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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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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
Inspired by lgx57
sqing   0
16 minutes ago
Source: Own
Let $ a,b>0. $ Prove that$$\dfrac{a^2}{ab+1}+\dfrac{b^3+2}{ab+b^2}\geq 2\sqrt{2}-1$$G

0 replies
sqing
16 minutes ago
0 replies
Find min
lgx57   2
N 27 minutes ago by sqing
Source: Own
Find min of $\dfrac{a^2}{ab+1}+\dfrac{b^2+2}{a+b}$
2 replies
+1 w
lgx57
Yesterday at 3:01 PM
sqing
27 minutes ago
Inequality
MathsII-enjoy   2
N 41 minutes ago by MathsII-enjoy
A interesting problem generalized :-D
2 replies
MathsII-enjoy
Yesterday at 1:59 PM
MathsII-enjoy
41 minutes ago
Sintetic geometry problem
ICE_CNME_4   5
N 44 minutes ago by Ianis
Source: Math Gazette Contest 2025
Let there be the triangle ABC and the points E ∈ (AC), F ∈ (AB), such that BE and CF are concurrent in O.
If {L} = AO ∩ EF and K ∈ BC, such that LK ⊥ BC, show that EKL = FKL.
5 replies
ICE_CNME_4
5 hours ago
Ianis
44 minutes ago
Solution needed ASAP
UglyScientist   9
N 44 minutes ago by MathsII-enjoy
$ABC$ is acute triangle. $H$ is orthocenter, $M$ is the midpoint of $BC$, $L$ is the midpoint of smaller arc $BC$. Point $K$ is on $AH$ such that, $MK$ is perpendicular to $AL$. Prove that: $HMLK$ is paralelogram(Synthetic sol needed).
9 replies
UglyScientist
Yesterday at 1:18 PM
MathsII-enjoy
44 minutes ago
AC, BF, DE concurrent
a1267ab   75
N an hour ago by NicoN9
Source: APMO 2020 Problem 1
Let $\Gamma$ be the circumcircle of $\triangle ABC$. Let $D$ be a point on the side $BC$. The tangent to $\Gamma$ at $A$ intersects the parallel line to $BA$ through $D$ at point $E$. The segment $CE$ intersects $\Gamma$ again at $F$. Suppose $B$, $D$, $F$, $E$ are concyclic. Prove that $AC$, $BF$, $DE$ are concurrent.
75 replies
a1267ab
Jun 9, 2020
NicoN9
an hour ago
FE on R+
AshAuktober   7
N an hour ago 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
an hour ago
2-adic Valuation Unbounded
tigerzhang   14
N an hour ago by GingerMan
Source: Own
For any nonzero integer, define $\nu_2(n)$ as the largest integer $k$ such that $2^k \mid n$. Find all integers $n$ that are not powers of $3$ such that the sequence $\nu_2\left(3^0-n\right),\nu_2\left(3^1-n\right),\nu_2\left(3^2-n\right),\ldots$ is unbounded.
14 replies
tigerzhang
Nov 5, 2021
GingerMan
an hour ago
Combinatorial Sum
P162008   2
N an hour ago by cazanova19921
Evaluate $\sum_{n=0}^{\infty} \frac{2^n + 1}{(2n + 1) \binom{2n}{n}}$
2 replies
P162008
Apr 24, 2025
cazanova19921
an hour ago
IMO 90/3 and IMO 00/5 cross-up
v_Enhance   60
N an hour ago by GingerMan
Source: USA TSTST 2018 Problem 8
For which positive integers $b > 2$ do there exist infinitely many positive integers $n$ such that $n^2$ divides $b^n+1$?

Evan Chen and Ankan Bhattacharya
60 replies
v_Enhance
Jun 26, 2018
GingerMan
an hour ago
square root problem
kjhgyuio   4
N an hour ago by aidan0626
........
4 replies
kjhgyuio
Yesterday at 4:48 AM
aidan0626
an hour ago
gcd (a^n+b,b^n+a) is constant
EthanWYX2009   81
N an hour ago by GingerMan
Source: 2024 IMO P2
Determine all pairs $(a,b)$ of positive integers for which there exist positive integers $g$ and $N$ such that
$$\gcd (a^n+b,b^n+a)=g$$holds for all integers $n\geqslant N.$ (Note that $\gcd(x, y)$ denotes the greatest common divisor of integers $x$ and $y.$)

Proposed by Valentio Iverson, Indonesia
81 replies
EthanWYX2009
Jul 16, 2024
GingerMan
an hour ago
Orthocenter
jayme   5
N 2 hours ago by Ianis
Dear Mathlinkers,

1. ABC an acuatangle triangle
2. H the orthcenter of ABC
3. DEF the orthic triangle of ABC
4. A* the midpoint of AH
5. X the point of intersection of AH and EF.

