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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.

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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
D1036 : Composition of polynomials
Dattier   1
N a few seconds ago by Dattier
Source: les dattes à Dattier
Find all $A \in \mathbb Q[x]$ with $\exists Q \in \mathbb Q[x], Q(A(x))= x^{2025!+2}+x^2+x+1$ and $\deg(A)>1$.
1 reply
Dattier
Yesterday at 1:52 PM
Dattier
a few seconds ago
number sequence contains every large number
mathematics2003   3
N a few seconds ago by sttsmet
Source: 2021ChinaTST test3 day1 P2
Given distinct positive integer $ a_1,a_2,…,a_{2020} $. For $ n \ge 2021 $, $a_n$ is the smallest number different from $a_1,a_2,…,a_{n-1}$ which doesn't divide $a_{n-2020}...a_{n-2}a_{n-1}$. Proof that every number large enough appears in the sequence.
3 replies
mathematics2003
Apr 13, 2021
sttsmet
a few seconds ago
IMO ShortList 2002, geometry problem 2
orl   28
N 4 minutes ago by ezpotd
Source: IMO ShortList 2002, geometry problem 2
Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \]
28 replies
orl
Sep 28, 2004
ezpotd
4 minutes ago
Arc Midpoints Form Cyclic Quadrilateral
ike.chen   56
N 11 minutes ago by ezpotd
Source: ISL 2022/G2
In the acute-angled triangle $ABC$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $AF$. The lines through $P$ parallel to $AC$ and $AB$ meet $BC$ at $D$ and $E$, respectively. Points $X \ne A$ and $Y \ne A$ lie on the circles $ABD$ and $ACE$, respectively, such that $DA = DX$ and $EA = EY$.
Prove that $B, C, X,$ and $Y$ are concyclic.
56 replies
ike.chen
Jul 9, 2023
ezpotd
11 minutes ago
Non-linear Recursive Sequence
amogususususus   4
N 22 minutes ago by GreekIdiot
Given $a_1=1$ and the recursive relation
$$a_{i+1}=a_i+\frac{1}{a_i}$$for all natural number $i$. Find the general form of $a_n$.

Is there any way to solve this problem and similar ones?
4 replies
amogususususus
Jan 24, 2025
GreekIdiot
22 minutes ago
Russian Diophantine Equation
LeYohan   2
N 24 minutes ago by RagvaloD
Source: Moscow, 1963
Find all integer solutions to

$\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3$.
2 replies
LeYohan
Yesterday at 2:59 PM
RagvaloD
24 minutes ago
PQ = r and 6 more conditions
avisioner   41
N 28 minutes ago by ezpotd
Source: 2023 ISL G2
Let $ABC$ be a triangle with $AC > BC,$ let $\omega$ be the circumcircle of $\triangle ABC,$ and let $r$ be its radius. Point $P$ is chosen on $\overline{AC}$ such taht $BC=CP,$ and point $S$ is the foot of the perpendicular from $P$ to $\overline{AB}$. Ray $BP$ mets $\omega$ again at $D$. Point $Q$ is chosen on line $SP$ such that $PQ = r$ and $S,P,Q$ lie on a line in that order. Finally, let $E$ be a point satisfying $\overline{AE} \perp \overline{CQ}$ and $\overline{BE} \perp \overline{DQ}$. Prove that $E$ lies on $\omega$.
41 replies
avisioner
Jul 17, 2024
ezpotd
28 minutes ago
Functional equation from R^2 to R
k.vasilev   19
N 29 minutes ago by megahertz13
Source: All-Russian Olympiad 2019 grade 10 problem 1
Each point $A$ in the plane is assigned a real number $f(A).$ It is known that $f(M)=f(A)+f(B)+f(C),$ whenever $M$ is the centroid of $\triangle ABC.$ Prove that $f(A)=0$ for all points $A.$
19 replies
k.vasilev
Apr 23, 2019
megahertz13
29 minutes ago
Functional equations in IMO TST
sheripqr   50
N 37 minutes ago by megahertz13
Source: Iran TST 1996
Find all functions $f: \mathbb R \to \mathbb R$ such that $$ f(f(x)+y)=f(x^2-y)+4f(x)y $$ for all $x,y \in \mathbb R$
50 replies
sheripqr
Sep 14, 2015
megahertz13
37 minutes ago
JBMO TST Bosnia and Herzegovina 2023 P4
FishkoBiH   1
N an hour ago by Maths_VC
Source: JBMO TST Bosnia and Herzegovina 2023 P4
Let $n$ be a positive integer. A board with a format $n*n$ is divided in $n*n$ equal squares.Determine all integers $n$3 such that the board can be covered in $2*1$ (or $1*2$) pieces so that there is exactly one empty square in each row and each column.
1 reply
FishkoBiH
Today at 1:38 PM
Maths_VC
an hour ago
IMO Shortlist 2009 - Problem C3
nsato   25
N an hour ago by popop614
Let $n$ be a positive integer. Given a sequence $\varepsilon_1$, $\dots$, $\varepsilon_{n - 1}$ with $\varepsilon_i = 0$ or $\varepsilon_i = 1$ for each $i = 1$, $\dots$, $n - 1$, the sequences $a_0$, $\dots$, $a_n$ and $b_0$, $\dots$, $b_n$ are constructed by the following rules: \[a_0 = b_0 = 1, \quad a_1 = b_1 = 7,\] \[\begin{array}{lll}
	a_{i+1} = 
	\begin{cases}
		2a_{i-1} + 3a_i, \\
		3a_{i-1} + a_i, 
	\end{cases} & 
        \begin{array}{l} 
                \text{if } \varepsilon_i = 0, \\  
                \text{if } \varepsilon_i = 1, \end{array} 
         & \text{for each } i = 1, \dots, n - 1, \\[15pt]
        b_{i+1}= 
        \begin{cases}
		2b_{i-1} + 3b_i, \\
		3b_{i-1} + b_i, 
	\end{cases} & 
        \begin{array}{l} 
                \text{if } \varepsilon_{n-i} = 0, \\  
                \text{if } \varepsilon_{n-i} = 1, \end{array} 
         & \text{for each } i = 1, \dots, n - 1.
	\end{array}\] Prove that $a_n = b_n$.

