Stay ahead of learning milestones! Enroll in a class over the summer!

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
|a^2-b^2-2abc|<2c implies abc EVEN!
tom-nowy   1
N 10 minutes ago by Tkn
Source: Own
Prove that if integers $a, b$ and $c$ satisfy $\left| a^2-b^2-2abc \right| <2c $, then $abc$ is an even number.
1 reply
tom-nowy
May 3, 2025
Tkn
10 minutes ago
Tricky inequality
Orestis_Lignos   28
N 24 minutes ago by MR.1
Source: JBMO 2023 Problem 2
Prove that for all non-negative real numbers $x,y,z$, not all equal to $0$, the following inequality holds

$\displaystyle \dfrac{2x^2-x+y+z}{x+y^2+z^2}+\dfrac{2y^2+x-y+z}{x^2+y+z^2}+\dfrac{2z^2+x+y-z}{x^2+y^2+z}\geq 3.$

Determine all the triples $(x,y,z)$ for which the equality holds.

Milan Mitreski, Serbia
28 replies
Orestis_Lignos
Jun 26, 2023
MR.1
24 minutes ago
JBMO Shortlist 2023 A1
Orestis_Lignos   5
N 32 minutes ago by MR.1
Source: JBMO Shortlist 2023, A1
Prove that for all positive real numbers $a,b,c,d$,

$$\frac{2}{(a+b)(c+d)+(b+c)(a+d)} \leq \frac{1}{(a+c)(b+d)+4ac}+\frac{1}{(a+c)(b+d)+4bd}$$
and determine when equality occurs.
5 replies
Orestis_Lignos
Jun 28, 2024
MR.1
32 minutes ago
square root problem
kjhgyuio   6
N 34 minutes ago by kjhgyuio
........
6 replies
kjhgyuio
May 3, 2025
kjhgyuio
34 minutes ago
9 ARML Location
deduck   38
N Today at 3:45 AM by shawnzeng
UNR -> Nevada
St Anselm -> New Hampshire
PSU -> Pennsylvania
WCU -> North Carolina


Put your USERNAME in the list ONLY IF YOU WANT TO!!!! !!!!!

I'm going to UNR if anyone wants to meetup!!! :D

Current List:
Iowa
UNR
PSU
St Anselm
WCU
38 replies
deduck
May 6, 2025
shawnzeng
Today at 3:45 AM
USAMO Medals
YauYauFilter   54
N Today at 3:26 AM by ohiorizzler1434
YauYauFilter
Apr 24, 2025
ohiorizzler1434
Today at 3:26 AM
Will I fail again
hashbrown2009   16
N Today at 3:24 AM by ohiorizzler1434
so this year I got 34 on JMO 772 774 and got docked 1 point from top honors + mop

I just got info that I pretty much cannot do math for the rest of summer due to family reasons, and the only time I have is winter break

do you guys think it's enough time to practice/grind to qualify mop through USAMO, or should I tell my parents to reschedule the stuff because I really want to make mop

(Note: I'm aiming for like 25+ on USAMO so at least silver but I'm not sure that's realistic given the circumstances i'm in)
16 replies
hashbrown2009
Yesterday at 1:54 PM
ohiorizzler1434
Today at 3:24 AM
Wizard101
El_Ectric   65
N Today at 3:13 AM by HamstPan38825
Source: USAMO 2016, Problem 6
Integers $n$ and $k$ are given, with $n\ge k\ge2$. You play the following game against an evil wizard.

The wizard has $2n$ cards; for each $i=1,\ldots,n$, there are two cards labeled $i$. Initially, the wizard places all cards face down in a row, in unknown order.

You may repeatedly make moves of the following form: you point to any $k$ of the cards. The wizard then turns those cards face up. If any two of the cards match, the game is over and you win. Otherwise, you must look away, while the wizard arbitrarily permutes the $k$ chosen cards and then turns them back face-down. Then, it is your turn again.

We say this game is winnable if there exist some positive integer $m$ and some strategy that is guaranteed to win in at most $m$ moves, no matter how the wizard responds.

For which values of $n$ and $k$ is the game winnable?
65 replies
El_Ectric
Apr 20, 2016
HamstPan38825
Today at 3:13 AM
SUMaC Residential vs. Ross
AwesomeDude10   9
N Today at 2:43 AM by Rong0625
Hi! I got into the SUMaC residential i program, and I also recently got off the Ross waitlist. I was wondering if anyone had any insight on which program is
1) More useful in furthering my mathematical knowledge (which has a better curriculum)
2) Since I'm a junior, which is more useful for college apps? (I know this is a little cringe)
Thanks!
9 replies
AwesomeDude10
May 6, 2025
Rong0625
Today at 2:43 AM
HCSSiM results
SurvivingInEnglish   63
N Yesterday at 11:52 PM by KevinChen_Yay
Anyone already got results for HCSSiM? Are there any point in sending additional work if I applied on March 19?
63 replies
SurvivingInEnglish
Apr 5, 2024
KevinChen_Yay
Yesterday at 11:52 PM
Sad Algebra
tastymath75025   46
N Yesterday at 11:40 PM by Ilikeminecraft
Source: 2019 USAMO 6, by Titu Andreescu and Gabriel Dospinescu
Find all polynomials $P$ with real coefficients such that $$\frac{P(x)}{yz}+\frac{P(y)}{zx}+\frac{P(z)}{xy}=P(x-y)+P(y-z)+P(z-x)$$holds for all nonzero real numbers $x,y,z$ satisfying $2xyz=x+y+z$.

