Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 INMO"

(Created page with "==Problem 1 \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{...")
 
(→‎Solution 1)
Line 3: Line 3:
 
\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 1==
 
==Solution 1==
−
 
−
\includegraphics[width=1.25\textwidth]{INMO 2024 P1.png}
 
−
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>
 
−
\text{Notice that} <cmath>\angle CBA = \angle FGA</cmath> because <math>CB \parallel EG</math>} \implies <math>\angle FAG =\angle FGA</math>
 
−
\:\text{or} \:<math>FA= FG</math>.\\
 
−
\text{Here F is the circumcentre of \traingle EAG becuase  F lies on the Perpendicular bisector of AG.}\\\\
 
−
\implies \text{<math>F</math> is the midpoint of <math>EG</math>} \implies \text{<math>FP</math> is the perpendicular bisector of <math>EG</math>.}\\
 
−
\text{This gives} \:<cmath>\angle EFP =90^\circ</cmath>.\\
 
−
\text{And because}<cmath>\angle EFP+\angle ECP=180^\circ</cmath>. \:\text{Points E,F,P,C are concyclic.}\\
 
−
\text{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 13:47, 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 1

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_INMO&oldid=218661"