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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
4 hours ago
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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jlacosta
4 hours ago
0 replies
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k a My Retirement & New Leadership at AoPS
rrusczyk   1571
N Mar 26, 2025 by SmartGroot
I write today to announce my retirement as CEO from Art of Problem Solving. When I founded AoPS 22 years ago, I never imagined that we would reach so many students and families, or that we would find so many channels through which we discover, inspire, and train the great problem solvers of the next generation. I am very proud of all we have accomplished and I’m thankful for the many supporters who provided inspiration and encouragement along the way. I'm particularly grateful to all of the wonderful members of the AoPS Community!

I’m delighted to introduce our new leaders - Ben Kornell and Andrew Sutherland. Ben has extensive experience in education and edtech prior to joining AoPS as my successor as CEO, including starting like I did as a classroom teacher. He has a deep understanding of the value of our work because he’s an AoPS parent! Meanwhile, Andrew and I have common roots as founders of education companies; he launched Quizlet at age 15! His journey from founder to MIT to technology and product leader as our Chief Product Officer traces a pathway many of our students will follow in the years to come.

Thank you again for your support for Art of Problem Solving and we look forward to working with millions more wonderful problem solvers in the years to come.

And special thanks to all of the amazing AoPS team members who have helped build AoPS. We’ve come a long way from here:IMAGE
1571 replies
rrusczyk
Mar 24, 2025
SmartGroot
Mar 26, 2025
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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
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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
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hard problem
Cobedangiu   0
40 minutes ago
Let $x,y,z>0$ and $xy+yz+zx=3$ : Prove that :
$\sum \ \frac{x}{y+z}\ge\sum \frac{1}{\sqrt{x+3}}$
0 replies
Cobedangiu
40 minutes ago
0 replies
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Ornaments and Christmas trees
Morskow   30
N 44 minutes ago by akliu
Source: Slovenia IMO TST 2018, Day 1, Problem 1
Let $n$ be a positive integer. On the table, we have $n^2$ ornaments in $n$ different colours, not necessarily $n$ of each colour. Prove that we can hang the ornaments on $n$ Christmas trees in such a way that there are exactly $n$ ornaments on each tree and the ornaments on every tree are of at most $2$ different colours.
30 replies
Morskow
Dec 17, 2017
akliu
44 minutes ago
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Informatics Competition (OTIS MOCK AIME 2025 II #11)
YaoAOPS   2
N an hour ago by akliu
Source: OTIS MOCK AIME 2025 II
At an informatics competition each student earns a score in $\{0, 1, \dots, 100\}$ on each of six problems, and their total score is the sum of the six scores (out of $600$). Given two students $A$ and $B$, we write $A \succ B$ if there are at least five problems on which $A$ scored strictly higher than $B$.

Compute the smallest integer $c$ such that the following statement is true: for every integer $n \ge 2$, given students $A_1$, \dots, $A_n$ satisfying $A_1 \succ A_2 \succ \dots \succ A_n$, the total score of $A_n$ is always at most $c$ points more than the total score of $A_1$.

Jiahe Liu
2 replies
1 viewing
YaoAOPS
Jan 22, 2025
akliu
an hour ago
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Prove that
abduqahhor_math   3
N an hour ago by rchokler
n^2+3n+5 is not divided to 121
3 replies
abduqahhor_math
Mar 31, 2025
rchokler
an hour ago
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Thanks u!
Ruji2018252   1
N an hour ago by alexheinis
Let $a_1,...,a_{2024}\in\mathbb{Z}$ and $1\leqslant a_1\leqslant a_2\leqslant ...\leqslant a_{2024}\leqslant 99$ and
\[P=a_1^2+a_2^2+...+a_{2024}^2-(a_1a_3+a_2a_4+...+a_{2022}a_{2024})\]Find maximum $P$
1 reply
Ruji2018252
Yesterday at 11:51 AM
alexheinis
an hour ago
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Number-erasing game with multiples of 3
p108771953   4
N an hour ago by conejita
Source: 2024 Mexican Mathematical Olympiad, Problem 6
Ana and Beto play on a blackboard where all integers from 1 to 2024 (inclusive) are written. On each turn, Ana chooses three numbers $a,b,c$ written on the board and then Beto erases them and writes one of the following numbers:
$$a+b-c, a-b+c, ~\text{or}~ -a+b+c.$$The game ends when only two numbers are left on the board and Ana cannot play. If the sum of the final numbers is a multiple of 3, Beto wins. Otherwise, Ana wins. ¿Who has a winning strategy?
4 replies
p108771953
Nov 6, 2024
conejita
an hour ago
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Another square grid :D
MathLuis   43
N an hour ago by akliu
Source: USEMO 2021 P1
Let $n$ be a fixed positive integer and consider an $n\times n$ grid of real numbers. Determine the greatest possible number of cells $c$ in the grid such that the entry in $c$ is both strictly greater than the average of $c$'s column and strictly less than the average of $c$'s row.

