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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
gcd (a^n+b,b^n+a) is constant
EthanWYX2009   86
N a minute ago by cubres
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
86 replies
+1 w
EthanWYX2009
Jul 16, 2024
cubres
a minute ago
Finding the minimal number of coins to pay without change
nAalniaOMliO   3
N 18 minutes ago by sami1618
Source: Belarusian-Iranian Friendly Competition 2025
In the magic land there are coins of all integer denominations from $1$ to $100$. Vlad has $n \geq 3$ coins, the sum of denominations of which is $200$. Find the minimal possible value of $n$ at which we can confidently say that Vlad is able to pay $100$ without change.
3 replies
+1 w
nAalniaOMliO
Jun 14, 2025
sami1618
18 minutes ago
Quadratic system
juckter   39
N 19 minutes ago by mudkip42
Source: Mexico National Olympiad 2011 Problem 3
Let $n$ be a positive integer. Find all real solutions $(a_1, a_2, \dots, a_n)$ to the system:

\[a_1^2 + a_1 - 1 = a_2\]\[ a_2^2 + a_2 - 1 = a_3\]\[\hspace*{3.3em} \vdots \]\[a_{n}^2 + a_n - 1 = a_1\]
39 replies
juckter
Jun 22, 2014
mudkip42
19 minutes ago
A geometry problem
Lttgeometry   0
25 minutes ago
Let triangle $ABC$ have $(w_A)$ as the $A$-mixtilinear incircle, and let $A'$ be the tangency point of $(w_A)$ with the circumcircle $(O)$. Let $AA''$ be a diameter of $(w_A)$. Define $B''$, $C''$ similarly. Prove that the lines $AA''$, $BB''$, and $CC''$ are concurrent.
0 replies
Lttgeometry
25 minutes ago
0 replies
Hard functional equation
Sardor   20
N 38 minutes ago by player-019
Source: IZHO2015.P3
Find all functions $ f\colon \mathbb{R} \to \mathbb{R} $ such that $ f(x^3+y^3+xy)=x^2f(x)+y^2f(y)+f(xy) $, for all $ x,y \in \mathbb{R} $.
20 replies
Sardor
Jan 13, 2015
player-019
38 minutes ago
P2 Cono Sur 2021
Leo890   10
N 41 minutes ago by lendsarctix280
Source: Cono Sur 2021 P2
Let $ABC$ be a triangle and $I$ its incenter. The lines $BI$ and $CI$ intersect the circumcircle of $ABC$ again at $M$ and $N$, respectively. Let $C_1$ and $C_2$ be the circumferences of diameters $NI$ and $MI$, respectively. The circle $C_1$ intersects $AB$ at $P$ and $Q$, and the circle $C_2$ intersects $AC$ at $R$ and $S$. Show that $P$, $Q$, $R$ and $S$ are concyclic.
10 replies
Leo890
Nov 30, 2021
lendsarctix280
41 minutes ago
Prove that $\angle FAC = \angle EDB$
micliva   32
N an hour ago by Fly_into_the_sky
Source: All-Russian Olympiad 1996, Grade 10, First Day, Problem 1
Points $E$ and $F$ are given on side $BC$ of convex quadrilateral $ABCD$ (with $E$ closer than $F$ to $B$). It is known that $\angle BAE = \angle CDF$ and $\angle EAF = \angle FDE$. Prove that $\angle FAC = \angle EDB$.

M. Smurov
32 replies
micliva
Apr 18, 2013
Fly_into_the_sky
an hour ago
Peru IMO TST 2023
diegoca1   1
N an hour ago by MathLuis
Source: Peru IMO TST 2023 D1 P2
Let $n$ be a positive integer. On an $n \times n$ board, players $A$ and $B$ take turns in a game. On each turn, a player selects an edge of the board (not on the board border) and makes a cut along that edge. If after the move the board is split into more than one piece, then that player loses the game.

Player $A$ moves first. Depending on the value of $n$, determine whether one of the players has a winning strategy.
1 reply
diegoca1
Yesterday at 8:03 PM
MathLuis
an hour ago
Purely projective statement
ChimkinGang   6
N 2 hours ago by axolotlx7
Source: Own
Let $ABCD$ be a quadrilateral with $S=AD\cap BC$, $T=AB\cap CD$, and $X=AC\cap BD$. Let $P$ be a point in the plane not on $TX$, $Q=BP\cap TX$, $R=SP\cap TX$, and $Q'$ be the point on $TX$ such that $(QQ';TX)=-1$. If $U=BD\cap PQ'$ and $V=AP\cap DR$, show that $U$, $V$, and $T$ are collinear.
6 replies
ChimkinGang
Jun 15, 2025
axolotlx7
2 hours ago
Rhombus EVAN
62861   72
N 2 hours ago by fearsum_fyz
Source: USA January TST for IMO 2017, Problem 2
Let $ABC$ be a triangle with altitude $\overline{AE}$. The $A$-excircle touches $\overline{BC}$ at $D$, and intersects the circumcircle at two points $F$ and $G$. Prove that one can select points $V$ and $N$ on lines $DG$ and $DF$ such that quadrilateral $EVAN$ is a rhombus.

Danielle Wang and Evan Chen
72 replies
62861
Feb 23, 2017
fearsum_fyz
2 hours ago
Probablity problem
AlanLG   1
N 2 hours ago by AlexCenteno2007
Source: Mathematics Regional Olympiad of Mexico Southeast 2019 P3
Eight teams are competing in a tournament all against all (every pair of team play exactly one time among them). There are not ties and both results of every game are equally probable. What is the probability that in the tournament every team had lose at least one game and won at least one game?
1 reply
AlanLG
Oct 23, 2021
AlexCenteno2007
2 hours ago
Four variables (5)
Nguyenhuyen_AG   1
N 2 hours ago by arqady
Let $a,\,b,\,c,\,d$ be non-negative real numbers, such that $a+b+c+d=4.$ Prove that
\[52  + 17(\sqrt a + \sqrt b + \sqrt c + \sqrt d)^2\geqslant 9(ab+ bc + ca + da  + db + dc)^2.\]hide
1 reply
Nguyenhuyen_AG
4 hours ago
arqady
2 hours ago
N-M where M,N two 5-digit ''consecutive'' palindromes
parmenides51   1
N 3 hours ago by AlexCenteno2007
Source: Mathematics Regional Olympiad of Mexico Center Zone 2018 P1
Let $M$ and $N$ be two positive five-digit palindrome integers, such that $M <N$ and there is no other palindrome number between them. Determine the possible values of $N-M$.
1 reply
parmenides51
Nov 13, 2021
AlexCenteno2007
3 hours ago
Peru Ibero TST 2022
diegoca1   1
N 3 hours ago by grupyorum
Source: Peru Ibero TST 2022 D2 P1
For every positive integer $m > 1$, let $p(m)$ be the largest prime number that divides $m$. For $m = 1$, define $p(1) = 1$.

a) Prove that there exists a positive integer $n$ such that the numbers $p(n - 2022)$, $p(n)$, and $p(n + 2022)$ are all less than $\frac{\sqrt{n}}{20}$.

b) Given a positive integer $N$, prove that there exists a positive integer $n$ such that the numbers $p(n - 2022)$, $p(n)$, and $p(n + 2022)$ are all less than $\frac{\sqrt{n}}{N}$.
1 reply
diegoca1
Today at 5:00 AM
grupyorum
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
This post has been edited 1 time. Last edited by alexgsi, Oct 20, 2023, 7:52 PM
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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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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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