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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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VERY HARD MATH PROBLEM!
slimshadyyy.3.60   29
N a minute ago by FunnyKoala17
Let a ≥b ≥c ≥0 be real numbers such that a^2 +b^2 +c^2 +abc = 4. Prove that
a+b+c+(√a−√c)^2 ≥3.
29 replies
slimshadyyy.3.60
Yesterday at 10:49 PM
FunnyKoala17
a minute ago
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Best bid given the other's bid
fancyfairy   0
2 minutes ago
Consider sets $N = \{1,2\}$ (player) and $M = \{a,b,c,d\}$ (items), with valuations given by matrix
$ V = \left[\begin{matrix} v_{1a} & v_{1b} & v_{1c} & v_{1d} \\ v_{2a} & v_{2b} & v_{2c} & v_{2d} \end{matrix}\right] $
where $v_{ij} \in [0, \infty]$.

A player’s total valuation is the sum of item values won, with half-value for half-won items.

Given player $N \backslash \{i\}$'s bid, player $i$'s bid $B_i$ is strictly better than another possible bid $B_i'$ of his if the total valuation derived from $B_i$ is strictly more than that derived from ${B_i}'$.

Each player simultaneously bids on items with a bid vector $B_i = [b^i_1, b^i_2, b^i_3, b^i_4]$ such that $b^i_j \in [0,1]$ and $\sum_{j \in M} b^i_j \leq 1$. The highest bidder wins an item; ties split ownership equally.

Given valuation matrix $V$, can we always find bid vectors $B_1$ and $B_2$ such that given $B_1$, the second player can not submit a bid strictly better than $B_2$ and vice-versa?
0 replies
1 viewing
fancyfairy
2 minutes ago
0 replies
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IMO 2016 Problem 1
quangminhltv99   151
N 5 minutes ago by Thelink_20
Source: IMO 2016
Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$. Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent.
151 replies
quangminhltv99
Jul 11, 2016
Thelink_20
5 minutes ago
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polynomial
hanzo.ei   1
N 9 minutes ago by alexheinis
Source: from a book
Given a polynomial \( P(x) \) satisfying \( P(0) = 0 \), \( P(1) = 1 \), and for \( n \) (\( n \geq 2, n \in \mathbb{N} \)) positive real numbers \( k_1, k_2, \dots, k_n \). Prove that there exists a strictly increasing sequence of real numbers \( (a_i)_{i=1}^{n} \subset (0,1) \) such that
\[ \sum_{i=1}^{n} \frac{k_i}{P'(a_i)} = \sum_{i=1}^{n} k_i. \]
1 reply
hanzo.ei
Today at 3:12 PM
alexheinis
9 minutes ago
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A complex integral
Gauler   1
N 6 hours ago by paxtonw
For $s=c+it$ with $|t|\le T$, prove that
$$\frac{1}{2\pi i}\int _{c-iT}^{c+iT}\frac{e^{su}}{s}ds=\frac{1}{2}+\frac{Tu}{\pi }+O\left(T^2u^2+\frac{c}{T}\right)$$for $T|u|\le 1$.
1 reply
Gauler
Today at 3:26 PM
paxtonw
6 hours ago
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Gheorghe Țițeica 2025 Grade 12 P4
AndreiVila   2
N Today at 12:06 PM by MS_asdfgzxcvb
Source: Gheorghe Țițeica 2025
Let $R$ be a ring. Let $x,y\in R$ such that $x^2=y^2=0$. Prove that if $x+y-xy$ is nilpotent, so is $xy$.

Janez Šter
2 replies
AndreiVila
Mar 28, 2025
MS_asdfgzxcvb
Today at 12:06 PM
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AB=BA if A-nilpotent
KevinDB17   1
N Today at 9:51 AM by RobertRogo
Let A,B 2 complex n*n matrices such that AB+I=A+B+BA
If A is nilpotent prove that AB=BA
1 reply
KevinDB17
Today at 9:37 AM
RobertRogo
Today at 9:51 AM
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Gheorghe Țițeica 2025 Grade 11 P2
AndreiVila   1
N Today at 9:39 AM by RobertRogo
Source: Gheorghe Țițeica 2025
Let $n\geq 2$ and $A,B\in\mathcal{M}_n(\mathbb{C})$ such that $$\{\text{rank}(A^k)\mid k\geq 1\}=\{\text{rank}(B^k)\mid k\geq 1\}.$$Prove that $\text{rank}(A^k)=\text{rank}(B^k)$ for all $k\geq 1$.

