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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!
[*]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]
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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
Where is the equality?
AndreiVila   2
N 28 minutes 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
28 minutes ago
Truth or lie at a table
SinaQane   7
N 35 minutes 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
+1 w
SinaQane
Jul 31, 2020
Oksutok
35 minutes ago
2 var inquality
sqing   4
N 36 minutes ago by sqing
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$$ a+b+ab\geq2$$
4 replies
sqing
Today at 4:06 AM
sqing
36 minutes ago
Functional inequality on N
BartSimpsons   21
N 37 minutes 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
37 minutes ago
A colouring game on a rectangular frame
Tintarn   0
42 minutes 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
42 minutes ago
0 replies
Find the value
sqing   5
N 42 minutes 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
42 minutes ago
A touching question on perpendicular lines
Tintarn   0
44 minutes 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
44 minutes ago
0 replies
The last nonzero digit of factorials
Tintarn   0
an hour ago
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 2
For each integer $n \ge 2$ we consider the last digit different from zero in the decimal expansion of $n!$. The infinite sequence of these digits starts with $2,6,4,2,2$. Determine all digits which occur at least once in this sequence, and show that each of those digits occurs in fact infinitely often.
0 replies
Tintarn
an hour ago
0 replies
Fridolin just can't get enough from jumping on the number line
Tintarn   0
an hour ago
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 1
Fridolin the frog jumps on the number line: He starts at $0$, then jumps in some order on each of the numbers $1,2,\dots,9$ exactly once and finally returns with his last jump to $0$. Can the total distance he travelled with these $10$ jumps be a) $20$, b) $25$?
0 replies
Tintarn
an hour ago
0 replies
D1016 : A strange result about the palindrom polynomials
Dattier   0
an hour ago
Source: les dattes à Dattier
Let $Q\in \{-1,1,0\}[x]$ with $Q(1)=0$.

