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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
Crossing ٍٍChords
matinyousefi   1
N 21 minutes ago by Trenod
Source: Iranian Combinatorics Olympiad 2020 P3
$1399$ points and some chords between them is given.
$a)$ In every step we can take two chords $RS,PQ$ with a common point other than $P,Q,R,S$ and erase exactly one of $RS,PQ$ and draw $PS,PR,QS,QR$ let $s$ be the minimum of chords after some steps. Find the maximum of $s$ over all initial positions.
$b)$ In every step we can take two chords $RS,PQ$ with a common point other than $P,Q,R,S$ and erase both of $RS,PQ$ and draw $PS,PR,QS,QR$ let $s$ be the minimum of chords after some steps. Find the maximum of $s$ over all initial positions.

Proposed by Afrouz Jabalameli, Abolfazl Asadi
1 reply
matinyousefi
Apr 24, 2020
Trenod
21 minutes ago
Nice NT with powers of two
oVlad   7
N 36 minutes ago by SimplisticFormulas
Source: Romania TST 2024 Day 1 P3
Let $n{}$ be a positive integer and let $a{}$ and $b{}$ be positive integers congruent to 1 modulo 4. Prove that there exists a positive integer $k{}$ such that at least one of the numbers $a^k-b$ and $b^k-a$ is divisible by $2^n.$

Cătălin Liviu Gherghe
7 replies
oVlad
Jul 31, 2024
SimplisticFormulas
36 minutes ago
Inequality in triangle
Nguyenhuyen_AG   0
an hour ago
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
\[\frac{1}{(a-4b)^2}+\frac{1}{(b-4c)^2}+\frac{1}{(c-4a)^2} \geqslant \frac{1}{ab+bc+ca}.\]
0 replies
Nguyenhuyen_AG
an hour ago
0 replies
D,E,F are collinear.
TUAN2k8   2
N an hour ago by TUAN2k8
Source: Own
Help me with this:
2 replies
TUAN2k8
May 28, 2025
TUAN2k8
an hour ago
Combinatorial identity
MehdiGolafshan   4
N an hour ago by watery
Let $n$ is a positive integer. Prove that
$$\sum_{k=0}^{n-1}\frac{1}{k+1}\binom{n-1}{k} = \frac{2^n-1}{n}.$$
4 replies
1 viewing
MehdiGolafshan
Jan 16, 2023
watery
an hour ago
JBMO Shortlist 2023 G7
Orestis_Lignos   7
N 2 hours ago by tilya_TASh
Source: JBMO Shortlist 2023, G7
Let $D$ and $E$ be arbitrary points on the sides $BC$ and $AC$ of triangle $ABC$, respectively. The circumcircle of $\triangle ADC$ meets for the second time the circumcircle of $\triangle BCE$ at point $F$. Line $FE$ meets line $AD$ at point $G$, while line $FD$ meets line $BE$ at point $H$. Prove that lines $CF, AH$ and $BG$ pass through the same point.
7 replies
Orestis_Lignos
Jun 28, 2024
tilya_TASh
2 hours ago
Reflected point lies on radical axis
Mahdi_Mashayekhi   5
N 2 hours ago by Mahdi_Mashayekhi
Source: Iran 2025 second round P4
Given is an acute and scalene triangle $ABC$ with circumcenter $O$. $BO$ and $CO$ intersect the altitude from $A$ to $BC$ at points $P$ and $Q$ respectively. $X$ is the circumcenter of triangle $OPQ$ and $O'$ is the reflection of $O$ over $BC$. $Y$ is the second intersection of circumcircles of triangles $BXP$ and $CXQ$. Show that $X,Y,O'$ are collinear.
5 replies
Mahdi_Mashayekhi
Apr 19, 2025
Mahdi_Mashayekhi
2 hours ago
Find the value
sqing   18
N 2 hours ago by Yiyj
Source: 2024 China Fujian High School Mathematics Competition
Let $f(x)=a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0,$ $a_i\in\{-1,1\} ,i=0,1,2,\cdots,6 $ and $f(2)=-53 .$ Find the value of $f(1).$
18 replies
sqing
Jun 22, 2024
Yiyj
2 hours ago
Number Theory
fasttrust_12-mn   14
N 3 hours ago by Namisgood
Source: Pan African Mathematics Olympiad P1
Find all positive intgers $a,b$ and $c$ such that $\frac{a+b}{a+c}=\frac{b+c}{b+a}$ and $ab+bc+ca$ is a prime number
14 replies
fasttrust_12-mn
Aug 15, 2024
Namisgood
3 hours ago
find question
mathematical-forest   5
N 3 hours ago by Jupiterballs
Are there any contest questions that seem simple but are actually difficult? :-D
5 replies
mathematical-forest
Thursday at 10:19 AM
Jupiterballs
3 hours ago
Own made functional equation
Primeniyazidayi   10
N 3 hours ago by Phat_23000245
Source: own(probably)
Find all functions $f:R \rightarrow R$ such that $xf(x^2+2f(y)-yf(x))=f(x)^3-f(y)(f(x^2)-2f(x))$ for all $x,y \in \mathbb{R}$
10 replies
Primeniyazidayi
May 26, 2025
Phat_23000245
3 hours ago
Tough inequality
TUAN2k8   4
N 3 hours ago by Phat_23000245
Source: Own
Let $n \ge 2$ be an even integer and let $x_1,x_2,...,x_n$ be real numbers satisfying $x_1^2+x_2^2+...+x_n^2=n$.
Prove that
$\sum_{1 \le i < j \le n} \frac{x_ix_j}{x_i^2+x_j^2+1} \ge \frac{-n}{6}$
4 replies
TUAN2k8
May 28, 2025
Phat_23000245
3 hours ago
Guess period of function
a1267ab   9
N 4 hours ago by HamstPan38825
Source: USA TST 2025
Let $n$ be a positive integer. Ana and Banana play a game. Banana thinks of a function $f\colon\mathbb{Z}\to\mathbb{Z}$ and a prime number $p$. He tells Ana that $f$ is nonconstant, $p<100$, and $f(x+p)=f(x)$ for all integers $x$. Ana's goal is to determine the value of $p$. She writes down $n$ integers $x_1,\dots,x_n$. After seeing this list, Banana writes down $f(x_1),\dots,f(x_n)$ in order. Ana wins if she can determine the value of $p$ from this information. Find the smallest value of $n$ for which Ana has a winning strategy.

