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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)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
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[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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
Calculus
youochange   4
N 3 minutes ago by youochange
Find the area enclosed by the curves $e^x,e^{-x},x^2+y^2=1$
4 replies
youochange
Yesterday at 2:38 PM
youochange
3 minutes ago
5-th powers is a no-go - JBMO Shortlist
WakeUp   9
N 11 minutes ago by Namisgood
Prove that there are are no positive integers $x$ and $y$ such that $x^5+y^5+1=(x+2)^5+(y-3)^5$.

Note
9 replies
WakeUp
Oct 30, 2010
Namisgood
11 minutes ago
IMO 2008, Question 1
orl   156
N 12 minutes ago by Siddharthmaybe
Source: IMO Shortlist 2008, G1
Let $ H$ be the orthocenter of an acute-angled triangle $ ABC$. The circle $ \Gamma_{A}$ centered at the midpoint of $ BC$ and passing through $ H$ intersects the sideline $ BC$ at points $ A_{1}$ and $ A_{2}$. Similarly, define the points $ B_{1}$, $ B_{2}$, $ C_{1}$ and $ C_{2}$.

Prove that the six points $ A_{1}$, $ A_{2}$, $ B_{1}$, $ B_{2}$, $ C_{1}$ and $ C_{2}$ are concyclic.

Author: Andrey Gavrilyuk, Russia
156 replies
orl
Jul 16, 2008
Siddharthmaybe
12 minutes ago
APMO 2015 P1
aditya21   63
N 13 minutes ago by Siddharthmaybe
Source: APMO 2015
Let $ABC$ be a triangle, and let $D$ be a point on side $BC$. A line through $D$ intersects side $AB$ at $X$ and ray $AC$ at $Y$ . The circumcircle of triangle $BXD$ intersects the circumcircle $\omega$ of triangle $ABC$ again at point $Z$ distinct from point $B$. The lines $ZD$ and $ZY$ intersect $\omega$ again at $V$ and $W$ respectively.
Prove that $AB = V W$

Proposed by Warut Suksompong, Thailand
63 replies
aditya21
Mar 30, 2015
Siddharthmaybe
13 minutes ago
AD=BE implies ABC right
v_Enhance   118
N 14 minutes ago by Siddharthmaybe
Source: European Girl's MO 2013, Problem 1
The side $BC$ of the triangle $ABC$ is extended beyond $C$ to $D$ so that $CD = BC$. The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$. Prove that, if $AD=BE$, then the triangle $ABC$ is right-angled.
118 replies
v_Enhance
Apr 10, 2013
Siddharthmaybe
14 minutes ago
Cyclic Quads and Parallel Lines
gracemoon124   17
N 15 minutes ago by Siddharthmaybe
Source: 2015 British Mathematical Olympiad?
Let $ABCD$ be a cyclic quadrilateral. Let $F$ be the midpoint of the arc $AB$ of its circumcircle which does not contain $C$ or $D$. Let the lines $DF$ and $AC$ meet at $P$ and the lines $CF$ and $BD$ meet at $Q$. Prove that the lines $PQ$ and $AB$ are parallel.
17 replies
gracemoon124
Aug 16, 2023
Siddharthmaybe
15 minutes ago
Problem 1 (First Day)
Valentin Vornicu   137
N 17 minutes ago by Siddharthmaybe
1. Let $ABC$ be an acute-angled triangle with $AB\neq AC$. The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$ respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$. Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the side $BC$.
137 replies
Valentin Vornicu
Jul 12, 2004
Siddharthmaybe
17 minutes ago
Concentric Circles
MithsApprentice   62
N 18 minutes ago by Siddharthmaybe
Source: USAMO 1998
Let ${\cal C}_1$ and ${\cal C}_2$ be concentric circles, with ${\cal C}_2$ in the interior of ${\cal C}_1$. From a point $A$ on ${\cal C}_1$ one draws the tangent $AB$ to ${\cal C}_2$ ($B\in {\cal C}_2$). Let $C$ be the second point of intersection of $AB$ and ${\cal C}_1$, and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects ${\cal C}_2$ at $E$ and $F$ in such a way that the perpendicular bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof, the ratio $AM/MC$.
62 replies
MithsApprentice
Oct 9, 2005
Siddharthmaybe
18 minutes ago
four points lie on a circle
pohoatza   77
N 23 minutes ago by Siddharthmaybe
Source: IMO Shortlist 2006, Geometry 2, AIMO 2007, TST 1, P2
Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic.

Proposed by Vyacheslev Yasinskiy, Ukraine
77 replies
pohoatza
Jun 28, 2007
Siddharthmaybe
23 minutes ago
Hard combi
EeEApO   6
N 30 minutes ago by aidan0626
In a quiz competition, there are a total of $100 $questions, each with $4$ answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?

Now, suppose my spouse and I move into a new home. Every year, we have twins. Starting at the age of $16$, each of our twin children also begins to have twins every year. If this pattern continues, how many years will it take for my family to grow large enough to have the required number of members to guarantee winning the quiz gift?
6 replies
EeEApO
May 8, 2025
aidan0626
30 minutes ago
The familiar right angle from the orthocenter
buratinogigle   0
44 minutes ago
Source: Own, HSGSO P6
Let $ABC$ be a triangle inscribed in a circle $\omega$ with orthocenter $H$ and altitude $BE$. Let $M$ be the midpoint of $AH$. Line $BM$ meets $\omega$ again at $P$. Line $PE$ meets $\omega$ again at $Q$. Let $K$ be the orthogonal projection of $E$ on the line $BC$. Line $QK$ meets $\omega$ again at $G$. Prove that $GA\perp GH$.
0 replies
buratinogigle
44 minutes ago
0 replies
This question just asks if you can factorise 12 factorial or not
Sadigly   4
N an hour ago by NicoN9
Source: Azerbaijan Junior MO 2025 P1
A teacher creates a fraction using numbers from $1$ to $12$ (including $12$). He writes some of the numbers on the numerator, and writes $\times$ (multiplication) between each number. Then he writes the rest of the numbers in the denominator and also writes $\times$ between each number. There is at least one number both in numerator and denominator. The teacher ensures that the fraction is equal to the smallest possible integer possible.

