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Difference between revisions of "2024 INMO"

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==Solution==
 
==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 \(\gcd(2,p) = 1\)

Latest revision as of 10:43, 14 December 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

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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