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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Jul 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
At least One pair with square of distance multiple of 2016
Johann Peter Dirichlet   5
N 11 minutes ago by ja.
Source: 38th Brazilian MO (2016) - First Day, Problem 2
Find the smallest number \(n\) such that any set of \(n\) ponts in a Cartesian plan, all of them with integer coordinates, contains two poitns such that the square of its mutual distance is a multiple of \(2016\).
5 replies
Johann Peter Dirichlet
Nov 23, 2016
ja.
11 minutes ago
A symmetric inequality in n variables (3)
Nguyenhuyen_AG   1
N 18 minutes ago by lbh_qys
Let $a_1,a_2,\ldots,a_n (n \geqslant 1)$ be non-negative real numbers. Prove that
\[\sum_{i=1}^n \frac{\displaystyle a_i^2 \left( \sum_{j=1}^n a_j - a_i \right)^2}{\displaystyle \sum_{j=1}^n a_j^2 - a_i^2} \geq \left(\sum_{i=1}^n a_i\right)^2 - \sum_{i=1}^n a_i^2.\]Assume all denominators are non-zero.
1 reply
+1 w
Nguyenhuyen_AG
3 hours ago
lbh_qys
18 minutes ago
shortlisted problems being used in undergraduate competition
enter16180   0
32 minutes ago
Hello, I am posting here to let know ( clarified after a post in College Math forum) that Problem 10 of Open Mathematical Olympiad for University Students ( OMOUS-2025) held at Ashgabat, Turkmenistan on 13-18 April, 2025 is found to be A6 Shortlisted Problems IMO-2024.
Following is discussion on College Math Forum
https://artofproblemsolving.com/community/c7h3551018_omous2025_team_competition_p10


Image of problem from competition for reference below.
0 replies
enter16180
32 minutes ago
0 replies
Four variables
Nguyenhuyen_AG   0
33 minutes ago
Let $a,\,b,\,c,\,d$ non-negative real numbers. Prove that
\[\frac{abc}{(a+b+c)^3}+\frac{bcd}{(b+c+d)^3}+\frac{cda}{(c+d+a)^3}+\frac{dab}{(d+a+b)^3} \leqslant \frac{(a+b+c+d)^2}{27(a^2+b^2+c^2+d^2)}.\]
0 replies
Nguyenhuyen_AG
33 minutes ago
0 replies
Functional equation
shactal   1
N an hour ago by Mathzeus1024
Source: Own
Hello, I found this functional equation that I can't solve, and I haven't got any hints. Could someone try and find the solution, it's actually quite difficult:
Find all continuous functions $f:\mathbb{R}\to \mathbb{R}$ such that, for all $x, y \in \mathbb{R} $,
$$
f(x + f(y)) + f(y + f(x)) = f(x \, f(y) + y \, f(x)) + f(x + y)$$Thank you.
1 reply
shactal
Yesterday at 11:15 PM
Mathzeus1024
an hour ago
Probability Inequality
EthanWYX2009   0
an hour ago
Source: 2024 June 谜之竞赛-5
Determine the minimum real number \(\lambda\) such that for any $2024$ real numbers \(a_1, a_2, \cdots, a_{2024}\) satisfying
\[\sum_{i=1}^{2024} a_i = 0,\quad\sum_{i=1}^{2024} a_i^2 = 1,\]there exists a non-empty subset \(I\) of \(\{1, 2, \cdots, 2024\}\) for which
\[\sum_{i\in I} a_i \leq \lambda \cdot \min\{|I|, 2024 - |I|\}.\]Proposed by Tianqin Li, High School Affiliated to Renmin University of China
0 replies
EthanWYX2009
an hour ago
0 replies
Elegant Geometry Problem
EthanWYX2009   0
an hour ago
Source: 2024 June 谜之竞赛-2
Let \( I \) be the incenter of \(\triangle ABC\). The incircle tangents to \( AC \), \( AB \) at \( E \), \( F \), respectively. Let \( EF \) intersect \( BC \) at \( P \). \(\odot BEP\) and \(\odot CFP\) intersect again at \( Q \). Let \( M \) be the midpoint of the arc \( BC \) of \(\odot ABC\). \(\odot MPQ\) intersects \(\odot ABC\) again at \( R \). Let \( H \) be the orthocenter of \(\triangle BIC\).

Prove that the intersection point of \( HR \) and \( QI \) lies on \(\odot MPQ\).

