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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.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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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
1996 St. Petersburg City Mathematical Olympiad
Sadece_Threv   2
N 12 minutes ago by reni_wee
Source: 1996 St. Petersburg City Mathematical Olympiad
Find all positive integers $n$ such that $3^{n-1}+5^{n-1}$ divides $3^{n}+5^{n}$
2 replies
Sadece_Threv
Jul 29, 2024
reni_wee
12 minutes ago
IMO 2010 Problem 5
mavropnevma   55
N 24 minutes ago by maromex
Each of the six boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$, $B_6$ initially contains one coin. The following operations are allowed

Type 1) Choose a non-empty box $B_j$, $1\leq j \leq 5$, remove one coin from $B_j$ and add two coins to $B_{j+1}$;

Type 2) Choose a non-empty box $B_k$, $1\leq k \leq 4$, remove one coin from $B_k$ and swap the contents (maybe empty) of the boxes $B_{k+1}$ and $B_{k+2}$.

Determine if there exists a finite sequence of operations of the allowed types, such that the five boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ become empty, while box $B_6$ contains exactly $2010^{2010^{2010}}$ coins.

Proposed by Hans Zantema, Netherlands
55 replies
mavropnevma
Jul 8, 2010
maromex
24 minutes ago
NT ineq: sum 1/a_i < (m+n)/m , {a_1,a_2,...,a_n} subset of {1,2,...,m}
parmenides51   1
N 41 minutes ago by DVDTSB
Source: 2006 MOP Homework Blue NT 6
Let $m$ and $n$ be positive integers with $m > n \ge 2$. Set $S =\{1,2,...,m\}$, and set $T = \{a_1,a_2,...,a_n\}$ is a subset of $S$ such that every element of $S$ is not divisible by any pair of distinct elements of $T$. Prove that
$$\frac{1}{a_1}+\frac{1}{a_2}+ ...+ \frac{1}{a_n} < \frac{m+n}{m}$$
1 reply
parmenides51
Apr 12, 2020
DVDTSB
41 minutes ago
Never 8
chess64   21
N 2 hours ago by reni_wee
Source: Canada 1970, Problem 10
Given the polynomial \[ f(x)=x^n+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots+a_{n-1}x+a_n \] with integer coefficients $a_1,a_2,\ldots,a_n$, and given also that there exist four distinct integers $a$, $b$, $c$ and $d$ such that \[ f(a)=f(b)=f(c)=f(d)=5, \] show that there is no integer $k$ such that $f(k)=8$.
21 replies
chess64
May 14, 2006
reni_wee
2 hours ago
Benelux fe
ErTeeEs06   11
N 5 hours ago by Pitchu-25
Source: BxMO 2025 P1
Does there exist a function $f:\mathbb{R}\to \mathbb{R}$ such that $$f(x^2+f(y))=f(x)^2-y$$for all $x, y\in \mathbb{R}$?
11 replies
ErTeeEs06
Apr 26, 2025
Pitchu-25
5 hours ago
My Unsolved FE on R+
ZeltaQN2008   4
N Today at 10:37 AM by mashumaro
Source: IDK
Give $a>0$. Find all funcitions $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that for all any $x,y\in (0,\infty):$
$$f(xf(y)+a)=yf(x+y+a)$$
4 replies
ZeltaQN2008
Today at 8:41 AM
mashumaro
Today at 10:37 AM
2015 solutions for quotient function!
raxu   49
N Today at 1:26 AM by blueprimes
Source: TSTST 2015 Problem 5
Let $\varphi(n)$ denote the number of positive integers less than $n$ that are relatively prime to $n$. Prove that there exists a positive integer $m$ for which the equation $\varphi(n)=m$ has at least $2015$ solutions in $n$.

Proposed by Iurie Boreico
49 replies
raxu
Jun 26, 2015
blueprimes
Today at 1:26 AM
My unsolved problem
ZeltaQN2008   1
N Yesterday at 4:46 PM by Adywastaken
Source: Belarus 2017
Find all funcition $f:(0,\infty)\rightarrow (0,\infty)$ such that for all any $x,y\in (0,\infty)$ :
$f(x+f(xy))=xf(1+f(y))$
1 reply
ZeltaQN2008
Yesterday at 3:47 PM
Adywastaken
Yesterday at 4:46 PM
IMO Shortlist 2011, Algebra 4
orl   18
N Yesterday at 3:54 PM by shanelin-sigma
Source: IMO Shortlist 2011, Algebra 4
Determine all pairs $(f,g)$ of functions from the set of positive integers to itself that satisfy \[f^{g(n)+1}(n) + g^{f(n)}(n) = f(n+1) - g(n+1) + 1\] for every positive integer $n$. Here, $f^k(n)$ means $\underbrace{f(f(\ldots f)}_{k}(n) \ldots ))$.

