Happy Memorial Day! Please note that AoPS Online is closed May 24-26th.

G
Topic
First Poster
Last Poster
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]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 1, 2025
0 replies
k i Peer-to-Peer Programs Forum
jwelsh   157
N Dec 11, 2023 by cw357
Many of our AoPS Community members share their knowledge with their peers in a variety of ways, ranging from creating mock contests to creating real contests to writing handouts to hosting sessions as part of our partnership with schoolhouse.world.

To facilitate students in these efforts, we have created a new Peer-to-Peer Programs forum. With the creation of this forum, we are starting a new process for those of you who want to advertise your efforts. These advertisements and ensuing discussions have been cluttering up some of the forums that were meant for other purposes, so we’re gathering these topics in one place. This also allows students to find new peer-to-peer learning opportunities without having to poke around all the other forums.

To announce your program, or to invite others to work with you on it, here’s what to do:

1) Post a new topic in the Peer-to-Peer Programs forum. This will be the discussion thread for your program.

2) Post a single brief post in this thread that links the discussion thread of your program in the Peer-to-Peer Programs forum.

Please note that we’ll move or delete any future advertisement posts that are outside the Peer-to-Peer Programs forum, as well as any posts in this topic that are not brief announcements of new opportunities. In particular, this topic should not be used to discuss specific programs; those discussions should occur in topics in the Peer-to-Peer Programs forum.

Your post in this thread should have what you're sharing (class, session, tutoring, handout, math or coding game/other program) and a link to the thread in the Peer-to-Peer Programs forum, which should have more information (like where to find what you're sharing).
157 replies
jwelsh
Mar 15, 2021
cw357
Dec 11, 2023
k i C&P posting recs by mods
v_Enhance   0
Jun 12, 2020
The purpose of this post is to lay out a few suggestions about what kind of posts work well for the C&P forum. Except in a few cases these are mostly meant to be "suggestions based on historical trends" rather than firm hard rules; we may eventually replace this with an actual list of firm rules but that requires admin approval :) That said, if you post something in the "discouraged" category, you should not be totally surprised if it gets locked; they are discouraged exactly because past experience shows they tend to go badly.
-----------------------------
1. Program discussion: Allowed
If you have questions about specific camps or programs (e.g. which classes are good at X camp?), these questions fit well here. Many camps/programs have specific sub-forums too but we understand a lot of them are not active.
-----------------------------
2. Results discussion: Allowed
You can make threads about e.g. how you did on contests (including AMC), though on AMC day when there is a lot of discussion. Moderators and administrators may do a lot of thread-merging / forum-wrangling to keep things in one place.
-----------------------------
3. Reposting solutions or questions to past AMC/AIME/USAMO problems: Allowed
This forum contains a post for nearly every problem from AMC8, AMC10, AMC12, AIME, USAJMO, USAMO (and these links give you an index of all these posts). It is always permitted to post a full solution to any problem in its own thread (linked above), regardless of how old the problem is, and even if this solution is similar to one that has already been posted. We encourage this type of posting because it is helpful for the user to explain their solution in full to an audience, and for future users who want to see multiple approaches to a problem or even just the frequency distribution of common approaches. We do ask for some explanation; if you just post "the answer is (B); ez" then you are not adding anything useful.

You are also encouraged to post questions about a specific problem in the specific thread for that problem, or about previous user's solutions. It's almost always better to use the existing thread than to start a new one, to keep all the discussion in one place easily searchable for future visitors.
-----------------------------
4. Advice posts: Allowed, but read below first
You can use this forum to ask for advice about how to prepare for math competitions in general. But you should be aware that this question has been asked many many times. Before making a post, you are encouraged to look at the following:
[list]
[*] Stop looking for the right training: A generic post about advice that keeps getting stickied :)
[*] There is an enormous list of links on the Wiki of books / problems / etc for all levels.
[/list]
When you do post, we really encourage you to be as specific as possible in your question. Tell us about your background, what you've tried already, etc.

