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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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Nordic 2025 P3
anirbanbz   8
N a minute ago by Primeniyazidayi
Source: Nordic 2025
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. Let $E$ and $F$ be points on the line segments $AC$ and $AB$ respectively such that $AEHF$ is a parallelogram. Prove that $\vert OE \vert = \vert OF \vert$.
8 replies
anirbanbz
Mar 25, 2025
Primeniyazidayi
a minute ago
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A number theory problem from the British Math Olympiad
Rainbow1971   11
N 22 minutes ago by Rainbow1971
Source: British Math Olympiad, 2006/2007, round 1, problem 6
I am a little surprised to find that I am (so far) unable to solve this little problem:

[quote]Let $n$ be an integer. Show that, if $2 + 2 \sqrt{1+12n^2}$ is an integer, then it is a perfect square.[/quote]

I set $k := \sqrt{1+12n^2}$. If $2 + 2 \sqrt{1+12n^2}$ is an integer, then $k (=\sqrt{1+12n^2})$ is at least rational, so that $1 + 12n^2$ must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs $(n, k)$ for which this happens are $(0,1), (2,7), (28,97)$ and $(390, 1351)$, and that, for every such pair $(n,k)$, the "next" such pair can be calculated as
$$ \begin{bmatrix} 7 & 2 \\ 24 & 7 \end{bmatrix} \begin{bmatrix} n \\ k \end{bmatrix} .$$The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?

Thank you.




11 replies
Rainbow1971
Yesterday at 8:39 PM
Rainbow1971
22 minutes ago
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f( - f (x) - f (y))= 1 -x - y , in Zxz
parmenides51   6
N 29 minutes ago by Chikara
Source: 2020 Dutch IMO TST 3.3
Find all functions $f: Z \to Z$ that satisfy $$f(-f (x) - f (y))= 1 -x - y$$for all $x, y \in Z$
6 replies
parmenides51
Nov 22, 2020
Chikara
29 minutes ago
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Hard limits
Snoop76   2
N an hour ago by maths_enthusiast_0001
$a_n$ and $b_n$ satisfies the following recursion formulas: $a_{0}=1, $ $b_{0}=1$, $ a_{n+1}=a_{n}+b_{n}$$ $ and $ $$ b_{n+1}=(2n+3)b_{n}+a_{n}$. Find $ \lim_{n \to \infty} \frac{a_n}{(2n-1)!!}$ $ $ and $ $ $\lim_{n \to \infty} \frac{b_n}{(2n+1)!!}.$
2 replies
Snoop76
Mar 25, 2025
maths_enthusiast_0001
an hour ago
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No more topics!
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Pick 3
chess64   2
N May 15, 2006 by ZetaX
Source: Canada 1970, Problem 7
Show that from any five integers, not necessarily distinct, one can always choose three of these integers whose sum is divisible by 3.
2 replies
chess64
May 14, 2006
ZetaX
May 15, 2006
J
Pick 3
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Reveal topic tags
number theory unsolved
number theory
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Source: Canada 1970, Problem 7
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chess64
4794 posts
chess64
#1 h  May 14, 2006, 11:40 PM • 2 Y
Y by Adventure10, Mango247
Show that from any five integers, not necessarily distinct, one can always choose three of these integers whose sum is divisible by 3.
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pavel kozlov
613 posts
pavel kozlov
#2 h  May 15, 2006, 5:40 AM • 1 Y
Y by Adventure10
It's very easy problem.
Let's replace our numbers on it's remainders modulo $3$.
If there is $\geq 3$ same numbers, their sum is divisible by $3$.
Else by Dirichlet's principle there is $3$ different numbers: $0,1,2$. Their sum is divisible by $3$ too, q.e.d
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ZetaX
7579 posts
ZetaX
#3 h  May 15, 2006, 10:00 AM • 3 Y
Y by myh2910, Adventure10, Mango247
Or just Erdös-Ginzburg-Ziv for $ n = 3$ ;)
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