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

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

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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
Hard Inequality
danilorj   0
8 minutes ago
Let $a, b, c > 0$ with $a + b + c = 1$. Prove that:
\[
\sqrt{a + (b - c)^2} + \sqrt{b + (c - a)^2} + \sqrt{c + (a - b)^2} \geq \sqrt{3},
\]with equality if and only if $a = b = c = \frac{1}{3}$.
0 replies
danilorj
8 minutes ago
0 replies
Orthocenter lies on circumcircle
whatshisbucket   89
N an hour ago by Mathandski
Source: 2017 ELMO #2
Let $ABC$ be a triangle with orthocenter $H,$ and let $M$ be the midpoint of $\overline{BC}.$ Suppose that $P$ and $Q$ are distinct points on the circle with diameter $\overline{AH},$ different from $A,$ such that $M$ lies on line $PQ.$ Prove that the orthocenter of $\triangle APQ$ lies on the circumcircle of $\triangle ABC.$

Proposed by Michael Ren
89 replies
whatshisbucket
Jun 26, 2017
Mathandski
an hour ago
Hard math inequality
noneofyou34   5
N an hour ago by JARP091
If a,b,c are positive real numbers, such that a+b+c=1. Prove that:
(b+c)(a+c)/(a+b)+ (b+a)(a+c)/(c+b)+(b+c)(a+b)/(a+c)>= Sqrt.(6(a(a+c)+b(a+b)+c(b+c)) +3
5 replies
noneofyou34
Sunday at 2:00 PM
JARP091
an hour ago
Interesting inequalities
sqing   0
an hour ago
Source: Own
Let $ a,b>0 $. Prove that
$$\frac{ab-1} {ab(a+b+2)} \leq \frac{1} {8}$$$$\frac{2ab-1} {ab(a+b+1)} \leq 6\sqrt 3-10$$
0 replies
sqing
an hour ago
0 replies
Inspired by SXJX (12)2022 Q1167
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b,c>0 $. Prove that$$\frac{kabc-1} {abc(a+b+c+8(2k-1))}\leq \frac{1}{16 }$$Where $ k>\frac{1}{2}.$
1 reply
sqing
Yesterday at 4:01 AM
sqing
2 hours ago
Algebra manipulation excercise
Marinchoo   3
N 2 hours ago by compoly2010
Source: 2007 Bulgarian Autumn Math Competition, Problem 9.2
Let $a$, $b$, $c$ be real numbers, such that $a+b+c=0$ and $a^4+b^4+c^4=50$. Determine the value of $ab+bc+ca$.
3 replies
Marinchoo
Mar 17, 2022
compoly2010
2 hours ago
Numbers on a circle
navi_09220114   2
N 2 hours ago by ja.
Source: TASIMO 2025 Day 1 Problem 1
For a given positive integer $n$, determine the smallest integer $k$, such that it is possible to place numbers $1,2,3,\dots, 2n$ around a circle so that the sum of every $n$ consecutive numbers takes one of at most $k$ values.
2 replies
navi_09220114
Yesterday at 11:35 AM
ja.
2 hours ago
Inspired by 2007 Bulgarian
sqing   0
2 hours ago
Source: Own
Let $a$, $b$, $c$ be real numbers such that $a+b+c=0$ and $a^2+b^2+c^2+a^4+b^4+c^4=2$. Prove that $$ab+bc+ca=\frac{1-\sqrt 5}{2}$$Let $a$, $b$, $c$ be real numbers such that $a+b+c=0$ and $ab+bc+ca+a^2+b^2+c^2+a^4+b^4+c^4=2$. Prove that $$ab+bc+ca=\frac{1-\sqrt{17}}{4}$$
0 replies
1 viewing
sqing
2 hours ago
0 replies
Gives typical russian combinatorics vibes
Sadigly   4
N 4 hours ago by lbd4203
Source: Azerbaijan Senior MO 2025 P3
You are given a positive integer $n$. $n^2$ amount of people stand on coordinates $(x;y)$ where $x,y\in\{0;1;2;...;n-1\}$. Every person got a water cup and two people are considered to be neighbour if the distance between them is $1$. At the first minute, the person standing on coordinates $(0;0)$ got $1$ litres of water, and the other $n^2-1$ people's water cup is empty. Every minute, two neighbouring people are chosen that does not have the same amount of water in their water cups, and they equalize the amount of water in their water cups.