Anthony Wang
9 replies
a1267ab
Dec 14, 2024
HamstPan38825
4 hours ago
Inequality with abc=1
tenplusten   11
N 4 hours ago by sqing
Source: JBMO 2011 Shortlist A7
$\boxed{\text{A7}}$ Let $a,b,c$ be positive reals such that $abc=1$.Prove the inequality $\sum\frac{2a^2+\frac{1}{a}}{b+\frac{1}{a}+1}\geq 3$
11 replies
tenplusten
May 15, 2016
sqing
4 hours ago
Three collinear points
andria   8
N Oct 8, 2020 by Ali3085
Source: Iranian third round geometry problem 4
Let $ABC$ be a triangle with incenter $I$. Let $K$ be the midpoint of $AI$ and $BI\cap \odot(\triangle ABC)=M,CI\cap \odot(\triangle ABC)=N$. points $P,Q$ lie on $AM,AN$ respectively such that $\angle ABK=\angle PBC,\angle ACK=\angle QCB$. Prove that $P,Q,I$ are collinear.
8 replies
andria
Sep 10, 2015
Ali3085
Oct 8, 2020
Three collinear points
G H J
G H BBookmark kLocked kLocked NReply
Source: Iranian third round geometry problem 4
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andria
824 posts
#1 • 3 Y
Y by Ali3085, Adventure10, Mango247
Let $ABC$ be a triangle with incenter $I$. Let $K$ be the midpoint of $AI$ and $BI\cap \odot(\triangle ABC)=M,CI\cap \odot(\triangle ABC)=N$. points $P,Q$ lie on $AM,AN$ respectively such that $\angle ABK=\angle PBC,\angle ACK=\angle QCB$. Prove that $P,Q,I$ are collinear.
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Luis González
4149 posts
#2 • 9 Y
Y by andria, ATimo, ali.agh, A_Math_Lover, Mathuzb, amar_04, hakN, Adventure10, Mango247
Let $L \equiv AI \cap BC,$ $D \equiv MN \cap BC,$ $Y \equiv AM \cap BC$ and $U \equiv BP \cap AI.$ As $MN$ is the perpendicular bisector of $AI$ (well-known), then $K \in MN.$ Since $BU,BL$ are the reflections of $BK,BA$ across $BI,$ we deduce that $(L,I,A,K)=(A,K,L,I)=(L,U,A,I)=B(L,U,A,I)=(Y,P,A,M)$ $\Longrightarrow$ $D(L,I,A,K)=D(Y,P,A,M)$ $\Longrightarrow$ $P \in DI.$ By similar reasoning $Q \in DI$ $\Longrightarrow$ $P,Q,I$ are collinear.
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buratinogigle
2401 posts
#3 • 2 Y
Y by Adventure10, Mango247
I have seen general problem

Let $ABC$ be a triangle inscribed in circle $(O)$ and $P,Q$ are two isogonal conjugate points. $PB,PC$ cut $(O)$ again at $M,N$. $QA$ cuts $MN$ at $K$. $L$ is isogonal conjugate of $K$. $LB,LC$ cut $AM,AN$ at $S,T$, resp. Prove that $S,Q,T$ are collinear.
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Luis González
4149 posts
#4 • 3 Y
Y by buratinogigle, Adventure10, Mango247
buratinogigle wrote:
I have seen general problem

Let $ABC$ be a triangle inscribed in circle $(O)$ and $P,Q$ are two isogonal conjugate points. $PB,PC$ cut $(O)$ again at $M,N.$ $QA$ cuts $MN$ at $K.$ $L$ is isogonal conjugate of $K.$ $LB,LC$ cut $AM,AN$ at $S,T,$ resp. Prove that $S,Q,T$ are collinear.

