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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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A lies on the radical axis of BQX and CPX
a_507_bc   35
N 19 minutes ago by jordiejoh
Source: APMO 2024 P1
Let $ABC$ be an acute triangle. Let $D$ be a point on side $AB$ and $E$ be a point on side $AC$ such that lines $BC$ and $DE$ are parallel. Let $X$ be an interior point of $BCED$. Suppose rays $DX$ and $EX$ meet side $BC$ at points $P$ and $Q$, respectively, such that both $P$ and $Q$ lie between $B$ and $C$. Suppose that the circumcircles of triangles $BQX$ and $CPX$ intersect at a point $Y \neq X$. Prove that the points $A, X$, and $Y$ are collinear.
35 replies
a_507_bc
Jul 29, 2024
jordiejoh
19 minutes ago
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sum(ab/4a^2+b^2) <= 3/5
truongphatt2668   3
N 29 minutes ago by Nguyenhuyen_AG
Source: I remember I read it somewhere
Let $a,b,c>0$. Prove that:
$$\dfrac{ab}{a^2+4b^2} + \dfrac{bc}{b^2+4c^2} + \dfrac{ca}{c^2+4a^2} \le \dfrac{3}{5}$$
3 replies
truongphatt2668
Monday at 1:23 PM
Nguyenhuyen_AG
29 minutes ago
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the epitome of olympiad nt
youlost_thegame_1434   30
N 42 minutes ago by Jupiterballs
Source: 2023 IMO Shortlist N3
For positive integers $n$ and $k \geq 2$, define $E_k(n)$ as the greatest exponent $r$ such that $k^r$ divides $n!$. Prove that there are infinitely many $n$ such that $E_{10}(n) > E_9(n)$ and infinitely many $m$ such that $E_{10}(m) < E_9(m)$.
30 replies
youlost_thegame_1434
Jul 17, 2024
Jupiterballs
42 minutes ago
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inequalities
Cobedangiu   5
N an hour ago by Cobedangiu
Source: own
$a,b>0$ and $a+b=1$. Find min P:
$P=\sqrt{\frac{1-a}{1+7a}}+\sqrt{\frac{1-b}{1+7b}}$
5 replies
Cobedangiu
Yesterday at 6:10 PM
Cobedangiu
an hour ago
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No more topics!
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Cyclic Quadrilateral, bisectors, points lie on perpendicular line
mruczek   5
N Sep 12, 2024 by War-Hammer
Source: 54 Polish MO 2003 Second Round - First Day Problem 2
The quadrilateral $ABCD$ is inscribed in the circle $o$. Bisectors of angles $DAB$ and $ABC$ intersect at point $P$, and bisectors of angles $BCD$ and $CDA$ intersect in point $Q$. Point $M$ is the center of this arc $BC$ of the circle $o$ which does not contain points $D$ and $A$. Point $N$ is the center of the arc $DA$ of the circle $o$, which does not contain points $B$ and $C$. Prove that the points $P$ and $Q$ lie on the line perpendicular to $MN$.
5 replies
mruczek
Feb 19, 2018
War-Hammer
Sep 12, 2024
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Cyclic Quadrilateral, bisectors, points lie on perpendicular line
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Reveal topic tags
geometry
cyclic quadrilateral
Poland
bisector
perpendicular
geometry unsolved
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Source: 54 Polish MO 2003 Second Round - First Day Problem 2
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mruczek
67 posts
mruczek
#1 h  Feb 19, 2018, 1:15 AM • 2 Y
Y by Adventure10, Rounak_iitr
The quadrilateral $ABCD$ is inscribed in the circle $o$. Bisectors of angles $DAB$ and $ABC$ intersect at point $P$, and bisectors of angles $BCD$ and $CDA$ intersect in point $Q$. Point $M$ is the center of this arc $BC$ of the circle $o$ which does not contain points $D$ and $A$. Point $N$ is the center of the arc $DA$ of the circle $o$, which does not contain points $B$ and $C$. Prove that the points $P$ and $Q$ lie on the line perpendicular to $MN$.
This post has been edited 4 times. Last edited by mruczek, Apr 21, 2018, 4:38 PM
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mruczek
67 posts
mruczek
#2 h  Mar 16, 2018, 5:14 PM • 2 Y
Y by Adventure10, Mango247
Bump, Bump
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coldheart361
26 posts
coldheart361
#3 h  Feb 2, 2020, 11:05 AM • 1 Y
Y by Adventure10
Can you give me hint?
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Radical_Axis
66 posts
Radical_Axis
#4 h  Feb 12, 2020, 2:41 PM • 2 Y
Y by coldheart361, Adventure10
coldheart361 wrote:
Can you give me hint?

Let $AD$ and $BC$ meet at $R$ .Then it's easy to see that $R,P,Q$ are collinear.
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coldheart361
26 posts
coldheart361
#5 h  Apr 3, 2020, 11:35 PM • 2 Y
Y by Mango247, Mango247
Thanks for the clue! I finally got it.
Let AD and BC meet at R, then P is be the incenter of $\triangle{ABR}$ and Q is the excenter of $\triangle{CDR}$, hence P, Q, and R lie on the bisector of $\angle{R}$.
Now let the tangent of circle at $M$ hit $PQ$ at $X$ and tangent of circle at $N$ hit $PQ$ at $Y$,
since $M$ and $N$ are midpoints of arc, we get $MX//AR$ and $NY//BR$,
Let $MX$ and $NY$ intersect at $Z$, then the bisector of $\angle{Z}$ is parallel to $PQR$, which is perpendicular to $MN$
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War-Hammer
670 posts
War-Hammer
#6 h  Sep 12, 2024, 3:49 AM
Y by
Also check Poland Second Round 2003 - Day 1 - Problem 2.
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