G
Topic
First Poster
Last Poster
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
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]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Sunday, Mar 2 - Jun 22
Friday, Mar 28 - Jul 18
Sunday, Apr 13 - Aug 10
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Tuesday, Mar 25 - Jul 8
Sunday, Apr 13 - Aug 10
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21


Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, Mar 23 - Jul 20
Monday, Apr 7 - Jul 28
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Sunday, Mar 16 - Jun 8
Wednesday, Apr 16 - Jul 2
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Monday, Mar 17 - Jun 9
Thursday, Apr 17 - Jul 3
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Sunday, Mar 2 - Jun 22
Wednesday, Apr 16 - Jul 30
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Tuesday, Mar 4 - Aug 12
Sunday, Mar 23 - Sep 21
Wednesday, Apr 23 - Oct 1
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Mar 16 - Sep 14
Tuesday, Mar 25 - Sep 2
Monday, Apr 21 - Oct 13
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Sunday, Mar 23 - Aug 3
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Friday, Apr 11 - Jun 27
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Sunday, Mar 16 - Aug 24
Wednesday, Apr 9 - Sep 3
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Wednesday, Mar 5 - May 21
Tuesday, Jun 10 - Aug 26

Calculus
Sunday, Mar 30 - Oct 5
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Sunday, Mar 23 - Jun 15
Wednesday, Apr 16 - Jul 2
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Friday, Apr 11 - Jun 27
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Tuesday, Mar 4 - May 20
Monday, Mar 31 - Jun 23
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Monday, Mar 24 - Jun 16
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Sunday, Mar 30 - Jun 22
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Tuesday, Mar 25 - Sep 2
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Sat & Sun, Apr 26 - Apr 27 (4:00 - 7:00 pm ET/1:00 - 4:00pm PT)
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
Mar 2, 2025
0 replies
Interesting inequalities
sqing   6
N 4 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 0 $ and $ a+b+c=3 $. Prove that
$$    \frac{a^2}{a^2+b+c+ \frac{3}{2}}+\frac{b^2}{b^2+c+a+\frac{3}{2}}+\frac{c^2}{c^2+a+b+\frac{3}{2}} \leq \frac{6}{7}$$Equality holds when $ (a,b,c)=(0,\frac{3}{2},\frac{3}{2}) $ or $ (a,b,c)=(0,0,3) .$
6 replies
1 viewing
sqing
Today at 3:12 AM
sqing
4 minutes ago
European Mathematical Cup 2016 senior division problem 2
steppewolf   6
N 16 minutes ago by SimplisticFormulas
For two positive integers $a$ and $b$, Ivica and Marica play the following game: Given two piles of $a$
and $b$ cookies, on each turn a player takes $2n$ cookies from one of the piles, of which he eats $n$ and puts $n$ of
them on the other pile. Number $n$ is arbitrary in every move. Players take turns alternatively, with Ivica going
first. The player who cannot make a move, loses. Assuming both players play perfectly, determine all pairs of
numbers $(a, b)$ for which Marica has a winning strategy.

