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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 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
angle wanted, right ABC, AM=CB , CN=MB
parmenides51   2
N 14 minutes ago by Tsikaloudakis
Source: 2022 European Math Tournament - Senior First + Grand League - Math Battle 1.3
In a right-angled triangle $ABC$, points $M$ and $N$ are taken on the legs $AB$ and $BC$, respectively, so that $AM=CB$ and $CN=MB$. Find the acute angle between line segments $AN$ and $CM$.
2 replies
parmenides51
Dec 19, 2022
Tsikaloudakis
14 minutes ago
Inspired by Czech-Polish-Slovak 2017
sqing   4
N 15 minutes ago by sqing
Let $x, y$ be real numbers. Prove that
$$\frac{(xy+1)(x + 2y)}{(x^2 + 1)(2y^2 + 1)} \leq \frac{3}{2\sqrt 2}$$$$\frac{(xy+1)(x +3y)}{(x^2 + 1)(3y^2 + 1)} \leq \frac{2}{ \sqrt 3}$$$$\frac{(xy - 1)(x + 2y)}{(x^2 + 1)(2y^2 + 1)} \leq \frac{1}{\sqrt 2}$$$$\frac{(xy - 1)(x + 3y)}{(x^2 + 1)(3y^2 + 1)} \leq \frac{\sqrt 3}{2}$$
4 replies
sqing
3 hours ago
sqing
15 minutes ago
interesting inequality
pennypc123456789   1
N 28 minutes ago by Quantum-Phantom
Let \( a,b,c \) be real numbers satisfying \( a+b+c = 3 \) . Find the maximum value of
\[P  = \dfrac{a(b+c)}{a^2+2bc+3} + \dfrac{b(a+c) }{b^2+2ca +3 } + \dfrac{c(a+b)}{c^2+2ab+3}.\]
1 reply
pennypc123456789
Yesterday at 9:47 AM
Quantum-Phantom
28 minutes ago
Jbmo 2015 problem 1
neverlose   11
N 34 minutes ago by MATHS_ENTUSIAST
Find all prime numbers $a,b,c$ and positive integers $k$ satisfying the equation \[a^2+b^2+16c^2 = 9k^2 + 1.\]

Proposed by Moldova
11 replies
neverlose
Jun 26, 2015
MATHS_ENTUSIAST
34 minutes ago
Colors over colors in a grid
FAA2533   1
N 36 minutes ago by Rohit-2006
Source: BdMO 2025 Higher Secondary P3
Two player are playing in an $100 \times 100$ grid. Initially the whole board is black. On $A$'s move, he selects $4 \times 4$ subgrid and color it white. On $B$'s move, he selects a $3 \times 3$ subgrid and colors it black. $A$ wants to make the whole board white. Can he do it?

Proposed by S M A Nahian
1 reply
FAA2533
Feb 8, 2025
Rohit-2006
36 minutes ago
AD and BC meet MN at P and Q
WakeUp   6
N 37 minutes ago by Nari_Tom
Source: Baltic Way 2005
Let $ABCD$ be a convex quadrilateral such that $BC=AD$. Let $M$ and $N$ be the midpoints of $AB$ and $CD$, respectively. The lines $AD$ and $BC$ meet the line $MN$ at $P$ and $Q$, respectively. Prove that $CQ=DP$.
6 replies
WakeUp
Dec 28, 2010
Nari_Tom
37 minutes ago
thank you
Piwbo   0
42 minutes ago
Let $p_n$ be the n-th prime number in increasing order for $n\geq 1$. Prove that there exists a sequence of distinct prime numbers $q_n$ satisfying $q_1+q_2+...+q_n=p_n$ for all $n\geq 1 $
0 replies
Piwbo
42 minutes ago
0 replies
IMO ShortList 2002, geometry problem 4
orl   24
N 44 minutes ago by Avron
Source: IMO ShortList 2002, geometry problem 4
Circles $S_1$ and $S_2$ intersect at points $P$ and $Q$. Distinct points $A_1$ and $B_1$ (not at $P$ or $Q$) are selected on $S_1$. The lines $A_1P$ and $B_1P$ meet $S_2$ again at $A_2$ and $B_2$ respectively, and the lines $A_1B_1$ and $A_2B_2$ meet at $C$. Prove that, as $A_1$ and $B_1$ vary, the circumcentres of triangles $A_1A_2C$ all lie on one fixed circle.
24 replies
orl
Sep 28, 2004
Avron
44 minutes ago
comb and nt mixed
MR.1   1
N an hour ago by MR.1
Source: own
in antarctica there is $n\geq3$ penguin and each one is numbered using numbers $1,2,\dots n$. penguin $i$ is called $mirza$ if $\binom{\underline i}{p_i}=1$ where $p_i$ is $i_{th}$ prime divisor of $n$ ($p_{kt+i}=p_{i}$(where k is number of $n$'s prime divisors). what is maximum ratio of $\frac{M}{n}$ where $M$ is number of $mirza$ in antarctica?($p_1\neq 1$)
1 reply
MR.1
Yesterday at 2:34 PM
MR.1
an hour ago
The best computer problems of the year
NT_G   9
N an hour ago by bin_sherlo
Source: https://t.me/NeuroGeometry
1. I is incenter of triangle $ABC$. $K$ is the midpoint of arc $BC$ not containing $A$. $L$ is the midpoint of arc $(BAC)$ $M$ is the midpoint of $IK$ and $N$ is the midpoint of $LM$.  $D$ is projection of $I$ onto $BC$. $S$ is the second intersection of $KD$ and $(ABC)$
Prove that $(ISD)$ is tangent to $(AMN)$.