Proposed by Titu Andreescu and Gabriel Dospinescu
46 replies
tastymath75025
Apr 18, 2019
Ilikeminecraft
Yesterday at 11:40 PM
Aime ll 2022 problem 5
Rook567   1
N Yesterday at 9:39 PM by clarkculus
I don’t understand the solution. I got 220 as answer. Why does it insist, for example two primes must add to the third, when you can take 2,19,19 or 2,7,11 which for drawing purposes is equivalent to 1,1,2 and 2,7,9?
1 reply
Rook567
Yesterday at 9:08 PM
clarkculus
Yesterday at 9:39 PM
MathILy 2025 Decisions Thread
mysterynotfound   41
N Yesterday at 9:11 PM by cappucher
Discuss your decisions here!
also share any relevant details about your decisions if you want
41 replies
mysterynotfound
Apr 21, 2025
cappucher
Yesterday at 9:11 PM
The answer of 2022 AIME II #5 is incorrect
minz32   5
N Yesterday at 8:14 PM by Rook567
Source: 2022 AIME II problem 5
The answer posted on the AOPS page for the 2022 AIME II problem 5 has some flaw in it and the final answer is incorrect.

The answer is trying to use the symmetric property for the a, b, c. However the numbers are not exactly cyclically symmetric in the problem.

For example, the solution works well with a = 20, it derives 4 different pairs of b and c for p2 is in (3, 5, 11, 17) as stated. However, for a = 19, only 3 p2 in the set works, since if the p2 = 17, we will have c = 0, which is not a possible vertex label. So only three pairs of b and c work for a= 19, which are (19, 17, 14), (19, 17, 12), (19, 17, 6).

Same for a = 18.

However, for a = 17, one more possibility joined. c can be bigger than a. So the total triple satisfying the problem is back to 4.

There are two more cases that the total number of working pairs of b, c is 3, when a = 5, and a = 4.

So the final answer for this problem is not 072. It should be 068.



5 replies
minz32
Nov 20, 2022
Rook567
Yesterday at 8:14 PM
two subsets with no fewer than four common elements.
micliva   40
N Apr 25, 2025 by Ilikeminecraft
Source: All-Russian Olympiad 1996, Grade 9, First Day, Problem 4
In the Duma there are 1600 delegates, who have formed 16000 committees of 80 persons each. Prove that one can find two committees having no fewer than four common members.

A. Skopenkov
40 replies
micliva
Apr 18, 2013
Ilikeminecraft
Apr 25, 2025
two subsets with no fewer than four common elements.
G H J
Source: All-Russian Olympiad 1996, Grade 9, First Day, Problem 4
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
micliva
172 posts
#1 • 4 Y
Y by Awwal, Littlelame, Adventure10, Mango247
In the Duma there are 1600 delegates, who have formed 16000 committees of 80 persons each. Prove that one can find two committees having no fewer than four common members.

A. Skopenkov
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Panoz93
61 posts
#2 • 7 Y
Y by vsathiam, Unsolved_cube, Aspiring_Mathletes, myh2910, Littlelame, Adventure10, Mango247
Consider a random couple of committees, and let $X$ be the number of delegates that join both of them. Consider $a_{d}$ the number of committees that include delegate ${d}$. Then the expected value of $X$ is
$\mathbb E [X]=\sum{d} \frac{\binom{a_{d}}{2}}{\binom{16000}{2}}\geq \frac{1600\binom{\frac{\sum{d} a_{d}}{1600}}{2}}{\binom{16000}{2}}=\frac{1600\binom{800}{2}}{\binom{16000}{2}}>3 $, and the conclusion follows.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
NewAlbionAcademy
910 posts
#3 • 2 Y
Y by Toinfinity, Adventure10
Even if there are only $78$ committees, I think this inequality holds! If this is the case, why did they ask for $16000$ when they could have asked for $80$?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
vsathiam
201 posts
#4 • 4 Y
Y by Aspiring_Mathletes, Littlelame, Adventure10, Mango247
Panoz93 wrote:
Consider a random couple of committees, and let $X$ be the number of delegates that join both of them. Consider $a_{d}$ the number of committees that include delegate ${d}$. Then the expected value of $X$ is
$\mathbb E [X]=\sum{d} \frac{\binom{a_{d}}{2}}{\binom{16000}{2}}\geq \frac{1600\binom{\frac{\sum{d} a_{d}}{1600}}{2}}{\binom{16000}{2}}=\frac{1600\binom{800}{2}}{\binom{16000}{2}}>3 $, and the conclusion follows.