Proposed by Bohan Zhang, Shanghai Minban Huayu Middle School
0 replies
EthanWYX2009
an hour ago
0 replies
Non-polynomial sequences satifying m+n|a_m+a_n?
TUAN2k8   0
an hour ago
Source: own
Consider a sequence of integers \((a_n)_{n>0}\) such that for every pair of distinct positive integers \((m, n)\), \(m + n\) is a divisor of \(a_m + a_n\).

a) Prove that \(a_n\) is divisible by \(n\) for every positive integer \(n\).

b) Does there exist a sequence \((a_n)_{n>0}\) that is not a polynomial in \(n\) (i.e., there does not exist a polynomial \(P(X) \in \mathbb{R}[X]\) such that \(a_n = P(n)\) for all \(n \in \mathbb{Z}_+\)) and satisfies the given condition?
0 replies
TUAN2k8
an hour ago
0 replies
Fraction Part Inequality
EthanWYX2009   0
an hour ago
Source: 2023 November 谜之竞赛-1
Let \( x \) be a real number.[list]
[*]Determine the maximum value of $ \left| \sum_{k=1}^{1012} \left(\{(2k-1)x\} - \{2kx\}\right) \right| $;
[*]Determine the maximum value of $\left| \sum_{k=1}^{1012} \left(\{kx\} - \{(k+1012)x\}\right) \right|$. [/list]
Proposed by Site Mu, Beijing 101 Middle School
0 replies
EthanWYX2009
an hour ago
0 replies
Tricky Geometry
zqy648   1
N an hour ago by EthanWYX2009
Source: 2023 October 谜之竞赛-5
Given triangle \( ABC \), let \( P \) be a moving point inside the triangle such that \( \angle ABP = \angle ACP \). \( BP \), \( CP \) intersect \( AO \) at \( E \), \( F \) respectively, where \( O \) is the circumcenter of \( \triangle ABC \). The circle with diameter \( AP \) meet the circumcircle of \( \triangle BPC \) at another point \( Q \).

Show that there exist two fixed circles tangent to the circumcircle of \( \triangle QEF \).

Created by Sheng Lu
IMAGE
1 reply
zqy648
Jul 19, 2025
EthanWYX2009
an hour ago
Help me prove these lemmas
dimi07   0
2 hours ago
In the name of God, the most Merciful, the most Compassionate.
Let $a,b,c$ $\in$ $\mathbb{Z}$,then prove the following
\[
a\mid c, b\mid c \implies lcm(a,b)\mid c.
\]And also prove that
\[
c\mid a,c\mid b \implies c\mid gcd(a,b)
\]And by the help of God I finish this question.
0 replies
dimi07
2 hours ago
0 replies
Come frome \frac {a^5+b^5} {a^3+b^3} \geq a^2+b^2-ab
sqing   3
N 2 hours ago by sqing
Source: Own
Let $a,b>0, $ Prove that $$ \frac {a^5+a^3b^2+a^2b^3+b^5} {(a^3+a^2b+ab^2+b^3)( a^2+ b^2-ab) }=1$$
3 replies
sqing
Today at 2:06 AM
sqing
2 hours ago
A symmetric inequality in n variables (2)
Nguyenhuyen_AG   2
N 3 hours ago by Nguyenhuyen_AG
Let $a_1,a_2,\ldots,a_n (n \geqslant 2)$ be non-negative real numbers. Prove that
\[\sum_{i=1}^n \frac{\displaystyle a_i^2 \left( \sum_{j=1}^n a_j - a_i \right)}{\displaystyle \sum_{j=1}^n a_j^2 - a_i^2} \geq \sum_{i=1}^n a_i.\]Assume all denominators are non-zero.
2 replies
Nguyenhuyen_AG
5 hours ago
Nguyenhuyen_AG
3 hours ago
Functional equation
proximo   13
N 3 hours ago by megarnie
Source: Belarusian Mathematical Olympiad 2017
Find all functions $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$, satisfying the following equation $$f(x+f(xy))=xf(1+f(y))$$for all positive $x$ and $y$
13 replies
proximo
Mar 31, 2017
megarnie
3 hours ago
angle relations in a convex ABCD given, double segment wanted
parmenides51   12
N May 15, 2025 by Nuran2010
Source: Iranian Geometry Olympiad 2018 IGO Intermediate p2
In convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ meet at the point $P$. We know that $\angle DAC = 90^o$ and $2 \angle ADB = \angle ACB$. If we have $ \angle DBC + 2 \angle ADC = 180^o$ prove that $2AP = BP$.

