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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
J
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A cyclic inequality
KhuongTrang   0
30 minutes ago
Source: own-CRUX
IMAGE
Link
0 replies
2 viewing
KhuongTrang
30 minutes ago
0 replies
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Image of f(a + b) - f(a) - f(b)
RootofUnityfilter   0
38 minutes ago
Source: my teacher
Let $\mathbb{Z}_{>0}$ be the set of all positive integers. Determine all subsets $\mathcal{S}$ of $\{2^0, 2^1, 2^2, \dots\}$ such that
$$\mathcal{S} = \{f(a + b) - f(a) - f(b) | a, b \in \mathbb{Z}_{>0}\}$$
0 replies
RootofUnityfilter
38 minutes ago
0 replies
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Complex + Radical Evaluation
Saucepan_man02   4
N 39 minutes ago by Mathzeus1024
Evaluate: (without calculators)
$$ (\sqrt{6 - 2 \sqrt{5}} + i \sqrt{2 \sqrt{5} + 10})^5 + (\sqrt{6 - 2 \sqrt{5}} - i \sqrt{2 \sqrt{5} + 10})^5$$
4 replies
Saucepan_man02
Mar 17, 2025
Mathzeus1024
39 minutes ago
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a+b+c=3 inequality
JK1603JK   0
43 minutes ago
Let $a,b,c\ge 0: a+b+c=3$ then prove
$$\color{black}{\sqrt{a+b+2c^{2}}+\sqrt{b+c+2a^{2}}+\sqrt{c+a+2b^{2}}\le 3\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}+3}.}$$
0 replies
JK1603JK
43 minutes ago
0 replies
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prove that $\angle Q L A=\angle M L A$
NJAX   3
N an hour ago by Baimukh
Source: 2nd Al-Khwarizmi International Junior Mathematical Olympiad 2024, Day2, Problem7
Two circles with centers $O_{1}$ and $O_{2}$ intersect at $P$ and $Q$. Let $\omega$ be the circumcircle of the triangle $P O_{1} O_{2}$; the circle $\omega$ intersect the circles centered at $O_{1}$ and $O_{2}$ at points $A$ and $B$, respectively. The point $Q$ is inside triangle $P A B$ and $P Q$ intersects $\omega$ at $M$. The point $E$ on $\omega$ is such that $P Q=Q E$. Let $M E$ and $A B$ meet at $L$, prove that $\angle Q L A=\angle M L A$.

