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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Jul 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
IMO 2025 Medal Cutoffs Prediction
GreenTea2593   12
N a minute ago by Polyquadratus
What are your prediction for IMO 2025 medal cutoffs?
12 replies
GreenTea2593
Today at 4:44 AM
Polyquadratus
a minute ago
Inspired by Nguyenhuyen_AG
sqing   3
N 7 minutes ago by sqing
Source: Own
Let $a, \, b, \, c\geq 0 .$ Prove that
$$\frac{4a}{b+c}+\frac{36b}{c+2a}+\frac{c}{a+3b} \geq 4$$$$  \frac{ 8a}{b+c}+\frac{65b}{c+2a}+\frac{c}{a+3b}  \geq 4\sqrt 2$$
3 replies
1 viewing
sqing
Today at 5:59 AM
sqing
7 minutes ago
geometry is mAJorly BACk
EpicBird08   10
N 10 minutes ago by Anzoteh
Source: ISL 2024/G6
Let $ABC$ be an acute triangle with $AB < AC$, and let $\Gamma$ be the circumcircle of $ABC$. Points $X$ and $Y$ lie on $\Gamma$ so that $XY$ and $BC$ meet on the external angle bisector of $\angle BAC$. Suppose that the tangents to $\Gamma$ at $x$ and $Y$ intersect at a point $T$ on the same side of $BC$ as $A$, and that $TX$ and $TY$ intersect $BC$ at $U$ and $V$, respectively. Let $J$ be the centre of the excircle of triangle $TUV$ opposite the vertex $T$.

Prove that $AJ$ bisects $\angle BAC$.
10 replies
EpicBird08
Today at 3:00 AM
Anzoteh
10 minutes ago
maximizing score
KevinYang2.71   3
N 26 minutes ago by Assassino9931
Source: ISL 2024 C1
Let $n$ be a positive integer. A class of $n$ students run $n$ races, in each of which they are ranked with no draws. A student is eligible for a rating $(a,\,b)$ for positive integers $a$ and $b$ if they come in the top $b$ places in at least $a$ of the races. Their final score is the maximum possible value of $a-b$ across all ratings for which they are eligible.

Find the maximum possible sum of all the scores of the $n$ students.
3 replies
+1 w
KevinYang2.71
Today at 3:00 AM
Assassino9931
26 minutes ago
orang NT
KevinYang2.71   14
N 29 minutes ago by star-1ord
Source: ISL 2024 N1
Find all positive integers $n$ with the following property: for all positive divisors $d$ of $n$, we have $d+1\mid n$ or $d+1$ is prime.
14 replies
KevinYang2.71
Today at 3:00 AM
star-1ord
29 minutes ago
Inspired by old results
sqing   0
30 minutes ago
Source: Own
Let $ a,b> 0,a+b+1=ab .$ Prove that
$$   \frac{1}{a^2+ab+1}+ \frac{1}{b^2+ab+1}  \leq \frac{2(7-4\sqrt 2)}{17}$$$$   \frac{a}{a^2+ 2 ab+1}+ \frac{b}{b^2+  2ab+1} \leq \frac{2\sqrt 2-1}{7}$$Let $ a,b> 0,a+b+2=ab .$ Prove that
$$   \frac{1}{a^2+2ab+1}+ \frac{1}{b^2+2ab+1}  \leq  \frac{2(13-6\sqrt 3)}{61}$$$$   \frac{a}{a^2+ ab+1}+ \frac{b}{b^2+ ab+1} \leq  \frac{2(5\sqrt 3-3)}{33}$$
0 replies
sqing
30 minutes ago
0 replies
Beetles on heights. If two on incircle, then third is also!
NO_SQUARES   2
N 34 minutes ago by LiTaotao
Source: Kvant 2024 no. 11-12 M2825; also from 'Sirius Mathematical Olympiad'
At the same time, three beetles with identical speeds began to crawl along the heights of an acute-angled non-isosceles triangle from its vertices. At some point, it turned out that the first and second beetles were on a circle inscribed in a triangle. Prove that at this moment the third beetle is also on this circle.
A. Kuznetsov
2 replies
NO_SQUARES
Jan 20, 2025
LiTaotao
34 minutes ago
Modula sequencia
EeEeRUT   2
N 35 minutes ago by Tkn
Source: thailand tstst 2025
Find all ordered pair of positive integers $(A,B)$ such that the sequence of non-negative integers $(a_n)$ with $a_0=A$ and $$a_{n+1}=(n+B)\;\mathrm{mod}\;(a_0+a_1+\cdots+a_{n}),\quad\text{for all integer }n\geqslant 0.$$is bounded.
Note that for any positive integers $a$ and $b$, define $a\;\mathrm{mod}\;b$ to be a unique integer $0\leqslant r<b$ such that $b$ divides $a-r$.
2 replies
EeEeRUT
Today at 12:57 AM
Tkn
35 minutes ago
quadrilateral geo with length conditions
OronSH   7
N 43 minutes ago by s27_SaparbekovUmar
Source: IMO Shortlist 2024 G1
Let $ABCD$ be a cyclic quadrilateral such that $AC<BD<AD$ and $\angle DBA<90^\circ$. Point $E$ lies on the line through $D$ parallel to $AB$ such that $E$ and $C$ lie on opposite sides of line $AD$, and $AC=DE$. Point $F$ lies on the line through $A$ parallel to $CD$ such that $F$ and $C$ lie on opposite sides of line $AD$, and $BD=AF$.