Actually, the absolute best way to get a helpful response is to take a few examples of problems that you tried to solve but couldn't, and explain what you tried on them / why you couldn't solve them. Here is a great example of a specific question.
-----------------------------
5. Publicity: use P2P forum instead
See https://artofproblemsolving.com/community/c5h2489297_peertopeer_programs_forum.
Some exceptions have been allowed in the past, but these require approval from administrators. (I am not totally sure what the criteria is. I am not an administrator.)
-----------------------------
6. Mock contests: use Mock Contests forum instead
Mock contests should be posted in the dedicated forum instead:
https://artofproblemsolving.com/community/c594864_aops_mock_contests
-----------------------------
7. AMC procedural questions: suggest to contact the AMC HQ instead
If you have a question like "how do I submit a change of venue form for the AIME" or "why is my name not on the qualifiers list even though I have a 300 index", you would be better off calling or emailing the AMC program to ask, they are the ones who can help you :)
-----------------------------
8. Discussion of random math problems: suggest to use MSM/HSM/HSO instead
If you are discussing a specific math problem that isn't from the AMC/AIME/USAMO, it's better to post these in Middle School Math, High School Math, High School Olympiads instead.
-----------------------------
9. Politics: suggest to use Round Table instead
There are important conversations to be had about things like gender diversity in math contests, etc., for sure. However, from experience we think that C&P is historically not a good place to have these conversations, as they go off the rails very quickly. We encourage you to use the Round Table instead, where it is much more clear that all posts need to be serious.
-----------------------------
10. MAA complaints: discouraged
We don't want to pretend that the MAA is perfect or that we agree with everything they do. However, we chose to discourage this sort of behavior because in practice most of the comments we see are not useful and some are frankly offensive.
[list] [*] If you just want to blow off steam, do it on your blog instead.
[*] When you have criticism, it should be reasoned, well-thought and constructive. What we mean by this is, for example, when the AOIME was announced, there was great outrage about potential cheating. Well, do you really think that this is something the organizers didn't think about too? Simply posting that "people will cheat and steal my USAMOO qualification, the MAA are idiots!" is not helpful as it is not bringing any new information to the table.
[*] Even if you do have reasoned, well-thought, constructive criticism, we think it is actually better to email it the MAA instead, rather than post it here. Experience shows that even polite, well-meaning suggestions posted in C&P are often derailed by less mature users who insist on complaining about everything.
[/list]
-----------------------------
11. Memes and joke posts: discouraged
It's fine to make jokes or lighthearted posts every so often. But it should be done with discretion. Ideally, jokes should be done within a longer post that has other content. For example, in my response to one user's question about olympiad combinatorics, I used a silly picture of Sogiita Gunha, but it was done within a context of a much longer post where it was meant to actually make a point.

On the other hand, there are many threads which consist largely of posts whose only content is an attached meme with the word "MAA" in it. When done in excess like this, the jokes reflect poorly on the community, so we explicitly discourage them.
-----------------------------
12. Questions that no one can answer: discouraged
Examples of this: "will MIT ask for AOIME scores?", "what will the AIME 2021 cutoffs be (asked in 2020)", etc. Basically, if you ask a question on this forum, it's better if the question is something that a user can plausibly answer :)
-----------------------------
13. Blind speculation: discouraged
Along these lines, if you do see a question that you don't have an answer to, we discourage "blindly guessing" as it leads to spreading of baseless rumors. For example, if you see some user posting "why are there fewer qualifiers than usual this year?", you should not reply "the MAA must have been worried about online cheating so they took fewer people!!". Was sich überhaupt sagen lässt, lässt sich klar sagen; und wovon man nicht reden kann, darüber muss man schweigen.
-----------------------------
14. Discussion of cheating: strongly discouraged
If you have evidence or reasonable suspicion of cheating, please report this to your Competition Manager or to the AMC HQ; these forums cannot help you.
Otherwise, please avoid public discussion of cheating. That is: no discussion of methods of cheating, no speculation about how cheating affects cutoffs, and so on --- it is not helpful to anyone, and it creates a sour atmosphere. A longer explanation is given in Seriously, please stop discussing how to cheat.
-----------------------------
15. Cutoff jokes: never allowed
Whenever the cutoffs for any major contest are released, it is very obvious when they are official. In the past, this has been achieved by the numbers being posted on the official AMC website (here) or through a post from the AMCDirector account.

You must never post fake cutoffs, even as a joke. You should also refrain from posting cutoffs that you've heard of via email, etc., because it is better to wait for the obvious official announcement. A longer explanation is given in A Treatise on Cutoff Trolling.
-----------------------------
16. Meanness: never allowed
Being mean is worse than being immature and unproductive. If another user does something which you think is inappropriate, use the Report button to bring the post to moderator attention, or if you really must reply, do so in a way that is tactful and constructive rather than inflammatory.
-----------------------------

Finally, we remind you all to sit back and enjoy the problems. :D

-----------------------------
(EDIT 2024-09-13: AoPS has asked to me to add the following item.)