Proposed by Ilya Bogdanov, Russia
25 replies
nsato
Jul 6, 2010
popop614
an hour ago
JBMO TST Bosnia and Herzegovina 2024 P4
FishkoBiH   1
N an hour ago by TopGbulliedU
Source: JBMO TST Bosnia and Herzegovina 2024 P4
Let $m$ and $n$ be natural numbers. Every one of the $m*n$ squares of the $m*n$ board is colored either black or white, so that no 2 neighbouring squares are the same color(the board is colored like in chess").In one step we can pick 2 neighbouring squares and change their colors like this:
- a white square becomes black;
-a black square becomes blue;
-a blue square becomes white.
For which $m$ and $n$ can we ,in a finite sequence of these steps, switch the starting colors from white to black and vice versa.
1 reply
FishkoBiH
6 hours ago
TopGbulliedU
an hour ago
Van der Corput Inequality
EthanWYX2009   1
N 2 hours ago by grupyorum
Source: en.wikipedia.org/wiki/Van_der_Corput_inequality
Let $V$ be a real or complex inner product space. Suppose that ${\displaystyle v,u_{1},\dots ,u_{n}\in V} $ and that ${\displaystyle \|v\|=1}.$ Then$${\displaystyle \displaystyle \left(\sum _{i=1}^{n}|\langle v,u_{i}\rangle |\right)^{2}\leq \sum _{i,j=1}^{n}|\langle u_{i},u_{j}\rangle |.}$$
1 reply
EthanWYX2009
Today at 3:36 AM
grupyorum
2 hours ago
two sequences of positive integers and inequalities
rmtf1111   52
N 2 hours ago by Kappa_Beta_725
Source: EGMO 2019 P5
Let $n\ge 2$ be an integer, and let $a_1, a_2, \cdots , a_n$ be positive integers. Show that there exist positive integers $b_1, b_2, \cdots, b_n$ satisfying the following three conditions:

$\text{(A)} \ a_i\le b_i$ for $i=1, 2, \cdots , n;$

$\text{(B)} \ $ the remainders of $b_1, b_2, \cdots, b_n$ on division by $n$ are pairwise different; and

$\text{(C)} \ $ $b_1+b_2+\cdots b_n \le n\left(\frac{n-1}{2}+\left\lfloor \frac{a_1+a_2+\cdots a_n}{n}\right \rfloor \right)$

(Here, $\lfloor x \rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)
52 replies
rmtf1111
Apr 10, 2019
Kappa_Beta_725
2 hours ago
Geo with unnecessary condition
egxa   8
N Apr 4, 2025 by ErTeeEs06
Source: Turkey Olympic Revenge 2024 P4
Let the circumcircle of a triangle $ABC$ be $\Gamma$. The tangents to $\Gamma$ at $B,C$ meet at point $E$. For a point $F$ on line $BC$ which is not on the segment $BC$, let the midpoint of $EF$ be $G$. Lines $GB,GC$ meet $\Gamma$ again at points $I,H$ respectively. Let $M$ be the midpoint of $BC$. Prove that the points $F,I,H,M$ lie on a circle.