Prove that, no matter what, the person standing on the coordinates $(x;y)$ will not have more than $\frac1{x+y+1}$ litres of water.
4 replies
Sadigly
May 8, 2025
lbd4203
4 hours ago
Product of Sum
shobber   4
N 4 hours ago by alexanderchew
Source: CGMO 2006
Given that $x_{i}>0$, $i = 1, 2, \cdots, n$, $k \geq 1$. Show that: \[\sum_{i=1}^{n}\frac{1}{1+x_{i}}\cdot \sum_{i=1}^{n}x_{i}\leq \sum_{i=1}^{n}\frac{x_{i}^{k+1}}{1+x_{i}}\cdot \sum_{i=1}^{n}\frac{1}{x_{i}^{k}}\]
4 replies
shobber
Aug 9, 2006
alexanderchew
4 hours ago
Prove that two different boards can be obtained
hectorleo123   1
N 4 hours ago by Joalro178
Source: 2014 Peru Ibero TST P2
Let $n\ge 4$ be an integer. You have two $n\times n$ boards. Each board contains the numbers $1$ to $n^2$ inclusive, one number per square, arbitrarily arranged on each board. A move consists of exchanging two rows or two columns on the first board (no moves can be made on the second board). Show that it is possible to make a sequence of moves such that for all $1 \le i \le n$ and $1 \le j \le n$, the number that is in the $i-th$ row and $j-th$ column of the first board is different from the number that is in the $i-th$ row and $j-th$ column of the second board.
1 reply
hectorleo123
Sep 15, 2023
Joalro178
4 hours ago
Italian WinterCamps test07 Problem4
mattilgale   90
N 4 hours ago by mathwiz_1207
Source: ISL 2006, G3, VAIMO 2007/5
Let $ ABCDE$ be a convex pentagon such that
\[ \angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE.
\]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$.

Proposed by Zuming Feng, USA
90 replies
mattilgale
Jan 29, 2007
mathwiz_1207
4 hours ago
Iran TST P8
TheBarioBario   8
N 5 hours ago by Mysteriouxxx
Source: Iranian TST 2022 problem 8
In triangle $ABC$, with $AB<AC$, $I$ is the incenter, $E$ is the intersection of $A$-excircle and $BC$. Point $F$ lies on the external angle bisector of $BAC$ such that $E$ and $F$ lieas on the same side of the line $AI$ and $\angle AIF=\angle AEB$. Point $Q$ lies on $BC$ such that $\angle AIQ=90$. Circle $\omega_b$ is tangent to $FQ$ and $AB$ at $B$, circle $\omega_c$ is tangent to $FQ$ and $AC$ at $C$ and both circles pass through the inside of triangle $ABC$. if $M$ is the Midpoint od the arc $BC$, which does not contain $A$, prove that $M$ lies on the radical axis of $\omega_b$ and $\omega_c$.

Proposed by Amirmahdi Mohseni
8 replies
TheBarioBario
Apr 2, 2022
Mysteriouxxx
5 hours ago
IMO 2010 Problem 6
mavropnevma   42
N 5 hours ago by awesomeming327.
Let $a_1, a_2, a_3, \ldots$ be a sequence of positive real numbers, and $s$ be a positive integer, such that
\[a_n = \max \{ a_k + a_{n-k} \mid 1 \leq k \leq n-1 \} \ \textrm{ for all } \ n > s.\]
Prove there exist positive integers $\ell \leq s$ and $N$, such that
\[a_n = a_{\ell} + a_{n - \ell} \ \textrm{ for all } \ n \geq N.\]

