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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
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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
J
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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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Sipnayan JHS 2021 F-9
PikaVee   1
N 5 minutes ago by PikaVee
Matt and Sai are playing a game of darts together. Matt has a slightly more accurate aim than Sai. In
fact, Matt can hit the bullseye 80% of the time while Sai can only hit it 60% of the time. They take turns
in playing and the first player is determined by a flip of a fair coin. If the probability that Sai scores the
first bullseye is given by $ \frac {a}{b} $ where a and b are relatively prime integers, what is b − a?
1 reply
PikaVee
23 minutes ago
PikaVee
5 minutes ago
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Standart looking FE
Kimchiks926   13
N 16 minutes ago by math-olympiad-clown
Source: Baltic Way 2022, Problem 5
Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(0)+1=f(1)$ and for any real numbers $x$ and $y$,
$$ f(xy-x)+f(x+f(y))=yf(x)+3 $$
13 replies
Kimchiks926
Nov 12, 2022
math-olympiad-clown
16 minutes ago
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A sharp one with 3 var (2)
mihaig   4
N 16 minutes ago by mihaig
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$a+b+c+\sqrt{abc}\geq4.$$
4 replies
mihaig
May 26, 2025
mihaig
16 minutes ago
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3 var inequality
SunnyEvan   11
N 17 minutes ago by mihaig
Let $ a,b,c \in R $ ,such that $ a^2+b^2+c^2=4(ab+bc+ca)$Prove that :$$ \frac{7-2\sqrt{14}}{48} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{7+2\sqrt{14}}{48} $$
11 replies
SunnyEvan
May 17, 2025
mihaig
17 minutes ago
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trigonometric inequality
MATH1945   12
N 18 minutes ago by mihaig
Source: ?
In triangle $ABC$, prove that $$sin^2(A)+sin^2(B)+sin^2(C) \leq \frac{9}{4}$$
12 replies
MATH1945
May 26, 2016
mihaig
18 minutes ago
J
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Prefix sums of divisors are perfect squares
CyclicISLscelesTrapezoid   38
N 18 minutes ago by maromex
Source: ISL 2021 N3
Find all positive integers $n$ with the following property: the $k$ positive divisors of $n$ have a permutation $(d_1,d_2,\ldots,d_k)$ such that for $i=1,2,\ldots,k$, the number $d_1+d_2+\cdots+d_i$ is a perfect square.
38 replies
1 viewing
CyclicISLscelesTrapezoid
Jul 12, 2022
maromex
18 minutes ago
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Low unsociable sets implies low chromatic number
62861   21
N 27 minutes ago by awesomeming327.
Source: IMO 2015 Shortlist, C7
In a company of people some pairs are enemies. A group of people is called unsociable if the number of members in the group is odd and at least $3$, and it is possible to arrange all its members around a round table so that every two neighbors are enemies. Given that there are at most $2015$ unsociable groups, prove that it is possible to partition the company into $11$ parts so that no two enemies are in the same part.

Proposed by Russia
21 replies
62861
Jul 7, 2016
awesomeming327.
27 minutes ago
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Sipnayan 2025 JHS F E-Spaghetti
PikaVee   2
N 33 minutes ago by PikaVee
There are two bins A and B which contain 12 balls and 24 balls, respectively. Each of these balls is marked with one letter: X, Y, or Z. In each bin, each ball is equally likely to be chosen. Randomly picking from bin A, the probability of choosing balls marked X and Y are $ \frac{1}{3} $ and $ \frac{1}{4} $, respectively. Randomly picking from bin B, the probability of choosing balls marked X and Y are $ \frac{1}{4} $ and $ \frac{1}{3} $, respectively. If the contents of the two bins are merged into one bin, what is the probability of choosing two balls marked X and Y from this bin?
2 replies
PikaVee
43 minutes ago
PikaVee
33 minutes ago
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100 Selected Problems Handout
Asjmaj   35
N 44 minutes ago by CBMaster
Happy New Year to all AoPSers!
 :clap2:

Here’s my modest gift to you all. Although I haven’t been very active in the forums, the AoPS community contributed to an immense part of my preparation and left a huge impact on me as a person. Consider this my way of giving back. I also want to take this opportunity to thank Evan Chen—his work has consistently inspired me throughout my olympiad journey, and this handout is no exception.



