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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. 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 events:
[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
[*]April 8th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS State Discussion
April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
[*]April 10th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MathILy and MathILy-Er Math Jam: Multibackwards Numbers
[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
jlacosta
Apr 2, 2025
0 replies
k i Suggestion Form
jwelsh   0
May 6, 2021
Hello!

Given the number of suggestions we’ve been receiving, we’re transitioning to a suggestion form. If you have a suggestion for the AoPS website, please submit the Google Form:
Suggestion Form

To keep all new suggestions together, any new suggestion threads posted will be deleted.

Please remember that if you find a bug outside of FTW! (after refreshing to make sure it’s not a glitch), make sure you’re following the How to write a bug report instructions and using the proper format to report the bug.

Please check the FTW! thread for bugs and post any new ones in the For the Win! and Other Games Support Forum.
0 replies
jwelsh
May 6, 2021
0 replies
k i Read me first / How to write a bug report
slester   3
N May 4, 2019 by LauraZed
Greetings, AoPS users!

If you're reading this post, that means you've come across some kind of bug, error, or misbehavior, which nobody likes! To help us developers solve the problem as quickly as possible, we need enough information to understand what happened. Following these guidelines will help us squash those bugs more effectively.

Before submitting a bug report, please confirm the issue exists in other browsers or other computers if you have access to them.

For a list of many common questions and issues, please see our user created FAQ, Community FAQ, or For the Win! FAQ.

What is a bug?
A bug is a misbehavior that is reproducible. If a refresh makes it go away 100% of the time, then it isn't a bug, but rather a glitch. That's when your browser has some strange file cached, or for some reason doesn't render the page like it should. Please don't report glitches, since we generally cannot fix them. A glitch that happens more than a few times, though, could be an intermittent bug.

If something is wrong in the wiki, you can change it! The AoPS Wiki is user-editable, and it may be defaced from time to time. You can revert these changes yourself, but if you notice a particular user defacing the wiki, please let an admin know.

The subject
The subject line should explain as clearly as possible what went wrong.

Bad: Forum doesn't work
Good: Switching between threads quickly shows blank page.

The report
Use this format to report bugs. Be as specific as possible. If you don't know the answer exactly, give us as much information as you know. Attaching a screenshot is helpful if you can take one.

Summary of the problem:
Page URL:
Steps to reproduce:
1.
2.
3.
...
Expected behavior:
Frequency:
Operating system(s):
Browser(s), including version:
Additional information:


If your computer or tablet is school issued, please indicate this under Additional information.

Example
Summary of the problem: When I click back and forth between two threads in the site support section, the content of the threads no longer show up. (See attached screenshot.)
Page URL: http://artofproblemsolving.com/community/c10_site_support
Steps to reproduce:
1. Go to the Site Support forum.
2. Click on any thread.
3. Click quickly on a different thread.
Expected behavior: To see the second thread.
Frequency: Every time
Operating system: Mac OS X
Browser: Chrome and Firefox
Additional information: Only happens in the Site Support forum. My tablet is school issued, but I have the problem at both school and home.

How to take a screenshot
Mac OS X: If you type ⌘+Shift+4, you'll get a "crosshairs" that lets you take a custom screenshot size. Just click and drag to select the area you want to take a picture of. If you type ⌘+Shift+4+space, you can take a screenshot of a specific window. All screenshots will show up on your desktop.

Windows: Hit the Windows logo key+PrtScn, and a screenshot of your entire screen. Alternatively, you can hit Alt+PrtScn to take a screenshot of the currently selected window. All screenshots are saved to the Pictures → Screenshots folder.

Advanced
If you're a bit more comfortable with how browsers work, you can also show us what happens in the JavaScript console.

In Chrome, type CTRL+Shift+J (Windows, Linux) or ⌘+Option+J (Mac).
In Firefox, type CTRL+Shift+K (Windows, Linux) or ⌘+Option+K (Mac).
In Internet Explorer, it's the F12 key.
In Safari, first enable the Develop menu: Preferences → Advanced, click "Show Develop menu in menu bar." Then either go to Develop → Show Error console or type Option+⌘+C.

