2024 INMO

==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

https://i.imgur.com/ivcAShL.png To Prove: Points E, F, P, C are concyclic

Observe: $\angle CAB=\angle CBA=\angle EGA$ $\angle ECB=\angle CEG=\angle EAB= 90^\circ$ Notice that $\angle CBA = \angle FGA$ because $CB \parallel EG$ $\Longrightarrow \angle FAG =\angle FGA \Longrightarrow FA= FG$ . Here F is the circumcentre of $\triangle EAG$ because $F$ lies on the Perpendicular bisector of AG $\Longrightarrow F$ is the midpoint of $EG \Longrightarrow FP$ is the perpendicular bisector of $EG$ . This gives $\angle EFP =90^\circ$ And because $\angle EFP+\angle ECP=180^\circ$ Points E, F, P, C are concyclic. Hence proven that the centre of the circumcircle of $\triangle EGB$ lies on the circumcircle of $\triangle ECF$ .

∼Lakshya Pamecha

Problem 3

Let p be an odd prime number and $a,b,c$ be integers so that the integers $a^{2023}+b^{2023}, b^{2024}+c^{2024}, c^{2025}+a^{2025}$ are all divisible by p. Prove that p divides each of $a,b,c$ .

Solution

$a^{2023} \equiv -b^{2023} \pmod{p}$ $a^{(2023 \cdot 2024 \cdot 2025)} \equiv b^{(2023 \cdot 2024 \cdot 2025)} \pmod{p}$ Similarly, $b^{(2023 \cdot 2024 \cdot 2025)} \equiv -c^{(2023 \cdot 2024 \cdot 2025)} \pmod{p}$ and lastly $a^{(2023 \cdot 2024 \cdot 2025)} \equiv c^{(2023 \cdot 2024 \cdot 2025)} \pmod{p}$ Using some equations, we get $2c^{(2023 \cdot 2024 \cdot 2025)} \equiv 0 \pmod{p}$ and by the question, $p$ is an odd prime, so we are done as $\gcd(2,p) = 1$

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2024 INMO

Solution

Problem 3

Solution