With 2025 drawing near, my High School Olympiad career will soon be over, so I want to share a compilation of the problems that I liked the most over the years and their respective detailed write-ups. Originally, I intended it just as a personal record, but I decided to give it some “textbook value” by not repeating the topics so that the selection would span many different approaches, adding hints, and including my motivations and thought process.

While IMHO it turned out to be quite instructive, I cannot call it a textbook by any means. I recommend solving it if you are confident enough and want to test your skills on miscellaneous, unordered, challenging, high-quality problems. Hints will allow you to not be stuck for too long, and the fully motivated solutions (often with multiple approaches) should help broaden your perspective. 



This is my first experience of writing anything in this format, and I’m not a writer by any means, so please forgive any mistakes or nonsense that may be written here. If you spot any typos, inconsistencies, or flawed arguments whatsoever (no one is immune :blush: ), feel free to DM me. In fact, I welcome any feedback or suggestions.

I left some authors/sources blank simply because I don’t know them, so if you happen to recognize where and by whom a problem originated, please let me know. And quoting the legend: “The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.” 



I’ll likely keep a separate file to track all the typos, and when there’s enough, I will update the main file. Some problems need polishing (at least aesthetically), and I also have more remarks to add.

This content is only for educational purposes and is not meant for commercial usage.



This is it! Good luck in 45^2, and I hope you enjoy working through these problems as much as I did!

Here's a link to Google Drive because of AoPS file size constraints: Selected Problems
35 replies
Asjmaj
Dec 31, 2024
CBMaster
44 minutes ago
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Inspired by SunnyEvan
sqing   3
N an hour ago by sqing
Source: Own
Let $ a,b,c $ be reals such that $ a^2+b^2+c^2=2(ab+bc+ca). $ Prove that$$ \frac{1}{12} \leq \frac{a^2b+b^2c+c^2a}{(a+b+c)^3} \leq \frac{5}{36} $$Let $ a,b,c $ be reals such that $ a^2+b^2+c^2=5(ab+bc+ca). $ Prove that$$ -\frac{1}{25} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{197}{675} $$
3 replies
sqing
May 17, 2025
sqing
an hour ago
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Centrally symmetric polyhedron
genius_007   0
an hour ago
Source: unknown
Does there exist a convex polyhedron with an odd number of sides, where each side is centrally symmetric?
0 replies
genius_007
an hour ago
0 replies
J
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2-var inequality
sqing   4
N an hour ago by sqing
Source: Own
Let $ a,b> 0 $ and $2a+2b+ab=5. $ Prove that
$$ \frac{a^2}{b^2}+\frac{1}{a^2}-a^2\geq 1$$$$ \frac{a^3}{b^3}+\frac{1}{a^3}-a^3\geq 1$$
4 replies
sqing
3 hours ago
sqing
an hour ago
J
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Combinatorial Game
Cats_on_a_computer   1
N an hour ago by Cats_on_a_computer

Let n>1 be odd. A row of n spaces is initially empty. Alice and Bob alternate moves (Alice first); on each turn a player may either
1. Place a stone in any empty space, or
2. Remove a stone from a non-empty space S, then (if they exist) place stones in the nearest empty spaces immediately to the left and to the right of S.

