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 | + | To Prove: Points E, F, P, C are concyclic |
− | |||
Observe: <cmath>\angle CAB=\angle CBA=\angle EGA</cmath> <cmath>\angle ECB=\angle CEG=\angle EAB= 90^\circ</cmath> | Observe: <cmath>\angle CAB=\angle CBA=\angle EGA</cmath> <cmath>\angle ECB=\angle CEG=\angle EAB= 90^\circ</cmath> |
Revision as of 14:12, 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
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
.