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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/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.

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0 replies
jlacosta
Mar 2, 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
<QBC =<PCB if BM = CN, <PMC = <MAB, <QNB = < NAC
parmenides51   1
N 6 minutes ago by dotscom26
Source: 2005 Estonia IMO Training Test p2
On the side BC of triangle $ABC$, the points $M$ and $N$ are taken such that the point $M$ lies between the points $B$ and $N$, and $| BM | = | CN |$. On segments $AN$ and $AM$, points $P$ and $Q$ are taken so that $\angle PMC = \angle  MAB$ and $\angle QNB = \angle NAC$. Prove that $\angle QBC = \angle PCB$.
1 reply
parmenides51
Sep 24, 2020
dotscom26
6 minutes ago
Bosnia and Herzegovina EGMO TST 2017 Problem 2
gobathegreat   2
N 18 minutes ago by anvarbek0813
Source: Bosnia and Herzegovina EGMO Team Selection Test 2017
It is given triangle $ABC$ and points $P$ and $Q$ on sides $AB$ and $AC$, respectively, such that $PQ\mid\mid BC$. Let $X$ and $Y$ be intersection points of lines $BQ$ and $CP$ with circumcircle $k$ of triangle $APQ$, and $D$ and $E$ intersection points of lines $AX$ and $AY$ with side $BC$. If $2\cdot DE=BC$, prove that circle $k$ contains intersection point of angle bisector of $\angle BAC$ with $BC$
2 replies
gobathegreat
Sep 19, 2018
anvarbek0813
18 minutes ago
Another NT FE
nukelauncher   58
N 29 minutes ago by andrewthenerd
Source: ISL 2019 N4
Find all functions $f:\mathbb Z_{>0}\to \mathbb Z_{>0}$ such that $a+f(b)$ divides $a^2+bf(a)$ for all positive integers $a$ and $b$ with $a+b>2019$.
58 replies
nukelauncher
Sep 22, 2020
andrewthenerd
29 minutes ago
Easiest Functional Equation
NCbutAN   7
N 31 minutes ago by InftyByond
Source: Random book
Find all functions $f: \mathbb R \to \mathbb R$ such that $$f(yf(x)+f(xy))=(x+f(x))f(y)$$Follows for all reals $x,y$.
7 replies
NCbutAN
Mar 2, 2025
InftyByond
31 minutes ago
Tennessee Math Tournament (TMT) Online 2025
TennesseeMathTournament   27
N 5 hours ago by sadas123
Hello everyone! We are excited to announce a new competition, the Tennessee Math Tournament, created by the Tennessee Math Coalition! Anyone can participate in the virtual competition for free.

The testing window is from March 22nd to April 5th, 2025. Virtual competitors may participate in the competition at any time during that window.

The virtual competition consists of three rounds: Individual, Bullet, and Team. The Individual Round is 60 minutes long and consists of 30 questions (AMC 10 level). The Bullet Round is 20 minutes long and consists of 80 questions (Mathcounts Chapter level). The Team Round is 30 minutes long and consists of 16 questions (AIME level). Virtual competitors may compete in teams of four, or choose to not participate in the team round.

To register and see more information, click here!

If you have any questions, please email connect@tnmathcoalition.org or reply to this thread!
27 replies
TennesseeMathTournament
Mar 9, 2025
sadas123
5 hours ago
OTIS Mock AIME 2025 airs Dec 19th
v_Enhance   38
N Today at 1:32 AM by smileapple
Source: https://web.evanchen.cc/mockaime.html
Satisfactory. Keep cooking.
IMAGE

Problems are posted at https://web.evanchen.cc/mockaime.html#current now!

Like last year, we're running the OTIS Mock AIME 2025 again, except this time there will actually be both a I and a II because we had enough problems to pull it off. However, the two versions will feel quite different from each other:

[list]
[*] The OTIS Mock AIME I is going to be tough. It will definitely be harder than the actual AIME, by perhaps 2 to 4 problems. But more tangibly, it will also have significant artistic license. Problems will freely assume IMO-style background throughout the test, and intentionally stretch the boundary of what constitutes an “AIME problem”.
[*] The OTIS Mock AIME II is meant to be more practically useful. It will adhere more closely to the difficulty and style of the real AIME. There will inevitably still be some more IMO-flavored problems, but they’ll appear later in the ordering.
[/list]
Like last time, all 30 problems are set by current and past OTIS students.

