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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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0 replies
jlacosta
Mar 2, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Flo0r functi0n
m4thbl3nd3r   5
N 4 minutes ago by pco
Find all positive integers such that $$n=\lfloor \sqrt{n}\rfloor^2+\lfloor \sqrt{n}\rfloor$$
5 replies
m4thbl3nd3r
an hour ago
pco
4 minutes ago
Hard T^T
Noname23   1
N 7 minutes ago by RagvaloD
<problem>
1 reply
Noname23
34 minutes ago
RagvaloD
7 minutes ago
At least 3 co-prime pairs
eezad3   0
7 minutes ago
You are given a random permutation of the first $n$ integers where $n>10$. Show that there exists at least three positions $i$ such that $gcd(p[i], p[i+1])=1$ (here $p[i]$ refers to the $i$-th position integer in the permutation)
0 replies
eezad3
7 minutes ago
0 replies
Suspicious Quadrilateral Geometry
YaoAOPS   4
N 8 minutes ago by Giabach298
Source: 2025 CTST P8
Let quadrilateral $A_1A_2A_3A_4$ be not cyclic and haves edges not parallel to each other.

Denote $B_i$ as the intersection of the tangent line at $A_i$ with respect to circle $A_{i-1}A_iA_{i+1}$ and the $A_{i+2}$-symmedian with respect to triangle $A_{i+1}A_{i+2}A_{i+3}$ and $C_i$ as the intersection of lines $A_iA_{i+1}$ and $B_iB_{i+1}$, where all indexes taken cyclically.

Prove that $C_1$, $C_2$, $C_3$, and $C_4$ are collinear.
4 replies
YaoAOPS
Mar 10, 2025
Giabach298
8 minutes ago
i need help
MR.1   0
35 minutes ago
Source: help
can you guys tell me problems about fe in $R+$(i know $R$ well). i want to study so if you guys have some easy or normal problems please send me
0 replies
MR.1
35 minutes ago
0 replies
Changeable polynomials, can they ever become equal?
mshtand1   4
N an hour ago by CHESSR1DER
Source: Ukrainian Mathematical Olympiad 2025. Day 2, Problem 11.5
Initially, two constant polynomials are written on the board: \(0\) and \(1\). At each step, it is allowed to add \(1\) to one of the polynomials and to multiply another one by the polynomial \(45x + 2025\). Can the polynomials become equal at some point?

Proposed by Oleksii Masalitin
4 replies
mshtand1
Yesterday at 12:47 AM
CHESSR1DER
an hour ago
Guessing polynomial by its maximum values on segments
NO_SQUARES   3
N an hour ago by pco
Source: Kvant 2025 no. 1 M2828 and The XIX Southern Mathematical Tournament
Maxim has guessed a polynomial $f(x)$ of degree $n$. Sasha wants to guess it (knowing $n$). During a turn, Sasha can name a certain segment $[a;b]$ and Maxim will give in response the maximum value of $f(x)$ on the segment $[a;b]$. Will Sasha be able to guess $f(x)$ in a finite number of steps?
M. Didin
3 replies
NO_SQUARES
Yesterday at 3:21 PM
pco
an hour ago
easy number theory
MuradSafarli   1
N an hour ago by Tuvshuu
\[
v_p(n!) \leq \frac{n}{p - 1}
\]
1 reply
MuradSafarli
2 hours ago
Tuvshuu
an hour ago
Inspired by Kazakhstan 2017
sqing   1
N an hour ago by sqing
Source: Own
Let $a,b,c\ge \frac{1}{2}$ and $a+b+c=2. $ Prove that
$$\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)\ge 1$$Let $a,b,c\ge \frac{1}{3}$ and $a+b+c=1. $ Prove that
$$\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)\ge 9$$
1 reply
+1 w
sqing
3 hours ago
sqing
an hour ago
Collinear geometry problem with incircle
ilovemath0402   1
N 2 hours ago by deraxenrovalo
Given acute $\triangle ABC$ not isosceles, the incircle $(I)$. $D,E,F$ is the intersection of $(I)$ with $BC,CA,AB$. $P$ is the projection of $D$ onto $EF$. $DP$ cut $(I)$ at the second point $K$. $L$ is the projection of $A$ onto $IK$. $(LEC), (LFB)$ cut $(I)$ at the second point $M,N$ respectively. Prove $M,N,P$ are collinear
1 reply
ilovemath0402
Jul 22, 2023
deraxenrovalo
2 hours ago
Wait wasn&#039;t it the reciprocal in the paper?
Supercali   6
N 3 hours ago by kes0716
Source: India TST 2023 Day 2 P1
Let $\mathbb{Z}_{\ge 0}$ be the set of non-negative integers and $\mathbb{R}^+$ be the set of positive real numbers. Let $f: \mathbb{Z}_{\ge 0}^2 \rightarrow \mathbb{R}^+$ be a function such that $f(0, k) = 2^k$ and $f(k, 0) = 1$ for all integers $k \ge 0$, and $$f(m, n) = \frac{2f(m-1, n) \cdot f(m, n-1)}{f(m-1, n)+f(m, n-1)}$$for all integers $m, n \ge 1$. Prove that $f(99, 99)<1.99$.

