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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
Divisibilty...
Sadigly   6
N 2 minutes ago by Sadigly
Source: Azerbaijan Junior MO 2025 P2
Find all $4$ consecutive even numbers, such that the square of their product divides the sum of their squares.
6 replies
Sadigly
34 minutes ago
Sadigly
2 minutes ago
how can I solve this FE
Jackson0423   4
N 2 minutes ago by Jackson0423

Let \( f : \mathbb{R} \to \mathbb{R} \) be a function that satisfies the following equation for all real numbers \( x \):
\[
f(x^2 + x + 3) + 2f(x^2 - 3x + 5) = 6x^2 - 10x + 17.
\]Find the value of \( f(100) \).
4 replies
Jackson0423
33 minutes ago
Jackson0423
2 minutes ago
IMO Genre Predictions
ohiorizzler1434   65
N 4 minutes ago by Oksutok
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
65 replies
ohiorizzler1434
May 3, 2025
Oksutok
4 minutes ago
k^2/p for k =1 to (p-1)/2
truongphatt2668   1
N 11 minutes ago by Double07
Let $p$ be a prime such that: $p = 4k+1$. Simplify:
$$\sum_{k=1}^{\frac{p-1}{2}}\begin{Bmatrix}\dfrac{k^2}{p}\end{Bmatrix}$$
1 reply
truongphatt2668
2 hours ago
Double07
11 minutes ago
Interesting inequality
imnotgoodatmathsorry   1
N 17 minutes ago by Bergo1305
Let $x,y,z > \frac{1}{2}$ and $x+y+z=3$.Prove that:
$\sqrt{x^3+y^3+3xy-1}+\sqrt{y^3+z^3+3yz-1}+\sqrt{z^3+x^3+3zx-1}+\frac{1}{4}(x+5)(y+5)(z+5) \le 60$
1 reply
imnotgoodatmathsorry
an hour ago
Bergo1305
17 minutes ago
every lucky set of values {a_1,a_2,..,a_n} satisfies a_1+a_2+...+a_n >n2^{n-1}
parmenides51   6
N 22 minutes ago by jonh_malkovich
Source: 2020 International Olympiad of Metropolises P3
Let $n>1$ be a given integer. The Mint issues coins of $n$ different values $a_1, a_2, ..., a_n$, where each $a_i$ is a positive integer (the number of coins of each value is unlimited). A set of values $\{a_1, a_2,..., a_n\}$ is called lucky, if the sum $a_1+ a_2+...+ a_n$ can be collected in a unique way (namely, by taking one coin of each value).
(a) Prove that there exists a lucky set of values $\{a_1, a_2, ..., a_n\}$ with $$a_1+ a_2+...+ a_n < n \cdot 2^n.$$(b) Prove that every lucky set of values $\{a_1, a_2,..., a_n\}$ satisfies $$a_1+ a_2+...+ a_n >n \cdot 2^{n-1}.$$
Proposed by Ilya Bogdanov
6 replies
parmenides51
Dec 19, 2020
jonh_malkovich
22 minutes ago
A strange NT problem
flower417477   0
25 minutes ago
Source: unknown
$p$ is a given prime number.$A=\{a_1,a_2,\cdots,a_{p-1}\}$ is a set which $\prod\limits_{i=1}^{p-1}a_i\equiv\frac{p-1}{2}\pmod p$.
Prove that there're at least $\frac{p-1}{2}$ non-empty subsets $B$ of $A$ such that $\sum\limits_{b\in B}b\equiv 1\pmod p$
0 replies
flower417477
25 minutes ago
0 replies
combi/nt
blug   0
44 minutes ago
Prove that every positive integer $n$ can be written in the form
$$n=2^{a_1}3^{b_1}+2^{a_2}3^{b_2}+..., $$where $a_i, b_j$ are non negative integers, such that
$$2^x3^y\nmid 2^z3^t$$for every $x, y, z, t$.
0 replies
blug
44 minutes ago
0 replies
Interesting inequalities
sqing   2
N an hour ago by sqing
Source: Own
Let $ a,b,c\geq 0 , a(b+c)=k.$ Prove that
$$\frac{1}{a+1}+\frac{2}{b+1}+\frac{1}{c+1}\geq  \frac{4\sqrt{k}-6}{ k-2}$$Where $5\leq  k\in N^+.$
Let $ a,b,c\geq 0 , a(b+c)=9.$ Prove that
$$\frac{1}{a+1}+\frac{2}{b+1}+\frac{1}{c+1}\geq \frac{6}{7}$$
2 replies
sqing
2 hours ago
sqing
an hour ago
Find smallest value of (x^2 + y^2 + z^2)/(xyz)
orl   7
N an hour ago by Bryan0224
Source: CWMO 2001, Problem 4
Let $ x, y, z$ be real numbers such that $ x + y + z \geq xyz$. Find the smallest possible value of $ \frac {x^2 + y^2 + z^2}{xyz}$.
