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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
Ring out the Old Year and ring in the New.
Kunihiko_Chikaya   1
N a minute ago by Mathzeus1024
Let $a,\ b,\ c$ be positive real numbers.

Prove that

$$\sqrt[3]{\left(\frac{a^{2022}-a}{b}+\frac{2021}{a^{\frac{a}{b}}}+1\right)\left(\frac{b^{2022}-b}{c}+\frac{2021}{b^{\frac{b}{c}}}+1\right)\left(\frac{c^{2022}-c}{a}+\frac{2021}{c^{\frac{c}{a}}}+1\right)}$$
$$\geq 2022.$$
Proposed by Kunihiko Chikaya/December 31, 2021
1 reply
Kunihiko_Chikaya
Dec 31, 2021
Mathzeus1024
a minute ago
Stronger inequality than an old result
KhuongTrang   21
N 4 minutes ago by arqady
Source: own, inspired
Problem. Find the best constant $k$ satisfying $$(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge \frac{9}{4}+k\cdot\frac{a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)}{(a+b+c)^{3}}$$holds for all $a,b,c\ge 0: ab+bc+ca>0.$
21 replies
+1 w
KhuongTrang
Aug 1, 2024
arqady
4 minutes ago
Inspired by Jackson0423
sqing   1
N 7 minutes ago by sqing
Source: Own
Let $ a, b, c>0 $ and $ a^2 + b^2 =c(a + b). $ Prove that
$$   \frac{b^2 +bc+ c^2}{ a(a +b+  c)} \geq 2\sqrt 3-3$$
1 reply
sqing
22 minutes ago
sqing
7 minutes ago
Combo problem
soryn   1
N 8 minutes ago by soryn
The school A has m1 boys and m2 girls, and ,the school B has n1 boys and n2 girls. Each school is represented by one team formed by p students,boys and girls. If f(k) is the number of cases for which,the twice schools has,togheter k girls, fund f(k) and the valute of k, for which f(k) is maximum.
1 reply
soryn
Today at 6:33 AM
soryn
8 minutes ago
Incredible combinatorics problem
A_E_R   2
N 12 minutes ago by quacksaysduck
Source: Turkmenistan Math Olympiad - 2025
For any integer n, prove that there are exactly 18 integer whose sum and the sum of the fifth powers of each are equal to the integer n.
2 replies
A_E_R
2 hours ago
quacksaysduck
12 minutes ago
What is the likelihood the last card left in the deck is black?
BEHZOD_UZ   1
N an hour ago by sami1618
Source: Yandex Uzbekistan Coding and Math Contest 2025
You have a deck of cards containing $26$ black and $13$ red cards. You pull out $2$ cards, one after another, and check their colour. If both cards are the same colour, then a black card is added to the deck. However, if the cards are of different colours, then a red card is used to replace them. Once the cards are taken out of the deck, they are not returned to the deck, and thus the number of cards keeps reducing. What is the likelihood the last card left in the deck is black?
1 reply
BEHZOD_UZ
an hour ago
sami1618
an hour ago
abc(a+b+c)=3, show that prod(a+b)>=8 [Indian RMO 2012(b) Q4]
Potla   31
N an hour ago by sqing
Let $a,b,c$ be positive real numbers such that $abc(a+b+c)=3.$ Prove that we have
\[(a+b)(b+c)(c+a)\geq 8.\]
Also determine the case of equality.
31 replies
Potla
Dec 2, 2012
sqing
an hour ago
AGI-Origin Solves Full IMO 2020–2024 Benchmark Without Solver (30/30) beat Alpha
AGI-Origin   10
N an hour ago by TestX01
Hello IMO community,

I’m sharing here a full 30-problem solution set to all IMO problems from 2020 to 2024.

Standard AI: Prompt --> Symbolic Solver (SymPy, Geometry API, etc.)

Unlike AlphaGeometry or symbolic math tools that solve through direct symbolic computation, AGI-Origin operates via recursive symbolic cognition.

AGI-Origin:
Prompt --> Internal symbolic mapping --> Recursive contradiction/repair --> Structural reasoning --> Human-style proof

It builds human-readable logic paths by recursively tracing contradictions, repairing structure, and collapsing ambiguity — not by invoking any external symbolic solver.

These results were produced by a recursive symbolic cognition framework called AGI-Origin, designed to simulate semi-AGI through contradiction collapse, symbolic feedback, and recursion-based error repair.

These were solved without using any symbolic computation engine or solver.
Instead, the solutions were derived using a recursive symbolic framework called AGI-Origin, based on:
- Contradiction collapse
- Self-correcting recursion
- Symbolic anchoring and logical repair

Full PDF: [Upload to Dropbox/Google Drive/Notion or arXiv link when ready]

This effort surpasses AlphaGeometry’s previous 25/30 mark by covering:
- Algebra
- Combinatorics
- Geometry
- Functional Equations

Each solution follows a rigorous logical path and is written in fully human-readable format — no machine code or symbolic solvers were used.

