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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.

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0 replies
jlacosta
May 1, 2025
0 replies
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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
J
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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
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2-var inequality
sqing   0
17 minutes ago
Source: Own
Let $ a,b\geq 0 $ and $ \frac{a}{b+2} +\frac{1}{a+2}+ \frac{b}{3} +ab\leq1. $ Prove that
$$ \frac{a}{2b+1}+\frac{1}{2a+1} +\frac{b}{3}\leq\frac{9\sqrt 2-1}{7}$$Let $ a,b \geq 0 $ and $\frac{a}{b+5} + \frac{b}{a+5} +\frac{1}{ab+5}\geq \frac{1}{2} . $ Prove that
$$ \frac{a}{2b+1}+\frac{1}{2a+1} +\frac{b}{3}\geq1$$
0 replies
sqing
17 minutes ago
0 replies
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help me ..
exoticc   1
N 32 minutes ago by jasperE3
Find all pairs of functions (f,g) : R->R that satisfy:
f(1)=2025;
g(f(x+y))+2x+y-1=f(x)+(2x+y)g(y) , ∀x,y∈R
1 reply
exoticc
3 hours ago
jasperE3
32 minutes ago
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Combi Algorithm/PHP/..
CatalanThinker   1
N 44 minutes ago by CatalanThinker
Source: Olympiad_Combinatorics_by_Pranav_A_Sriram
5. [Czech and Slovak Republics 1997]
Each side and diagonal of a regular n-gon (n ≥ 3) is colored blue or green. A move consists of choosing a vertex and
switching the color of each segment incident to that vertex (from blue to green or vice versa). Prove that regardless of the initial coloring, it is possible to make the number of blue segments incident to each vertex even by following a sequence of moves. Also show that the final configuration obtained is uniquely determined by the initial coloring.
1 reply
CatalanThinker
Today at 5:47 AM
CatalanThinker
44 minutes ago
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(Theorem, Lemma, Result, ...) Marathon
Oksutok   0
an hour ago
1.(Świerczkowski's Theorem) A tetrahedral chain is a sequence of regular tetrahedra where any two consecutive tetrahedra are glued together face to face. Show that there is no closed tetrahedral chain.
0 replies
Oksutok
an hour ago
0 replies
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cute geo
Royal_mhyasd   4
N 2 hours ago by Royal_mhyasd
Source: own(?)
Let $\triangle ABC$ be an acute triangle and $I$ it's incenter. Let $A'$, $B'$ and $C'$ be the projections of $I$ onto $BC$, $AC$ and $AB$ respectively. $BC \cap B'C' = \{K\}$ and $Y$ is the projection of $A'$ onto $KI$. Let $M$ be the middle of the arc $BC$ not containing $A$ and $T$ the second intersection of $A'M$ and the circumcircle of $ABC$. If $N$ is the midpoint of $AI$, $TY \cap IA' = \{P\}$, $BN \cap PC' = \{D\}$ and $CN \cap PB' =\{E\}$, prove that $NEPD$ is cyclic.
PS i'm not sure if this problem is actually original so if it isn't someone please tell me so i can change the source (if that's possible)
4 replies
Royal_mhyasd
5 hours ago
Royal_mhyasd
2 hours ago
J
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Easy Number Theory
jj_ca888   13
N 2 hours ago by cursed_tangent1434
Source: SMO 2020/1
The sequence of positive integers $a_0, a_1, a_2, \ldots$ is recursively defined such that $a_0$ is not a power of $2$, and for all nonnegative integers $n$:

(i) if $a_n$ is even, then $a_{n+1} $ is the largest odd factor of $a_n$
(ii) if $a_n$ is odd, then $a_{n+1} = a_n + p^2$ where $p$ is the smallest prime factor of $a_n$

Prove that there exists some positive integer $M$ such that $a_{m+2} = a_m $ for all $m \geq M$.

