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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
Radical center and perspective center
Lsway   3
N 16 minutes ago by flower417477
Source: Own
(2017.3.17) Given $\triangle ABC$ with circumcircle $\Gamma $,orthocenter $ H$.
Let $D,E,F$ be three points such taht $AE,AF$ are isogonal line WRT $ \angle BAC$, $BF,BD $ are isogonal line WRT $ \angle ABC$, $CD,CE$ are isogonal line WRT $ \angle ACB$.
Let $\Gamma _1,\Gamma _2,\Gamma _3$ be the pedal circle of $D,E,F$ WRT $\triangle ABC$ , respectively.
Let $T $ be the radical center of $\Gamma _1,\Gamma _2,\Gamma _3$ , $ l_1,l_2,l_3$ be the radical axis of $\Gamma _1 $ and $\Gamma $ , $\Gamma _2$ and $\Gamma $,$\Gamma _3$ and $\Gamma $, respectively. $l_2  \cap  l_3=X,l_1 \cap l_3=Y,l_1 \cap  l_2=Z. $ Then $AX,BY,CZ,HT$ are collinear.
P.S. I discovered this conclusion when trying a problem from my friend.
3 replies
Lsway
Mar 18, 2017
flower417477
16 minutes ago
symetric 3 vars 2 conditions
perfect_square   1
N 17 minutes ago by lbh_qys
Let $a,b,c \ge 0$ which satisfy:
$  \begin{cases}
      a^4+b^4+c^4 =3 \\
      a^5+b^5+c^5 =3.1
     \end{cases}$
Prove that: $ a^6+b^6+c^6 < 4$
1 reply
perfect_square
38 minutes ago
lbh_qys
17 minutes ago
Game theory in 2025???
Iveela   3
N 18 minutes ago by flower417477
Source: IMSC 2025 P3
Alice and Bob play a game on a $K_{2026}$. They take turns with Alice playing first. Initially all edges are uncolored. On Alice's turn she chooses any uncolored edge and colors it red. On Bob's turn, he chooses 1, 2, or 3 uncolored edges and colors them blue. The game ends once all edges have been colored. Let $r$ and $b$ be the number of vertices of the largest red and blue clique, respectively. Bob wins if at the end of the game $b > r$. Show that Bob has a winning strategy.

Note: $K_n$ denotes the complete graph with $n$ vertices and where there is an edge between any pair of vertices. A red (or blue, respectively) clique refers to a complete subgraph of which all the edges are red (or blue, respectively).
3 replies
Iveela
Jul 5, 2025
flower417477
18 minutes ago
D1052 : How it's possible ?
Dattier   11
N 18 minutes 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
11 replies
Dattier
Jul 15, 2025
Dattier
18 minutes ago
Solve the equation
Tip_pay   3
N 24 minutes ago by Tip_pay
Is there a nice way to replace this equation?

$$\dfrac{2x}{x^2-4x+2}+\dfrac{3x}{x^2+x+2}+\dfrac{5}{4}=0$$
If there is no suitable replacement, then how can you solve the equation without raising it to the 4th degree?
3 replies
Tip_pay
38 minutes ago
Tip_pay
24 minutes ago
Polynomial with prime
Eeightqx   4
N 40 minutes ago by navid
Source: 2025 China South East Mathematical Olympiad Grade10 P8
Find all positive integer pairs $(p,\,m)$ where $p$ is a prime such that the polynomial
$$f(x)=x^{2m}-px^m+p^2$$can be written as the product of two rational coefficient polynomial whose degree less than $2m$.
4 replies
Eeightqx
5 hours ago
navid
40 minutes ago
Bonza functions
KevinYang2.71   71
N 44 minutes ago by supremitas.mathematica
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 $n$.

