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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
35 minutes ago
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following events:
[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
[*]April 8th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS State Discussion
April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
[*]April 10th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MathILy and MathILy-Er Math Jam: Multibackwards Numbers
[*]April 16th (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
35 minutes ago
0 replies
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H
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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Prove that there are no tuples $(x, y, z)$ sastifying $x^2+y^2-z^2=xyz-2$
Anabcde   2
N 4 minutes ago by IMUKAT
Prove that there are no tuples $(x, y, z) \in \mathbb{Z}^3$ sastifying $x^2+y^2-z^2=xyz-2$
2 replies
Anabcde
3 hours ago
IMUKAT
4 minutes ago
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Old problem :(
Drakkur   0
14 minutes ago
Let a, b, c be positive real numbers. Prove that
$$\dfrac{1}{\sqrt{a^2+bc}}+\dfrac{1}{\sqrt{b^2+ca}}+\dfrac{1}{\sqrt{c^2+ab}}\le \sqrt{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)$$
0 replies
Drakkur
14 minutes ago
0 replies
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Functional equations
hanzo.ei   6
N 25 minutes ago by hanzo.ei
Source: Greekldiot
Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$
6 replies
+2 w
hanzo.ei
Mar 29, 2025
hanzo.ei
25 minutes ago
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Polynomial
EtacticToe   2
N an hour ago by yuribogomolov
Source: Own
Let $f(x)$ be a monic polynomial with integer coefficient. And suppose there exist 4 distinct integer $a,b,c,d$ such that $f(a)=…=f(d)=5$.

Find all $k$ such that $f(k)=8$
2 replies
EtacticToe
Dec 14, 2024
yuribogomolov
an hour ago
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No more topics!
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2020 IGO Elementary P3
Gaussian_cyber   5
N Dec 18, 2022 by parmenides51
Source: 7th Iranian Geometry Olympiad (Elementary) P3
According to the figure, three equilateral triangles with side lengths $a,b,c$ have one
common vertex and do not have any other common point. The lengths $x, y$, and $z$ are defined as
in the figure. Prove that $3(x+y+z)>2(a+b+c)$.
Proposed by Mahdi Etesamifard
5 replies
Gaussian_cyber
Nov 4, 2020
parmenides51
Dec 18, 2022
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2020 IGO Elementary P3
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Reveal topic tags
geometry
IGO
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Source: 7th Iranian Geometry Olympiad (Elementary) P3
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Gaussian_cyber
162 posts
Gaussian_cyber
#1 h  Nov 4, 2020, 10:51 AM • 2 Y
Y by jhu08, HWenslawski
According to the figure, three equilateral triangles with side lengths $a,b,c$ have one
common vertex and do not have any other common point. The lengths $x, y$, and $z$ are defined as
in the figure. Prove that $3(x+y+z)>2(a+b+c)$.
Proposed by Mahdi Etesamifard
Attachments:
This post has been edited 1 time. Last edited by Gaussian_cyber, Nov 4, 2020, 6:29 PM
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dungnguyentien
105 posts
dungnguyentien
#2 h  Nov 4, 2020, 7:24 PM • 3 Y
Y by buratinogigle, wasikgcrushedbi, jhu08
One solution for this problem.
Attachments:
This post has been edited 1 time. Last edited by dungnguyentien, Nov 4, 2020, 7:25 PM
Reason: .
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buratinogigle
2321 posts
buratinogigle
#3 h  Nov 5, 2020, 4:56 PM • 2 Y
Y by dungnguyentien, jhu08
I like this problem the best in 7th IGO. See the picture for an easy generalization with cut and jigsaw proof.
Attachments:
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dungnguyentien
105 posts
dungnguyentien
#4 h  Nov 5, 2020, 6:36 PM • 2 Y
Y by buratinogigle, jhu08
buratinogigle wrote:
I like this problem the best in 7th IGO. See the picture for an easy generalization with cut and jigsaw proof.

Dear my teacher, I absolutely agree with you.
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wasikgcrushedbi
52 posts
wasikgcrushedbi
#5 h  Sep 15, 2021, 5:13 AM • 1 Y
Y by jhu08
dungnguyentien wrote:
One solution for this problem.

It would be $x+y+z=y+(z+x)>y+t>2a$ instead of $2c$ right?
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parmenides51
30629 posts
parmenides51
#6 h  Dec 18, 2022, 8:25 AM
Y by
According to the figure, three equilateral triangles with side lengths $a,b,c$ have one common vertex and do not have any other common point. The lengths $x, y$, and $z$ are defined as in the figure. Prove that $3(x+y+z)>2(a+b+c)$.
Proposed by Mahdi Etesamifard
https://cdn.artofproblemsolving.com/attachments/6/8/e3179c696b8b01c6b779dc536d72d1eaa2d556.png

Click to reveal hidden text
This post has been edited 1 time. Last edited by parmenides51, Dec 18, 2022, 9:27 AM
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