Proposed by Andrew Wen
13 replies
jj_ca888
Aug 28, 2020
cursed_tangent1434
2 hours ago
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3^n + 61 is a square
VideoCake   26
N 2 hours ago by maromex
Source: 2025 German MO, Round 4, Grade 11/12, P6
Determine all positive integers \(n\) such that \(3^n + 61\) is the square of an integer.
26 replies
VideoCake
Monday at 5:14 PM
maromex
2 hours ago
J
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Nice Collinearity
oVlad   10
N 2 hours ago by Trenod
Source: KöMaL A. 831
In triangle $ABC$ let $F$ denote the midpoint of side $BC$. Let the circle passing through point $A$ and tangent to side $BC$ at point $F$ intersect sides $AB$ and $AC$ at points $M$ and $N$, respectively. Let the line segments $CM$ and $BN$ intersect in point $X$. Let $P$ be the second point of intersection of the circumcircles of triangles $BMX$ and $CNX$. Prove that points $A, F$ and $P$ are collinear.

Proposed by Imolay András, Budapest
10 replies
oVlad
Oct 11, 2022
Trenod
2 hours ago
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Hardest in ARO 2008
discredit   29
N 2 hours ago by JARP091
Source: ARO 2008, Problem 11.8
In a chess tournament $ 2n+3$ players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least $ n$ next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
29 replies
discredit
Jun 11, 2008
JARP091
2 hours ago
J
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Macedonian Mathematical Olympiad 2019 problem 1
Lukaluce   5
N 2 hours ago by AylyGayypow009
In an acute-angled triangle $ABC$, point $M$ is the midpoint of side $BC$ and the centers of the $M$- excircles of triangles $AMB$ and $AMC$ are $D$ and $E$, respectively. The circumcircle of triangle $ABD$ intersects line $BC$ at points $B$ and $F$. The circumcircle of triangle $ACE$ intersects line $BC$ at points $C$ and $G$. Prove that $BF\hspace{0.25mm} = \hspace{0.25mm} CG$ .
5 replies
Lukaluce
Apr 20, 2019
AylyGayypow009
2 hours ago
J
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Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   9
N 2 hours ago by JARP091
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
9 replies
OgnjenTesic
May 22, 2025
JARP091
2 hours ago
J
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Number of real roots
girishpimoli   2
N 3 hours ago by HAL9000sk
Number of real roots of

$\displaystyle 2\sin(\theta)\cos(3\theta)\sin(5\theta)=-1$
2 replies
girishpimoli
Yesterday at 5:35 PM
HAL9000sk
3 hours ago
J
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Inequality
lgx57   1
N 5 hours ago by sqing
$a,b,c \in \mathbb{R}^{+}$,$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1$. Prove that
$$a^abc+b^bac+c^cab \ge 27(ab+bc+ca)$$
1 reply
lgx57
5 hours ago
sqing
5 hours ago
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Pls help I'm stuck
UtkarshSri   3
N 5 hours ago by lbh_qys
So I was solving this problem
Cbrt(x+1) - cbrt(x-1) = pow((x²-1),1/6)


Here, the square root is defined in the domain x≤-1 OR x≥1
I substituted a= cbrt(x+1) and b= cbrt(x-1)
And then the equation turned out to be
a-b= sqrt(ab)
Here, since the RHS is +ve, the LHS must be too, so a>b (Done so as to prevent extraneous roots)
Now, I can write the domain in terms of a and b as ab≥0
Squaring, we get a²+b²-3ab=0
On solving we get a/b = (3±sqrt(5))/2
Now, we know that a>b from the above condition, so the only root is a/b= (3+sqrt(5))/2 and on substituting for x, and applying componendo and dividendo after cubing, I get x= sqrt(5)/2