What is this positive integer, which is also the value of the fraction?
4 replies
Sadigly
Friday at 7:34 AM
NicoN9
an hour ago
Divisibility NT
reni_wee   0
an hour ago
Source: Iran 1998
Suppose that $a$ and $b$ are natural numbers such that
$$p = \frac{b}{4}\sqrt{\frac{2a-b}{2a+b}}$$is a prime number. Find all possible values of $a$,$b$,$p$.
0 replies
reni_wee
an hour ago
0 replies
Six variables
Nguyenhuyen_AG   0
an hour ago
Let $a,\,b,\,c,\,x,\,y,\,z$ be six positive real numbers. Prove that
$$\frac{a}{b+c} \cdot \frac{y+z}{x} + \frac{b}{c+a} \cdot \frac{z+x}{y} + \frac{c}{a+b} \cdot \frac{x+y}{z} \geqslant 2+\sqrt{\frac{8abc}{(a+b)(b+c)(c+a)}}.$$
0 replies
Nguyenhuyen_AG
an hour ago
0 replies
find f
ali666   5
N Apr 30, 2025 by Blackbeam999
find all valued functions $f$ such that for all real $x,y$:
$f(x-y)=f(x)f(y)$
5 replies
ali666
Aug 19, 2006
Blackbeam999
Apr 30, 2025
find f
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G H BBookmark kLocked kLocked NReply
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ali666
352 posts
#1 • 2 Y
Y by Adventure10, Mango247
find all valued functions $f$ such that for all real $x,y$:
$f(x-y)=f(x)f(y)$
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Rust
5049 posts
#2 • 2 Y
Y by Adventure10, Mango247
One solution is f(x)=0.
y=0 give f(x)(1-f(0))=0. If exist x, suth that $f(x)\not =0$, then f(0)=1.
x=y give $f(y)^{2}=1$ and x=2y give f(y)(1-f(2y))=0.
Therefore only 2 solutions:
1. $f(x)=0 \ \forall x\in R,$
2. $f(x)=1 \ \forall x\in R.$
Z K Y
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ali666
352 posts
#3 • 2 Y
Y by Adventure10, Mango247
Rust wrote:
One solution is f(x)=0.
y=0 give f(x)(1-f(0))=0. If exist x, suth that $f(x)\not =0$, then f(0)=1.
x=y give $f(y)^{2}=1$ and x=2y give f(y)(1-f(2y))=0.
Therefore only 2 solutions:
1. $f(x)=0 \ \forall x\in R,$
2. $f(x)=1 \ \forall x\in R.$
thank Rust.
a similar problem : http://www.mathlinks.ro/Forum/viewtopic.php?t=97604
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Blackbeam999
24 posts
#4
Y by
Rust wrote:
One solution is f(x)=0.
y=0 give f(x)(1-f(0))=0. If exist x, suth that $f(x)\not =0$, then f(0)=1.
x=y give $f(y)^{2}=1$ and x=2y give f(y)(1-f(2y))=0.
Therefore only 2 solutions:
1. $f(x)=0 \ \forall x\in R,$
2. $f(x)=1 \ \forall x\in R.$

How can I prove that there is no a,b exist such that f(a)=1 , f(b)= -1?
Z K Y
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jasperE3
11320 posts
#5 • 1 Y
Y by Blackbeam999
Blackbeam999 wrote:
Rust wrote:
One solution is f(x)=0.
y=0 give f(x)(1-f(0))=0. If exist x, suth that $f(x)\not =0$, then f(0)=1.
x=y give $f(y)^{2}=1$ and x=2y give f(y)(1-f(2y))=0.
Therefore only 2 solutions:
1. $f(x)=0 \ \forall x\in R,$
2. $f(x)=1 \ \forall x\in R.$

How can I prove that there is no a,b exist such that f(a)=1 , f(b)= -1?

Because if $f(a)=-1$ for some $a$ then from $f(x)(1-f(2x))=0$ we have:
$$f\left(\frac a2\right)(1-f(a))=0$$and so $f\left(\frac a2\right)=0$, but this contradicts $f\left(\frac a2\right)\in\{-1,1\}$.
Z K Y
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Blackbeam999
24 posts
#6
Y by
jasperE3 wrote:
Blackbeam999 wrote:
Rust wrote:
One solution is f(x)=0.
y=0 give f(x)(1-f(0))=0. If exist x, suth that $f(x)\not =0$, then f(0)=1.
x=y give $f(y)^{2}=1$ and x=2y give f(y)(1-f(2y))=0.
Therefore only 2 solutions:
1. $f(x)=0 \ \forall x\in R,$
2. $f(x)=1 \ \forall x\in R.$

How can I prove that there is no a,b exist such that f(a)=1 , f(b)= -1?

Because if $f(a)=-1$ for some $a$ then from $f(x)(1-f(2x))=0$ we have:
$$f\left(\frac a2\right)(1-f(a))=0$$and so $f\left(\frac a2\right)=0$, but this contradicts $f\left(\frac a2\right)\in\{-1,1\}$.

Oh thanks
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