Summer is a great time to explore cool problems to keep your skills sharp!  Schedule a class today!

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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
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Prealgebra 2 Self-Paced

Prealgebra 2
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Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
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Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
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Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
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Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
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Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
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Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
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Wednesday, Jun 25 - Dec 10
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Intermediate Counting & Probability
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Intermediate Number Theory
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Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
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Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
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Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
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Intermediate Programming with Python
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USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
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Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
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Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 1, 2025
0 replies
k i Peer-to-Peer Programs Forum
jwelsh   157
N Dec 11, 2023 by cw357
Many of our AoPS Community members share their knowledge with their peers in a variety of ways, ranging from creating mock contests to creating real contests to writing handouts to hosting sessions as part of our partnership with schoolhouse.world.

To facilitate students in these efforts, we have created a new Peer-to-Peer Programs forum. With the creation of this forum, we are starting a new process for those of you who want to advertise your efforts. These advertisements and ensuing discussions have been cluttering up some of the forums that were meant for other purposes, so we’re gathering these topics in one place. This also allows students to find new peer-to-peer learning opportunities without having to poke around all the other forums.

To announce your program, or to invite others to work with you on it, here’s what to do:

1) Post a new topic in the Peer-to-Peer Programs forum. This will be the discussion thread for your program.

2) Post a single brief post in this thread that links the discussion thread of your program in the Peer-to-Peer Programs forum.

Please note that we’ll move or delete any future advertisement posts that are outside the Peer-to-Peer Programs forum, as well as any posts in this topic that are not brief announcements of new opportunities. In particular, this topic should not be used to discuss specific programs; those discussions should occur in topics in the Peer-to-Peer Programs forum.

Your post in this thread should have what you're sharing (class, session, tutoring, handout, math or coding game/other program) and a link to the thread in the Peer-to-Peer Programs forum, which should have more information (like where to find what you're sharing).
157 replies
jwelsh
Mar 15, 2021
cw357
Dec 11, 2023
k i C&P posting recs by mods
v_Enhance   0
Jun 12, 2020
The purpose of this post is to lay out a few suggestions about what kind of posts work well for the C&P forum. Except in a few cases these are mostly meant to be "suggestions based on historical trends" rather than firm hard rules; we may eventually replace this with an actual list of firm rules but that requires admin approval :) That said, if you post something in the "discouraged" category, you should not be totally surprised if it gets locked; they are discouraged exactly because past experience shows they tend to go badly.
-----------------------------
1. Program discussion: Allowed
If you have questions about specific camps or programs (e.g. which classes are good at X camp?), these questions fit well here. Many camps/programs have specific sub-forums too but we understand a lot of them are not active.
-----------------------------
2. Results discussion: Allowed
You can make threads about e.g. how you did on contests (including AMC), though on AMC day when there is a lot of discussion. Moderators and administrators may do a lot of thread-merging / forum-wrangling to keep things in one place.
-----------------------------
3. Reposting solutions or questions to past AMC/AIME/USAMO problems: Allowed
This forum contains a post for nearly every problem from AMC8, AMC10, AMC12, AIME, USAJMO, USAMO (and these links give you an index of all these posts). It is always permitted to post a full solution to any problem in its own thread (linked above), regardless of how old the problem is, and even if this solution is similar to one that has already been posted. We encourage this type of posting because it is helpful for the user to explain their solution in full to an audience, and for future users who want to see multiple approaches to a problem or even just the frequency distribution of common approaches. We do ask for some explanation; if you just post "the answer is (B); ez" then you are not adding anything useful.

You are also encouraged to post questions about a specific problem in the specific thread for that problem, or about previous user's solutions. It's almost always better to use the existing thread than to start a new one, to keep all the discussion in one place easily searchable for future visitors.
-----------------------------
4. Advice posts: Allowed, but read below first
You can use this forum to ask for advice about how to prepare for math competitions in general. But you should be aware that this question has been asked many many times. Before making a post, you are encouraged to look at the following:
[list]
[*] Stop looking for the right training: A generic post about advice that keeps getting stickied :)
[*] There is an enormous list of links on the Wiki of books / problems / etc for all levels.
[/list]
When you do post, we really encourage you to be as specific as possible in your question. Tell us about your background, what you've tried already, etc.