Is it true that $\exists Q \in \mathbb Z[x]$ palindrom with $Q | P$ ?
0 replies
Dattier
an hour ago
0 replies
EMC last problem
GreekIdiot   1
N an hour ago by Tintarn
Source: EMC 2024 P4, Ioannis Galamatis
Find all functions $f: \mathbb{R_+} \rightarrow \mathbb{R_+}$ such that $f(x+yf(x))=xf(y+1) \: \forall\: x, \:y \in \mathbb{R_+}$.
Note: With the symbol $\mathbb{R_+}$ we denote the set of positive real numbers.
1 reply
GreekIdiot
Yesterday at 7:02 PM
Tintarn
an hour ago
Functional equation with a parameter
Tintarn   8
N an hour ago by NicoN9
Source: Baltic Way 2024, Problem 1
Let $\alpha$ be a non-zero real number. Find all functions $f: \mathbb{R}\to\mathbb{R}$ such that
\[
xf(x+y)=(x+\alpha y)f(x)+xf(y)
\]for all $x,y\in\mathbb{R}$.
8 replies
Tintarn
Nov 16, 2024
NicoN9
an hour ago
D1015 : A strange EF for polynomials
Dattier   1
N an hour ago by Dattier
Source: les dattes à Dattier
Find all $P \in \mathbb R[x,y]$ with $P \not\in \mathbb R[x] \cup \mathbb R[y]$ and $\forall g,f$ homeomorphismes of $\mathbb R$, $P(f,g)$ is an homoemorphisme too.
1 reply
Dattier
Yesterday at 8:37 PM
Dattier
an hour ago
Functional Equations Marathon March 2025
Levieee   19
N an hour ago by Jupiterballs
1. before posting another problem please try your best to provide the solution to the previous solution because we don't want a backlog of many problems
2.one is welcome to send functional equations involving calculus (mainly basic real analysis type of proofs) as long it is of the form $\text{"find all functions:"}$
19 replies
Levieee
Today at 1:03 AM
Jupiterballs
an hour ago
IMO 2014 Problem 1
Amir Hossein   131
N Today at 2:09 AM by eg4334
Let $a_0 < a_1 < a_2 < \dots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n\geq 1$ such that
\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]
Proposed by Gerhard Wöginger, Austria.
131 replies
Amir Hossein
Jul 8, 2014
eg4334
Today at 2:09 AM
IMO 2014 Problem 1
G H J
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Amir Hossein
5452 posts
#1 • 36 Y
Y by gobathegreat, test20, Kunihiko_Chikaya, wiseman, quangminhltv99, Davi-8191, Tawan, SRKTK, Wizard_32, myh2910, donotoven, megarnie, math31415926535, son7, ImSh95, mathmax12, Stuart111, Anulick, Lamboreghini, Adventure10, Mango247, ProMaskedVictor, Aopamy, D_S, ItsBesi, cubres, and 10 other users
Let $a_0 < a_1 < a_2 < \dots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n\geq 1$ such that
\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]
Proposed by Gerhard Wöginger, Austria.
This post has been edited 1 time. Last edited by v_Enhance, Nov 5, 2023, 5:17 PM
Reason: missing < sign
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manuel153
324 posts
#2 • 12 Y
Y by Amir Hossein, TheMathLife, son7, ImSh95, Lamboreghini, Adventure10, Mango247, cubres, and 4 other users
There should be some fixed point theorem in the background...
And perhaps some contraction?
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Fedor Petrov
520 posts
#3 • 141 Y
Y by Amir Hossein, manuel153, capttawish, chaotic_iak, alibez, Ygg, shaiephraim, CatalinBordea, test20, numbertheorist17, Mathematicalx, mgalushka, jakab922, lfetahu, professordad, MexicOMM, randomasdf97, Niosha, biomathematics, ssk9208, ZacPower123, cohesive-aggregate, mihaith, B101099, Popescu, v_Enhance, quangminhltv99, Kezer, anantmudgal09, TheDarkPrince, Tommy2000, A_Math_Lover, Swag00, xilias, hwl0304, jt314, JasperL, rocketscience, propali_mat, Tawan, tarzanjunior, InCtrl, futurestar, Wizard_32, Anar24, ThisIsASentence, sunfishho, samoha, Pluto1708, Arhaan, Kayak, richrow12, MelonGirl, TheMathLife, AlastorMoody, alijahan, mathleticguyyy, Reef334, Kanep, leilei80, TwilightZone, Illuzion, hellomath010118, Aryan-23, green_leaf, Limerent, OlympusHero, Pitagar, aops29, xymking, MathBoy23, kevinmathz, 554183, math31415926535, samrocksnature, hsiangshen, megarnie, Siddharth03, mo.s.k14142, ZHEKSHEN, asdf334, Tafi_ak, L567, myh2910, Wizard0001, ali3985, son7, chystudent1-_-, Aimingformygoal, ImSh95, Quidditch, 407420, hieuan682, skyguy88, Elyson, trying_to_solve_br, PERA2008, GeoMetrix, Danielzh, ihatemath123, melowmolly, Mogmog8, EpicBird08, Schur-Schwartz, OronSH, Lamboreghini, zhenghua, Adventure10, golue3120, khina, Mango247, bobthegod78, Aopamy, Stuffybear, DroneChaudhary, Deadline, vrondoS, sabkx, Springles, Sedro, Math_legendno12, Wild, akliu, aidan0626, alexanderhamilton124, NicoN9, ESAOPS, Scilyse, cubres, jason02, Funcshun840, MS_asdfgzxcvb, and 9 other users
Denote $b_n=(a_n-a_{n-1})+\dots+(a_n-a_1)$. Then $b_1=0$, $b_n$ increases and we are searching for $n$ such that $b_n<a_0\leq b_{n+1}$. It clearly exists and is unique.
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mavropnevma
15142 posts
#4 • 23 Y
Y by FlakeLCR, anantmudgal09, azmath333, DrMath, Tawan, HamstPan38825, son7, ImSh95, khina, Lamboreghini, Amir Hossein, Adventure10, Mango247, cubres, MS_asdfgzxcvb, and 8 other users
Of course, the fact the sequence $(a_n)_{n\geq 1}$ has integer terms is irrelevant; all that matters is that the sequence $(b_n)_{n\geq 1}$ is unbounded.