Details are written out at https://web.evanchen.cc/mockaime.html, but to highlight important info:
[list]
[*] Free, obviously. Anyone can participate.
[*]Both tests will release sometime Dec 19th. You can do either/both.
[*]If you'd like to submit for scoring, you should do so by January 20th at 23:59 Pacific time (same deadline for both). Please hold off on public spoilers before then.
[*]Solutions, statistics, and maybe some high scores will be published shortly after that.
[/list]
Feel free to post questions, hype comments, etc. in this thread.
38 replies
v_Enhance
Dec 6, 2024
smileapple
Today at 1:32 AM
How to get better at AMC 10
Dream9   6
N Today at 1:31 AM by sadas123
I'm nearly in high school now but only average like 75 on AMC 10 sadly. I want to get better so I'm doing like the first 11 questions of previous AMC 10's almost every day because I also did previous years for AMC 8. Is there any specific way to get better scores and understand more difficult problems past AMC 8? I have almost no trouble with AMC 8 problem given enough time (like 23-24 right with enough time).
6 replies
Dream9
Yesterday at 1:17 AM
sadas123
Today at 1:31 AM
An FE. Who woulda thunk it?
nikenissan   112
N Today at 1:04 AM by Marcus_Zhang
Source: 2021 USAJMO Problem 1
Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that for positive integers $a$ and $b,$ \[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\]
112 replies
nikenissan
Apr 15, 2021
Marcus_Zhang
Today at 1:04 AM
AIME score for college apps
Happyllamaalways   56
N Yesterday at 4:45 PM by Countmath1
What good colleges do I have a chance of getting into with an 11 on AIME? (Any chances for Princeton)

Also idk if this has weight but I had the highest AIME score in my school.
56 replies
Happyllamaalways
Mar 13, 2025
Countmath1
Yesterday at 4:45 PM
MIT Beaverworks Summer Institute
PowerOfPi_09   0
Yesterday at 4:30 PM
Hi! I was wondering if anyone here has completed this program, and if so, which track did you choose? Do rising juniors have a chance, or is it mainly rising seniors that they accept? Also, how long did it take you to complete the prerequisites?
Thanks!
0 replies
PowerOfPi_09
Yesterday at 4:30 PM
0 replies
k HOT TAKE: MIT SHOULD NOT RELEASE THEIR DECISIONS ON PI DAY
alcumusftwgrind   8
N Yesterday at 10:13 AM by maxamc
rant lol

Imagine a poor senior waiting for their MIT decisions just to have their hopes CRUSHED on 3/14 and they can't even celebrate pi day...

and even worse, this year's pi day is special because this year is a very special number...

8 replies
alcumusftwgrind
Yesterday at 2:11 AM
maxamc
Yesterday at 10:13 AM
rows are DERANGED and a SOCOURGE to usajmo .
GrantStar   26
N Yesterday at 6:00 AM by joshualiu315
Source: USAJMO 2024/4
Let $n \geq 3$ be an integer. Rowan and Colin play a game on an $n \times n$ grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is orderly if: [list] [*]no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and [*]no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. [/list] In terms of $n$, how many orderly colorings are there?

Proposed by Alec Sun
26 replies
GrantStar
Mar 21, 2024
joshualiu315
Yesterday at 6:00 AM
Geo equals ABsurdly proBEMatic
ihatemath123   73
N Yesterday at 5:38 AM by joshualiu315
Source: 2024 USAMO Problem 5, JMO Problem 6
Point $D$ is selected inside acute $\triangle ABC$ so that $\angle DAC = \angle ACB$ and $\angle BDC = 90^{\circ} + \angle BAC$. Point $E$ is chosen on ray $BD$ so that $AE = EC$. Let $M$ be the midpoint of $BC$.

Show that line $AB$ is tangent to the circumcircle of triangle $BEM$.

Proposed by Anton Trygub
73 replies
ihatemath123
Mar 21, 2024
joshualiu315
Yesterday at 5:38 AM
average FE
KevinYang2.71   74
N Yesterday at 4:55 AM by joshualiu315
Source: USAJMO 2024/5
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy
\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]for all $x,y\in\mathbb{R}$.