Proposed by Mehmet Can Baştemir
8 replies
egxa
Aug 6, 2024
ErTeeEs06
Apr 4, 2025
Geo with unnecessary condition
G H J
G H BBookmark kLocked kLocked NReply
Source: Turkey Olympic Revenge 2024 P4
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egxa
211 posts
#1 • 2 Y
Y by SerdarBozdag, Rounak_iitr
Let the circumcircle of a triangle $ABC$ be $\Gamma$. The tangents to $\Gamma$ at $B,C$ meet at point $E$. For a point $F$ on line $BC$ which is not on the segment $BC$, let the midpoint of $EF$ be $G$. Lines $GB,GC$ meet $\Gamma$ again at points $I,H$ respectively. Let $M$ be the midpoint of $BC$. Prove that the points $F,I,H,M$ lie on a circle.

Proposed by Mehmet Can Baştemir
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sami1618
913 posts
#3
Y by
Consider the linear function $f(X)=\mathbb{P}(X,\Gamma)-\mathbb{P}(X,G)$. Then $$\mathbb{P}(G,\Gamma)=f(G)=\frac{f(E)+f(F)}{2}=\frac{BE^2+BF \cdot CF -EG^2-FG^2}{2}=$$$$\frac{(BM^2+BF\cdot CF+EM^2)-2EG^2}{2}=\frac{(FM^2+EM^2)-2EG^2}{2}=\frac{EF^2-2EG^2}{2}=EG$$Thus inverting about $G$ with radius $EG$ maps $F$, $I$, $H$, and $M$ all to line $BC$, finishing the problem.
Attachments:
This post has been edited 1 time. Last edited by sami1618, Aug 6, 2024, 1:45 PM
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SerdarBozdag
892 posts
#4 • 4 Y
Y by alexanderhamilton124, bin_sherlo, egxa, ehuseyinyigit
$\angle FME = 90, GF = GE$ implies $GF = GM = GE$. Note that $G$ is on the radical axis of the point circle $(E)$ and $(ABC)$ because it lies on the $E$-midline of the triangle $EBC$. This gives
$GF^2 = GM^2 = GB \cdot GI = GH \cdot GC$ and because $B,C \in FM$; $F,I,M,H,G$ are cyclic.
This post has been edited 2 times. Last edited by SerdarBozdag, Aug 6, 2024, 2:23 PM
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hukilau17
288 posts
#5 • 1 Y
Y by ehuseyinyigit
We complex bash. Let $\Gamma$ be the unit circle (and let $j$, rather than $i$, denote the coordinate of $I$), so that
$$|a|=|b|=|c|=1$$$$e=\frac{2bc}{b+c}$$$$m=\frac{b+c}2$$$$\overline{f} = \frac{b+c-f}{bc}$$$$g = \frac{e+f}2 = \frac{2bc+bf+cf}{2(b+c)}$$$$\overline{g} = \frac{\frac2{bc}+\frac{b+c-f}{b^2c}+\frac{b+c-f}{bc^2}}{2\left(\frac1b+\frac1c\right)} = \frac{b^2+4bc+c^2-bf-cf}{2bc(b+c)}$$$$j = \frac{b-g}{b\overline{g}-1} = \frac{\frac{2b^2-bf-cf}{2(b+c)}}{\frac{b^2+2bc-c^2-bf-cf}{2c(b+c)}} = \frac{c(2b^2-bf-cf)}{b^2+2bc-c^2-bf-cf}$$$$h = \frac{c-g}{c\overline{g}-1} = \frac{\frac{2c^2-bf-cf}{2(b+c)}}{\frac{-b^2+2bc+c^2-bf-cf}{2b(b+c)}} = \frac{b(2c^2-bf-cf)}{-b^2+2bc+c^2-bf-cf}$$$$j-f = \frac{bf^2+cf^2-b^2f-3bcf+2b^2c}{b^2+2bc-c^2-bf-cf} = \frac{(b-f)(2bc-bf-cf)}{b^2+2bc-c^2-bf-cf}$$$$h-f = \frac{bf^2+cf^2-c^2f-3bcf+2bc^2}{-b^2+2bc+c^2-bf-cf} = \frac{(c-f)(2bc-bf-cf)}{-b^2+2bc+c^2-bf-cf}$$$$j-m = \frac{-b^3+b^2c-bc^2+c^3+b^2f-c^2f}{2(b^2+2bc-c^2-bf-cf)} = \frac{(c-b)(b^2+c^2-bf-cf)}{2(b^2+2bc-c^2-bf-cf)}$$$$h-m = \frac{b^3-b^2c+bc^2-c^3-b^2f+c^2f}{2(-b^2+2bc+c^2-bf-cf)} = \frac{(b-c)(b^2+c^2-bf-cf)}{2(-b^2+2bc+c^2-bf-cf)}$$Thus
$$\frac{(j-f)(h-m)}{(h-f)(j-m)} = -\frac{b-f}{c-f}$$which is real since $F$ lies on line $BC$. $\blacksquare$
