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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
c^a + a = 2^b
Havu   11
N a few seconds ago by ilikemath247365
Find $a, b, c\in\mathbb{Z}^+$ such that $a,b,c$ coprime, $a + b = 2c$ and $c^a + a = 2^b$.
11 replies
Havu
May 10, 2025
ilikemath247365
a few seconds ago
Angles in a triangle with integer cotangents
Stear14   2
N 5 minutes ago by Stear14
In a triangle $ABC$, the point $M$ is the midpoint of $BC$ and $N$ is a point on the side $BC$ such that $BN:NC=2:1$. The cotangents of the angles $\angle BAM$, $\angle MAN$, and $\angle NAC$ are positive integers $k,m,n$.
(a) Show that the cotangent of the angle $\angle BAC$ is also an integer and equals $m-k-n$.
(b) Show that there are infinitely many possible triples $(k,m,n)$, some of which consisting of Fibonacci numbers.
2 replies
Stear14
May 21, 2025
Stear14
5 minutes ago
Parallelograms and concyclicity
Lukaluce   33
N 17 minutes ago by HamstPan38825
Source: EGMO 2025 P4
Let $ABC$ be an acute triangle with incentre $I$ and $AB \neq AC$. Let lines $BI$ and $CI$ intersect the circumcircle of $ABC$ at $P \neq B$ and $Q \neq C$, respectively. Consider points $R$ and $S$ such that $AQRB$ and $ACSP$ are parallelograms (with $AQ \parallel RB, AB \parallel QR, AC \parallel SP$, and $AP \parallel CS$). Let $T$ be the point of intersection of lines $RB$ and $SC$. Prove that points $R, S, T$, and $I$ are concyclic.
33 replies
Lukaluce
Apr 14, 2025
HamstPan38825
17 minutes ago
IMO Shortlist 2013, Number Theory #4
lyukhson   30
N 19 minutes ago by Martin2001
Source: IMO Shortlist 2013, Number Theory #4
Determine whether there exists an infinite sequence of nonzero digits $a_1 , a_2 , a_3 , \cdots $ and a positive integer $N$ such that for every integer $k > N$, the number $\overline{a_k a_{k-1}\cdots a_1 }$ is a perfect square.
30 replies
lyukhson
Jul 10, 2014
Martin2001
19 minutes ago
v_p of factorials
TacH   9
N an hour ago by cursed_tangent1434
Source: InfinityDots MO 2 Problem 1
Determine whether there exists a finite set $S$ of primes such that for all positive integers $m$, there exists a positive integer $n$ and prime $p\in S$ such that $p^m\mid n!$ but $p^{m+1}\nmid n!$.

Proposed by TacH
9 replies
TacH
Apr 9, 2018
cursed_tangent1434
an hour ago
How Many Rooks can be Removed?
bluecarneal   10
N an hour ago by quantam13
Source: Fall 2005 Tournament of Towns Junior A-Level #3
Originally, every square of $8 \times 8$ chessboard contains a rook. One by one, rooks which attack an odd number of others are removed. Find the maximal number of rooks that can be removed. (A rook attacks another rook if they are on the same row or column and there are no other rooks between them.)

(6 points)
10 replies
bluecarneal
Mar 25, 2015
quantam13
an hour ago
silk road angle chasing , perpendiculars given, equal angles wanted
parmenides51   7
N 2 hours ago by Rayvhs
Source: SRMC 2019 P1
The altitudes of the acute-angled non-isosceles triangle $ ABC $ intersect at the point $ H $. On the segment $ C_1H $, where $ CC_1 $ is the altitude of the triangle, the point $ K $ is marked. Points $ L $ and $ M $ are the feet of perpendiculars from point $ K $ on straight lines $ AC $ and $ BC $, respectively. The lines $ AM $ and $ BL $ intersect at $ N $. Prove that $ \angle ANK = \angle HNL $.
7 replies
parmenides51
Jul 16, 2019
Rayvhs
2 hours ago
Find the value
sqing   5
N 2 hours ago by sqing
Source: Own
Let $ a,b $ be real numbers such that $ (a^2 + b^2) (a + 1) (b + 1) =  a ^ 3 + b ^ 3 =2 $. Find the value of $ a b .$