Advertising paid program or service: never allowed

Per the AoPS Terms of Service (rule 5h), general advertisements are not allowed.

While we do allow advertisements of official contests (at the MAA and MATHCOUNTS level) and those run by college students with at least one successful year, any and all advertisements of a paid service or program is not allowed and will be deleted.
0 replies
v_Enhance
Jun 12, 2020
0 replies
k i Stop looking for the "right" training
v_Enhance   50
N Oct 16, 2017 by blawho12
Source: Contest advice
EDIT 2019-02-01: https://blog.evanchen.cc/2019/01/31/math-contest-platitudes-v3/ is the updated version of this.

EDIT 2021-06-09: see also https://web.evanchen.cc/faq-contest.html.

Original 2013 post
50 replies
v_Enhance
Feb 15, 2013
blawho12
Oct 16, 2017
inequality
danilorj   0
4 minutes ago
Suppose $a,b,c$ are positive real numbers and $a+b+c=2$. Show that

$(1+a^2)(1+b^2)(1+c^2)\geq 3$.
0 replies
danilorj
4 minutes ago
0 replies
Parallelogram in the Plane
Taco12   8
N 11 minutes ago by lpieleanu
Source: 2023 Canada EGMO TST/2
Parallelogram $ABCD$ is given in the plane. The incircle of triangle $ABC$ has center $I$ and is tangent to diagonal $AC$ at $X$. Let $Y$ be the center of parallelogram $ABCD$. Show that $DX$ and $IY$ are parallel.
8 replies
+1 w
Taco12
Feb 10, 2023
lpieleanu
11 minutes ago
Combinatorial
|nSan|ty   7
N 27 minutes ago by SomeonecoolLovesMaths
Source: RMO 2007 problem
How many 6-digit numbers are there such that-:
a)The digits of each number are all from the set $ \{1,2,3,4,5\}$
b)any digit that appears in the number appears at least twice ?
(Example: $ 225252$ is valid while $ 222133$ is not)
[weightage 17/100]
7 replies
|nSan|ty
Oct 10, 2007
SomeonecoolLovesMaths
27 minutes ago
pairs (m, n) such that a fractional expression is an integer
cielblue   0
an hour ago
Find all pairs $(m,\ n)$ of positive integers such that $\frac{m^3-mn+1}{m^2+mn+2}$ is an integer.
0 replies
cielblue
an hour ago
0 replies
Perfect squares: 2011 USAJMO #1
v_Enhance   227
N 2 hours ago by ray66
Find, with proof, all positive integers $n$ for which $2^n + 12^n + 2011^n$ is a perfect square.
227 replies
v_Enhance
Apr 28, 2011
ray66
2 hours ago
Mustang Math Recruitment is Open!
MustangMathTournament   0
2 hours ago
The Interest Form for joining Mustang Math is open!

Hello all!

We're Mustang Math, and we are currently recruiting for the 2025-2026 year! If you are a high school or college student and are passionate about promoting an interest in competition math to younger students, you should strongly consider filling out the following form: https://link.mustangmath.com/join. Every member in MM truly has the potential to make a huge impact, no matter your experience!

About Mustang Math

Mustang Math is a nonprofit organization of high school and college volunteers that is dedicated to providing middle schoolers access to challenging, interesting, fun, and collaborative math competitions and resources. Having reached over 4000 U.S. competitors and 1150 international competitors in our first six years, we are excited to expand our team to offer our events to even more mathematically inclined students.

PROJECTS
We have worked on various math-related projects. Our annual team math competition, Mustang Math Tournament (MMT) recently ran. We hosted 8 in-person competitions based in Washington, NorCal, SoCal, Illinois, Georgia, Massachusetts, Nevada and New Jersey, as well as an online competition run nationally. In total, we had almost 900 competitors, and the students had glowing reviews of the event. MMT International will once again be running later in August, and with it, we anticipate our contest to reach over a thousand students.

