Difference between revisions of "2024 INMO"
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==Solution== | ==Solution== | ||
| − | + | \(a^{2023} \equiv -b^{2023} \pmod{p}\) | |
| + | |||
\(a^{(2023 \cdot 2024 \cdot 2025)} \equiv b^{(2023 \cdot 2024 \cdot 2025)} \pmod{p}\) | \(a^{(2023 \cdot 2024 \cdot 2025)} \equiv b^{(2023 \cdot 2024 \cdot 2025)} \pmod{p}\) | ||
| + | |||
Similarly, | Similarly, | ||
| + | |||
\(b^{(2023 \cdot 2024 \cdot 2025)} \equiv -c^{(2023 \cdot 2024 \cdot 2025)} \pmod{p}\) | \(b^{(2023 \cdot 2024 \cdot 2025)} \equiv -c^{(2023 \cdot 2024 \cdot 2025)} \pmod{p}\) | ||
| + | |||
and lastly | and lastly | ||
| + | |||
\(a^{(2023 \cdot 2024 \cdot 2025)} \equiv c^{(2023 \cdot 2024 \cdot 2025)} \pmod{p}\) | \(a^{(2023 \cdot 2024 \cdot 2025)} \equiv c^{(2023 \cdot 2024 \cdot 2025)} \pmod{p}\) | ||
| + | |||
Using some equations, we get | Using some equations, we get | ||
| + | |||
\(2c^{(2023 \cdot 2024 \cdot 2025)} \equiv 0 \pmod{p}\) | \(2c^{(2023 \cdot 2024 \cdot 2025)} \equiv 0 \pmod{p}\) | ||
| + | |||
and by the question, \(p\) is an odd prime, so we are done as \(\gcd(2,p) = 1\) | 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 
, \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: ![\[\angle CAB=\angle CBA=\angle EGA\]](http://latex.artofproblemsolving.com/8/0/4/804eea72b35117cbeb648bd29eb74cd965fc3103.png)
![\[\angle ECB=\angle CEG=\angle EAB= 90^\circ\]](http://latex.artofproblemsolving.com/7/2/2/722712b65d8613d3797ea63c4f56a91c07bc74bb.png)
Notice that ![\[\angle CBA = \angle FGA\]](http://latex.artofproblemsolving.com/d/2/a/d2a5044184909fe92d8505f40511b9321c91a798.png)
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 ![\[\angle EFP =90^\circ\]](http://latex.artofproblemsolving.com/0/8/f/08ffa237cee2e795c91719df6617a5e70836a3a0.png)
And because ![\[\angle EFP+\angle ECP=180^\circ\]](http://latex.artofproblemsolving.com/7/9/e/79ec6002f0be9c7e10f9e7c7b5fc805c0bc90903.png)
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 ![\[a^{2023}+b^{2023}, b^{2024}+c^{2024}, c^{2025}+a^{2025}\]](http://latex.artofproblemsolving.com/2/3/3/23366379c5800a35a8a898097ea72ba86a02ebf1.png)
are all divisible by p. Prove that p divides each of .
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 \(\gcd(2,p) = 1\)