Prove that the perpendicular bisectors of segments $BC$ and $EF$ intersect on the circumcircle of $ABCD$.

Proposed by Mykhailo Shtandenko, Ukraine
7 replies
OronSH
Today at 3:13 AM
s27_SaparbekovUmar
43 minutes ago
I am [not] a parallelogram
peppapig_   12
N an hour ago by s27_SaparbekovUmar
Source: ISL 2024/G4
Let $ABCD$ be a quadrilateral with $AB$ parallel to $CD$ and $AB<CD$. Lines $AD$ and $BC$ intersect at a point $P$. Point $X$ distinct from $C$ lies on the circumcircle of triangle $ABC$ such that $PC=PX$. Point $Y$ distinct from $D$ lies on the circumcircle of triangle $ABD$ such that $PD=PY$. Lines $AX$ and $BY$ intersect at $Q$.

Prove that $PQ$ is parallel to $AB$.

Fedir Yudin, Mykhailo Shtandenko, Anton Trygub, Ukraine
12 replies
peppapig_
Today at 3:00 AM
s27_SaparbekovUmar
an hour ago
IMO 2025 Comments
Primeniyazidayi   2
N an hour ago by OGMATH
What do you think about IMO 2025 exam problems and IMO 2024 Shortlist problems?
2 replies
Primeniyazidayi
2 hours ago
OGMATH
an hour ago
D1052 : How it's possible ?
Dattier   1
N an hour ago by Dattier
Source: les dattes à Dattier
Is it true for all $n$ natural integer : $(E\times 29^n \mod F) \mod 3\neq 0$ ?

E=163999081217965835070356295641931525591357567735624606830696386586994976172813816839262877
50922205042989559280142547393636527346272001339597483126086699049357460700008119117240578043
46281799731794620614941989125738298381362079843446150841376016501310942563338531951229469261
017554376486801


F=666165351866558215458553224258271230186695252725433417706426521946436303489813878149680284
24137540935869820330945301911165461108917069581547697978809314332789769417680417107249591988
71252061235265894516611712110379162326930843580773177773789047845826833190774483296276708089
431095213731040922452939281280
1 reply
Dattier
Yesterday at 4:02 PM
Dattier
an hour ago
I miss Turbo
sarjinius   16
N an hour ago by mofumofu
Source: 2025 IMO P6
Consider a $2025\times2025$ grid of unit squares. Matilda wishes to place on the grid some rectangular tiles, possibly of different sizes, such that each side of every tile lies on a grid line and every unit square is covered by at most one tile.

