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 | + | https://i.imgur.com/ivcAShL.png |
− | + | 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> | ||
− | Notice that <cmath>\angle CBA = \angle FGA</cmath> because <math>CB \parallel EG</math> <math>\Longrightarrow \angle FAG =\angle FGA | + | Notice that <cmath>\angle CBA = \angle FGA</cmath> because <math>CB \parallel EG</math> <math>\Longrightarrow \angle FAG =\angle FGA \Longrightarrow FA= FG</math>. |
− | + | Here F is the circumcentre of <math>\triangle EAG</math> because <math>F</math> lies on the Perpendicular bisector of AG | |
− | + | <math>\Longrightarrow F</math> is the midpoint of <math>EG \Longrightarrow FP</math> is the perpendicular bisector of <math>EG</math>. | |
− | + | This gives <cmath>\angle EFP =90^\circ</cmath> | |
− | This gives<cmath>\angle EFP =90^\circ</cmath> | + | 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> | ||
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>. | ||
+ | |||
+ | ∼Lakshya Pamecha | ||
+ | ==Problem 3== | ||
+ | Let p be an odd prime number and <math>a,b,c</math> be integers so that the integers <cmath>a^{2023}+b^{2023}, b^{2024}+c^{2024}, c^{2025}+a^{2025}</cmath> are all divisible by p. Prove that p divides each of <math>a,b,c</math>. | ||
+ | |||
+ | ==Solution== |
Latest revision as of 13:28, 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
https://i.imgur.com/ivcAShL.png 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 .
∼Lakshya Pamecha
Problem 3
Let p be an odd prime number and be integers so that the integers are all divisible by p. Prove that p divides each of .