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Difference between revisions of "2024 INMO"

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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>.
  
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∼Lakshya Pamecha
 
 
==Problem 3==
 
==Problem 3==
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Let p be an odd prime number and a,b,c 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>.
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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>.
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==Solution==
 
==Solution==
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If <math>p\vert{a}</math> \Rightarrow <math>p\vert a^{2023}</math> and <math> p \vert a^{2023}+b^{2023}</math> \Rightarrow p\vert b  \Rightarrow <math>p\vert b^{2024}</math> and <math>p\vert b^{2024}+c^{2024}</math> \Rightarrow <math>p\vert c</math>.\\
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\(a^{2023} \equiv -b^{2023} \pmod{p}\)
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Therefore, if <math>p</math> divides one of <math>a,b,c</math> it will divide all of them.\\
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Assume that <math>p</math> does not divide <math>a, b </math>or <math>c</math>
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\(a^{(2023 \cdot 2024 \cdot 2025)} \equiv b^{(2023 \cdot 2024 \cdot 2025)} \pmod{p}\)
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Set
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<math></math> a^{2023} &\equiv k \pmod{p} \Rightarrow b^{2023} \equiv -k \pmod{p} \\ b^{2024} &\equiv -bk \pmod{p} \Rightarrow c^{2024} \equiv kb \pmod{p}\\ c^{2025} &\equiv kbc \pmod{p}\Rightarrow a^{2025} \equiv -kbc \pmod{p}\\$<math>
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Similarly,
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<cmath>\Rightarrow \boxed{a^2 &\equiv -bc \pmod{p}}</cmath>
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Now we see that
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\(b^{(2023 \cdot 2024 \cdot 2025)} \equiv -c^{(2023 \cdot 2024 \cdot 2025)} \pmod{p}\)
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</math><math>(a^{2023})^2 &\equiv (b^{2023})^2 \pmod{p}\\ (-bc)^{2023} &\equiv (b^2)^{2023} \pmod{p}\\ \Rightarrow -c^{2023} &\equiv b^{2023} \pmod{p}\; \text{and} \; b^{4048} \equiv c^{4048}\pmod{p}\$</math>
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\text{So},
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and lastly
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<cmath>\boxed{b^2 &\equiv c^2 \pmod{p}}</cmath>
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This gives us to 2 cases:\\
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\(a^{(2023 \cdot 2024 \cdot 2025)} \equiv c^{(2023 \cdot 2024 \cdot 2025)} \pmod{p}\)
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Case I:
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<cmath>b-c \equiv 0 \pmod{p}
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Using some equations, we get
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    \Rightarrow b^{2024} \equiv c^{2024} \pmod{p}
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    \Rightarrow 2b^{2024} &\equiv 0 \pmod{p} \Rightarrow p\vert b </cmath>
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\(2c^{(2023 \cdot 2024 \cdot 2025)} \equiv 0 \pmod{p}\)
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Case II:
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<cmath>b+c \equiv 0\pmod{p} \Rightarrow -bc \equiv c^2 \pmod{p} \Rightarrow a^2 \equiv c^2 \pmod{p} \\\Rightarrow a \equiv c \pmod{p} \;\;\text{OR} \;\; a \equiv -c \pmod{p}</cmath>
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and by the question, \(p\) is an odd prime, so we are done as \(\gcd(2,p) = 1\)
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On checking for both cases we get <math>p\vert{a}</math> which implies <math>p\vert{b}</math> and <math>p\vert{c}</math>.
 

Latest revision as of 06:29, 10 August 2025

==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 $\triangle ECF$.

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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