7 replies
orl
Dec 27, 2008
Bryan0224
an hour ago
easy substitutions for a functional in reals
Circumcircle   9
N an hour ago by Bardia7003
Source: Kosovo Math Olympiad 2025, Grade 11, Problem 2
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ with the property that for every real numbers $x$ and $y$ it holds that
$$f(x+yf(x+y))=f(x)+f(xy)+y^2.$$
9 replies
Circumcircle
Nov 16, 2024
Bardia7003
an hour ago
writing words around circle, two letters
jasperE3   1
N an hour ago by pi_quadrat_sechstel
Source: VJIMC 2000 2.2
If we write the sequence $\text{AAABABBB}$ along the perimeter of a circle, then every word of the length $3$ consisting of letters $A$ and $B$ (i.e. $\text{AAA}$, $\text{AAB}$, $\text{ABA}$, $\text{BAB}$, $\text{ABB}$, $\text{BBB}$, $\text{BBA}$, $\text{BAA}$) occurs exactly once on the perimeter. Decide whether it is possible to write a sequence of letters from a $k$-element alphabet along the perimeter of a circle in such a way that every word of the length $l$ (i.e. an ordered $l$-tuple of letters) occurs exactly once on the perimeter.
1 reply
jasperE3
Jul 27, 2021
pi_quadrat_sechstel
an hour ago
Arithmetic Sequence of Products
GrantStar   19
N an hour ago by OronSH
Source: IMO Shortlist 2023 N4
Let $a_1, \dots, a_n, b_1, \dots, b_n$ be $2n$ positive integers such that the $n+1$ products
\[a_1 a_2 a_3 \cdots a_n, b_1 a_2 a_3 \cdots a_n, b_1 b_2 a_3 \cdots a_n, \dots, b_1 b_2 b_3 \cdots b_n\]form a strictly increasing arithmetic progression in that order. Determine the smallest possible integer that could be the common difference of such an arithmetic progression.
19 replies
GrantStar
Jul 17, 2024
OronSH
an hour ago
Inequality Involving Complex Numbers with Modulus Less Than 1
tom-nowy   0
2 hours ago
Let $x,y,z$ be complex numbers such that $|x|<1, |y|<1,$ and $|z|<1$.
Prove that $$ |x+y+z|^2 +3>|xy+yz+zx|^2+3|xyz|^2 .$$
0 replies
tom-nowy
2 hours ago
0 replies
Inequality with permutations
Vlados021   9
N Dec 14, 2024 by Seungjun_Lee
Source: 2019 Belarus Team Selection Test 7.3
Given a positive integer $n$, determine the maximal constant $C_n$ satisfying the following condition: for any partition of the set $\{1,2,\ldots,2n \}$ into two $n$-element subsets $A$ and $B$, there exist labellings $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$ of $A$ and $B$, respectively, such that
$$
(a_1-b_1)^2+(a_2-b_2)^2+\ldots+(a_n-b_n)^2\ge C_n.
$$
(B. Serankou, M. Karpuk)
9 replies
Vlados021
Sep 2, 2019
Seungjun_Lee
Dec 14, 2024
Inequality with permutations
G H J
G H BBookmark kLocked kLocked NReply
Source: 2019 Belarus Team Selection Test 7.3
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Vlados021
184 posts
#1 • 4 Y
Y by brokendiamond, Ab_Rin, Adventure10, farhad.fritl
Given a positive integer $n$, determine the maximal constant $C_n$ satisfying the following condition: for any partition of the set $\{1,2,\ldots,2n \}$ into two $n$-element subsets $A$ and $B$, there exist labellings $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$ of $A$ and $B$, respectively, such that
$$
(a_1-b_1)^2+(a_2-b_2)^2+\ldots+(a_n-b_n)^2\ge C_n.
$$
(B. Serankou, M. Karpuk)
Z K Y
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brokendiamond
349 posts
#3 • 1 Y
Y by Adventure10
How to solve this problem ???
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XbenX
590 posts
#5 • 1 Y
Y by Adventure10
We're trying to maximize $$C_n=\sum_{i=1}^{n}(a_i-b_i)^2=\sum_{i=1}^{2n}i^2-2\sum_{i=1}^{n}a_ib_i=\frac{2n(2n+1)(4n+1)}{6}-2\sum_{i=1}^{n}a_ib_i$$, so we need to minimize $S:=\sum_{i=1}^{n}a_ib_i$.