Actually, the absolute best way to get a helpful response is to take a few examples of problems that you tried to solve but couldn't, and explain what you tried on them / why you couldn't solve them. Here is a great example of a specific question.
-----------------------------
5. Publicity: use P2P forum instead
See https://artofproblemsolving.com/community/c5h2489297_peertopeer_programs_forum.
Some exceptions have been allowed in the past, but these require approval from administrators. (I am not totally sure what the criteria is. I am not an administrator.)
-----------------------------
6. Mock contests: use Mock Contests forum instead
Mock contests should be posted in the dedicated forum instead:
https://artofproblemsolving.com/community/c594864_aops_mock_contests
-----------------------------
7. AMC procedural questions: suggest to contact the AMC HQ instead
If you have a question like "how do I submit a change of venue form for the AIME" or "why is my name not on the qualifiers list even though I have a 300 index", you would be better off calling or emailing the AMC program to ask, they are the ones who can help you :)
-----------------------------
8. Discussion of random math problems: suggest to use MSM/HSM/HSO instead
If you are discussing a specific math problem that isn't from the AMC/AIME/USAMO, it's better to post these in Middle School Math, High School Math, High School Olympiads instead.
-----------------------------
9. Politics: suggest to use Round Table instead
There are important conversations to be had about things like gender diversity in math contests, etc., for sure. However, from experience we think that C&P is historically not a good place to have these conversations, as they go off the rails very quickly. We encourage you to use the Round Table instead, where it is much more clear that all posts need to be serious.
-----------------------------
10. MAA complaints: discouraged
We don't want to pretend that the MAA is perfect or that we agree with everything they do. However, we chose to discourage this sort of behavior because in practice most of the comments we see are not useful and some are frankly offensive.
[list] [*] If you just want to blow off steam, do it on your blog instead.
[*] When you have criticism, it should be reasoned, well-thought and constructive. What we mean by this is, for example, when the AOIME was announced, there was great outrage about potential cheating. Well, do you really think that this is something the organizers didn't think about too? Simply posting that "people will cheat and steal my USAMOO qualification, the MAA are idiots!" is not helpful as it is not bringing any new information to the table.
[*] Even if you do have reasoned, well-thought, constructive criticism, we think it is actually better to email it the MAA instead, rather than post it here. Experience shows that even polite, well-meaning suggestions posted in C&P are often derailed by less mature users who insist on complaining about everything.
[/list]
-----------------------------
11. Memes and joke posts: discouraged
It's fine to make jokes or lighthearted posts every so often. But it should be done with discretion. Ideally, jokes should be done within a longer post that has other content. For example, in my response to one user's question about olympiad combinatorics, I used a silly picture of Sogiita Gunha, but it was done within a context of a much longer post where it was meant to actually make a point.

On the other hand, there are many threads which consist largely of posts whose only content is an attached meme with the word "MAA" in it. When done in excess like this, the jokes reflect poorly on the community, so we explicitly discourage them.
-----------------------------
12. Questions that no one can answer: discouraged
Examples of this: "will MIT ask for AOIME scores?", "what will the AIME 2021 cutoffs be (asked in 2020)", etc. Basically, if you ask a question on this forum, it's better if the question is something that a user can plausibly answer :)
-----------------------------
13. Blind speculation: discouraged
Along these lines, if you do see a question that you don't have an answer to, we discourage "blindly guessing" as it leads to spreading of baseless rumors. For example, if you see some user posting "why are there fewer qualifiers than usual this year?", you should not reply "the MAA must have been worried about online cheating so they took fewer people!!". Was sich überhaupt sagen lässt, lässt sich klar sagen; und wovon man nicht reden kann, darüber muss man schweigen.
-----------------------------
14. Discussion of cheating: strongly discouraged
If you have evidence or reasonable suspicion of cheating, please report this to your Competition Manager or to the AMC HQ; these forums cannot help you.
Otherwise, please avoid public discussion of cheating. That is: no discussion of methods of cheating, no speculation about how cheating affects cutoffs, and so on --- it is not helpful to anyone, and it creates a sour atmosphere. A longer explanation is given in Seriously, please stop discussing how to cheat.
-----------------------------
15. Cutoff jokes: never allowed
Whenever the cutoffs for any major contest are released, it is very obvious when they are official. In the past, this has been achieved by the numbers being posted on the official AMC website (here) or through a post from the AMCDirector account.