Proposed by Carl Schildkraut
74 replies
KevinYang2.71
Mar 21, 2024
joshualiu315
Yesterday at 4:55 AM
IMO ShortList 2002, combinatorics problem 5
orl   18
N Jun 10, 2023 by awesomeming327.
Source: IMO ShortList 2002, combinatorics problem 5
Let $r\geq2$ be a fixed positive integer, and let $F$ be an infinite family of sets, each of size $r$, no two of which are disjoint. Prove that there exists a set of size $r-1$ that meets each set in $F$.
18 replies
orl
Sep 28, 2004
awesomeming327.
Jun 10, 2023
IMO ShortList 2002, combinatorics problem 5
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO ShortList 2002, combinatorics problem 5
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orl
3647 posts
#1 • 3 Y
Y by Inequality_forever, Adventure10, Mango247
Let $r\geq2$ be a fixed positive integer, and let $F$ be an infinite family of sets, each of size $r$, no two of which are disjoint. Prove that there exists a set of size $r-1$ that meets each set in $F$.
Attachments:
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orl
3647 posts
#2 • 3 Y
Y by Inequality_forever, Adventure10, Mango247
Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions :)
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dzeta
129 posts
#3 • 4 Y
Y by Tawan, Inequality_forever, Adventure10, Mango247
Just an idea. Let $A_{1},A_{2},...,A_{n} \in F$. There is $X \in F$ which is not a subset of $\bigcup_{i=1}^{n}{A_{i}}$ (since $F$ is infinite), so $X'=X-\bigcup_{i=1}^{n}{A_{i}}$ has at most $r-1$ elements and also has noneempty intersection with all $A_{i}$. So, I've proved that for all finite subfamilies $F' \subset F$, there is a set with $r-1$ elements which meets all elements in $F'$.
Can you see a nice way to finish it from this point? I think I found one, but it's quite boring, so I won't post it now.
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yilmazk
1 post
#4 • 6 Y
Y by ValidName, Inequality_forever, Adventure10, Sross314, Mango247, Stuffybear
:) consider the union of all sets in the family, call it X. For any set B with at most r-1 elements, we can find a set in the family, A, outside B (not intersecting with B).

Now, take any set in the family, there is an element ,a1, in that set, that is contained in infinitely many sets in the family. Consider this infinite family, I1. Take a set ,A1, in the family but outside (not containing) a1,now there is an element a2 in A1 that is contained in infinitely many sets in the family I1, lets call it I2. Inductively, outside of the constructed set {a1, a2, ... a(r-1)} there is a set in the family, with an element ar that is in infinitely many sets in the family I(r-1), lets call it Ir, thus we have infinitely many sets(Ir family of sets) each contains a1...ar we reach a contradiction, as sets in the family are all distinct...
I think this does it.. it's a lovely problem indeed..
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Raúl
69 posts
#5 • 3 Y
Y by Inequality_forever, Adventure10, Mango247
What if we put $F_1=\{r_1, r_2\}, F_2=\{r_2, r_3\}$ and $F_a=\{r_3, r_1\}$, $a\geq 3$. It contradicts the problem :o
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goldeneagle
240 posts
#6 • 4 Y
Y by Tawan, Inequality_forever, Adventure10, Mango247
Raúl wrote:
What if we put $F_1=\{r_1, r_2\}, F_2=\{r_2, r_3\}$ and $F_a=\{r_3, r_1\}$, $a\geq 3$. It contradicts the problem :o
the problem said " let be an infinite family of sets" .....it's a set and cannot have two similar element!!

i think i solved it but i'm not sure..!!
let $k \in F$ so every other element of $F$ has a subset of $k$ as it's intersection with $k$ so because the subset's of $K$ are finite so there are infinitely set of $F$ such that their interesction with $K$ is a subset of $K$ like $A_1$ and $K=A_1 \cup X_1$ ....so we can show them like $\{A_1 \cup X_1 , A_1 \cup Y_1,A_1 \cup Y_2,A_1 \cup Y_3,.... \} $and $X_1 \cap Y_i = \emptyset $ ...also $\mid A_1\mid \leq r-1 $ (because $F$ is a set and it's elements are different!)
if we repeat this steps for $Y_1,Y_2,Y_3,...$ and do this $r+1$ times ...i mean that each of $Y_2,Y_3,Y_4,... $ have intersection with a subset of $Y_1$ (maybe empty!) so infinitely of them have intersection with a subset of $Y_1$ call it $A_2$ that $ Y_1= A_2 \cup X_2$ so we show them as : $A_1 \cup A_2 \cup X_2,  A_1 \cup A_2 \cup Z_1 ,A_1 \cup A_2 \cup Z_2 ,......$ and so $X_2 \cap Z_i = \emptyset $ and $\mid A_1 \cup A_2 \mid \leq r-1$ now repeat this for$ Z_1 , Z_2,Z_3,.....$

then we arrive to $A_1 \cup X_1 , A_1 \cup A_2 \cup X_2 , A_1 \cup A_2 \cup A_3 \cup X_3 ,..., A_1 \cup A_2 \cup....A_{r+1} \cup X_{r+1} $ and we konw that $X_i \cap X_j =\emptyset$
$P= A_1 \cup ...A_{r+1} $ thjat because of the last step and difference between the elements of $F$ so $\mid P \mid \leq r-1$ and if $T \in F , T \cap P=\emptyset  \Rightarrow   T \cap X_i $ is nonempty and because
1. $x_i$ are disjiont
2.$\mid P \mid \leq r-1 \Rightarrow \mid X_i \mid \geq 1 $
so $\mid T \mid \geq r+1$ this is contradiction!!
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yugrey
2326 posts
#7 • 8 Y
Y by Tawan, freefrom_i, huricane, ValidName, Inequality_forever, Supercali, IAmTheHazard, Adventure10
Lemma: Say all elements of an infinite family of (finite) sets $F$ intersect an infinite bunch $G$ of distinct $r$ sets (any $f$ in $F$ and any $g$ in $G$ intersect). Then $F$ also intersects an $r-1$ set. This lemma kills the problem because we just take $G=F$ and we are done.

