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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
Something nice
KhuongTrang   32
N 4 minutes ago by arqady
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
32 replies
KhuongTrang
Nov 1, 2023
arqady
4 minutes ago
Infimum of decreasing sequence b_n/n^2
a1267ab   35
N 18 minutes ago by shendrew7
Source: USA Winter TST for IMO 2020, Problem 1 and TST for EGMO 2020, Problem 3, by Carl Schildkraut and Milan Haiman
Choose positive integers $b_1, b_2, \dotsc$ satisfying
\[1=\frac{b_1}{1^2} > \frac{b_2}{2^2} > \frac{b_3}{3^2} > \frac{b_4}{4^2} > \dotsb\]and let $r$ denote the largest real number satisfying $\tfrac{b_n}{n^2} \geq r$ for all positive integers $n$. What are the possible values of $r$ across all possible choices of the sequence $(b_n)$?

Carl Schildkraut and Milan Haiman
35 replies
a1267ab
Dec 16, 2019
shendrew7
18 minutes ago
IMO Genre Predictions
ohiorizzler1434   52
N 37 minutes ago by justaguy_69
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
52 replies
ohiorizzler1434
May 3, 2025
justaguy_69
37 minutes ago
My Unsolved FE in R+
ZeltaQN2008   0
39 minutes ago
Source: Ho Chi Minh TST 2017 - 2018
Find all functions $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that for all any $x,y\in (0,\infty):$
$$f(1+xf(y))=yf(x+y)$$
0 replies
ZeltaQN2008
39 minutes ago
0 replies
\sqrt{a^2+b^2+2}+\sqrt{b^2+c^2+2 }+\sqrt{c^2+a^2+2}\ge 6
parmenides51   19
N 2 hours ago by NicoN9
Source: JBMO Shortlist 2017 A1
Let $a, b, c$ be positive real numbers such that $a + b + c + ab + bc + ca + abc = 7$. Prove
that $\sqrt{a^2 + b^2 + 2 }+\sqrt{b^2 + c^2 + 2 }+\sqrt{c^2 + a^2 + 2 } \ge 6$ .
19 replies
parmenides51
Jul 25, 2018
NicoN9
2 hours ago
Inspired by Austria 2025
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b\geq 0 ,a,b\neq 1$ and $  a^2+b^2=1. $ Prove that$$   (a + b ) \left( \frac{a}{(b -1)^2} + \frac{b}{(a - 1)^2} \right) \geq 12+8\sqrt 2$$
1 reply
sqing
2 hours ago
sqing
2 hours ago
Concurrency from isogonal Mittenpunkt configuration
MarkBcc168   17
N 4 hours ago by Ilikeminecraft
Source: Fake USAMO 2020 P3
Let $\triangle ABC$ be a scalene triangle with circumcenter $O$, incenter $I$, and incircle $\omega$. Let $\omega$ touch the sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ at points $D$, $E$, and $F$ respectively. Let $T$ be the projection of $D$ to $\overline{EF}$. The line $AT$ intersects the circumcircle of $\triangle ABC$ again at point $X\ne A$. The circumcircles of $\triangle AEX$ and $\triangle AFX$ intersect $\omega$ again at points $P\ne E$ and $Q\ne F$ respectively. Prove that the lines $EQ$, $FP$, and $OI$ are concurrent.

Proposed by MarkBcc168.
17 replies
MarkBcc168
Apr 28, 2020
Ilikeminecraft
4 hours ago
Inequality with a,b,c
GeoMorocco   7
N 4 hours ago by lele0305
Source: Morocco Training
Let $   a,b,c   $ be positive real numbers such that : $   ab+bc+ca=3   $ . Prove that : $$\frac{\sqrt{1+a^2}}{1+ab}+\frac{\sqrt{1+b^2}}{1+bc}+\frac{\sqrt{1+c^2}}{1+ca}\ge \sqrt{\frac{3(a+b+c)}{2}}$$
7 replies
GeoMorocco
Apr 11, 2025
lele0305
4 hours ago
Property of the divisors of k^3 - 2
Scilyse   2
N 5 hours ago by Assassino9931
Source: KoMaL A. 892
Given two integers, $k$ and $d$ such that $d$ divides $k^3 - 2$. Show that there exists integers $a$, $b$, $c$ satisfying $d = a^3 + 2b^3 + 4c^3 - 6abc$.

