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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
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]
Our full course list for upcoming classes is below:
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0 replies
jlacosta
Mar 2, 2025
0 replies
inequalities
Cobedangiu   1
N 21 minutes ago by Natrium
Source: own
$a,b>0$ and $a+b=1$. Find min P:
$P=\sqrt{\frac{1-a}{1+7a}}+\sqrt{\frac{1-b}{1+7b}}$
1 reply
Cobedangiu
2 hours ago
Natrium
21 minutes ago
Is this FE solvable?
Mathdreams   0
an hour ago
Find all $f:\mathbb{R} \rightarrow \mathbb{R}$ such that \[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]for all reals $x$ and $y$.
0 replies
+1 w
Mathdreams
an hour ago
0 replies
OFM2021 Senior P1
medhimdi   0
an hour ago
Let $a_1, a_2, a_3, \dots$ and $b_1, b_2, b_3, \dots$ be two sequences of integers such that $a_{n+2}=a_{n+1}+a_n$ and $b_{n+2}=b_{n+1}+b_n$ for all $n\geq1$. Suppose that $a_n$ divides $b_n$ for an infinity of integers $n\geq1$. Prove that there exist an integer $c$ such that $b_n=ca_n$ for all $n\geq1$
0 replies
medhimdi
an hour ago
0 replies
Hard NT problem
tiendat004   2
N an hour ago by avinashp
Given two odd positive integers $a,b$ are coprime. Consider the sequence $(x_n)$ given by $x_0=2,x_1=a,x_{n+2}=ax_{n+1}+bx_n,$ $\forall n\geq 0$. Suppose that there exist positive integers $m,n,p$ such that $mnp$ is even and $\dfrac{x_m}{x_nx_p}$ is an integer. Prove that the numerator in its simplest form of $\dfrac{m}{np}$ is an odd integer greater than $1$.
2 replies
tiendat004
Aug 15, 2024
avinashp
an hour ago
No more topics!
Concurrency and collinearity
bozzio   15
N Aug 4, 2013 by XmL
Source: BMO tst 2012- italy
ABC is a triangle and $\Gamma$ the circumcircle of ABC. $A_0$ , $B_0$ , $C_0$ are the midpoints of the sides BC, AC and AB. D is the feet of the altitude from A and $D_0$ is the projection of $A_0$ on $B_0 C_0$ . G is the centroid of ABC. $\omega$ is the circumference which passes through $B_0$ $C_0$ and is tangent to $\Gamma$ in P (different from A). Show that:
1) $B_0 C_0$ and the tangents to $\Gamma$ in A and P are concurrent;
2) P,D,G,$D_0$ are collinear.
15 replies
bozzio
Jan 3, 2013
XmL
Aug 4, 2013
Concurrency and collinearity
G H J
Source: BMO tst 2012- italy
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bozzio
59 posts
#1 • 2 Y
Y by Adventure10, Mango247
ABC is a triangle and $\Gamma$ the circumcircle of ABC. $A_0$ , $B_0$ , $C_0$ are the midpoints of the sides BC, AC and AB. D is the feet of the altitude from A and $D_0$ is the projection of $A_0$ on $B_0 C_0$ . G is the centroid of ABC. $\omega$ is the circumference which passes through $B_0$ $C_0$ and is tangent to $\Gamma$ in P (different from A). Show that:
1) $B_0 C_0$ and the tangents to $\Gamma$ in A and P are concurrent;
2) P,D,G,$D_0$ are collinear.
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bozzio
59 posts
#2 • 2 Y
Y by Adventure10, Mango247
Is it possible to prove the first point by inversion? I thought that thoose 3 lines are collinear iff the point of intersection of the tangents (that I call N) is on B_0C_0. The tangents are the polars of P and A with respects to $\Gamma$ , so if N is on B_0C_0 than the pole of B_0C_0 belongs to the line AP. How to prove that the pole of B_0C_0 belongs to ABC? (D_0 is the inverse of the pole with respect to $\Gamma$)
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Wolstenholme
543 posts
#3 • 2 Y
Y by Adventure10, Mango247
I haven't solved the problem, but I noticed that if we extend BB_0 and CC_0 to hit the circle at X, Y respectively, then XY also passes through the intersection of the tangents to the circle in A and P.
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Wolstenholme
543 posts
#4 • 2 Y
Y by Adventure10, Mango247
Also you can easily prove that D, G, D_0 are collinear by a simple coord bash but I dont know how to handle P.
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Burii
137 posts
#5 • 2 Y
Y by Adventure10, Mango247
hints:
If $F$ is image of $D$ under the homothety with center $G$ and scale $-2$, that collinearity of points $P$, $D$ and $G$ is equivalent condition: $\angle CFP=\angle CFD$, which is equivalent this:$ \angle CAP=\angle BAM$, where $M$ is foot $F$ on line $BC$. To prove this use inwersion with center at $A$ and any radius.
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Wolstenholme
543 posts
#6 • 2 Y
Y by Adventure10, Mango247
I've seen this problem in a problem set entitled "pascal and desargue" so is it possible to use these theorems?
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Burii
137 posts
#7 • 2 Y
Y by Adventure10, Mango247
My solution does not involve these theorems;)
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Wolstenholme
543 posts
#8 • 1 Y
Y by Adventure10
When you say homothety with scale -2 does that mean a dilation with scale 2 or does it mean reflect the point and extend it by a scale of 2?
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Burii
137 posts
#9 • 1 Y
Y by Adventure10
Ohh sorry, of course dilation;)
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Wolstenholme
543 posts
#10 • 1 Y
Y by Adventure10
Ok cool, cause my picture was looking WIERD!
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Wolstenholme
543 posts
#11 • 1 Y
Y by Adventure10
Wouldn't it be just the midpoint of GD?
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Burii
137 posts
#12 • 2 Y
Y by Adventure10, Mango247
I will post my solution maybe tomorrow;)
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Wolstenholme
543 posts
#13 • 1 Y
Y by Adventure10
Does anyone have a solution to this problem?
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Wolstenholme
543 posts
#14 • 2 Y
Y by Adventure10, Mango247
Is there anyway we can prove b) with a)?
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XmL
552 posts
#15 • 2 Y
Y by Adventure10, Mango247
Please delete this post. :oops:
This post has been edited 2 times. Last edited by XmL, Aug 4, 2013, 6:26 AM
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XmL
552 posts
#16 • 2 Y
Y by Adventure10, Mango247
a) It's easy to prove that circles $(AC_0B_0)$ and $\Gamma$ are tangent. Thus $C_0B_0$, tangents to $(ABC)$ at $A,P$ are basically that pairwise radical axex of the three circles $(AC_0B_0)$, $\Gamma$, $\omega$, which we know are supposed to concur.



b) Since $X$ is on the perpendicular bisector of $AD$, we have $AX=DX=XP$, which means that the circumcenter of $ADP$ is $X$ and we have $\angle APD=\angle AXD_0$(*)
Let $O$ be the circumcenter of $ABC$. Since $\angle XD_0O=\angle XAO=\angle XPD_0=90$, thus $A,X,P,O,D_0$ are concyclic, which means that $\angle APD_0=\angle AXD_0=\angle APD$(*), hence $P,D,D_0$are collinear.
Since $\frac {AG}{GA_0}=2=\frac {AD}{D_0A_0}$, thus $D,G,D_0$ are collinear, hence $P,D,G,D_0$ are collinear

some interesting stuff I found about this diagram: Let $A'$ be the point on $\Gamma$ such that $AA'\parallel BC$, thus $A'$ is in $PD$. This is very easy to prove.
I will post more tomorrow,it's time to sleep :(
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