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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.
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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 15 minutes ago by pco
Find all positive integers such that $$n=\lfloor \sqrt{n}\rfloor^2+\lfloor \sqrt{n}\rfloor$$
5 replies
m4thbl3nd3r
2 hours ago
pco
15 minutes ago
Hard T^T
Noname23   1
N 18 minutes ago by RagvaloD
<problem>
1 reply
Noname23
an hour ago
RagvaloD
18 minutes ago
At least 3 co-prime pairs
eezad3   0
18 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
18 minutes ago
0 replies
Suspicious Quadrilateral Geometry
YaoAOPS   4
N 19 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
19 minutes ago
i need help
MR.1   0
an hour 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
an hour 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 2 hours ago by Tuvshuu
\[
v_p(n!) \leq \frac{n}{p - 1}
\]
1 reply
MuradSafarli
2 hours ago
Tuvshuu
2 hours ago
Inspired by Kazakhstan 2017
sqing   1
N 2 hours 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
sqing
3 hours ago
sqing
2 hours 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 4 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
4 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
Tiling a grid
a_507_bc   4
N Sep 16, 2024 by ESCmath
Source: Singapore Open MO 2023 Round 2 2023 P2
A grid of cells is tiled with dominoes such that every cell is covered by exactly one domino. A subset $S$ of dominoes is chosen. Is it true that at least one of the following 2 statements is false?
(1) There are $2022$ more horizontal dominoes than vertical dominoes in $S$.
(2) The cells covered by the dominoes in $S$ can be tiled completely and exactly by $L$-shaped tetrominoes.
4 replies
a_507_bc
Jul 1, 2023
ESCmath
Sep 16, 2024
Tiling a grid
G H J
G H BBookmark kLocked kLocked NReply
Source: Singapore Open MO 2023 Round 2 2023 P2
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a_507_bc
676 posts
#1
Y by
A grid of cells is tiled with dominoes such that every cell is covered by exactly one domino. A subset $S$ of dominoes is chosen. Is it true that at least one of the following 2 statements is false?
(1) There are $2022$ more horizontal dominoes than vertical dominoes in $S$.
(2) The cells covered by the dominoes in $S$ can be tiled completely and exactly by $L$-shaped tetrominoes.
Z K Y
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qwedsazxc
167 posts
#2 • 1 Y
Y by Ngoyh
Routine coloring problem :)
Assume that it is possible for $S$ have $2022$ more horizontal dominoes than vertical dominoes, and the cells covered by the dominoes in $S$ can be fully tiled only with L-tetrominoes.
Let the number of vertical dominoes be $n$. Thus there are $2022+n$ horizontal dominoes and we need $1011+n$ L-tetrominoes to tile $S$ completely.
We color each row black and white alternating, and we consider the difference of black and white cells in $S$.
Vertical dominoes have one black and one white cells; they make up no difference in the number of black and white cells.
Horizontal dominoes are either fully black or white. Let $m$ be the number of white horizontal dominoes. Then there are $2(m-(n+2022-m)=4m-2n-4044$ more white cells than black.
When $S$ is completely tiled with $n+1011$ L-tetrominoes, each L-tetromino has two more white cells than black, or the other way around. Let $k$ be the number of L-tetrominoes with more white cells. Then there are $2(k-(n+1011-k)=4k-2n-2022$ more white cells than black.
Here $4m-2n-4044=4k-2n-2022$, and $2(m-k)=1011$; $2|1011$. This is a contradiction. Therefore the two statements can't be both true.
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beansenthusiast505
26 posts
#3
Y by
AAAAAAAAAAAAAAAAAAAA i'm so dumb why didnt i see this in smo

We define horizontal parity as a colouring of each row with alternate colours (i.e. odd rows black even rows white) and vertical parity to be its vertical colouring (odd columns black even columns white)

Each horizontal domino decreases horizontal parity by 2 and keeps vertical parity constant, vice versa for the vertical domino. For each L-tetromino, it decreases both horizontal and vertical parity by 2.

