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

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Pisano period
JARP091   0
a minute ago
Source: At the time of writing this problem i do not know the source if any
Let $F_n$ denote the $n$th Fibonacci number, defined by the recurrence:
\[
F_0 = 0,\quad F_1 = 1,\quad \text{and} \quad F_{n+2} = F_{n+1} + F_n \quad \text{for } n \geq 0.
\]For a fixed modulus $m \in \mathbb{N}$, define the set
\[
S_m = \left\{ (i, j) \in \mathbb{N}^2 : F_i \cdot F_j \equiv 1 \pmod{m} \right\}.
\]Prove that $|S_m| = \infty$ if and only if $5 \nmid m$.

Note: Its a very beautiful problem
0 replies
JARP091
a minute ago
0 replies
real functional equation
DottedCaculator   20
N 3 minutes ago by megahertz13
Source: OMMC POTM February 2022
Find all functions $f:\mathbb R \to \mathbb R$ (from the set of real numbers to itself) where$$f(x-y)+xf(x-1)+f(y)=x^2$$for all reals $x,y.$

Proposed by cj13609517288
20 replies
+1 w
DottedCaculator
Nov 2, 2023
megahertz13
3 minutes ago
Simson lines on OH circle
DVDTSB   3
N 16 minutes ago by Funcshun840
Source: Romania TST 2025 Day 2 P4
Let \( ABC \) and \( DEF \) be two triangles inscribed in the same circle, centered at \( O \), and sharing the same orthocenter \( H \ne O \). The Simson lines of the points \( D, E, F \) with respect to triangle \( ABC \) form a non-degenerate triangle \( \Delta \).
Prove that the orthocenter of \( \Delta \) lies on the circle with diameter \( OH \).

Note. Assume that the points \( A, F, B, D, C, E \) lie in this order on the circle and form a convex, non-degenerate hexagon.

Proposed by Andrei Chiriță
3 replies
DVDTSB
May 13, 2025
Funcshun840
16 minutes ago
Twisted easy Number Theory
reni_wee   1
N 26 minutes ago by reni_wee
Source: ONTCP Example 2.1.1.
Let $m$ and $n$ be positive integers posessing the following property: the equation
$$\gcd(11k-1, m) = \gcd(11k-1 , n)$$holds for all positive integers $k$. Prove that $m = 11^rn$ for some integer $r$.
1 reply
reni_wee
38 minutes ago
reni_wee
26 minutes ago
graph thory
o.k.oo   0
an hour ago
There are 10 people at a party. None of the 3 friends of each person are friends with each other. What is the maximum number of friends at this party?
0 replies
o.k.oo
an hour ago
0 replies
IMO Shortlist 2012, Geometry 8
lyukhson   33
N an hour ago by awesomeming327.
Source: IMO Shortlist 2012, Geometry 8
Let $ABC$ be a triangle with circumcircle $\omega$ and $\ell$ a line without common points with $\omega$. Denote by $P$ the foot of the perpendicular from the center of $\omega$ to $\ell$. The side-lines $BC,CA,AB$ intersect $\ell$ at the points $X,Y,Z$ different from $P$. Prove that the circumcircles of the triangles $AXP$, $BYP$ and $CZP$ have a common point different from $P$ or are mutually tangent at $P$.

Proposed by Cosmin Pohoata, Romania
33 replies
lyukhson
Jul 29, 2013
awesomeming327.
an hour ago
Consecutive squares are floors
ICE_CNME_4   11
N 2 hours ago by ICE_CNME_4

Determine how many positive integers \( n \) have the property that both
\[
\left\lfloor \sqrt{2n - 1} \right\rfloor \quad \text{and} \quad \left\lfloor \sqrt{3n + 2} \right\rfloor
\]are consecutive perfect squares.
11 replies
ICE_CNME_4
Yesterday at 1:50 PM
ICE_CNME_4
2 hours ago
Finding all possible $n$ on a strange division condition!!
MathLuis   10
N 2 hours ago by justaguy_69
Source: Bolivian Cono Sur Pre-TST 2021 P1
Find the sum of all positive integers $n$ such that
$$\frac{n+11}{\sqrt{n-1}}$$is an integer.
10 replies
1 viewing
MathLuis
Nov 12, 2021
justaguy_69
2 hours ago
IMO 2012 P5
mathmdmb   123
N 2 hours ago by SimplisticFormulas
Source: IMO 2012 P5
Let $ABC$ be a triangle with $\angle BCA=90^{\circ}$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK=BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$. Let $M$ be the point of intersection of $AL$ and $BK$.

