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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
Inspired by youthdoo
sqing   0
6 minutes ago
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$$ab+bc+ca\leq 10$$
0 replies
1 viewing
sqing
6 minutes ago
0 replies
Sharygin 2025 CR P2
Gengar_in_Galar   4
N 7 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
7 minutes ago
2 var inquality
sqing   5
N 9 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$$ a+b+ab\geq2$$
5 replies
sqing
Today at 4:06 AM
ionbursuc
9 minutes ago
100 Selected Problems Handout
Asjmaj   32
N 12 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
Asjmaj
Dec 31, 2024
John_Mgr
12 minutes ago
No more topics!
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
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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
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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
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#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
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#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
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#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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