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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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Japanese Triangles
pikapika007   67
N 7 minutes ago by quantam13
Source: IMO 2023/5
Let $n$ be a positive integer. A Japanese triangle consists of $1 + 2 + \dots + n$ circles arranged in an equilateral triangular shape such that for each $i = 1$, $2$, $\dots$, $n$, the $i^{th}$ row contains exactly $i$ circles, exactly one of which is coloured red. A ninja path in a Japanese triangle is a sequence of $n$ circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. Here is an example of a Japanese triangle with $n = 6$, along with a ninja path in that triangle containing two red circles.
IMAGE
In terms of $n$, find the greatest $k$ such that in each Japanese triangle there is a ninja path containing at least $k$ red circles.
67 replies
pikapika007
Jul 9, 2023
quantam13
7 minutes ago
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Three Nagel points collinear
jayme   2
N 18 minutes ago by jayme
Dear Marthlinkers,

1. ABCD a square
2. M a point on the segment CD sothat MA < MB
3. Nm, Na, Nb the Nagel’s points of the triangles MAD, ADM, BCM.

Prove : Nm, Na and Nb are collinear.

Sincerely
Jean-Louis
2 replies
jayme
Mar 31, 2025
jayme
18 minutes ago
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D1018 : Can you do that ?
Dattier   1
N 29 minutes ago by Dattier
Source: les dattes à Dattier
We can find $A,B,C$, such that $\gcd(A,B)=\gcd(C,A)=\gcd(A,2)=1$ and $$\forall n \in \mathbb N^*, (C^n \times B \mod A) \mod 2=0 $$.

For example :

$C=20$
$A=47650065401584409637777147310342834508082136874940478469495402430677786194142956609253842997905945723173497630499054266092849839$

$B=238877301561986449355077953728734922992395532218802882582141073061059783672634737309722816649187007910722185635031285098751698$

Can you find $A,B,C$ such that $A>3$ is prime, $C,B \in (\mathbb Z/A\mathbb Z)^*$ with $o(C)=(A-1)/2$ and $$\forall n \in \mathbb N^*, (C^n \times B \mod A) \mod 2=0 $$?
1 reply
Dattier
Mar 24, 2025
Dattier
29 minutes ago
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n-gon function
ehsan2004   9
N an hour ago by AshAuktober
Source: Romanian IMO Team Selection Test TST 1996, problem 1
Let $ f: \mathbb{R}^2 \rightarrow \mathbb{R} $ be a function such that for every regular $ n $-gon $ A_1A_2 \ldots A_n $ we have $ f(A_1)+f(A_2)+\cdots +f(A_n)=0 $. Prove that $ f(x)=0 $ for all reals $ x $.
9 replies
ehsan2004
Sep 13, 2005
AshAuktober
an hour ago
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Kazakhstan National Olympiad 2015 Final Round, 9grade, P3
zhumazhenis   1
N Mar 14, 2015 by Luis González
Source: Kazakhstan National Olympiad 2015 Final Round, 9grade, P3
Given right trangle with $\angle{C}=90^{\circ}$. Inscribed and escribed circles of $ABC$ are tangent to side $BC$ at points $A_1$ and $A_2$. Similarly, we define the points $B_1$ and $B_2$. Prove that the segments $A_1B_2$ and $B_1A_2$ intersect at an altitude drawn from vertex $C$ of the triangle $ABC$.
1 reply
zhumazhenis
Mar 14, 2015
Luis González
Mar 14, 2015
J
Kazakhstan National Olympiad 2015 Final Round, 9grade, P3
G H J
Reveal topic tags
geometry
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G H BBookmark kLocked kLocked NReply
Source: Kazakhstan National Olympiad 2015 Final Round, 9grade, P3
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zhumazhenis
94 posts
zhumazhenis
#1 h  Mar 14, 2015, 1:58 PM • 2 Y
Y by Adventure10, Mango247
Given right trangle with $\angle{C}=90^{\circ}$. Inscribed and escribed circles of $ABC$ are tangent to side $BC$ at points $A_1$ and $A_2$. Similarly, we define the points $B_1$ and $B_2$. Prove that the segments $A_1B_2$ and $B_1A_2$ intersect at an altitude drawn from vertex $C$ of the triangle $ABC$.
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Luis González
4145 posts
Luis González
#2 h  Mar 14, 2015, 3:53 PM • 2 Y
Y by Adventure10, Mango247
B-excircle touches $AB$ at $Y.$ $D$ is the foot of the C-altitude and $A_1B_2$ cuts $CD$ at $X.$ Using standard triangle notation, we get

$\frac{YA}{YB} \cdot \frac{BA_1}{A_1C} \cdot \frac{CB_2}{B_2A}=\frac{s-c}{s} \cdot \frac{s-b}{s-c} \cdot \frac{s-a}{s-c}=\frac{(s-a) \cdot (s-b)}{s \cdot (s-c)}=1.$

Thus, by Menelaus' theorem $A_1,B_2,Y$ are collinear $\Longrightarrow$ $\widehat{CXA_1}=\widehat{AB_2Y}+\widehat{ACD}=\tfrac{1}{2}\widehat{A}+90^{\circ}-\widehat{A}=90^{\circ}-\tfrac{1}{2}\widehat{A}=90^{\circ}-\tfrac{1}{2}\widehat{XCA_1}$ $\Longrightarrow$ $\triangle CA_1X$ is C-isosceles $\Longrightarrow$ $CX=CA_1=s-c.$ Similarly $B_1A_2$ will cut $CD$ at the same point $X.$
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