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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
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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.
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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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IGO 2022 advanced/free P1
Tafi_ak   11
N 8 minutes ago by jordiejoh
Source: Iranian Geometry Olympiad 2022 P1 Advanced, Free
Four points $A$, $B$, $C$ and $D$ lie on a circle $\omega$ such that $AB=BC=CD$. The tangent line to $\omega$ at point $C$ intersects the tangent line to $\omega$ at $A$ and the line $AD$ at $K$ and $L$. The circle $\omega$ and the circumcircle of triangle $KLA$ intersect again at $M$. Prove that $MA=ML$.

Proposed by Mahdi Etesamifard
11 replies
Tafi_ak
Dec 13, 2022
jordiejoh
8 minutes ago
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Interesting inequalities
sqing   4
N 18 minutes ago by lbh_qys
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) .$
4 replies
+1 w
sqing
2 hours ago
lbh_qys
18 minutes ago
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Interesting inequalities
sqing   1
N 32 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 0 $ and $ a+b+c=3 $. Prove that
$$ \frac{a^2+1}{a^2+b+c-\frac{1}{2}}+\frac{b^2+1}{b^2+c+a-\frac{1}{2}}+\frac{c^2+1}{c^2+a+b-\frac{1}{2}} \leq \frac{12}{5}$$$$ \frac{a^2+2}{a^2+b+c+\frac{1}{2}}+\frac{b^2+2}{b^2+c+a+\frac{1}{2}}+\frac{c^2+2}{c^2+a+b+\frac{1}{2}} \leq \frac{18}{7}$$Equality holds when $ (a,b,c)=(0,\frac{3}{2},\frac{3}{2}) $ or $ (a,b,c)=(1,1,1) .$
1 reply
1 viewing
sqing
39 minutes ago
sqing
32 minutes ago
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euler line and midpoint tangents
Thelink_20   3
N 41 minutes ago by hectorleo123
Source: My problem
Let the medians of a triangle $\Delta ABC$ intersect its circumcircle $\Gamma$ at $N_A, N_B, N_C$. The tangets to $\Gamma$ from $N_A,N_B,N_C$ determine a triangle $\Delta X_AX_BX_C$, where $X_A$ is relative to $A$ and so on. Prove that lines $AX_A,BX_B,CX_C$ are concurrent at a point $P$ and that $P$ belongs to the euler line of $\Delta ABC$.
Remark:Click to reveal hidden text

3 replies
Thelink_20
Jan 24, 2025
hectorleo123
41 minutes ago
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inequalities
Cobedangiu   4
N 2 hours ago by sqing
Source: own
$a,b>0$ and $a+b=1$. Find min P:
$P=\sqrt{\frac{1-a}{1+7a}}+\sqrt{\frac{1-b}{1+7b}}$
4 replies
Cobedangiu
Yesterday at 6:10 PM
sqing
2 hours ago
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Problem 4: ISL 2008/G3 Constructed Four Times
ike.chen   25
N 2 hours ago by Yiyj1
Source: USEMO 2022/4
Let $ABCD$ be a cyclic quadrilateral whose opposite sides are not parallel. Suppose points $P, Q, R, S$ lie in the interiors of segments $AB, BC, CD, DA,$ respectively, such that $$\angle PDA = \angle PCB, \text{ } \angle QAB = \angle QDC, \text{ } \angle RBC = \angle RAD, \text{ and } \angle SCD = \angle SBA.$$Let $AQ$ intersect $BS$ at $X$, and $DQ$ intersect $CS$ at $Y$. Prove that lines $PR$ and $XY$ are either parallel or coincide.

