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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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FE over R
IAmTheHazard   19
N an hour ago by Bardia7003
Source: ELMO Shortlist 2024/A3
Find all functions $f : \mathbb{R}\to\mathbb{R}$ such that for all real numbers $x$ and $y$,
$$f(x+f(y))+xy=f(x)f(y)+f(x)+y.$$
Andrew Carratu
19 replies
IAmTheHazard
Jun 22, 2024
Bardia7003
an hour ago
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Find values of $a b+a c+b c$
NJAX   11
N an hour ago by Baimukh
Source: 2nd Al-Khwarizmi International Junior Mathematical Olympiad 2024, Day2, Problem6
Let $a, b, c$ be distinct real numbers such that $a+b+c=0$ and $$ a^{2}-b=b^{2}-c=c^{2}-a. $$Evaluate all the possible values of $a b+a c+b c$.

Proposed by Nguyen Anh Vu, Vietnam
11 replies
NJAX
May 31, 2024
Baimukh
an hour ago
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Final fool geometry
giangtruong13   0
an hour ago
Let $ABC$ be a pointed triangle and altitudes $AD, BE, CF$. Prove that: $$S_{DEF}=(sin^2A*sin^2B+sin^2C-2)S_{ABC}$$
0 replies
giangtruong13
an hour ago
0 replies
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Interesting config
TheUltimate123   37
N an hour ago by E50
Source: ELMO 2023/4
Let \(ABC\) be an acute scalene triangle with orthocenter \(H\). Line \(BH\) intersects \(\overline{AC}\) at \(E\) and line \(CH\) intersects \(\overline{AB}\) at \(F\). Let \(X\) be the foot of the perpendicular from \(H\) to the line through \(A\) parallel to \(\overline{EF}\). Point \(B_1\) lies on line \(XF\) such that \(\overline{BB_1}\) is parallel to \(\overline{AC}\), and point \(C_1\) lies on line \(XE\) such that \(\overline{CC_1}\) is parallel to \(\overline{AB}\). Prove that points \(B\), \(C\), \(B_1\), \(C_1\) are concyclic.

Proposed by Luke Robitaille
37 replies
TheUltimate123
Jun 26, 2023
E50
an hour ago
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No more topics!
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complex bash oops
megahertz13   2
N Mar 30, 2025 by lpieleanu
Source: PUMaC Finals 2016 A3
On a cyclic quadrilateral $ABCD$, let $M$ and $N$ denote the midpoints of $\overline{AB}$ and $\overline{CD}$. Let $E$ be the projection of $C$ onto $\overline{AB}$ and let $F$ be the reflection of $N$ over the midpoint of $\overline{DE}$. Assume $F$ lies in the interior of quadrilateral $ABCD$. Prove that $\angle BMF = \angle CBD$.
2 replies
megahertz13
Nov 5, 2024
lpieleanu
Mar 30, 2025
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complex bash oops
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geometry
complex bash
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Source: PUMaC Finals 2016 A3
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megahertz13
3180 posts
megahertz13
#1 h  Nov 5, 2024, 2:33 AM • 1 Y
Y by endless_abyss
On a cyclic quadrilateral $ABCD$, let $M$ and $N$ denote the midpoints of $\overline{AB}$ and $\overline{CD}$. Let $E$ be the projection of $C$ onto $\overline{AB}$ and let $F$ be the reflection of $N$ over the midpoint of $\overline{DE}$. Assume $F$ lies in the interior of quadrilateral $ABCD$. Prove that $\angle BMF = \angle CBD$.
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maths_enthusiast_0001
132 posts
maths_enthusiast_0001
#2 h  Dec 19, 2024, 2:45 PM
Y by
Toss the figure on the complex plane with circumcircle of $ABCD$ as the unit circle.
Thus we have, $\boxed{A=a,B=b,C=c,D=d,M=\frac{a+b}{2},N=\frac{c+d}{2}}$. Also, $E=\frac{1}{2}\left(a+b+c-\frac{ab}{c}\right)$.
Now clearly $\square FEND$ is a parallelogram thus, $F+N=E+D$ and we get, $\boxed{F=\frac{1}{2}\left(a+b+d-\frac{ab}{c}\right)}$. Now we have, $\angle BMF=arg\left(\frac{(b-a)c}{(cd-ab)}\right)$ and $\angle CBD=arg\left(\frac{c-b}{d-b}\right)$.
Claim: $\angle BMF=\angle CBD \iff arg\left(\frac{(b-a)c}{(cd-ab)}\right)=arg\left(\frac{c-b}{d-b}\right) \iff \left(\frac{\frac{(b-a)c}{(cd-ab)}}{\frac{c-b}{d-b}}\right) \in \mathbb{R}$
Proof: Let $\alpha=\left(\frac{\frac{(b-a)c}{(cd-ab)}}{\frac{c-b}{d-b}}\right)$ then we want $\alpha=\overline{\alpha}$. Now,
$$\alpha=\left(\frac{\frac{(b-a)c}{(cd-ab)}}{\frac{c-b}{d-b}}\right)=\left(\frac{(b-d)(a-b)c}{(ab-cd)(b-c)}\right)$$$$ \implies \overline{\alpha}=\left(\frac{(\frac{1}{b}-\frac{1}{d})(\frac{1}{a}-\frac{1}{b})\frac{1}{c}}{(\frac{1}{ab}-\frac{1}{cd})(\frac{1}{b}-\frac{1}{c})}\right)=\left(\frac{\frac{(d-b)(b-a)}{adcb^{2}}}{\frac{(cd-ab)(c-b)}{ab^{2}c^{2}d}}\right)=\left(\frac{c(b-d)(a-b)}{(ab-cd)(b-c)}\right)=\alpha$$as desired. Thus $\angle{BMF}=\angle{CBD}$. ($\mathcal{QED}$) $\blacksquare$
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lpieleanu
2853 posts
lpieleanu
#3 h  Mar 30, 2025, 12:22 AM
Y by
Solution
This post has been edited 1 time. Last edited by lpieleanu, Mar 30, 2025, 12:27 AM
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