We do this by induction. First we do $r=2$. Note that there is some element that appears infinitely many times in $G$, else we have infinitely many disjoint sets in $G$ and all elements of $F$ must intersect all of them and have infinite size, contradiction.

Say it is true for $r=k$. Now we look at $r=k+1$. All elements of the infinite family of $F$ intersect each of an infinite family $G$ of $k+1$ sets. Now some element must show up infinitely many times in $G$ by the exact same argument as for the $r=2$ case. Call this element $1$. Say $H$ is the infinite subfamily of $G$ with $1$'s, and $I$ is the infinite family of $k$ elements formed by removing the $1$'s from the sets in $G$. Now, every member of $F$ that lacks a $1$ must intersect every element of $I$. Hence, by induction some set of $k-1$ elements meets everything in $F$ without a $1$. Then take the union of $1$ and this set, and we have a set of $k$ elements that meets everything in $F$.

The lemma is proved by induction, and the statement follows.
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theflowerking
92 posts
#8 • 4 Y
Y by Tawan, Inequality_forever, Adventure10, Mango247
Let a number $k \le r$ be "good" if, for any $k$ things, only a finite number of the sets of $F$ contain these $k$ things. Clearly $r$ is good since the sets are distinct. Now, assume $k$ is good, and $k \ge 2$.

Then take any $k-1$ things, and assume an infinite number of sets of $F$ contain these $k-1$ things. Call these sets "good" sets, and let $X$ be the set of the $k-1$ things. Consider one of these good sets, $X \cup \{ t_1, ..., t_{r-k+1} \}$. By the goodness of $k$, only finitely many good sets contain one of the $t_i$. So take one good set without any $t_i$, say $X \cup \{t_1', ..., t_{r-k+1}'\}$. By the goodness of $k$, only finitely many good sets contain one of the $t_i$ or $t_i'$, so take one good set without any of those. Continue until I have $r+1$ good sets that only share elements in $X$. Let them be $G_1 \cap X, G_2 \cup X, ..., G_{r+1} \cup X$.

If every set intersects $X$, then the problem is proven. Otherwise, there exists a set in $F$ that doesn't intersect $X$. Then it must intersect $G_1, ..., G_{r+1}$. But all the $G$'s are coprime, so this set has cardinality $r+1$, contradiction!

Thus, $k-1$ is good. By induction, $1$ is good.

Take any set of $r$ things $a_1,...,a_r$. Finite sets contain $a_i$ for any $i$, and thus finite sets intersect $\{a_1,...,a_r\}$. But all sets of $F$ intersect $\{a_1,...,a_r\}$, contradiction!
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ABCDE
1963 posts
#9 • 3 Y
Y by Inequality_forever, Adventure10, Mango247
For a set $A\in F$ and a subset $B$ of $A$, let $S(A,B)$ denote the set of sets $T\in F$ with $A\cap T=B$.

Let $|B|=1$ is a subset of $A\in F$. If $S(A,B)$ is empty, then we take all elements in $A$ except for the element in $B$ as our set of size $r-1$ and this clearly intersects all sets in $F$. Hence, we may suppose that $S(A,B)$ is nonempty for all such sets $A$ and $B$.

Now, we will prove that $S(A,B)$ is finite for all $B\subseteq A\in F$. We will strong induct on $|B|$. Clearly, the statement is true for $|B|=r$, since $S(A,B)=\{A\}$. Now, suppose that the statement is true for $|B|=k,k+1,\ldots,r$. We will now prove it for all $|B|=k-1$. Consider a set $C\in F$ such that $|A\cap C|=1$ and $B\cap C\cap A=\emptyset$. Such a $C$ exists by the previous paragraph, and let $C'$ be the subset of $C$ consisting of all elements of $C$ except the element it shares with $A$. Now, suppose that infinitely many sets in $F$ intersect $A$ at $B$. Let $T$ be the set of those sets. Now, all sets in $T$ intersect $C'$, so since there are infinitely many of them, some subset $X$ of $C'$ is the set of intersection infinitely many times. Now, take some $D$ such that $D\cap C'=X$, and note that infinitely many sets intersect $D$ at $B\cup X$. But since $|B\cup X|=|B|+|X|\geq k$, we have a contradiction by strong induction.