Let $ a,b $ be real numbers such that $ (a^2 + b^2) (a + 1) (b + 1) = 2 $ and $ a ^ 3 + b ^ 3 = 1 $. Find the value of $ a + b .$
5 replies
1 viewing
sqing
Yesterday at 2:29 PM
sqing
2 hours ago
2-var inequality
sqing   0
2 hours ago
Source: Own
Let $ a,b\geq  0 ,a+b+ab=2.$ Prove that
$$ (a^2+\frac{27}{5}ab+b^2)(a+1)(b+1) \leq 12 $$$$ (a^2+\frac{11}{2}ab+b^2)(a+1)(b+1) \leq 45(2-\sqrt 3) $$
0 replies
sqing
2 hours ago
0 replies
circumcenter of ARS lies on AD
Melid   1
N 2 hours ago by Acrylic3491
Source: own
In triangle $ABC$, let $D$ be a point on arc $BC$ of circle $ABC$ which doesn't contain $A$. $AD$ and $BC$ intersect at $E$. Let $P$ and $Q$ be the reflection of $E$ about to $AB$ and $AC$, respectively. $PD$ intersects $AB$ at $R$, and $QD$ intersects $AC$ at $S$. Prove that circumcenter of triangle $ARS$ lies on $AD$.
1 reply
Melid
Today at 9:30 AM
Acrylic3491
2 hours ago
2-var inequality
sqing   10
N 3 hours ago by sqing
Source: Own
Let $ a,b> 0 ,a^3+ab+b^3=3.$ Prove that
$$ (a+b)(a+1)(b+1) \leq 8$$$$ (a^2+b^2)(a+1)(b+1) \leq 8$$Let $ a,b> 0 ,a^3+ab(a+b)+b^3=3.$ Prove that
$$ (a+b)(a+1)(b+1) \leq \frac{3}{2}+\sqrt[3]{6}+\sqrt[3]{36}$$
10 replies
sqing
Yesterday at 1:35 PM
sqing
3 hours ago
Inspired by Czech-Polish-Slovak 2024
sqing   1
N 3 hours ago by sqing
Source: Own
Let $ a,b,c\geq 0, (a+1)(b+ c )=2025.$ Prove that$$ a+b^2+c\geq \frac{355}{4}$$Let $ a,b,c\geq 0, (a-1)(b+ c )=2025.$ Prove that$$ a+b^2+c\geq \frac{364}{4}$$Let $ a,b,c\geq 0, (a+ 1)(b- c )=2025.$ Prove that$$ a+b^2+c\geq \frac{135 \sqrt[3]{90}-2}{2}$$
1 reply
sqing
3 hours ago
sqing
3 hours ago
FE i created on bijective function with x≠y
benjaminchew13   8
N 3 hours ago by benjaminchew13
Source: own (probably)
Find all bijective functions $f:\mathbb{R}\to \mathbb{R}$ such that $$(x-y)f(x+f(f(y)))=xf(x)+f(y)^{2}$$for all $x,y\in \mathbb{R}$ such that $x\neq y$.
8 replies
benjaminchew13
5 hours ago
benjaminchew13
3 hours ago
Sum of divisors
Kimchiks926   3
N 3 hours ago by math-olympiad-clown
Source: Baltic Way 2022, Problem 17
Let $n$ be a positive integer such that the sum of its positive divisors is at least $2022n$. Prove that $n$ has at least $2022$ distinct prime factors.
3 replies
Kimchiks926
Nov 12, 2022
math-olympiad-clown
3 hours ago
Isogonal lines at the intersection of two circles
anantmudgal09   12
N Apr 29, 2025 by Tony_stark0094
Source: RMO Delhi 2016, P3
Two circles $C_1$ and $C_2$ intersect each other at points $A$ and $B$. Their external common tangent (closer to $B$) touches $C_1$ at $P$ and $C_2$ at $Q$. Let $C$ be the reflection of $B$ in line $PQ$. Prove that $\angle CAP=\angle BAQ$.
12 replies
anantmudgal09
Oct 11, 2016
Tony_stark0094
Apr 29, 2025
Isogonal lines at the intersection of two circles
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G H BBookmark kLocked kLocked NReply
Source: RMO Delhi 2016, P3
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anantmudgal09
1980 posts
#1 • 2 Y
Y by Adventure10, Rounak_iitr
Two circles $C_1$ and $C_2$ intersect each other at points $A$ and $B$. Their external common tangent (closer to $B$) touches $C_1$ at $P$ and $C_2$ at $Q$. Let $C$ be the reflection of $B$ in line $PQ$. Prove that $\angle CAP=\angle BAQ$.
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ThE-dArK-lOrD
4071 posts
#2 • 2 Y
Y by Adventure10, Mango247
Just angle-chasing.
Let $\angle{PAB}=x,\angle{QAB}=y\implies \angle{BPQ}=\angle{CPQ}=x,\angle{BQP}=\angle{CQP}=y$.
So $A,P,Q,C$ concyclic and thus $\angle{BAQ}=y=\angle{CQP}=\angle{CAP}$.
This post has been edited 1 time. Last edited by ThE-dArK-lOrD, Feb 10, 2019, 1:00 PM
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WizardMath
2487 posts
#3 • 3 Y
Y by amar_04, Adventure10, Mango247
The A-Humpty point lemma.
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jayme
9801 posts
#4 • 1 Y
Y by Adventure10
Dear Mathlinkers,
a link with

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=9076.

can be helpfull.