It'll look something like this:
IMAGE
3 replies
slester
Apr 9, 2015
LauraZed
May 4, 2019
k i Community Safety
dcouchman   0
Jan 18, 2018
If you find content on the AoPS Community that makes you concerned for a user's health or safety, please alert AoPS Administrators using the report button (Z) or by emailing sheriff@aops.com . You should provide a description of the content and a link in your message. If it's an emergency, call 911 or whatever the local emergency services are in your country.

Please also use those steps to alert us if bullying behavior is being directed at you or another user. Content that is "unlawful, harmful, threatening, abusive, harassing, tortuous, defamatory, vulgar, obscene, libelous, invasive of another's privacy, hateful, or racially, ethnically or otherwise objectionable" (AoPS Terms of Service 5.d) or that otherwise bullies people is not tolerated on AoPS, and accounts that post such content may be terminated or suspended.
0 replies
dcouchman
Jan 18, 2018
0 replies
No More than √㏑x㏑㏑x Digits
EthanWYX2009   3
N 25 minutes ago by MathisWow
Source: 2024 April 谜之竞赛-3
Let $f(x)\in\mathbb Z[x]$ have positive integer leading coefficient. Show that there exists infinte positive integer $x,$ such that the number of digit that doesn'r equal to $9$ is no more than $\mathcal O(\sqrt{\ln x\ln\ln x}).$

Created by Chunji Wang, Zhenyu Dong
3 replies
EthanWYX2009
Mar 24, 2025
MathisWow
25 minutes ago
Sum of points' powers
Suntafayato   2
N an hour ago by fadhool
Given 2 circles $\omega_1, \omega_2$, find the locus of all points $P$ such that $\mathcal{P}ow(P, \omega_1) + \mathcal{P}ow(P, \omega_2) = 0$ (i.e: sum of powers of point $P$ with respect to the two circles $\omega_1, \omega_2$ is zero).
2 replies
Suntafayato
Mar 24, 2020
fadhool
an hour ago
IMO ShortList 2001, number theory problem 3
orl   10
N an hour ago by OronSH
Source: IMO ShortList 2001, number theory problem 3
Let $ a_1 = 11^{11}, \, a_2 = 12^{12}, \, a_3 = 13^{13}$, and $ a_n = |a_{n - 1} - a_{n - 2}| + |a_{n - 2} - a_{n - 3}|, n \geq 4.$ Determine $ a_{14^{14}}$.
10 replies
orl
Sep 30, 2004
OronSH
an hour ago
Existence of a solution of a diophantine equation
syk0526   4
N an hour ago by ihategeo_1969
Source: North Korea Team Selection Test 2013 #6
Show that $ x^3 + x+ a^2 = y^2 $ has at least one pair of positive integer solution $ (x,y) $ for each positive integer $ a $.
4 replies
syk0526
May 17, 2014
ihategeo_1969
an hour ago
k How to delete a private forum you created
Platinum_Dragon   2
N Yesterday at 9:36 PM by jlacosta
Is this possible? thank you
2 replies
Platinum_Dragon
Yesterday at 4:26 PM
jlacosta
Yesterday at 9:36 PM
This has been going on for a while now, can anyone else relate?
jmr2010   2
N Yesterday at 3:02 AM by valisaxieamc
Most of the time when I type in something for the tags or search for a user, the AoPS suggestion box pops up, and most of the time, when I click the suggestion, the box just disappears, meaning the automatic system usually never works
2 replies
jmr2010
Apr 29, 2025
valisaxieamc
Yesterday at 3:02 AM
k How to remove tags from a PM after you've removed yourself from it
Platinum_Dragon   2
N Apr 29, 2025 by Platinum_Dragon
Is it possible? Because it's kind of annoying to have a whole bunch of tags that stick around forever.