Proposed by Lorenzo Sarria, Colombia
71 replies
KevinYang2.71
Jul 15, 2025
supremitas.mathematica
44 minutes ago
Strange board
truongphatt2668   0
an hour ago
A 8x8 chess board is called "weird" when the black and white squares on the board are arranged in those rules:
i) The number of black squares in each column is equal to each other;
ii) The number of black squares in each row is different from each other.

Find the maximum numbers of the couple of squares on the board such in each couple, they share one common edge and have different colors?

An example of a chess board:
0 replies
truongphatt2668
an hour ago
0 replies
Iran TST P5
TheBarioBario   5
N an hour ago by megarnie
Source: Iranian TST 2022 problem 5
Find all $C\in \mathbb{R}$ such that every sequence of integers $\{a_n\}_{n=1}^{\infty}$ which is bounded from below and for all $n\geq 2$ satisfy $$0\leq a_{n-1}+Ca_n+a_{n+1}<1$$is periodic.

Proposed by Navid Safaei
5 replies
TheBarioBario
Apr 2, 2022
megarnie
an hour ago
Prove Equidistance
Eeightqx   3
N an hour ago by X.Luser
Source: 2025 China South East Mathematical Olympiad Grade10 P3
Two circles $\omega_1$ and $\omega_2$ intersects at one point $T$. The common tangent further from $T$ is $XY$ where $X\in\omega_1$ and $Y\in\omega_2$. A line parallel to $XY$ cuts $\omega_1,\,\omega_2$ at four points $A,\,C,\,B,\,D$. $\odot(TAC)$ cuts $\odot(TBD)$ at $P$. If $TA\perp TD$, show that the distance from $X$ to $PC$ is equal to the distance from $Y$ to $PB$.
3 replies
Eeightqx
Yesterday at 5:28 AM
X.Luser
an hour ago
A chair set
truongphatt2668   0
an hour ago
During an extracurricular activity, there are 100 chairs arranged into two rows facing each other, with each row containing 50 chairs. The teacher assigns 100 students to sit on the 100 chairs, with one student per chair.

(a) Suppose each student only knows the person sitting directly opposite or the person sitting next to them. What is the minimum number of students that need to be selected such that every student knows at least one selected student?

(b) Suppose one student leaves their seat. The teacher then repeatedly performs the following operation: pick any student who is sitting in a row with at least one empty seat and move them to an empty seat. This operation can be performed multiple times. Prove that after a finite number of such operations, the teacher cannot move any student to a seat where they are sitting directly opposite their original seat.
0 replies
truongphatt2668
an hour ago
0 replies
2025 Greece IMO TST P3
brainfertilzer   19
N 2 hours ago by AN1729
Source: 2024 ISL G3
Let $ABCDE$ be a convex pentagon and let $M$ be the midpoint of $AB$. Suppose that segment $AB$ is tangent to the circumcircle of triangle $CME$ at $M$ and that $D$ lies on the circumircles of $AME$ and $BMC$. Lines $AD$ and $ME$ interesect at $K$, and lines $BD$ and $MC$ intersect at $L$. Points $P$ and $Q$ lie on line $EC$ so that $\angle PDC = \angle EDQ = \angle ADB$.

Prove that lines $KP, LQ,$ and $MD$ are concurrent.
19 replies
brainfertilzer
Jul 16, 2025
AN1729
2 hours ago
Recursive sequence satisfies inequality
orl   4
N 2 hours ago by math-olympiad-clown
Source: IMO Shortlist 1996, A3
Let $ a > 2$ be given, and starting $ a_0 = 1, a_1 = a$ define recursively:

\[ a_{n+1} = \left(\frac{a^2_n}{a^2_{n-1}} - 2 \right) \cdot a_n.\]

