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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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Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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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
Polynomial application with complex number
RenheMiResembleRice   1
N 6 minutes ago by Mathzeus1024
$P\left(x\right)=128x^{4}-32x^{2}+1$

By examining the roots of P(x), find the exact value of $\sin\left(\frac{\pi}{8}\right)\sin\left(\frac{3\pi}{8}\right)$
1 reply
RenheMiResembleRice
32 minutes ago
Mathzeus1024
6 minutes ago
square geometry bisect $\angle ESB$
GorgonMathDota   12
N 7 minutes ago by AshAuktober
Source: BMO SL 2019, G1
Let $ABCD$ be a square of center $O$ and let $M$ be the symmetric of the point $B$ with respect to point $A$. Let $E$ be the intersection of $CM$ and $BD$, and let $S$ be the intersection of $MO$ and $AE$. Show that $SO$ is the angle bisector of $\angle ESB$.
12 replies
GorgonMathDota
Nov 8, 2020
AshAuktober
7 minutes ago
Number of modular sequences with different residues
PerfectPlayer   1
N 24 minutes ago by Z4ADies
Source: Turkey TST 2025 Day 3 P9
Let \(n\) be a positive integer. For every positive integer $1 \leq k \leq n$ the sequence ${\displaystyle {\{ a_{i}+ki\}}_{i=1}^{n }}$ is defined, where $a_1,a_2, \dots ,a_n$ are integers. Among these \(n\) sequences, for at most how many of them does all the elements of the sequence give different remainders when divided by \(n\)?
1 reply
1 viewing
PerfectPlayer
6 hours ago
Z4ADies
24 minutes ago
D1010 : How it is possible ?
Dattier   13
N an hour ago by Dattier
Source: les dattes à Dattier
Is it true that$$\forall n \in \mathbb N^*, (24^n \times B \mod A) \mod 2 = 0 $$?

A=1728400904217815186787639216753921417860004366580219212750904
024377969478249664644267971025952530803647043121025959018172048
336953969062151534282052863307398281681465366665810775710867856
720572225880311472925624694183944650261079955759251769111321319
421445397848518597584590900951222557860592579005088853698315463
815905425095325508106272375728975

B=2275643401548081847207782760491442295266487354750527085289354
965376765188468052271190172787064418854789322484305145310707614
546573398182642923893780527037224143380886260467760991228567577
953725945090125797351518670892779468968705801340068681556238850
340398780828104506916965606659768601942798676554332768254089685
307970609932846902
13 replies
Dattier
Mar 10, 2025
Dattier
an hour ago
Interesting inequality
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b\geq 2  . $ Prove that
$$ (a^2-1)(b^2-1) (1-ab)+\frac{27}{8}a^2b^2\leq 27$$$$ (a^2-1)(b^2-1)(1-a^2b^2 )+\frac{81}{4}a^2b^2     \leq 189$$$$ (a^2-1)(b^2-1)(1-a^2b^2 )+ 162  ab  \leq 513$$$$  (a^2-1)(b^2-1) (1-a^2b^2 )+21 a^2b^2\leq \frac{3219}{16}$$$$ (a^2-1)(b^2-1) (1-ab)+\frac{27}{8}a^2b^2\leq\frac{415+61\sqrt{61}}{18}$$
1 reply
sqing
2 hours ago
sqing
an hour ago
Minimal Grouping in a Complete Graph
swynca   1
N an hour ago by swynca
Source: 2025 Turkey TST P1
In a complete graph with $2025$ vertices, each edge has one of the colors $r_1$, $r_2$, or $r_3$. For each $i = 1,2,3$, if the $2025$ vertices can be divided into $a_i$ groups such that any two vertices connected by an edge of color $r_i$ are in different groups, find the minimum possible value of $a_1 + a_2 + a_3$.
1 reply
1 viewing
swynca
4 hours ago
swynca
an hour ago
Nice FE as the First Day Finale
swynca   1
N an hour ago by swynca
Source: 2025 Turkey TST P3
Find all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for all $x,y \in \mathbb{R}-\{0\}$,
$$ f(x) \neq 0 \text{ and } \frac{f(x)}{f(y)} + \frac{f(y)}{f(x)} - f \left( \frac{x}{y}-\frac{y}{x} \right) =2 $$
1 reply
swynca
4 hours ago
swynca
an hour ago
Cn/lnn bound for S
EthanWYX2009   0
2 hours ago
Source: 2025 March 谜之竞赛-2
Prove that there exists an constant $C,$ such that for all integer $n\ge 2$ and a subset $S$ of $[n],$ satisfy $a\mid\tbinom ab$ for all $a,b\in S,$ $a>b,$ then $|S|\le \frac{Cn}{\ln n}.$