thank you
2 replies
Platinum_Dragon
Apr 29, 2025
Platinum_Dragon
Apr 29, 2025
k Reaper....
Happycat2   22
N Apr 29, 2025 by jlacosta
Can someone explain what the reaper is this time? I'm sorry but I don't know what "Rapper ear error a pear perrier ear ape ea games" means.
22 replies
Happycat2
Apr 27, 2025
jlacosta
Apr 29, 2025
k Reaper Description
Jwenslawski   1
N Apr 29, 2025 by LostInBali
What does the Reaper game description mean?

"rapper eaE error a Sear perriTr ear apeIea games"

Is it in another language?
1 reply
Jwenslawski
Apr 29, 2025
LostInBali
Apr 29, 2025
k blog glitch?
ngelyy   18
N Apr 29, 2025 by Blossom_tree_17
RESOLVED
18 replies
ngelyy
Apr 29, 2025
Blossom_tree_17
Apr 29, 2025
k Hieroglyph (and Petroglyph?) on Post #1
awesomeming327.   10
N Apr 28, 2025 by bpan2021
Steps to Reproduce:
1. Click this link
2. Scroll up.
Behavior: it should show two copies of a certain "post #0" above post #1.
Frequency: 100%

Mac OS, Arc Browser (which runs on Chromium)

Note: you can also click on "view topic" on the N7 of the 2023 ISL
10 replies
awesomeming327.
Apr 27, 2025
bpan2021
Apr 28, 2025
k something glitched?
JohannIsBach   10
N Apr 27, 2025 by valisaxieamc
hi. my aops whenever i go 2 the website the website is like way to big and i cant access anything? i am on via an ingognito window but its kind of inconvenient not sure if its just my laptop or the website. i cant recreat it on any other device? plz help [/resolved]
10 replies
JohannIsBach
Apr 25, 2025
valisaxieamc
Apr 27, 2025
k alcumus not working? [RESOLVED]
JohannIsBach   3
N Apr 27, 2025 by MathDolphin95
4 some reason alcumus isnt working??? in the 1st attachment there isnt any "play, profile, ...etc."
and in the second pic i clicked the "play" button, but its just a blank screen??? i tried refreshing it didnt work????
3 replies
JohannIsBach
Apr 27, 2025
MathDolphin95
Apr 27, 2025
Is this just my device? Typing glitch
pearlhaas   4
N Apr 26, 2025 by valisaxieamc
This started today.
When I tried to type in the AoPS homework section (as well as forums), where you put your answers into the slots, I found that I couldn’t? This might only be my device but it was delaying my work and I just thought I’d share.
I refreshed multiple times, but it only works sometimes.
Sometimes, the blinking blue bar shows but I can’t type, and other times, the blue bar doesn’t show at all.
I am able to type this forum up since this was one of the few times that my device let me type (for more, I have an iPad. Not sure what generation, but it’s a pretty recent version, I believe).
If this is a problem with my device, I apologize for wasting your time :)
Just wanted to make sure I could get this notice out if others have the same problem.
4 replies
pearlhaas
Apr 25, 2025
valisaxieamc
Apr 26, 2025
Functions
Potla   23
N Apr 25, 2025 by Ilikeminecraft
Source: 0
Find all functions $ f: \mathbb{R}\longrightarrow \mathbb{R}$ such that
\[f(x+y)+f(y+z)+f(z+x)\ge 3f(x+2y+3z)\]
for all $x, y, z \in \mathbb R$.
23 replies
Potla
Feb 21, 2009
Ilikeminecraft
Apr 25, 2025
Functions
G H J
G H BBookmark kLocked kLocked NReply
Source: 0
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Potla
1886 posts
#1 • 2 Y
Y by Adventure10, HWenslawski
Find all functions $ f: \mathbb{R}\longrightarrow \mathbb{R}$ such that
\[f(x+y)+f(y+z)+f(z+x)\ge 3f(x+2y+3z)\]
for all $x, y, z \in \mathbb R$.
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arcsin1.01
27 posts
#2 • 2 Y
Y by Adventure10, HWenslawski
Hint 1
Hint 2
Big hint (do not open before you give up)
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addictedtomath
108 posts
#3 • 2 Y
Y by Adventure10, HWenslawski
solution
Z K Y
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popcorn1
1098 posts
#4
Y by
$1 + 2 = 3$ again?!

