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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.

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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!
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0 replies
jlacosta
Mar 2, 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
Stronger than Iran 96, not completely symmetric
MeoMayBe   2
N a few seconds ago by CHESSR1DER
Source: Own
Let a, b, c\geq 0. Prove that
(bc+ca+ab)\left[\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}+\frac{1}{(a+b)^{2}}\right]\geq\frac{9}{4}+\frac{2abc(b-c)^{2}}{(a+b)^{2}(a+c)^{2}(b+c)}.
PS. 2 is also the best constant.
PS (2). Sorry again, I cannot type LaTeX since new members cannot share image.
2 replies
MeoMayBe
May 14, 2023
CHESSR1DER
a few seconds ago
Equilateral triangle geo
MathSaiyan   0
2 minutes ago
Source: PErA 2025/3
Let \( ABC \) be an equilateral triangle with circumcenter \( O \). Let \( X \) and \( Y \) be two points on segments \( AB \) and \( AC \), respectively, such that \( \angle XOY = 60^\circ \). If \( T \) is the reflection of \( O \) with respect to line \( XY \), prove that lines \( BT \) and \( OY \) are parallel.
0 replies
MathSaiyan
2 minutes ago
0 replies
Inspired by youthdoo
sqing   1
N 3 minutes ago by sqing
Source: Own
Let $ a,b,c $ be real numbers such that $      \frac{3}{a^2+6}+\frac{1}{b^2+2}+\frac{3}{c^2+6}=1. $ Prove that$$ab+bc+ca\leq 7$$Let $ a,b,c $ be real numbers such that $   \frac{2}{a^2+4}+\frac{3}{b^2+6}+\frac{2}{c^2+4}=1. $ Prove that$$ab+bc+ca\leq 7$$Let $ a,b,c $ be real numbers such that $    \frac{3}{a^2+6}+\frac{2}{b^2+4}+\frac{3}{c^2+6}=1. $ Prove that$$ab+bc+ca\leq 8$$Let $ a,b,c $ be real numbers such that $  \frac{3}{a^2+6}+\frac{4}{b^2+8}+\frac{3}{c^2+6}=1. $ Prove that$$ab+bc+ca\leq 10$$
1 reply
1 viewing
sqing
18 minutes ago
sqing
3 minutes ago
Clean number theory
MathSaiyan   0
6 minutes ago
Source: PErA 2025/2
Let $m$ be a positive integer. We say that a positive integer $x$ is $m$-good if $a^m$ divides $x$ for some integer $a > 1$. We say a positive integer $x$ is $m$-bad if it is not $m$-good.
(a) Is it true that for every positive integer $n$ there exist $n$ consecutive $m$-bad positive integers?
(b) Is it true that for every positive integer $n$ there exist $n$ consecutive $m$-good positive integers?
0 replies
MathSaiyan
6 minutes ago
0 replies
Many-solutions combogeo
MathSaiyan   1
N 6 minutes ago by Speedysolver1
Source: PErA 2025/1
Let $S$ be a set of at least three points of the plane in general position. Prove that there exists a non-intersecting polygon whose vertices are exactly the points of $S$.
1 reply
MathSaiyan
9 minutes ago
Speedysolver1
6 minutes ago
2 var inquality
sqing   6
N 10 minutes ago by ionbursuc
Source: Own
Let $ a ,  b\geq 0 $ and $ \frac{1}{a^2+1}+\frac{1}{b^2+1}\le   \frac{3}{2}. $ Show that$$ a+b+ab\geq1$$Let $ a ,  b\geq 0 $ and $ \frac{1}{a^2+1}+\frac{1}{b^2+1}\le   \frac{5}{6}. $ Show that$$ a+b+ab\geq2$$
6 replies
sqing
Today at 4:06 AM
ionbursuc
10 minutes ago
Sharygin 2025 CR P2
Gengar_in_Galar   4
N 19 minutes ago by FKcosX
Source: Sharygin 2025
Four points on the plane are not concyclic, and any three of them are not collinear. Prove that there exists a point $Z$ such that the reflection of each of these four points about $Z$ lies on the circle passing through three remaining points.
Proposed by:A Kuznetsov
4 replies
Gengar_in_Galar
Mar 10, 2025
FKcosX
19 minutes ago
100 Selected Problems Handout
Asjmaj   32
N 23 minutes ago by John_Mgr
Happy New Year to all AoPSers!
 :clap2:

Here’s my modest gift to you all. Although I haven’t been very active in the forums, the AoPS community contributed to an immense part of my preparation and left a huge impact on me as a person. Consider this my way of giving back. I also want to take this opportunity to thank Evan Chen—his work has consistently inspired me throughout my olympiad journey, and this handout is no exception.