The proof to this generalization is very similar to what I did in my previous solution. Letting $D \equiv MN \cap BC,$ $Y \equiv AM \cap BC,$ $U \equiv AQ \cap BC$ and $ V \equiv AP \cap BC,$ we get $B(Y,S,A,M)=B(V,L,A,P)=(A,K,U,Q)=(U,Q,A,K)$ $\Longrightarrow$ $D(Y,S,A,M)=D(U,Q,A,K)$ $\Longrightarrow$ $S \in DQ$ and likewise $T \in DQ$ $\Longrightarrow$ $S,Q,T$ are collinear.
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K.N
532 posts
#5 • 2 Y
Y by Adventure10, Mango247
Hello
Please give some other solutions to see different ideas
I have got that $BP,CQ$ interest each other on $AI$ At a point like $S$ and we can easily get that $(I_{a}ISK)=-1$
Can this be useful?!
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H.HAFEZI2000
328 posts
#8 • 2 Y
Y by Adventure10, Mango247
we'll solve the problem using bary coordinates. let $\triangle ABC$ be the refrence triangle and let $k'=(QC,BP)$ and literally $K,K'$ are isogonal conjugates and we have that: $K=(2a+b+c:b:c) \implies K'=(\frac{a^2}{2a+b+c}:b:c)$ and now we can easily calculate $Q$ from cevians $CK',AN$ and already know that: $N=(a:b:\frac{-c^2}{a+b}), M=(a:\frac{-b^2}{a+c}:c) \implies Q=(\frac{a^2}{2a+b+c}:b:\frac{-c^2}{a+b}), P=(\frac{a^2}{2a+b+c}:\frac{-b^2}{a+c}:c)$ and it's pretty much straightforward to show

$\begin{vmatrix}
\frac{a^2}{2a+b+c} & b &\frac{-c^2}{a+b} \\ 
a & b &c \\ 
\frac{a^2}{2a+b+c} & \frac{-b^2}{a+c} & c \\ 
\end{vmatrix}=0$
This post has been edited 1 time. Last edited by H.HAFEZI2000, Aug 5, 2018, 10:39 AM
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AlastorMoody
2125 posts
#9 • 2 Y
Y by Adventure10, Mango247
Iran MO Round 3 2015 G4 wrote:
Let $ABC$ be a triangle with incenter $I$. Let $K$ be the midpoint of $AI$ and $BI\cap \odot(\triangle ABC)=M,CI\cap \odot(\triangle ABC)=N$. points $P,Q$ lie on $AM,AN$ respectively such that $\angle ABK=\angle PBC,\angle ACK=\angle QCB$. Prove that $P,Q,I$ are collinear.
Solution (with Aryan-23): Let $MN \cap BC=S$, $AN \cap BC=E$ & $AI$ $\cap$ $BC$ $=$ $U$. Let $K'$ be the isogonal conjugate of $K$ WRT $\Delta ABC$
$$(A,Q;N,E) \overset{C}{=} (A,K';I,U) =(U,K;I,A)=(A,I;K,U)$$Hence, $S,P,I$ collinear and similarly, $Q,I,S$ collinear $\qquad \blacksquare$
This post has been edited 2 times. Last edited by Luis González, Oct 6, 2019, 2:47 AM
Reason: Unhiding solution
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Arefe
65 posts
#11 • 1 Y
Y by AlastorMoody
Let $B' , C'$ be points on $BK , CK$ such that $AB'||BI$ and $AC'||CI$ , so $CK=CK'$ and $BK=BK'$ then $BC'||CB'$ and we can get $\angle{ABC'}+\angle{ACB'}=\angle{BAC}$ $(1)$
It's easy to check that $P , B'$ are isogonal conjucate and $Q , C'$ are , too .
If $ S $ is intersection of $QB , CP$ , we can check easily from $(1)$ that $ S $ is on the circle of $ABC$ .
So with pascal on $NCSBMA$ we can get $I , P , Q $ are collinear $\blacksquare$
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Ali3085
214 posts
#12 • 1 Y
Y by Stuffybear
isogonality lemma kills it :D
Let $MN \cap BC =R$
note that $(CK,CQ) , (CA,CR)$ are pairs of isogonal line wrt $\angle BCA$
since $N=AQ \cap KR $ we have that $RQ \cap AK$ is isogonal to $AN$ so $RQ \cap AK ,A,N,I$ are collinear
so $I=RQ \cap AK$
similarly $I=PQ \cap AK$
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