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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.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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0 replies
jlacosta
Mar 2, 2025
0 replies
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D1018 : Can you do that ?
Dattier   1
N 2 minutes 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 minutes ago
J
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n-gon function
ehsan2004   9
N 19 minutes 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
+1 w
ehsan2004
Sep 13, 2005
AshAuktober
19 minutes ago
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Iran geometry
Dadgarnia   37
N 23 minutes 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
23 minutes ago
J
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Inequatity
mrmath2006   13
N 33 minutes 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
33 minutes ago
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perimeter inequality in convex hegaxon , 1/2 \le s/(u+v) \le 1
parmenides51   1
N Jan 12, 2020 by tk1
Source: Austrian Regional Competition For Advanced Students 2002 p3
In the convex $ABCDEF$ (has all interior angles less than $180^o$) with the perimeter $s$ the triangles $ACE$ and $BDF$ have perimeters $u$ and $v$ respectively.
a) Show the inequalities $\frac{1}{2} \le \frac{s}{u+v}\le 1$
b) Check whether $1$ is replaced by a smaller number or $1/2$ by a larger number can the inequality remains valid for all convex hexagons.
1 reply
parmenides51
Jan 11, 2020
tk1
Jan 12, 2020
J
perimeter inequality in convex hegaxon , 1/2 \le s/(u+v) \le 1
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Reveal topic tags
geometry
perimeter
inequalities
convex
hexagon
geometric inequality
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Source: Austrian Regional Competition For Advanced Students 2002 p3
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parmenides51
30629 posts
parmenides51
#1 h  Jan 11, 2020, 8:44 PM • 1 Y
Y by Adventure10
In the convex $ABCDEF$ (has all interior angles less than $180^o$) with the perimeter $s$ the triangles $ACE$ and $BDF$ have perimeters $u$ and $v$ respectively.
a) Show the inequalities $\frac{1}{2} \le \frac{s}{u+v}\le 1$
b) Check whether $1$ is replaced by a smaller number or $1/2$ by a larger number can the inequality remains valid for all convex hexagons.
Z K Y
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tk1
937 posts
tk1
#2 h  Jan 12, 2020, 1:03 AM • 1 Y
Y by Adventure10
(a) $AB+BC\ge AC,$ etc. Adding, we get $s\ge u.$ Similarly $s\ge v,$ and therefore $s\ge (u+v)/2.$

Let $G,\,H$ be the intersection points of $AC$ with $BF,\,BD,$ respectively. Let $I,\,J$ be the intersection points of $CE$ with $BD,\,DF,$ respectively. Let $K,\,L$ be the intersection points of $EA$ with $FD,\,FB,$ respectively. Then $AB\le AG+GB,$ etc. Adding we get $s\le u+v-(GH+HI+IJ+JK+KL)\le u+v.$

(b) At the limit where, for example, $B$ is on $AC$ and close to $A,$ $D$ is on $CE$ and close to $C,$ and finally $F$ is on $AE$ and close to $E,$ we get the left equality. At the limit where points $B,\,C,\,E,\,F$ collapse into one, then we get the right equality. For the case of a regular hexagon of side $a$, we get $s=6a,$ and $u=v=3\sqrt{3}a,$ which leads to $\frac{s}{u+v}=\frac{\sqrt{3}}{3}.$
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