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k a My Retirement & New Leadership at AoPS
rrusczyk   1571
N Mar 26, 2025 by SmartGroot
I write today to announce my retirement as CEO from Art of Problem Solving. When I founded AoPS 22 years ago, I never imagined that we would reach so many students and families, or that we would find so many channels through which we discover, inspire, and train the great problem solvers of the next generation. I am very proud of all we have accomplished and I’m thankful for the many supporters who provided inspiration and encouragement along the way. I'm particularly grateful to all of the wonderful members of the AoPS Community!

I’m delighted to introduce our new leaders - Ben Kornell and Andrew Sutherland. Ben has extensive experience in education and edtech prior to joining AoPS as my successor as CEO, including starting like I did as a classroom teacher. He has a deep understanding of the value of our work because he’s an AoPS parent! Meanwhile, Andrew and I have common roots as founders of education companies; he launched Quizlet at age 15! His journey from founder to MIT to technology and product leader as our Chief Product Officer traces a pathway many of our students will follow in the years to come.

Thank you again for your support for Art of Problem Solving and we look forward to working with millions more wonderful problem solvers in the years to come.

And special thanks to all of the amazing AoPS team members who have helped build AoPS. We’ve come a long way from here:IMAGE
1571 replies
rrusczyk
Mar 24, 2025
SmartGroot
Mar 26, 2025
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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
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0 replies
jlacosta
Mar 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
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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
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Olympiad Geometry problem-second time posting
kjhgyuio   2
N 6 minutes ago by ND_
Source: smo problem
In trapezium ABCD,AD is parallel to BC and points E and F are midpoints of AB and DC respectively. If
Area of AEFD/Area of EBCF =√3 + 1/3-√3 and the area of triangle ABD is √3 .find the area of trapezium ABCD
2 replies
kjhgyuio
Today at 1:03 AM
ND_
6 minutes ago
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D1010 : How it is possible ?
Dattier   13
N 12 minutes ago by Dattier
Source: les dattes à Dattier
Is it true that$$\forall n \in \mathbb N^*, (24^n \times B \mod A) \mod 2 = 0 $$?

A=1728400904217815186787639216753921417860004366580219212750904
024377969478249664644267971025952530803647043121025959018172048
336953969062151534282052863307398281681465366665810775710867856
720572225880311472925624694183944650261079955759251769111321319
421445397848518597584590900951222557860592579005088853698315463
815905425095325508106272375728975

B=2275643401548081847207782760491442295266487354750527085289354
965376765188468052271190172787064418854789322484305145310707614
546573398182642923893780527037224143380886260467760991228567577
953725945090125797351518670892779468968705801340068681556238850
340398780828104506916965606659768601942798676554332768254089685
307970609932846902
13 replies
Dattier
Mar 10, 2025
Dattier
12 minutes ago
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Japanese Triangles
pikapika007   67
N 30 minutes ago by quantam13
Source: IMO 2023/5
Let $n$ be a positive integer. A Japanese triangle consists of $1 + 2 + \dots + n$ circles arranged in an equilateral triangular shape such that for each $i = 1$, $2$, $\dots$, $n$, the $i^{th}$ row contains exactly $i$ circles, exactly one of which is coloured red. A ninja path in a Japanese triangle is a sequence of $n$ circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. Here is an example of a Japanese triangle with $n = 6$, along with a ninja path in that triangle containing two red circles.
IMAGE
In terms of $n$, find the greatest $k$ such that in each Japanese triangle there is a ninja path containing at least $k$ red circles.
67 replies
pikapika007
Jul 9, 2023
quantam13
30 minutes ago
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Three Nagel points collinear
jayme   2
N 40 minutes ago by jayme
Dear Marthlinkers,

1. ABCD a square
2. M a point on the segment CD sothat MA < MB
3. Nm, Na, Nb the Nagel’s points of the triangles MAD, ADM, BCM.

Prove : Nm, Na and Nb are collinear.

Sincerely
Jean-Louis
2 replies
jayme
Mar 31, 2025
jayme
40 minutes ago
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D1018 : Can you do that ?
Dattier   1
N an hour ago by Dattier
Source: les dattes à Dattier
We can find $A,B,C$, such that $\gcd(A,B)=\gcd(C,A)=\gcd(A,2)=1$ and $$\forall n \in \mathbb N^*, (C^n \times B \mod A) \mod 2=0 $$.

For example :

$C=20$
$A=47650065401584409637777147310342834508082136874940478469495402430677786194142956609253842997905945723173497630499054266092849839$

$B=238877301561986449355077953728734922992395532218802882582141073061059783672634737309722816649187007910722185635031285098751698$

Can you find $A,B,C$ such that $A>3$ is prime, $C,B \in (\mathbb Z/A\mathbb Z)^*$ with $o(C)=(A-1)/2$ and $$\forall n \in \mathbb N^*, (C^n \times B \mod A) \mod 2=0 $$?
1 reply
Dattier
Mar 24, 2025
Dattier
an hour ago
J
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n-gon function
ehsan2004   9
N an hour ago by AshAuktober
Source: Romanian IMO Team Selection Test TST 1996, problem 1
Let $ f: \mathbb{R}^2 \rightarrow \mathbb{R} $ be a function such that for every regular $ n $-gon $ A_1A_2 \ldots A_n $ we have $ f(A_1)+f(A_2)+\cdots +f(A_n)=0 $. Prove that $ f(x)=0 $ for all reals $ x $.
9 replies
ehsan2004
Sep 13, 2005
AshAuktober
an hour ago
J
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Iran geometry
Dadgarnia   37
N an hour ago by amirhsz
Source: Iranian TST 2018, first exam day 2, problem 4
Let $ABC$ be a triangle ($\angle A\neq 90^\circ$). $BE,CF$ are the altitudes of the triangle. The bisector of $\angle A$ intersects $EF,BC$ at $M,N$. Let $P$ be a point such that $MP\perp EF$ and $NP\perp BC$. Prove that $AP$ passes through the midpoint of $BC$.

