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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]
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0 replies
jlacosta
Mar 2, 2025
0 replies
inequalities
Cobedangiu   8
N 15 minutes ago by xytunghoanh
problem
8 replies
3 viewing
Cobedangiu
Mar 31, 2025
xytunghoanh
15 minutes ago
April Fools Geometry
awesomeming327.   4
N 20 minutes ago by avinashp
Let $ABC$ be an acute triangle with $AB<AC$, and let $D$ be the projection from $A$ onto $BC$. Let $E$ be a point on the extension of $AD$ past $D$ such that $\angle BAC+\angle BEC=90^\circ$. Let $L$ be on the perpendicular bisector of $AE$ such that $L$ and $C$ are on the same side of $AE$ and
\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]Let the reflection of $D$ across $AB$ and $AC$ be $W$ and $Y$, respectively. Let $X\in AW$ and $Z\in AY$ such that $\angle XBE=\angle ZCE=90^\circ$. Let $EX$ and $EZ$ intersect the circumcircles of $EBD$ and $ECD$ at $J$ and $K$, respectively. Let $LB$ and $LC$ intersect $WJ$ and $YK$ at $P$ and $Q$. Let $PQ$ intersect $BC$ at $F$. Prove that $FB/FC=DB/DC$.
4 replies
awesomeming327.
Yesterday at 2:52 PM
avinashp
20 minutes ago
Olympiad Geometry problem-second time posting
kjhgyuio   4
N 27 minutes ago by ND_
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
4 replies
kjhgyuio
Today at 1:03 AM
ND_
27 minutes ago
Prove that there are no tuples $(x, y, z)$ sastifying $x^2+y^2-z^2=xyz-2$
Anabcde   0
30 minutes ago
Prove that there are no tuples $(x, y, z) \in \mathbb{Z}^3$ sastifying $x^2+y^2-z^2=xyz-2$
0 replies
Anabcde
30 minutes ago
0 replies
No more topics!
cyclic quadrilateral
AndrewTom   17
N Mar 11, 2023 by yofro
Source: BrMO 2014/2015, Round 1, question 5
Let $ABCD$ be a cyclic quadrilateral. Let $F$ be the midpoint of the arc $AB$ of its circumcircle which does not contain $C$ or $D$. Let the lines $DF$ and $AC$ meet at $P$ and the lines $CF$ and $BD$ meet at $Q$. Prove that the lines $PQ$ and $AB$ are parallel.

See here: http://www.bmoc.maths.org/home/bmo1-2015.pdf
17 replies
AndrewTom
Nov 29, 2014
yofro
Mar 11, 2023
cyclic quadrilateral
G H J
Source: BrMO 2014/2015, Round 1, question 5
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AndrewTom
12750 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ABCD$ be a cyclic quadrilateral. Let $F$ be the midpoint of the arc $AB$ of its circumcircle which does not contain $C$ or $D$. Let the lines $DF$ and $AC$ meet at $P$ and the lines $CF$ and $BD$ meet at $Q$. Prove that the lines $PQ$ and $AB$ are parallel.

See here: http://www.bmoc.maths.org/home/bmo1-2015.pdf
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TelvCohl
2312 posts
#2 • 3 Y
Y by AndrewTom, Adventure10, Mango247
My solution:

Easy to see the tangent passing through $ F $ is parallel to $ AB $ ,
so from Pascal theorem ( for $ FFDBAC $ ) we get $ PQ \parallel AB $ .

Q.E.D
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AndrewTom
12750 posts
#3 • 2 Y
Y by Adventure10, Mango247
How do we prove that the tangent passing through $F$ is parallel to $AB$?
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TelvCohl
2312 posts
#4 • 3 Y
Y by AndrewTom, Adventure10, Mango247
AndrewTom wrote:
How do we prove that the tangent passing through $F$ is parallel to $AB$?

Just notice $ \triangle FAB $ is an isosceles triangle ( $ FA=FB $ ) :)
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AndrewTom
12750 posts
#5 • 2 Y
Y by Adventure10, Mango247
Thanks, TelvCohl.

Is it obvious that if the arc lengths $AF$ and $FB$ are equal then so are the chord lengths $AF$ and $FB$ are also equal? How would we prove this?
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cobbler
2180 posts
#6 • 6 Y
Y by disneyalice710, Adventure10, Mango247, and 3 other users
No need for Pascal's theorem; remember, BMO1 problems are designed to be solved using very rudimentary tools.

Note: For brevity, I will employ the symbol < in place of $\angle$.

Refer to attached diagram.
Since arc AF=FB we have (angles subtending same arc) <ADF=<FDB=<FCA=<BCF=a (say). Furthermore letting arcAD=2m and arcBC=2n we have (angles at circumference = half angles at center) <BDC=<BAC=n and <ACD=<ABD=m. Also (angles subtending same arc) <DAC=<DFC=<DBC=x (say). Therefore, since ABCD is cyclic we have <DAB+<BCD=180, i.e. m+n+2a+x=180. We want to prove that <QPC=<BAC=n and <PQD=<ABD=m.

Claim: It suffices to prove that quadrilateral PQCD is cyclic.
Proof: If PQCD is cyclic, it would follow that <QPC+<CPD+<QCD=180, and plugging the known values of <CPD and <QCD, and solving for <QPC, yields <QPC=n. Similarly for <PQD.

