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k a June Highlights and 2025 AoPS Online Class Information
jlacosta   0
Jun 2, 2025
Congratulations to all the mathletes who competed at National MATHCOUNTS! If you missed the exciting Countdown Round, you can watch the video at this link. Are you interested in training for MATHCOUNTS or AMC 10 contests? How would you like to train for these math competitions in half the time? We have accelerated sections which meet twice per week instead of once starting on July 8th (7:30pm ET). These sections fill quickly so enroll today!

[list][*]MATHCOUNTS/AMC 8 Basics
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[*]AMC 10 Problem Series[/list]
For those interested in Olympiad level training in math, computer science, physics, and chemistry, be sure to enroll in our WOOT courses before August 19th to take advantage of early bird pricing!

Summer camps are starting this 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 a transformative 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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0 replies
jlacosta
Jun 2, 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
J
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Non-classical FE
M11100111001Y1R   0
an hour ago
Source: Iran TST 2025 Test 2 Problem 3
Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for all $x, y >0$ we have:
$$f(f(f(xy))+x^2)=f(y)(f(x)-f(x+y))$$
0 replies
M11100111001Y1R
an hour ago
0 replies
J
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Calculus
Bob96   0
an hour ago
Given a sequence \{a_n\} of positive real numbers that decrease to 0, and define f(x)=\sum_{n=1}^\infty a_n^n x^n. If \sum_{n=1}^\infty a_n diverges, prove that \int_1^\infty \frac{\ln f(x)}{x^2}dx diverges
0 replies
Bob96
an hour ago
0 replies
J
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Most accurate rounding
popcorn1   24
N an hour ago by HamstPan38825
Source: USA Team Selection Test for IMO 2024, Problem 1
Find the smallest constant $C > 1$ such that the following statement holds: for every integer $n \geq 2$ and sequence of non-integer positive real numbers $a_1, a_2, \dots, a_n$ satisfying $$\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} = 1,$$it's possible to choose positive integers $b_i$ such that
(i) for each $i = 1, 2, \dots, n$, either $b_i = \lfloor a_i \rfloor$ or $b_i = \lfloor a_i \rfloor + 1$, and
(ii) we have $$1 < \frac{1}{b_1} + \frac{1}{b_2} + \cdots + \frac{1}{b_n} \leq C.$$(Here $\lfloor \bullet \rfloor$ denotes the floor function, as usual.)

Merlijn Staps
24 replies
popcorn1
Dec 11, 2023
HamstPan38825
an hour ago
J
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Concurrence
LiamChen   1
N an hour ago by phonghatemath
Source: MOP1998
Problem:
1 reply
LiamChen
3 hours ago
phonghatemath
an hour ago
J
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Merlijn Has 100 Coins
tastymath75025   40
N an hour ago by peace09
Source: USA TSTST 2019 Problem 4
Consider coins with positive real denominations not exceeding 1. Find the smallest $C>0$ such that the following holds: if we have any $100$ such coins with total value $50$, then we can always split them into two stacks of $50$ coins each such that the absolute difference between the total values of the two stacks is at most $C$.

Merlijn Staps
40 replies
tastymath75025
Jun 25, 2019
peace09
an hour ago
J
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IMO ShortList 1998, algebra problem 1
orl   39
N an hour ago by Siddharthmaybe
Source: IMO ShortList 1998, algebra problem 1
Let $a_{1},a_{2},\ldots ,a_{n}$ be positive real numbers such that $a_{1}+a_{2}+\cdots +a_{n}<1$. Prove that

\[ \frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}. \]
39 replies
orl
Oct 22, 2004
Siddharthmaybe
an hour ago
J
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2013 Japan MO Finals P5
parkjungmin   0
an hour ago
2013 Japan MO Finals
0 replies
parkjungmin
an hour ago
0 replies
J
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Might be slightly generalizable
Rijul saini   6
N 2 hours ago by WLOGQED1729
Source: India IMOTC Day 3 Problem 1
Let $ABC$ be an acute angled triangle with orthocenter $H$ and $AB<AC$. Let $T(\ne B,C, H)$ be any other point on the arc $\stackrel{\LARGE\frown}{BHC}$ of the circumcircle of $BHC$ and let line $BT$ intersect line $AC$ at $E(\ne A)$ and let line $CT$ intersect line $AB$ at $F(\ne A)$. Let the circumcircles of $AEF$ and $ABC$ intersect again at $X$ ($\ne A$). Let the lines $XE,XF,XT$ intersect the circumcircle of $(ABC)$ again at $P,Q,R$ ($\ne X$). Prove that the lines $AR,BC,PQ$ concur.
6 replies
Rijul saini
Yesterday at 6:39 PM
WLOGQED1729
2 hours ago
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Prove that circumcircle of the triangle TEF tangent to (O), (K), (L)
centos6   8
N 2 hours ago by ihategeo_1969
Let $(O)$ be the circumcircle of the triangle $\triangle ABC$. $A’$ be the antipode of $A$ in $(O)$. Angle bisector of angle $\angle A$ meets $BC$ and $A-Mixtilinear$ at $D$ and $E$. Let $N$ be the midpoind of the arc $BAC$. $ T = A’E \cap (O), T \neq A’, F = AD \cap NT$. $(K)$ and $(L)$ be the Thebault circles of the cevian $AD$. Prove that circumcircle of the triangle $\triangle TEF$ tangent to $(O)$, $(K)$ and $(L)$.