2. Circle $w$ with center $I$ is incircle of triangle $ABC$. $D$ is projection of $I$ onto $BC$ Points $E, F$ on segments $BI$, $CI$ are following condition: $IE = IF$. $M$ is the midpoint of $EF$. $P$ is the second intersection of $(IMD)$ and $w$. $Q$ is the second intersection of $AP$ and $(IMD)$.
Prove, that $I,E,F,Q$ are concyclic.

I am grateful to Savva Chuev for making the diagrams.
9 replies
NT_G
Dec 31, 2023
bin_sherlo
an hour ago
geometry problem
Medjl   4
N an hour ago by tigerBoss101
Source: Netherlands TST for IMO 2017 day 3 problem 1
A circle $\omega$ with diameter $AK$ is given. The point $M$ lies in the interior of the circle, but not on $AK$. The line $AM$ intersects $\omega$ in $A$ and $Q$. The tangent to $\omega$ at $Q$ intersects the line through $M$ perpendicular to $AK$, at $P$. The point $L$ lies on $\omega$, and is such that $PL$ is tangent to $\omega$ and $L\neq Q$.
Show that $K, L$, and $M$ are collinear.
4 replies
Medjl
Feb 1, 2018
tigerBoss101
an hour ago
IMO ShortList 2002, geometry problem 3
orl   71
N an hour ago by Avron
Source: IMO ShortList 2002, geometry problem 3
The circle $S$ has centre $O$, and $BC$ is a diameter of $S$. Let $A$ be a point of $S$ such that $\angle AOB<120{{}^\circ}$. Let $D$ be the midpoint of the arc $AB$ which does not contain $C$. The line through $O$ parallel to $DA$ meets the line $AC$ at $I$. The perpendicular bisector of $OA$ meets $S$ at $E$ and at $F$. Prove that $I$ is the incentre of the triangle $CEF.$
71 replies
orl
Sep 28, 2004
Avron
an hour ago
Perpendicularity
April   31
N an hour ago by Tsikaloudakis
Source: CGMO 2007 P5
Point $D$ lies inside triangle $ABC$ such that $\angle DAC = \angle DCA = 30^{\circ}$ and $\angle DBA = 60^{\circ}$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF = 2FC$. Prove that $DE \perp EF$.
31 replies
April
Dec 28, 2008
Tsikaloudakis
an hour ago
China Mathematical Olympiad 1986 problem3
jred   4
N an hour ago by alexanderhamilton124
Source: China Mathematical Olympiad 1986 problem3
Let $Z_1,Z_2,\cdots ,Z_n$ be complex numbers satisfying $|Z_1|+|Z_2|+\cdots +|Z_n|=1$. Show that there exist some among the $n$ complex numbers such that the modulus of the sum of these complex numbers is not less than $1/6$.
4 replies
jred
Jan 17, 2014
alexanderhamilton124
an hour ago
Geometry
youochange   8
N Apr 7, 2025 by RANDOM__USER
m:}
Let $\triangle ABC$ be a triangle inscribed in a circle, where the tangents to the circle at points $B$ and $C$ intersect at the point $P$. Let $M$ be a point on the arc $AC$ (not containing $B$) such that $M \neq A$ and $M \neq C$. Let the lines $BC$ and $AM$ intersect at point $K$. Let $P'$ be the reflection of $P$ with respect to the line $AM$. The lines $AP'$ and $PM$ intersect at point $Q$, and $PM$ intersects the circumcircle of $\triangle ABC$ again at point $N$.

