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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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Polynomial
EtacticToe   2
N 16 minutes ago by yuribogomolov
Source: Own
Let $f(x)$ be a monic polynomial with integer coefficient. And suppose there exist 4 distinct integer $a,b,c,d$ such that $f(a)=…=f(d)=5$.

Find all $k$ such that $f(k)=8$
2 replies
+1 w
EtacticToe
Dec 14, 2024
yuribogomolov
16 minutes ago
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Is this FE solvable?
Mathdreams   2
N 21 minutes ago by Mathdreams
Find all $f:\mathbb{R} \rightarrow \mathbb{R}$ such that \[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]for all reals $x$ and $y$.
2 replies
Mathdreams
Yesterday at 6:58 PM
Mathdreams
21 minutes ago
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Prove that there are no tuples $(x, y, z)$ sastifying $x^2+y^2-z^2=xyz-2$
Anabcde   1
N 21 minutes ago by giangtruong13
Prove that there are no tuples $(x, y, z) \in \mathbb{Z}^3$ sastifying $x^2+y^2-z^2=xyz-2$
1 reply
Anabcde
2 hours ago
giangtruong13
21 minutes ago
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inequalities hard
Cobedangiu   3
N 22 minutes ago by sqing
problem
3 replies
1 viewing
Cobedangiu
Mar 31, 2025
sqing
22 minutes ago
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No more topics!
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CM bisects <BCD, extension of side and diagonal of parallelogram, BK = DL
parmenides51   1
N May 29, 2022 by FornaxA
Source: 2022 SPbU Qualifying round, grades 10-11 p8 - Saint Petersburg State University School Olympiad
On the extension of side $AB$ of parallelogram $ABCD$ beyond point $B$ , the point $K$ is marked, and the point $L $is marked on the extension of the side $AD$ beyond the point $D$. It turned out that $BK = DL$. Segments $BL$ and $DK$ intersect at point $M$ . Prove that $CM$ is the bisector of angle $\angle BCD$.
1 reply
parmenides51
May 29, 2022
FornaxA
May 29, 2022
J
CM bisects <BCD, extension of side and diagonal of parallelogram, BK = DL
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Reveal topic tags
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parallelogram
angle bisector
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Source: 2022 SPbU Qualifying round, grades 10-11 p8 - Saint Petersburg State University School Olympiad
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parmenides51
30629 posts
parmenides51
#1 h  May 29, 2022, 9:30 AM • 1 Y
Y by FornaxA
On the extension of side $AB$ of parallelogram $ABCD$ beyond point $B$ , the point $K$ is marked, and the point $L $is marked on the extension of the side $AD$ beyond the point $D$. It turned out that $BK = DL$. Segments $BL$ and $DK$ intersect at point $M$ . Prove that $CM$ is the bisector of angle $\angle BCD$.
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FornaxA
7 posts
FornaxA
#2 h  May 29, 2022, 11:48 AM
Y by
In triangle BCD, we have: $\ \displaystyle \frac{\sin{\widehat{DCM}}}{\sin{\widehat{MCB}}} \cdot \frac{\sin{\widehat{CBL}}}{\sin{\widehat{LBD}}} \cdot \frac{\sin{\widehat{BDK}}}{\sin{\widehat{KDC}}}=1, \ $ so $\ \displaystyle \frac{\sin{\widehat{MCB}}}{\sin{\widehat{DCM}}} = \frac{\sin{\widehat{CBL}}}{\sin{\widehat{KDC}}} \cdot \frac{\sin{\widehat{BDK}}/BK}{\sin{\widehat{DBL}}/DL} = \frac{\sin{\widehat{CBL}}}{\sin{\widehat{KDC}}} \cdot \frac{\sin{\widehat{DKB}}/BD}{\sin{\widehat{DLB}/DB}} = \frac{\sin{\widehat{CBL}}}{\sin{\widehat{KDC}}} \cdot \frac{\sin{\widehat{DKB}}}{\sin{\widehat{DLB}}}, \ $ but, because ABCD is a parallelogram, $\ \widehat{DKB}= \widehat{KDC}, \ $ and $\ \widehat{DLB}= \widehat{LBC}, \ $ and so $\ \displaystyle \frac{\sin{\widehat{MCB}}}{\sin{\widehat{DCM}}}= \frac{\sin{\widehat{CBL}}}{\sin{\widehat{KDC}}} \cdot \frac{\sin{\widehat{KDC}}}{\sin{\widehat{CBL}}}=1. \ $ Therefore, $\ \sin{\widehat{MCB}} = \sin{\widehat{DCM}}. \ $ The sum of these angles is less than $\ 180^{\circ}, \ $ so they are equal and CM bisects the angle $\ \widehat{DCB}$.
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