Determine the minimum number of tiles Matilda needs to place so that each row and each column of the grid has exactly one unit square that is not covered by any tile.
16 replies
sarjinius
Today at 3:20 AM
mofumofu
an hour ago
Bonza functions
KevinYang2.71   43
N an hour ago by PHSH
Source: 2025 IMO P3
Let $\mathbb{N}$ denote the set of positive integers. A function $f\colon\mathbb{N}\to\mathbb{N}$ is said to be bonza if
\[
f(a)~~\text{divides}~~b^a-f(b)^{f(a)}
\]for all positive integers $a$ and $b$.

Determine the smallest real constant $c$ such that $f(n)\leqslant cn$ for all bonza functions $f$ and all positive integers $n$.
43 replies
+1 w
KevinYang2.71
Yesterday at 3:38 AM
PHSH
an hour ago
Hard Function
johnlp1234   11
N May 17, 2025 by GreekIdiot
f:R+--->R+:
f(x^3+f(y))=y+(f(x))^3
11 replies
johnlp1234
Jul 7, 2020
GreekIdiot
May 17, 2025
Hard Function
G H J
G H BBookmark kLocked kLocked NReply
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johnlp1234
35 posts
#1 • 1 Y
Y by alexey_phenichniy
f:R+--->R+:
f(x^3+f(y))=y+(f(x))^3
Z K Y
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TuZo
19351 posts
#2
Y by
johnlp1234 wrote:
$f:R+--->R+$:
$f(x^3+f(y))=y+(f(x))^3$
Z K Y
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Aritra12
1026 posts
#3 • 3 Y
Y by abhradeep12, CatsMeow12, Wisphard
johnlp1234 wrote:
f:R+--->R+:
f(x^3+f(y))=y+(f(x))^3

Can you please say what is the question,please dont post incomplete problems
please read this post's point number c Rules
This post has been edited 3 times. Last edited by Aritra12, Jul 7, 2020, 5:06 PM
Z K Y
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SomeUser221104
144 posts
#4 • 1 Y
Y by abhradeep12
johnlp1234 wrote:
Find all function $f:\mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that :
$$f(x^3+f(y))=y+(f(x))^3$$

FTFY
Z K Y
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jasperE3
11432 posts
#5
Y by
johnlp1234 wrote:
f:R+--->R+:
f(x^3+f(y))=y+(f(x))^3

Let $P(x,y)$ be the assertion $f\left(x^3+f(y)\right)=y+f(x)^3$.
$P(f(x),y^3+f(z))\Rightarrow f\left(f(x)^3+y+f(z)^3\right)=f(f(x))^3+y^3+f(z)$
Swapping $x,z$, we get that $f(f(x))^3=f(x)+c$ for all $x>0$ where $c\in\mathbb R$ is some constant, that is, $f(x)=\sqrt[3]{x+c}$ for all $x\in f(\mathbb R^+)$.
$P(1,x)\Rightarrow f(1+f(x))=x+f(1)^3$ so $\left(f(1)^3,\infty\right)\subseteq f(\mathbb R^+)$ and $f(x)=\sqrt[3]{x+c}$ for all $x>f(1)^3$.