Notice that in fact the precise condition for the thesis to hold true is that for the sequence $(\delta_n)_{n\geq 1}$, given by $\displaystyle \delta_n = a_{n+1}-a_n > 0$ for all $n\geq 1$,
to have $\displaystyle \sum_{k=1}^{\infty} k\delta_k > a_0$; then there will exist $m\geq 1$ so that $\displaystyle b_{m+1} = \sum_{k=1}^{m} k\delta_k > a_0$.
If the sequence $(\delta_n)_{n\geq 1}$ is so that $\displaystyle \sum_{k=1}^{\infty} k\delta_k \leq a_0$, then $\displaystyle b_{m+1} = \sum_{k=1}^{m} k\delta_k < a_0$ and the thesis holds no more; such exemple is $\delta_n = \dfrac {a_0}{n 2^n}$.
This post has been edited 2 times. Last edited by mavropnevma, Jul 8, 2014, 2:21 PM
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ngv
251 posts
#5 • 16 Y
Y by Amir Hossein, rkm0959, Tawan, son7, ImSh95, Lamboreghini, Adventure10, cubres, and 8 other users
Amir Hossein wrote:
Let $a_0 < a_1 < a_2 \ldots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n\geq 1$ such that
\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]
Key Lemma:
If $S_i=S_{i-1}+u_i$ is a cumulative sequence with $S_0=0$ and $u_i>0$ for all $i$, then for every positive integer $k$, there is a unique positive integer $n$ so that, $S_n < k\leq S_{n+1}$.
Proof:
Almost obvious. But if you still need one, take $n$ to be the smallest positive integer so that $S_{n+1}\geq k$. Then, we must have $S_{n}<k$. If not, it would contradict the minimality of $n$.
Now, all we have to do is find an appropriate $\{u\}$ and $\{S\}$. Take, $S_n=\sum_{i=1}^na_i$, which is clearly increasing. We need, $S_n+a_0>na_n$ and $S_{n}+a_0\leq na_{n+1}$. Clearly this want us to consider the sequence $T_n=na_{n+1}-S_{n}$. It can be written as $T_n=\sum_{i=1}^nc_i=T_{i-1}+c_i$ where $c_i=a_{n+1}-a_i$. Obviously, $c_i>0$ for all $i$. Thus, $\{T\}$ is a cumulative sequence, and therefore, such a $n$ always exists no matter what the choice of $a_0$ is as long as it is a positive integer.

Remark: Something similar can not be said if it is replaced by non-negative integers. Because then the sequence $\{S\}$ would not be strictly increasing. Also, the crucial idea seems kind of trivial to me if someone has ever done "Binary Search" or something like that in an algorithm course/in CSE, whatever.
This post has been edited 1 time. Last edited by ngv, Jul 8, 2014, 1:28 PM
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Davi Medeiros
118 posts
#6 • 27 Y
Y by Amir Hossein, CatalinBordea, Torus121, DominicanAOPSer, randomasdf97, InCtrl, Tawan, A_Math_Lover, Lukaluce, Illuzion, megarnie, son7, ImSh95, mathmax12, naonaoaz, Danielzh, Lamboreghini, ihatemath123, Adventure10, cubres, and 7 other users
The old $\Delta$ trick, again!

Let $ a_n = a_0 + \Delta_1 + \Delta_2 + ... + \Delta_n $ for each $ n \ge 1$. Notice that $ \Delta_n \ge 1 $

Now, the condition $a_n < \frac{a_0 + ... +a_n}{n} \le a_{n+1}$ can be rewritten as:

\[ a_0 + \Delta_1 + \Delta_2 + ... + \Delta_n < \frac{(n+1)a_0 + n\Delta_1 +(n-1)\Delta_2+ ... + \Delta_n}{n} \le \]
\[ \le a_0+\Delta_1 + ... +\Delta_n + \Delta_{n+1} \]

After a little calculation, one can have that $\Delta_2 +2\Delta_3 +...+(n-1)\Delta_n < a_0 \le \Delta_2 +2\Delta_3 +...+(n-1)\Delta_n +n\Delta_{n+1}$

Now, defining $f: \mathbb{N} \rightarrow \mathbb{Z}$ by $f(1) = 0$ and $f(n)=\Delta_2 + 2\Delta_3 +.. +(n-1)\Delta_n$, for all $n \ge 2$, we see that $f$ is increasing and unbounded ($f(n+1)-f(n)=n\Delta_{n+1} \ge n$).

Now, it remains to prove that exists a unique $n\in \mathbb{N}$ such that $f(n)<a_0<f(n+1)$ This is clear, since $f(1)=0<a_0$ and $f$ is estrictely increasing and unbounded. Problem Solved!
This post has been edited 2 times. Last edited by Davi Medeiros, Jul 8, 2014, 1:51 PM
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andreiromania
52 posts
#7 • 17 Y
Y by Amir Hossein, InCtrl, Tawan, A_Math_Lover, megarnie, son7, ImSh95, Lamboreghini, Adventure10, sabkx, cubres, and 6 other users
Fedor Petrov wrote:
Denote $b_n=(a_n-a_{n-1})+\dots+(a_n-a_1)$. Then $b_1=0$, $b_n$ increases and we are searching for $n$ such that $b_n<a_0\leq b_{n+1}$. It clearly exists and is unique.

