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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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Interesting inequalities
sqing   0
a minute ago
Source: Own
Let $ a,b,c> 0 $ and $ a+b+c=3 $. Prove that
$$ \frac{a^2+1}{a^2+b+c-\frac{1}{2}}+\frac{b^2+1}{b^2+c+a-\frac{1}{2}}+\frac{c^2+1}{c^2+a+b-\frac{1}{2}} \leq \frac{12}{5}$$Equality holds when $ (a,b,c)=(0,\frac{3}{2},\frac{3}{2}) $ or $ (a,b,c)=(1,1,1) .$
0 replies
1 viewing
sqing
a minute ago
0 replies
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euler line and midpoint tangents
Thelink_20   3
N 3 minutes ago by hectorleo123
Source: My problem
Let the medians of a triangle $\Delta ABC$ intersect its circumcircle $\Gamma$ at $N_A, N_B, N_C$. The tangets to $\Gamma$ from $N_A,N_B,N_C$ determine a triangle $\Delta X_AX_BX_C$, where $X_A$ is relative to $A$ and so on. Prove that lines $AX_A,BX_B,CX_C$ are concurrent at a point $P$ and that $P$ belongs to the euler line of $\Delta ABC$.
Remark:Click to reveal hidden text

3 replies
Thelink_20
Jan 24, 2025
hectorleo123
3 minutes ago
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inequalities
Cobedangiu   4
N an hour ago by sqing
Source: own
$a,b>0$ and $a+b=1$. Find min P:
$P=\sqrt{\frac{1-a}{1+7a}}+\sqrt{\frac{1-b}{1+7b}}$
4 replies
Cobedangiu
Yesterday at 6:10 PM
sqing
an hour ago
J
H
Interesting inequalities
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b,c> 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) .$
1 reply
1 viewing
sqing
an hour ago
sqing
an hour ago
J
No more topics!
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Concurrent perpendiculars in a rectangle
Steff9   3
N Jun 3, 2021 by JustKeepRunning
Source: JBMO Shortlist 2013, G6
Let $P$ and $Q$ be the midpoints of the sides $BC$ and $CD$, respectively in a rectangle $ABCD$. Let $K$ and $M$ be the intersections of the line $PD$ with the lines $QB$ and $QA$, respectively, and let $N$ be the intersection of the lines $PA$ and $QB$. Let $X$, $Y$ and $Z$ be the midpoints of the segments $AN$, $KN$ and $AM$, respectively. Let $\ell_1$ be the line passing through $X$ and perpendicular to $MK$, $\ell_2$ be the line passing through $Y$ and perpendicular to $AM$ and $\ell_3$ the line passing through $Z$ and perpendicular to $KN$. Prove that the lines $\ell_1$, $\ell_2$ and $\ell_3$ are concurrent.
3 replies
Steff9
Jun 11, 2017
JustKeepRunning
Jun 3, 2021
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Concurrent perpendiculars in a rectangle
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Reveal topic tags
geometry
rectangle
perpendicular lines
concurrency
midpoints
L
G H BBookmark kLocked kLocked NReply
Source: JBMO Shortlist 2013, G6
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Steff9
58 posts
Steff9
#1 h  Jun 11, 2017, 10:38 AM • 3 Y
Y by celilcelil, Adventure10, Mango247
Let $P$ and $Q$ be the midpoints of the sides $BC$ and $CD$, respectively in a rectangle $ABCD$. Let $K$ and $M$ be the intersections of the line $PD$ with the lines $QB$ and $QA$, respectively, and let $N$ be the intersection of the lines $PA$ and $QB$. Let $X$, $Y$ and $Z$ be the midpoints of the segments $AN$, $KN$ and $AM$, respectively. Let $\ell_1$ be the line passing through $X$ and perpendicular to $MK$, $\ell_2$ be the line passing through $Y$ and perpendicular to $AM$ and $\ell_3$ the line passing through $Z$ and perpendicular to $KN$. Prove that the lines $\ell_1$, $\ell_2$ and $\ell_3$ are concurrent.
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celilcelil
164 posts
celilcelil
#2 h  Apr 21, 2021, 8:04 PM
Y by
We need to prove that XM² - XK² + KZ² - ZQ² + QY² - YM² = 0.
In triangle AMN MX is median. Then we can find value of XM² depending on the values of AM, AN, MN. Similarly we can find values of XK, KZ, YM from triangles AKN, AKM, MKN, respectively.
After rewrite it in the equation we find that we need to prove : AM² - KN² = 4( QZ² - QY²).
Let AZ = ZM = a, QM = b, NY = NK = C, QK = d. Rewrite it in the equation we find that
b(b + 2a) = d(d + 2c), which means AMKN is cyclic.
Then we need to prove that AMKN is cyclic.
Note that PA = PD.
<DAQ = <CBQ.
<PAD = <PDA = 90° - <PDC = <DPC. Then,
<BKP = <DPC - <CBQ = <PAD - <DAQ = <PAQ.
We are done!!!


I think it was very nice problem for JBMO.
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Steff9
58 posts
Steff9
#3 h  Apr 21, 2021, 10:17 PM
Y by
See Property 10.4.3 from the book A Beautiful Journey Through Olympiad Geometry attached below.
Attachments:
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JustKeepRunning
2958 posts
JustKeepRunning
#4 h  Jun 3, 2021, 9:10 PM
Y by
The problem reduces to the following claim

Claim: $ANKM$ is cyclic

Proof: Angle chase by letting $\angle ADP=\theta$ and $\angle DAQ=\alpha,$ and you get that $\angle ANQ=\angle QMK=180^{\circ}-\theta=\alpha$.

Once this is done, we know that the lines are concurrent, and in fact, by lemma above, the point of concurrency is even know. It is the reflection of the circumcenter of $ANKM$ over the intersection of the diagonals of its varigon gram.
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