In our classes, we teach students math in fun and engaging math lessons and help them discover the beauty of mathematics. Our aspiring tech team is working on a variety of unique projects like our website and custom test platform. We also have a newsletter, which, combined with our social media presence, helps to keep the mathematics community engaged with cool puzzles, tidbits, and information about the math world! Our design team ensures all our merch and material is aesthetically pleasing.

Some highlights of this past year include 1000+ students in our classes, AMC10 mock with 150+ participants, our monthly newsletter to a subscriber base of 6000+, creating 8 designs for 800 pieces of physical merchandise, as well as improving our custom website (mustangmath.com, 20k visits) and test-taking platform (comp.mt, 6500+ users).

Why Join Mustang Math?

As a non-profit organization on the rise, there are numerous opportunities for volunteers to share ideas and suggest projects that they are interested in. Through our organizational structure, members who are committed have the opportunity to become a part of the leadership team. Overall, working in the Mustang Math team is both a fun and fulfilling experience where volunteers are able to pursue their passion all while learning how to take initiative and work with peers. We welcome everyone interested in joining!

More Information

To learn more, visit https://link.mustangmath.com/RecruitmentInfo. If you have any questions or concerns, please email us at contact@mustangmath.com.

https://link.mustangmath.com/join
0 replies
MustangMathTournament
2 hours ago
0 replies
Zsigmondy's theorem
V0305   2
N 3 hours ago by Andyluo
Is Zsigmondy's theorem allowed on the IMO, and is it allowed on the AMC series of proof competitions (e.g. USAJMO, USA TSTST)?
2 replies
V0305
3 hours ago
Andyluo
3 hours ago
Moving P(o)in(t)s
bobthegod78   71
N 3 hours ago by ray66
Source: USAJMO 2021/4
Carina has three pins, labeled $A, B$, and $C$, respectively, located at the origin of the coordinate plane. In a move, Carina may move a pin to an adjacent lattice point at distance $1$ away. What is the least number of moves that Carina can make in order for triangle $ABC$ to have area 2021?

(A lattice point is a point $(x, y)$ in the coordinate plane where $x$ and $y$ are both integers, not necessarily positive.)
71 replies
bobthegod78
Apr 15, 2021
ray66
3 hours ago
Jane street swag package? USA(J)MO
arfekete   45
N 4 hours ago by vsarg
Hey! People are starting to get their swag packages from Jane Street for qualifying for USA(J)MO, and after some initial discussion on what we got, people are getting different things. Out of curiosity, I was wondering how they decide who gets what.
Please enter the following info:

- USAMO or USAJMO
- Grade
- Score
- Award/Medal/HM
- MOP (yes or no, if yes then color)
- List of items you got in your package

I will reply with my info as an example.
45 replies
arfekete
May 7, 2025
vsarg
4 hours ago
Proof Writing Help
gulab_jamun   3
N Today at 3:01 PM by maromex
Ok so like, i'm working on proofs, and im prolly gonna use this page for any questions. My question as of now is what can I cite? Like for example, if for a question I use Evan Chen's fact 5, in my proof do I have to prove fact 5 all over again or can i say "this result follows from Evan Chen's fact 5"?
3 replies
gulab_jamun
Yesterday at 2:45 PM
maromex
Today at 3:01 PM
MAN IS KID
DrMath   136
N Today at 11:00 AM by lakshya2009
Source: USAMO 2017 P3, Evan Chen
Let $ABC$ be a scalene triangle with circumcircle $\Omega$ and incenter $I$. Ray $AI$ meets $\overline{BC}$ at $D$ and meets $\Omega$ again at $M$; the circle with diameter $\overline{DM}$ cuts $\Omega$ again at $K$. Lines $MK$ and $BC$ meet at $S$, and $N$ is the midpoint of $\overline{IS}$. The circumcircles of $\triangle KID$ and $\triangle MAN$ intersect at points $L_1$ and $L_2$. Prove that $\Omega$ passes through the midpoint of either $\overline{IL_1}$ or $\overline{IL_2}$.