Sincerely
Jean-Louis
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Anar24
475 posts
#5 • 2 Y
Y by Adventure10, Mango247
ThE-dArK-lOrD wrote:
Just angle-chasing, we let $\angle{PAB}=x,\angle{QAB}=y\Rightarrow \angle{BPQ}=\angle{CPQ}=x,\angle{BQP}=\angle{CQP}=y$
So $A,P,Q,C$ concyclic, we get $\angle{BAQ}=y=\angle{CQP}=\angle{CAP}$, we are done.
ThE-dArK-lOrD wrote:
Just angle-chasing, we let $\angle{PAB}=x,\angle{QAB}=y\Rightarrow \angle{BPQ}=\angle{CPQ}=x,\angle{BQP}=\angle{CQP}=y$
So $A,P,Q,C$ concyclic, we get $\angle{BAQ}=y=\angle{CQP}=\angle{CAP}$, we are done.

can you explain why A,P,Q and C are concyclic?
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Anar24
475 posts
#6 • 1 Y
Y by Adventure10
WizardMath wrote:
The A-Humpty point lemma.

can you show me what is that lemma,i couldn't find the source!!!
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WizardMath
2487 posts
#7 • 2 Y
Y by Adventure10, Mango247
Actually, the name is not that well known, but anyways, here's the link:
https://drive.google.com/file/d/0B3gLVLnxtyRvejlpenRmZGh6SDQ/view
See the one in the section on Symmedian related properties.
This post has been edited 1 time. Last edited by WizardMath, Oct 11, 2016, 1:56 PM
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Ayushakj
240 posts
#8 • 2 Y
Y by Adventure10, Mango247
please add figure here.
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Euler149
389 posts
#10 • 3 Y
Y by hansu, Adventure10, Mango247
http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYS8yLzQ1ZTcyZDUyOTE1YTE0MDM2ZGRjZmZiZDE4NzM2N2RmNjM2OGY1LmpwZWc=&rn=aW1hZ2UuanBlZw==
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kk108
2649 posts
#11 • 1 Y
Y by Adventure10
anantmudgal09 wrote:
Two circles $C_1$ and $C_2$ intersect each other at points $A$ and $B$. Their external common tangent (closer to $B$) touches $C_1$ at $P$ and $C_2$ at $Q$. Let $C$ be the reflection of $B$ in line $PQ$. Prove that $\angle CAP=\angle BAQ$.

What is meant by the reflection of a point in a line ??
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AlastorMoody
2125 posts
#12 • 2 Y
Y by Adventure10, Mango247
Very easy to proceed, after noticing that $B$ is the $A-\text{HM}$ point in $\Delta APQ$, Anyways, here's a solution with just angle chasing,

$\angle PCQ=\angle PBQ=180^{\circ}-\angle BPQ - \angle BQP=180^{\circ}-\angle BAP-\angle BAQ=180^{\circ}-A$ $\implies$ $APCQ$ is cyclic, $\angle BAQ$ $=$ $\angle BQP$ $=$ $\angle PQC$ $=$ $\angle PAC$
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Jishnu4414l
155 posts
#13
Y by
Here's a solution that uses only angle chasing.
Let $\angle BPQ=\theta$, and $\angle BQP=\phi$.
Construct the lines $CP$ and $CQ$.
We have by congruency, $\angle PQB=\angle PQC=\phi$.
Similarly, we have $\angle QPB-\angle QPC=\theta$.
Now because $PQ$ is tangent to $\mathbf{C_1}$ and $\mathbf{C_2}$ at $P$ and $Q$ respectively, we have $\angle PAB=\theta$ and $\angle BAQ=\phi$.
But $\angle PCQ=180^{\circ}-\theta-\phi$.
So, we have that points $A,P,C,Q$ are concyclic.
Now, we have $\angle PQC=\angle PAC=\angle CAP=\phi$, and $\angle BAQ=\phi$ too.
Our proof is thus complete.
This post has been edited 1 time. Last edited by Jishnu4414l, Dec 14, 2023, 4:02 PM
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Tony_stark0094
69 posts
#14
Y by
let $BC \cap PQ = D$ and we know $DB*DA=DP^2=DP*DQ$ $\implies$ $PACQ$ is cyclic
so $\angle CAP = \angle CQP = \angle BQP = \angle BAQ$ hence done
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