Proposed by Morteza Saghafiyan, Iran
42 replies
mavropnevma
Jul 8, 2010
awesomeming327.
5 hours ago
IMO problem 1
iandrei   77
N Apr 23, 2025 by YaoAOPS
Source: IMO ShortList 2003, combinatorics problem 1
Let $A$ be a $101$-element subset of the set $S=\{1,2,\ldots,1000000\}$. Prove that there exist numbers $t_1$, $t_2, \ldots, t_{100}$ in $S$ such that the sets \[ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100  \] are pairwise disjoint.
77 replies
iandrei
Jul 14, 2003
YaoAOPS
Apr 23, 2025
IMO problem 1
G H J
Source: IMO ShortList 2003, combinatorics problem 1
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Math4Life7
1703 posts
#69
Y by
We can see that for $a_x, a_y \in A$. We want no pairs such that $a_x - a_y = t_i - t_j$.

We choose our $t_i$ greedily and by induction. Firstly we choose $t_1 =1$.

For our induction step we let $t_{n+1}$ be the first value such that $t_{n+1} - t_j \neq a_x - a_y$ for all $j \in [1, n]$. We can see that there are $\binom{101}{2}$ values of $a_x - a_y$, and there are $n$ values of $j$. Thus the number of values that cannot be obtained by $t_{100}$ is bounded above by $\binom{101}{2} \cdot 100 \leq 10^6$. $\blacksquare$
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ihatemath123
3448 posts
#70
Y by
The "disjoint" condition is satisfied iff $a_w + t_x \neq a_y + t_z$ for all valid $(w,x,y,z)$; in other words, $|a_w - a_y| \neq |t_x - t_z|$ for all valid $w, x, y$ and $z$. There are up to $\binom{101}{2} = 5050$ distinct absolute differences between two elements of $A$.

Consider just arithmetic sequences $t_1, t_2, \dots, t_{100}$ with common difference $d$, where $1 \leq d \leq 10101$. The possible values of $|t_x - t_z|$ are $\{ d, 2d, \dots, 99d \}$; as $d$ varies from $1$ to $10101$, there must be some (in fact, at least $5051$) values of $d$ for which none of $\{ d, 2d, \dots, 99d \}$ are the absolute difference between any two elements of $A$, so we just pick $d$ to be any of those values.
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mannshah1211
652 posts
#71
Y by
Let $k$ be the largest size of such a set $t$. Let $t_1, t_2, \cdots, t_k$ be the numbers, and let $A_1, A_2, \cdots, A_{101}$ be the numbers. Let $n$ be the number of numbers of the form $x - y + t_j$, where $x, y  \in A$. Then, we must have the inequality $10^6 - k \le n,$ since if $10^6 - k > n,$ we have at least one more valid choice to add to $t$. But, in turn, $n \le 101 \cdot 100 \cdot k$, which gives the desired inequality.

EDIT: This is incorrect. I'll edit in the correct solution sometime soon.
EDIT 2: Done.
This post has been edited 3 times. Last edited by mannshah1211, Jan 17, 2024, 9:53 AM
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eibc
600 posts
#72
Y by
Suppose that we pick $n$ numbers $t_1, t_2, \ldots, t_n$ satisfying the problem conditions for maximal $n$. Then for $x \in S$ and $x \neq t_k$ for some $k$, we must have $x - t_{\ell} = a_i - a_j$ or $x = t_{\ell} + a_i - a_j$ for some integers $1 \le \ell \le n$ and $1 \le i \neq j \le 100$. For a fixed $t_\ell$, there are $101 \cdot 100 = 10100$ ways to choose $(a_i, a_j)$, which is enough to imply that $10100n \ge 10^6 - n$, or $n \ge 100$.
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RGB
19 posts
#73
Y by
We can prove this by contradiction. Suppose there is a $101$-element subset $A$ of $S = \{1,2,\ldots,1000000\}$ such that, for any choice of $100$ numbers $t_1, t_2, \ldots, t_{100}$ from $S$, there exist indices $i$ and $j$ ($1 \leq i < j \leq 100$) such that $A_i$ and $A_j$ have at least one common element.

Let $B_1, B_2, \ldots, B_{100}$ be the sets $A_1, A_2, \ldots, A_{100}$ respectively. For each $j$, the set $B_j$ can be expressed as $B_j = A + t_j = \{x + t_j \mid x \in A\}$.