Cristi Săvescu
1 reply
AndreiVila
Mar 28, 2025
RobertRogo
Today at 9:39 AM
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Integral
Martin.s   2
N Today at 9:14 AM by Entrepreneur
Show that
$$\int_{0}^{\infty} \log \left(\frac{1+a \sin^{2} bx}{1-a \sin^{2} bx} \right) \frac{1}{x^{2}} \ dx = \frac{\pi b}{2} \left( \sqrt{1+a} - \sqrt{1-a} \right)$$
2 replies
Martin.s
Dec 9, 2023
Entrepreneur
Today at 9:14 AM
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Positiion vectors of complex numbers
MetaphysicalWukong   1
N Today at 8:55 AM by MetaphysicalWukong
I cant even understand the question. Can someone help me?

1 reply
MetaphysicalWukong
Today at 8:54 AM
MetaphysicalWukong
Today at 8:55 AM
J
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Prove the inequality, very tight...
Butterfly   0
Today at 8:54 AM
Prove $$xe^x-x^3\sqrt{x+1}-\ln x>0.$$
0 replies
Butterfly
Today at 8:54 AM
0 replies
J
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Gheorghe Țițeica 2025 Grade 11 P1
AndreiVila   3
N Today at 7:23 AM by Levieee
Source: Gheorghe Țițeica 2025
Find all continuous functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x+y)=f(x+f(y))$ for all $x,y\in\mathbb{R}$.
3 replies
AndreiVila
Mar 28, 2025
Levieee
Today at 7:23 AM
J
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Gheorghe Țițeica 2025 Grade 12 P2
AndreiVila   1
N Today at 5:40 AM by Alphaamss
Source: Gheorghe Țițeica 2025
Let $f:[0,1]\rightarrow\mathbb{R}$ be a continuous function. Prove that $$\int_0^{\pi/2}f(\sin(2x))\sin x\, dx = \int_0^{\pi/2} f(\cos^2 x)\cos x\, dx.$$
1 reply
AndreiVila
Mar 28, 2025
Alphaamss
Today at 5:40 AM
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something like MVT
mqoi_KOLA   8
N Today at 4:20 AM by Alphaamss
If $F$ is a continuous function on $[0,1]$ such that $F(0) = F(1)$, then there exists a $c \in (0,1)$ such that:

\[ F(c) = \frac{1}{c} \int_0^c F(x) \,dx \]
8 replies
mqoi_KOLA
Yesterday at 11:37 AM
Alphaamss
Today at 4:20 AM
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J
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k Factorials, Sum of Powers, Diophantine Equations,
John_Mgr   3
N Mar 27, 2025 by John_Mgr
Source: Nepal NMO 2025 p4
Find all the pairs of positive integers $n$ and $x$ such that: \[1^n+2^n+3^n+\cdots +n^n=x!\]
$Petko$ $Lazarov, Bulgaria$
3 replies
John_Mgr
Mar 15, 2025
John_Mgr
Mar 27, 2025
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Factorials, Sum of Powers, Diophantine Equations,
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Reveal topic tags
number theory
nmo
Diophantine equation
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G H BBookmark kLocked kLocked NReply
Source: Nepal NMO 2025 p4
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John_Mgr
58 posts
John_Mgr
#1 h  Mar 15, 2025, 12:59 PM
Y by
Find all the pairs of positive integers $n$ and $x$ such that: \[1^n+2^n+3^n+\cdots +n^n=x!\]
$Petko$ $Lazarov, Bulgaria$
This post has been edited 1 time. Last edited by John_Mgr, Mar 15, 2025, 2:18 PM
Reason: positive integers
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RedFireTruck
4221 posts
RedFireTruck
#2 h  Mar 15, 2025, 1:51 PM
Y by
Solved with Alex-131.

When n=1, we get solutions (1,0) and (1,1). We claim these are the only ones. Assume n>=2.

Note that x>n.

Taking mod 2 gives 2 | 1+0+1+... so 4 | n.

When 2^k | n, then mod 2^k gives 2^k | 1+0+1+... so 2^(k+1) | n.

Therefore, no such n>=2 exists.
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AshAuktober
948 posts
AshAuktober
#3 h  Mar 15, 2025, 3:43 PM
Y by
Hi, kindly note that this is the main post vying for being added to contest collections right now.
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John_Mgr
58 posts
John_Mgr
#4 h  Mar 27, 2025, 2:49 AM
Y by
As @above mention heres the original one https://artofproblemsolving.com/community/c6h3528150p34284772,
How can i delete this??
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