Prove : X is the orthocenter of A*BC.

Sincerely
Jean-Louis
5 replies
jayme
Mar 25, 2015
Ianis
2 hours ago
positive integers forming a perfect square
cielblue   1
N 2 hours ago by aaravdodhia
Find all positive integers $n$ such that $2^n-n^2+1$ is a perfect square.
1 reply
cielblue
Friday at 8:25 PM
aaravdodhia
2 hours ago
hard problem
Cobedangiu   18
N Apr 15, 2025 by giangtruong13
problem
18 replies
Cobedangiu
Mar 27, 2025
giangtruong13
Apr 15, 2025
hard problem
G H J
G H BBookmark kLocked kLocked NReply
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Cobedangiu
66 posts
#1 • 1 Y
Y by PikaPika999
problem
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Cobedangiu
66 posts
#2 • 1 Y
Y by PikaPika999
no one? .
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Alex-131
5378 posts
#3 • 1 Y
Y by PikaPika999
Wrong solution in-english class, will check out later
This post has been edited 5 times. Last edited by Alex-131, Mar 27, 2025, 4:54 PM
Reason: r
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ND_
51 posts
#4 • 1 Y
Y by PikaPika999
When is equality achieved? a=b=c doesn't satisfy the initial condition. I think minimum is 10.
This post has been edited 2 times. Last edited by ND_, Mar 27, 2025, 4:50 PM
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Nuran2010
82 posts
#7 • 2 Y
Y by PikaPika999, TunarHasanzade
https://artofproblemsolving.com/community/c6h1834400p23600351
Also posted here
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Cobedangiu
66 posts
#8 • 1 Y
Y by PikaPika999
ND_ wrote:
When is equality achieved? a=b=c doesn't satisfy the initial condition. I think minimum is 10.

with equality occurring when $a=2, b=1,c=1$
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ND_
51 posts
#10 • 1 Y
Y by PikaPika999
\( b+c \ge 2\sqrt{bc} \iff \frac{b+c}{bc} \ge \frac{4}{b+c} \iff \) $\frac{1}{b} + \frac{1}{c} \ge \frac{2}{b+c} + \frac{2}{b+c}$
\( \Rightarrow (a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \ge \left(a+\frac{b+c}{2}+\frac{b+c}{2}\right)\left(\frac{1}{a} + \frac{2}{b+c} + \frac{2}{b+c}\right) \)

Hence, minimum occurs at \( b=c \). Plugging onto original equation, we get \( 2 + \left(\frac{b}{a}\right)^3=5\left(\frac{b}{a}\right) \), or \( b=2a \). So, minimum of 10 occurs at \( (2b, b, b) \).
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EaZ_Shadow
1265 posts
#11 • 1 Y
Y by PikaPika999
I used Cauchy and got $(a+b+c)(\frac1a+\frac1b+\frac1c)\geq9$ so i think 10 is the min, and I dont think 9 is achievable.
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xytunghoanh
33 posts
#12 • 1 Y
Y by PikaPika999
Cobedangiu wrote:
problem

cmath?
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Cobedangiu
66 posts
#13 • 2 Y
Y by xytunghoanh, PikaPika999
xytunghoanh wrote:
Cobedangiu wrote:
problem

cmath?
yes, but this inequality is from someone else, just took it back from that person.
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xytunghoanh
33 posts
#14 • 1 Y
Y by PikaPika999
I can help you to solve this. Check pm for Vietnamese solution.
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Cobedangiu
66 posts
#15 • 2 Y
Y by xytunghoanh, PikaPika999
xytunghoanh wrote:
I can help you to solve this. Check pm for Vietnamese solution.
I know how to do it
This post has been edited 1 time. Last edited by Cobedangiu, Mar 27, 2025, 5:27 PM
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sqing
41975 posts
#16 • 1 Y
Y by PikaPika999
Cobedangiu wrote:
problem
Let $a,b,c>0 $ and $a^3+b^3+c^3=5abc.$ Prove that$$10 \leq (a+b+c)\left(\frac {1}{a}+\frac {1}{b}+\frac {1}{c}\right)\leq5+4\sqrt{2}$$https://artofproblemsolving.com/community/c6h3455063p33365029
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InftyByond
205 posts
#17 • 2 Y
Y by Maximilian113, PikaPika999
k this is SUPER hard
why is inequality not satisfied by a=b=c?
suggests something funny like calculus