Since finitely many sets intersect at $A$ at each one of its subsets, there are finitely many sets in $F$, a contradiction. Hence, the only possibility is the one described in the second paragraph, and we have already found the set of size $r-1$, so we're done.
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v_Enhance
6857 posts
#10 • 8 Y
Y by tenplusten, Ankoganit, huricane, Inequality_forever, v4913, DanDumitrescu, Adventure10, Mango247
I'll prove the generalized HMMT formulation:
HMMT 2016 Team Round #9 wrote:
Fix positive integers $r>s$, and let $\mathcal F$ be an infinite family of sets, each of size $r$, no two of which share fewer than $s$ elements. Prove that there exists a set of size $r-1$ that shares at least $s$ elements with each set in $F$.

We consider the following algorithm. Initially, let $X = \varnothing$ and $\mathcal G = \mathcal F$. Then, perform the following:
  • If there exists $x \notin X$ which lies in infinitely elements of $\mathcal G$, then add $x$ to $X$ and remove all sets in $\mathcal G$ not containing $x$.
  • Otherwise, stop.
The end result is an infinite sub-family $\mathcal G$, such that each $S \in \mathcal G$ contains every element of $X$, while $|\mathcal G| = \infty$ still.

We claim that this $X$ works. Evidently, $|X| \le r-1$.

We can take a subset $G_1$, \dots, $G_r$ of sets in $\mathcal G$ such that $G_i \cap G_j = X$ for $i \neq j$, since each element of $G_i \setminus X$ is in finitely many guys. In particular, $|X| \ge s$.

Crude picture:
[asy] 	size(6cm); 	draw( (0,0)--(3,0)--(3,1)--(0,1)--cycle ); 	label("$X$", (3,1), dir(45)); 	dot( (0.5,0.5) ); 	dot( (1.5,0.5) ); 	dot( (2.5,0.5) ); 	dot( (-0.5,1.2), blue ); 	dot( (-1.2,1.2), blue ); 	dot( (-0.5,0.6), red ); 	dot( (-1.2,0.6), red ); 	dot( (-0.5,-0.2), heavygreen ); 	dot( (-1.2,-0.2), heavygreen ); [/asy]
($G_1$ is blue plus $X$, et cetera)

Then any $S \in \mathcal F$ must meet each $G_i$ in at least $s$ places. It thus has to meet $X$ in at least $s$ places, since otherwise it would have to meet each $G_i$ in at least one element outside of $X$, but those are disjoint.
This post has been edited 1 time. Last edited by v_Enhance, Mar 27, 2017, 12:54 AM
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mathadventurer
54 posts
#11 • 3 Y
Y by Inequality_forever, Adventure10, Mango247
Suppose a set $X \in F$ has elements $\{1, 2, \cdots n\}$. For each other set in $F$, its intersection with $X$ is a subset of $X$. Now consider all the subsets of $X$, there is a finite number of them. There are infinitely many sets in $F$, so there exists some set $Y \subset X$ with such that the set $S$ containing all sets in $F$ with precisely $Y$ as the intersection with $X$ is infinite.

First suppose the size of $Y$ is less than $n-1$. If all the subsets in $F$ meet $Y$, then we are done. Otherwise, take a set $Z$ in $F$ that is disjoint with $Y$. For each set in $S$, they must also intersect with $Z$ at another element not in $Y$. Let $Y_1$ be the set containing $Y$ and this element. Since $|S|$ is infinite, there exists an infinite set $S_1 \in S$ such that all the elements in $S_1$ contain $Y_1$.

Now do the same thing again to get an infinite set $S_2$ such that all sets in $S_2$ contain $Y_2$ where $|Y_2| > |Y_1|$. We can continue to do so until we get to some infinite set $S_i$ such that all sets in $S_i$ contain $Y_i$ where $|Y_i| = n-1$ (For the case when $|Y| = n-1$, we can just jump to this step and let $Y_i = Y$).

But now there can't exist another set in $F-S_i$ which doesn't meet $Y_i$ as otherwise an infinite subset of $S_i$ must contain the same $n$ elements and so we're done.
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tastymath75025
3223 posts
#12 • 4 Y
Y by yugrey, Inequality_forever, IAmTheHazard, Adventure10
Here is a solution with StoryScene that only works when $F$ is countable. yugrey says this should be generalizable to all $F$ with transfinite induction, but I'm not knowledgeable enough on the subject to verify this.