Show that $MK=ML$.

Proposed by Josef Tkadlec, Czech Republic
123 replies
mathmdmb
Jul 11, 2012
SimplisticFormulas
2 hours ago
Fixed line
TheUltimate123   14
N 2 hours ago by amirhsz
Source: ELMO Shortlist 2023 G4
Let \(D\) be a point on segment \(PQ\). Let \(\omega\) be a fixed circle passing through \(D\), and let \(A\) be a variable point on \(\omega\). Let \(X\) be the intersection of the tangent to the circumcircle of \(\triangle ADP\) at \(P\) and the tangent to the circumcircle of \(\triangle ADQ\) at \(Q\). Show that as \(A\) varies, \(X\) lies on a fixed line.

Proposed by Elliott Liu and Anthony Wang
14 replies
TheUltimate123
Jun 29, 2023
amirhsz
2 hours ago
Computing functions
BBNoDollar   7
N 2 hours ago by ICE_CNME_4
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )
7 replies
BBNoDollar
May 18, 2025
ICE_CNME_4
2 hours ago
RMO 2024 Q2
SomeonecoolLovesMaths   14
N 2 hours ago by Adywastaken
Source: RMO 2024 Q2
For a positive integer $n$, let $R(n)$ be the sum of the remainders when $n$ is divided by $1,2, \cdots , n$. For example, $R(4) = 0 + 0 + 1 + 0 = 1,$ $R(7) = 0 + 1 + 1 + 3 + 2 + 1 + 0 = 8$. Find all positive integers such that $R(n) = n-1$.
14 replies
SomeonecoolLovesMaths
Nov 3, 2024
Adywastaken
2 hours ago
Decimal functions in binary
Pranav1056   3
N 3 hours ago by ihategeo_1969
Source: India TST 2023 Day 3 P1
Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that $f(x) + y$ and $f(y) + x$ have the same number of $1$'s in their binary representations, for any $x,y \in \mathbb{N}$.
3 replies
Pranav1056
Jul 9, 2023
ihategeo_1969
3 hours ago
Beautiful numbers in base b
v_Enhance   21
N 3 hours ago by Martin2001
Source: USEMO 2023, problem 1
A positive integer $n$ is called beautiful if, for every integer $4 \le b \le 10000$, the base-$b$ representation of $n$ contains the consecutive digits $2$, $0$, $2$, $3$ (in this order, from left to right). Determine whether the set of all beautiful integers is finite.

Oleg Kryzhanovsky
21 replies
v_Enhance
Oct 21, 2023
Martin2001
3 hours ago
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
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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
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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
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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
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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 $
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grupyorum
1428 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.
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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.
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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. $\blacksquare$
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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.
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Grizzy
920 posts
#10 • 1 Y
Y by Mango247
Solution
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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.
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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
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IAmTheHazard
5003 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.
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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
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primesarespecial
364 posts
#15
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This is just Corradi's Lemma...
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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.
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HamstPan38825
8868 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
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IAmTheHazard
5003 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.
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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 $\blacksquare$
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ike.chen
1162 posts
#21
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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
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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
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RedFireTruck
4243 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.
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megarnie
5610 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
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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.
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pi271828
3371 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.
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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. $\square$
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OronSH
1748 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.
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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 $\square$
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peppapig_
280 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.
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Shreyasharma
683 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
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cj13609517288
1922 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
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dolphinday
1328 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.
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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!
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shendrew7
799 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
\[\sum_{d \in S} \frac{\binom{c(d)}{2}}{\binom{16000}{2}} \ge 1600 \cdot \frac{\binom{800}{2}}{\binom{16000}{2}} = 80 \cdot \frac{799}{15999} > 3.\]
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blueprimes
356 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.
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Mathandski
772 posts
#37
Y by
Iconic problem
Attachments:
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lpieleanu
3002 posts
#38
Y by
Solution
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eg4334
636 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.
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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
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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 $\frac{1600{800\choose 2}}{{16000\choose 2}} > 3$. $\blacksquare$
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de-Kirschbaum
202 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.
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Ilikeminecraft
658 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.
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