Now, consider the union of all $B_j$'s for $j = 1, 2, \ldots, 100$:

\[ B = B_1 \cup B_2 \cup \ldots \cup B_{100} \]
Since $B_j = A + t_j$, the union $B$ is essentially the set of all elements in $S$ that can be obtained by adding an element from $A$ to any of the $100$ chosen $t_j$'s.

Now, by our assumption, there must exist two indices $i$ and $j$ ($1 \leq i < j \leq 100$) such that $B_i \cap B_j \neq \emptyset$. This implies that there exists an element $x$ in $A$ such that $x + t_i = x + t_j$, which means $t_i = t_j$.

However, this contradicts our assumption that any choice of $100$ numbers $t_1, t_2, \ldots, t_{100}$ must yield distinct sets $B_j$. Therefore, our assumption must be false, and there must be a $101$-element subset $A$ with the desired property.
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dolphinday
1328 posts
#74
Y by
Keep on choosing $t_i$ until we get stuck(we can't pick any more $t$) for some $i = k$. This implies that for all $t \in S$, it has already been picked or that for some $a$, $b$, $t_i \in S$ we have $t = b + t_i - a$. So then there are at most $k \cdot 10100$($101$ choices for $b$ and $100$ for $a$) values for $t$, so $k \geq 100$ as desired.
This post has been edited 1 time. Last edited by dolphinday, Feb 25, 2024, 7:21 PM
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bjump
1030 posts
#75 • 1 Y
Y by megarnie
badly worded solution
Choose $t_i$'s greedily then for all $a$, $b$ in $A$, and $1 \le j \le 100$ we cannot pick $t_j+a-b$, or $t_i-b+a$ in the list of $t$'s in the future. So for every new $t$ added we eliminate at most $101 \cdot 100 = 10100$ more $t$'s. So we can pick at least $\left \lceil \frac{10^6}{10100} \right \rceil = 100$ different $t$'s before we could be forced to stop, finishing the problem.
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RedFireTruck
4226 posts
#76
Y by
we basically just dont want any absolute difference in $A$ to also be an absolute difference in $t$

there are at most $\binom{101}{2}=5050$ distinct absolute differences in $A$

this means that every time we pick a $t_i$, at most $1+5050\cdot2=10101$ elements get eliminated from $S$ as possibilities for other $t_i$

if we pick $t_1=1$, then only at most $5051$ elements get eliminated from $S$ the first time

since $5051+98\cdot 10101=989898+5051<995000$, we can pick $t_{100}$, as desired.
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ezpotd
1279 posts
#77
Y by
Observe the condition is equivalent to the difference between no two $t_i$ being the difference between two $a_i$. Choose $t_i$, for each $t_i$ we automatically eliminate out of candidacy for future $t_i$ all the numbers that are a different of two $A_i$ from $t_i$, which is at most $2 \cdot \binom{101}{2}$, where the $2$ is for positve/negative, and we also eliminate the chosen number itself, for a total for $10101$ eliminations per chosen $t_i$. Now any number not eliminated can be added to $t_i$ and the condition will still be satisfied, so add the first $99$ numbers and see that we have eliminated $999999$ numbers, and there is one number remaining and we see that we can just add this and done.
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joshualiu315
2534 posts
#78
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Suppose we have a working set: $A = \{a_1, a_2, \dots, a_{101} \}$. Note that $A_i$ and $A_j$ are disjoint if and only if

\[t_i-t_j \notin \{a_p-a_q\vert 1\le p \ne q \le 101\}.\]
In order to get such a set, we will proceed by a greedy algorithm. Assume that we have already picked $n$ elements in $T = \{t_1, \dots, t_n\}$ such that $T$ is as large as possible. That means every $i \in \{1,2,\dots,10^6\}$ is either in $T$ or satisfies

\[i = t_i+b-a,\]
for some $t_i \in T$ and $a,b \in A$.