Let $F=\{S_1,S_2,\dots\}$. For any $t$ and any $1\le i\le r$ let $X_i(t)$ be the set of all sets of size $i$ which meet each of $S_1,S_2,\dots, S_t$. Note that $X_r(t)$ is always infinite, since every element of $F$ lies in it. Let $f(t)$ be the smallest $i$ so that $X_i(t)\neq \emptyset$. Since $X_i (t+1)\subseteq X_i(t)$ for all $t$, it's clear that $f(t)$ is nondecreasing, but $f(t)$ only takes values in $\{1,2,\dots, r\}$, so it eventually becomes some constant $k$.

Now I claim that if $f(t)=k$, then $|X_k(t)|$ must be finite. Suppose otherwise, so $X_k(t)$ is infinite. Then there exists some $S\in X_k(t)$ so that $S$ isn't in $S_1\cup S_2\cup \dots \cup S_t$, so $S$ contains some element $e$ not in any of the $S_i$ for $1\le i\le t$. Now consider $S' = S\backslash e$, which has size $k-1$ and meets $S_1,S_2,\dots, S_t$. This implies $X_{k-1}(t)\neq \emptyset$, contradiction.

Now since $|X_k(t)|$ is finite but nonzero for all large $t$ and $X_k(t+1)\subseteq X_k(t)$, it follows $X_k(t)$ eventually becomes constant for $t\ge T$. Now take any $S\in X_k(T)$; it follows that $S$ meets all sets in $F$, and since $X_r(t)$ is infinite, we have $k\le r-1$, so $|S|\le r-1$ as desired.
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yugrey
2326 posts
#13 • 3 Y
Y by Inequality_forever, Adventure10, Mango247
The transfinite induction that tastymath references is rather straightforward. In any case, there is something else very interesting about this problem.

Amusingly in my research I am reading a paper by mathematician (and IMO 1971 gold medalist) Peter Frankl (https://www.renyi.hu/~pfrankl/2017-4.pdf) that asks what the right bound on $F$ is (it is known that we need $F$ to be bigger than $r^r/2^r$ and it cannot be bigger than $r^r/(1+o(1))^r$). My solution in this thread is reminiscent of some of the lemmas in this paper (in particular the ones about $A$ and $G$).
This post has been edited 2 times. Last edited by yugrey, Oct 6, 2019, 5:41 AM
Reason: referenced tastymath's post
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dgrozev
2459 posts
#14 • 2 Y
Y by Adventure10, Mango247
orl wrote:
Let $r\geq2$ be a fixed positive integer, and let $F$ be an infinite family of sets, each of size $r$, no two of which are disjoint. Prove that there exists a set of size $r-1$ that meets each set in $F$.

We denote the family of sets by $\mathcal{F}$. Let $X:=\bigcup_{F\in\mathcal{F}}F$ be the underlying set. $X$ is infinite set, since $\mathcal{F}$ is infinite. For some $x_1\in X$ we choose a set $F_1$ of $k-1$ elements such that $\{x_1\}\cup F_1\in \mathcal{F}$. Then we choose $x_2\in X, x_2\notin \{x_1\}\cup F_1$ and a set $F_2$ with $\{x_2\}\cup F_2\in \mathcal{F}$.

Proceeding this way, we obtain (disjoint) sets $ U_i:=\{x_i\}, i\geq 1$, and sets of $r-1$ elements $F_i, i\geq 1$ such that $U_i\cup F_i \in \mathcal{F}, i=1,2,\dots$ and $\left(\bigcup_{i} U_i\right) \bigcap \left(\bigcup_i F_i\right)=\emptyset$.
Consider $X':=\bigcup_i F_i$. If $X'$ is infinite we run the same process again and obtain mutually disjoint sets (we use the same notations again) $U_i,i\geq 1$, each one consisting of two elements and sets $F_i,i\ge 1$, each one of $r-2$ elements, such that $U_i\cup F_i\in\mathcal{F}$ and $\left(\bigcup_{i} U_i\right) \bigcap \left(\bigcup_i F_i\right)=\emptyset$.

In case $X':=\bigcup_i F_i$ is infinite, we proceed with this process further until we come to a moment we have disjoint sets $U_i,i\in\mathbb{N}$ each with $k,k<n$ elements and sets $F_i$, each with $r-k$ elements, such that $U_i\cup F_i\in \mathcal{F}$ and $\left(\bigcup_{i} U_i\right) \bigcap \left(\bigcup_i F_i\right)=\emptyset$ and $ X':=\bigcup_{i}F_i$ is finite. (Such moment certainly will come, since otherwise we finally get $F_i=\emptyset$ and disjoint set of $U_i$ of $r$ elements, $U_i\in\mathcal{F}$, which contradicts that any two sets in $\mathcal{F}$ meet.)

