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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 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
(Theorem, Lemma, Result, ...) Marathon
Oksutok   0
6 minutes 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
6 minutes ago
0 replies
cute geo
Royal_mhyasd   4
N an hour 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
4 hours ago
Royal_mhyasd
an hour ago
Easy Number Theory
jj_ca888   13
N an hour 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
an hour ago
3^n + 61 is a square
VideoCake   26
N an hour 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
an hour ago
Nice Collinearity
oVlad   10
N an hour 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
an hour ago
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
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
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
DE is tangent to a fixed circle whose radius is half the radius of (O)
parmenides51   1
N 2 hours ago by TigerOnion
Source: 2017 Saudi Arabia JBMO Training Tests 2
Let $ABC$ be a triangle inscribed in circle $(O)$ such that points $B, C$ are fixed, while $A$ moves on major arc $BC$ of $(O)$. The tangents through $B$ and $C$ to $(O)$ intersect at $P$. The circle with diameter $OP$ intersects $AC$ and $AB$ at $D$ and $E$, respectively. Prove that $DE$ is tangent to a fixed circle whose radius is half the radius of $(O)$.
1 reply
parmenides51
May 28, 2020
TigerOnion
2 hours ago
Orthocorrespondent of P on Euler line
Luis González   1
N 2 hours ago by AuroralMoss
Let $O,G$ and $K$ be the circumcenter, centroid and symmedian point of $\triangle ABC,$ respectively. $P$ is an arbitrary point on Euler line $OG.$ Show that the orthocorrespondent of $P$ WRT $\triangle {ABC}$ falls on $GK.$
1 reply
Luis González
Feb 8, 2025
AuroralMoss
2 hours ago
JBMO Shortlist 2023 N3
Orestis_Lignos   9
N 2 hours ago by Just1
Source: JBMO Shortlist 2023, N3
Let $A$ be a subset of $\{2,3, \ldots, 28 \}$ such that if $a \in A$, then the residue obtained when we divide $a^2$ by $29$ also belongs to $A$.

Find the minimum possible value of $|A|$.
9 replies
Orestis_Lignos
Jun 28, 2024
Just1
2 hours ago
polygon's area doesn't add much when combined with its centric symmetry
mathematics2003   7
N 2 hours ago by sttsmet
Source: 2021ChinaTST test4 day2 P2
Find the smallest real $\alpha$, such that for any convex polygon $P$ with area $1$, there exist a point $M$ in the plane, such that the area of convex hull of $P\cup Q$ is at most $\alpha$, where $Q$ denotes the image of $P$ under central symmetry with respect to $M$.
7 replies
mathematics2003
Apr 14, 2021
sttsmet
2 hours ago
a^2-bc square implies 2a+b+c composite
v_Enhance   41
N 2 hours ago by cursed_tangent1434
Source: ELMO 2009, Problem 1
Let $a,b,c$ be positive integers such that $a^2 - bc$ is a square. Prove that $2a + b + c$ is not prime.

Evan o'Dorney
41 replies
v_Enhance
Dec 31, 2012
cursed_tangent1434
2 hours ago
IMO Shortlist 2008, Geometry problem 2
April   42
N 2 hours ago by s27_SaparbekovUmar
Source: IMO Shortlist 2008, Geometry problem 2, German TST 2, P1, 2009
Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE = \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$.

Proposed by Charles Leytem, Luxembourg
42 replies
April
Jul 9, 2009
s27_SaparbekovUmar
2 hours ago
Find all pairs (x; y)
sqing   5
N Apr 18, 2025 by EVKV
Source: 7th European Mathematical Cup , Junior Category, Q2
Find all pairs $ (x; y) $ of positive integers such that
$$xy | x^2 + 2y -1.$$
5 replies
sqing
Dec 25, 2018
EVKV
Apr 18, 2025
Find all pairs (x; y)
G H J
G H BBookmark kLocked kLocked NReply
Source: 7th European Mathematical Cup , Junior Category, Q2
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sqing
42473 posts
#1 • 4 Y
Y by rightways, Illuzion, Adventure10, Mango247
Find all pairs $ (x; y) $ of positive integers such that
$$xy | x^2 + 2y -1.$$
Z K Y
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TheDarkPrince
3042 posts
#2 • 1 Y
Y by Adventure10
sqing wrote:
Find all pairs $ (x; y) $ of positive integers such that
$$xy | x^2 + 2y -1.$$

