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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.

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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
interesting functional equation
tabel   1
N 6 minutes ago by GioOrnikapa
Source: random romanian contest
Determine all functions \( f : (0, \infty) \to (0, \infty) \) that satisfy the functional equation:
\[
f(f(x)(1 + y)) = f(x) + f(xy), \quad \forall x, y > 0.
\]
1 reply
tabel
an hour ago
GioOrnikapa
6 minutes ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   4
N 11 minutes ago by GreenTea2593
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
4 replies
OgnjenTesic
May 22, 2025
GreenTea2593
11 minutes ago
pairs (m, n) such that a fractional expression is an integer
cielblue   2
N 28 minutes ago by cielblue
Find all pairs $(m,\ n)$ of positive integers such that $\frac{m^3-mn+1}{m^2+mn+2}$ is an integer.
2 replies
cielblue
Yesterday at 8:38 PM
cielblue
28 minutes ago
Sociable set of people
jgnr   23
N 37 minutes ago by quantam13
Source: RMM 2012 day 1 problem 1
Given a finite number of boys and girls, a sociable set of boys is a set of boys such that every girl knows at least one boy in that set; and a sociable set of girls is a set of girls such that every boy knows at least one girl in that set. Prove that the number of sociable sets of boys and the number of sociable sets of girls have the same parity. (Acquaintance is assumed to be mutual.)

(Poland) Marek Cygan
23 replies
jgnr
Mar 3, 2012
quantam13
37 minutes ago
diophantine equation
m4thbl3nd3r   0
41 minutes ago
Find all positive integers $n,k$ such that $$5^{2n+1}-5^n+1=k^2$$
0 replies
m4thbl3nd3r
41 minutes ago
0 replies
A geometry problem
Lttgeometry   1
N an hour ago by Funcshun840
Triangle $ABC$ has two isogonal conjugate points $P$ and $Q$. The circle $(BPC)$ intersects circle $(AP)$ at $R \neq P$, and the circle $(BQC)$ intersects circle $(AQ)$ at $S\neq Q$. Prove that $R$ and $S$ are isogonal conjugates in triangle $ABC$.
Note: Circle $(AP)$ is the circle with diameter $AP$, Circle $(AQ)$ is the circle with diameter $AQ$.
1 reply
Lttgeometry
Today at 4:03 AM
Funcshun840
an hour ago
Inequality em981
oldbeginner   15
N an hour ago by sqing
Source: Own
Let $a, b, c>0, a+b+c=3$. Prove that
\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]
15 replies
oldbeginner
Sep 22, 2016
sqing
an hour ago
Functional equation
shobber   19
N 2 hours ago by Unique_solver
Source: Canada 2002
Let $\mathbb N = \{0,1,2,\ldots\}$. Determine all functions $f: \mathbb N \to \mathbb N$ such that
\[ xf(y) + yf(x) = (x+y) f(x^2+y^2)  \]
for all $x$ and $y$ in $\mathbb N$.
19 replies
shobber
Mar 5, 2006
Unique_solver
2 hours ago
Prove the inequality
Butterfly   0
2 hours ago
Let $a,b,c$ be real numbers such that $a+b+c=3$. Prove $$a^3b+b^3c+c^3a\le \frac{9}{32}(63+5\sqrt{105}).$$
0 replies
Butterfly
2 hours ago
0 replies
Functional equation
shactal   1
N 3 hours ago by ariopro1387
Let $f:\mathbb R\to \mathbb R$ a function satifying $$f(x+2xy) = f(x) + 2f(xy)$$for all $x,y\in \mathbb R$.
If $f(1991)=a$, then what is $f(1992)$, the answer is in terms of $a$.
1 reply
shactal
4 hours ago
ariopro1387
3 hours ago
interesting diophantiic fe in natural numbers
skellyrah   5
N 3 hours ago by skellyrah
Find all functions \( f : \mathbb{N} \to \mathbb{N} \) such that for all \( m, n \in \mathbb{N} \),
\[
mn + f(n!) = f(f(n))! + n \cdot \gcd(f(m), m!).
\]
5 replies
skellyrah
Yesterday at 8:01 AM
skellyrah
3 hours ago
Non-linear Recursive Sequence
amogususususus   3
N 3 hours ago by SunnyEvan
Given $a_1=1$ and the recursive relation
$$a_{i+1}=a_i+\frac{1}{a_i}$$for all natural number $i$. Find the general form of $a_n$.

Is there any way to solve this problem and similar ones?
3 replies
amogususususus
Jan 24, 2025
SunnyEvan
3 hours ago
Inspired by 2025 Beijing
sqing   6
N 3 hours ago by sqing
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
6 replies
sqing
Yesterday at 4:56 PM
sqing
3 hours ago
Serbian selection contest for the IMO 2025 - P4
OgnjenTesic   2
N 4 hours ago by sqing-inequality-BUST
Source: Serbian selection contest for the IMO 2025
For a permutation $\pi$ of the set $A = \{1, 2, \ldots, 2025\}$, define its colorfulness as the greatest natural number $k$ such that:
- For all $1 \le i, j \le 2025$, $i \ne j$, if $|i - j| < k$, then $|\pi(i) - \pi(j)| \ge k$.
What is the maximum possible colorfulness of a permutation of the set $A$? Determine how many such permutations have maximal colorfulness.