Proposed by Evan Chen
136 replies
DrMath
Apr 19, 2017
lakshya2009
Today at 11:00 AM
have you done DCX-Russian?
GoodMorning   83
N Today at 7:30 AM by ray66
Source: 2023 USAJMO Problem 3
Consider an $n$-by-$n$ board of unit squares for some odd positive integer $n$. We say that a collection $C$ of identical dominoes is a maximal grid-aligned configuration on the board if $C$ consists of $(n^2-1)/2$ dominoes where each domino covers exactly two neighboring squares and the dominoes don't overlap: $C$ then covers all but one square on the board. We are allowed to slide (but not rotate) a domino on the board to cover the uncovered square, resulting in a new maximal grid-aligned configuration with another square uncovered. Let $k(C)$ be the number of distinct maximal grid-aligned configurations obtainable from $C$ by repeatedly sliding dominoes. Find the maximum value of $k(C)$ as a function of $n$.

Proposed by Holden Mui
83 replies
GoodMorning
Mar 23, 2023
ray66
Today at 7:30 AM
Titu Factoring Troll
GoodMorning   77
N Today at 7:19 AM by ray66
Source: 2023 USAJMO Problem 1
Find all triples of positive integers $(x,y,z)$ that satisfy the equation
$$2(x+y+z+2xyz)^2=(2xy+2yz+2zx+1)^2+2023.$$
77 replies
GoodMorning
Mar 23, 2023
ray66
Today at 7:19 AM
Need help with combi problems
JARP091   4
N Today at 4:19 AM by JARP091
I want to create a problem set of some of the hardest combi problems that are yet to appear in any contest. Can anyone help me out? Also can anyone give me some tips to create combi problems.
4 replies
JARP091
Yesterday at 5:43 PM
JARP091
Today at 4:19 AM
IMO problem 1
iandrei   77
N Apr 23, 2025 by YaoAOPS
Source: IMO ShortList 2003, combinatorics problem 1
Let $A$ be a $101$-element subset of the set $S=\{1,2,\ldots,1000000\}$. Prove that there exist numbers $t_1$, $t_2, \ldots, t_{100}$ in $S$ such that the sets \[ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100  \] are pairwise disjoint.
77 replies
iandrei
Jul 14, 2003
YaoAOPS
Apr 23, 2025
IMO problem 1
G H J
Source: IMO ShortList 2003, combinatorics problem 1
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
iandrei
138 posts
#1 • 11 Y
Y by Davi-8191, Wizard_32, aops5234, Adventure10, Sprites, THEfmigm, megarnie, Mango247, and 3 other users
Let $A$ be a $101$-element subset of the set $S=\{1,2,\ldots,1000000\}$. Prove that there exist numbers $t_1$, $t_2, \ldots, t_{100}$ in $S$ such that the sets \[ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100  \] are pairwise disjoint.
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Anonymous
334 posts
#2 • 8 Y
Y by ValidName, lahmacun, Adventure10, myh2910, Mango247, and 3 other users
Assume that we have already found t_1, t_2, ..., t_k (k<=99) and are searching for t_{k+1}. We can not take t_{k+1} only of form
t_i+a_j-a_l, where 1<=i<=k and a_j and a_l are elements of A.
So, we have k*101*100 forbidden values of t_{k+1}, which correspond to a_j<>a_l, and k forbidden values which correspond to a_j=a_l. So, at most 99*101*100+99=1000000-1 forbidden values, and at least 1 admissible.
Z K
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Alison
264 posts
#3 • 6 Y
Y by A_Math_Lover, biomathematics, Adventure10, Mango247, and 2 other users
Notice that if instead of taking an arbitrary t_{k+1}, you always take the smallest t_{k+1} that is not forbidden, you will only have to make sure that t_{k+1} is distinct from all t_i+a_j-a_l with a_j>a_l. This is true because the values with a_j<a_l are all <=t_k, and so have been forbidden at some earlier step.

This gives you ceiling(1000000/5051) = 198 t_i's.

However, Fedor's proof also shows that you can choose 100 t_i's such that the sets are parwise disjoint even when taken mod 10^6.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Anonymous
334 posts
#4 • 6 Y
Y by ValidName, Adventure10, Mango247, and 3 other users
Let A = {a1 < a2 < ... < a101}.

Draw an undirected graph and put a vertex between i and j iff the sets (A+i) and (A+j) are disjoint . The graph will have 10^6 vertices . For an arbitrary vertex x to be joined with y , x-y must not be one of the numbers ai - aj (i<>j) , which are 101*100 numbers . So the vertex x has degree at least 10^6 - 101*100 .