Lemma. The minimum of $S$ is achieved when none of the pairs $(a_i,b_i)$ are both greater than $n$.

Proof. Assume not, then there are four numbers $i,j,k,l \in \{1,2,3,\dots,n\}$ such that $(n+i)(n+j)+lk$ appears in $S$, but we can replace these numbers with $(n+i)l+(n+j)k$ and decrease $S$ because $(n+i)(n+j-l)\geq k(n+j-l)$. $\blacksquare$

So, we can let $A=\{1,2,\dots ,n\}$ and $B=\{n+1,n+2,\dots ,2n\}$ and by rearrangement inequality we get that $$S\geq \sum_{i=1}^{n} i(2n+1-i)=\frac{n(n+1)(2n+1)}{3}$$.
Hence, we have $C_n=\frac{2n(2n+1)(4n+1)}{6}-2S\leq \frac{n(2n+1)(2n-1)}{3}$.
This post has been edited 1 time. Last edited by XbenX, Jan 29, 2020, 1:10 PM
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somartino
8 posts
#6 • 2 Y
Y by Aegaertargaryen1102, YOUsername
I don't quite understand. We have to find the maximum value of the sum for each partition and then the smallest of those, don't we?
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indulged
69 posts
#8
Y by
XbenX wrote:
We're trying to maximize $$C_n=\sum_{i=1}^{n}(a_i-b_i)^2=\sum_{i=1}^{2n}i^2-2\sum_{i=1}^{n}a_ib_i=\frac{2n(2n+1)(4n+1)}{6}-2\sum_{i=1}^{n}a_ib_i$$, so we need to minimize $S:=\sum_{i=1}^{n}a_ib_i$.

Lemma. The minimum of $S$ is achieved when none of the pairs $(a_i,b_i)$ are both greater than $n$.

Proof. Assume not, then there are four numbers $i,j,k,l \in \{1,2,3,\dots,n\}$ such that $(n+i)(n+j)+lk$ appears in $S$, but we can replace these numbers with $(n+i)l+(n+j)k$ and decrease $S$ because $(n+i)(n+j-l)\geq k(n+j-l)$. $\blacksquare$