Now, $X'$ being finite means there exist finitely many subsets of $X'$ with $r-k$ elements, implying infinitely many sets among $F_i,i\geq 1$ coincide with some $F_0\subset X', |F_0|=r-k$.
We claim the elements in $F_0$ satisfy the requirement. Indeed, suppose there exists $F\in\mathcal{F}, F\cap F_0=\emptyset$. But $U_i\cup F_0\in \mathcal{F}$ for infinitely many $i$, and since $U_i$ are disjoint, some $U_i$ does not meet $F$, thus $U_i\cup F_0$ does not meet $F$, contradiction.

Remark. The fact exploited is that the family $\mathcal{F}$ is too large. There is no need $\mathcal{F}$ to be infinite. One can go through and estimate how large should be $|\mathcal{F}|$, if finite, in order to apply the same arguments.
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dgrozev
2459 posts
#15 • 3 Y
Y by tastymath75025, Adventure10, Mango247
tastymath75025 wrote:
Here is a solution with StoryScene that only works when $F$ is countable. yugrey says this should be generalizable to all $F$ with transfinite induction, but I'm not knowledgeable enough on the subject to verify this.

Let $F=\{S_1,S_2,\dots\}$. For any $t$ and any $1\le i\le r$ let $X_i(t)$ be the set of all sets of size $i$ which meet each of $S_1,S_2,\dots, S_t$. Note that $X_r(t)$ is always infinite, since every element of $F$ lies in it. Let $f(t)$ be the smallest $i$ so that $X_i(t)\neq \emptyset$. Since $X_i (t+1)\subseteq X_i(t)$ for all $t$, it's clear that $f(t)$ is nondecreasing, but $f(t)$ only takes values in $\{1,2,\dots, r\}$, so it eventually becomes some constant $k$.

Now I claim that if $f(t)=k$, then $|X_k(t)|$ must be finite. Suppose otherwise, so $X_k(t)$ is infinite. Then there exists some $S\in X_k(t)$ so that $S$ isn't in $S_1\cup S_2\cup \dots \cup S_t$, so $S$ contains some element $e$ not in any of the $S_i$ for $1\le i\le t$. Now consider $S' = S\backslash e$, which has size $k-1$ and meets $S_1,S_2,\dots, S_t$. This implies $X_{k-1}(t)\neq \emptyset$, contradiction.

Now since $|X_k(t)|$ is finite but nonzero for all large $t$ and $X_k(t+1)\subseteq X_k(t)$, it follows $X_k(t)$ eventually becomes constant for $t\ge T$. Now take any $S\in X_k(T)$; it follows that $S$ meets all sets in $F$, and since $X_r(t)$ is infinite, we have $k\le r-1$, so $|S|\le r-1$ as desired.
Nice! No need for transfinite induction in case the family is uncountable. But, if you use it, it usually is made by indexing the sets in $F$ with ordinals, like $\{S_{\alpha}:\alpha <\beta\}$, where $\beta$ is some ordinal. In case $\beta$ is countable, one can arrange $S_{\alpha},\alpha <\beta$ in a sequence and apply the same argument. But, if $\beta$ is uncountable, for example the first uncountable ordinal $\omega_1$ (when you take the limit, and $\omega_1+1$,....etc.), one should use some other argument like the one used below. Not that it is of any difficulty, but it makes the induction pointless.

Suppose $\mathcal{F}$ is uncountable family satisfying the condition. As you proved, for any countable subfamily $F\subset \mathcal{F}, F=\{S_1,S_2,\dots\}$, there exists a finite collection $X_k(T),k\leq r-1$ of sets with $k$ elements (eventually depending on $F$ ), as described in post N12. We simply denote $X_k(T)$ as $X_k$ or rather $X_k(F)$ (to mark it may depend on $F$).
Let $\ell$ me the maximal possible $k$ in $X_k(F)$ when $F$ varies. Note that for any two countable subfamilies $F_1,F_2\subset \mathcal{F}$ with $X_{\ell}(F_1)$ and $X_{\ell}(F_2)$, $X_{\ell}(F_1)\cap X_{\ell}(F_2)\neq \emptyset$. Indeed, $F_1\cup F_2$ is also countable, and arranging those sets in a row it easily follows $X_{\ell}(F_1\cup F_2)=X_{\ell}(F_1)\cap X_{\ell}(F_2)$.