Proposed by Navilarekallu Tejaswi
6 replies
Supercali
Jul 9, 2023
kes0716
3 hours ago
About old Inequality
perfect_square   0
3 hours ago
Source: Arqady
This is: $a,b,c>0$ which satisfy $abc=1$
Prove that: $ \frac{a+b+c}{3} \ge \sqrt[10]{\frac{a^3+b^3+c^3}{3}}$
By $  uvw $ method, I can assum $b=c=x,a=\frac{1}{x^2}$
But I can't prove:
$ \frac{2x+\frac{1}{x^2}}{3} \ge \sqrt[10]{ \frac{2x^3+ \frac{1}{x^6}}{3}} $
Is there an another way?
0 replies
perfect_square
3 hours ago
0 replies
inquality
karasuno   1
N 3 hours ago by sqing
The real numbers $x,y,z \ge \frac{1}{2}$ are given such that $x^{2}+y^{2}+z^{2}=1$. Prove the inequality $$(\frac{1}{x}+\frac{1}{y}-\frac{1}{z})(\frac{1}{x}-\frac{1}{y}+\frac{1}{z})\ge 2 .$$
1 reply
karasuno
4 hours ago
sqing
3 hours ago
Number Theory
karasuno   0
4 hours ago
Solve the equation $$n!+10^{2014}=m^{4}$$in natural numbers m and n.
0 replies
karasuno
4 hours ago
0 replies
x | (y + 1), y | (z + 1) and z | (x + 1)
parmenides51   7
N Jun 18, 2024 by Gryphos
Source: 2021 Austrian Regional Competition For Advanced Students p4
Determine all triples $(x, y, z)$ of positive integers satisfying $x | (y + 1)$, $y | (z + 1)$ and $z | (x + 1)$.

(Walther Janous)
7 replies
parmenides51
May 31, 2021
Gryphos
Jun 18, 2024
x | (y + 1), y | (z + 1) and z | (x + 1)
G H J
G H BBookmark kLocked kLocked NReply
Source: 2021 Austrian Regional Competition For Advanced Students p4
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parmenides51
30627 posts
#1 • 3 Y
Y by chessgocube, HWenslawski, Mango247
Determine all triples $(x, y, z)$ of positive integers satisfying $x | (y + 1)$, $y | (z + 1)$ and $z | (x + 1)$.

(Walther Janous)
This post has been edited 2 times. Last edited by parmenides51, May 31, 2021, 10:54 PM
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Mahdi.sh
73 posts
#2 • 2 Y
Y by HWenslawski, Mango247
Just multiplieng the conditions
Z K Y
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Mahdi.sh
73 posts
#4
Y by
$xyz\mid (x+1)(y+1)(z+1)=xyz+xy+yz+zx+x+y+z+1\rightarrow xyz\mid xy+yz+zx+x+y+z+1\rightarrow xyz\leq xy+yz+zx+x+y+z+1$...
This post has been edited 1 time. Last edited by Mahdi.sh, May 31, 2021, 11:25 PM
Z K Y
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Mathmick51
478 posts
#5
Y by
First , if $x=1$ or $y=1$ or $z=1$.
WLOG; $x=1$
$z\mid 2$
$z=1$ or $2$.
$z=1$ then $y\mid 2$
$y=1$ or $2$
$z=2$ then $y\mid 3$
$y=1$ or $3$
Therefore, $(x,y,z)=(1,1,1),(1,2,1),(1,3,1)$ and their permutation.
Now, if $x,y,z\geq 2$
Mahdi.sh wrote:
$xyz\mid (x+1)(y+1)(z+1)=xyz+xy+yz+zx+x+y+z+1\rightarrow xyz\mid xy+yz+zx+x+y+z+1\rightarrow xyz\leq xy+yz+zx+x+y+z+1$...

From Last Inequality that @above show.
$yz+y+z+1\geq x(yz-y-z-1) \geq 2yz-2y-2z-2$
$yz-3y-3z-3\leq 0$
$(y-3)(z-3)\leq 12$
and then case work
Z K Y
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Arslan
268 posts
#6
Y by
Mathmick51 wrote:
First , if $x=1$ or $y=1$ or $z=1$.
WLOG; $x=1$
$z\mid 2$
$z=1$ or $2$.
$z=1$ then $y\mid 2$
$y=1$ or $2$
$z=2$ then $y\mid 3$
$y=1$ or $3$
Therefore, $(x,y,z)=(1,1,1),(1,2,1),(1,3,1)$ and their permutation.
Now, if $x,y,z\geq 2$
Mahdi.sh wrote:
$xyz\mid (x+1)(y+1)(z+1)=xyz+xy+yz+zx+x+y+z+1\rightarrow xyz\mid xy+yz+zx+x+y+z+1\rightarrow xyz\leq xy+yz+zx+x+y+z+1$...