You must never post fake cutoffs, even as a joke. You should also refrain from posting cutoffs that you've heard of via email, etc., because it is better to wait for the obvious official announcement. A longer explanation is given in A Treatise on Cutoff Trolling.
-----------------------------
16. Meanness: never allowed
Being mean is worse than being immature and unproductive. If another user does something which you think is inappropriate, use the Report button to bring the post to moderator attention, or if you really must reply, do so in a way that is tactful and constructive rather than inflammatory.
-----------------------------

Finally, we remind you all to sit back and enjoy the problems. :D

-----------------------------
(EDIT 2024-09-13: AoPS has asked to me to add the following item.)

Advertising paid program or service: never allowed

Per the AoPS Terms of Service (rule 5h), general advertisements are not allowed.

While we do allow advertisements of official contests (at the MAA and MATHCOUNTS level) and those run by college students with at least one successful year, any and all advertisements of a paid service or program is not allowed and will be deleted.
0 replies
v_Enhance
Jun 12, 2020
0 replies
k i Stop looking for the "right" training
v_Enhance   50
N Oct 16, 2017 by blawho12
Source: Contest advice
EDIT 2019-02-01: https://blog.evanchen.cc/2019/01/31/math-contest-platitudes-v3/ is the updated version of this.

EDIT 2021-06-09: see also https://web.evanchen.cc/faq-contest.html.

Original 2013 post
50 replies
v_Enhance
Feb 15, 2013
blawho12
Oct 16, 2017
Nice Collinearity
oVlad   10
N 2 minutes ago by Trenod
Source: KöMaL A. 831
In triangle $ABC$ let $F$ denote the midpoint of side $BC$. Let the circle passing through point $A$ and tangent to side $BC$ at point $F$ intersect sides $AB$ and $AC$ at points $M$ and $N$, respectively. Let the line segments $CM$ and $BN$ intersect in point $X$. Let $P$ be the second point of intersection of the circumcircles of triangles $BMX$ and $CNX$. Prove that points $A, F$ and $P$ are collinear.

Proposed by Imolay András, Budapest
10 replies
1 viewing
oVlad
Oct 11, 2022
Trenod
2 minutes ago
cute geo
Royal_mhyasd   1
N 10 minutes ago by Royal_mhyasd
Source: own(?)
Let $\triangle ABC$ be an acute triangle and $I$ it's incenter. Let $A'$, $B'$ and $C'$ be the projections of $I$ onto $BC$, $AC$ and $AB$ respectively. $BC \cap B'C' = \{K\}$ and $Y$ is the projection of $A'$ onto $KI$. Let $M$ be the middle of the arc $BC$ not containing $A$ and $T$ the second intersection of $A'M$ and the circumcircle of $ABC$. If $N$ is the midpoint of $AI$, $TY \cap IA' = \{P\}$, $BN \cap PC' = \{D\}$ and $CN \cap PB' =\{E\}$, prove that $NEPD$ is cyclic.
PS i'm not sure if this problem is actually original so if it isn't someone please tell me so i can change the source (if that's possible)
1 reply
+1 w
Royal_mhyasd
3 hours ago
Royal_mhyasd
10 minutes ago
Hardest in ARO 2008
discredit   29
N 12 minutes ago by JARP091
Source: ARO 2008, Problem 11.8
In a chess tournament $ 2n+3$ players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least $ n$ next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
29 replies
discredit
Jun 11, 2008
JARP091
12 minutes ago
Macedonian Mathematical Olympiad 2019 problem 1
Lukaluce   5
N 15 minutes ago by AylyGayypow009
In an acute-angled triangle $ABC$, point $M$ is the midpoint of side $BC$ and the centers of the $M$- excircles of triangles $AMB$ and $AMC$ are $D$ and $E$, respectively. The circumcircle of triangle $ABD$ intersects line $BC$ at points $B$ and $F$. The circumcircle of triangle $ACE$ intersects line $BC$ at points $C$ and $G$. Prove that $BF\hspace{0.25mm} = \hspace{0.25mm} CG$ .
5 replies
Lukaluce
Apr 20, 2019
AylyGayypow009
15 minutes ago
Do you need to attend mop
averageguy   5
N Yesterday at 5:55 PM by babyzombievillager
So I got accepted into a summer program and already paid the fee of around $5000 dollars. It's for 8 weeks (my entire summer) and it's in person. I have a few questions
1. If I was to make MOP this year am I forced to attend?
2.If I don't attend the program but still qualify can I still put on my college application that I qualified for MOP or can you only put MOP qualifier if you actually attend the program.
5 replies
averageguy
Mar 5, 2025
babyzombievillager
Yesterday at 5:55 PM
2014 amc 10 a problem 23
Rook567   3
N Yesterday at 4:13 PM by Rook567
Why do solutions assume 30 60 90 triangles?
If you assume 45 45 90 you get 5/6 as answer, don’t you?
3 replies
Rook567
Monday at 7:31 PM
Rook567
Yesterday at 4:13 PM
Inequality with a^2+b^2+c^2+abc=4
cn2_71828182846   72
N Yesterday at 4:12 PM by endless_abyss
Source: USAMO 2001 #3
Let $a, b, c \geq 0$ and satisfy \[ a^2+b^2+c^2 +abc = 4 . \] Show that \[ 0 \le ab + bc + ca - abc \leq 2. \]
72 replies
cn2_71828182846
Jun 27, 2004
endless_abyss
Yesterday at 4:12 PM
Special Points on $BC$
tenniskidperson3   40
N Yesterday at 3:43 PM by dipinsubedi
Source: 2013 USAMO Problem 6
Let $ABC$ be a triangle. Find all points $P$ on segment $BC$ satisfying the following property: If $X$ and $Y$ are the intersections of line $PA$ with the common external tangent lines of the circumcircles of triangles $PAB$ and $PAC$, then \[\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.\]
40 replies
tenniskidperson3
May 1, 2013
dipinsubedi
Yesterday at 3:43 PM
Alcumus vs books
UnbeatableJJ   12
N Yesterday at 2:59 PM by pingpongmerrily
If I am aiming for AIME, then JMO afterwards, is Alcumus adequate, or I still need to do the problems on AoPS books?

