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a My Retirement & New Leadership at AoPS
rrusczyk   1398
N a minute ago by ExcitablePorcupine48
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
1398 replies
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rrusczyk
Monday at 6:37 PM
ExcitablePorcupine48
a minute ago
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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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Slightly weird points which are not so weird
Pranav1056   12
N 5 minutes ago by ihategeo_1969
Source: India TST 2023 Day 4 P1
Suppose an acute scalene triangle $ABC$ has incentre $I$ and incircle touching $BC$ at $D$. Let $Z$ be the antipode of $A$ in the circumcircle of $ABC$. Point $L$ is chosen on the internal angle bisector of $\angle BZC$ such that $AL = LI$. Let $M$ be the midpoint of arc $BZC$, and let $V$ be the midpoint of $ID$. Prove that $\angle IML = \angle DVM$
12 replies
Pranav1056
Jul 9, 2023
ihategeo_1969
5 minutes ago
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Combinatoric's comeback
giangtruong13   1
N 5 minutes ago by wassupevery1
Source: Vietnam TST IMO 2025 P5
Given $n$x$n$ square board has the row and column numbered from $1$ to $n$, square in $ith$ row and $jth$ column get symbolized by square $(i,j)$ . Subset $A$ of squares on the board is called "good" subset if two random squares $({x}_1, y), ({x}_2, y)$ belong to $A$ satisfy that the squares $(u,v)$ with $ {x}_1 <u \leq {x}_2,v<y$ or ${x}_1 \leq u <{x}_2, v>y$ are not belong to $A$. Find the minimum number of "good" distinct subsets such that each square on the board belongs to only one subset
1 reply
giangtruong13
15 minutes ago
wassupevery1
5 minutes ago
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The three lines AA', BB' and CC' meet on the line IO
WakeUp   44
N 5 minutes ago by ihategeo_1969
Source: Romanian Master Of Mathematics 2012
Let $ABC$ be a triangle and let $I$ and $O$ denote its incentre and circumcentre respectively. Let $\omega_A$ be the circle through $B$ and $C$ which is tangent to the incircle of the triangle $ABC$; the circles $\omega_B$ and $\omega_C$ are defined similarly. The circles $\omega_B$ and $\omega_C$ meet at a point $A'$ distinct from $A$; the points $B'$ and $C'$ are defined similarly. Prove that the lines $AA',BB'$ and $CC'$ are concurrent at a point on the line $IO$.

(Russia) Fedor Ivlev
44 replies
+1 w
WakeUp
Mar 3, 2012
ihategeo_1969
5 minutes ago
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Nice problemm
hanzo.ei   0
11 minutes ago

Consider the sequence $(a_n)$ defined as follows:
\[ a_1 = \frac{\sqrt{6}}{3}, \quad a_{n+1} = a_n + \frac{1}{3a_n}, \quad \forall n \in \mathbb{N}. \]
a, Prove that
\[ 0 \le a_n \sqrt{6} - 2\sqrt{n} \le \frac{1}{4\sqrt{n}} \Bigl( 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \Bigr), \quad \forall n \in \mathbb{N}. \]b,For each $n \in \mathbb{N}$, define
\[ b_n = \frac{3a_n^2 - 2n - 1}{\ln(n+1)}. \]Compute the limit $\displaystyle \lim_{n \to \infty} b_n.$
0 replies
hanzo.ei
11 minutes ago
0 replies
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Gaussian integral
soruz   3
N Today at 8:25 AM by Mathzeus1024
Exist a method of calculation for $ \int e^{-x^2}\,dx $, with help of $ e^{i \phi}=cos \phi + i sin \phi $ and Moivre's formula.
3 replies
soruz
Oct 20, 2013
Mathzeus1024
Today at 8:25 AM
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Limit conundrum
MetaphysicalWukong   4
N Today at 7:42 AM by MetaphysicalWukong
Source: UNSW
Why is the last statement not true? And how do we know the selected option is true?
4 replies
MetaphysicalWukong
Yesterday at 8:00 AM
MetaphysicalWukong
Today at 7:42 AM
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Finding supremum of a weird function
pokoknyaakuimut   4
N Today at 6:56 AM by MihaiT
Find $\text{sup}\{2^{2x}+2^{\frac{1}{2x}}:x\in\mathbb{R}, x<0\}$. Easy to guess that the answer is $1$, but I haven't found the reason yet. :(
4 replies
pokoknyaakuimut
Feb 14, 2025
MihaiT
Today at 6:56 AM
J
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real analysis
ay19bme   3
N Yesterday at 8:46 PM by ay19bme
...........................
3 replies
ay19bme
Yesterday at 4:19 PM
ay19bme
Yesterday at 8:46 PM
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Integration Bee Kaizo
Calcul8er   50
N Yesterday at 7:10 PM by Shikhar_
Hey integration fans. I decided to collate some of my favourite and most evil integrals I've written into one big integration bee problem set. I've been entering integration bees since 2017 and I've been really getting hands on with the writing side of things over the last couple of years. I hope you'll enjoy!
50 replies
Calcul8er
Mar 2, 2025
Shikhar_
Yesterday at 7:10 PM
J
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Putnam 1950 B1
centslordm   2
N Yesterday at 6:19 PM by KAME06
In each of $n$ houses on a straight street are one or more boys. At what point should all the boys meet so that the sum of the distances that they walk is as small as possible?
2 replies
centslordm
May 25, 2022
KAME06
Yesterday at 6:19 PM
J
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Another integral
Martin.s   2
N Yesterday at 12:43 PM by MS_asdfgzxcvb


