Difference between revisions of "2024 INMO"
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∼Lakshya Pamecha | ∼Lakshya Pamecha | ||
| + | ==Problem 3== | ||
| + | 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>. | ||
| + | ==Solution== | ||
| + | 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>.\\ | ||
| + | Therefore, if <math>p</math> divides one of <math>a,b,c</math> it will divide all of them.\\ | ||
| + | Assume that <math>p</math> does not divide <math>a, b </math>or <math>c</math> | ||
| + | Set | ||
| + | <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> | ||
| + | <cmath>\Rightarrow \boxed{a^2 &\equiv -bc \pmod{p}}</cmath> | ||
| + | Now we see that | ||
| + | </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> | ||
| + | \text{So}, | ||
| + | <cmath>\boxed{b^2 &\equiv c^2 \pmod{p}}</cmath> | ||
| + | This gives us to 2 cases:\\ | ||
| + | Case I: | ||
| + | <cmath>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 </cmath> | ||
| + | Case II: | ||
| + | <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> | ||
| + | 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>. | ||
Revision as of 14:27, 25 April 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 a,b,c 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
If 
\Rightarrow 
and 
\Rightarrow p\vert b \Rightarrow 
and 
\Rightarrow 
.\\
Therefore, if 
divides one of it will divide all of them.\\
Assume that does not divide 
or 
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:
![\[b+c \equiv 0\pmod{p} \Rightarrow -bc \equiv c^2 \pmod{p} \Rightarrow a^2 \equiv c^2 \pmod{p} \\\Rightarrow a \equiv c \pmod{p} \;\;\text{OR} \;\; a \equiv -c \pmod{p}\]](http://latex.artofproblemsolving.com/a/d/9/ad95cc4a36d0f801f4f747b834ff93c5e2fd64a4.png)
On checking for both cases we get which implies 
and 
.