Now, my question is, since the equation is even i.e f(x)=f(-x), it's symmetric about the y-axis, so x= -sqrt(5)/2 must also satisfy the equation, and indeed it does, so where did I made some irreversible step that caused this loss of the root?
when I plugged the equation to wolframalpha, it showed that the "principle root" has the former solⁿ only where's for a "real valued root", it has both.I also found that on solving the equation
b-a= sqrt(ab), I get the latter root.But I mean where did that extra -ve sign come from?
I'm familiar that f(x)=sqrt(g(x)) is equivalent to f²(x)=g(x) AND f(x)≥0 AND g(x)≥0 but the g(x)≥0 condition is not necessary as f²(x) is already ≥0
and also that abs(f(x))=abs(g(x)) is equivalent to f²(x)=g²(x)

PS: Pardon me I don't know latex well, I'm tried to make this as comprehensive as possible and made sure there's no loss of equivalence in any step.
I hope someone clarifies this for me, it's really bothering me for a while.
3 replies
UtkarshSri
Yesterday at 9:57 AM
lbh_qys
5 hours ago
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BABBAGE'S THEOREM EXTENSION
Mathgloggers   4
N May 6, 2025 by Mathgloggers
A few days ago I came across. this interesting result is someone interested in proving this.

$\boxed{\sum_{k=1}^{p-1} \frac{1}{k} \equiv \sum_{k=p+1}^{2p-1} \frac{1}{k} \equiv \sum_{k=2p+1}^{3p-1}\frac{1}{k} \equiv.....\sum_{k=p(p-1)+1}^{p^2-1}\frac{1}{k} \equiv 0(mod p^2)}$
4 replies
Mathgloggers
Apr 29, 2025
Mathgloggers
May 6, 2025
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BABBAGE'S THEOREM EXTENSION
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Reveal topic tags
number theory
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Mathgloggers
91 posts
Mathgloggers
#1 h  Apr 29, 2025, 12:18 PM • 1 Y
Y by orange0707
A few days ago I came across. this interesting result is someone interested in proving this.

$\boxed{\sum_{k=1}^{p-1} \frac{1}{k} \equiv \sum_{k=p+1}^{2p-1} \frac{1}{k} \equiv \sum_{k=2p+1}^{3p-1}\frac{1}{k} \equiv.....\sum_{k=p(p-1)+1}^{p^2-1}\frac{1}{k} \equiv 0(mod p^2)}$
This post has been edited 1 time. Last edited by Mathgloggers, Apr 29, 2025, 12:22 PM
Reason: D
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orange0707
2 posts
orange0707
#2 h  May 6, 2025, 3:46 AM • 1 Y
Y by Mathgloggers
Is this an extension of Wolstenholme's theorem?
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Mathgloggers
91 posts
Mathgloggers
#3 h  May 6, 2025, 4:18 AM
Y by
Yes a sort of because that connects with $\frac{1}{k} (mod p^2) $ congruences but this also follows from Babbages theorem which is:
$ \binom{aP}{bP} \equiv 0 (mod p) $ which act a sort of proof for Wolstenholme's theorem.
So when I was working I saw the congruences of of the other part because at $\equiv 0 (mod p) $ modulo inverse does not exist and I came up with this.
This post has been edited 1 time. Last edited by Mathgloggers, May 6, 2025, 4:19 AM
Reason: m
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orange0707
2 posts
orange0707
#4 h  May 6, 2025, 6:50 AM • 1 Y
Y by Mathgloggers
This is really an interesting idea. When I saw your conclusion, I happened to be studying the method of constructing polynomials and then proving Wolstenholme's theorem by using Lagrange's theorem and Fermat's little theorem, and found that it could also be used here. Additionally, may I ask when new users can upload images?Can complex mathematical formulas be directly entered using a LaTeX editor?
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Mathgloggers
91 posts
Mathgloggers
#5 h  May 6, 2025, 7:11 AM
Y by
Yes you can upload images through the pin image icon the extreme right of head bar where you send solution and yes you can upload complex mathematical formula .
I would recommend you the "LATEX FORUMS " on aops you can check that to form your formulas
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