Prove that the point $Q$ lies on the circumcircle of $\triangle ANK$.
8 replies
youochange
Apr 6, 2025
RANDOM__USER
Apr 7, 2025
Geometry
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youochange
160 posts
#1 • 2 Y
Y by PikaPika999, Rounak_iitr
m:}
Let $\triangle ABC$ be a triangle inscribed in a circle, where the tangents to the circle at points $B$ and $C$ intersect at the point $P$. Let $M$ be a point on the arc $AC$ (not containing $B$) such that $M \neq A$ and $M \neq C$. Let the lines $BC$ and $AM$ intersect at point $K$. Let $P'$ be the reflection of $P$ with respect to the line $AM$. The lines $AP'$ and $PM$ intersect at point $Q$, and $PM$ intersects the circumcircle of $\triangle ABC$ again at point $N$.

Prove that the point $Q$ lies on the circumcircle of $\triangle ANK$.
This post has been edited 1 time. Last edited by youochange, Apr 6, 2025, 11:28 AM
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youochange
160 posts
#2 • 1 Y
Y by PikaPika999
Bump :first:
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youochange
160 posts
#3 • 1 Y
Y by PikaPika999
Helpmmmmmmme
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Double07
72 posts
#4 • 3 Y
Y by PikaPika999, RANDOM__USER, youochange
Try complex bashing:

Take $(ABC)$ to be the unit circle and WLOG suppose $m=1$.

Denote by $S=AP\cap(ABC), S\neq A$ and $R=AP'\cap (ABC), R\neq A$.

Since $P'$ is the reflection of $P$ over $AM\implies \widehat{SAM}=\widehat{MAR}\implies M$ is the midpoint of arc $\widehat{RS}$, so $r\cdot s=m^2=1\implies r=\frac{1}{s}=\overline{s}$.

Compute $p=\frac{2bc}{b+c}$ and $s=\frac{ab+ac-2bc}{2a-b-c}$, so $r=\frac{b+c-2a}{2bc-ab-ac}$.

Since $M, N, P$ are collinear and $|m|=|n|=1\implies -mn=\frac{p-m}{\overline{p}-\overline{m}}\implies n=\frac{2bc-b-c}{b+c-2}$.

$Q=AR\cap MN\implies q=\frac{ar(m+n)-mn(a+r)}{ar-mn}=\frac{a(b+c-2a)(2bc-2)-(2bc-b-c)(2abc-a^2b-a^2c+b+c-2a)}{a(b+c-2a)(b+c-2)-(2bc-b-c)(2bc-ab-ac)}=$
$=\frac{(2a+2bc-ab-ac-b-c)(ab+ac+2a-2abc-b-c)}{2(a-bc)(2a+2bc-ab-ac-b-c)}=\frac{ab+ac+2a-2abc-b-c}{2(a-bc)}$.

Compute $K=AM\cap BC\implies k=\frac{am(b+c)-bc(a+m)}{an-bc}=\frac{ab+ac-bc-abc}{a-bc}$.

Now, to prove that $A, N, Q, K$ are concyclic, we need to prove that $\frac{a-n}{a-q}\cdot\frac{k-q}{k-n}\in\mathbb{R}$.