\[ I = \int_{0}^{\frac{1}{\sqrt{3}}} \frac{u \arctan(u)}{(1 - u^2) \sqrt{1 - 2 u^2}} \, du \]
2 replies
Martin.s
Mar 9, 2025
MS_asdfgzxcvb
Yesterday at 12:43 PM
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Some integrals and sums(series)
Martin.s   1
N Yesterday at 12:09 PM by Entrepreneur
Source: Inspired from silver08
I saw Silver's post, so I thought I'd share some integrals and sums as well.


It's Christmas!!! (or boxing day.)


\begin{align*} 1. & \quad \int_{0}^{1} \frac{K(-x) - E(-x)}{x \sqrt{x+1}} \ln\left(\frac{1-x}{1+x}\right) \, dx = \frac{\pi - 4 \ln(2)}{4\sqrt{\pi}} \cdot \Gamma^2\left(\frac{1}{4}\right) \\ & \text{where:} \\ & \quad E(x) = \int_{0}^{1} \frac{\sqrt{1 - t^2 x}}{\sqrt{1 - t^2}} \, dt, \quad K(x) = \int_{0}^{1} \frac{1}{\sqrt{1 - t^2} \sqrt{1 - t^2 x}} \, dt. \end{align*}
\begin{align*} 2. & \quad I = \int_{0}^{\infty} \frac{1}{1+x} \ln\left(\prod_{k=1}^{\infty}\left(1 + e^{-(2k+1)\sqrt{x}}\right) \prod_{k=1}^{\infty}\left(1 + e^{-(2k+1)\pi^2\sqrt{x}}\right)\right) \, dx \end{align*}
\begin{align*} 3. & \quad \Omega = \sum_{n=1}^{\infty} (-1)^{n-1} \left(\frac{a n + b}{n(n+1)}\right)^3 H_n^2 \\ & \text{where: } H_n = \sum_{k=1}^{n} \frac{1}{k} \quad \text{(Harmonic numbers)}, \quad a, b \in \mathbb{R}. \end{align*}
\begin{align*} 4. & \quad \int_{0}^{\infty} \frac{\ln\left(\sqrt{z^4 + z^2 + 1}\right) - \ln(z)}{z^{10} + 1} \cdot \frac{z^2 + 1}{z^4 + z^2 + 1} \, dz \end{align*}
\begin{align*} 5. & \quad \int_{0}^{\frac{\pi}{2}} \frac{\left(c(a_1 - a_2 \sin^2 x)(b_1 - b_2 \cos^2 x)\right)}{\left(\alpha_1 + \beta_1 \sin^2 x\right)\left(\alpha_2 + \beta_2 \cos^2 x\right)} \, dx \end{align*}
\begin{align*} 6. & \quad \Omega = \int_{0}^{\infty} e^{-(a+b+c)x} \prod_{n=1}^{\infty}\left(1 + \frac{(a-b)^2 x^2}{n^2}\right) \, dx, \quad a, b \in \mathbb{R}^{+}, \, 0 \leq c \in \mathbb{R}. \end{align*}

