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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
1 viewing
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
True Generalization of 2023 CGMO T7
EthanWYX2009   0
6 minutes ago
Source: aops.com/community/c6h3132846p28384612
Given positive integer $n.$ Let $x_1,\ldots ,x_n\ge 0$ and $x_1x_2\cdots x_n\le 1.$ Show that
\[\sum_{k=1}^n\frac{1}{1+\sum_{j\neq k}x_j}\le\frac n{1+(n-1)\sqrt[n]{x_1x_2\cdots x_n}}.\]
0 replies
+1 w
EthanWYX2009
6 minutes ago
0 replies
Interesting inequality
sqing   0
8 minutes ago
Source: Own
Let $ a,b,c\geq \frac{1}{3}$ and $ a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+8  . $ Prove that
$$ ab+bc +ca\leq 17+2\sqrt{73}$$Let $ a,b,c\geq \frac{1}{2}$ and $ a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+8  . $ Prove that
$$ ab+bc +ca\leq \frac{469+115\sqrt{17}}{32}$$Let $ a,b,c\geq \frac{1}{5}$ and $ a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+8  . $ Prove that
$$ ab+bc +ca\leq \frac{569+34\sqrt{281}}{25}$$
0 replies
sqing
8 minutes ago
0 replies
Not homogenous, messy inequality
Kimchiks926   10
N 18 minutes ago by Marcus_Zhang
Source: Latvian TST for Baltic Way 2019 Problem 1
Prove that for all positive real numbers $a, b, c$ with $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} =1$ the following inequality holds:
$$3(ab+bc+ca)+\frac{9}{a+b+c} \le \frac{9abc}{a+b+c} + 2(a^2+b^2+c^2)+1$$
10 replies
Kimchiks926
May 29, 2020
Marcus_Zhang
18 minutes ago
Interesting inequality
sqing   0
22 minutes ago
Source: Own
Let $ a,b,c\geq 2.$ Prove that
$$ (a+1)(b+1)(c +1)-\frac{9}{4}abc\leq 9$$$$ (a+1)(b+1)(c +1)-\frac{23}{10}abc\leq\frac{43}{5}$$
0 replies
1 viewing
sqing
22 minutes ago
0 replies
USA 97 [1/(b^3+c^3+abc) + ... >= 1/(abc)]
Maverick   45
N an hour ago by Marcus_Zhang
Source: USAMO 1997/5; also: ineq E2.37 in Book: Inegalitati; Authors:L.Panaitopol,V. Bandila,M.Lascu
Prove that, for all positive real numbers $ a$, $ b$, $ c$, the inequality
\[ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc}
\]
holds.
45 replies
1 viewing
Maverick
Sep 12, 2003
Marcus_Zhang
an hour ago
The prime inequality learning problem
orl   137
N 2 hours ago by Marcus_Zhang
Source: IMO 1995, Problem 2, Day 1, IMO Shortlist 1995, A1
Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Prove that
\[ \frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}.
\]
137 replies
orl
Nov 9, 2005
Marcus_Zhang
2 hours ago
hard ............ (2)
Noname23   2
N 3 hours ago by mathprodigy2011
problem
2 replies
Noname23
Yesterday at 5:10 PM
mathprodigy2011
3 hours ago
Abelkonkurransen 2025 3a
Lil_flip38   5
N 3 hours ago by ariopro1387
Source: abelkonkurransen
Let \(ABC\) be a triangle. Let \(E,F\) be the feet of the altitudes from \(B,C\) respectively. Let \(P,Q\) be the projections of \(B,C\) onto line \(EF\). Show that \(PE=QF\).
5 replies
Lil_flip38
Yesterday at 11:14 AM
ariopro1387
3 hours ago
Inequality by Po-Ru Loh
v_Enhance   54
N 3 hours ago by Marcus_Zhang
Source: ELMO 2003 Problem 4
Let $x,y,z \ge 1$ be real numbers such that \[ \frac{1}{x^2-1} + \frac{1}{y^2-1} + \frac{1}{z^2-1} = 1. \] Prove that \[ \frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} \le 1. \]
54 replies
v_Enhance
Dec 29, 2012
Marcus_Zhang
3 hours ago
Problem 5
Functional_equation   14
N 3 hours ago by ali123456
Source: Azerbaijan third round 2020(JBMO Shortlist 2019 N6)
$a,b,c$ are non-negative integers.
Solve: $a!+5^b=7^c$

Proposed by Serbia
14 replies
Functional_equation
Jun 6, 2020
ali123456
3 hours ago
a^12+3^b=1788^c
falantrng   6
N 4 hours ago by ali123456
Source: Azerbaijan NMO 2024. Junior P3
Find all the natural numbers $a, b, c$ satisfying the following equation:
$$a^{12} + 3^b = 1788^c$$.
6 replies
falantrng
Jul 8, 2024
ali123456
4 hours ago
stuck on a system of recurrence sequence
Nonecludiangeofan   0
4 hours ago
Please guys help me solve this nasty problem that i've been stuck for the past month:
Let \( (a_n) \) and \( (b_n) \) be two sequences defined by:
\[
a_{n+1} = \frac{1 + a_n + a_n b_n}{b_n} \quad \text{and} \quad b_{n+1} = \frac{1 + b_n + a_n b_n}{a_n}
\]for all \( n \ge 0 \), with initial values \( a_0 = 1 \) and \( b_0 = 2 \).

Prove that:
\[
a_{2024} < 5.
\]
(btw am still not comfortable with system of recurrence sequences)
0 replies
Nonecludiangeofan
4 hours ago
0 replies
A huge group of children compare their heights
Tintarn   5
N 4 hours ago by InCtrl
Source: All-Russian MO 2024 9.8
$1000$ children, no two of the same height, lined up. Let us call a pair of different children $(a,b)$ good if between them there is no child whose height is greater than the height of one of $a$ and $b$, but less than the height of the other. What is the greatest number of good pairs that could be formed? (Here, $(a,b)$ and $(b,a)$ are considered the same pair.)
Proposed by I. Bogdanov
5 replies
Tintarn
Apr 22, 2024
InCtrl
4 hours ago
Iran Inequality
mathmatecS   15
N 5 hours ago by Marcus_Zhang
Source: Iran 1998
When $x(\ge1),$ $y(\ge1),$ $z(\ge1)$ satisfy $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2,$ prove in equality.
$$\sqrt{x+y+z}\ge\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$$
15 replies
mathmatecS
Jun 11, 2015
Marcus_Zhang
5 hours ago
JBMO Shortlist 2023 C1
Orestis_Lignos   6
N Mar 18, 2025 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
Mar 18, 2025
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?
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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
Z K Y
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jasperE3
11100 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
1043 posts
#9
Y by
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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