Very ingenuous approach.However not all students may find such an idea so I decided to give a few ideas that one can find from trial and error in order to build a solution.
I will refer to $a_n <\frac{a_0+a_1+a_2+\cdots+a_n}{n}$ as "the LHS for n" and to $\frac{a_0+a_1+a_2+\cdots+a_n}{n}\leq a_{n+1}$ as "the RHS for n".

1)First,we prove that at most one n can satisfy the given inequality.If we suppose there are more than 1 n's,we choose the minimal one,and find out that the RHS for n turns into the opposite of the LHS for n+1,which by induction turns into the opposite of the LHS for every m>n.
2)Then,we prove that,given a particular n,one cannot have the opposite of the RHS for n,and then the opposite for the LHS for n+1.
3)This being proven,we then assume that for n=1 we have the opposite of the RHS(the LHS being obviously true,if the RHS were to be true too we would have found our n).If there would be an n for which the opposite for the LHS for n would be true,then there would have been an m<n for which both the LHS for m AND the RHS for m would be true,otherwise there would be an m for which the opposite of the RHS for m is true and then the opposite of the LHS for m+1 is true,which contradicts 2).
4)From 3) we conjecture that if there would not exist an n for which both the LHS and the RHS are true,the RHS must be false for all n,however with a bit of manipulation this turns into $a_0+a_1>a_n$ for all n,which is obviously false.
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Matija
53 posts
#8 • 14 Y
Y by Amir Hossein, Tawan, A_Math_Lover, son7, ImSh95, Adventure10, Mango247, cubres, and 6 other users
The same idea written differently. Let $S_n$ be the statement $ a_n <\frac{a_0+a_1+a_2+\cdots+a_{n-1}}{n-1} $ for $n\geq 2$. Then the statement of the problem is equivalent to existence of a unique $n$ such that $S_n$ and not $S_{n+1}$.

$S_1$ should be phrased as $0<a_0$ which obviously holds. $S_2, ..., S_n$ imply together that $a_n < a_0+a_1$, so there must be $n$ for which $S_{n+1}$ does not hold. But if $S_m$ does not hold, then $S_{m+1}$ also does not hold, so there can be only one integer with the sought property.
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colinhy
751 posts
#9 • 24 Y
Y by Ygg, Amir Hossein, Stefan4024, DominicanAOPSer, Kezer, Tawan, strategos21, megarnie, JG666, son7, ImSh95, hellnish, Adventure10, ike.chen, cubres, and 9 other users
Hmm another pretty similar proof, just worded differently:

Proof

Rather straightforward, even for a #1.
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gavrilos
233 posts
#10 • 12 Y
Y by Amir Hossein, Tawan, son7, ImSh95, Adventure10, Mango247, cubres, and 5 other users
If colinhy's proof is correct,mine is correct too (it's the same).A rather easy problem I have to say.In my opinion easier than last year's one.
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dibyo_99
487 posts
#11 • 14 Y
Y by Tawan, Amir Hossein, son7, ImSh95, Adventure10, Mango247, cubres, and 7 other users
We are required to show the existence of a unique intger $n \ge 1$ such that \[ na_n < \sum_{i=0}^n a_i \le na_{n+1} \Longleftrightarrow \sum_{i=1}^{n-1} (a_n - a_i) < a_0 \le \sum_{i=1}^n (a_{n+1}-a_i) \]Note that the sequence $\left( \sum_{i=1}^{n-1} (a_n - a_i) \right)_n$ is strictly increasing (since $a_n$ is strictly increasing, each of the differences also increase), has the first term $0$ and is also obviously unbounded. Therefore, there must exist some term of the sequence strictly smaller than $a_0$ such that the next term is greater (weakly) than $a_0$.
This post has been edited 1 time. Last edited by dibyo_99, Jul 8, 2014, 2:38 PM
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tchebytchev
584 posts
#12 • 15 Y
Y by Tawan, Amir Hossein, Rustem-E303, son7, ImSh95, rstenetbg, Adventure10, Mango247, cubres, and 6 other users
Suppose that for every $n$ we have \[a_{n+1} < \frac{a_0+...+a_n}{n}\]
for $n=1$, $a_2 < a_0+a_1$. For $n=2$, $2a_3 <a_0+a_1+a_2<2(a_0+a_1)$ so $a_3<a_0+a_1$ and by induction $a_n<a_0+a_1$ which it is a contradiction because we have an infinite sequence of positive integer and bounded !