Created by Yuxing Ye
0 replies
2 viewing
EthanWYX2009
2 hours ago
0 replies
Natural function and cubelike expression
sarjinius   2
N 2 hours ago by Kaimiaku
Source: Philippine Mathematical Olympiad 2025 P8
Let $\mathbb{N}$ be the set of positive integers. Find all functions $f : \mathbb{N} \to \mathbb{N}$ such that for all $m, n \in \mathbb{N}$, \[m^2f(m) + n^2f(n) + 3mn(m + n)\]is a perfect cube.
2 replies
sarjinius
Mar 9, 2025
Kaimiaku
2 hours ago
hard problem
Noname23   3
N 2 hours ago by Noname23
problem
3 replies
Noname23
Sunday at 4:57 PM
Noname23
2 hours ago
Roots, bounding and other delusions
anantmudgal09   28
N 2 hours ago by kes0716
Source: INMO 2021 Problem 6
Let $\mathbb{R}[x]$ be the set of all polynomials with real coefficients. Find all functions $f: \mathbb{R}[x] \rightarrow \mathbb{R}[x]$ satisfying the following conditions:

[list]
[*] $f$ maps the zero polynomial to itself,
[*] for any non-zero polynomial $P \in \mathbb{R}[x]$, $\text{deg} \, f(P) \le 1+ \text{deg} \, P$, and
[*] for any two polynomials $P, Q \in \mathbb{R}[x]$, the polynomials $P-f(Q)$ and $Q-f(P)$ have the same set of real roots.
[/list]

Proposed by Anant Mudgal, Sutanay Bhattacharya, Pulkit Sinha
28 replies
anantmudgal09
Mar 7, 2021
kes0716
2 hours ago
Inspired by my own results
sqing   5
N 2 hours ago by sqing
Source: Own
Let $ a ,  b  $ be reals such that $ a+b+ab=1. $ Show that$$ 1-\frac{1 }{\sqrt2}\le \frac{1}{a^2+1}+\frac{1}{b^2+1}\le 1+\frac{1 }{\sqrt2} $$Let $ a ,  b\geq 0 $ and $ a+b+ab=1. $ Show that$$ \frac{3}{2}\le \frac{1}{a^2+1}+\frac{1}{b^2+1}\le 1+\frac{1 }{\sqrt2} $$
5 replies
sqing
Yesterday at 8:32 AM
sqing
2 hours ago
Polygon formed by the edges of an infinite chessboard
AlperenINAN   1
N 2 hours ago by AlperenINAN
Source: Turkey TST 2025 P5
Let $P$ be a polygon formed by the edges of an infinite chessboard, which does not intersect itself. Let the numbers $a_1,a_2,a_3$ represent the number of unit squares that have exactly $1,2\text{ or } 3$ edges on the boundary of $P$ respectively. Find the largest real number $k$ such that the inequality $a_1+a_2>ka_3$ holds for each polygon constructed with these conditions.
1 reply
AlperenINAN
4 hours ago
AlperenINAN
2 hours ago
Interesting inequality
sqing   5
N 2 hours ago by sqing
Source: Own
Let $ a,b\geq 2  . $ Prove that
$$(a^2-1)(b^2-1) -6ab\geq-15$$$$(a^2-1)(b^2-1)  -7ab\geq  -\frac{58}{3}$$$$(a^3-1)(b^3-1)  -\frac{21}{4}a^2b^2\geq -35$$$$(a^3-1)(b^3-1)  -6a^2b^2\geq-\frac{2391}{49}$$
5 replies
sqing
6 hours ago
sqing
2 hours ago
JBMO Shortlist 2023 C1
Orestis_Lignos   6
N Today at 12:28 AM by zhenghua
Source: JBMO Shortlist 2023, C1
Given is a square board with dimensions $2023 \times 2023$, in which each unit cell is colored blue or red. There are exactly $1012$ rows in which the majority of cells are blue, and exactly $1012$ columns in which the majority of cells are red.