So, we can let $A=\{1,2,\dots ,n\}$ and $B=\{n+1,n+2,\dots ,2n\}$ and by rearrangement inequality we get that $$S\geq \sum_{i=1}^{n} i(2n+1-i)=\frac{n(n+1)(2n+1)}{3}$$.
Hence, we have $C_n=\frac{2n(2n+1)(4n+1)}{6}-2S\leq \frac{n(2n+1)(2n-1)}{3}$.
Oh,you did it in a wrong place ,you need to find where S max is,not the miniment。But your method is right ,and i think the ture result is n。
This post has been edited 1 time. Last edited by indulged, May 7, 2021, 2:22 PM
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Takeya.O
769 posts
#9
Y by
$n$ is even: $\frac{13n^{3}-4n}{12}$
$n$ is odd: $\frac{13n^{3}-n}{12}$
This post has been edited 1 time. Last edited by Takeya.O, Jun 24, 2023, 7:12 AM
Reason: error
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Takeya.O
769 posts
#10
Y by
I don't know the time when the sum is minimized.
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dragon-YBW
40 posts
#11
Y by
use karamata ineq.
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dragon-YBW
40 posts
#12
Y by
Consruction:
if n=2k+1,take A={1,2,...,k,3k+2,...,4k+2} and B={k+1,...,3k+1}
if n=2k,take A={1,...,k,3k+1,...,4k} and B={k+1,...,3k}
Proof:
let c(k)=|a(k)-b(k)|,rearrange c(k) as d1≥d2≥...≥dn
if n=2k+1,we can prove that {dk}>{3k+1,3k,...,k+1}
if n=2k,we have {dk}>{3k-1,3k-1,3k-3,3k-3,...,k+1,k+1} (nontrivial but not hard)
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Seungjun_Lee
526 posts
#13
Y by
I don't know why there isn't any full solution for this great problem! Nice and elegant integer inequality from 2019 RMMSL. XD

Let $s_i = a_i + b_i$. Since $\sum{a_i^2} + \sum{b_i^2}$ is constant, we only need to find the maximum value of $\sum_{i = 1}^{n} a_ib_i$ where $A$ and $B$ are partitions of $[2n]$, and $|A| = |B| = n$. This is simply maximizing $\sum_{i = 1}^{n} s_i^2$.

We claim that $|s_i - s_j| \le n$ for any $i < j$ both in the set $\{ 1, 2, \cdots, n\}$. This is because $|s_i - s_j| = |a_i - a_j + b_i - b_j|$. As $a_i \ge a_1 + i - 1$ and $a_j \le a_n - n + j$, we have $a_i - a_j \ge a_1 - a_n + (n-1) + (i - j)$. Also, $b_i - b_j \ge j - i$. Therefore, we have that $|s_i - s_j| \le n$. Now, we forget about our original definition of $s_i$, and just focus on the fact that $s_i$ are integers that satisfies $|s_i - s_j| \le n$ with $\sum_{i = 1}^{n} s_i = n(2n+1)$. As $|s_i - s_j| \le n$, we can set an integer $t$ so that $s_i \in [t, t+n]$ for any $i$. As the function $x \mapsto x^2$ is convex, we can force for some $s_i < s_j$, changing $(s_i, s_j)$ into $(s_i - 1, s_j + 1)$ makes the value $\sum_{i = 1}^{n} s_i^2$ larger. If there are two $i, j$ such that $a < s_i < s_j < a+n$, then we can force $s_i = a$ or $s_j = a+n$ so that the sum gets larger. Hence, we can see that only one value can lie in the interval $(t, t+n)$. The remaining job is just tedious calculation, which I will omit. :pilot:

Now, we proved that (I actually did the calculation) $\sum s_i^2$ is maximized when the sequence of $s_i$ consists of $\lfloor n/2 \rfloor$ number of $\left \lfloor \dfrac{3n}{2} \right \rfloor$ and $\lceil n/2 \rceil$ number of $\left \lceil \dfrac{5n}{2} \right \rceil$. Then, we can easily construct $A$ and $B$ so that $s_i$ becomes the prementioned sequence. Hence, we can calculate $C_n$, which turns out to be$$\sum_{i=1}^{n}{(a_i - b_i)^2} \ge \dfrac{2n(2n+1)(4n+1)}{3} - \left \lfloor \dfrac{n}{2} \right \rfloor \cdot \left \lfloor \dfrac{3n}{2} \right \rfloor^2 - \left \lceil \dfrac{n}{2} \right \rceil \cdot \left \lceil \dfrac{5n}{2} \right \rceil^2$$
This post has been edited 1 time. Last edited by Seungjun_Lee, Dec 14, 2024, 2:41 AM
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