$a-n=\frac{ab+ac+b+c-2a-2bc}{b+c-2}$

$a-q=\frac{2a^2-ab-ac-2a+b+c}{2(a-bc)}=\frac{(a-1)(2a-b-c)}{2(a-bc)}$

$k-q=\frac{ab+ac+b+c-2a-2bc}{2(a-bc)}$

$k-n=\frac{(ab+ac-bc-abc)(b+c-2)-(2bc-b-c)(a-bc)}{(a-bc)(b+c-2)}=\frac{(b-1)(c-1)(2bc-ab-ac)}{(a-bc)(b+c-2)}$

So $\frac{a-n}{a-q}\cdot\frac{k-q}{k-n}=\frac{(a-bc)(ab+ac+b+c-2a-2bc)^2}{(a-1)(b-1)(c-1)(2a-b-c)(2bc-ab-ac)}$, which is real by conjugating, so we're done.
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RANDOM__USER
6 posts
#5 • 2 Y
Y by PikaPika999, youochange
Hmm, very interesting problem, sadly I only have minor results that might be useful. :(

Claim 1: If \(D\) is the midpoint of \(BC\), then \(ADNK\) is cyclic.
Proof: We intersect \(AN\) with \(BC\) at a point \(F\). Then because \(BCMN\) is harmonic (the tangents from \(B\) and \(C\) intersect on \(NM\)) it must be that if we project this harmonic quad from \(A\) onto \(BC\) that \((B,C;F,K)=-1\). Now using a very well known property of harmonic sets, we know that \(BF \cdot FC = FD \cdot FK\). However, due to PoP we know that \(BF \cdot FC = AF \cdot FN\), thus \(AF \cdot FN = FD \cdot FK\), meaning that, indeed \(ADNK\) is cyclic. \(\square\)

Now for another cool observation,

Claim 2: The problem is equivelent to showing that \(PDQP'\) is cyclic.
Proof: Assume \(PDQP'\) is cyclic, then \(\angle{DQA} = \angle{P'PD}\). If \(X = PP' \cup AK\), then \(\angle{PXK} = \frac{\pi}{2}\) and \(\angle{PDK} = \frac{\pi}{2}\). Thus \(\angle{DPP'} = \angle{DKA}\) and thus \(\angle{DKA} = \angle{DQA}\) which means that \(AQKD\) is cyclic. Taking into account the result that \(ADNK\) is cyclic, we obtain that \(ANKQ\) is cyclic. \(\square\)

And finally the last observation I think is note worthy is the following,

Claim 3: If \(E\) is the intersection of \(AP\) and \((ABC)\), then \(E, N, K\) and \(E,F,M\) are colinear.
Proof: Quite trivial through harmonics and projective ideas. \(\square\)
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lolsamo
10 posts
#6 • 2 Y
Y by RANDOM__USER, youochange
Person above is just done, $\angle QAK=\angle PAK=\angle KNM=\angle KNQ$, as desired
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Captainscrubz
50 posts
#7 • 2 Y
Y by RANDOM__USER, youochange
ig the simplest solution take $\sqrt bc$ inversion
Suppose $X$ is a point in the plane let after inversion it be $X'$
forgive me cuz I used $P'$ as a point after inversion of $P$ while there was $P'$ in the problem :P

So $P \rightarrow P'$ where $P'$ will be the $A-$Humpty point
$M'$ will be a random point on $\overrightarrow{CB}$
$N'=(M'AP')\cap BC$ ,$P''=$ reflection of $P'$ in $AM'$ , $Q'=AP''\cap (M'AP')$ and $K'=AM'\cap (ABC)$
We need to prove that $Q'-K'-N'$
Let $E$ be the reflection of $P'$ in $BC$ see that $E$ will lie on $(ABC)$

We will use phantom points here
Let $K^*=AM'\cap EN'$
and Let $D$ be the midpoint of $BC$
So-
$$\angle EN'D=\angle DN'P'=\angle M'AP'$$$$\implies (AK^*N'D)$$$$\implies \angle AK^*E =\angle AK^*N'=\angle ADC=\angle ABC+ \angle BAD=\angle EBC$$$$\therefore K^*\equiv K'$$$$\therefore \angle M'N'K'=\angle K'AP'=\angle M'AQ'$$$$\blacksquare$$
Attachments:
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SimplisticFormulas
89 posts
#8 • 2 Y
Y by RANDOM__USER, youochange
unless im seriously mistaken, the simplest solution is using projective
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RANDOM__USER
6 posts
#9 • 1 Y
Y by youochange
Yea, that seems to be the correct solution! That is essentially my solution in addition to the comment that I somehow didn't notice to finish of my solution :)
This post has been edited 1 time. Last edited by RANDOM__USER, Apr 7, 2025, 7:39 AM
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