Proposed by Iman Maghsoudi, Hooman Fattahi
37 replies
Dadgarnia
Apr 8, 2018
amirhsz
an hour ago
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Inequatity
mrmath2006   13
N an hour ago by KhuongTrang
Given $a,b,c>0$ & $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=10$. Prove that
$(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\ge \frac{27}{2}$
13 replies
mrmath2006
Jan 5, 2016
KhuongTrang
an hour ago
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weird 3-var cyclic ineq
RainbowNeos   1
N 2 hours ago by lbh_qys
Given $a,b,c\geq 0$ with $a+b+c=1$ and at most one of them being zero, show that
\[\frac{1}{\max(a^2,b)}+\frac{1}{\max(b^2,c)}+\frac{1}{\max(c^2,a)}\geq\frac{27}{4}\]
1 reply
+1 w
RainbowNeos
Mar 28, 2025
lbh_qys
2 hours ago
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Tangent Spheres and Tangents to Spheres
Math-Problem-Solving   2
N 2 hours ago by Math-Problem-Solving
Source: 2002 British Mathematical Olympiad Round 2
Prove this.
2 replies
Math-Problem-Solving
6 hours ago
Math-Problem-Solving
2 hours ago
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Find all sequences satisfying two conditions
orl   33
N 2 hours ago by AshAuktober
Source: IMO Shortlist 2007, C1, AIMO 2008, TST 1, P1
Let $ n > 1$ be an integer. Find all sequences $ a_1, a_2, \ldots a_{n^2 + n}$ satisfying the following conditions:
\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n; \]

\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n. \]
Author: Dusan Dukic, Serbia
33 replies
orl
Jul 13, 2008
AshAuktober
2 hours ago
J
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Interesting inequalities
sqing   6
N 2 hours ago by sqing
Source: Own
Let $ a,b,c\geq 0 $ and $ a+b+c=3 $. Prove that
$$ \frac{a^2}{a^2+b+c+ \frac{3}{2}}+\frac{b^2}{b^2+c+a+\frac{3}{2}}+\frac{c^2}{c^2+a+b+\frac{3}{2}} \leq \frac{6}{7}$$Equality holds when $ (a,b,c)=(0,\frac{3}{2},\frac{3}{2}) $ or $ (a,b,c)=(0,0,3) .$
6 replies
sqing
Today at 3:12 AM
sqing
2 hours ago
J
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European Mathematical Cup 2016 senior division problem 2
steppewolf   6
N 2 hours ago by SimplisticFormulas
For two positive integers $a$ and $b$, Ivica and Marica play the following game: Given two piles of $a$
and $b$ cookies, on each turn a player takes $2n$ cookies from one of the piles, of which he eats $n$ and puts $n$ of
them on the other pile. Number $n$ is arbitrary in every move. Players take turns alternatively, with Ivica going
first. The player who cannot make a move, loses. Assuming both players play perfectly, determine all pairs of
numbers $(a, b)$ for which Marica has a winning strategy.

Proposed by Petar Orlić
6 replies
steppewolf
Dec 31, 2016
SimplisticFormulas
2 hours ago
J
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inequalities hard
Cobedangiu   1
N 3 hours ago by Cobedangiu
problem
1 reply
Cobedangiu
Monday at 11:45 AM
Cobedangiu
3 hours ago
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P inside ABC such that 3 < ABP = 3 < ACP = < ABC + < ACB
Arne   1
N Jan 9, 2005 by Armo
Source: Poland 97
Let P be a point inside triangle ABC such that 3<ABP = 3<ACP = <ABC + <ACB. Prove that AB/(AC + PB) = AC/(AB + PC).
1 reply
Arne
Jan 9, 2005
Armo
Jan 9, 2005
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P inside ABC such that 3 < ABP = 3 < ACP = < ABC + < ACB
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Reveal topic tags
geometry proposed
geometry
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G H BBookmark kLocked kLocked NReply
Source: Poland 97
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Arne
3660 posts
Arne
#1 h  Jan 9, 2005, 2:41 PM • 2 Y
Y by Adventure10, Mango247
Let P be a point inside triangle ABC such that 3<ABP = 3<ACP = <ABC + <ACB. Prove that AB/(AC + PB) = AC/(AB + PC).
Z K Y
The post below has been deleted. Click to close.
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Armo
373 posts
Armo
#2 h  Jan 9, 2005, 9:25 PM • 2 Y
Y by Adventure10, Mango247
Let $CP$ cuts $AB$ at $F$ and $BP$ cuts $AC$ at $E$. Simple angle chasing gives that $\angle{FBP}=\angle{BPF}=\angle{EPC}=\angle{PCE}$ so $BF=FP, PE=EC$. After this: $AC+PB=AE+EC+PB=AE+PE+PB=AE+BE$, similarly we obtain $AB+PC=AF+FC$. So we have to prove that $\frac{AB}{AE+BE}=\frac{AC}{AF+FC}$ and this is obvious as triangles $ABE$ and $ACF$ are similar.
I hope my solution is right (I am too lazy to check it).
:)
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