I got AMC 23 this year, and never took amc 10 before. If I master the alcumus of intermediate algebra (making all of the bars blue). How likely I can qualify for AIME 2026?
12 replies
UnbeatableJJ
Apr 23, 2025
pingpongmerrily
Yesterday at 2:59 PM
Zsigmondy's theorem
V0305   17
N Yesterday at 7:22 AM by whwlqkd
Is Zsigmondy's theorem allowed on the IMO, and is it allowed on the AMC series of proof competitions (e.g. USAJMO, USA TSTST)?
17 replies
V0305
May 24, 2025
whwlqkd
Yesterday at 7:22 AM
USAJMO problem 3: Inequality
BOGTRO   106
N Monday at 11:12 PM by Learning11
Let $a,b,c$ be positive real numbers. Prove that $\frac{a^3+3b^3}{5a+b}+\frac{b^3+3c^3}{5b+c}+\frac{c^3+3a^3}{5c+a} \geq \frac{2}{3}(a^2+b^2+c^2)$.
106 replies
BOGTRO
Apr 24, 2012
Learning11
Monday at 11:12 PM
what the yap
KevinYang2.71   31
N Monday at 3:26 PM by Blast_S1
Source: USAMO 2025/3
Alice the architect and Bob the builder play a game. First, Alice chooses two points $P$ and $Q$ in the plane and a subset $\mathcal{S}$ of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair $A,\,B$ of cities, they are connected with a road along the line segment $AB$ if and only if the following condition holds:
[center]For every city $C$ distinct from $A$ and $B$, there exists $R\in\mathcal{S}$ such[/center]
[center]that $\triangle PQR$ is directly similar to either $\triangle ABC$ or $\triangle BAC$.[/center]
Alice wins the game if (i) the resulting roads allow for travel between any pair of cities via a finite sequence of roads and (ii) no two roads cross. Otherwise, Bob wins. Determine, with proof, which player has a winning strategy.

Note: $\triangle UVW$ is directly similar to $\triangle XYZ$ if there exists a sequence of rotations, translations, and dilations sending $U$ to $X$, $V$ to $Y$, and $W$ to $Z$.
31 replies
KevinYang2.71
Mar 20, 2025
Blast_S1
Monday at 3:26 PM
Mustang Math Recruitment is Open!
MustangMathTournament   3
N Monday at 3:21 PM by Puzzlebooks206
The Interest Form for joining Mustang Math is open!

Hello all!

We're Mustang Math, and we are currently recruiting for the 2025-2026 year! If you are a high school or college student and are passionate about promoting an interest in competition math to younger students, you should strongly consider filling out the following form: https://link.mustangmath.com/join. Every member in MM truly has the potential to make a huge impact, no matter your experience!

About Mustang Math

Mustang Math is a nonprofit organization of high school and college volunteers that is dedicated to providing middle schoolers access to challenging, interesting, fun, and collaborative math competitions and resources. Having reached over 4000 U.S. competitors and 1150 international competitors in our first six years, we are excited to expand our team to offer our events to even more mathematically inclined students.