What is the maximal possible side length of the largest monochromatic square?
6 replies
Orestis_Lignos
Jun 28, 2024
zhenghua
Today at 12:28 AM
JBMO Shortlist 2023 C1
G H J
G H BBookmark kLocked kLocked NReply
Source: JBMO Shortlist 2023, C1
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Orestis_Lignos
555 posts
#1 • 2 Y
Y by ItsBesi, lian_the_noob12
Given is a square board with dimensions $2023 \times 2023$, in which each unit cell is colored blue or red. There are exactly $1012$ rows in which the majority of cells are blue, and exactly $1012$ columns in which the majority of cells are red.

What is the maximal possible side length of the largest monochromatic square?
Z K Y
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Assassino9931
1197 posts
#3
Y by
Answer

Bound

Example
Z K Y
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crezk
899 posts
#4 • 1 Y
Y by ehuseyinyigit
generalization for $(2n+1)^2$ is $n$
Z K Y
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ItsBesi
136 posts
#5
Y by
Finally did a combi :)

Answer: $2011$

Bounding:

Let $X$ be the length of the largest monochromatic square.

FTSOC assume $X \geq 2012$

Assume that the color in the monochromatic square is blue so if $X \geq 2012$ note that because of the square there should be $1012$ rows which are blue and also $1012$ columns that are also blue hence there are $2023-1012=1011$ columns that the majority of cells are red but this is a contradiction by our condition . $\rightarrow \leftarrow$

Hence $X <2012 \iff X \leq 2011$

Now we just show that $X=2011$ works:

Construction
Attachments:
This post has been edited 1 time. Last edited by ItsBesi, Jan 29, 2025, 10:47 AM
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Mhremath
66 posts
#6
Y by
easy
Let's say this square AZ Square

now Let's AZ square side be $A$
$\textcolor{red}{Claim:}$ There is no way that $A\geq 1012$
proof:
that's easy to show according to the our AZ square includes oly the same colored unit cells
and we know that majority of the rows and columns $A>1012$ is Absurd
and$A=1012$ contradict majority
then A=1011 is our answer and there is no any contradiction

and here is some cells
This post has been edited 1 time. Last edited by Mhremath, Jan 30, 2025, 8:28 PM
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jasperE3
11096 posts
#8
Y by
Orestis_Lignos wrote:
Given is a square board with dimensions $2023 \times 2023$, in which each unit cell is colored blue or red. There are exactly $1012$ rows in which the majority of cells are blue, and exactly $1012$ columns in which the majority of cells are red.

What is the maximal possible side length of the largest monochromatic square?
Call a row or column red-dominated if it has at least $c$ red cells (id est, the majority of the cells are red), and define blue-dominated similarly. Note that each row and column is either red-dominated or blue-dominated since each row and column has an odd number $2c+1$ of cells.

We will show that the answer is $1011$, by proving the following generalized problem for special case $c=1011$.
generalized problem wrote:
Let $c\in\mathbb N$. Given is a square board with dimensions $(2c+1)\times(2c+1)$, in which each unit cell is colored blue or red. There are exactly $c+1$ blue-dominated rows, and exactly $c+1$ red-dominated columns.