Tilek Askerbekov
25 replies
ike.chen
Oct 23, 2022
Yiyj1
2 hours ago
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Inspired by old results
sqing   8
N 2 hours ago by sqing
Source: Own
Let $ a,b,c> 0 $ and $ abc=1 $. Prove that
$$\frac1{a^2+a+k}+\frac1{b^2+b+k}+\frac1{c^2+c+k}\geq \frac{3}{k+2}$$Where $ 0<k \leq 1.$
8 replies
sqing
Monday at 1:42 PM
sqing
2 hours ago
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function
REGNA   2
N 2 hours ago by jasperE3
find all $f : \mathbb{R^+}\rightarrow \mathbb{R^+}$ such that :
$f(x+3f(y))=f(x)+f(y)+2y)$
2 replies
REGNA
Mar 19, 2023
jasperE3
2 hours ago
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sum(ab/4a^2+b^2) <= 3/5
truongphatt2668   3
N 2 hours ago by Nguyenhuyen_AG
Source: I remember I read it somewhere
Let $a,b,c>0$. Prove that:
$$\dfrac{ab}{a^2+4b^2} + \dfrac{bc}{b^2+4c^2} + \dfrac{ca}{c^2+4a^2} \le \dfrac{3}{5}$$
3 replies
truongphatt2668
Monday at 1:23 PM
Nguyenhuyen_AG
2 hours ago
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Inspired by old results
sqing   2
N 2 hours ago by sqing
Source: Own
Let $ a,b $ be reals such that $ a^2+b^2 +ab =1.$ Prove that $$9\geq (a^2-a+1)(b^2-b+1) (a^2-ab+b^2) \geq \frac{19-8\sqrt 3}{27}$$Let $ a,b $ be reals such that $ a^3+b^3 -2ab =1.$ Prove that $$ (a^2-a+1)(b^2-b+1) (a^2-ab+b^2) \geq 1$$
2 replies
sqing
2 hours ago
sqing
2 hours ago
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Painting Beads on Necklace
amuthup   45
N 3 hours ago by maromex
Source: 2021 ISL C2
Let $n\ge 3$ be a fixed integer. There are $m\ge n+1$ beads on a circular necklace. You wish to paint the beads using $n$ colors, such that among any $n+1$ consecutive beads every color appears at least once. Find the largest value of $m$ for which this task is $\emph{not}$ possible.