Proposed by Iman Maghsoudi
12 replies
parmenides51
Sep 19, 2018
Nuran2010
May 15, 2025
angle relations in a convex ABCD given, double segment wanted
G H J
G H BBookmark kLocked kLocked NReply
Source: Iranian Geometry Olympiad 2018 IGO Intermediate p2
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parmenides51
30655 posts
#1 • 4 Y
Y by Maths_Guy, Adventure10, Mango247, ItsBesi
In convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ meet at the point $P$. We know that $\angle DAC = 90^o$ and $2 \angle ADB = \angle ACB$. If we have $ \angle DBC + 2 \angle ADC = 180^o$ prove that $2AP = BP$.

Proposed by Iman Maghsoudi
This post has been edited 1 time. Last edited by parmenides51, Sep 20, 2018, 9:22 AM
Reason: Proposed
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TheDarkPrince
3042 posts
#2 • 4 Y
Y by Maths_Guy, Pluto1708, myh2910, Adventure10
Drop feet from $C$ on $BP$. Angle conditions give congruency and we are done.
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MarkBcc168
1631 posts
#3 • 2 Y
Y by Adventure10, Mango247
Let $Q$ be the point on $AD$ such that $QC=QD$. Then the condition $\angle DBC+2\angle ADC=180^{\circ}$ implies $BCDQ$ is concyclic. Thus $\angle QCB=\angle ADB$ or $\angle QCP = \angle QDP$, which implies $PC=PD$. Now let $R$ be the point on ray $PA$ such that $AR=AP$. Hence $BCDR$ is cyclic, implying $PB=PR$ so we are done.
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achen29
561 posts
#4 • 2 Y
Y by Adventure10, Mango247
One can just angle chase and $PC=PD$, and through obtaining all angles in terms of one variable $x$ use sine rule on $\triangle APD $ and $\triangle BPC$.
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Math-wiz
6107 posts
#5 • 2 Y
Y by Adventure10, DEKT
$Let$ $\angle BAD=x^{\circ}$.
$Simple$ $angle$ $chasing$ $gives$ $\angle ACB=2x$, $\angle CDP=\angle DCP=\frac{90-x}{2}$.

$So,$ $PD=PC$

$In$ $\triangle APD$,
$\sin x=\frac{AP}{PD}\implies AP=PD\sin x$

$In$ $\triangle PCB$, $by$ $Sine$ $rule,$
$\frac{\sin 2x}{\sin 90-x}=\frac{BP}{PC}\implies \frac{BP}{2}=PC\sin x$

$(As$ $\frac{\sin 2x}{\sin 90-x}=\frac{2\sin x\cos x}{\cos x}=2\sin x)$

$But,$ $PC=PD\implies PC\sin x=PD\sin x$

$Substituting$ $our$ $results,$

$\frac{BP}{2}=AP\implies 2AP=BP$
This post has been edited 2 times. Last edited by Math-wiz, Aug 27, 2019, 1:24 PM
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checkmated
6 posts
#6 • 1 Y
Y by Mango247
Why am I getting this:
$\angle$ ADP = 90- $\angle$ APD = 90- $\angle$ BPC; as $\angle$ PCB = 2 $\angle$ ADP, $\implies$ $\angle$ PBC = 180-( $\angle$ BPC + $\angle$ PCB) =90- $\angle$ ADP $\implies$ 180-2( $\angle$ ADP + $\angle$ PDC) =90- $\angle$ ADP $\implies$ $\angle$ ADP +2 $\angle$ PDC =90 $\implies$ $\angle$ ACD = $\angle$ PDC $\implies$ $\angle$ DPC = 180-2 $\angle$ PDC $\implies$ $\angle$ APC = $\angle$ ADP + 180 -2 $\angle$ PDC =90. But A, P and C are on a straight line!!!
This post has been edited 2 times. Last edited by checkmated, Oct 30, 2020, 5:30 AM
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VicKmath7
1392 posts
#7 • 2 Y
Y by Mango247, Mango247
Sketch. Note that $PC=CB$ by angle chase. If $M$ is midpoint of $PB$, then triangles $APD$ and $PMC$ are congruent, because of $PD=PC$ (angle chase). So we're done.
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cursed_noob
38 posts
#8 • 1 Y
Y by UndefinedUser
Pretty simple by angle chasing , as far as I did. Denote $ \angle ADP=x$. So, $ \angle APD= \angle BPC = 90^{\circ}-x$ and $ \angle PCB= 2x$. This yields, $$ \angle PBC= 90^{\circ}-x \\ \implies \angle PBC= \angle BPC \\ \implies BC= PC $$Now, drawing a perpendicular from $C$ to point $M$ of $PB$ concludes $ \triangle PCM \sim \triangle PAD$. But we need to show their congruence.We can do this if we can prove $PD=PC$. How can we do that? Denote a point $Q$ on $AC$ such that $ \angle QDA=\angle ADC$ .Hence, $$\triangle QDA \cong \triangle ADC \implies QD=DC$$. Since $\triangle QDC \sim \triangle PDC$ , we can conclude $PD=PC$ . Thus we proved the congruence of $ \triangle PCM$ and $\triangle PAD$ which implies $PA=PM \implies PA=\frac{PB}{2} \implies 2PA=PB$ $\blacksquare$
This post has been edited 3 times. Last edited by cursed_noob, Oct 6, 2021, 2:54 PM
Reason: Edited this time for LaTeX
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Mahdi_Mashayekhi
705 posts
#9
Y by
Let CH be perpendicular to BP.
∠DBC + 2∠ADC = 180 ---> ∠DBC + 2∠ADB + 2∠PDC = ∠DBC + ∠BCP + 2∠PDC so ∠BPC = 2∠PDC and it means DPC is isosceles.
∠DAC = 90 = ∠DHC ---> AHCD is cyclic ---> ∠HCP = ∠PDA = ∠BCP/2 so CH is angle bisector of ∠BCP and it means BCP is isosceles.
so we have BP = 2HP and now it's easy to show triangles HPC and APD are congruent so HP = AP.
BP = 2HP = 2AP so we're Done.
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lian_the_noob12
173 posts
#10 • 1 Y
Y by Rounak_iitr
$\color{magenta} \boxed{\textbf{SOLUTION}}$

