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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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Procedure involving colored numbers
Ciobi_   0
a minute ago
Source: Romania NMO 2025 9.1
Let $N \geq 1$ be a positive integer. There are two numbers written on a blackboard, one red and one blue. Initially, both are 0. We define the following procedure: at each step, we choose a nonnegative integer $k$ (not necessarily distinct from the previously chosen ones), and, if the red and blue numbers are $x$ and $y$ respectively, we replace them with $x+k+1$ and $y+k^2+2$, which we color blue and red (in this order). We keep doing this procedure until the blue number is at least $N$.
Determine the minimum value of the red number at the end of this procedure.
0 replies
Ciobi_
a minute ago
0 replies
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One nice inequlity 1
prof.   0
4 minutes ago
If $a,b,c$ are positiv real number, such that $a^2+b^2+c^2=3abc$ prove inequality $$\frac{a}{b^2c^2}+\frac{b}{c^2a^2}+\frac{c}{a^2b^2}\ge\frac{9}{a+b+c}.$$
0 replies
1 viewing
prof.
4 minutes ago
0 replies
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2 var inquality
sqing   0
9 minutes ago
Source: Own
Let $ a,b\geq 0 $ and $ 2a+2b+ab=5 $. Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+ \frac{1}{a+b}+\frac{9ab}{20(a+b+ab)} \geq \frac{33}{20}$$$$\frac{1}{a+1}+\frac{1}{b+1}+ \frac{1}{a+b}+\frac{31ab}{50(a+b+ab)} \geq \frac{59}{35}$$$$\frac{1}{a+1}+\frac{1}{b+1}+ \frac{1}{a+b}+\frac{0.616327 ab}{a+b+ab} \geq \frac{59}{35}$$
0 replies
sqing
9 minutes ago
0 replies
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Poland Inequalities
wangzishan   6
N 11 minutes ago by AshAuktober
Leta,b,c are postive real numbers,proof that $ \frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\geq1$
6 replies
wangzishan
Apr 23, 2009
AshAuktober
11 minutes ago
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No more topics!
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easy geometry..
ts0_9   1
N Mar 28, 2014 by MrRTI
Source: Kazakhstan National Olympiad 2014 P1 D1 9 grade
In a triangle $ABC$ a point $I$ — the inscribed circle center, and $w$ — described circle. Lines $BI$ and $CI$ cross $w$ respectively in points $B_1$ and $C_1$, and line $B_1C_1$ intersect lines $AB$ and $AC$ in points $C_2$ and $B_2$, respectively. Let $w_1$— triangle circumscribed circle $IB_1C_1$ and lines $IB_2$ and $IC_2$ cross $w_1$ 1 in points $M$ and $N$, respectively. Prove that $BC_2*B_2C = B_2M*C_2N$
1 reply
ts0_9
Mar 27, 2014
MrRTI
Mar 28, 2014
J
easy geometry..
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Reveal topic tags
geometry
rhombus
circumcircle
power of a point
radical axis
geometry proposed
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G H BBookmark kLocked kLocked NReply
Source: Kazakhstan National Olympiad 2014 P1 D1 9 grade
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ts0_9
134 posts
ts0_9
#1 h  Mar 27, 2014, 4:15 PM • 2 Y
Y by Adventure10, Mango247
In a triangle $ABC$ a point $I$ — the inscribed circle center, and $w$ — described circle. Lines $BI$ and $CI$ cross $w$ respectively in points $B_1$ and $C_1$, and line $B_1C_1$ intersect lines $AB$ and $AC$ in points $C_2$ and $B_2$, respectively. Let $w_1$— triangle circumscribed circle $IB_1C_1$ and lines $IB_2$ and $IC_2$ cross $w_1$ 1 in points $M$ and $N$, respectively. Prove that $BC_2*B_2C = B_2M*C_2N$
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MrRTI
191 posts
MrRTI
#2 h  Mar 28, 2014, 4:00 PM • 2 Y
Y by Adventure10, Mango247
Hmmm, this is weird, I got $BC_{2} = C_2N$ and $B_{2}C =B_{2}M$.
Notice that $B_{1}$ and $C_{1}$ are the midpoints of the arc $BC$ and $CA$, so $\angle{B_{2}B_{1}I} = \angle{B_{2}CI}$, so $B_{2}B_{1}CI$ is cyclic, then, $\angle{AB_{2}C_{2}} = \angle{CB_{2}B_{1}} = \angle{CIB_{1}} = \frac{\angle{B}}{2} + \frac{\angle{C}}{2} = 90^{\circ} - \frac{\angle{A}}{2}$. But clearly, $\angle{AI,B_{1}C_{1}} = 180^{\circ} - \frac{\angle{A}}{2} - \angle{B_{1}AC} - \angle{AB_{1}C_{1}} = 180^{\circ} - \frac{\angle{A}+\angle{B}+\angle{C}}{2} = 90^{\circ}$.
So, $AB_{2}IC_{2}$ is a rhombus.

Now, because $B_{2}$ is on the radical axis $B_{1}C_{1}$, then it has an equal power w.r.t both $\omega$ and $\omega_{1}$, so $BC_{2} \cdot C_{2}A = NC_{2} \cdot C_{2}I$. So, $AIBN$ is cyclic. But, $AB_{2}C_{2}I$ is a rhombus, so we must have $BC_{2} = C_{2}N$. Working analogously, we get $CB_{2} = B_{2}M$. Thus, the problem has solved.
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