Proposed by Bojan Bašić, Serbia
18 replies
orl
Jul 11, 2012
shanelin-sigma
Yesterday at 3:54 PM
IMO 2011 Problem 5
orl   84
N Yesterday at 12:57 PM by alexanderhamilton124
Let $f$ be a function from the set of integers to the set of positive integers. Suppose that, for any two integers $m$ and $n$, the difference $f(m) - f(n)$ is divisible by $f(m- n)$. Prove that, for all integers $m$ and $n$ with $f(m) \leq f(n)$, the number $f(n)$ is divisible by $f(m)$.

Proposed by Mahyar Sefidgaran, Iran
84 replies
orl
Jul 19, 2011
alexanderhamilton124
Yesterday at 12:57 PM
Functional Equation
Keith50   2
N Yesterday at 3:49 AM by jasperE3
Source: Own
Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that \[f(x+f(x)+2f(y))+f(2f(x)-y)=4x+f(y)\]holds for all reals $x$ and $y$.
2 replies
Keith50
Jun 24, 2021
jasperE3
Yesterday at 3:49 AM
n-term Sequence
MithsApprentice   14
N Yesterday at 2:32 AM by chenghaohu
Source: USAMO 1996, Problem 4
An $n$-term sequence $(x_1, x_2, \ldots, x_n)$ in which each term is either 0 or 1 is called a binary sequence of length $n$. Let $a_n$ be the number of binary sequences of length $n$ containing no three consecutive terms equal to 0, 1, 0 in that order. Let $b_n$ be the number of binary sequences of length $n$ that contain no four consecutive terms equal to 0, 0, 1, 1 or 1, 1, 0, 0 in that order. Prove that $b_{n+1} = 2a_n$ for all positive integers $n$.
14 replies
MithsApprentice
Oct 22, 2005
chenghaohu
Yesterday at 2:32 AM
FE on R+
AshAuktober   7
N Yesterday at 1:37 AM by GingerMan
Source: 2007 MOP
(Note I couldn't find a post w/ this from AoPS search so I'm posting, please do tell if there exists a post.)

Solve over positive real numbers the functional equation
\[ f\left( f(x) y + \frac xy \right) = xyf(x^2+y^2). \]
7 replies
AshAuktober
Sep 2, 2024
GingerMan
Yesterday at 1:37 AM
Property of a function
Ritangshu   0
May 3, 2025
Let \( f(x, y) = xy \), where \( x \geq 0 \) and \( y \geq 0 \).
Prove that the function \( f \) satisfies the following property:

\[
f\left( \lambda x + (1 - \lambda)x',\; \lambda y + (1 - \lambda)y' \right) > \min\{f(x, y),\; f(x', y')\}
\]
for all \( (x, y) \ne (x', y') \) and for all \( \lambda \in (0, 1) \).

0 replies
Ritangshu
May 3, 2025
0 replies
Angle bisectors
bgn   3
N Aug 11, 2020 by Arefe
Source: Iran MO 3rd round 2016 finals - Geometry P3
Given triangle $\triangle ABC$ and let $D,E,F$ be the foot of angle bisectors of $A,B,C$ ,respectively.
$M,N$ lie on $EF$ such that $AM=AN$. Let $H$ be the foot of $A$-altitude on $BC$.
Points $K,L$ lie on $EF$ such that triangles $\triangle AKL, \triangle HMN$ are correspondingly similiar (with the given order of vertices) such that $AK \not\parallel HM$ and $AK \not\parallel HN$.

Show that: $DK=DL$
3 replies
bgn
Sep 5, 2016
Arefe
Aug 11, 2020
Angle bisectors
G H J
G H BBookmark kLocked kLocked NReply
Source: Iran MO 3rd round 2016 finals - Geometry P3
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bgn
178 posts
#1 • 2 Y
Y by doxuanlong15052000, Adventure10
Given triangle $\triangle ABC$ and let $D,E,F$ be the foot of angle bisectors of $A,B,C$ ,respectively.
$M,N$ lie on $EF$ such that $AM=AN$. Let $H$ be the foot of $A$-altitude on $BC$.
Points $K,L$ lie on $EF$ such that triangles $\triangle AKL, \triangle HMN$ are correspondingly similiar (with the given order of vertices) such that $AK \not\parallel HM$ and $AK \not\parallel HN$.