With 2025 drawing near, my High School Olympiad career will soon be over, so I want to share a compilation of the problems that I liked the most over the years and their respective detailed write-ups. Originally, I intended it just as a personal record, but I decided to give it some “textbook value” by not repeating the topics so that the selection would span many different approaches, adding hints, and including my motivations and thought process.

While IMHO it turned out to be quite instructive, I cannot call it a textbook by any means. I recommend solving it if you are confident enough and want to test your skills on miscellaneous, unordered, challenging, high-quality problems. Hints will allow you to not be stuck for too long, and the fully motivated solutions (often with multiple approaches) should help broaden your perspective. 



This is my first experience of writing anything in this format, and I’m not a writer by any means, so please forgive any mistakes or nonsense that may be written here. If you spot any typos, inconsistencies, or flawed arguments whatsoever (no one is immune :blush: ), feel free to DM me. In fact, I welcome any feedback or suggestions.

I left some authors/sources blank simply because I don’t know them, so if you happen to recognize where and by whom a problem originated, please let me know. And quoting the legend: “The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.” 



I’ll likely keep a separate file to track all the typos, and when there’s enough, I will update the main file. Some problems need polishing (at least aesthetically), and I also have more remarks to add.

This content is only for educational purposes and is not meant for commercial usage.



This is it! Good luck in 45^2, and I hope you enjoy working through these problems as much as I did!

Here's a link to Google Drive because of AoPS file size constraints: Selected Problems
32 replies
1 viewing
Asjmaj
Dec 31, 2024
John_Mgr
23 minutes ago
Where is the equality?
AndreiVila   2
N an hour ago by MihaiT
Source: Romanian District Olympiad 2025 9.3
Determine all positive real numbers $a,b,c,d$ such that $a+b+c+d=80$ and $$a+\frac{b}{1+a}+\frac{c}{1+a+b}+\frac{d}{1+a+b+c}=8.$$
2 replies
AndreiVila
Mar 8, 2025
MihaiT
an hour ago
Truth or lie at a table
SinaQane   7
N an hour ago by Oksutok
Source: 239 2019 S4
There are $n>1000$ people at a round table. Some of them are knights who always tell the truth, and the rest are liars who always tell lies. Each of those sitting said the phrase: “among the $20$ people sitting clockwise from where I sit there are as many knights as among the $20$ people seated counterclockwise from where I sit”. For what $n$ could this happen?
7 replies
SinaQane
Jul 31, 2020
Oksutok
an hour ago
Functional inequality on N
BartSimpsons   21
N an hour ago by Tony_stark0094
Source: European Mathematical Cup 2017 Problem 1
Find all functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that the inequality $$f(x)+yf(f(x))\le x(1+f(y))$$holds for all positive integers $x, y$.

Proposed by Adrian Beker.
21 replies
BartSimpsons
Dec 27, 2017
Tony_stark0094
an hour ago
A colouring game on a rectangular frame
Tintarn   0
an hour ago
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 4
For integers $m,n \ge 3$ we consider a $m \times n$ rectangular frame, consisting of the $2m+2n-4$ boundary squares of a $m \times n$ rectangle.

Renate and Erhard play the following game on this frame, with Renate to start the game. In a move, a player colours a rectangular area consisting of a single or several white squares. If there are any more white squares, they have to form a connected region. The player who moves last wins the game.

