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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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Inclusion Exclusion Principle
chandru1   1
N 13 minutes ago by onofre.campos
How does one prove the identity $$1=\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}2^{n-k}$$This easy via the binomial theorem for the quantity is just $(2-1)^{k}$, but how do we arrive at this using the I-E-P?
1 reply
chandru1
Dec 4, 2020
onofre.campos
13 minutes ago
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inequalities
Cobedangiu   3
N 39 minutes ago by Nguyenhuyen_AG
Source: own
$a,b>0$ and $a+b=1$. Find min P:
$P=\sqrt{\frac{1-a}{1+7a}}+\sqrt{\frac{1-b}{1+7b}}$
3 replies
Cobedangiu
Yesterday at 6:10 PM
Nguyenhuyen_AG
39 minutes ago
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Olympiad Geometry problem-second time posting
kjhgyuio   0
39 minutes ago
Source: smo problem
In trapezium ABCD,AD is parallel to BC and points E and F are midpoints of AB and DC respectively. If
Area of AEFD/Area of EBCF =√3 + 1/3-√3 and the area of triangle ABD is √3 .find the area of trapezium ABCD
0 replies
kjhgyuio
39 minutes ago
0 replies
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Inspired by old results
sqing   7
N 3 hours ago by SunnyEvan
Source: Own
Let $ a,b,c> 0 $ and $ abc=1 $. Prove that
$$\frac1{a^2+a+k}+\frac1{b^2+b+k}+\frac1{c^2+c+k}\geq \frac{3}{k+2}$$Where $ 0<k \leq 1.$
7 replies
sqing
Monday at 1:42 PM
SunnyEvan
3 hours ago
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No more topics!
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Supplementary Angles ???
utkarshgupta   2
N Jul 5, 2014 by jlammy
Source: BrMO 2005 R2
Let $ABC$ be a triangle with $AC > AB$. The point $X$ lies on the side $BA$ extended through $A$, and the point $Y$ lies on the side $CA$ in such a way that $BX = CA$ and $CY = BA$. The line $XY$ meets the perpendicular bisector of side $BC$ at $P$. Show that
$\angle BPC + \angle BAC = 180$.
2 replies
utkarshgupta
Jul 5, 2014
jlammy
Jul 5, 2014
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Supplementary Angles ???
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Reveal topic tags
geometry
circumcircle
rhombus
geometric transformation
reflection
perpendicular bisector
geometry proposed
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G H BBookmark kLocked kLocked NReply
Source: BrMO 2005 R2
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utkarshgupta
2280 posts
utkarshgupta
#1 h  Jul 5, 2014, 2:34 PM • 2 Y
Y by Adventure10, Mango247
Let $ABC$ be a triangle with $AC > AB$. The point $X$ lies on the side $BA$ extended through $A$, and the point $Y$ lies on the side $CA$ in such a way that $BX = CA$ and $CY = BA$. The line $XY$ meets the perpendicular bisector of side $BC$ at $P$. Show that
$\angle BPC + \angle BAC = 180$.
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Luis González
4145 posts
Luis González
#2 h  Jul 5, 2014, 5:33 PM • 1 Y
Y by Adventure10
Let $M$ be the midpoint of the arc $BC$ of the circumcircle $\odot(ABC),$ not containing $A.$ $AM$ and $XY$ cut $BC$ at $D$ and $Z,$ respectively. By Menelaus' theorem for $\triangle ABC$ cut by $\overline{XYZ},$ we get

$\frac{BZ}{ZC}=\frac{YA}{CY} \cdot \frac{XB}{XA}=\frac{AC-AB}{CY} \cdot \frac{XB}{AC-AB}=\frac{AC}{AB}=\frac{CD}{DB},$

which means that $D$ and $Z$ are symmetric WRT the midpoint of $\overline{BC},$ i.e. $\triangle MDZ$ is M-isosceles. Since $AX=AY=AC-AB,$ then $XYZ \parallel AM$ $\Longrightarrow$ $PZMD$ is a rhombus, i.e. $P$ is reflection of $M$ on $BC$ $\Longrightarrow$ $\widehat{BPC}=\widehat{BMC}=180^{\circ}-\widehat{BAC}.$
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jlammy
1099 posts
jlammy
#3 h  Jul 5, 2014, 9:03 PM • 2 Y
Y by Adventure10, Mango247
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=575260
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=327416
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=482947
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=470946
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=254072&p=1389325
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=72694&p=421058

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