PROJECTS
We have worked on various math-related projects. Our annual team math competition, Mustang Math Tournament (MMT) recently ran. We hosted 8 in-person competitions based in Washington, NorCal, SoCal, Illinois, Georgia, Massachusetts, Nevada and New Jersey, as well as an online competition run nationally. In total, we had almost 900 competitors, and the students had glowing reviews of the event. MMT International will once again be running later in August, and with it, we anticipate our contest to reach over a thousand students.

In our classes, we teach students math in fun and engaging math lessons and help them discover the beauty of mathematics. Our aspiring tech team is working on a variety of unique projects like our website and custom test platform. We also have a newsletter, which, combined with our social media presence, helps to keep the mathematics community engaged with cool puzzles, tidbits, and information about the math world! Our design team ensures all our merch and material is aesthetically pleasing.

Some highlights of this past year include 1000+ students in our classes, AMC10 mock with 150+ participants, our monthly newsletter to a subscriber base of 6000+, creating 8 designs for 800 pieces of physical merchandise, as well as improving our custom website (mustangmath.com, 20k visits) and test-taking platform (comp.mt, 6500+ users).

Why Join Mustang Math?

As a non-profit organization on the rise, there are numerous opportunities for volunteers to share ideas and suggest projects that they are interested in. Through our organizational structure, members who are committed have the opportunity to become a part of the leadership team. Overall, working in the Mustang Math team is both a fun and fulfilling experience where volunteers are able to pursue their passion all while learning how to take initiative and work with peers. We welcome everyone interested in joining!

More Information

To learn more, visit https://link.mustangmath.com/RecruitmentInfo. If you have any questions or concerns, please email us at contact@mustangmath.com.

https://link.mustangmath.com/join
3 replies
MustangMathTournament
May 24, 2025
Puzzlebooks206
Monday at 3:21 PM
[TEST RELEASED] OMMC Year 5
DottedCaculator   173
N Monday at 3:01 PM by drhong
Test portal: https://ommc-test-portal-2025.vercel.app/

Hello to all creative problem solvers,

Do you want to work on a fun, untimed team math competition with amazing questions by MOPpers and IMO & EGMO medalists? $\phantom{You lost the game.}$
Do you want to have a chance to win thousands in cash and raffle prizes (no matter your skill level)?

Check out the fifth annual iteration of the

Online Monmouth Math Competition!

Online Monmouth Math Competition, or OMMC, is a 501c3 accredited nonprofit organization managed by adults, college students, and high schoolers which aims to give talented high school and middle school students an exciting way to develop their skills in mathematics.

Our website: https://www.ommcofficial.org/

This is not a local competition; any student 18 or younger anywhere in the world can attend. We have changed some elements of our contest format, so read carefully and thoroughly. Join our Discord or monitor this thread for updates and test releases.

How hard is it?

We plan to raffle out a TON of prizes over all competitors regardless of performance. So just submit: a few minutes of your time will give you a great chance to win amazing prizes!

How are the problems?

You can check out our past problems and sample problems here:
https://www.ommcofficial.org/sample
https://www.ommcofficial.org/2022-documents
https://www.ommcofficial.org/2023-documents
https://www.ommcofficial.org/ommc-amc

How will the test be held?/How do I sign up?

Solo teams?

Test Policy

Timeline:
Main Round: May 17th - May 24th
Test Portal Released. The Main Round of the contest is held. The Main Round consists of 25 questions that each have a numerical answer. Teams will have the entire time interval to work on the questions. They can submit any time during the interval. Teams are free to edit their submissions before the period ends, even after they submit.