Furthermore, no move may produce a position that has appeared earlier. The player loses when they cannot make a legal move.
Assuming optimal play, which move(s) can Alice make on her first turn?
1 reply
Cats_on_a_computer
2 hours ago
Cats_on_a_computer
an hour ago
J
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A game of digits and seventh powers
v_Enhance   28
N an hour ago by blueprimes
Source: Taiwan 2014 TST3 Quiz 1, P2
Alice and Bob play the following game. They alternate selecting distinct nonzero digits (from $1$ to $9$) until they have chosen seven such digits, and then consider the resulting seven-digit number by concatenating the digits in the order selected, with the seventh digit appearing last (i.e. $\overline{A_1B_2A_3B_4A_6B_6A_7}$). Alice wins if and only if the resulting number is the last seven decimal digits of some perfect seventh power. Please determine which player has the winning strategy.
28 replies
v_Enhance
Jul 18, 2014
blueprimes
an hour ago
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4 variables with quadrilateral sides 2
mihaig   6
N May 2, 2025 by mihaig
Source: Own
Let $a,b,c,d\geq0$ satisfying
$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}=2.$$Prove
$$\left(a+b+c+d-2\right)^2+8\geq3\left(abc+abd+acd+bcd\right).$$
6 replies
mihaig
Apr 29, 2025
mihaig
May 2, 2025
J
4 variables with quadrilateral sides 2
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Reveal topic tags
inequalities
geometry
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G H BBookmark kLocked kLocked NReply
Source: Own
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mihaig
7388 posts
mihaig
#1 h  Apr 29, 2025, 8:47 PM • 1 Y
Y by arqady
Let $a,b,c,d\geq0$ satisfying
$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}=2.$$Prove
$$\left(a+b+c+d-2\right)^2+8\geq3\left(abc+abd+acd+bcd\right).$$
Z K Y
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mihaig
7388 posts
mihaig
#2 h  Apr 30, 2025, 6:29 PM
Y by
See, of course,
https://artofproblemsolving.com/community/c6t243f6h3555859_4_variables_with_quadrilateral_sides
Z K Y
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mihaig
7388 posts
mihaig
#3 h  May 1, 2025, 4:45 PM
Y by
Let $a,b,c,d\geq0$ satisfying
$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}=2.$$Prove or disprove
$$\left(a+b+c+d-3\right)^2+7\geq2\left(abc+abd+acd+bcd\right).$$
Z K Y
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arqady
30260 posts
arqady
#4 h  May 1, 2025, 5:35 PM
Y by
mihaig wrote:
Let $a,b,c,d\geq0$ satisfying
$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}=2.$$Prove or disprove
$$\left(a+b+c+d-3\right)^2+7\geq2\left(abc+abd+acd+bcd\right).$$
It's wrong.
Try $(a,b,c,d)=\left(\frac{1}{2},\frac{5}{4},\frac{5}{4},\frac{5}{4}\right).$
Z K Y
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arqady
30260 posts
arqady
#5 h  May 1, 2025, 5:36 PM
Y by
mihaig wrote:
Let $a,b,c,d\geq0$ satisfying
$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}=2.$$Prove
$$\left(a+b+c+d-2\right)^2+8\geq3\left(abc+abd+acd+bcd\right).$$
Nice! There is a nice solution.
Z K Y
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mihaig
7388 posts
mihaig
#6 h  May 2, 2025, 9:21 AM
Y by
arqady wrote:
mihaig wrote:
Let $a,b,c,d\geq0$ satisfying
$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}=2.$$Prove or disprove
$$\left(a+b+c+d-3\right)^2+7\geq2\left(abc+abd+acd+bcd\right).$$
It's wrong.
Try $(a,b,c,d)=\left(\frac{1}{2},\frac{5}{4},\frac{5}{4},\frac{5}{4}\right).$

Bravo!
Z K Y
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mihaig
7388 posts
mihaig
#7 h  May 2, 2025, 3:30 PM
Y by
arqady wrote:
mihaig wrote:
Let $a,b,c,d\geq0$ satisfying
$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}=2.$$Prove
$$\left(a+b+c+d-2\right)^2+8\geq3\left(abc+abd+acd+bcd\right).$$
Nice! There is a nice solution.

Thank you!
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