Proposed by Petar Orlić
6 replies
steppewolf
Dec 31, 2016
SimplisticFormulas
16 minutes ago
inequalities hard
Cobedangiu   1
N 29 minutes ago by Cobedangiu
problem
1 reply
Cobedangiu
Monday at 11:45 AM
Cobedangiu
29 minutes ago
Something nice
KhuongTrang   25
N 32 minutes ago by KhuongTrang
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
25 replies
KhuongTrang
Nov 1, 2023
KhuongTrang
32 minutes ago
geometry incentre config
Tony_stark0094   1
N 2 hours ago by Tony_stark0094
In a triangle $\Delta ABC$ $I$ is the incentre and point $F$ is defined such that $F \in AC$ and $F \in \odot BIC$
prove that $AI$ is the perpendicular bisector of $BF$
1 reply
Tony_stark0094
Yesterday at 4:09 PM
Tony_stark0094
2 hours ago
geometry
Tony_stark0094   1
N 2 hours ago by Tony_stark0094
Consider $\Delta ABC$ let $\omega_1$ and $\omega_2$ be the circles passing through $A,B$ and $A,C$ respectively such that $BC$ is tangent to $\omega_1$ and $\omega_2$ define $R$ to be a point such that it lies on both the circles $\omega_1$ and $\omega_2$ prove that $HR$ and $AR$ are perpendicular.
1 reply
Tony_stark0094
2 hours ago
Tony_stark0094
2 hours ago
one very nice!
MihaiT   1
N 4 hours ago by MihaiT
Given $m_1$ weights, each weighing $k_1$ and another $m_2$ weights with $k_2$ each. Write a algorithm that determines the ways in which a scale can be balanced with a weight $X$ on the left pan, and display the number of possible solutions. (The weights can be placed on both pans and the program starts with the numbers $m_1,k_1,m_2,k_2,X$. What will be displayed after three successive runs: 5,2,5,1,4 | 5,2,5,1,11 | 5,2,5,1,20?

One answer is possible:
a)10;5;0;
b)20;7;0;
c)20;7;1;
d)10;10;0;
e)10;7;0;
f)20;5;0,
1 reply
MihaiT
Mar 31, 2025
MihaiT
4 hours ago
Geo Mock #4
Bluesoul   1
N 4 hours ago by Sedro
Consider acute triangle $ABC$ with orthocenter $H$. Extend $AH$ to meet $BC$ at $D$. The angle bisector of $\angle{ABH}$ meets the midpoint of $AD$. If $AB=10, BH=6$, compute the area of $\triangle{ABC}$.
1 reply
Bluesoul
Yesterday at 7:03 AM
Sedro
4 hours ago
An inequality
JK1603JK   2
N 6 hours ago by lbh_qys
Let a,b,c\ge 0: a+b+c=3 then prove \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le \frac{27}{2}\cdot\frac{1}{2ab+2bc+2ca+3}.
2 replies
JK1603JK
Today at 3:11 AM
lbh_qys
6 hours ago
Geo Mock #3
Bluesoul   2
N Yesterday at 11:31 PM by mathprodigy2011
Consider square $ABCD$ with side length of $5$. The point $P$ is selected on the diagonal $AC$ such that $\angle{BPD}=135^{\circ}$. Denote the circumcenters of $\triangle{BPA}, \triangle{APD}$ as $O_1,O_2$. Find the length of $O_1O_2$
2 replies
Bluesoul
Yesterday at 7:02 AM
mathprodigy2011
Yesterday at 11:31 PM
Complex + Radical Evaluation
Saucepan_man02   3
N Yesterday at 4:45 PM by SmartHusky
Evaluate: (without calculators)
$$ (\sqrt{6 - 2 \sqrt{5}} + i \sqrt{2 \sqrt{5} + 10})^5 + (\sqrt{6 - 2 \sqrt{5}} - i \sqrt{2 \sqrt{5} + 10})^5$$
3 replies
Saucepan_man02
Mar 17, 2025
SmartHusky
Yesterday at 4:45 PM
Geo Mock #2
Bluesoul   1
N Yesterday at 4:36 PM by Sedro
Consider convex quadrilateral $ABCD$ such that $AB=6, BC=10, \angle{ABC}=90^{\circ}$. Denote the midpoints of $AD,CD$ as $M,N$ respectively, compute the area of $\triangle{BMN}$ given the area of $ABCD$ is $50$.
1 reply
Bluesoul
Yesterday at 6:59 AM
Sedro
Yesterday at 4:36 PM
Geo Mock #1
Bluesoul   1
N Yesterday at 4:30 PM by Sedro
Consider the rectangle $ABCD$ with $AB=4$. Point $E$ lies inside the rectangle such that $\triangle{ABE}$ is equilateral. Given that $C,E$ and the midpoint of $AD$ are on the same line, compute the length of $BC$.
1 reply
Bluesoul
Yesterday at 6:58 AM
Sedro
Yesterday at 4:30 PM
Inequalities (Please help me!!!)
yt12   6
N Yesterday at 4:16 PM by lamhihi1234
Let $a,b,c$ be reals with $a+b+c=1$and $ a,b,c \ge \frac{-3}{ 4}$. Prove that
$$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{ c^2+1} \le \frac{9}{ 10}$$
6 replies
yt12
Mar 4, 2023
lamhihi1234
Yesterday at 4:16 PM
<NHP=<AHQ wanted, 2 equilateral triangles exterior to a right triangle
parmenides51   2
N Mar 27, 2022 by parmenides51
Source: 2017 Romania JBMO TST4 p4
Let $ABC$ be a right triangle, with the right angle at $A$. The altitude from $A$ meets $BC$ at $H$ and $M$ is the midpoint of the hypotenuse $[BC]$. On the legs, in the exterior of the triangle, equilateral triangles $BAP$ and $ACQ$ are constructed. If $N$ is the intersection point of the lines $AM$ and $PQ$, prove that the angles $\angle NHP$ and $\angle AHQ$ are equal.