So, we take minimal $X_{\ell}(F)$ when $F$ varies (but providing in $X_k(F)$ we have $k=\ell$). Denote it with $X(F_0)$. I claim, all sets in $X(F_0)$ (which consist of exactly $\ell<r$ elements) do the job. Indeed let $S_0\in X(F_0)$. Take any set $S\in\mathcal{F}$. We may consider $F:=\{S\}\cup F_0$. Since $X_\ell(F)\subset X_{\ell}(F_0)$ and $X_{\ell}(F_0)$ is the minimal one, we get $X_\ell(F)=X_\ell(F_0)$ or $S\cap S_0\neq \emptyset$. $\blacksquare$

Remark. The idea in post #12 may be applied in slightly different way, working only with finite sets. For any collection $F$ of finite sets $S_1,S_2,\dots,S_n$, let $X(F)$ denotes the collection of all sets of size at most $r-1$ that meet any set in $F$, and there aren't two sets in $X(F)$ one is the subset of the other. In the same way, as you proved it in #12, there exist finite collection $F$ with $X(F)$ of finite size. So, you take a minimal $X(F)$, under inclusion, and further apply the same method as above.
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yugrey
2326 posts
#16
Y by
See https://arxiv.org/abs/2010.02541 for new bounds on this problem.
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mathaddiction
308 posts
#17
Y by
Suppose on the contrary that such set does not exist.

Claim. For each $1\leq n\leq r-1$ there exists some $\{a_1,...,a_n\}$ which is contained in an infinite number of sets in $F$.
Proof. Induct on $n$. For $n=1$, suppose $S_0\in F$. Then every $S$ in $F$ intersect $S_0$, hence by pigeonhole principle, there exists infinitely many sets intersecting the same element in $S_0$, so we are done.

Now for the general case $n$ we can find some $\{a_1,...,a_{n-1}\}$ contained in an infinite number of sets $S_1,S_2,...$ in $F$. If this set intersects every other set in $F$ then we are done. Otherwise consider a set $S_0$ not intersecting it. Then each of the $S_1,S_2,...$ intersect $S_0$. By pigeonhole principle an infinite number of them intersect at the same element $a_n$, hence taking $\{a_1,...,a_n\}$ we are done. $\blacksquare$

Now we can take $T=\{a_1,...,a_{r-1}\}$ satisfying the condition in the Claim, being contained in sets $S_1=T\cap \{b_1\},S_2=T\cap\{b_2\} ,...$, where the $b's$ are all distinct. Suppose on the contrary that some set $S_0$ in $F$ do not intersect $\{a_1,...,a_{r-1}\}$. Then it must contain all $b_1,b_2,...$, contradiction.
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Rg230403
222 posts
#18 • 2 Y
Y by L567, starchan
FTSOC, assume problem statement is false.

Let $S$ be the set of largest size which is a subset of infinitely many elements of $F$. If there are many such $S$, then pick any of them. Now, we claim $|S|=r-1$.
Proof: Assume not, clearly $|S|<r$. Now, we can find $f\in F$ such that $f\cap S=\Phi$(else, problem statement follows). Thus, there must be some element $t\in f$, such that $t$ is in infinitely many of the sets that contain $S$. Thus, $S\cup \{t\}$ would be a larger set satisfying problem conditions.

Thus, we have found an $r-1$ set which is a subset of infinitely many sets in $F$. But, observe that we should still be able to apply the above algorithm if problem statement is false. But, we cannot have $|S|=r$. Thus, we are done.
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awesomeming327.
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Clearly, there must exist at least one element $x$ that appears in infinitely many sets $F$. Otherwise, consider any $S\in F$. For each $x\in S$, finitely many sets $T\in F$ have the property that $x\in T$. However, this means that only finitely many sets $T\in F$ can meet $S$, so infinitely many sets are disjoint from it, contradiction.

Now, let $U$ be the maximal set, selected from a greedy algorithm, such that $U$ is a subset of infinitely many $S\in F$. Clearly, $|U|\le r-1$. Let $F_0$ be the maximal subfamily of $F$ such that $U\subset S$ for all $S\in F_0$. Since $U$ is maximal, for any $V_1,V_2,\dots, V_n\in F_0$, there is infinitely many $W\in F_0$ such that $V_i$ and $W$ only intersect at $U$ for all $i$. Otherwise, there is infinitely many $W\in F_0$ such that $V_i\setminus U$ and $W\setminus U$ aren't disjoint for some $I$. Since the union of $V_i$ is finite, though, there must exist a particular element that appears in infinitely many $W\in F_0$, contradiction.

Thus, let $V_1,V_2,\dots,V_{r+1}$ all be mutually disjoint except at $U$. For all $S\in F$, $S$ and $V_i$ are not disjoint for all $i$. If $S$ is disjoint from $U$, then it must reach one element from each $V_i\setminus U$, but that's $r+1$ distinct elements, absurd.
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