Probably typo, but it should be

$\mathbb E [X]=\sum_{d} \frac{\binom{a_{d}}{2}}{\binom{16000}{2}}\geq \frac{1600\binom{\frac{\sum a_{d}}{1600}}{2}}{\binom{16000}{2}}= \frac{1600\binom{\frac{16000 \cdot 80}{1600}}{2}}{\binom{16000}{2}} = \frac{1600\binom{800}{2}}{\binom{16000}{2}}>3 $
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
grupyorum
1418 posts
#6 • 3 Y
Y by Aspiring_Mathletes, Illuzion, Mango247
For instructive purposes, I'll post a non-probabilistic proof of this problem, which employs double counting. In some sense, it is the same steps but interpreted differently.

For each $i\in\{1,2,\dots,1600\}$, denote by $n_i$ the number of committees delegate $i$ belongs to. Thus we immediately have $\sum_{i=1}^{1600}n_i=16000\cdot 80$. We now count all triples of form $(i,C_1,C_2)$, where $i$ is a delegate, and $C_1,C_2$ are two distinct committees such that $i\in C_1\cap C_2$. Clearly, there are $\sum_i \binom{n_i}{2}$ such triples. In particular, using the convexity of $x\mapsto \binom{x}{2}$, together with Jensen's we further get $\sum_i \binom{n_i}{2}\geqslant 1600  \binom{\bar{N}}{2}$, where $\bar{N}=\frac{1}{1600}\sum_i n_i = 800$. Now the ratio
$$
R \triangleq \frac{\sum_i n_i}{\binom{16000}{2}},
$$is larger than $3$, using the estimate above. Thus by Pigeonhole principle, there is a pair of committees belonging to at least four triples, each with a distinct first coordinate, finishing the proof.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pad
1671 posts
#7 • 2 Y
Y by Aspiring_Mathletes, mijail
Let $T$ be the number of triples Suppose every two committees have $\le 3$ people in common. Let $T$ be the number of triples $(i,C_1,C_2)$, where person $i$ is part of both $C_1$ and $C_2$. Choose the person $i$. Suppose he is in $x_i$ committees. Then
\[ T = \binom{x_1}{2} + \cdots+\binom{x_{1600}}{2} \ge 1600\binom{\frac{x_1+\cdots+x_{1600}}{1600}}{2} = 1600 \binom{80\cdot 16000/1600}{2} = 1600\binom{800}{2}. \]Choose $C_1,C_2$ in $\tbinom{16000}{2}$ ways. These have at most 3 people in common, so
\[ T \le 3\binom{16000}{2}. \]However, $1600\tbinom{800}{2} > 3\tbinom{16000}{2}$, contradiction.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Vitriol
113 posts
#8
Y by
literally the same asdfjkk :(

Number the delegates $1$ to $1600$. Let $a_k$ be the number of committees person $k$ is in. Notice that $a_1 + a_2 + \dots + a_{1600} = 16000 \cdot 80$.

Now notice that for each $k$, there exist $\binom{a_k}{2}$ pairs of conmittees both containing person $k$, hence each pair of committees has a $\frac{\binom{a_k}{2}}{\binom{16000}{2}}$ chance of containing person $k$.

It follows that, over all pairs of committees,
\begin{align*}
    \mathbb{E} [\text{number of common members}] &= \sum_{k=1}^{1600} \mathbb{P} [\text{person } k \text{ is in both}] \\
    &= \frac1{\binom{16000}{2}} \sum_{k=1}^{1600} \binom{a_k}{2} \\
    &\ge \frac1{\binom{16000}{2}} \cdot 1600 \binom{800}{2} \\
    &= 80 \cdot \frac{799}{15999} = 4 - \varepsilon,
\end{align*}where the inequality is by Jensen's. Therefore there exist one pair of committees having at least four common members. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathlogician
1051 posts
#9
Y by
We use the probabilistic method. Choose $2$ random committees, and let $\mathbb{E}[X]$ be the expected number of delegates on both the committees. We prove that $\mathbb{E}[X] >3$.