Let $u>\max\left\{f(1)^3,8647\right\}$ be sufficiently large, then $u^3+f(x)>f(1)^3$ for all $x>0$ and $u>f(1)^3$, so:
$P(u,x)\Rightarrow\sqrt[3]{u^3+f(x)+c}=x+u+c$
So $f$ is a cubic polynomials, however since $f(x)=\sqrt[3]{x+c}$ for all $x>f(1)^3$ it cannot be a cubic polynomial. Hence no solutions.
Z K Y
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GreekIdiot
322 posts
#6
Y by
But $f(x)=x$ clearly satisfies...
Z K Y
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GreekIdiot
322 posts
#7
Y by
Let $P(x,y)$ denote the assertion
Fixing $x$ and varying $y$ we see that $f$ is surjective
Let $f(a)=f(b)$ for some $a$, $b$ then
$P(x,a)-P(x,b) \implies f(x^3+f(a))-f(x^3-f(b))=a-b \implies a=b$ thus $f$ is bijective
To be continued
Ι found $f(f(x))=x \: \forall \: x \in \mathbb{R_+}$ using the substitution jasper chose above. Too lazy to writeup.
This post has been edited 1 time. Last edited by GreekIdiot, May 17, 2025, 9:49 AM
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ektorasmiliotis
120 posts
#8
Y by
surjective in R+ or for values greater than (f(c))^3 ?
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GreekIdiot
322 posts
#9
Y by
I didnt use surjectivity anyways so it shouldnt matter...
$f$ is non constant thus picking $x$ to minimize $f(x)$ we get that $f$ is surjective on interval $(min^3\{f\}, \infty)$
This post has been edited 1 time. Last edited by GreekIdiot, May 17, 2025, 4:57 PM
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maromex
283 posts
#10
Y by
Actually $f$ is surjective because it's an involution.

Should I post about this FE here instead of the identical https://artofproblemsolving.com/community/c6h2179422?
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maromex
283 posts
#11
Y by
Hi. $P(f(x), y^3+ f(z)) : f(f(x)^3 + z + f(y)^3) = f(f(x))^3 + y^3 + f(z).$
And therefore by switching $x, y$ and comparing,$$f(f(x))^3 = x^3 + c.$$$P(f(x), y) : f(f(x)^3 + f(y)) = y + x^3 + c$
$P(x, f(y)) : f(x^3 + f(f(y))) = f(y) + f(x)^3$
which implies $f(x^3 + \sqrt[3]{y^3 + c}) = f(y) + f(x)^3$
Take function of both sides, and substitute into $P(f(x), y)$, get $f(f(x^3 + \sqrt[3]{y^3 + c})) = y + x^3 + c$
Cube both sides and apply some stuff, now we have$$(x^3 + \sqrt[3]{y^3 + c})^3 + c = (x^3 + y + c)^3$$for all $x, y$. This is equivalent to $(x^3 + y + c)^3 - (x^3 + \sqrt[3]{y^3 + c})^3 = c$. Fix a $y$ and let $a = y + c$ and $b = \sqrt[3]{y^3 + c}$. Both $a$ and $b$ are positive. Then we have $(x^3 + a)^3 - (x^3 + b)^3 = c$. When we factor, we get$$(a - b)((x^3 + a)^2 + (x^3 + a)(x^3 + b) + (x^3 + b)^2) = c.$$Notice that, when increasing $x$, the LHS gets arbitrarily large (and therefore not constant) if $a - b > 0$, and gets arbitrarily small (and therefore not constant) if $a - b < 0$. The only possibility is $a - b = 0$ and therefore $c = 0$. Therefore,$$f(f(x))^3 = x^3 \implies f(f(x)) = x.$$This implies $f$ is bijective. If we take $t=x^3 + f(y)$, then $f(t) > y = f(f(y))$. Now $t$ can be anything greater than $f(y)$. Because $f$ is surjective, $f(y)$ can be anything and $f$ is strictly increasing.

If, for some $x$ we have $f(x) < x$, then $x = f(f(x)) > f(x)$ contradicting the fact that $f$ is strictly increasing. If, for some $x$ we have $f(x) > x$, then $f(x) > f(f(x)) = x$ which also contradicts $f$ strictly increasing. Therefore $$f(x) = x$$for all $x$, and it works.
This post has been edited 7 times. Last edited by maromex, May 17, 2025, 8:16 PM
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GreekIdiot
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maromex wrote:
Actually $f$ is surjective because it's an involution.

Yeah you are right I forgot I proved that $f(f(x))=x$ :blush:
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