Proposed by Pavle Martinović
2 replies
OgnjenTesic
May 22, 2025
sqing-inequality-BUST
4 hours ago
Problem 1
SpectralS   146
N Apr 23, 2025 by YaoAOPS
Given triangle $ABC$ the point $J$ is the centre of the excircle opposite the vertex $A.$ This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G.$ Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC.$ Prove that $M$ is the midpoint of $ST.$

(The excircle of $ABC$ opposite the vertex $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)

Proposed by Evangelos Psychas, Greece
146 replies
SpectralS
Jul 10, 2012
YaoAOPS
Apr 23, 2025
Problem 1
G H J
G H BBookmark kLocked kLocked NReply
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SpectralS
5 posts
#1 • 17 Y
Y by buratinogigle, Mualpha7, Amir Hossein, iarnab_kundu, Davi-8191, math1181, AlastorMoody, A-Thought-Of-God, jhu08, megarnie, HWenslawski, ImSh95, Adventure10, Tastymooncake2, ItsBesi, Rounak_iitr, and 1 other user
Given triangle $ABC$ the point $J$ is the centre of the excircle opposite the vertex $A.$ This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G.$ Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC.$ Prove that $M$ is the midpoint of $ST.$

(The excircle of $ABC$ opposite the vertex $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)

Proposed by Evangelos Psychas, Greece
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FelixD
588 posts
#2 • 20 Y
Y by UraharaKisuke, buratinogigle, AndrewTom, biomathematics, karitoshi, Mathasocean, A-Thought-Of-God, translate, jhu08, yshk, ImSh95, Adventure10, Mango247, shafikbara48593762, Tastymooncake2, and 5 other users
If I'm not mistaken, it's easy to see that $\angle LFJ=\alpha/2$ and so $AFJL$ is cyclic. But $\angle JLA=90$ and so $\angle AFJ=90$. Thus, $AB=BS$ and hence $MS=AK$. Similarly, $MT=AL$, but $AK=AL$, so we are done.
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AndreiAndronache
88 posts
#3 • 7 Y
Y by jhu08, ImSh95, Adventure10, Tastymooncake2, and 3 other users
Otherwise: $m(\angle{JFL})=m(\angle{JAL})=\frac{A}{2}\Rightarrow A,F,J,L$ are concyclic $\Rightarrow S,J,M,F$ are concyclic $\Rightarrow m(\angle{JST})=m(\angle{JTS})=\frac{A}{2}\Rightarrow JS=JT\Rightarrow SM=MT$.
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WakeUp
1347 posts
#4 • 9 Y
Y by buratinogigle, jhu08, metricpaper, ImSh95, Adventure10, Tastymooncake2, and 3 other users
Slightly different start to FelixD's approach:

Let $P,Q$ be the midpoints of $MK,ML$. Clearly $FP\perp GP$ and $GQ\perp FQ$ so $FPQG$ is cyclic. Thus $\angle FGP=\angle FQP=\angle MQP=\angle MLK$ ($PQ||KL$ of course). Thus $FGLK$ is cyclic. Angle-chasing shows that this angle is actually $\frac{B}{2}$, which is equal to $\angle FJK$ and so $J$ also lies on this circle. But $AJ$ is clearly the diameter of $(JKL)$, so the points $A,F,K,J,L,G$ are concyclic.
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Concyclicboy
49 posts
#5 • 2 Y
Y by jhu08, Adventure10
Yes, we only need that <MFB= <A/2 and then AFJL cyclic then <AFJ=<ALJ=90 then <SFJ=<SMJ=90 then SFMJ cyclic with <MSJ=<A/2. Similar to <JTM=<A/2
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Zhero
2043 posts
#6 • 7 Y
Y by buratinogigle, yugrey, jhu08, Adventure10, Tastymooncake2, and 2 other users
By a simple angle chase, $\angle FBM = \angle ABJ = 90 + \angle B/2$ and $\angle FMB = \angle AJB = \angle C / 2$, so $\triangle BFM \sim \triangle BAJ$. Thus, $\triangle BFA \sim \triangle BMJ$ (since $BF/BA = BM/BJ$ and $\angle FBA = \angle MBJ = 90 - \angle B / 2$.) Since $BM \perp MJ$, we have $BF \perp AF$. Since $\angle ABF = \angle FBS = 90 - \angle B/2$, we have $BS = BA$. Similarly, $CA = CT$, so $SM = (s-c) + c = s = (s-b) + b = MT$.
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golem24
92 posts
#7 • 19 Y
Y by yugrey, amar_04, ETS1331, Aryan-23, jhu08, mistakesinsolutions, Adventure10, Mango247, MS_Kekas, Tastymooncake2, Funcshun840, and 8 other users
The solutions are smaller than the problem statement.
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Kunihiko_Chikaya
14514 posts
#8 • 7 Y
Y by AceOfDiamonds, buratinogigle, jhu08, yshk, Adventure10, Mango247, Tastymooncake2
Here is the attached figure.
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Orin
22 posts
#9 • 3 Y
Y by buratinogigle, Adventure10, Mango247
It is well-known that $\angle BJC=90^{\circ}-\frac {1}{2}\angle A$
now $\angle LJC=\angle MJC=90^{\circ}-\angle MCJ=90^{\circ}-(90^{\circ}- \frac {1}{2}\angle C)=\frac {1}{2} \angle C$
so $\angle FJL=\angle BJC+\angle LJC=90^{\circ} -\frac {1}{2} \angle A+\frac {1}{2} \angle C$....(1)
now $\angle MLJ=90^{\circ}- \angle LJC=90^{\circ}-\frac {1}{2} \angle C$....(2)
so from (1),(2) we get in $\triangle LJF,\angle LFJ=\frac {1}{2} \angle A$ which inplies $AFJL$ is cyclic.
since $\angle JLA=90^{\circ}$ we get $AFJ=90^{\circ}$
now $\angle FAB=\angle FAL- \angle A=\angle \frac {1}{2} \angle B$
and $\angle FSB=90^{\circ}-\angle FBS=90^{\circ}-\angle CBJ=\angle \frac {1}{2} \angle B$
hence $\angle FAB=\angle FSB$
so $\triangle FBS \cong \triangle FBA$[ASA] implying $AB=BS$
now $MS=MB+BS=BK+AB=AK$
simimlarly $MT=AL$
but $AK=AL$[tangent to the excircle from $A$]
hence $MS=MT$ which means $M$ is the midpoint of $ST$
P.S.:Edited only to correct a typo.
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math_genie
41 posts
#10 • 2 Y
Y by Adventure10, Mango247
Its quite easy to see that
$A,K,J,G$ is cyclic. (1)
$A,F,J,L$ is cyclic. (2)

Since $\angle AKJ = 90^{\circ}$, this implies that $\angle AGJ = 90^{\circ}$ from (1)
Similarly $\angle AFJ = 90^{\circ}$.
Hence we get
$A,F,K,J$ is cyclic (3)
$A,G,L,J$ is cyclic (4)

Thus from (1),(2),(3),(4) we get that all six points $A,G,F,J,K,L$ to be cyclic.
Now, note that from triangle ABS, $AB = BS = c$.
Similarly, $AC = CT = b$.
Thus $SM = SB + BM = c + (s-c) = s$
Similarly $TM = TC + CM = b + (s-b) = s$.
Hence proved.
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sjaelee
485 posts
#11 • 2 Y
Y by Adventure10 and 1 other user
Solution
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Neerjhor
12 posts
#12 • 1 Y
Y by Adventure10
It was really easy. Mine is almost similar as everyone. I proved that $AFJL$ and $AGJK$ are cyclic, which concludes that $AB=BS$ and $AC=CT$. Then $BM=s-c$ and $CM=s-b$. Now, we add these and $MS=MT$. :-)

This was quite easy for IMO 1 I guess. :-)
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pi37
2079 posts
#13 • 1 Y
Y by Adventure10
Slightly different approach
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guysfromKIROVOGRAD
3 posts
#14 • 2 Y
Y by Adventure10, Mango247
WakeUp wrote:
Slightly different start to FelixD's approach:

Let $P,Q$ be the midpoints of $MK,ML$. Clearly $FP\perp GP$ and $GQ\perp FQ$ so $FPQG$ is cyclic. Thus $\angle FGP=\angle FQP=\angle MQP=\angle MLK$ ($PQ||KL$ of course). Thus $FGLK$ is cyclic. Angle-chasing shows that this angle is actually $\frac{B}{2}$, which is equal to $\angle FJK$ and so $J$ also lies on this circle. But $A$ is clearly the diameter of $(JKL)$, so the points $A,F,K,J,L,G$ are concyclic.

Cool solution, bro. But ... too many letters as for me. Ты ещё не задолбался точечки отмечать? ( Use russian translator, bro)
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#15 • 2 Y
Y by Adventure10, Mango247
Notice that $MK$ and $ML$ are parallel to the internal bisectors of $B$ and $C$ respectively. Then $MK \perp BJ$, $ML \perp CJ$, so $M$ is the orthocenter of $JFG$. Since $JM \perp BC$ from tangency, and $JM \perp FG$ from the orthocenter, $BC \parallel FG$. Now we can prove that $MF \parallel AG$ and $MG \parallel AF$ with any of the methods above, or by considering the excenters $I_b, I_c$, the fact tht $I_bI_c$ is antiparallel to $BC$ and a short angle chase. With this, $FMG$ is the medial triangle of $AST$ and we are done.
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