First, note that there are $\binom{16000}{2}$ pairs of committees. Let $a_i$ be the number of committees delegate $i$ is on, for $1 \leq i \leq 1600$. Furthermore, the sum of all the $a_i$ must be equal to $16000 \cdot 80.$ Now we have that$$\sum_{i = 1}^{1600} \binom{a_i}{2} \geq 1600 \binom{800}{2}$$and computing $\mathbb{E}[X]$, we find that$$\mathbb{E}[X] = \frac{\sum_{i = 1}^{1600} \binom{a_i}{2}}{\binom{16000}{2}} \geq \frac{1600 \binom{800}{2}}{\binom{16000}{2}} = \frac{80 \cdot 799}{15999} > 3$$as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Grizzy
920 posts
#10 • 1 Y
Y by Mango247
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
brianzjk
1201 posts
#11
Y by
Let $c_i$ be the number of committees the $i$th member is in. Then,
\[\sum_{i=0}^{1600}c_i=16000\cdot80\]But Jensens tells us
\[\frac{\sum \binom{c_i}{2}}{\binom{16000}{2}}\geq \frac{1600\cdot \binom{80}{2}}{\binom{16000}{2}}\]A simple calculation tells us the RHS is greater than 3, so there exist two committees with at least four common members, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
HrishiP
1346 posts
#12
Y by
Let the number of committees the $i^{\text{th}}$ delegate is on be $a_i.$ We will calculate the expected number of committees that any two delegates are both on.

We see that $\sum_i^{1600} a_i = 16000\times 80.$ Let $\mathbb{E}[X]$ be the expected number of delegates on any two randomly chosen committees. Then, we can easily compute $\mathbb{E}[X]$ to get
$$\mathbb{E}[X]=\frac{\sum \tbinom{a_i}{2}}{\tbinom{16000}{2}} \ge \frac{1600\tbinom{800}{2}}{\tbinom{16000}{2}} = \frac{80 \times 799}{15999} > 3,$$Where we use Jensen's Inequality. Because the expected value is greater than $3$, there must certainly exist two committees with at least $4$ coinciding members.

Edit: 1100$^{\text{th}}$ post!
This post has been edited 1 time. Last edited by HrishiP, Oct 29, 2020, 7:02 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
IAmTheHazard
5001 posts
#13 • 1 Y
Y by centslordm
Let the delegates be numbered $1,2,\ldots,1600$, and let delegate $i$ be in $a_i$ committees. Clearly:
$$\sum_{i=1}^{1600} a_i=16000\cdot 80,$$so on average each delegate is a member of $800$ committees.
Now pick two (distinct) committees randomly. Let $X$ be the number of delegates that are members of both. Then:
\begin{align*}
\mathbb{E}[X]&=\sum_{i=1}^{1600} \frac{\binom{a_i}{2}}{\binom{16000}{2}}\\
&\geq \frac{1600 \binom{800}{2}}{\binom{16000}{2}}\\
&=\frac{80(799)}{15999}\\
&>3,
\end{align*}hence there must be some pair of committees with at least $4$ common delegates, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Sprites
478 posts
#14
Y by
micliva wrote:
In the Duma there are 1600 delegates, who have formed 16000 committees of 80 persons each. Prove that one can find two committees having no fewer than four common members.

A. Skopenkov
Order the committees from $1,2,3,........,16,000$.
Define the committee set $=C(j)$ for each member to be the set $[a_1,a_2,......,a_X]$ as the committee's in which an member is in and denote $a_d=|C(d)|$
Then $$\mathbb{E} \left( \mid \max_{1 \le j,k \le 16000 } C(j) \displaystyle \cap  C(k) \mid\right) \geq \frac{1600\binom{\frac{\sum{d} a_{d}}{1600}}{2}}{\binom{16000}{2}}=\frac{16000\binom{800}{2}}{\binom{16000}{2}}>3$$,hence we are done.$\blacksquare$
This post has been edited 2 times. Last edited by Sprites, Oct 8, 2021, 9:58 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
primesarespecial
364 posts
#15
Y by
This is just Corradi's Lemma...
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
SPHS1234
466 posts
#16
Y by
ARO1996 wrote:
In the Duma there are $1600(=y)$ delegates, who have formed $16000(=k)$ committees of $80(=x)$ persons each. Prove that one can find two committees having no fewer than four common members.

A. Skopenkov
Assume FTSOC $|A_i \cap A_j |\leq 3 $.
Let $a_i$ be the number of committees the $i_{th}$ is in.Then $\sum {a_i}=xk$.Count triplets $\{A_i,A_j,D\}$ such that the delegate $D$ is in both $A_i$ and $A_j$.Denote this sum $T$.Then $$3\binom{k}{2}\geq T =\sum {\binom{a_i}{2}} \geq y\binom{\frac{xk}{y}}{2} $$this gives that $3 \times 15999 \geq 80\times 799$, a contradiction.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
HamstPan38825
8861 posts
#17
Y by
Old Incorrect Solution, for storage

Because there are $16000 \cdot 80$ total appearances of any delegate in any committee, we expect to see each delegate in 800 committees. Thus, there are a total of ${800 \choose 2} \cdot 1600$ delegates in the intersection of any two committees. As there are ${16000 \choose 2}$ committees, the average size of intersection is $${800 \choose 2} \cdot \frac{1600}{{16000 \choose 2}} = \frac{63920}{15999} > 3.$$As a result, there must exist two committees whose intersection contains four or more members