The answer is $f(x)=c$ for all reals $c$; it's easy to see that these solutions work. Let $P(x,y,z)$ denote the given assertion.

Setting $P(x,0,0)$ gives $3f(x) \leq 2f(x) + f(0)$, so $f(x) \leq f(0)$ for all reals $x$. Setting $P(\frac{x}{2},\frac{x}{2},-\frac{x}{2})$ gives $3f(0) \leq 2f(0) + f(x)$, so $f(0) \leq f(x)$ for all reals $x$. Therefore $f(x) \leq f(0) \leq f(x)$, or $f$ is constant.

Remark. I really like this FE because both steps are motivated: you get the first from just doing stuff and you wonder, ``huh, how do I use the inequality?'' Then the solution is natural.
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MrOreoJuice
594 posts
#5 • 2 Y
Y by Mango247, Mango247
Really cute :)
As usual let $P(x,y,z)$ denote the given assertion.
  • $P(x,0,0) \implies f(x) \le f(0)$ for all $x$.
  • $P\left( \dfrac{-3x}{2} , \dfrac{3x}{2} , \dfrac {-x}{2} \right) \implies 2f(0) \le f(-2x) + f(x)$ which is also $\le 2f(0) \implies f(-2x) + f(x) = 2f(0)$.
  • $P(-x , 2x , -x) \implies f(x) \ge f(0)$.
So $f(x) = c$ for constant $c$.

Edit: wait bruh how did I miss the more natural substitution shown above.
This post has been edited 1 time. Last edited by MrOreoJuice, Sep 3, 2021, 10:21 AM
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jasperE3
11281 posts
#6
Y by
$P(x,0,0)\Rightarrow f(x)\le f(0)$
$P(x,x,-x)\Rightarrow f(2x)\ge f(0)\Rightarrow f(x)\ge f(0)\Rightarrow\boxed{f(x)=c}$ for some $c\in\mathbb R$, which works.
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554183
484 posts
#7
Y by
$P(y,y,-y) \implies f(2y)+f(0)+f(0) \geq 3f(0) \implies f(x) \geq f(0) \forall x$
$P(-y,y,-y) \implies f(0)+f(0)+f(-2y) \geq 3f(-2y) \implies f(0) \geq f(-2y) \implies f(0) \geq f(x) \forall x$
Therefore, $f(x)=c$ where $c$ is a constant for all $x$
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HamstPan38825
8857 posts
#8
Y by
Only constant functions work.

Summing this inequality cyclically, $$f(x+2y+3z) + f(y+2z+3x)+f(z+2x+3y) \leq f(x+y) + f(y+z) + f(z+x).$$On the other hand, substituting $P(2x+y, 2y+z, 2z+x)$, we obtain exactly the reverse inequality. As a result, equality must hold everywhere. Now, setting $y=z=0$, $$3f(x) = f(x) + 2f(0) \implies f(x) = f(0)$$is constant, as needed.
This post has been edited 1 time. Last edited by HamstPan38825, Mar 25, 2023, 2:15 AM
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Mogmog8
1080 posts
#9 • 1 Y
Y by centslordm
We claim the only solution is $f(x)=c$, where $c$ is a real constant. Note that this satisfies the given conditions. Let $P(x,y,z)$ denote the given assertion and $P(x,0,0)$ yields $f(x)\le f(0)$. Then, $P(-z,-z,z)$ yields $f(0)\le f(-2z)$ so $f(x)\ge f(0)$. Hence, $f(x)=f(0)$ so we are done. $\square$
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joshualiu315
2533 posts
#10
Y by
Plugging in $(x,y,z)=(x,0,0)$ to get

