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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.
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[*]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
Cobedangiu   5
N 20 minutes ago by Cobedangiu
Source: own
$a,b>0$ and $a+b=1$. Find min P:
$P=\sqrt{\frac{1-a}{1+7a}}+\sqrt{\frac{1-b}{1+7b}}$
5 replies
Cobedangiu
Yesterday at 6:10 PM
Cobedangiu
20 minutes ago
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Proving ∠BHF=90
BarisKoyuncu   17
N 34 minutes ago by jordiejoh
Source: IGO 2021 Advanced P1
Acute-angled triangle $ABC$ with circumcircle $\omega$ is given. Let $D$ be the midpoint of $AC$, $E$ be the foot of altitude from $A$ to $BC$, and $F$ be the intersection point of $AB$ and $DE$. Point $H$ lies on the arc $BC$ of $\omega$ (the one that does not contain $A$) such that $\angle BHE=\angle ABC$. Prove that $\angle BHF=90^\circ$.
17 replies
BarisKoyuncu
Dec 30, 2021
jordiejoh
34 minutes ago
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Trapezium inscribed in a circle
shivangjindal   27
N 40 minutes ago by andrewthenerd
Source: Balkan Mathematics Olympiad 2014 - Problem-3
Let $ABCD$ be a trapezium inscribed in a circle $\Gamma$ with diameter $AB$. Let $E$ be the intersection point of the diagonals $AC$ and $BD$ . The circle with center $B$ and radius $BE$ meets $\Gamma$ at the points $K$ and $L$ (where $K$ is on the same side of $AB$ as $C$). The line perpendicular to $BD$ at $E$ intersects $CD$ at $M$. Prove that $KM$ is perpendicular to $DL$.

Greece - Silouanos Brazitikos
27 replies
shivangjindal
May 4, 2014
andrewthenerd
40 minutes ago
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Tangent Spheres and Tangents to Spheres
Math-Problem-Solving   0
an hour ago
Source: 2002 British Mathematical Olympiad Round 2
Prove this.
0 replies
Math-Problem-Solving
an hour ago
0 replies
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Sum of any 3 sides of a quad div by 4th=>some 2 are equal
YESMAths   7
N Apr 20, 2022 by sanyalarnab
Source: RMO (Mumbai Region) 2015 P8
The length of each side of a convex quadrilateral $ABCD$ is a positive integer. If the sum of the lengths of any three sides is divisible by the length of the remaining side then prove that some two sides of the quadrilateral have the same length.
7 replies
YESMAths
Dec 6, 2015
sanyalarnab
Apr 20, 2022
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Sum of any 3 sides of a quad div by 4th=>some 2 are equal
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Reveal topic tags
inequalities
number theory
geometry
geometry proposed
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G H BBookmark kLocked kLocked NReply
Source: RMO (Mumbai Region) 2015 P8
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YESMAths
829 posts
YESMAths
#1 h  Dec 6, 2015, 1:54 PM • 2 Y
Y by Adventure10, Mango247
The length of each side of a convex quadrilateral $ABCD$ is a positive integer. If the sum of the lengths of any three sides is divisible by the length of the remaining side then prove that some two sides of the quadrilateral have the same length.
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libra8
240 posts
libra8
#2 h  Dec 6, 2015, 2:31 PM • 2 Y
Y by Adventure10, Mango247
assume a< b< c< d. then as d divides a+b+c, and a+b+c < 3d, and a+b+c > d (triangle inequality), a+b+c = 2d. perform similarly with c, b, a.
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AnonymousBunny
339 posts
AnonymousBunny
#3 h  Dec 6, 2015, 2:31 PM • 2 Y
Y by Adventure10, Mango247
Does this work?

Click to reveal hidden text

edit: sniped by few seconds :P
This post has been edited 2 times. Last edited by AnonymousBunny, Dec 6, 2015, 2:33 PM
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libra8
240 posts
libra8
#4 h  Dec 6, 2015, 2:32 PM • 2 Y
Y by Adventure10, Mango247
yeah. same.
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aditya21
717 posts
aditya21
#5 h  Dec 6, 2015, 2:45 PM • 2 Y
Y by Adventure10, Mango247
yeah that's a nice and easy problem.
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trimo
62 posts
trimo
#6 h  Nov 3, 2019, 6:06 AM • 1 Y
Y by Adventure10
Same as Q. No 62 in the book "Maths Problem Book 1" by Kin Y. Li.
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guptaamitu1
656 posts
guptaamitu1
#7 h  Oct 13, 2020, 6:29 PM
Y by
AnonymousBunny wrote:
Does this work?

Click to reveal hidden text

edit: sniped by few seconds :P

This solution is wrong as there are could be some multiples of 3a which are less than a and have more larger values, for example, 20a/21 etc
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sanyalarnab
928 posts
sanyalarnab
#8 h  Apr 20, 2022, 4:02 AM
Y by
ok ig there is no complete solutions. So here it goes:
Assume the sides are $a,b,c,d$ and that they are distinct. Then we can get an ordering. WLOG we assume $$a<b<c<d$$.
Let $b+c+d=am_1$
$a+c+d=bm_2$
$b+a+d=cm_3$
$b+c+a=dm_4$
From the quadrilateral inequality we get $m_4>1$
Also $dm_4 < d+d+d$
Hence we get $1<m_4<3 \implies m_4=2$
Also $a(m_1+1)=b(m_2+1)=c(m_3+1)=3d$
$\implies m_1>m_2>m_3>2$
$\implies m_1 \ge 5,m_2 \ge 4, m_3 \ge 3$
Hence $3(a+b+c+d)=am_1+bm_2+cm_3+dm_4 \ge 5a+4b+3c+2d$
$\implies d \ge 2a+b$....(*)
Also $(d+a+b)+(a+b+c) = cm_3+dm_4 \ge 3c+2d$
$\implies d \le 2a+2b-2c < 2a+2c-2c=2a < 2a+b$ which contradicts (*).
Hence our assumption for $a,b,c,d$ to be distinct was wrong, completing the proof.
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