This can be rephrased more "rigorously" by rewriting the expected 800 number in terms of sums, but I'm not really bothered to do so. It does avoid the subtleties given in the previous solution by computing delegate-intersection pairs via the delegates themselves, not from the committees. The previous solution fails because the number of committees any delegate is in is not fixed; the vice-versa case however is, which validates our EV computations.
This post has been edited 3 times. Last edited by HamstPan38825, Jun 12, 2022, 4:12 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
IAmTheHazard
5001 posts
#18 • 6 Y
Y by HamstPan38825, insertionsort, channing421, Dansman2838, eibc, NicoN9
After a (retrospectively) surprisingly long discussion on Discord we have finally figured out why any solution that claims the expected number to be exactly 4 is wrong: for a given delegate, the probability that they are in a randomly selected committee is not $\frac{80}{1600}=\frac{1}{20}$, but instead depends on the number of committees that specific delegate is in. The probability is different if they're in every committee or in only one.
Of course, when laid out like this it seems rather obvious, but I think it's difficult to notice that it's wrong. Without knowledge of proofs that (implicitly) show that an EV of $4-\varepsilon$ is achievable, I wouldn't have batted an eye at the erroneous solutions.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
th1nq3r
146 posts
#19 • 1 Y
Y by rstenetbg
I am practicing my probabilistic method, as I observe everyone else here is also.

Arbitrarily choose two committees. Let $X$ be the number of people in the two chosen committees, and define $X_i$ to be $1$ if person $i$ is in both of the chosen committees and $0$ otherwise. Let $n_i$ denote the number of committees delegate $i$ is a member of. Then we have that $X = X_1 + X_2 + \cdots X_{1600}$, and also $\mathbb{E}[X_i] = \mathbb{P}(X_i = 1) = \frac{\binom{n_i}{2}}{\binom{16000}{2}}$. Finally we have that $\sum_{i = 1}^{1600} n_i = 16000 \cdot 80$, and so the average amount of committees delegate $i$ is a member of is $16000 \cdot 80/1600 = 800$. Thus we have that

\begin{align*}
\mathbb{E} &= \sum_{i = 1}^{1600} \binom{n_i}{2}/\binom{16000}{2} \\
&= \frac{\binom{n_1}{2} + \binom{n_2}{2} + \cdots \binom{n_{1600}}{2}}{\binom{16000}{2}} \\
&\geq 1600 \frac{\binom{\frac{n_1 + n_2 + \cdots + n_{1600}}{1600}}{2}}{\binom{16000}{2}} \\
&= 1600 \frac{\binom{800}{2}}{\binom{16000}{2}} \\
&> 3.9952.
\end{align*}However $X$ must be an integer so $\mathbb{E}[X] \geq 4$, and we are done $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ike.chen
1162 posts
#21
Y by
Let the delegates be $D_1, D_2, \ldots, D_{1600}$ and the committees be $C_1, C_2, \ldots, C_{16000}$. In addition, we define $S$ as the number of triplets $(D_i, C_j, C_k)$ such that $D_i$ belongs to both $C_j$ and $C_k$.

