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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.

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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
D1036 : Composition of polynomials
Dattier   1
N 31 minutes ago by Dattier
Source: les dattes à Dattier
Find all $A \in \mathbb Q[x]$ with $\exists Q \in \mathbb Q[x], Q(A(x))= x^{2025!+2}+x^2+x+1$ and $\deg(A)>1$.
1 reply
Dattier
Yesterday at 1:52 PM
Dattier
31 minutes ago
number sequence contains every large number
mathematics2003   3
N 31 minutes ago by sttsmet
Source: 2021ChinaTST test3 day1 P2
Given distinct positive integer $ a_1,a_2,…,a_{2020} $. For $ n \ge 2021 $, $a_n$ is the smallest number different from $a_1,a_2,…,a_{n-1}$ which doesn't divide $a_{n-2020}...a_{n-2}a_{n-1}$. Proof that every number large enough appears in the sequence.
3 replies
mathematics2003
Apr 13, 2021
sttsmet
31 minutes ago
IMO ShortList 2002, geometry problem 2
orl   28
N 35 minutes ago by ezpotd
Source: IMO ShortList 2002, geometry problem 2
Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \]
28 replies
orl
Sep 28, 2004
ezpotd
35 minutes ago
Arc Midpoints Form Cyclic Quadrilateral
ike.chen   56
N 42 minutes ago by ezpotd
Source: ISL 2022/G2
In the acute-angled triangle $ABC$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $AF$. The lines through $P$ parallel to $AC$ and $AB$ meet $BC$ at $D$ and $E$, respectively. Points $X \ne A$ and $Y \ne A$ lie on the circles $ABD$ and $ACE$, respectively, such that $DA = DX$ and $EA = EY$.
Prove that $B, C, X,$ and $Y$ are concyclic.
56 replies
ike.chen
Jul 9, 2023
ezpotd
42 minutes ago
Non-linear Recursive Sequence
amogususususus   4
N an hour ago by GreekIdiot
Given $a_1=1$ and the recursive relation
$$a_{i+1}=a_i+\frac{1}{a_i}$$for all natural number $i$. Find the general form of $a_n$.

Is there any way to solve this problem and similar ones?
4 replies
amogususususus
Jan 24, 2025
GreekIdiot
an hour ago
Russian Diophantine Equation
LeYohan   2
N an hour ago by RagvaloD
Source: Moscow, 1963
Find all integer solutions to

$\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3$.
2 replies
LeYohan
Yesterday at 2:59 PM
RagvaloD
an hour ago
PQ = r and 6 more conditions
avisioner   41
N an hour ago by ezpotd
Source: 2023 ISL G2
Let $ABC$ be a triangle with $AC > BC,$ let $\omega$ be the circumcircle of $\triangle ABC,$ and let $r$ be its radius. Point $P$ is chosen on $\overline{AC}$ such taht $BC=CP,$ and point $S$ is the foot of the perpendicular from $P$ to $\overline{AB}$. Ray $BP$ mets $\omega$ again at $D$. Point $Q$ is chosen on line $SP$ such that $PQ = r$ and $S,P,Q$ lie on a line in that order. Finally, let $E$ be a point satisfying $\overline{AE} \perp \overline{CQ}$ and $\overline{BE} \perp \overline{DQ}$. Prove that $E$ lies on $\omega$.
41 replies
avisioner
Jul 17, 2024
ezpotd
an hour ago
Functional equation from R^2 to R
k.vasilev   19
N an hour ago by megahertz13
Source: All-Russian Olympiad 2019 grade 10 problem 1
Each point $A$ in the plane is assigned a real number $f(A).$ It is known that $f(M)=f(A)+f(B)+f(C),$ whenever $M$ is the centroid of $\triangle ABC.$ Prove that $f(A)=0$ for all points $A.$
19 replies
k.vasilev
Apr 23, 2019
megahertz13
an hour ago
Functional equations in IMO TST
sheripqr   50
N an hour ago by megahertz13
Source: Iran TST 1996
Find all functions $f: \mathbb R \to \mathbb R$ such that $$ f(f(x)+y)=f(x^2-y)+4f(x)y $$ for all $x,y \in \mathbb R$
50 replies
sheripqr
Sep 14, 2015
megahertz13
an hour ago
nice ecuation
MihaiT   1
N an hour ago by Hello_Kitty
Find real values $m$ , s.t. ecuation: $x+1=me^{|x-1|}$ have 2 real solutions .
1 reply
MihaiT
Today at 2:03 PM
Hello_Kitty
an hour ago
IMO Shortlist 2009 - Problem C3
nsato   25
N 2 hours ago by popop614
Let $n$ be a positive integer. Given a sequence $\varepsilon_1$, $\dots$, $\varepsilon_{n - 1}$ with $\varepsilon_i = 0$ or $\varepsilon_i = 1$ for each $i = 1$, $\dots$, $n - 1$, the sequences $a_0$, $\dots$, $a_n$ and $b_0$, $\dots$, $b_n$ are constructed by the following rules: \[a_0 = b_0 = 1, \quad a_1 = b_1 = 7,\] \[\begin{array}{lll}
	a_{i+1} = 
	\begin{cases}
		2a_{i-1} + 3a_i, \\
		3a_{i-1} + a_i, 
	\end{cases} & 
        \begin{array}{l} 
                \text{if } \varepsilon_i = 0, \\  
                \text{if } \varepsilon_i = 1, \end{array} 
         & \text{for each } i = 1, \dots, n - 1, \\[15pt]
        b_{i+1}= 
        \begin{cases}
		2b_{i-1} + 3b_i, \\
		3b_{i-1} + b_i, 
	\end{cases} & 
        \begin{array}{l} 
                \text{if } \varepsilon_{n-i} = 0, \\  
                \text{if } \varepsilon_{n-i} = 1, \end{array} 
         & \text{for each } i = 1, \dots, n - 1.
	\end{array}\] Prove that $a_n = b_n$.