Proposed by Csongor Beke and László Bence Simon, Cambridge
2 replies
Scilyse
Jan 13, 2025
Assassino9931
5 hours ago
Centroid, altitudes and medians, and concyclic points
BR1F1SZ   1
N 5 hours ago by sami1618
Source: Austria National MO Part 1 Problem 2
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)
1 reply
BR1F1SZ
Yesterday at 9:45 PM
sami1618
5 hours ago
Nordic 2025 P3
anirbanbz   8
N 5 hours ago by lksb
Source: Nordic 2025
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. Let $E$ and $F$ be points on the line segments $AC$ and $AB$ respectively such that $AEHF$ is a parallelogram. Prove that $\vert OE \vert = \vert OF \vert$.
8 replies
anirbanbz
Mar 25, 2025
lksb
5 hours ago
another functional inequality?
Scilyse   32
N 6 hours ago by ihategeo_1969
Source: 2023 ISL A4
Let $\mathbb R_{>0}$ be the set of positive real numbers. Determine all functions $f \colon \mathbb R_{>0} \to \mathbb R_{>0}$ such that \[x \big(f(x) + f(y)\big) \geqslant \big(f(f(x)) + y\big) f(y)\]for every $x, y \in \mathbb R_{>0}$.
32 replies
Scilyse
Jul 17, 2024
ihategeo_1969
6 hours ago
Mount Inequality erupts in all directions!
BR1F1SZ   1
N 6 hours ago by sami1618
Source: Austria National MO Part 1 Problem 1
Let $a$, $b$ and $c$ be pairwise distinct nonnegative real numbers. Prove that
\[
(a + b + c) \left( \frac{a}{(b - c)^2} + \frac{b}{(c - a)^2} + \frac{c}{(a - b)^2} \right) > 4.
\](Karl Czakler)
1 reply
BR1F1SZ
Yesterday at 9:41 PM
sami1618
6 hours ago
Division involving difference of squares
BR1F1SZ   1
N 6 hours ago by grupyorum
Source: Austria National MO Part 1 Problem 4
Determine all integers $n$ that can be written in the form
\[
n = \frac{a^2 - b^2}{b},
\]where $a$ and $b$ are positive integers.

(Walther Janous)
1 reply
BR1F1SZ
Yesterday at 9:50 PM
grupyorum
6 hours ago
FE solution too simple?
Yiyj1   9
N Apr 23, 2025 by jasperE3
Source: 101 Algebra Problems from the AMSP
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the equality $$f(f(x)+y) = f(x^2-y)+4f(x)y$$holds for all pairs of real numbers $(x,y)$.

My solution

I feel like my solution is too simple. Is there something I did wrong or something I missed?
9 replies
Yiyj1
Apr 9, 2025
jasperE3
Apr 23, 2025
FE solution too simple?
G H J
G H BBookmark kLocked kLocked NReply
Source: 101 Algebra Problems from the AMSP
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Yiyj1
1265 posts
#1
Y by
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the equality $$f(f(x)+y) = f(x^2-y)+4f(x)y$$holds for all pairs of real numbers $(x,y)$.

My solution

I feel like my solution is too simple. Is there something I did wrong or something I missed?
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InterLoop
280 posts
#2 • 1 Y
Y by Yiyj1
You cannot immediately "cancel" the $f$ without further conclusions.

For example $f(3) = f(2) = 1$ is possible for a function - this does not mean that $3 = 2$.
The property that leads you to $f(a) = f(b) \implies a = b$ or the "cancellation" of $f$ is called injectivity. You have to prove the function is injective first before cancellation.

