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

If $p\vert{a}$ \Rightarrow $p\vert a^{2023}$ and $p \vert a^{2023}+b^{2023}$ \Rightarrow p\vert b \Rightarrow $p\vert b^{2024}$ and $p\vert b^{2024}+c^{2024}$ \Rightarrow $p\vert c$ .\\ Therefore, if $p$ divides one of $a,b,c$ it will divide all of them.\\ Assume that $p$ does not divide $a, b$ or $c$ Set $$ (Error compiling LaTeX. Unknown error_msg) 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}\$$<cmath>\Rightarrow \boxed{a^2 &\equiv -bc \pmod{p}}</cmath> Now we see that$ (Error compiling LaTeX. Unknown error_msg)$(a^{2023})^2 &\equiv (b^{2023})^2 \pmod{p}\\ (-bc)^{2023} &\equiv (b^2)^{2023} \pmod{p}\\ \Rightarrow -c^{2023} &\equiv b^{2023} \pmod{p}\; \text{and} \; b^{4048} \equiv c^{4048}\pmod{p}$ (Error compiling LaTeX. Unknown error_msg) \text{So},

\[\boxed{b^2 &\equiv c^2 \pmod{p}}\] (Error compiling LaTeX. Unknown error_msg)

This gives us to 2 cases:\\ Case I:

\[b-c \equiv 0 \pmod{p}
    \Rightarrow b^{2024} \equiv c^{2024} \pmod{p}
    \Rightarrow 2b^{2024} &\equiv 0 \pmod{p} \Rightarrow p\vert b\] (Error compiling LaTeX. Unknown error_msg)

Case II: $b+c \equiv 0\pmod{p} \Rightarrow -bc \equiv c^2 \pmod{p} \Rightarrow a^2 \equiv c^2 \pmod{p} \\\Rightarrow a \equiv c \pmod{p} \;\;\text{OR} \;\; a \equiv -c \pmod{p}$ On checking for both cases we get $p\vert{a}$ which implies $p\vert{b}$ and $p\vert{c}$ .

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2024 INMO

Solution

Problem 3

Solution