Determine all pairs $(m,n)$ for which Renate has a winning strategy.
0 replies
Tintarn
an hour ago
0 replies
Find the value
sqing   5
N an hour ago by sqing
Source: Own
Let $a,b,c$ be distinct real numbers such that $ \frac{a^2}{(a-b)^2}+ \frac{b^2}{(b-c)^2}+ \frac{c^2}{(c-a)^2} =1. $ Find the value of $\frac{a}{a-b}+ \frac{b}{b-c}+ \frac{c}{c-a}.$
Let $a,b,c$ be distinct real numbers such that $\frac{a^2}{(b-c)^2}+ \frac{b^2}{(c-a)^2}+ \frac{c^2}{(a-b)^2}=2. $ Find the value of $\frac{a}{b-c}+ \frac{b}{c-a}+ \frac{c}{a-b}.$
Let $a,b,c$ be distinct real numbers such that $\frac{(a+b)^2}{(a-b)^2}+ \frac{(b+c)^2}{(b-c)^2}+ \frac{(c+a)^2}{(c-a)^2}=2. $ Find the value of $\frac{a+b}{a-b}+\frac{b+c}{b-c}+ \frac{c+a}{c-a}.$
5 replies
sqing
Today at 5:02 AM
sqing
an hour ago
A touching question on perpendicular lines
Tintarn   0
an hour ago
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 3
Let $k$ be a semicircle with diameter $AB$ and midpoint $M$. Let $P$ be a point on $k$ different from $A$ and $B$.

The circle $k_A$ touches $k$ in a point $C$, the segment $MA$ in a point $D$, and additionally the segment $MP$. The circle $k_B$ touches $k$ in a point $E$ and additionally the segments $MB$ and $MP$.

Show that the lines $AE$ and $CD$ are perpendicular.
0 replies
Tintarn
an hour ago
0 replies
Can this sequence be bounded?
darij grinberg   66
N Today at 2:34 AM by shendrew7
Source: German pre-TST 2005, problem 4, ISL 2004, algebra problem 2
Let $a_0$, $a_1$, $a_2$, ... be an infinite sequence of real numbers satisfying the equation $a_n=\left|a_{n+1}-a_{n+2}\right|$ for all $n\geq 0$, where $a_0$ and $a_1$ are two different positive reals.

Can this sequence $a_0$, $a_1$, $a_2$, ... be bounded?

Proposed by Mihai Bălună, Romania
66 replies
darij grinberg
Jan 19, 2005
shendrew7
Today at 2:34 AM
Can this sequence be bounded?
G H J
G H BBookmark kLocked kLocked NReply
Source: German pre-TST 2005, problem 4, ISL 2004, algebra problem 2
The post below has been deleted. Click to close.
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darij grinberg
6555 posts
#1 • 8 Y
Y by Davi-8191, Adventure10, jhu08, megarnie, HWenslawski, ImSh95, Mango247, and 1 other user
Let $a_0$, $a_1$, $a_2$, ... be an infinite sequence of real numbers satisfying the equation $a_n=\left|a_{n+1}-a_{n+2}\right|$ for all $n\geq 0$, where $a_0$ and $a_1$ are two different positive reals.

Can this sequence $a_0$, $a_1$, $a_2$, ... be bounded?

Proposed by Mihai Bălună, Romania
This post has been edited 1 time. Last edited by djmathman, Sep 27, 2015, 2:12 PM
Z K Y
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Valentin Vornicu
7301 posts
#2 • 6 Y
Y by jhu08, megarnie, Adventure10, AlphaBetaGammaOmega, ImSh95, Mango247
Yes, actually this is a Romanian proposal by Mihai Bălună. Have fun with this nice problem.
Z K Y
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Peter Scholze
644 posts
#3 • 5 Y
Y by Adventure10, jhu08, megarnie, ImSh95, Mango247
? yes?
there are thousands of proofs given(in particular, by darij and me), that this sequence is always unbounded.
Z K Y
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darij grinberg
6555 posts
#4 • 4 Y
Y by jhu08, Adventure10, ImSh95, Mango247
Well, it's not exactly that I didn't give a proof, but it wasn't quite correct either... So, the problem is not too trivial...

darij
Z K Y
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zhaobin
2382 posts
#5 • 4 Y
Y by Adventure10, jhu08, ImSh95, Mango247
I think a_i must be position.

