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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
2 hours ago
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following events:
[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
[*]April 8th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS State Discussion
April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
[*]April 10th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MathILy and MathILy-Er Math Jam: Multibackwards Numbers
[*]April 16th (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
1 viewing
jlacosta
2 hours ago
0 replies
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
Thanks u!
Ruji2018252   1
N 2 minutes ago by alexheinis
Let $a_1,...,a_{2024}\in\mathbb{Z}$ and $1\leqslant a_1\leqslant a_2\leqslant ...\leqslant a_{2024}\leqslant 99$ and
\[P=a_1^2+a_2^2+...+a_{2024}^2-(a_1a_3+a_2a_4+...+a_{2022}a_{2024})\]Find maximum $P$
1 reply
Ruji2018252
Yesterday at 11:51 AM
alexheinis
2 minutes ago
Number-erasing game with multiples of 3
p108771953   4
N 4 minutes ago by conejita
Source: 2024 Mexican Mathematical Olympiad, Problem 6
Ana and Beto play on a blackboard where all integers from 1 to 2024 (inclusive) are written. On each turn, Ana chooses three numbers $a,b,c$ written on the board and then Beto erases them and writes one of the following numbers:
$$a+b-c, a-b+c, ~\text{or}~ -a+b+c.$$The game ends when only two numbers are left on the board and Ana cannot play. If the sum of the final numbers is a multiple of 3, Beto wins. Otherwise, Ana wins. ¿Who has a winning strategy?
4 replies
p108771953
Nov 6, 2024
conejita
4 minutes ago
Another square grid :D
MathLuis   43
N 12 minutes ago by akliu
Source: USEMO 2021 P1
Let $n$ be a fixed positive integer and consider an $n\times n$ grid of real numbers. Determine the greatest possible number of cells $c$ in the grid such that the entry in $c$ is both strictly greater than the average of $c$'s column and strictly less than the average of $c$'s row.

Proposed by Holden Mui
43 replies
1 viewing
MathLuis
Oct 30, 2021
akliu
12 minutes ago
Number theory
Maaaaaaath   0
25 minutes ago
Let $m$ be a positive integer . Prove that there exists infinitely many pairs of positive integers $(x,y)$ such that $\gcd(x,y)=1$ and :

$$xy  |  x^2+y^2+m$$
0 replies
Maaaaaaath
25 minutes ago
0 replies
No more topics!
projection of foot of altitude on line, is incenter of another triangle
parmenides51   1
N Sep 14, 2018 by LuisXender
Source: Kazakhstan NMO 2016 grade IX P4
In the triangle $ABC$ from the largest angle $C$, $CH$ let be the altitude. Lines $HM$ and $HN$, are altitudes of triangles $ACH$ and $BCH$ respectively, $HP$ and $HQ$ are bisectors of triangles $AMH$ and $BNH$. Let $R$ be the foot of the perpendicular from the point $H$ on the line $PQ$ . Prove that $R$ is the intersection point of the angle bisectors of the triangle $MNH$ .
1 reply
parmenides51
Sep 13, 2018
LuisXender
Sep 14, 2018
projection of foot of altitude on line, is incenter of another triangle
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Source: Kazakhstan NMO 2016 grade IX P4
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parmenides51
30629 posts
#1 • 1 Y
Y by Adventure10
In the triangle $ABC$ from the largest angle $C$, $CH$ let be the altitude. Lines $HM$ and $HN$, are altitudes of triangles $ACH$ and $BCH$ respectively, $HP$ and $HQ$ are bisectors of triangles $AMH$ and $BNH$. Let $R$ be the foot of the perpendicular from the point $H$ on the line $PQ$ . Prove that $R$ is the intersection point of the angle bisectors of the triangle $MNH$ .
This post has been edited 2 times. Last edited by parmenides51, Dec 15, 2022, 9:09 PM
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LuisXender
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#2 • 2 Y
Y by Adventure10, Mango247
It is easy to see $CP=CH=CQ$, thats because $\angle AHM=\angle HCP$ and $\angle PHC= 90-\frac{\angle HCP}{2}$. So $C$ is the center of the circumcircle of $PHQ$ therefore $\angle HQP= \frac{\angle HCP}{2}=\frac{\angle HNM}{2}$, but $HQNR$ is cyclic so $\angle HNR=\angle HQR=\angle HQP$. Therefore $NR$ bisects $\angle HNM$. Do the analogous for $MR$ and done.
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