From Last Inequality that @above show.
$yz+y+z+1\geq x(yz-y-z-1) \geq 2yz-2y-2z-2$
$yz-3y-3z-3\leq 0$
$(y-3)(z-3)\leq 12$
and then case work

Are you sure that $(1,3,1)$ is a solution?
Z K Y
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Arslan
268 posts
#7
Y by
After realizing my mistake, let me present my humble and lengthy solution.

$\left\{\begin{array}{cc}x|y+1\\y|z+1\\z|x+1\end{array}\right.$ $\implies$ $xyz|(x+1)(y+1)(z+1)$ $\implies$ $xyz|xyz+xy+yz+zx+x+y+z+1$ $\implies$ $xyz|xy+yz+zx+x+y+z+1$.
So, we are looking for all triples of positive integers such that
$\frac{xy+yz+zx+x+y+z+1}{xyz}\in\mathbb{N}$ or $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xyz}\in\mathbb{N}$.
Let define a function on the set of natural numbers: $f(x,y,z):=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xyz}$.
I will not waste a time on the case when any two of the three are equal.
So, assume that $x,y$ and $z$ are pairwisely different.
Since $f(x,y,z)$ is a symmetric function (not the original conditions of the problem!), we can assume wlog $x<y<z$.

$x\ge4.$
Then $y\ge5, z\ge6.$
$0<f(x,y,z)\le\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot4}+\frac{1}{4\cdot5\cdot6}=\frac{90}{120}<1$.
So, in this case we don't get any solutions.

$x=3.$
Then $y\ge4, z\ge5.$
$0<f(x,y,z)\le\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot3}+\frac{1}{3\cdot4\cdot5}=1$.
So, we get only one solution: $(x,y,z)=(3,4,5)$ with its permutations.
However, $(x,y,z)=(4,3,5)$ and its cyclic permutations satisfy the original problem.

$x=2.$
Then $y\ge3, z\ge4.$
$0<f(2,y,z)\le\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot2}+\frac{1}{2\cdot3\cdot4}=\frac{36}{24}<2$.
So, the only possibility is $f(2,y,z)=1$.
$\implies$ $\frac{1}{2}+\frac{1}{y}+\frac{1}{z}+\frac{1}{2y}+\frac{1}{yz}+\frac{1}{z\cdot2}+\frac{1}{2yz}=1$
$\implies$ $(y-3)(z-3)=12=1\cdot12=2\cdot6=3\cdot4$ $\implies$ $(y,z)=(4,15),(5,9),(6,7)$. However, none of these satisfies the original problem.

$x=1.$
Then $y\ge2, z\ge3.$
$1<f(1,y,z)\le\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot1}+\frac{1}{1\cdot2\cdot3}=\frac{18}{6}=3$.
So, the only possibilities are $f(1,y,z)=2$ and $f(1,y,z)=3$.
$f(1,y,z)=2$.
$\implies$ $\frac{1}{1}+\frac{1}{y}+\frac{1}{z}+\frac{1}{1\cdot y}+\frac{1}{yz}+\frac{1}{z\cdot1}+\frac{1}{1\cdot yz}=2$
$\implies$ $(y-2)(z-2)=6=1\cdot6=2\cdot3$ $\implies$ $(y,z)=(3,8),(4,5)$. However, none of these satisfies the original problem.
$f(1,y,z)=3$.
$\implies$ $\frac{1}{1}+\frac{1}{y}+\frac{1}{z}+\frac{1}{1\cdot y}+\frac{1}{yz}+\frac{1}{z\cdot1}+\frac{1}{1\cdot yz}=3$
$\implies$ $(y-1)(z-1)=2=1\cdot2$ $\implies$ $(y,z)=(2,3)$.
After checking, only $(3,2,1)$ and its cyclic permutations satisfy the original problem.
So, all solutions are $(1,1,1), (1,2,1), (4,3,5), (3,2,1)$ and their cyclic permutations.
$\blacksquare$
This post has been edited 3 times. Last edited by Arslan, Jul 2, 2024, 6:17 PM
Reason: Typo
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Pal702004
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#8
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All solutions, with max $z$ are:

$(x,y,z)=(1,1,1);(1,1,2);(2,1,3);(4,3,5)$
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Gryphos
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Arslan wrote:
Now we can assume wlog $x<y<z$...

This is not true since the condition is not symmetric in $x,y,z$, only cyclic. However, one can assume that either $x<y<z$ or $x>y>z$, if the three numbers are distinct. In the second case $x | y+1$ and $x \ge y+1 \Longrightarrow x=y+1$. Similarly, $y=z+1$. Then $z \mid x+1 = z+3$, hence $z \mid 3 \Longrightarrow z \in \{1, 3\}$. This gives the two additional solutions $(5,4,3)$, $(3,2,1)$ (and their cyclic permutations).
This post has been edited 1 time. Last edited by Gryphos, Jun 18, 2024, 9:32 AM
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