Now suppose $n$ vertical dominoes and $n+2022$ horizontal dominoes. We know that if this set $S$ can be supposedly tiled with L-tetrominoes, it can be tiled with $n+1011$ of them. Then the horizontal parity is $2n+4044$ (mod 4) $=$ $2n$ (mod 4) and the vertical parity is also $2n$ (mod 4) if tiled by the horizontal and vertical dominoes. If tiled by L-tetrominoes though, the horizontal and vertical parities would be $2n+2022$ (mod 4). Since 2022 $\equiv$ 2 (mod 4), contradiction and at least 1 statement has to be false.
Z K Y
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bxiao31415
2 posts
#4 • 1 Y
Y by asdia
Solution with just 1 coloring:

Let $(i,j)$ be colored $0$ if $i$ and $j$ are even, $1$ if $i$ is odd and $j$ is even, $3$ if $i$ is even and $j$ is odd, and $2$ if both $i$ and $j$ are odd.

Assume for contradiction that both (1) and (2) can be true. Each horizontal domino covers squares such that the sum of numbers on them is $1 \pmod 4$ and each vertical domino covers squares such that the sum of numbers on them is $-1 \pmod 4$, so the total sum of numbers on covered squares is $2022 \equiv 2 \pmod 4$. But each L-shape, which can be tiled with $1$ horizontal and $1$ vertical domino, contributes $0 \pmod 4$, so we have $0 \equiv 2 \pmod 4$, a contradiction.
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ESCmath
3 posts
#5 • 3 Y
Y by BR1F1SZ, kiyoras_2001, Akeyla
This problem was proposed by me, along with Q4. Solution is attached below.

Also, here is the story of how I came up with this problem. It is quite interesting.

As we all know, all Singaporean males have to go through 2 years of National Service, and like most others I did mine in the army. When I was doing my Basic Military Training (BMT), there was once I had to fall out of a conduct due to injury, and so I was sitting on the side while the rest were doing exercises on the parade square. Being bored with nothing to do, I decided why not try to come up with a problem. So I looked around to try to find inspiration.

It was then I noticed that the parade square was tiled with bricks in the following pattern:
https://fastly.4sqi.net/img/general/200x200/40417116_IiHFPsBl595cttYLNmkrKt0r3qsEZQGqNkuoMftEUso.jpg
(Basically alternating sets of 2 horizontal dominoes by 2 vertical dominoes)

Noticing this pattern, I decided why not try to come up with a combi problem that involved this particular tiling of dominoes? So I got to work exploring this particular pattern of dominoes to try and find nice properties I could create a problem around.

The exact details behind how L-shaped tetrominoes came into the picture has largely been lost to time, but I think it was because of observing how a pair of adjacent horizontal and vertical dominoes can be paired to form an L-shaped tetromino, and so I thought whether it is possible to expand on this idea and create sets of dominoes which can be replaced by L-shaped tetrominoes in this manner. Then I realised there were sets of dominoes which can be tiled by L-shaped tetrominoes but this one-to-one domino-pair-to-tetromino mapping doesn't exist (e.g. 4x2 rectangle with 2 horizontal dominoes on the left and 2 vertical dominoes on the right). Also, if such a one-to-one mapping exists, it basically guarantees that the number of horizontal and vertical tetrominoes in the set are the same, so I thought about exploring sets where the number of horizontal and vertical tetrominoes are not the same, but it can still be tiled by L-shaped tetrominoes.

The exact details of what happened after that have also been lost to time, but I think I did some trial and error and found that #horizontal dominoes - #vertical dominoes has to be divisible by 4, so I went about exploring why values not divisible by 4 do not work. When exploring the case where the difference is 2 mod 4, I found a solution that made it a nice problem to work out. So eventually, that was the problem I proposed.

Notice that until now I have been using the tiling from the parade square, and in fact in the original version of the problem I specified that to be the tiling used for the infinite grid. It was only afterwards that I realised that the solution worked for all tilings, not just this one, so I removed the condition and generalised it to all tilings, which gives the final version of the problem you see above.

Edit: I found an older version of the problem, attached below. If you want you can try it, but it doesn't differ that meaningfully from the original problem.
Attachments:
My Problem 4.pdf (87kb)
This post has been edited 1 time. Last edited by ESCmath, Sep 16, 2024, 6:44 PM
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