Proposed by Ilya Bogdanov, Russia
25 replies
nsato
Jul 6, 2010
popop614
2 hours ago
Linear algebra problem
Feynmann123   1
N 5 hours ago by Etkan
Let A \in \mathbb{R}^{n \times n} be a matrix such that A^2 = A and A \neq I and A \neq 0.

Problem:
a) Show that the only possible eigenvalues of A are 0 and 1.
b) What kind of matrix is A? (Hint: Think projection.)
c) Give a 2×2 example of such a matrix.
1 reply
Feynmann123
Today at 9:33 AM
Etkan
5 hours ago
Different Paths Probability
Qebehsenuef   6
N 5 hours ago by Etkan
Source: OBM
A mouse initially occupies cage A and is trained to change cages by going through a tunnel whenever an alarm sounds. Each time the alarm sounds, the mouse chooses any of the tunnels adjacent to its cage with equal probability and without being affected by previous choices. What is the probability that after the alarm sounds 23 times the mouse occupies cage B?
6 replies
Qebehsenuef
Apr 28, 2025
Etkan
5 hours ago
Linear algebra
Feynmann123   6
N Today at 1:09 PM by OGMATH
Hi everyone,

I was wondering whether when I tried to compute e^(2x2 matrix) and got the expansions of sinx and cosx with the method of discounting the constant junk whether it plays any significance. I am a UK student and none of this is in my School syllabus so I was just wondering…


6 replies
Feynmann123
Yesterday at 6:44 PM
OGMATH
Today at 1:09 PM
fof=f' (open question for me)
blang   3
N Jan 1, 2010 by blang
Hi !

An open question for me :

Find all $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that $ f \circ f =f'$ .

Thanks !
3 replies
blang
Dec 31, 2009
blang
Jan 1, 2010
fof=f' (open question for me)
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blang
165 posts
#1 • 3 Y
Y by Adventure10 and 2 other users
Hi !

An open question for me :

Find all $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that $ f \circ f =f'$ .

Thanks !
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fedja
6920 posts
#2 • 6 Y
Y by pieater314159, Adventure10, Mango247, and 3 other users
Nice to have you back!

This is going to be long and technical, so I'm not even sure I'll be able to post the entire thing at once. OK, let's start at least :).

Step 1: $ f\in C^1$. Indeed, $ f$ muct be at least continuous, whence the derivative is continuous due to the equation. Actually, you can get $ C^\infty$ this way.

Step 2: $ f$ cannot have a fixed point $ a > 0$. Indeed, then we would have $ f(x)\ge a$ for all $ x\ge a$ (you cannot cross the level $ a$ down because every time you are at that level the derivative is $ a > 0$), whence $ f'(x)\ge a$ when $ x\ge a$, whence $ f(x)\ge a + a(x - a)$ for $ x\ge a$, whence $ f'(x) = f(f(x))\ge a + a^2(x - a)$ for $ x > a$, whence $ f(x)\ge a + c(x - a)^2$, so $ f'(x)\ge a + c(f(x) - a)^2$, but this inequality leads to a blow-up in finite time as $ x$ increases.

Step 3: $ f$ has a fixed point. Indeed, otherwise we have either $ f(x) > x$ for all $ x$, or $ f(x) < x$ for all $ x$. The first option will result in $ f'(x) > x$, in $ f(x) > \frac 12 x^2$, and, finally in $ f'(x)\ge \frac 12 f(x)^2$ and a blow-up in finite time. The second option will result in $ f'(x) < 0$ for all $ x < 0$, which is incompatible with going to $ - \infty$ at $ - \infty$.

Step 4: Suppose that $ f$ has $ 0$ as a fixed point. Then we have
\[ f(x) = \int_0^x f(f(t))\,dt.\]
Take the largest interval $ ( - a,a)$ with the property that $ |f\circ f|\le 1$ on $ [ - a,a]$ (possibly the entire real line). Now, we have $ |f(x)|\le |x|$ on $ ( - a,a)$ whence $ f(x)\in( - a,a)$ for $ x\in ( - a,a)$ and $ |f'(x)| = |f(f(x))|\le |f(x)|$ on $ ( - a,a)$. Now, Gronwall's lemma implies that $ f\equiv 0$ on $ ( - a,a)$, so if $ a = + \infty$, we are done and if $ a < + \infty$, then $ a$ can be increased contrdicting the choice of $ a$.