Final Round: May 26th - May 28th
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173 replies
DottedCaculator
Apr 26, 2025
drhong
Monday at 3:01 PM
Circumcircle excircle chaos
CyclicISLscelesTrapezoid   25
N Apr 22, 2025 by bin_sherlo
Source: ISL 2021 G8
Let $ABC$ be a triangle with circumcircle $\omega$ and let $\Omega_A$ be the $A$-excircle. Let $X$ and $Y$ be the intersection points of $\omega$ and $\Omega_A$. Let $P$ and $Q$ be the projections of $A$ onto the tangent lines to $\Omega_A$ at $X$ and $Y$ respectively. The tangent line at $P$ to the circumcircle of the triangle $APX$ intersects the tangent line at $Q$ to the circumcircle of the triangle $AQY$ at a point $R$. Prove that $\overline{AR} \perp \overline{BC}$.
25 replies
CyclicISLscelesTrapezoid
Jul 12, 2022
bin_sherlo
Apr 22, 2025
Circumcircle excircle chaos
G H J
G H BBookmark kLocked kLocked NReply
Source: ISL 2021 G8
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CyclicISLscelesTrapezoid
372 posts
#1 • 8 Y
Y by GioOrnikapa, LoloChen, PHSH, crazyeyemoody907, bjump, v4913, Rounak_iitr, Funcshun840
Let $ABC$ be a triangle with circumcircle $\omega$ and let $\Omega_A$ be the $A$-excircle. Let $X$ and $Y$ be the intersection points of $\omega$ and $\Omega_A$. Let $P$ and $Q$ be the projections of $A$ onto the tangent lines to $\Omega_A$ at $X$ and $Y$ respectively. The tangent line at $P$ to the circumcircle of the triangle $APX$ intersects the tangent line at $Q$ to the circumcircle of the triangle $AQY$ at a point $R$. Prove that $\overline{AR} \perp \overline{BC}$.
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ltf0501
191 posts
#2 • 4 Y
Y by PRMOisTheHardestExam, kyou46, Mango247, Rounak_iitr
Let $I_a$ the $A$-excenter of triangle $ABC$, $S$ be the second intersection of the line $PX$ and $\omega$, and $T$ be a point on the line $PX$ such that $\measuredangle TAI_a = \measuredangle I_aAS$. We show that $\triangle ASI_a \stackrel{+}{\sim} \triangle AIT$.

Let $K$ be the second intersection of the line $I_aX$ and $\omega$. By Euler's formula we have
$$
I_aX \times I_aK = 2Rr_a,
$$where $R$ and $r_a$ are the radius of $\omega$ and $\Omega_A$, respectively. Therefore, $I_aK = 2R = KS$. Let $M$ be the midpoint of $I_aS$, then $\measuredangle KMS = 90^{\circ} = \measuredangle KXS$, so $M$ lies on $\omega$.

Let $A'$ be the symmetric point of $A$ w.r.t $M$, then
$$
\measuredangle I_aA'A = \measuredangle SAM = \measuredangle MAX = \measuredangle I_aSX.
$$Hence there exists a point $T'$ on the line $SX$ such that $\triangle I_aAA' \stackrel{+}{\sim} I_aT'S$. Then we have
$$
\frac{SA}{SI_a} = \frac{I_aA'}{I_aS} = \frac{I_aA}{I_aT'},
$$and
$$
\measuredangle ASI_a = \measuredangle A'I_aS = \measuredangle A'I_aA - \measuredangle SI_aA = \measuredangle SI_aT' - \measuredangle SI_aA = \measuredangle AI_aT'.
$$This implies that $\triangle ASI_a \stackrel{+}{\sim} \triangle AI_aT'$. Hence $\measuredangle SAI_a = \measuredangle I_aAT'$ and $T = T'$.

Let $\ell$ be a tangent line of $\Omega_A$ different from $BC$ but parallel to $BC$, and $B', C'$ be the intersection of $\ell$ and $AB, AC$, respectively. Consider the transformation $\varphi$ which is the composition of the inversion $\mathcal{I}(A, AI_a^2)$ and the symmetry with respect to $AI_a$. It is known that
$\varphi(B) = C'$ and $\varphi(C) = B'$, so $\varphi(ABC) = \ell$. Besides, by the conclusion above we have $\varphi(S) = T$, so $T$ lies on $\ell$.

Finally, let $R'$ be the projection of $A$ onto $\ell$, and $O$ be the circumcenter of $ABC$, then by angle chasing we have
$$
\measuredangle R'PT = \measuredangle R'AT = \measuredangle SAO = \measuredangle KXA = \measuredangle PAX,
$$which implies that $R'P$ is tangent to $\odot(APX)$. Similarly $R'Q$ is tangent to $\odot(AQY)$. Hence $R' = R$ and the statement is proved.
This post has been edited 1 time. Last edited by ltf0501, Jul 16, 2022, 5:43 AM
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tastymath75025
3223 posts
#3 • 2 Y
Y by GuvercinciHoca, Kingsbane2139
Fun problem :) Here is a solution quite different from the official one, using Poncelet’s porism and Simson lines.