Proposed by Holden Mui
43 replies
MathLuis
Oct 30, 2021
akliu
an hour ago
J
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Number theory
Maaaaaaath   0
2 hours ago
Let $m$ be a positive integer . Prove that there exists infinitely many pairs of positive integers $(x,y)$ such that $\gcd(x,y)=1$ and :

$$xy | x^2+y^2+m$$
0 replies
Maaaaaaath
2 hours ago
0 replies
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Only consecutive terms are coprime
socrates   35
N 2 hours ago by zuat.e
Source: 7th RMM 2015, Problem 1
Does there exist an infinite sequence of positive integers $a_1, a_2, a_3, . . .$ such that $a_m$ and $a_n$ are coprime if and only if $|m - n| = 1$?
35 replies
socrates
Feb 28, 2015
zuat.e
2 hours ago
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D1019 : Dominoes 2*1
Dattier   4
N 2 hours ago by Dattier
I have a 9*9 grid like this one:

IMAGE

We choose 5 white squares on the lower triangle, 5 black squares on the upper triangle and one on the diagonal, which we remove from the grid.
Like for example here:

IMAGE

Can we completely cover the grid remove from these 11 squares with 2*1 dominoes like this one:

IMAGE
4 replies
Dattier
Mar 26, 2025
Dattier
2 hours ago
J
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inequalities hard
Cobedangiu   4
N 2 hours ago by Cobedangiu
problem
4 replies
Cobedangiu
Mar 31, 2025
Cobedangiu
2 hours ago
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Find pair of m and n numbers
abduqahhor_math   5
N 3 hours ago by abduqahhor_math
(m+6:n)=4*(m;n)
5 replies
abduqahhor_math
Mar 31, 2025
abduqahhor_math
3 hours ago
J
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geometry incentre config
Tony_stark0094   2
N 3 hours ago by LearnMath_105
In a triangle $\Delta ABC$ $I$ is the incentre and point $F$ is defined such that $F \in AC$ and $F \in \odot BIC$
prove that $AI$ is the perpendicular bisector of $BF$
2 replies
Tony_stark0094
Yesterday at 4:09 PM
LearnMath_105
3 hours ago
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Geo Mock #5
Bluesoul   1
N 4 hours ago by Sedro
Consider triangle $ABC$ with $AB=13, BC=14, AC=15$. Denote the orthocenter of $\triangle{ABC}$ as $H$, the intersection of $(BHC)$ and $AC$ as $P\neq C$. Compute the length of $AP$.
1 reply
Bluesoul
Yesterday at 7:03 AM
Sedro
4 hours ago
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J
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Polynomial optimization problem
ReticulatedPython   2
N Today at 12:38 PM by Mathzeus1024
Let $$p(x)=-ax^4+x^3$$, where $a$ is a real number. Prove that for all positive $a$, $$p(x) \le \frac{27}{256a^3}.$$
2 replies
ReticulatedPython
Mar 31, 2025
Mathzeus1024
Today at 12:38 PM
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Polynomial optimization problem
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Reveal topic tags
algebra
polynomial
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ReticulatedPython
550 posts
ReticulatedPython
#1 h  Mar 31, 2025, 2:51 PM
Y by
Let $$p(x)=-ax^4+x^3$$, where $a$ is a real number. Prove that for all positive $a$, $$p(x) \le \frac{27}{256a^3}.$$
This post has been edited 1 time. Last edited by ReticulatedPython, Mar 31, 2025, 3:03 PM
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Lankou
1366 posts
Lankou
#2 h  Monday at 7:09 PM
Y by
The function has four real zeros: $0$ of multiplicity $3$ and $\frac{1}{a}$ of multiplicity $1$. The extremities of the graph both approache the $-\infty$ so the function has an absolute maximum between $0$ and $\frac{1}{a}$
$p'(x)=x^3(-ax+1)$
The graph has a turning point at $x=\frac {3}{4a}$
At this point the maximum is $=\frac {27}{256a^3}$
Hence $p(x) \le \frac{27}{256a^3}.$
This post has been edited 1 time. Last edited by Lankou, Monday at 7:11 PM
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Mathzeus1024
775 posts
Mathzeus1024
#4 h  Today at 12:38 PM
Y by
Taking $a \in \mathbb{R}^{+}$, the derivative of $p(x)$ equal to zero yields the critical values:

$p'(x) = -4ax^3+3x^2 = x^2(3-4ax) = 0 \Rightarrow x = 0, \frac{3}{4a}$ (i).

A second derivative check of (i) shows that:

$p''(x) = -12ax^2+6x \Rightarrow p''(0) = 0$ (an inflection point) and $p''\left(\frac{3}{4a}\right) = -\frac{9}{4a} < 0$ (a global maximum).

Thus, $p_{MAX} = p\left(\frac{3}{4a}\right) = \textcolor{red}{\frac{27}{256a^3}} \le \frac{27}{256a^3}$.

QED
This post has been edited 2 times. Last edited by Mathzeus1024, Today at 12:43 PM
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