If each delegate $D_i$ belongs to $x_i$ committees, then we know \[ \sum_{i=1}^{1600} x_i = 16000 \cdot 80 \]and \[ S = \sum_{i=1}^{1600} \binom{x_i}{2} = \frac{1}{2} \left(\sum_{i=1}^{1600} x_i(x_i - 1) \right) = \frac{1}{2} \left(\left( \sum_{i=1}^{1600} x_i^2 \right) - 16000 \cdot 80 \right) \]\[ \ge \frac{1}{2} \left( \frac{\left( x_1 + x_2 + \ldots + x_{1600} \right)^2}{1600} - 16000 \cdot 80 \right) \]\[ = \frac{1}{2} \left( 160000 \cdot 80^2 - 16000 \cdot 80 \right) = 8000 \cdot 80 \cdot 799 \]where the inequality follows from Cauchy-Schwarz. Now, we have \[ \frac{S}{\binom{16000}{2}} \ge \frac{8000 \cdot 80 \cdot 799}{\binom{16000}{2}} = \frac{80 \cdot 799}{15999} > 3. \]Thus, the average number of shared members between two committees is greater than $3$. The desired result follows easily. $\blacksquare$
This post has been edited 4 times. Last edited by ike.chen, Jun 18, 2024, 6:22 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mogmog8
1080 posts
#22 • 1 Y
Y by centslordm
We count triplets $(A,X,Y)$ such that $A$ is a person in both committees $X$ and $Y$. Suppose person $A_i$ is in $B_i$ committees, noting $\sum B_i=16000\cdot 80$ Counting by committees, this sum is $\sum_{X\neq Y}|X\cap Y|$. Counting by person, this sum is \[\sum_{i=1}^{1600}\binom{B_i}{2}\ge 1600\binom{\sum B_i/1600}{2}=1600\binom{800}{2}\]Hence, the expected value of $|X\cap Y|$ is \[\frac{1600\binom{800}{2}}{\binom{16000}{2}}>3\]so at least one of $|X\cap Y|$ is at least $4$. $\square$
This post has been edited 1 time. Last edited by Mogmog8, Apr 3, 2023, 2:08 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
RedFireTruck
4223 posts
#23
Y by
Let the $n$th delegate be in $a_n$ committees. The total number of common members over all pairs of committees is $$\sum_{i=1}^{1600}\binom{a_i}{2}\ge 1600\binom{\frac{\sum_{i=1}^{1600}a_i}{1600}}{2}=1600\binom{\frac{80\cdot 16000}{1600}}{2}=1600\binom{800}{2}$$and there are $\binom{16000}{2}$ pairs of committees. Since $3\binom{16000}{2}<1600\binom{800}{2}$, some pair of committees must have at least $4$ common members.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
megarnie
5606 posts
#24
Y by
Suppose that the $n$th delegate was in $a_n$ committees. We see that $\sum_{i=1}^{1600} a_i = 16000\cdot 80$. For any two committees the expected value of the number of common members is \[ \frac{ \sum_{i=1}
^{1600} \binom{a_i}{2} }{\binom{16000}{2}}  \ge \frac{1600 \cdot \binom{800}{2}}{\binom{16000}{2}} = \frac{1600 \cdot 800 \cdot 799}{16000 \cdot 15999} > \frac{48000}{15999} > 3 ,\]by Jensen. The largest integer greater than this is $4$, so we must have two committees having at least $4$ common members.
This post has been edited 1 time. Last edited by megarnie, Aug 9, 2023, 6:31 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
eibc
600 posts
#25
Y by
Name the delegates $1, 2, \ldots, 1600$, and let $x_n$ be the number of committies $n$ is in. Then we have
$$\sum_{i = 1}^{1600} x_i = 16000 \cdot 80.$$Now, let $S$ be the number of (unordered) triplets $(C_i, C_j, k)$, where $C_i$, $C_j$ are two committees and $k$ belongs to both of them. Then if every two committees have at most $3$ common members, we have
$$S \le 3\binom{16000}{2}$$But counting $S$ in terms of the $x_i$, from Jensen we get
$$S = \sum_{i = 1}^{1600} \binom{x_i}{2} \ge 1600\binom{800}{2} > 3 \binom{16000}{2} \ge S,$$contradiction.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pi271828
3369 posts
#26
Y by
Let $c_i$ denote the number of committees the $i$th person is in. Note that the average of all $c_i$ is $800$. The expected value of delegates in the intersection of two random committees is $$\sum_{i = 1}^{1600} \frac{{c_i \choose 2}}{{16000 \choose 2}} \ge \frac{1600 \cdot {800 \choose 2}}{{16000 \choose 2}}$$where the last step is true by Jensen's. A quick computation confirms this is greater than 3, so there must exist two committees that have at least four delegates in common.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
joshualiu315
2534 posts
#27
Y by
Let $a_i$ be the number of committees the $i$th delegate is in. We have $\sum a_i=16000 \cdot 80$. For any two committees, the expected number of commmon members is $\frac{\sum \binom{a_i}{2}}{\binom{16000}{2}}$.

Now, let $f(x)=\binom{x}{2}=\frac{x(x-1)}{2}$. Clearly, $f(x)$ is convex so we apply Jensen's to get

\[\sum \binom{a_i}{2}=\sum f(a_i) \ge 1600 f \left(\frac{\sum a_i}{1600} \right)=1600f(800)\]
Thus, we have