Proposed by Amir Parsa Hoseini Nayeri, Iran
3 replies
NJAX
May 31, 2024
Baimukh
an hour ago
J
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Three Nagel points collinear
jayme   3
N an hour ago by buratinogigle
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
3 replies
jayme
Mar 31, 2025
buratinogigle
an hour ago
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How many acute triangles can a right isosceles triangle be divided into?
truongphatt2668   3
N an hour ago by truongphatt2668
Source: I don't know
How many acute triangles can a right isosceles triangle be divided into?
3 replies
truongphatt2668
Yesterday at 2:48 PM
truongphatt2668
an hour ago
J
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sum(ab/4a^2+b^2) <= 3/5
truongphatt2668   4
N an hour ago by truongphatt2668
Source: I remember I read it somewhere
Let $a,b,c>0$. Prove that:
$$\dfrac{ab}{a^2+4b^2} + \dfrac{bc}{b^2+4c^2} + \dfrac{ca}{c^2+4a^2} \le \dfrac{3}{5}$$
4 replies
truongphatt2668
Monday at 1:23 PM
truongphatt2668
an hour ago
J
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Olympiad Geometry problem-second time posting
kjhgyuio   2
N an hour 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_
an hour ago
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D1010 : How it is possible ?
Dattier   13
N an hour 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
an hour ago
J
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Japanese Triangles
pikapika007   67
N 2 hours 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
2 hours ago
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D1018 : Can you do that ?
Dattier   1
N 2 hours 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
2 hours ago
J
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Inequality
JK1603JK   1
N 3 hours ago by lbh_qys
Let $a,b,c\ge 0: a+b+c=3$ then prove \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2}\cdot\frac{abc}{ab+bc+ca}\ge \frac{5}{3}.$$
1 reply
JK1603JK
3 hours ago
lbh_qys
3 hours ago
J
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Inequality
JK1603JK   1
N 4 hours ago by lbh_qys
Prove that 9ab\left(a-b+c\right)+9bc\left(b-c+a\right)+9ca\left(c-a+b\right)\ge \left(a+b+c\right)^{3},\ \ a\ge 0\ge b\ge c: a+b+c\le 0.
1 reply
JK1603JK
4 hours ago
lbh_qys
4 hours ago
J
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J
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2016 Chandigarh RMO Geometry box tools are on table as in figure, angle find
parmenides51   7
N Oct 29, 2024 by cyberhacker
Two of the Geometry box tools are placed on the table as shown. Determine the angle $\angle ABC$
IMAGE
7 replies
parmenides51
Aug 9, 2019
cyberhacker
Oct 29, 2024
J
2016 Chandigarh RMO Geometry box tools are on table as in figure, angle find
G H J
Reveal topic tags
geometry
angles
L
G H BBookmark kLocked kLocked NReply
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parmenides51
30629 posts
parmenides51
#1 h  Aug 9, 2019, 10:44 AM • 3 Y
Y by Adventure10, Mango247, Rounak_iitr
Two of the Geometry box tools are placed on the table as shown. Determine the angle $\angle ABC$
https://2.bp.blogspot.com/--DWVwVQJgMM/XU1OK08PSUI/AAAAAAAAKfs/dgZeYwiYOrQJE4eKQT5s13GQdBEHPqy9QCK4BGAYYCw/s1600/prmo%2B16%2BChandigarh%2Bp7.png
This post has been edited 2 times. Last edited by parmenides51, Aug 21, 2019, 9:54 PM
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gamerrk1004
1519 posts
gamerrk1004
#2 h  Aug 9, 2019, 10:46 AM • 2 Y
Y by Adventure10, Mango247
Is the answer $10$ ?
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APX007
50 posts
APX007
#3 h  Aug 9, 2019, 10:46 AM • 2 Y
Y by Adventure10, Mango247
yes i got the same
Z K Y
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APX007
50 posts
APX007
#4 h  Aug 9, 2019, 11:02 AM • 2 Y
Y by Adventure10, Mango247
extend line $C$ to meet $AB$ at $D$
making line $C$ as transversal
$ \angle ADC = 130^{o}$
USING EXTERIOR ANGLE PROPERTY $\angle DCA + \angle CAD = 130^{o}$
$ \angle DCA = 180^{o} - 60^{o}$ = $120^{o}$
THEREFORE $ \angle ABC = 10^{o}$ :jump:
This post has been edited 3 times. Last edited by APX007, Aug 9, 2019, 11:06 AM
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RishiNandha_M
531 posts
RishiNandha_M
#5 h  Oct 14, 2019, 3:18 PM • 1 Y
Y by Adventure10
This was RMO???
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Math-wiz
6107 posts
Math-wiz
#6 h  Oct 14, 2019, 3:19 PM • 1 Y
Y by Adventure10
RishiNandha_M wrote:
This was RMO???

Actually its confusing. It is between PRMO and RMO
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Siddharthmaybe
106 posts
Siddharthmaybe
#7 h  Oct 29, 2024, 8:06 AM
Y by
Let the set square vertices be CMN where angle M= 90

extend MC too meet AB at K then angle BKC = 50 degree since the ruler opposite sides are parallel
angle KCB = 120 (exterior angle property)
so in triange KCB -> angle ABC = angle KBC = 10 by angle sum property
Z K Y
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cyberhacker
400 posts
cyberhacker
#8 h  Oct 29, 2024, 1:58 PM
Y by
pentagon ABC(intersection of line c and the ruler top side) total angle is 540, subtract 180+50+300, and you get 10 as the final answer
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