But a simple angle chase shows that <DPC=<DQC=a+x, and the result follows.
Attachments:
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djmathman
7936 posts
#7 • 3 Y
Y by cobbler, Adventure10, and 1 other user
All these solutions are too complicated :P

Note that $\angle PDQ=\tfrac12\widehat{FB}=\tfrac12\widehat{AF}=\angle PCQ$, whence quadrilateral $DPQC$ is cyclic. Now $PQ$ and $AB$ are both antiparallel to $CD$, so they must be parallel to each other. Done. (This is often referred to as the converse of Reim's theorem).
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AndrewTom
12750 posts
#8 • 2 Y
Y by Adventure10, Mango247
Thanks. Are there other ways of answering this question?

For example, can we prove this using vectors?
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v_Enhance
6870 posts
#9 • 3 Y
Y by AndrewTom, Adventure10, Mango247
AndrewTom wrote:
For example, can we prove this using vectors?
If you really want to, complex numbers works. For let $f=1$ and $a,b,c,d$ have their usual meanings, but noting $ab=1$. Using standard formulas gives \[ p = \frac{da+dc-ac-acd}{d-ac} \text{ and } q = \frac{c+acd-ad-cd}{ac-d} \] so \[ p-q = \frac{c-ac}{d-ac} \] which is easily verified to be pure imaginary.

However, I still think the Pascal solution is the "morally correct" solution. :)
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jayme
9775 posts
#10 • 3 Y
Y by AndrewTom, Adventure10, Mango247
Dear Mathlinkers,
all the proofs are in connection with a personnal point of view...
I like the Pascal's theorem,
and also the angle technic when it is used with art...
Sincerely
Jean-Louis
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disneyalice710
20 posts
#11 • 3 Y
Y by AndrewTom, Adventure10, Mango247
In my point of view, I think Thales theorem is enough beautiful to use, just notice that $APD$ is similar to $BQC$, then we have: $AP$/$BQ$= $AD$/$BC$. Moreover, triangle $AXD$ and $BXC$ are similar ($X$ is the intersection of $AC$ and $BD$), which implies $AD$/$BC$= $AX$/$XB$. So, $AX$/$BX$= $AP$/$BQ$ and we have $PQ$//$AB$_
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Dexenberg
143 posts
#12 • 4 Y
Y by AndrewTom, disneyalice710, Adventure10, Mango247
disneyalice710 wrote:
In my point of view, I think Thales theorem is enough beautiful to use, just notice that $APD$ is similar to $BQC$, then we have: $AP$/$BQ$= $AD$/$BC$. Moreover, triangle $AXD$ and $BXC$ are similar ($X$ is the intersection of $AC$ and $BD$), which implies $AD$/$BC$= $AX$/$XB$. So, $AX$/$BX$= $AP$/$BQ$ and we have $PQ$//$AB$_
Or $\dfrac{BQ}{QX}=\dfrac{BC}{CX}=\dfrac{DA}{DX}=\dfrac{AP}{PX}$ using that $(DP$ and $(CQ$ are bisectors and $AXD$ and $BXC$ are similar too...
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bigs
14 posts
#13 • 2 Y
Y by Adventure10, Mango247
Let G be the midpoint of arc DC. Let the lines AG and DB meet at R and the lines BG and AC meet at S. Prove that PQRS is cyclic.
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PROF65
2016 posts
#14 • 2 Y
Y by Adventure10, Mango247
Angle chase yields $\widehat{DPC}=\widehat{DAC}+\widehat{ADF},\widehat{DQC}=\widehat{DBC}+\widehat{BCF}$ then $ PQDC$ is cyclic.Let $ D',C'$ points of intersections of $ DF$ and $AB,CF$ and $AB$
$\widehat{FD'C'}=\widehat{DFA}+\widehat{FAB},\widehat{FCD}=\widehat{FCA}+\widehat{ACD}$ thus $C'D'CD$ is cyclic therefore by angle chase or Reim's theorem the result follows
WCP
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bigs
14 posts
#15 • 2 Y
Y by Adventure10, Mango247
Reim's theorem? I mean thx.
I see the angles but I don't know Reim's theory.
Maybe (use your notation) A'B'C'D' is also cyclic? And we can find some more quadrangles...
What's the reason?
This post has been edited 1 time. Last edited by bigs, Feb 7, 2016, 2:10 PM
Reason: Thx
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Davidng
69 posts
#16 • 2 Y
Y by Adventure10, Mango247
My solution is somewhat simple: We know ABCD is a cyclic, thus angle ABD=ACD (1), we also have PQCD is a cylic since angle PDQ=PCQ, therefore angle PQD=PCD (2). From (1) and (2) angle ABD=PQD, therefore AB//PQ
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ninhduccuong
12 posts
#17 • 2 Y
Y by Adventure10, Mango247
because $F$ is midpoint of arc $AB$ so arc $FA$ = arc $FB$
hence $\widehat{ADF}$=$\widehat{FDB}$=$\widehat{ACF}$=$\widehat{FCB}$
so $PQCD$ is cyclic
so $\widehat{QPC}$=$\widehat{QDC}$ (1)
$ABCD$ is cyclic so $\widehat{BAC}$=$\widehat{BDC}$ (2)
from (1)(2) hence $\widehat{BAC}$=$\widehat{QPC}$
$PQ // AB$
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yofro
3146 posts
#18
Y by
Note $\frac{\widehat{CD}-\widehat{AF}}{2}=\angle CPD = \frac{\widehat{CD}-\widehat{BF}}{2} = \angle CQD\implies (CQPD)$, so done by Reim's.
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