BTW,I am intrested in this problem.can you post the solution.
Z K Y
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Xell
40 posts
#6 • 4 Y
Y by Adventure10, jhu08, ImSh95, Mango247
First, some basic facts: all the terms of the sequence are positive, two consecutive terms are always different, we have $u_{n+2}=u_{n+1}-u_{n}$ or $u_{n+2}=u_{n+1}+u_{n}$, but if $u_{n}>u_{n+1}$, then necessarily, $u_{n+2}=u_{n+1}+u_{n}$.

I will then prove the following result: if $u_{n}<u_{n+1}$ for some integer n, then for some $m>n$ we have $u_{n+1}<u_{m}$ and $u_{n}+u_{n+1} < u_{m+1}$.

Property 1: let a,b be two positive real numbers with $a<b$ and suppose that $u_{n}=b-a$,$u_{n+1}=2kb-(2k-1)a$ for some integer k>0. If $u_{n+2}=u_{n+1}-u_{n}$ then we have $u_{n+2}<u_{n+1}$ so, from a previous remark, we have necessarily, $u_{n+3}=u_{n+1}+u_{n+2}$. We check that in that case, $u_{n+2}>b$ and $u_{n+3}>a+b$.

Property 2: let a,b be two positive real numbers with $a<b$ and suppose that $u_{n}=2kb-(2k-1)a$,$u_{n+1}=(2k+1)b-2ka$ for some integer k>0. If $u_{n+2}=u_{n+1}+u_{n}$, then We check that $u_{n+1}> b$ and $u_{n+2}> a+b$.

Suppose now that $a=u_{n}<u_{n+1}=b$. If $u_{n+2}=a+b$, the result is proved. So suppose that $u_{n+2}=b-a$. Then necessarily, $u_{n+3}=u_{n+2}+u_{n+1}=2b-a$. Property 1 (with k=1) shows that the result is proved if $u_{n+4}=u_{n+3}-u_{n+2}$, so suppose that $u_{n+4}=u_{n+3}+u_{n+2}=3b-2a$. Property 2 (with k=1) proves the result if $u_{n+5}=u_{n+4}+u_{n+3}$ so suppose that $u_{n+5}=u_{n+4}-u_{n+3}=b-a$.$u_{n+4}>u_{n+5}$, so $u_{n+6}=u_{n+5}+u_{n+4}=4b-3a$. And so on (= induction) we show that the result is proved by using property 1 and 2 unless we have the following sequence:
a,b,b-a,2b-a,3b-2a,b-a,4b-3a,5b-4a,...,b-a,2kb-(2k-1)a,(2k+1)b-2ka,b-a,...
But in that last case, 2kb-(2k-1)a, can be as high as we want, for example higher than b, and (2k+1)b-2ka higher than a+b.
The result is therefore proved.

Now, there are always consecutive terms such that $u_{n}<u_{n+1}$ (if $u_{n}>u_{n+1}$ for some n, then $u_{n+2}=u_{n}+u_{n+1}>u_{n+1}$). Name $a=u_{n}<u_{n+1}=b$. Then an easy induction using the result shows that one can find terms higher than $k(a+b)$ for every integer k, which shows that such a sequence can not be bounded.
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k2c901_1
146 posts
#7 • 3 Y
Y by Adventure10, jhu08, ImSh95
This problem was used as problem 1 of the final exam of the 3rd TST of Taiwan 2005.
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indybar
398 posts
#8 • 5 Y
Y by ultralako, Adventure10, jhu08, ImSh95, Mango247
I'm not really clear with Xell's solution. Can anybody elaborate?
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bomb
365 posts
#9 • 3 Y
Y by Adventure10, jhu08, ImSh95
Wasnt this posted before???

Bomb
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Philip_Leszczynski
327 posts
#10 • 4 Y
Y by Adventure10, jhu08, ImSh95, Mango247
Since $a_n = |a_{n+1}-a_{n+2}|$, we see that all elements of $\{a_n\}$ are nonnegative.
We first show that all elements of $\{a_n\}$ are positive.