\[\frac{\sum \binom{a_i}{2}}{\binom{16000}{2}} \ge \frac{1600 \cdot \binom{800}{2}}{\binom{16000}{2}}>3\]
meaning we must have two committees having at least $4$ common members. $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
OronSH
1738 posts
#28 • 2 Y
Y by megarnie, The_Great_Learner
Let the $i$th person be in $a_i$ committees. If we choose two random distinct committees, the expected number of delegates in both committees is then $\frac{\sum_{i=1}^{1600} a_i^2-a_i}{16000 \cdot 15999}.$ However, we have that $\sum_{i=1}^{1600} a_i=16000 \cdot 80=1280000,$ and by QM-AM we get $\sum_{i=1}^{1600} a_i^2-a_i \ge 1022720000.$ Then we compute $\frac{1022720000}{16000 \cdot 15999}>3,$ so there must be a pair with at least $4$ common members.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ex-center
27 posts
#29
Y by
Let the $i^{\text{th}}$ delegate be in $a_i$ committees. Note that the total number of intersections of committees is given by and by jensens is greater than
$$\sum_{i=1}^{1600}\binom{a_i}{2}\geq 1600\binom{800}{2}$$The average number of intersections between $2$ committees over all pairs is thus greater than
$$\frac{1600\binom{800}{2}}{\binom{16000}{2}} > 3$$Hence there must exists $2$ committees with an intersection of at least $4$ people $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
peppapig_
281 posts
#30
Y by
Let there be $1600$ delegates, label them $d_1$, $d_2$, $\dots$, $d_{1600}$, and let delegate $d_i$ for some integer $1\leq i\leq1600$ be in $a_i$ different groups. Let $F(g_1,g_2)$ of two delegate groups count how many delegates are in both $g_1$ and $g_2$. Over all unordered pairs of different groups $(g_1,g_2)$, let $S$ be the sum of all $F(g_1,g_2)$'s. Note that
\[S=\sum_{i=1}^{1600}\binom{a_i}{2},\]and since there are $16000$ groups with $80$ people each, we also know that $a_1+a_2+\dots+a_{1600}=16000(80)$. Therefore, by Jensen-Karamata's, we have that
\[S=\sum_{i=1}^{1600}\binom{a_i}{2}\geq 1600*\binom{\frac{16000(80)}{1600}}{2}=1600\binom{800}{2}.\]Therefore the total average of the number of intersections between any two groups of delegates is
\[\frac{1600\binom{800}{2}}{\binom{16000}{2}}=\frac{1600(800)(799)}{16000(15999)}=\frac{80(799)}{15999}>\frac{80(600)}{15999}=\frac{48000}{15999}>3.\]By probabilistic method, this gives us that there must be at least one pair of two different groups such that there are at least $4$ different delegates that are members of both groups.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Shreyasharma
682 posts
#31
Y by
Let delegate $i$ be denoted by $a_i$ and committee $i$ be denoted by $C_i$. Now assume that person $a_i$ appears in $x_i$ committees. Then let us count groups of the form $(C_i, C_j, a_k)$ such that $a_k$ appears in $C_i$ and $C_j$ where $i \neq j$. Clearly this quantity is just,
\begin{align*}
\sum_{i=1}^{1600} \binom{c_i}{2} \geq  1600 \cdot \binom{800}{2}
\end{align*}Now note that in total there are $\binom{16000}{2}$ pairs of committees $(C_i, C_j)$. Thus if we find that, \begin{align*}
1600 \cdot \binom{800}{2} \geq 3 \cdot \binom{16000}{2},
\end{align*}then we would be done by pigeonhole as then at least one committee pair, $(C_i, C_j)$ would have $4$ triples of the form $(C_i, C_j, a_k)$ for $4$ distinct values of $k$ . However this is true so we are done.
This post has been edited 1 time. Last edited by Shreyasharma, Nov 20, 2023, 1:41 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cj13609517288
1915 posts
#32
Y by
Double count (delegate,(unordered pair of committees)) pairs. Delegates have $800$ committees on average(by global counting (delegate,committee) pairs), so by Jensen the number of these pairs is at least $1600\cdot\binom{800}{2}$. Therefore, some pair of committees has
\[\left\lceil \frac{1600\cdot\binom{800}{2}}{\binom{16000}{2}} \right\rceil=4\]common members. $\blacksquare$
This post has been edited 1 time. Last edited by cj13609517288, Dec 8, 2023, 5:23 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
dolphinday
1324 posts
#33
Y by
Let $d_i$ be the $i$th delegate. Then $x_i$ is the number of committees that delegate $d_i$ is in.
Then the expected value of shared members of two random committees is
\[\frac{\sum_{i=1}^{1600}\binom{x_i}{2}}{\binom{16000}{2}}\].
Then since $\binom{x}{2}$ is convex, by Jensen's we have
\[\frac{\sum_{i=1}^{1600}\binom{x_i}{2}}{\binom{16000}{2}} \geq \frac{1600 \cdot \binom{800}{2}}{\binom{16000}{2}} = \frac{1600\cdot800\cdot799}{16000\cdot15999}\]. This is approximately less than \[80 \cdot \frac{1}{20} = 4\]Since the expected value is less than $4$, and the number of common committee members is whole, there has to be at least one pair of committees that has $4$ common members.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
asdf334
7585 posts
#34 • 1 Y
Y by GeoKing
Label delegates $1,2,\dots,1600$.

Let $d_1,\dots, d_{1600}$ be the number of committees each delegate is a part of. Consider counting pairs $(c,d)$ of committees and delegates. Counting by committees then delegates we find
\[d_1+d_2+\dots+d_{1600}=16000\cdot 80.\]
Now let's count pairs $((c_1,c_2),d)$ of two committees and one delegate. Counting by delegates, we have
\[\binom{d_1}{2}+\binom{d_2}{2}+\dots+\binom{d_{1600}}{2}\ge 1600\binom{16000\cdot 80\div 1600}{2}=A\]and there are also $\binom{16000}{2}=B$ pairs of committees. Hence it suffices to show that $\frac{A}{B}>3$ to finish by Pigeonhole.