But this means that the graph has at least 10^6(10^6 - 101*100)/2 edges. The problem asks to prove that there is an 100-clique in the graph. But by Turan's theorem , there is a k-clique in a graph with n vertices iff the number of edges is strictly greater than :

M(n,k) = (k-2)/(k-1) * (n^2 - r^2)/2 + r*(r-1)/2

where we have taken r to be the remainder of n when divided by k-1.

In our case n=10^6 , k=100 and r=1. A simple calculation shows that the number of vertices is greater than M(10^6,100)+1 and thus we are done.

P.S. : in an IMO paper , should one prove Turan's theorem or not ? I guess so ..
Z K
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Anonymous
334 posts
#5 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Fedor Petrov wrote:
Assume that we have already found t_1, t_2, ..., t_k (k<=99) and are searching for t_{k+1}. We can not take t_{k+1} only of form
t_i+a_j-a_l, where 1<=i<=k and a_j and a_l are elements of A.
So, we have k*101*100 forbidden values of t_{k+1}, which correspond to a_j<>a_l, and k forbidden values which correspond to a_j=a_l. So, at most 99*101*100+99=1000000-1 forbidden values, and at least 1 admissible.
so it is right to assume that for every K there exists a number such that \lim \lambda
Z K
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
me@home
2349 posts
#6 • 4 Y
Y by Adventure10, Mango247, and 2 other users
Fedor Petrov wrote:
Assume that we have already found $\{t_k\}_{1=k}^{99}$ and are searching for $t_{k+1}$. We can not take $t_{k+1}$ only of form
$t_i+a_j-a_l \ | \ 1\leq i\leq k \ \ a_j, a_l \in A$.
So, we have $k*101*100$ forbidden values of $t_{k+1}$, which correspond to $a_j<>a_l$, and $k$ forbidden values which correspond to $a_j=a_l$. So, at most $99*101*100+99=1000000-1$ forbidden values, and at least $1$ admissible.
Alison wrote:
Notice that if instead of taking an arbitrary $t_{k+1}$, you always take the smallest $t_{k+1}$ that is not forbidden, you will only have to make sure that $t_{k+1}$ is distinct from all $t_i+a_j-a_l \ | \ a_j>a_l$. This is true because the values with $a_j<a_l$ are all $\leq t_k$, and so have been forbidden at some earlier step.

This gives you $\lceil 1000000/5051 \rceil = 198 t_i's$.

However, Fedor's proof also shows that you can choose $100 t_i's$ such that the sets are parwise disjoint even when taken $mod 10^6$.
Guest wrote:
Let $A = {a_1 < a_2 < ... < a_{101}}$.

Draw an undirected graph and put a vertex between $i, j$ iff the sets $(A+i)$ and $(A+j)$ are disjoint . The graph will have $10^6$ vertices . For an arbitrary vertex $x$ to be joined with $y$ , $x-y$ must not be one of the numbers $ai - aj (i<>j)$ , which are $101*100$ numbers . So the vertex $x$ has degree at least $10^6 - 101*100$ .

But this means that the graph has at least $10^6(10^6 - 101*100)/2$ edges. The problem asks to prove that there is an $100-clique$ in the graph. But by Turan's theorem , there is a $k-clique$ in a graph with $n$ vertices iff the number of edges is strictly greater than :

$M(n,k) = (k-2)/(k-1) * (n^2 - r^2)/2 + r*(r-1)/2$

where we have taken $r$ to be the remainder of $n$ when divided by $k-1$.

In our case $n=10^6 , k=100 , r=1$. A simple calculation shows that the number of vertices is greater than $M(10^6,100)+1$ and thus we are done.

P.S. : in an IMO paper , should one prove Turan's theorem or not ? I guess so ..
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
msecco
154 posts
#7 • 4 Y
Y by Adventure10, Mango247, and 2 other users
This problem has been proposed by Carlos Gustavo Tamm de Araujo Moreira, from Brazil.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Bugi
1857 posts
#8 • 3 Y
Y by Adventure10 and 2 other users
http://www.artofproblemsolving.com/Wiki/index.php/IMO_Problems_and_Solutions

It already says so in the Wiki. If you know an author which isn't listed there, please contribute!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Dragonboy
38 posts
#9 • 3 Y
Y by Adventure10, Mango247, and 1 other user
May be my solution is wrong but it seems to me that we just need $|S|\geq 99\binom{101}{2}+100$. Please help me if you find any bug in my solution
SOLUTION
Click to reveal hidden text
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mahi
52 posts
#10 • 4 Y
Y by Adventure10, Mango247, and 2 other users
Dragonboy wrote:
Make a set $S_{i+1}\subset S_i$ such that $S_{i+1}$ doesn't contain $t_{i+1}$ and any element $K$ satisfying $K-t_{i+1}=|x-y|$ for any distinct $x,y\in A$

