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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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inequalities hard
Cobedangiu   2
N 17 minutes ago by IceyCold
problem
2 replies
1 viewing
Cobedangiu
Mar 31, 2025
IceyCold
17 minutes ago
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Find the probability
ali3985   0
26 minutes ago
Let $A$ be a set of Natural numbers from $1$ to $N$.
Now choose $k$ ($k \geq 3$) distinct elements from this set.

What is the probability of these numbers to be an increasing geometric progression ?
0 replies
ali3985
26 minutes ago
0 replies
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IMOC 2017 G2 , (ABC) <= (DEF) . perpendiculars related
parmenides51   7
N 31 minutes ago by AshAuktober
Source: https://artofproblemsolving.com/community/c6h1740077p11309077
Given two acute triangles $\vartriangle ABC, \vartriangle DEF$. If $AB \ge DE, BC \ge EF$ and $CA \ge FD$, show that the area of $\vartriangle ABC$ is not less than the area of $\vartriangle DEF$
7 replies
parmenides51
Mar 20, 2020
AshAuktober
31 minutes ago
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The Sums of Elements in Subsets
bobaboby1   3
N 31 minutes ago by bobaboby1
Given a finite set \( X = \{x_1, x_2, \ldots, x_n\} \), and the pairwise comparison of the sums of elements of all its subsets (with the empty set defined as having a sum of 0), which amounts to \( \binom{2}{2^n} \) inequalities, these given comparisons satisfy the following three constraints:

1. The sum of elements of any non-empty subset is greater than 0.
2. For any two subsets, removing or adding the same elements does not change their comparison of the sums of elements.
3. For any two disjoint subsets \( A \) and \( B \), if the sums of elements of \( A \) and \( B \) are greater than those of subsets \( C \) and \( D \) respectively, then the sum of elements of the union \( A \cup B \) is greater than that of \( C \cup D \).

The question is: Does there necessarily exist a positive solution \( (x_1, x_2, \ldots, x_n) \) that satisfies all these conditions?
3 replies
bobaboby1
Mar 12, 2025
bobaboby1
31 minutes ago
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Inequality
JK1603JK   1
N 5 hours ago by lbh_qys
Let $a,b,c\ge 0: a+b+c=3$ then prove \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2}\cdot\frac{abc}{ab+bc+ca}\ge \frac{5}{3}.$$
1 reply
JK1603JK
6 hours ago
lbh_qys
5 hours ago
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Inequality
JK1603JK   1
N Today at 8:38 AM by lbh_qys
Prove that 9ab\left(a-b+c\right)+9bc\left(b-c+a\right)+9ca\left(c-a+b\right)\ge \left(a+b+c\right)^{3},\ \ a\ge 0\ge b\ge c: a+b+c\le 0.
1 reply
JK1603JK
Today at 8:27 AM
lbh_qys
Today at 8:38 AM
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Solve the equetion
yt12   4
N Today at 8:00 AM by lgx57
Solve the equetion:$\sin 2x+\tan x=2$
4 replies
yt12
Mar 31, 2025
lgx57
Today at 8:00 AM
J
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Inequalities
sqing   2
N Today at 7:41 AM by sqing
Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=11.$ Prove that
$$a+ab+abc\leq\frac{49}{6}$$Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=10.$ Prove that
$$a+ab+abc\leq\frac{169}{24}$$Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=14.$ Prove that
$$a+ab+abc\leq\frac{63+5\sqrt 5}{6}$$Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=32.$ Prove that
$$a+ab+abc\leq48+\frac{64\sqrt{2}}{3}$$
2 replies
sqing
Yesterday at 2:59 PM
sqing
Today at 7:41 AM
J
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geometry incentre config
Tony_stark0094   1
N Today at 7:38 AM 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
Today at 7:38 AM
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geometry
Tony_stark0094   1
N Today at 7:36 AM 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
Today at 7:04 AM
Tony_stark0094
Today at 7:36 AM
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one very nice!
MihaiT   1
N Today at 5:28 AM 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
Today at 5:28 AM
J
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School Math Problem
math_cool123   1
N Today at 5:12 AM by MathPerson12321
Find all ordered pairs of nonzero integers $(a, b)$ that satisfy $$(a^2+b)(a+b^2)=(a-b)^3.$$
1 reply
math_cool123
Today at 5:03 AM
MathPerson12321
Today at 5:12 AM
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Geo Mock #4
Bluesoul   1
N Today at 4:44 AM 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
Today at 4:44 AM
J
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An inequality
JK1603JK   2
N Today at 3:24 AM 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
Today at 3:24 AM
J
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J
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3 out of n points exist always which an angle < = \pi / n
parmenides51   2
N Jul 27, 2018 by NikoIsLife
Source: Brazilian Mathematical Olympiad 1985 P2
Given $n$ points in the plane, show that we can always find three which give an angle $\le \pi / n$.
2 replies
parmenides51
Jul 27, 2018
NikoIsLife
Jul 27, 2018
J
3 out of n points exist always which an angle < = \pi / n
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Reveal topic tags
geometry
combinatorial geometry
angles
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Source: Brazilian Mathematical Olympiad 1985 P2
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parmenides51
30629 posts
parmenides51
#1 h  Jul 27, 2018, 6:39 AM • 2 Y
Y by Adventure10, Mango247
Given $n$ points in the plane, show that we can always find three which give an angle $\le \pi / n$.
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TuZo
19351 posts
TuZo
#2 h  Jul 27, 2018, 8:12 AM • 1 Y
Y by Adventure10
If the points determine a convex n-gone, we have:
We assume the opposite: all the angles there are $>\frac{\pi }{n}$. Draw the diagonals. In every vertex we have $(n-2)$ interior disjoint angle. So the in on of the vertex,the sum of angles is $>(n-2)\frac{\pi }{n}$, so the sum of all interior angles of the polygon is $>n(n-2)\frac{\pi }{n}=(n-2)\pi $, and this is a contradiction.
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NikoIsLife
9657 posts
NikoIsLife
#3 h  Jul 27, 2018, 9:22 AM • 3 Y
Y by leibnitz, Adventure10, Mango247
Kind of hard to explain, but I'll try.
This post has been edited 6 times. Last edited by NikoIsLife, Jul 27, 2018, 9:33 AM
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