From given condition,
$$\angle DBC+2(\angle ADC)=180°\implies \angle DBC+2\angle ADB+2\angle BDC=180° \implies \angle DBC+\angle ACB+\angle BDC+\angle BDC=180° \implies 180°-\angle ACD+\angle BDC=180°$$$$\implies \angle ACD=\angle BDC$$
Let, $P'$ be the reflection of $P$ over $AD$
So, $$\angle BDP'=\angle PDP'=2\angle ADP=2\alpha$$From given condition,
$$\angle BCP'=\angle ACB=2\angle ADB=2\alpha$$
So, $P'BCD$ is cyclic $$\implies \angle PBP'=\angle DBP'=\angle DCP'=\angle DCP=\angle CDP=\angle BDC=\angle BP'C=\angle PP'B \implies BP=PP'=2AP \blacksquare$$
This post has been edited 1 time. Last edited by lian_the_noob12, Sep 1, 2023, 8:46 PM
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alexgsi
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Solution
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ali123456
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I will give a sketch of my solution:
(1) Consider $(D1)$ the bisector of angle $ACB$
(2) Notice that angle $LCA=LDA$
(3) $ALCD$ is cyclic
(4) $DLC=90$
(5) Notice $L$ is the midpoint of $PB$ so the problem becomes trying to prove that $P L=PA$
(6) $LCP=LCB$ let it be y and let $ALD=x$
(7) And now comes the brilliant part by noticing that : $180=APD+2ADC=APD+2ADP+2PDC=90+ADP+2PDC=90+ADC+PDC=90+ADC+90-LCD$ then $ADC=LCD=x$ then you can see that $PAL=PLA=x$
And conclude
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Nuran2010
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#13 • 1 Y
Y by FarrukhBurzu
My (ugly) solution:
Let $\angle ADB=\alpha$.So,$\angle ACB=2\alpha$.Then,$\angle APD=\angle BPC=\angle PBC=90-\alpha$ is satisfied.So, call $|BC|=|PC|=a$.Now,we must have $90-\alpha+2\alpha+2 \angle BDC=180^{\circ}$.So,we get $\angle PDC=\angle PCD=\frac{90-\alpha}{2}$.So,$|BC|=|PC|=|PD|=a$.Call $|AP|=b,|BP|=c$,If we find $c=2b$,problem is finished.Now,Pythagorean theorem in $\triangle APD$ gives:$|AD|=\sqrt{a^2-b^2}$.Pythagorean again in $\triangle ACD$ gives:$|DC|=\sqrt{2a^2+2ab}$.One last step,Stewart in $\triangle BDC$ gives:$c \cdot (\sqrt{2a^2+2ab})^2+a^3=ac(a+c)+a^2(a+c)$.Clearing gives $2abc=ac^2$ and this is equivalent to $2b=c$,we're done.
This post has been edited 1 time. Last edited by Nuran2010, May 15, 2025, 9:21 AM
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