There are at most $|T| \cdot |A| \cdot (|A|-1) = 10100n$ values of $i$. Thus,

\[n+10100n \ge 10^6 \implies n>99,\]
as desired. $\blacksquare$
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Saucepan_man02
1346 posts
#79
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Here we go:
Consider the set $T$ to have $t_1, \cdots, t_n$ currently. Let $A = \{ a_1, a_2, \cdots, a_{101} \}$. Let set $D$ denote the pair-wise differences of elements of set $A$.
Then, if we cant add another element $t \in S$ to $S$, then we must have: $t = t_k + d$ for some $d \in D$. Note that: $|D| \le 101*100$. Thus: $$101*100*n > 10^6 \implies n \ge 100$$and we are done.
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Maximilian113
575 posts
#80
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Let the elements of $A$ be $a_1, a_2, \cdots a_{101},$ and let the elements of $T$ be $t_1, t_2, \cdots, t_n.$ Since each $x+t_j, x\in A$ is distinct, then clearly for all $p, q, r, s$ $$|t_p-t_q| \neq |a_r-a_s|.$$Set $t_1=1.$ Clearly since there are at most $\binom{101}{2}=5050$ pairwise differences among the elements of $A,$ every time we add a number to the set $T$ we reduce the possible number of potential numbers to add to $T$ by at most $2 \cdot 5050 + 1 = 10101.$ (we account for both sides, and the number itself) Therefore, after choosing $99$ numbers, there are at least $$10^6 - 10101\cdot 99=1$$possible number to add to the set $T,$ so thus it is possible to have $100$ elements in $T.$ QED
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megahertz13
3183 posts
#81
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First choose a random $t_1\in S$.

We will prove that if we have already chosen distinct $t_1,t_2,\dots,t_k$ satisfying the condition, then we can choose a $t_{k+1}$ such that the disjoint condition is satisfied. This will finish the problem.

If there are any intersections, there exists a $1\le j\le k$ and $x,y\in A$ satisfying $$t_j+x=t_{k+1}+y\implies t_{k+1}=t_j+x-y.$$However, since $t_j\ne t_{k+1}$, we have $x\ne y$. Then, there are $$k\cdot 101\cdot 100=10100k$$ways to choose the right-hand side of the equation $t_{k+1}=t_j+x-y$. Since $t_{k+1}$ is not equal to any of $t_1, t_2,\dots,t_k$, there are at most $10101k$ numbers that $t_{k+1}$ cannot be. However, $$10101k\le 999999<|S|,$$so there always exists a $t_{k+1}$.
This post has been edited 1 time. Last edited by megahertz13, Nov 26, 2024, 5:05 PM
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ihategeo_1969
238 posts
#82
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Say $t_1$, $\dots$, $t_k$ are max such subsets.

So adding $A_{k+1}$ has an element common with one of $A_1$, $A_2$, $\dots$, $A_k$. Hence for any $T \in S/\{t_1\dots,t_k\}$, we have \[a_\ell+T=a_j+t_i\]Fix RHS and see that $a_\ell$ takes $101-1$ values and so \[101k \ge \frac{10^6-k}{100} \implies k>99 \iff k \ge 100\]And done.
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YaoAOPS
1541 posts
#83
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Let $D = A - A$ be the elements which can be expressed as the difference of two elements in the list. Then $|S| \le 101 \cdot 100 + 1 = 10101$ since $0 \in D$ is over counted $100$ times. We now select each $t_i$ iteratively. Note that if $t_i$ is selected, then any subsequent $t_j$ selected can't be in $t_i + D$ for the sets to be pairwise disjoint, and furthermore this is sufficient. As such, we may select $t_i$ from $S \setminus \bigcup_{j=1}^{i-1} (t_j + D)$ while preserving the condition. For each $i$, we have that
\begin{align*}
	\left|S \setminus \bigcup_{j=1}^{i-1} (t_j + D)\right| &\ge
	1000000 - \sum_{j=1}^{99} \left|(t_j + D)\right|  \\
    &\ge 1000000 - 99 \cdot 10101 = 1.
\end{align*}so we may choose $t_{100}$ as well to finish.
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