\[3f(x) \le 2f(x)+f(0)\]\[\implies f(x) \le f(0)\]
Plugging in $(x,y,z)=(x/2,x/2,-x/2)$ gives

\[3f(0) \le f(x)+2f(0)\]\[\implies f(x) \ge f(0)\]
Thus, $f(x)=f(0)$, so our answer is $\boxed{f(x)=c, \ c \in \mathbb{R}}$.
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shendrew7
794 posts
#11
Y by
Substitute $(2k,0,0)$ and $(k,k-k)$ to find
\[f(2k) \leq f(0), \quad f(2k) \ge f(0).\]
Hence the constant $f(0)$ is the only value $f$ can take, so $\boxed{f(x) = c, \quad c \in \mathbb{R}}$, which evidently works. $\blacksquare$
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kamatadu
479 posts
#12
Y by
$f\equiv c$ for some constant $c$ is the solution.

$P(x,0,0) \implies 3f(x) \le f(x) + f(0) + f(x) \implies f(x) \le f(0)$.

$P(-x,-x,x) \implies 3f(-x + 2(-x) + 3x) \le f(-2x) + f(0) + f(0) \implies 3f(0) \le f(-2x) + 2f(0) \implies f(-2x) \ge f(0)$.
Now changing $-2x \rightarrow x$, we get that $f(x) \ge f(0)$.

Thus $f(x) = f(0)$. :yoda:
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dolphinday
1324 posts
#14
Y by
Letting $y = z = 0$ gives $f(x) \leq f(0)$.Then $z = -x = -y$ gives $3f(0) \leq f(2x) + 2f(0)$ so $f(0) \leq f(2x)$, which is only possible if $f(x) = f(0) = c$.
This post has been edited 1 time. Last edited by dolphinday, Mar 6, 2024, 8:40 PM
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eibc
600 posts
#15
Y by
The answer is $f \equiv c$ for any real constant $c$, which works. Let $P(x, y, z)$ denote the given assertion. From $P(x, 0, 0)$ we have $3f(x) \le 2f(x) + f(0)$, so $f(x) \le f(0)$. From $P(\tfrac{x}{2}, \tfrac{x}{2}, -\tfrac{x}{2})$ we have $3f(0) \le f(x) + 2f(0)$, so $f(x) \ge f(0)$. Hence $f(x) = f(0)$ for all $x$, which implies the solution set.
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pikapika007
298 posts
#16
Y by
The answer is $f(x) = c$ only - clearly this works. Now
\[ P(x, 0, 0) \implies 3f(x) \le 2f(x) + f(0) \implies f(x) \le f(0) \]so
\[ 3f(x+2y+3z) \le f(x+y) + f(y+z) + f(z+x) \le 3f(0). \]Now $P(a, a, -a)$ implies that
\[ 3f(0) \le f(2a) + 2f(0) \le 3f(0) \]so $f(0) \le f(a)$ for all $a$, and we're done.
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Jndd
1416 posts
#17
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The answer is $f(x)=c$ for $c\in \mathbb{R}$, and it is easy to see that this satisfies our inequality.