Note that
\[A=1600\cdot 400\cdot 799\]\[B=8000\cdot 15999\]\[\frac{A}{B}=\frac{80\cdot 799}{15999}>3\iff 80\cdot 799>3\cdot 15999\]which is obviously true.
Done!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
shendrew7
795 posts
#35
Y by
Let $S$ be the set of delegates, and $c(d)$ be the number of committees the delegate is in. Note that
\[\sum_{d \in S} c(d) = 16000 \cdot 80 \quad \text{and} \sum_{d \in S} \quad \binom{c(d)}{2}\]
represents the sum of the amount of overlap between any two committees. Thus, for any two randomly selected committees, the expected number of delegates shared is
\[\sum_{d \in S} \frac{\binom{c(d)}{2}}{\binom{16000}{2}} \ge 1600 \cdot \frac{\binom{800}{2}}{\binom{16000}{2}} = 80 \cdot \frac{799}{15999} > 3.\]
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
blueprimes
353 posts
#36
Y by
Suppose each delegate $1 \le i \le 1600$ belongs to $c_i$ commitees. We will choose two commitees at random, the expected number of common members by linearity of expectation is
$$\dfrac{1}{\binom{16000}{2}}\sum_{i = 1}^{1600} \dbinom{c_i}{2} \ge \dfrac{1600}{\binom{16000}{2}} \binom{\sum_{i = 1}^{1600} c_i / 1600}{2}$$Now it is easy to see that $\sum_{i = 1}^{1600} c_i = 16000 \cdot 80 \implies \sum_{i = 1}^{1600} c_i / 1600 = 800$. Therefore it is sufficient to show that
$$\frac{1600 \binom{800}{2}}{\binom{16000}{2}} = \frac{1600 \cdot 800 \cdot 799}{16000 \cdot 15999} = \frac{80 \cdot 799}{15999} > 3$$which implies that two commitees can be chosen with at least $4$ common members, as wanted.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mathandski
757 posts
#37
Y by
Iconic problem
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lpieleanu
3001 posts
#38
Y by
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
eg4334
637 posts
#39
Y by
Let the delegates $d_1, d_2, \dots d_{1600}$ be in $x_1, x_2, \dots, x_{1600}$ committes, respectively. The key is to count the number of common members by each member then divide by $\binom{16000}{2}$. The expected number of common members from an arbitrary pair of committes is then $\frac{\sum \binom{x_i}{2}}{\binom{16000}{2}}$ and also $\sum x_i = 16000 \cdot 80$. By Jensens we have $$\frac{\sum \binom{x_i}{2}}{\binom{16000}{2}} \geq \frac{1600 \cdot \binom{800}{2}}{\binom{16000}{2}} \approx 3.99$$, so one pair of committes must have at least four members. This really is the epitome of an easy global problem.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Maximilian113
575 posts
#40
Y by
Let the number of committees the delegates join be $x_1, x_2, \cdots, x_{1600}.$ Then for some pair of committees, we count the expected number of delegates that joined both of them.

First, observe that $\sum x_i = 80 \cdot 16000.$ Thus by Jensen's the expected value is $$E=\sum \frac{\binom{x_i}{2}}{\binom{16000}{2}} \geq \frac{1600 \cdot \binom{800}{2}}{\binom{16000}{2}} > 3.$$QED
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
gladIasked
648 posts
#41
Y by
Let $a_i$ be the number of committees that the $i$-th delegate is in. The probability that a delegate is in two arbitrarily selected committees is $\frac{a_i(a_i-1)}{16000(15999)}$. Summing over all $i$ gives us the expected number of people in both of the committees: $\frac{{a_1\choose 2}+{a_2\choose 2}+\dots+{a_{1600}\choose 2}}{16000(15999)}$. Because $a_1+a_2+\dots+a_{1600}=80(16000)$, Jensen allows us to determine the lower bound on the expression to be $\frac{1600{800\choose 2}}{{16000\choose 2}} > 3$. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
de-Kirschbaum
199 posts
#42
Y by
Let the committees formed be $C_1, \ldots, C_{160000}$. We will count the number $T$ of unordered pairs of $(C_i, C_j, x)$ where person $x$ is in committees $C_i, C_j$. Suppose for sake of contradiction all pairs of committees have at most $3$ common members. Then $$T \leq 3\binom{16000}{2}$$
Now let $a_i$ be the number of commitees person $x_i$ is in. Then by Jensen's $$T=\sum_{i=1}^{1600} \binom{a_i}{2} \geq 1600\binom{800}{2}$$
So $1600\binom{800}{2}\leq 3\binom{16000}{2}$ which is untrue.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ilikeminecraft
623 posts
#43
Y by
Let $a_i$ be the number of committees the $i^{\text{th}}$ member is in. Thus, we have that $\sum a_i = 16000 \cdot 80.$ We also have that
\begin{align*}
  \sum \frac{\binom{a_i}{2}}{\binom{16000}{2}} & \geq \frac{1600\cdot\binom{\sum a_i / 1600}{2}}{\binom{16000}{2}} \\
  & = \frac{1600\binom{800}{2}}{\binom{16000}{2}} \\
  & = \frac{80\cdot799}{1599} \\
  & > 3
\end{align*}Thus, we are done.
Z K Y
N Quick Reply
G
H
=
a