Suppose for the sake of contradiction that there exists an $i \ge 0$ such that $a_{i+2}=0$.
(Obviously $a_0,a_1 \ne 0$ since this is given.)
$a_i = |a_{i+1} - a_{i+2}|$, so $a_i = a_{i+1}$. Let $x=a_i=a_{i+1}$.
We will show that for any $j \ge 0$, the set $\{a_k|j \le k \le i_1 \}$ has only two distinct elements: $x$ and $0$.
For the base case we have $j=i$. We will induct backwards on $j$, from $i$ to $0$.
By the inductive hypothesis, $a_{j+1} \epsilon \{0,x\}$, $a_{j+2} \epsilon \{0,x\}$.
Since $a_j = |a_{j+1} - a_{j+2}|$, $a_j \epsilon \{|0-0|,|0-x|,|x-0|,|x-x|\} = \{0,x\}$.
(Remember that $x$ is nonnegative.)
This completes the induction. So $a_0 \epsilon \{0,x\}$ and $a_1 \epsilon \{0,x\}$.
Since we are given that $a_0$ and $a_1$ are positive, $a_0=a_1=x$. (If $x=0$, then we have our contradiction.)
But we are also given that $a_0$ and $a_1$ are different, so we have our contradiction.
Therefore, all elements of $\{a_n\}$ are positive, and as a corollary, no two consecutive elements of $\{a_n\}$ are equal.

We will now show that $a_0$ cannot be the largest element of $\{a_n\}$.
Suppose for the sake of contradiction that it is.
Let $a_0 = x/k$, and let $a_1 = 1/k$. Then $k>0$ and $x>1$.
$ka_0 = x$
$ka_1 = 1$
$ka_2 = b$
$a_0 = |a_1 - a_2|$, so $ka_0 = k|a_1 - a_2| = |ka_1 - ka_2| = |1-b|$.
Then $x=|1-b|$. So $x=1-b$ or $x=b-1$. The former is impossible since it would imply that $b=1-x$, which would make $b$ negative.
So $x=b-1$, and $b=x+1$. $b>x$. So $a_2 = b/k > x/k = a_0$. This contradicts the assumption that $a_0$ is the largest element of $\{a_n\}$.

We will now suppose, for the sake of contradiction, that $a_{i+1}$, for $i \ge 0$, is the largest element of $\{a_n\}$.
Let $a_0 = 1/k$, and let $a_1 = x/k$. Then $k>0$ and $x>1$.
$ka_i = 1$
$ka_{i+1} = x$
$ka_{i+2} = x-1$ (It can't be $1-x$, since this is negative.)
$ka_{i+3} = b$
$ka_{i+1} = k|a_{i+2} - a_{i+3}| = |ka_{i+2} - ka_{i+3}| = |x-1-b|$
$x=|x-1-b|$
The option $x=x-1-b$ is impossible, since it implies $b=-1$.
Then $x=b+1-x$, and $2x=b+1$, and $b=2x-1$.
Suppose $2x-1<x$. Then $x-1<0$, so $x<1$. Contradiction.
So $a_{i+3} = b/k > x/k = a_{i+1}$. This contradicts our assumption that $a_{i+1}$ is the largest of $\{a_n\}$.

We conclude that $\{a_n\}$ does not have a largest element, and it is therefore not bounded.

QED.
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darktreb
732 posts
#11 • 4 Y
Y by Adventure10, jhu08, ImSh95, Mango247
This was also the first problem of the first MOP test for the Black group (USAMO winners) in 2005.
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Valentin Vornicu
7301 posts
#12 • 4 Y
Y by Adventure10, jhu08, ImSh95, Mango247
darktreb wrote:
This was also the first problem of the first MOP test for the Black group (USAMO winners) in 2005.
Also in the Romanian TST in 2005 :) Spread problem :D
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nipolo
7 posts
#13 • 4 Y
Y by Adventure10, jhu08, ImSh95, Mango247
What a easy problem????

There is no bounded.It is unlimited or it has at least 2 limits
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ZetaX
7579 posts
#14 • 4 Y
Y by Adventure10, jhu08, ImSh95, Assassino9931
Philip_Leszczynski wrote:
[...]
We conclude that $\{a_n\}$ does not have a largest element, and it is therefore not bounded.

QED.
You can't make that assumption, e.g the sequence $a_i = 1- \frac{1}{i}$ doesn't have a largest element and is bounded.
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me@home
2349 posts
#15 • 4 Y
Y by Adventure10, jhu08, ImSh95, Mango247
I'm guessing nipolo's brief "proof" is incorrect? Or if it is correct, can someone elaborate?
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