Miguel Ochoa Sanchez and Leonard Giugiuc
2 replies
parmenides51
May 16, 2020
parmenides51
Mar 27, 2022
<NHP=<AHQ wanted, 2 equilateral triangles exterior to a right triangle
G H J
G H BBookmark kLocked kLocked NReply
Source: 2017 Romania JBMO TST4 p4
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
parmenides51
30629 posts
#1
Y by
Let $ABC$ be a right triangle, with the right angle at $A$. The altitude from $A$ meets $BC$ at $H$ and $M$ is the midpoint of the hypotenuse $[BC]$. On the legs, in the exterior of the triangle, equilateral triangles $BAP$ and $ACQ$ are constructed. If $N$ is the intersection point of the lines $AM$ and $PQ$, prove that the angles $\angle NHP$ and $\angle AHQ$ are equal.

Miguel Ochoa Sanchez and Leonard Giugiuc
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ak_47
121 posts
#2
Y by
Let $AH$ meet $PQ$ at $K$
$\angle MAP = \angle MAB + 60 = B+60 \implies \angle NAP = 180-B-60=120-B$
Similarly, $\angle NAQ=120-C$

Hence, $\frac{PN}{NQ} = \frac{AP}{AQ}.\frac{sin(120-B)}{sin(120-C)}$

$\angle HAP = \angle HAB + 60 = C+60 \implies \angle KAP = 180-C-60=120-C$
Similarly, $\angle KAQ=120-B$

Hence, $\frac{PK}{KQ} = \frac{AP}{AQ}.\frac{sin(120-C)}{sin(120-B)}$

Therefore, $\frac{PN}{NQ}.\frac{PK}{KQ} = \frac{AP^2}{AQ^2}$

$\angle HAP=\angle HCQ = C+60$

$\frac {HC}{CQ}=\frac{HC}{AC}=cos C$

$\frac {AH}{AP}=\frac{AH}{AB}=cos C \implies \frac {HC}{CQ}=\frac {AH}{AP}$

Hence, $\triangle HAP \sim \triangle HCQ \implies \frac{AP}{CQ} = \frac{PH}{HQ} \implies \frac{AP}{AQ}=\frac{PH}{HQ}$

Therefore, $\frac{AP^2}{AQ^2}=\frac{PH^2}{HQ^2} \implies \frac{PN}{NQ}.\frac{PK}{KQ}=\frac{PH^2}{HQ^2}$

Therefore, $HK, HN$ are isogonal conjugates through $H$ in $\triangle PHQ$
Hence, $\angle NHP = \angle KHQ \implies \angle NHP = \angle AHQ$
This post has been edited 3 times. Last edited by ak_47, May 16, 2020, 10:26 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
parmenides51
30629 posts
#3
Y by
solved also here
Z K Y
N Quick Reply
G
H
=
a