Step 1: Relating $X,Y$ to the common tangents of $\omega,\Omega_A$.

First recall Poncelet’s Porism, which says for any $P\in \omega$, if we take $P_1\neq P_2\in \omega$ with $PP_1,PP_2$ tangent to $\Omega_A$, then $P_1P_2$ is tangent to $\Omega_A$ as well. By taking the limit as $P\to X$, we find that the tangent to $\Omega_A$ at $X$ meets $\omega$ at some point $X_1$ lying on a common tangent $\ell_X$ between $\omega,\Omega_A$. Define $Y_1, \ell_Y$ similarly. For reference, in my diagram $X,\ell_Y$ are closer to $B$ and $Y, \ell_X$ are closer to $C$.

Step 2: Getting rid of $(APX), (AQY)$.

Let $P_1\in \ell_X$ with $PP_1$ tangent to $(APX)$. Then $\angle PP_1X_1 = 180^{\circ} -\angle PX_1P_1 - \angle P_1PX_1 = \angle XAX_1 - \angle XAP = \angle PAX_1$, so $P_1\in (APX_1)$ and therefore $AP_1 \perp \ell_X$. Similarly we can define $Q_1$ as the projection of $A$ onto $\ell_Y$ and we only need to show $R=PP_1\cap QQ_1$ satisfies $AR\perp BC$.

Step 3: Reduction to Simson lines.

Define $P_2\in \ell_X$ and $P_3\in XX_1$ with $P_2P_3||BC$ and $AX_1P_2P_3$ cyclic. Similarly define $Q_2,Q_3$. Then the Simson line of $A$ with respect to $X_1P_2P_3$ passes through $P_1,P$, and the projection of $A$ onto $P_2P_3$ and similarly for $Y_1Q_2Q_3$. Therefore, if we can show the lines $P_2P_3,Q_2Q_3$ are the same line $\ell || BC$, it will follow that $P_1P\cap Q_1Q$ is the projection of $A$ onto $\ell$, which will finish the problem.

Step 4: Identifying $P_3,Q_3$.

By Reim’s theorem on $(AX_1P_2P_3), \omega$ with lines $AP_3$ and $X_1P_2$, we know that line $AP_3$ meets $\omega$ at some $X_1’$ with $X_1X_1’||BC$. It follows that if we define $P_3’ = AP_3\cap BC$, $P_3’$ is the inverse of $X_1$ in $\sqrt{bc}$-inversion. Therefore, if $I,I_A$ are the incenter and $A$-excenter of $\triangle ABC$, we know $AP_3’I\sim AI_AX_1$ so $\angle AP_3’I = \angle AI_AX_1$. Since $AI_A$ bisects the angle between $AB,AC$ and $I_AX_1$ bisects the angle between $X_1X, \ell_X$, we have \[\angle (I_AA, I_AX_1) = \frac{1}{2} \angle (AB, XX_1) + \frac{1}{2} \angle (AC, \ell_X)= \frac{1}{4} (\stackrel{\frown}{AX_1} - \stackrel{\frown}{BX} ) +\]\[\frac{1}{4} (\stackrel{\frown}{AX_1} - \stackrel{\frown}{CX_1})= \frac{1}{4} \stackrel{\frown}{AX_1} + \frac{1}{4} ( \stackrel{\frown}{AX_1} - \stackrel{\frown}{XX_1’}) = \frac{1}{2}\angle (AP_3, BC) + \frac{1}{2} \angle (AP_3, P_3X).\]It follows that reflecting $BC$ over $P_3’I$ gives a line parallel to $P_3X$; since this line is tangent to the incircle $(I)$ of $\triangle ABC$, it follows that the homothety sending $(I)$ to $\Omega_A$ also sends $P_3’$ to $P_3$, hence $P_3$ lies on the other tangent to $\Omega_A$ parallel to $BC$. Similarly $Q_3$ lies on this line as well, so $P_2,P_3,Q_2,Q_3$ lie on a common line as desired, and we are done.
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ABCDE
1963 posts
#5 • 1 Y
Y by LoloChen
Let $O$ and $I_A$ be the centers of $\omega$ and $\Omega_A$, $P',Q',R'$ be the reflections of $A$ across $P,Q,R$, $X',Y'$ be on $\omega$ with $XX',YY'$ tangent to $\Omega_A$, $Z=XP'\cap YQ'$, and $O'$ be the circumcenter of $P'Q'Z$. Then, we have that $P',Q'$ are the reflections of $A$ across $XX,YY'$ and $P'R'\perp P'X,Q'R'\perp Q'Y$, so $R'$ is the reflection of $Z$ over $O'$.