Thus, we may assume that $ f( - a) = - a$ for some $ a > 0$. We will show that for each $ a > 0$, there is a unique function $ f$ satisfying the equation and this initial condition. Needless to say, I have no hope for an explicit elementary formula for it.

Step 5: Similarly to Step 2, we can show that $ f(x)\le - a$ for $ x\ge - a$ and $ f(x)\ge - a$ for $ x\le - a$. Denoting $ g(t) = - f( - a + t) - a$ and $ h(t) = f( - a - t) + a$, we get the system
\[ g'(t) = a - h(g(t)),\qquad h'(t) = a + g(h(t))\]
with the initial data $ g(0) = h(0) = 0$ to solve in non-negative smooth functions on $ [0, + \infty)$.

Step 6: $ h$ is increasing and $ h(t)\ge at$. That is obvious.

Step 7: $ g$ is increasing and tends to a finite positive limit as $ t\to + \infty$ equal to the unique solution $ L\le 1$ of the equation $ h(L) = a$. This is not so obvious but pretty standard. The key is the Lemma in the next step that should be applied to the solution $ g$ and the constant solution $ L$ to show that $ g(t)\le L$ everywhere after which everything becomes clear.

Step 8:
Lemma: If $ F\ge G$ are two smooth functions on and $ f,g$ are two smooth solutions of $ f' = F(f)$, $ g' = G(g)$ with $ f(0)\ge g(0)$, then $ f\ge g$ on $ [0, + \infty)$.

Proof: Replace $ F$ by $ F_\delta = F + \delta$ with $ \delta > 0$ and consider $ f_\delta$ solving $ f_\delta' = F_\delta(f_\delta), f_\delta(0) = f(0) + \delta$. Then the graph of $ f_\delta$ cannot cross or even touch that of $ g$ because at the first crossing point we would have $ f_\delta' - g'\ge\delta > 0$. So, $ f_\delta > g$. But $ f_\delta\to f$ as $ \delta\to 0$ by the classical stability theorem.

Now we have come to the real meat but I'll have to interrupt my typing for at least half an hour. I'll continue as soon as I find free time.
This post has been edited 1 time. Last edited by fedja, Feb 15, 2010, 2:54 PM
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fedja
6920 posts
#3 • 3 Y
Y by pieater314159, Adventure10, and 1 other user
Uphh... Back to the proof. We have the system $ g'=a-h(g),h'=a+g(h)$. We want to solve it iteratively. Start with $ g$, find $ h$ from the second equation, find new $ g$ from the first equation for that $ h$, and so on. Everything will be fine if we are able to introduce some "distance" between possible functions $ g$, in which this single loop $ g\mapsto h\mapsto g$ is contractive. So, here goes

Step 9: Our possible functions $ g$ are $ C^1$, increasing, have positive derivative at $ 0$ and tend to some limit not greater than $ 1$ at $ +\infty$. This describes our space. For each pair $ g,g_1$ of such functions, there exists the least $ \lambda\ge 1$ such that $ \lambda^{-1}g\le g_1\le \lambda g$. This $ \lambda$ (or, more precisely, $ \lambda -1$ will measure how close $ g$ and $ g_1$ are. Note that it is not a distance but the fixed point theorem still holds in this strange space.

Step 10: Suppose that $ g$ and $ g_1$ are $ \lambda$-close, then
$ h(\mu^{-1}t) \le h_1(t)\le h(\mu t)$ with $ \mu-1=\frac{\lambda-1}{1+a}$. Indeed, we know that $ h$ solves $ h'=a+g(h)$, so $ H(t)=H(\mu t)$ solves
\[ \begin{aligned}
H'=\mu a+\mu g(H)&= a+ a(\mu-1)+\mu g(H)\ge a+ [a(\mu-1)+\mu] g(H)
\\
&=
a+\lambda g(H)\ge a+g_1(H)\,,
\end{aligned}\]
(we used that $ 0\le g\le 1$) so, by Lemma, $ H\ge h_1$. The other inequality is obtained by exchanging the roles of $ g$ and $ g_1$.

Step 11: Now it remains to show that the above inequality for $ h$ and $ h_1$ implies that new $ g$ and $ g_1$ are $ \mu$-close. Again, consider $ G=\mu g$. It solves
\[ G'=\mu(a-h(\mu^{-1}G))_+\ge (a-h_1(G))_+\]
and the lemma does the trick again (note that we can safely drop the values of $ h,h_1$ exceeding $ a$ in our differential equations for $ g,g_1$ because we never get to that area.

The end! :lol:
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blang
165 posts
#4 • 2 Y
Y by Adventure10, Mango247
Thanks, fedja ! :)
I will study your solution.
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