Prove that the maximal possible side length of the largest monochromatic square is $c$.
For the construction, we assign coordinates $(i,j)$ to each cell, where $1\le i\le2c+1$ is the column number and $1\le j\le2c+1$ is the row number. This is formally defined recursively by the following:
  1. $(1,1)$ is the bottom-left-most cell in the board
  2. if $1\le i\le2c+1$ and $1\le j\le2c$ then the cell $(i,j+1)$ is the cell directly above the cell $(i,j)$
  3. if $1\le i\le2c$ and $1\le j\le2c+1$ then the cell $(i+1,j)$ is the cell directly to the right of the cell $(i,j)$
We use the following coloring: color cell $(i,j)$ blue if $1\le i,j\le2c+1$ and one of the following mutually exclusive conditions are met:
  1. $i=j\le c+1$
  2. $i\ge c+2$ and $j\le c+1$
Color it red otherwise.

For example, this is the coloring for $c=6$:
https://i.ibb.co/rGsvcrFJ/pixil-frame-0-2.png

This works because:
  • there is a monochromatic blue $c\times c$ square in the bottom-right of the board, at coordinates $c+2\le i\le2c+1$, $1\le j\le c+1$ (as well as another blue $c\times c$ square one unit above that, and $c+2$ red $c\times c$ squares along the top of the board)
  • there are $c+1$ blue-dominated rows at coordinates $1\le j\le c+1$, since each of these rows will have $1$ blue cell at $i\le c+1$ that follow condition $A$ above, and $2c-(c+2)+1=c-1$ blue cells at $i\ge c+2$ following condition $B$
  • there are $c$ red-dominated rows (meaning that there cannot be more than $c+1$ blue-dominated rows) at coordinates $c+2\le j\le2c+1$, since they don't follow conditions $A$ or $B$
  • there are $c+1$ red-dominated columns at coordinates $1\le i\le c+1$, since each of these columns will have $1$ blue cell following condition $A$ and no other blue cells
  • there are $c$ blue-dominated columns (meaning that there cannot be more than $c+1$ red-dominated rows) at coordinates $c+2\le i\le2c+1$, since each of these columns have $c+1$ blue cells following condition $B$


Now suppose that it is possible to have a monochromatic $(c+1)\times(c+1)$ square.

If the square were red, then each of the $c+1$ rows of the board that intersect the square would have $c+1$ red cells, and therefore be red-dominated. But since the problem statement requires us to have at least $c+1$ blue-dominated rows, this would require us to have $(c+1)+(c+1)>2c+1$ rows, contradiction.

Likewise, if the square were blue, each of the $c+1$ columns that intersect the square would be blue-dominated, which is impossible as the problem requires exactly $c+1$ red-dominated columns.

So it is impossible to have a monochromatic $(c+1)\times(c+1)$ square in such a $(2c+1)\times(2c+1)$ board, and the largest possible such monochromatic square is indeed $c\times c$.
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zhenghua
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Let's generalize this by turning $2023\rightarrow 2n+1$ where $n>1$. Then there must be $n+1$ rows where the majority of the cells are blue, and exactly $n+1$ columns where the majority of the cells are red.

We claim the answer is a monochromatic square of size $n.$ The below configuration works for any $n>1$.
[asy]

fill((0,0)--(0,7)--(7,7)--(7,0)--cycle, blue);
fill((0,0)--(1,0)--(1,7)--(0,7)--cycle, red);
fill((4,0)--(7,0)--(7,3)--(4,3)--cycle, red);
fill((4,3)--(5,3)--(5,4)--(4,4)--cycle, red);
fill((5,4)--(6,4)--(6,5)--(5,5)--cycle, red);
fill((6,5)--(7,5)--(7,6)--(6,6)--cycle, red);

add(grid(7,7));

draw((4,0)--(7,0)--(7,3)--(4,3)--cycle, green);

[/asy]

Now we prove that any monochromatic square with a size more than $n$ violates one of the conditions. WLOG, let the color of the square be red and have side length $n+1$. Now note that $n+1>\frac{2n+1}{2}$ so there can be a maximum of $n$ columns with a majority of blue squares. However, this contradicts since we need exactly $n+1$ columns with a majority of blue squares. Thus, contradiction. Therefore the answer is $n$.

This problem is when $n=1011$ so the answer is $\boxed{1011}$.

$\mathbb{Q.E.D.}$
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