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ohhh
48 posts
#6
Y by
Fix triangle $ABC$ and move the point $F$ along the line $BC$ (Yes, ignore the "which is not on the segment" condition).
$degF = 1 \implies degG  = 1 \implies degH = degI =1$
Notice that the condition $(IMGF)$ cyclic has degree at most $degGI + degMI + degMF +deg GF= 1 + 1 + 0 + 1= 3$.
So $4$ points are enough!
$1: F = B$
$2: F = M$
$3: F$ such that $G = M_{bc}$
$4: F$ such that $G = M_a$
Okay! Similarly, we can prove $(HMGF)$ and consequently $(IMHGF)$, as desired.
This post has been edited 3 times. Last edited by ohhh, Aug 7, 2024, 12:41 AM
Reason: Typos
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Anancibedih
18 posts
#7 • 2 Y
Y by ehuseyinyigit, Ege_Saribass
Let $T$ be the point that satisfies $FBET$ parallelogram. $\angle{TFB}=\angle{EBC}=\angle{BCE},FC||TE\Rightarrow \hspace{1mm}\text{FTEC is cyclic}$ $\angle{TEC}+\angle{FCE}=180^\circ=\angle{TEC}+\angle{TIC}\Rightarrow \text{I lies on FTEC circle}$. So $FB\cdot BC=2FB\cdot BM=IB\cdot BT=2BG\cdot IB\Rightarrow \text{FGMI is cyclic}$. We know that $FG=GM$ so $\angle{FIG}=\angle {GIM}=\angle{FMG}$ so $\triangle {GMB}\sim \triangle {GIM}\Rightarrow GM^2=GB\cdot GI=GH\cdot GC$ so $\angle{GMF}=\angle{MHC}$ and we get $\angle{GIM}+\angle{GHM}=180^\circ$ so $H$ lies on $FGMI$ circle. We're done.
This post has been edited 3 times. Last edited by Anancibedih, Aug 21, 2024, 2:01 PM
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khanhnx
1618 posts
#8 • 1 Y
Y by MS_asdfgzxcvb
We have $BC$ is polar of $E$ WRT $\Gamma$ so $\bigodot(EF)$ is orthogonal to $\Gamma$. From this, we have $\mathcal{I}^{k = GM^2}_G: \Gamma \longleftrightarrow \Gamma$. Hence $\mathcal{I}^{k = GM^2}_G: F \longleftrightarrow F, M \longleftrightarrow M, B \longleftrightarrow I, C \longleftrightarrow H$. But $B, C, M, F$ are collinear then $G, I, H, F, M$ lie on a circle
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ehuseyinyigit
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#10
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The most natural way is constructing parallelograms. Let $C'$ and $B'$ be reflection of the points $C$ and $B$ wrt to $G$ respectively. Then $\angle BHC=\angle BAC=\angle BCE=\angle BFC'$ implies $BHC'F$ is cyclic and thus $FHMG$ (Since $BC.CF=2MC.CF=C'C.CH=2GC.CH$). On the other hand, $BE=EC$ implies $CEB'F$ is cyclic-isosceles trapezoid. Furthermore $\angle BHC=\angle CIB'=\angle CEB'$ gives $B'EICF$ is cyclic. Then PoP gives $BC.BF=2BM.BF=BI.BB'=2BI.BG$ implies $FGMI$ is cyclic. The result follows from $(FGMI)$ and $(FHMG)$ as desired.
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ErTeeEs06
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$G$ is on the radax of $(ABC)$ and pointcircle $E$, since that radax is the midparallel in triangle $EBC$. So inversion with radius $GE$ around $G$ sends $H$ to $C$ and $I$ to $B$. Now since $G$ is midpoint of $EF$ and $\angle EMF=90^\circ$ we see that $GE=GM=GF$ so $M$ and $F$ remain fixed. So the images of $F, I, H, M$ are collinear so $F, I, H, M$ are concyclic.
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