Another example is simply the fact that you have not "excluded" the solution $f(x) \equiv 0$ from the equation $f(f(x)) =f(x^2)$ in any way - so $f(x) = x^2$ is wrong for that function as well. (thus $f(x) \equiv 0$ is not injective)
This post has been edited 2 times. Last edited by InterLoop, Apr 9, 2025, 3:39 AM
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Yiyj1
1265 posts
#3
Y by
InterLoop wrote:
You cannot immediately "cancel" the $f$ without further conclusions.

For example $f(3) = f(2) = 1$ is possible for a function - this does not mean that $3 = 2$.
The property that leads you to $f(a) = f(b) \implies a = b$ or the "cancellation" of $f$ is called injectivity. You have to prove the function is injective first before cancellation.

ahh ic. I'll try to prove the injectivity. ty!
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AshAuktober
1004 posts
#4
Y by
This is in fact from Iran TST.
Z K Y
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davichu
8 posts
#5
Y by
Clearly, $f(x)\equiv0$ is a trivial solution, from now on, we assume it is not the case
Let $P(x,y)$ denote the assertion $f(f(x)+y) = f(x^2-y)+4f(x)y$
$$P(x,-f(x))\rightarrow f(0)=f(x^2+f(x))-4f(x)^2$$$$P(x,x^2)\rightarrow f(x^2+f(x))=f(0)+4f(x)x^2$$Adding these two together we get:
$4f(x)^2=4f(x)x^2\rightarrow f(x)^2=f(x)x^2$
Since $f(x)\neq0$,we can divide by $f(x)$ on both sides to get $f(x)=x^2$
so the only solutions are $f(x)\equiv0$ and $f(x)=x^2\forall x \in \mathbb{R}$
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Primeniyazidayi
93 posts
#6
Y by
davichu wrote:
Since $f(x)\neq0$,we can divide by $f(x)$ on both sides to get $f(x)=x^2$

You must at first prove that $f(x) =0 \text{ iff } x=0$(or simply avoid pointwise trap).
This post has been edited 1 time. Last edited by Primeniyazidayi, Apr 22, 2025, 11:12 AM
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Primeniyazidayi
93 posts
#7
Y by
The finish for @2above(hopefully correct):We will avoid pointwise trap.We of course have $f(0) =0$.Let $f(t) =0$ for $t \neq 0$.$P(t,y)$ gives $f(y) =f(t^2-y)$.Take some $u$ such that $f(u) =u^2 \neq 0$.Then we have $u^2=t^2(t^2-2u) +u^2$ or $u=\frac{t^2}{2}$.But $P(0, x) $ gives that $f$ is even which means $\frac{t^2}{2}=-\frac{t^2}{2}$ or $t=0$, contradiction. Thus we are done.
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ariopro1387
14 posts
#8
Y by
Let $P(x,y)$ be the assertion of the problem.
$P(x,\frac{x^2-f(x)}{2});$ $\frac{x^2-f(x)}{2}.f(x) = 0$
$\forall x \in \mathbb{R}$
1. $f(x)\equiv0$
2. $f(x)=x^2$
we have to just check that both won't happen:
if $f(x_{1}) = 0:$
$P(x_{1},y);$ $f(y) = f(x_{1}^2-y)$
then by changing $y$ value we get that $x_{1} = 0$ or $f(x)\equiv C$ (Just $C=0$ works).
This post has been edited 1 time. Last edited by ariopro1387, Apr 22, 2025, 4:06 PM
Reason: edit
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lksb
171 posts
#9 • 1 Y
Y by Yiyj1
one-liner
This post has been edited 1 time. Last edited by lksb, Apr 22, 2025, 7:15 PM
Reason: typo
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jasperE3
11293 posts
#10
Y by
lksb wrote:
one-liner

pointwise trap
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