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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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Inspired by old results
sqing   7
N 2 hours ago by SunnyEvan
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.$
7 replies
1 viewing
sqing
Monday at 1:42 PM
SunnyEvan
2 hours ago
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Modular Arithmetic and Integers
steven_zhang123   3
N 2 hours ago 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
2 hours ago
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Polynomials and their shift with all real roots and in common
Assassino9931   4
N 2 hours ago by Assassino9931
Source: Bulgaria Spring Mathematical Competition 2025 11.4
We call two non-constant polynomials friendly if each of them has only real roots, and every root of one polynomial is also a root of the other. For two friendly polynomials \( P(x), Q(x) \) and a constant \( C \in \mathbb{R}, C \neq 0 \), it is given that \( P(x) + C \) and \( Q(x) + C \) are also friendly polynomials. Prove that \( P(x) \equiv Q(x) \).
4 replies
1 viewing
Assassino9931
Mar 30, 2025
Assassino9931
2 hours ago
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2025 Caucasus MO Seniors P7
BR1F1SZ   2
N 2 hours ago by sami1618
Source: Caucasus MO
From a point $O$ lying outside the circle $\omega$, two tangents are drawn touching $\omega$ at points $M$ and $N$. A point $K$ is chosen on the segment $MN$. Let points $P$ and $Q$ be the midpoints of segments $KM$ and $OM$ respectively. The circumcircle of triangle $MPQ$ intersects $\omega$ again at point $L$ ($L \neq M$). Prove that the line $LN$ passes through the centroid of triangle $KMO$.
2 replies
BR1F1SZ
Mar 26, 2025
sami1618
2 hours ago
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No more topics!
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equilateral triangle
AndrewTom   1
N Oct 14, 2014 by Tsikaloudakis
In the acute-angled triangle $ ABC$, $ CF$ is an altitude, with $ F$ on $ AB$, and $ BM$ is a median with $ M$ on $ CA$. Given that $ BM=CF$ and $ \angle MBC = \angle FCA$, prove that the triangle $ ABC$ is equilateral.
1 reply
AndrewTom
Jan 23, 2010
Tsikaloudakis
Oct 14, 2014
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equilateral triangle
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Reveal topic tags
geometry unsolved
geometry
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AndrewTom
12750 posts
AndrewTom
#1 h  Jan 23, 2010, 7:28 PM • 2 Y
Y by Adventure10, Mango247
In the acute-angled triangle $ ABC$, $ CF$ is an altitude, with $ F$ on $ AB$, and $ BM$ is a median with $ M$ on $ CA$. Given that $ BM=CF$ and $ \angle MBC = \angle FCA$, prove that the triangle $ ABC$ is equilateral.
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Tsikaloudakis
974 posts
Tsikaloudakis
#2 h  Oct 14, 2014, 11:55 AM • 3 Y
Y by AndrewTom, kaparos, Adventure10
$\begin{gathered} C\hat FM = F\hat CM = C\hat BM = \varphi \mathop {}\limits^{} \Rightarrow \mathop {}\limits_{} \hfill \\ B,C,M,F\mathop {}\limits_{}^{} are\mathop {}\limits_{}^{} concyclyc\mathop {}\limits_{}^{} \Rightarrow \hfill \\ C\hat MB = {90^o}\mathop {}\limits^{} \Rightarrow \mathop {}\limits_{} BC = BA \hfill \\ \left. \begin{gathered} BM = CF\mathop {}\limits_{}^{} \hfill \\ \hat M = \hat F = {90^o}\mathop {}\limits_{}^{} \hfill \\ C\hat BM = F\hat CA = \varphi \hfill \\ \end{gathered} \right\}\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} tr.BCM = tr.FCA\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} AC = BC = AB \hfill \\ \end{gathered} $
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