IMAGE
8 replies
+1 w
centos6
Nov 30, 2018
ihategeo_1969
2 hours ago
J
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What the isogonal conjugate on IO
reni_wee   1
N 2 hours ago by Funcshun840
Source: buratinogigle
Given a triangle $ABC$ with incircle $(I)$ tangent to $BC, CA, AB$ at points $D, E, F$, respectively. Let $P$ be a point such that its isogonal conjugate lies on the line $OI$ (where $O$ is the circumcenter and $I$ the incenter of $ABC$). The line $PA$ intersects segments $DE$ and $DF$ at points $M_a$ and $N_a$, respectively, such that the circle with diameter $M_a N_a$ meets $BC$ at points $P_a$ and $Q_a$.

1) Prove that the circle $(AP_a Q_a)$ is tangent to the incircle $(I)$ at some point $X$.

2) Similarly define points $Y, Z$ corresponding to vertices $B, C$. Prove that the lines $AX, BY, CZ$ are concurrent.
1 reply
reni_wee
4 hours ago
Funcshun840
2 hours ago
J
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Maybe a old inequality?
pxchg1200   5
N 2 hours ago by teomihai
Source: Tran Quoc Anh
Let $ a,b,c>0$ with $ a+b+c=3 $ prove that:
\[ (a^2+1)(b^2+1)(c^2+1)\geq (a+1)(b+1)(c+1) \]



Click to reveal hidden text
5 replies
pxchg1200
Mar 6, 2012
teomihai
2 hours ago
J
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linear algebra
ay19bme   1
N 2 hours ago by ay19bme
........
1 reply
ay19bme
Today at 11:17 AM
ay19bme
2 hours ago
J
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Another integral limit
RobertRogo   3
N Today at 11:59 AM by Bayram_Turayew
Source: "Traian Lalescu" student contest 2025, Section A, Problem 3
Let $f \colon [0, \infty) \to \mathbb{R}$ be a function differentiable at 0 with $f(0) = 0$. Find
$$\lim_{n \to \infty} \frac{1}{n} \int_{2^n}^{2^{n+1}} f\left(\frac{\ln x}{x}\right) dx$$
3 replies
RobertRogo
May 9, 2025
Bayram_Turayew
Today at 11:59 AM
J
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Derivatives on a Functional Equation
Kingofmath101   1
N Today at 11:28 AM by Mathzeus1024
Let $g$ be a smooth function on $\mathbb{R}$ where $g^{(n)}(0)$ for all $n \in \mathbb{N}^+$ and $g$ satisfies

$$g(x) = xg(x^2 - 4)$$
for all $x \in \mathbb{R}$. Prove that $g^{(n)}(-2) = g^{(n)}(2) = 0$ for all $n \in \mathbb{N}$.

1 reply
Kingofmath101
Jun 4, 2017
Mathzeus1024
Today at 11:28 AM
J
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J
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Can cos(√2 t) be expressed as a polynomial in cost?
tom-nowy   2
N May 7, 2025 by wh0nix
Source: Question arising while viewing https://artofproblemsolving.com/community/c51293h3562250
Can $\cos ( \sqrt{2}\, t )$ be expressed as a polynomial in $\cos t$ with real coefficients?
2 replies
tom-nowy
May 6, 2025
wh0nix
May 7, 2025
J
Can cos(√2 t) be expressed as a polynomial in cost?
G H J
Reveal topic tags
real analysis
trigonometry
L
G H BBookmark kLocked kLocked NReply
Source: Question arising while viewing https://artofproblemsolving.com/community/c51293h3562250
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tom-nowy
134 posts
tom-nowy
#1 h  May 6, 2025, 7:10 AM
Y by
Can $\cos ( \sqrt{2}\, t )$ be expressed as a polynomial in $\cos t$ with real coefficients?
This post has been edited 2 times. Last edited by tom-nowy, May 6, 2025, 9:25 AM
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Aiden-1089
306 posts
Aiden-1089
#2 h  May 6, 2025, 7:17 AM • 1 Y
Y by tom-nowy
The answer is no since $\cos( \frac{\pi}{2} ) = \cos( \frac{3\pi}{2} ) =0$ but $\cos( \sqrt{2} \cdot \frac{\pi}{2} ) \neq \cos( \sqrt{2} \cdot \frac{3\pi}{2} )$.
This post has been edited 1 time. Last edited by Aiden-1089, May 6, 2025, 7:17 AM
Reason: typo
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wh0nix
27 posts
wh0nix
#3 h  May 7, 2025, 4:54 AM • 1 Y
Y by tom-nowy
Alternative solution: We can prove that $\cos (\sqrt{2}x)=P(\cos x)$ it is not possible. We can se that for $t=\frac{1}{\sqrt{2}}\left( \frac{\pi }{2}+2k\pi \right)$ we have $\cos (\sqrt{2}t)=0$ but $\cos t\ne 0$ because $\sqrt{2}\ne \frac{p}{q}$ and it is not period for $\cos t$ . So result that the polynomial $P$ have infinite many roots, contradiction.
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