Show that: $DK=DL$
Z K Y
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drmzjoseph
445 posts
#2 • 2 Y
Y by Adventure10, Mango247
Nice problem

General problem

Given triangle $\triangle ABC$ and let $D$ be the foot of angle bisectors of $A$. If $X \in AD, BX \cap AC \equiv E, CX \cap AB \equiv F$
$M,N$ lie on $EF$ such that $AM=AN$. Let $H$ be the foot of $A$-altitude on $BC$.
Points $K,L$ lie on $EF$ such that triangles $\triangle AKL, \triangle HMN$ are correspondingly similiar (with the given order of vertices) such that $AK \not\parallel HM$ and $AK \not\parallel HN$.

Show that: $DK=DL$

Proof

$Y \equiv EF \cap BC, R \equiv AH \cap EF$ let $W,Z$ the proyections of $A,D$ at $EF$ respectively. Denote $P$ a point at $ZD$ such that $\odot (APD)$ is tangent to $AZ$, and $Q \equiv EF \cap AP$

Is clear $\angle AYF =\angle ADP =\angle PAZ \Rightarrow AZ^2=ZY.ZQ=ZP.ZD \Rightarrow YQPD$ is cyclical $\Rightarrow \angle ZYD=\angle APZ \Rightarrow \triangle AZP \cup Q \sim \triangle HWA \cup R \Rightarrow \frac{AR}{RH}=\frac{PQ}{QA}$ and $\angle ARN=\angle MQA$

Denote $K'$ and $L'$ points at $EF$ such that $AN \parallel PK'$ and $AM \parallel PL' \Rightarrow AMN \cup H \sim PK'L' \cup A \Rightarrow \triangle HMN \sim \triangle AK'L' \Rightarrow K' \equiv K$ and $L \equiv L'$ is evident that $KP=PL \Rightarrow DK=DL$
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nguyenhaan2209
111 posts
#3 • 3 Y
Y by top1csp2020, Adventure10, Mango247
First we see that K,L can be construct uniquely by let the symmetric of H wrt EF be H' and draw two parallel line from A to H'M, H'N we have KL. The same idea as IMO 2018 G6, we will reconstruct it. Let the altitude of A,D and BC meet EF at U,V,W and P lies on DU such that UPA~UAD then UP.UD=UA^2 and UDA=UWA=UAP so UW.UQ=UD.UP gives us WQPD cyclic. Cause AVH=AUP=180-AWD and APU=UAD=VWH=VAR so AUP ~ HVA and PUQ=RVA=90 then AR/RH=PQ/QA + ARN=KQA. Re-construct K,L as PK//AN, PL//AM so AMN-H~PKL-A so AKL~HMN. (Q.E.D)
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Arefe
65 posts
#4
Y by
If we get $ T , P , Y $ foot of $ A , D , Y $ with $ EF$ and $Q , X$ intersection of $AH , AD$ with $EF$ .
Get $N$ the point on $TE$ and $L' , K'$ reflection of $L , K$ about $T$ .
$HN||AL' , HM||AK' \rightarrow \frac{NQ}{NT+TL}=\frac{NQ}{ML}=\frac{HQ}{HA}=\frac{MQ}{NK}=\frac{2NT-NQ}{NT+TK}$
$\rightarrow NT+TL=\frac{AH}{HQ}.NQ , NT+TK=\frac{AH}{HQ}.(2NT-NQ) \rightarrow |TL-TK|=2.\frac{AH}{HQ}.|NQ-NT| = 2.\frac{AH}{HQ}.QT$
so the problem is equal $\leftrightarrow PL=PK \leftrightarrow |TL-TK|=2TP \leftrightarrow TP=\frac{AH}{HQ}.QT $
If $D'$ intersection of $EF , BC$ we can see that $\angle{DAD'}=90$ so $ATHD' , PDD'A$ are cyclic .
$\frac{AH}{HQ}.QT=AH.\frac{QT}{QA}.\frac{AQ}{QH}=AH . \sin{BD'F} . \frac{\sin{AD'F}}{\sin{BD'F}}.\frac{AD'}{D'H} = AH . \sin{AD'F} . \frac{AD}{AH} = \sin{AD'F} . AD = TP $ and we're done $\blacksquare$
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