Carl Schildkraut, USA
45 replies
amuthup
Jul 12, 2022
maromex
3 hours ago
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Inclusion Exclusion Principle
chandru1   1
N 3 hours ago by onofre.campos
How does one prove the identity $$1=\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}2^{n-k}$$This easy via the binomial theorem for the quantity is just $(2-1)^{k}$, but how do we arrive at this using the I-E-P?
1 reply
chandru1
Dec 4, 2020
onofre.campos
3 hours ago
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Olympiad Geometry problem-second time posting
kjhgyuio   0
4 hours ago
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
0 replies
kjhgyuio
4 hours ago
0 replies
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Modular Arithmetic and Integers
steven_zhang123   3
N Yesterday at 10:50 PM by steven_zhang123
Integers \( n, a, b \in \mathbb{Z}^+ \) satisfies \( n + a + b = 30 \). If \( \alpha < b, \alpha \in \mathbb{Z^+} \), find the maximum possible value of $\sum_{k=1}^{\alpha} \left \lfloor \frac{kn^2 \bmod a }{b-k} \right \rfloor $.
3 replies
steven_zhang123
Mar 28, 2025
steven_zhang123
Yesterday at 10:50 PM
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Problem 1(Geometry)
Erken   6
N Sep 22, 2007 by cefer
Source: SRMC 2002
Let $ \triangle ABC$ be a triangle with incircle $ \omega(I,r)$and circumcircle $ \zeta(O,R)$.Let $ l_{a}$ be the angle bisector of $ \angle BAC$.Denote $ P=l_{a}\cap\zeta$.Let $ D$ be the point of tangency $ \omega$ with $ [BC]$.Denote $ Q=PD\cap\zeta$.Show that $ PI=QI$ if $ PD=r$.
6 replies
Erken
Sep 16, 2007
cefer
Sep 22, 2007
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Problem 1(Geometry)
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Reveal topic tags
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circumcircle
angle bisector
geometry proposed
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Source: SRMC 2002
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Erken
1363 posts
Erken
#1 h  Sep 16, 2007, 4:05 PM • 1 Y
Y by Adventure10
Let $ \triangle ABC$ be a triangle with incircle $ \omega(I,r)$and circumcircle $ \zeta(O,R)$.Let $ l_{a}$ be the angle bisector of $ \angle BAC$.Denote $ P=l_{a}\cap\zeta$.Let $ D$ be the point of tangency $ \omega$ with $ [BC]$.Denote $ Q=PD\cap\zeta$.Show that $ PI=QI$ if $ PD=r$.
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cefer
293 posts
cefer
#2 h  Sep 22, 2007, 11:44 AM • 3 Y
Y by Adventure10, Mango247, and 1 other user
From angles $ \angle DQB =\angle PQB =\angle PAB =\angle PAC =\angle PBC =\angle PBD$ So circumcircle of $ \bigtriangleup QDB$ is tangent to $ PB$. From tangent teorem and $ PB = PI$ (it is well-known) we get $ PD\cdot PQ = PB^{2}= PI^{2}\Rightarrow$ circumcircle of $ \bigtriangleup DIQ$ ıs tangent to $ PI\Rightarrow\angle IQP =\angle IQD =\angle DIP =\angle IPD =\angle IPQ\Rightarrow IP = IQ$
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Erken
1363 posts
Erken
#3 h  Sep 22, 2007, 11:55 AM • 2 Y
Y by Adventure10, Mango247
Dear,cefer,Did you take participant in this olympiad,i mean not exactly in 2002's,but in SRMC at all.Thank you for your nice solution.
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cefer
293 posts
cefer
#4 h  Sep 22, 2007, 12:30 PM • 1 Y
Y by Adventure10
I participated in 2005-2006, but i dont know my teachers sent our results or not. I think in 2006 they were late for sendıng answers.
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Erken
1363 posts
Erken
#5 h  Sep 22, 2007, 12:39 PM • 2 Y
Y by Adventure10, Mango247
Where must your teacher sent your results?I suppose that in your country,as in mine,APMO and SRMC come together,APMO results are sent to Korea,but i don't know about SRMC.. :maybe:
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Virgil Nicula
7054 posts
Virgil Nicula
#6 h  Sep 22, 2007, 1:21 PM • 3 Y
Y by Adventure10, Mango247, and 1 other user
Very nice problem and the Cefer's proof ! Here is another proof.
Erken wrote:
Let $ ABC$ be a triangle with the incircle $ \omega(I,r)$ and the circumcircle $ \Omega (O,R)$. Let $ l_{a}$ be the angle bisector of the angle $ \widehat{BAC}$. Denote $ P = l_{a}\cap\Omega$. Let $ D$ be the point of tangency of the circle $ \omega$ with the side $ [BC]$. Denote the second intersection $ Q\in PD\cap\Omega$. Show that $ PD = r\implies IQ = IP$.
Proof. Denote the second intersection $ A'\in AO\cap\Omega$. Since $ \widehat{DIP}\equiv\widehat{IAA'}$ and the power $ p_{\Omega}(I)$ of the point $ I$ w.r.t. the circle $ \Omega$ is $ -p_{\Omega}(I) = AI\cdot IP = 2Rr = ID\cdot AA'$, i.e. $ \frac{ID}{IP}=\frac{AI}{AA'}$ results $ \triangle IDP\sim\triangle AIA'$. Thus $ \widehat{IPD}\equiv\widehat{AA'I}$ what means that the point $ \boxed{\ Q\in PD\cap IA'\ }$. Therefore, $ \widehat{DQI}\equiv\widehat{IAA'}\equiv\widehat{DIP}$. In conclusion, $ PD = r$ $ \Longleftrightarrow$ $ PD = DI$ $ \Longleftrightarrow$ $ \widehat{DPI}\equiv\widehat{DIP}$ $ \Longleftrightarrow$ $ \widehat{DPI}\equiv\widehat{DQI}$ $ \Longleftrightarrow$ $ IP = IQ$.

Remark. $ \boxed{\ PD = r\Longleftrightarrow IP = IQ\Longleftrightarrow IO\perp IA\Longleftrightarrow b+c = 2a\ }$.
This post has been edited 4 times. Last edited by Virgil Nicula, Sep 22, 2007, 3:16 PM
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cefer
293 posts
cefer
#7 h  Sep 22, 2007, 1:34 PM • 2 Y
Y by Adventure10, Mango247
Erken wrote:
Where must your teacher sent your results?I suppose that in your country,as in mine,APMO and SRMC come together,APMO results are sent to Korea,but i don't know about SRMC.. :maybe:

I think they send SRMC results to Turkey
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