Show that for all integers $ k > 0,$ we have: $ \sum^k_{i = 0} \frac{1}{a_i} < \frac12 \cdot (2 + a - \sqrt{a^2-4}).$
4 replies
orl
Aug 9, 2008
math-olympiad-clown
2 hours ago
Peru IMO TST 2024
diegoca1   5
N 2 hours ago by zhoujef000
Source: Peru IMO TST 2024 D2 P3
Determine the minimum possible value of \( x^2 + y^2 + z^2 \), where \( x, y, z \) are positive real numbers satisfying:
\[
    \begin{cases} 
    x + y + z = 5, \\ 
    \frac{1} {x} + \frac{1} {y} + \frac{1} {z} = 2.
    \end{cases}
    \]
5 replies
diegoca1
Jul 25, 2025
zhoujef000
2 hours ago
4 variables
Nguyenhuyen_AG   11
N May 23, 2025 by arqady
Let $a,\,b,\,c,\,d$ are non-negative real numbers and $0 \leqslant k \leqslant \frac{2}{\sqrt{3}}.$ Prove that
$$a^2+b^2+c^2+d^2+kabcd \geqslant k+4+(k+2)(a+b+c+d-4).$$hide
11 replies
Nguyenhuyen_AG
Dec 21, 2020
arqady
May 23, 2025
4 variables
G H J
G H BBookmark kLocked kLocked NReply
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Nguyenhuyen_AG
3422 posts
#1 • 1 Y
Y by Mango247
Let $a,\,b,\,c,\,d$ are non-negative real numbers and $0 \leqslant k \leqslant \frac{2}{\sqrt{3}}.$ Prove that
$$a^2+b^2+c^2+d^2+kabcd \geqslant k+4+(k+2)(a+b+c+d-4).$$hide
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mihaig
7553 posts
#2 • 1 Y
Y by Mango247
Nice. I actually have a more complex problem, which involves parameters and $n$ variables. But since I never made it public, I only congratulate you for getting a same result as I did, independently.
This post has been edited 1 time. Last edited by mihaig, Dec 21, 2020, 9:54 AM
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Nguyenhuyen_AG
3422 posts
#3 • 3 Y
Y by Mango247, Mango247, Mango247
mihaig wrote:
Nice. I actually have a more complex problem, which involves parameters and $n$ variables. But since I never made it public, I only congratulate you for getting a same result as I did, independently.
I don't know your idea. I found this while prove an old problem

If $a,\,b,\,c,\,d \geqslant 0$ and $a^2+b^2+c^2+d^2+abcd=5$ then $$a+b+c+d \leqslant 4.$$
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mihaig
7553 posts
#4
Y by
Of course you don't know my idea. As I said, I'm glad you found a result similar to my result. No matter how. Coincidences happen.
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mihaig
7553 posts
#5
Y by
In addition, we also have this. But it's just an example, it's far from being the complete result.

Find $k\in\mathbb{R}$ such that
$$\sum_{i=1}^{5}{a_i^2}+k\prod_{i=1}^{5}{a_i}+4k+5\geq\left(k+2\right)\sum_{i=1}^{5}{a_i}$$holds for all $a_1,\ldots,a_5\geq0.$
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mihaig
7553 posts
#6
Y by
Bump to these beauties
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DipoleOfMonorak
235 posts
#7
Y by
mihaig wrote:
In addition, we also have this. But it's just an example, it's far from being the complete result.

Find $k\in\mathbb{R}$ such that
$$\sum_{i=1}^{5}{a_i^2}+k\prod_{i=1}^{5}{a_i}+4k+5\geq\left(k+2\right)\sum_{i=1}^{5}{a_i}$$holds for all $a_1,\ldots,a_5\geq0.$

Let $f(a_1,a_2,a_3,a_4,a_5)=\sum_{i=1}^{5}{a_i^2}+k\prod_{i=1}^{5}{a_i}+4k+5-\left(k+2\right)\sum_{i=1}^{5}{a_i},$ then