Let $n_0$ be an integer such that \[a_{n_0+1} \geq  \frac{a_0+...+a_{n_0}}{n_0}\]
we have $a_1<a_0+a_1$ if $a_0+a_1 \leq a_2$ we get the existence, if not then $a_2 <\frac{a_0+a_1+a_2}{2}$ and again if $\frac{a_0+a_1+a_2}{2} \leq a_3$ we have done and if not we have $a_3 <\frac{a_0+a_1+a_2+a_3}{3}$ and we continue until reaching the integer $n_0$ for which we have
\[a_{n_0} < \frac{a_0+...+a_{n_0}}{n_0} \leq a_{n_0+1}\]

The uniqueness is obvious because if $\frac{a_0+...+a_n}{n} \leq a_{n+1}$ then $\frac{a_0+...+a_n+a_{n+1}}{n+1} \leq a_{n+1}$ and since the sequence is strictly increasing we show easily that for any $m>n+1$ we have $\frac{a_0+...+a_m}{m} < a_m$ so the property can't be satisfied for $m> n_0$
This post has been edited 1 time. Last edited by tchebytchev, Jul 8, 2014, 4:58 PM
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Matematika
139 posts
#13 • 14 Y
Y by Tawan, Amir Hossein, son7, ImSh95, Lamboreghini, Adventure10, Mango247, cubres, and 6 other users
As the problem is already solved in quite a few ways I would like to comment on the possible generalizations.

First question is necessity of the weird integer condition.

The answer if I am not mistaken is that we can't remove it altogether as if we choose:

$a_0=\frac{\pi^2}{6}$

and $a_{n+1}=a_n+\frac{1}{(n+1)^3}$ for $n\geq 0$ should provide a counterexample for the problem without the "integer" condition.

Of course the uniqueness proof as shown above works regardless but the problem breaks at existence condition.

If we denote by $b_n=a_{n+1}-a_n$ as someone already did above if $a_0>nb_n+\cdots +2b_2+b_1$ there is no $n$ as described in the problem.


About the problem.
I believe it is a good problem for problem 1 which should really be as easy as possible while still requiring some thought. I personally dislike it due to a weird imposing of integer numbers on an obviously algebraic problem.
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AnonymousBunny
339 posts
#14 • 15 Y
Y by Tawan, Amir Hossein, son7, ImSh95, Adventure10, Mango247, cubres, and 8 other users
We rewrite the condition as
\[na_n < \displaystyle \sum_{i=0}^{n} a_i < na_{n+1} \implies \begin{cases} \displaystyle \sum_{i=1}^{n} (a_n - a_i) < a_0 \\
\sum_{i=1}^{n+1} (a_{n+1} - a_i) \geq a_0 \end{cases}.\]
Note that for all $n,$
\[\left( \displaystyle \sum_{i=1}^{n+1} (a_{n+1} - a_i) \right) - \left( \displaystyle \sum_{i=1}^{n} (a_n - a_i) \right) = n a_{n+1} - na_n > 0,\]
which implies $\displaystyle \sum_{i=1}^{n} \left( a_n - a_i \right)$ is monotonically increasing. Obviously $\displaystyle \sum_{i=1}^{n} (a_n-a_i)$ is unbounded from above. At $n=1,$ it equals zero. Now let $n$ be the largest integer such that $\displaystyle \sum_{i=1}^{n} \left( a_n - a_i \right) <a_0.$ We have that $a_{n+1} \geq a_0$ by our hypothesis, so this is our desired $n$ which is clearly unique. $\blacksquare$
This post has been edited 2 times. Last edited by AnonymousBunny, Jul 8, 2014, 5:04 PM
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MathD
204 posts
#15 • 14 Y
Y by tim9099xxzz, Tawan, Amir Hossein, son7, ImSh95, Adventure10, Mango247, cubres, and 6 other users
Would someone be so kind and check my proof?
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