/a/d2a5044184909fe92d8505f40511b9321c91a798.png)
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 ![\[\angle EFP =90^\circ\]](http://latex.artofproblemsolving.com/0/8/f/08ffa237cee2e795c91719df6617a5e70836a3a0.png)
And because ![\[\angle EFP+\angle ECP=180^\circ\]](http://latex.artofproblemsolving.com/7/9/e/79ec6002f0be9c7e10f9e7c7b5fc805c0bc90903.png)
Points E, F, P, C are concyclic.
Hence proven that the centre of the circumcircle of 
lies on the circumcircle of 
.
∼Lakshya Pamecha