Solution: We have $x^2 + 2y - 1\equiv 0 \pmod x\implies 2y - 1\equiv 0 \pmod x$ or $2y - 1 = xz$ for some positive integer $z$. Now we have $y\mid x+z$ or \[\frac{xz+1}{2}\mid x+z\implies xz+1\le 2(x+z) \implies (x-2)(z-2)\le 3.\]For $x = 1$, $\frac{z+1}{2}\mid z+1$ which is true for any odd natural $z$. For $x = 3$, we have $z \le 5$ which on checking gives $z = 1$ or $z = 5$. For $x = 5$, $z = 3$, and for $z=1$, $x$ is any odd. so we have solutions: \[(x,y) = (1,k),(3,2),(3,8),(5,8),(2k-1,k) ~~\text{for any positive  integer } k.\]
This post has been edited 2 times. Last edited by TheDarkPrince, Dec 25, 2018, 2:58 PM
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Bikey
131 posts
#3 • 2 Y
Y by danepale, Adventure10
TheDarkPrince wrote:
sqing wrote:
Find all pairs $ (x; y) $ of positive integers such that
$$xy | x^2 + 2y -1.$$

Solution: We have $x^2 + 2y - 1\equiv 0 \pmod x\implies 2y - 1\equiv 0 \pmod x$ or $2y - 1 = xz$ for some positive integer $z$. Now we have $y\mid x+z$ or \[\frac{xz+1}{2}\mid x+z\implies xz+1\le 2(x+z) \implies (x-2)(z-2)\le 3.\]For $x = 1$, $\frac{z+1}{2}\mid z+1$ which is true for any odd natural $z$. For $x = 3$, we have $z \le 5$ which on checking gives $z = 1$ or $z = 5$. For $x = 5$, $z = 3$, so we have solutions: \[(x,y) = (1,k),(3,2),(3,8),(5,8) ~~\text{for any positive  integer } k.\]
O.k but check the pair $(2k-1,k)$
Z K Y
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Daniil02
53 posts
#4 • 2 Y
Y by itslumi, Adventure10
Let $x^2 + 2y - 1 = kxy$
$
D = (ky)^2 - 8y + 4 = s^2 \implies 4( (\frac{ky}{2})^2 -2y + 1) = s^2 \implies (\frac{ky}{2})^2 -2y + 1 = k^2
$

If $s$ bigger than y, $(ky)^2 - s^2 \le (y+1)^2 - y^2 = 2y + 1$ so $s\le y-1$ and $(\frac{ky}{2})^2 -2y + 1) \le (y-1)^2 \implies k \le 2$.
If $k = 2$ we get solutions $(1;k)$ and $(2k-1; k)$. If $k = 1$ than notice that
$y^2 - 8y + 4 \le (y - 5)^2$ and $(y - 6)^2 \le y^2 - 8y + 4$ for all $y$ more than 8. After checking small cases we get $(3;8)$ and $(5;8)$.
This post has been edited 4 times. Last edited by Daniil02, Jun 15, 2019, 8:10 AM
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Ibrahim_K
62 posts
#6 • 2 Y
Y by Tellocan, Aoxz
Since $x|x^2+2y-1 \implies x|2y-1 \implies 2y-1=xk(1) \implies xy|x^2 + xk \implies y|x+k \implies x+k=yt(2)$
By $(1)$ we have $y=\frac{xk+1}{2}$ putting in $(2)$ we get $2(x+k)=t(xk+1)$.If $t \ge 4$ it is obvious to see that
$LHS<RHS$.Thus $t=1,2,3$

1. $t=1$
Then $2x+2k=xk+1 \implies (x-2)(k-2)=3 \implies x=3,k=5 ; x=5,k=3 \implies (x,y)=(3,8),(5,8)$

2.$t=2$
Then $x+k=xk+1 \implies (x-1)(k-1)=0$ if $x=1 \implies (x,y)=(1,n)$ , if $k=1 \implies (x,y)=(2n-1,n)$

3.$t=3$
Then $2x+2k=3xk+3 \implies \frac{2}{k} + \frac {2}{x} = 3 + \frac{3}{xk}$ if $x,k \ge 2$ then $LHS<RHS$
Thus at least one of them is equal to $1$.After checking we will not get new solution...

Hence $(x,y)=(3,8),(5,8),(1,n),(2n-1,n)$ , so we are done :-D
This post has been edited 1 time. Last edited by Ibrahim_K, Nov 22, 2022, 6:48 AM
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EVKV
71 posts
#7
Y by
2y-1=xk
y$x^2$ -1 <x
So k≤2x
putting in the value of y
xk<xk +1≤2x+k≤4x
k≤4
Now k≠4,2 as 2y-1 is odd
k=3 => x≤5 which then can be bashed
k=1 => (2a-1,a) for all naturals a
Z K Y
N Quick Reply
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