I would greatly appreciate any feedback on the solution structure, logic clarity, or symbolic methodology.

Thank you!

— AGI-Origin Team
AGI-Origin.com
10 replies
AGI-Origin
6 hours ago
TestX01
an hour ago
FE solution too simple?
Yiyj1   6
N an hour ago by Primeniyazidayi
Source: 101 Algebra Problems from the AMSP
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the equality $$f(f(x)+y) = f(x^2-y)+4f(x)y$$holds for all pairs of real numbers $(x,y)$.

My solution

I feel like my solution is too simple. Is there something I did wrong or something I missed?
6 replies
Yiyj1
Apr 9, 2025
Primeniyazidayi
an hour ago
Two very hard parallel
jayme   5
N 2 hours ago by jayme
Source: own inspired by EGMO
Dear Mathlinkers,

1. ABC a triangle
2. D, E two point on the segment BC so that BD = DE= EC
3. M, N the midpoint of ED, AE
4. H the orthocenter of the acutangle triangle ADE
5. 1, 2 the circumcircle of the triangle DHM, EHN
6. P, Q the second point of intersection of 1 and BM, 2 and CN
7. U, V the second points of intersection of 2 and MN, PQ.

Prove : UV is parallel to PM.

Sincerely
Jean-Louis
5 replies
jayme
Yesterday at 12:46 PM
jayme
2 hours ago
Number theory
XAN4   1
N 2 hours ago by NTstrucker
Source: own
Prove that there exists infinitely many positive integers $x,y,z$ such that $x,y,z\ne1$ and $x^x\cdot y^y=z^z$.
1 reply
XAN4
Apr 20, 2025
NTstrucker
2 hours ago
R+ FE with arbitrary constant
CyclicISLscelesTrapezoid   25
N 3 hours ago by DeathIsAwe
Source: APMO 2023/4
Let $c>0$ be a given positive real and $\mathbb{R}_{>0}$ be the set of all positive reals. Find all functions $f \colon \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that \[f((c+1)x+f(y))=f(x+2y)+2cx \quad \textrm{for all } x,y \in \mathbb{R}_{>0}.\]
25 replies
CyclicISLscelesTrapezoid
Jul 5, 2023
DeathIsAwe
3 hours ago
Combo with cyclic sums
oVlad   1
N 4 hours ago by ja.
Source: Romania EGMO TST 2017 Day 1 P4
In $p{}$ of the vertices of the regular polygon $A_0A_1\ldots A_{2016}$ we write the number $1{}$ and in the remaining ones we write the number $-1.{}$ Let $x_i{}$ be the number written on the vertex $A_i{}.$ A vertex is good if \[x_i+x_{i+1}+\cdots+x_j>0\quad\text{and}\quad x_i+x_{i-1}+\cdots+x_k>0,\]for any integers $j{}$ and $k{}$ such that $k\leqslant i\leqslant j.$ Note that the indices are taken modulo $2017.$ Determine the greatest possible value of $p{}$ such that, regardless of numbering, there always exists a good vertex.
1 reply
oVlad
Yesterday at 1:41 PM
ja.
4 hours ago
Incircle of a triangle is tangent to (ABC)
amar_04   11
N 4 hours ago by Nari_Tom
Source: XVII Sharygin Correspondence Round P18
Let $ABC$ be a scalene triangle, $AM$ be the median through $A$, and $\omega$ be the incircle. Let $\omega$ touch $BC$ at point $T$ and segment $AT$ meet $\omega$ for the second time at point $S$. Let $\delta$ be the triangle formed by lines $AM$ and $BC$ and the tangent to $\omega$ at $S$. Prove that the incircle of triangle $\delta$ is tangent to the circumcircle of triangle $ABC$.
11 replies
amar_04
Mar 2, 2021
Nari_Tom
4 hours 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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G H BBookmark kLocked kLocked NReply
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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
76 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
7 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. \(\square\)
Attachments:
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lolsamo
11 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
57 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-
$$\angle EN'D=\angle DN'P'=\angle M'AP'$$$$\implies (AK^*N'D)$$$$\implies \angle AK^*E =\angle AK^*N'=\angle ADC=\angle ABC+ \angle BAD=\angle EBC$$$$\therefore K^*\equiv K'$$$$\therefore \angle M'N'K'=\angle K'AP'=\angle M'AQ'$$$$\blacksquare$$
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This post has been edited 2 times. Last edited by Captainscrubz, Apr 7, 2025, 4:47 AM
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SimplisticFormulas
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#8 • 2 Y
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unless im seriously mistaken, the simplest solution is using projective
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RANDOM__USER
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#9 • 1 Y
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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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