Plugging in $y=z=0$, we get $3f(x)\leq 2f(x) + f(0)$, which gives $f(x)\leq f(0)$ for all $x$. Then, plugging in $x=y=-z$, we get $3f(0)\leq f(-2z)+2f(0)$, so $f(0)\leq f(-2z)$, giving $f(x)\geq f(0)$ for all $x$. Since $f(x)\geq f(0)\geq f(x)$ for all $x$, we must have $f(x)=f(0)$, as desired.
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Markas
105 posts
#18
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Let y = z = 0. Now we plug that in and we get that $3f(x) \leq f(x) + f(0) + f(x)$ $\Rightarrow$ $f(x) \leq f(0)$. Now let x = -z and y = -z. After we plug that in, we get $3f(0) \leq f(-2z) + f(0) + f(0)$ $\Rightarrow$ $f(0) \leq f(-2z)$. By the two substitutions we did we get $f(x) \leq f(0) \leq f(-2z)$ and we can choose x = -2z $\Rightarrow$ we now have $f(x) \leq f(0) \leq f(x)$ $\Rightarrow$ $f(0) = f(x)$ $\Rightarrow$ f is constant. Now we only need to check this which is obviously true since $3c \leq c + c + c$. We are ready since we proved that $f(x) = c$ is the only solution and it works.
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balllightning37
388 posts
#19
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Neat!

$P(x,0,0)$ implies $3f(x)\leq 2f(x)+f(0)$ or $f(x)\leq f(0)$ for all $x$.

Then, $P(x,x,-x)$ implies $3f(0)\leq f(x)+2f(0)$ or $f(0)\leq f(x)$ for all $x$. motivation

Combining, these two, we get that $f(x)=c$ for some constant $c$, which of course works.
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AshAuktober
1000 posts
#20
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Very kawaii, as expected from a Russian problem!
The only function that works is the constant function. It works as then the inequation becomes $3c \ge 3c$ which is true. We now prove it is the only such function.
Let $P(x, y, z)$ note the given assertion.

Claim 1: $f(x) \le f(0)$ for all $x \in \mathbb{R}$.
Proof: $P(x, 0, 0)$ $$\implies 2f(x) + f(0) \ge 3f(x) \implies f(x) \le f(0)$$as desired. $\square$

Claim 2: $f(x) \ge f(0)$ for all $x \in \mathbb{R}$.
Proof: $P\left(\frac x2, \frac x2, -\frac x2 \right)$ $$\implies 2f(0) + f(x) \ge 3f(0) \implies f(x) \ge f(0),$$as desired. $\square$

Combining the above two claims, $f(x) = f(0)$, and thus $f$ is indeed a constant function. $\blacksquare$
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eg4334
637 posts
#21
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Let $x+y=a, y+z=b, x+z=c$ so it rewrites into $$f(a)+f(b)+f(c) \geq 3f(2b+c)$$. Set $b=0, a=c$ to get $2f(a)+f(0) \geq 3f(a) \implies f(0) \geq f(a)$ And then $b=c=0$ to get $f(a) \geq f(0)$. Therefore $f(a)=f(0)$ so $\boxed{f(x) \equiv c}$
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blueprimes
344 posts
#22
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We claim the answer is $f(x) = c$ for a constant $c$. Obviously it works. Now the assertion $y = z = 0$ yields $f(x) \ge f(0)$ for all $x$, whereas $(x, y, z) \mapsto \left(\dfrac{x}{2}, \dfrac{x}{2}, - \dfrac{x}{2} \right)$ yields $f(x) \le f(0)$ for all $x$. So $f(x) = f(0)$ and $f$ is constant as wanted.
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Marcus_Zhang
980 posts
#23
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Almost a one liner
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Maximilian113
573 posts
#24
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Let $P(x, y, z)$ denote the assertion. Then $$P(-x, -x, x) \implies 3f(0) \leq f(-2x)+2f(0) \implies f(x) \geq f(0) \, \forall x \in \mathbb R.$$Also, $$P(-x, x, -x) \implies 3f(-2x) \leq 2f(0)+f(-2x) \implies f(x) \leq f(0) \, \forall x \in \mathbb R.$$It follows that $f(x)=c$ for some constant $c,$ and this clearly satisfies the assertion.
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Ilikeminecraft
609 posts
#25
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I claim that $f(x) = c$ for some constant $c\in\mathbb R$. This clearly works.

Plug in $x = y = -z$ to get that $f(0) \leq f(2x),$ and then $x = -y = z$ to get that $f(2x) \leq f(0).$ Thus, $f(x)$ is constant.
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