$$8. \int_{0}^{1} \frac{\tan^{-1}(x)}{1-x} \ln\left(\frac{1}{2} \left(\frac{1}{\sqrt{x}} + \sqrt{x}\right)\right) \, dx = \frac{\ln(2)}{4} - C - \frac{\pi^3}{192} + \frac{\pi}{32} (\ln(2))^2.$$

$$9. \int_{0}^{\frac{\pi}{4}} \int_{0}^{\frac{\pi}{4}} \frac{\ln^{2n}(\sin x) \sum_{k=0}^{\infty} \sum_{j=0}^{2n-1} \binom{j}{k-1} \left( \frac{\ln(\sec x)}{\ln(\sin x)} \right)^k}{\cot x \left( \cos^2 y + \tan x \cos y \sin y \right)} \, dy \, dx, \quad n \in \mathbb{Z}^+. $$


\[ \text{Find: } 10. \sum_{n=1}^\infty \sum_{m=-\infty}^\infty \frac{1}{n^p m^2 (m^2 + 1)^3 (n+1)^q}, \quad 2 \leq p, q \in \mathbb{Z}, \]\[ 11. \sum_{n=1}^\infty \sum_{m=-\infty}^\infty \frac{(-1)^{n+m}}{n^p m^2 (m^2 + 1)^3 (n+1)^q}, \quad 2 \leq p, q \in \mathbb{Z}. \]

\[12. \int_{0}^{\frac{\pi}{4}} \frac{\sin x}{\cos(2x) + 2} \tan^{-1}\left(\frac{\cos x \cot(2x)}{\sqrt{2}}\right) dx = \frac{5\pi^2}{48\sqrt{2}} - \frac{\pi}{4\sqrt{2}} \cos^{-1}\left(\frac{1}{3}\right). \]
1 reply
Martin.s
Dec 26, 2024
Entrepreneur
Yesterday at 12:09 PM
J
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An integral
gaussiemann144   1
N Yesterday at 10:10 AM by vanstraelen
Given $\alpha, \beta$-
$\alpha = \int_0^1 xe^{\frac{x^2-1}{2}} \cos(x) dx \quad \beta = \int_1^{3/2} e^{2(x^2-2x)} \sqrt{1-\cos(4x-4)} dx$
Find- $$\frac{\alpha - \cos(1) + e^{-1/2}}{\beta}$$
1 reply
gaussiemann144
Monday at 8:15 PM
vanstraelen
Yesterday at 10:10 AM
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Ahlfors 3.3.1.2
centslordm   3
N Yesterday at 9:14 AM by Mathzeus1024
If \[T_1 z = \frac{z + 2}{z + 3}, \qquad T_2 z = \frac z{z + 1},\]find $T_1 T_2z, \,T_2 T_1z$ and ${T_1}^{-1} T_2 z.$
3 replies
centslordm
Jan 8, 2025
Mathzeus1024
Yesterday at 9:14 AM
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Integral inequality
ohiorizzler1434   1
N Dec 20, 2024 by rudransh61
Given functions f(x) and g(x) non-negative such that g(x) is decreasing, show $\int f'(x)g(x) \geq f(x)g(x)$.
1 reply
ohiorizzler1434
Dec 20, 2024
rudransh61
Dec 20, 2024
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Integral inequality
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ohiorizzler1434
730 posts
ohiorizzler1434
#1 h  Dec 20, 2024, 12:11 PM
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Given functions f(x) and g(x) non-negative such that g(x) is decreasing, show $\int f'(x)g(x) \geq f(x)g(x)$.
This post has been edited 2 times. Last edited by ohiorizzler1434, Dec 20, 2024, 12:15 PM
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rudransh61
21 posts
rudransh61
#2 h  Dec 20, 2024, 12:51 PM
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If $g(x)$ is decreasing so $g'(x)<0$

So , $(f(x)g(x))'=f'(x)g(x)+f(x)g'(x) \le f'(x)g(x)$

Now integrate both side as both functions are always $>0$,

$f(x)g(x) \le \int{f'(x)g(x)}$
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