edit: ok im back with a fake solve!!!!!!!
Solve:
Expand $$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$to get $$\sum (1+\frac{a}{b}+\frac{b}{a})$$into which we now substitute $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$.
this now gives us $$\sum (1+x+\frac{1}{x})$$which taking the double derivative gives that this function is convex.
so we can apply Karamata (smoothing kind of) to give us that equality case is where two of $x, y, z$ are equal and consequently two of $a, b, c$ equal.
note that scaling doesn't change anything in the inequality so you homogenize by setting idk $a+2c=4$ ;) and then going back into the condition
we should get $$a=2, b=1, c=1$$must be the minimum and consequently the minimum value is achieved at 10????????????
This post has been edited 1 time. Last edited by InftyByond, Apr 6, 2025, 5:34 AM
Reason: fake proof issues die hard
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Cobedangiu
66 posts
#18 • 1 Y
Y by PikaPika999
InftyByond wrote:
k this is SUPER hard
why is inequality not satisfied by a=b=c?
suggests something funny like calculus

edit: ok im back with a fake solve!!!!!!!
Solve:
Expand $$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$to get $$\sum (1+\frac{a}{b}+\frac{b}{a})$$into which we now substitute $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$.
this now gives us $$\sum (1+x+\frac{1}{x})$$which taking the double derivative gives that this function is convex.
so we can apply Karamata (smoothing kind of) to give us that equality case is where two of $x, y, z$ are equal and consequently two of $a, b, c$ equal.
note that scaling doesn't change anything in the inequality so you homogenize by setting idk $a+2c=4$ ;) and then going back into the condition
we should get $$a=2, b=1, c=1$$must be the minimum and consequently the minimum value is achieved at 10????????????

The problem has been solved ;) , but the solution will not be posted here :P
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Nguyenhuyen_AG
3320 posts
#19 • 3 Y
Y by PikaPika999, anduran, truongphatt2668
Cobedangiu wrote:
problem
See here: https://nguyenhuyenag.wordpress.com/2025/04/06/inequality-43/
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InftyByond
205 posts
#20 • 1 Y
Y by PikaPika999
Cobedangiu wrote:
InftyByond wrote:
k this is SUPER hard
why is inequality not satisfied by a=b=c?
suggests something funny like calculus

edit: ok im back with a fake solve!!!!!!!
Solve:
Expand $$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$to get $$\sum (1+\frac{a}{b}+\frac{b}{a})$$into which we now substitute $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$.
this now gives us $$\sum (1+x+\frac{1}{x})$$which taking the double derivative gives that this function is convex.
so we can apply Karamata (smoothing kind of) to give us that equality case is where two of $x, y, z$ are equal and consequently two of $a, b, c$ equal.
note that scaling doesn't change anything in the inequality so you homogenize by setting idk $a+2c=4$ ;) and then going back into the condition
we should get $$a=2, b=1, c=1$$must be the minimum and consequently the minimum value is achieved at 10????????????

The problem has been solved ;) , but the solution will not be posted here :P

K so i looked at the solve
Its pretty nice and slick with no calculus
Is my alternate solve wrong then? not surprised
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DKI
32 posts
#21
Y by
we have
\[
(a+b-c)(a+b-2c+2\sqrt2c)(a+b-2c-2\sqrt2c)=4(a^3+b^3+c^3-5abc)-(a-b)^2(3a+3b+5c)
\]hence $a+b\ge c$, and similarly $a+c\ge b$, $b+c\ge a$,
\[
T=(a+b+c)(\frac1a+\frac1b+\frac1c)=10+\frac{a^3+b^3+c^3-5abc+(a+b-c)(a+c-b)(b+c-a)}{abc}\ge10
\]
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giangtruong13
140 posts
#22
Y by
See this https://artofproblemsolving.com/community/c6h1834400p23632179 for more information
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N Quick Reply
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