In here, you are just striking out the elements such that $K-t_{i+1}=|x-y|$, but the case remains where $K-t_{i+1}=-|x-y|$, which can also satisfy $y+t_i=x+t_j$ instead of $y+t_j=x+t_i$. So the limit is (about) doubled and it reaches near $10^6$, which can be achieved with a little more strict bounding.

Although, nice approach with algorithmic way :)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Dragonboy
38 posts
#11 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Mahi wrote:
Dragonboy wrote:
Make a set $S_{i+1}\subset S_i$ such that $S_{i+1}$ doesn't contain $t_{i+1}$ and any element $K$ satisfying $K-t_{i+1}=|x-y|$ for any distinct $x,y\in A$

In here, you are just striking out the elements such that $K-t_{i+1}=|x-y|$, but the case remains where $K-t_{i+1}=-|x-y|$, which can also satisfy $y+t_i=x+t_j$ instead of $y+t_j=x+t_i$. So the limit is (about) doubled and it reaches near $10^6$, which can be achieved with a little more strict bounding.

Although, nice approach with algorithmic way :)
I think i have mentioned that $t_{i+1}$ is the smallest element in $S_i$ and any other element $K$ in $S_i$ is greater than $t_{i+1}$.So, there is no $K$ satisfying $K-t_{i+1}=-|x-y|$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mahi
52 posts
#12 • 4 Y
Y by Adventure10, Mango247, and 2 other users
Yes, if you choose $t_i$'s in increasing sequence, then it reduces to $t_i-t_j=y-x$ where $i>j$ which implies $y>x$ and thus the strategy is optimized by two :)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Dragonboy
38 posts
#13 • 4 Y
Y by Adventure10, Mango247, and 2 other users
Mahi wrote:
Yes, if you choose $t_i$'s in increasing sequence, then it reduces to $t_i-t_j=y-x$ where $i>j$ which implies $y>x$ and thus the strategy is optimized by two :)
I'm not understanding what you're saying. Let me make it more clear for you (As much as i can)
After choosing all $t_i$ by the algorithm, for the sake of contradiction , let's assume there exists $t_i>t_j$ and $x>y$ such that $t_i-t_j=x-y$.
But It's not possible since we've banished such $t_i$ when we choose $t_{j+1}$ (According to algorithm).
Is it clear now?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mahi
52 posts
#14 • 4 Y
Y by Adventure10, Mango247, and 2 other users
In my last post, I just shared my opinion about the algorithm. I understood it earlier. It was clear to me after I noticed the part "greatest in the set $S_i$". Thanks for your concern.
By the way, something similar was also told by Alison in a previous post.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Wolstenholme
543 posts
#15 • 3 Y
Y by Adventure10 and 2 other users
Let each element of $ S $ be the vertex of a graph where two vertices $ u, v $ are connected by an edge if and only if the sets $ A + u $ and $ A + v $ are disjoint. Consider an arbitrary vertex $ v $. Since the $ |A + v| = 101 $ the maximum number of vertices $ w $ such that sets $ A + v $ and $ A + w $ are not disjoint is $ 100 * 101 $. Therefore every vertex of the graph has degree at least $ 10^6 - 100*101 - 1. $ Therefore the graph has at least $ \frac{10^6(10^6 - 100*101 - 1)}{2} $ edges. It suffices to show that this graph contains $ K_{100} $ as a subgraph.

Now, by Turan's Theorem, the maximum number of edges a graph with $ 10^6 $ vertices that does not contain $ K_{100} $ may contain is obtained when the graph is a complete $ 99 $-partite graph with $ 98 $ independent sets of size $ 10101 $ and $ 1 $ independent set of size $ 10102 $. It is easy to compute that this graph has $ \binom{98}{2}10101^2 + 98 \cdot 10101 \cdot 10102 = 494949494949 $ edges. But since $ \frac{10^6(10^6 - 100*101 - 1)}{2} = 494949500000 > 494949494949 $ we have the desired result.
Z K Y
G
H
=
a