$f(x,x,x,1,1)=(x-1)^2(kx+2k+3)\geq0\Rightarrow k\geq0$

$f(x,x,x,x,0)=4x^2-4(k+2)x+4k+5\geq0\Rightarrow k\in[-1..1]$

so can only be $k\in[0..1].$

Let $\sum_{i=1}^{5}{a_i}=s$ then

let
$g(a_1,a_2,a_3,a_4,a_5,~k,s)=\dfrac{1}{s^3}\sum_{i=1}^{5}{a_i^2}\left(\sum_{i=1}^{5}{a_i}\right)^3 + k\prod_{i=1}^{5}{a_i} + \left(\dfrac{4k+5}{s^5}-\dfrac{k+2}{s^4} \right)\left(\sum_{i=1}^{5}{a_i}\right)^5$

it remains to make sure that

$g\left(x,x+y,x+y+z,x+y+z+t,x+y+z+t+w,~\dfrac{1}{1+p},\dfrac{5}{q+1}\right),$

$g\left(x,x+y,x+y+z,x+y+z+t,x+y+z+t+w,~\dfrac{1}{1+p},5+\dfrac{1}{q+1}\right),$

$g\left(x,x+y,x+y+z,x+y+z+t,x+y+z+t+w,~\dfrac{1}{1+p},q+6\right)$

have only positive coefficients.
Is the particular case of your inequality true for all $n?$

$\prod_{i=1}^{n}{a_i}=1,~~~\sum_{i=1}^{n}{(a_i-1)(a_i-2)}\geq0$
This post has been edited 1 time. Last edited by DipoleOfMonorak, Aug 9, 2023, 4:07 PM
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mihaig
7553 posts
#8
Y by
No. There are a few general cases more complex and difficult than the above problems.
EDIT: I figured what you mean. The inequality posed by you can be easily proved (or disproved) using standard techniques.
But the inequality I proposed is not true for all $n.$
I provided counterexample a few years ago. I don't remember when or where.
This post has been edited 3 times. Last edited by mihaig, Aug 9, 2023, 5:26 PM
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DipoleOfMonorak
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#9
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Сounterexample?
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arqady
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Nguyenhuyen_AG wrote:
Let $a,\,b,\,c,\,d$ are non-negative real numbers and $0 \leqslant k \leqslant \frac{2}{\sqrt{3}}.$ Prove that
$$a^2+b^2+c^2+d^2+kabcd \geqslant k+4+(k+2)(a+b+c+d-4).$$hide
Smoothing helps.
Let $a+c=constant.$ Thus, we have a linear inequality of $ac$.
The same things we can say about $b+d$.
Let $d=0$.

Thus, $$a^2+b^2+c^2+d^2+kabcd-k-4-(k+2)(a+b+c+d-4)=a^2+b^2+c^2-(k+2)(a+b+c)+3k+4=$$$$=\sum_{cyc}\left(a-\frac{k+2}{2}\right)^2+3k+4-\frac{3(k+2)^2}{4}\geq0.$$But for $b=a$ and $d=c$ we obtain:
$$2(a^2+c^2)+ka^2c^2-k-4-(k+2)(2a+2c-4)\geq0,$$which is true because
$$2(a^2+c^2)+ka^2c^2-k-4-(k+2)(2a+2c-4)\geq$$$$\geq2(a^2+c^2)+2kac-k-k-4-(k+2)(2a+2c-4)\geq$$$$\geq2\cdot\frac{(k+2)}{4}(a+c)^2-2k-4-(k+2)(2a+2c-4)=$$$$=\frac{k+2}{2}(a+c-2)^2\geq0.$$
This post has been edited 1 time. Last edited by arqady, May 21, 2025, 10:07 AM
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Butterfly
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why you can make $d=0$?
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arqady
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Butterfly wrote:
why you can make $d=0$?
Since if $c+d=const$, the inequality is linear inequality of $cd$, which says that it's enough to understand, what happens for an extremal value of $cd$.
If $cd$ is minimally, so it's enough to assume $d=0$.
If $cd$ is maximally, so it's enough to assume $c=d$.
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