Fix the circumcircle and excircle of $ABC$, and vary triangle $ABC$ so that they have the same circumcircle and excircle, which is possible by Poncelet's Porism. Varying $A$ at a fixed angular velocity, we note that $P'$ and $Q'$ vary at the opposite angular velocity on fixed circles as they are reflections of $A$ across fixed lines. Since $\angle P'XY$ and $\angle Q'YX$ vary at the same velocity, $Z$ lies on a fixed circle through $X$ and $Y$ and varies at the same angular velocity as $P'$ and $Q'$. Because $\angle P'ZQ'$ is fixed, $P'O'Q'$ is similar to a fixed triangle, and hence $O'$ also lies on a fixed circle and moves at the same angular velocity. Finally, as $R'$ is the reflection of $Z$ over $O'$, the same is true for $R'$: it lies on a fixed circle and moves at an angular velocity opposite to that of $A$.

We now apply complex numbers. Let $\omega$ be the unit circle, $a$ be $A$, $j$ be the center of $\Gamma_A$, $r$ be $R'$. By the previous paragraph, we have that there exist constants $u$ and $v$ with $r=u+\frac va$. The condition that $AR=AR'\perp BC$ is equivalent to $AI_A$ bisecting $\angle R'AO$, or $\frac{P(a)}{Q(a)}=\frac{a(a-r)}{(a-j)^2}\in\mathbb R\iff P(a)\overline Q(a)=\overline P(a)Q(a)$, where $P$ and $Q$ are fixed quadratics. To verify that this is true for all $a$ on the unit circle, it suffices to check five cases.

When $A$ lies on the symmetry axis of $\omega$ and $\Omega_A$, the result follows by symmetry.

When $A=X$, we have that $B=C=X'$ so $P'=X$. Hence, $XP'$ is the reflection of the tangent at $X$ to $\omega$ over $XX'$. Then, it is clear that $AR'=XP'\parallel OX'\perp BC$.

When $A=X'$, we have that $BC=XX'$ so $P'=X'$. Then $AR'=X'R'\perp P'X=BC$.
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CANBANKAN
1301 posts
#6 • 2 Y
Y by trying_to_solve_br, JingheZhang
But you've only checked 3 cases, not 5

Actually if $|a|=1$ and $P(1/a)Q(a)=P(a)Q(1/a)$ where $P,Q$ are quadratics, the two polynomials have the same "leading term" so $a^2(P(1/a)Q(a)-P(a)Q(1/a))$ should be a poly of degree 3? So does that mean you check 4 cases?
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ABCDE
1963 posts
#7 • 1 Y
Y by LoloChen
Looks like I accidentally left out the last sentence while copy pasting:

The cases $A=Y$ and $A=Y'$ follow analogously, so we are done.

I’m not sure how many cases are actually needed beyond five being sufficient. There is also actually a sixth trivial case which comes from the other point on the symmetry axis.
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duanby
76 posts
#8
Y by
The problem is equivalent to
sin(AXY-2AIaY)/sin(AYX-2AIaX) = (AY/AX)*( cos(AYIa)/cos(AXIa) )^2 ---- (*)
Let A' be the antipodal of A, U,V be the second intersection of IaX,IaY and (ABC)
Then IaU=IaV=2R
Therefore UV, tangent at A' wrt (ABC),perpendicular bisector of AIa are concurrent
Then use Menelaus's theorem we can get(*)
This post has been edited 2 times. Last edited by duanby, Jul 13, 2022, 1:02 PM
Reason: restate
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MarkBcc168
1595 posts
#9 • 1 Y
Y by GuvercinciHoca
Very beautiful and very challenging geometry! Here is my solution.

Let the tangents to $\Omega_A$ at $X$ and $Y$ meet each other at $T$ and meet $\omega$ again at $X_1,Y_1$. Let $AT$ meet $\omega$ again at $K$. The main lemma is the following.
Quote:
Lines $KY_1$, $XX_1$, $BC$ are concurrent at $U$. Similarly, lines $KX_1$, $YY_1$, $BC$ are concurrent at $V$.

To prove this lemma, we define more points. Let $N$ be the midpoint of arc $BAC$. Let $\Omega_A$ touch $BC,CA,AB$ at $D,E,F$.
Claim: $X_1Y_1, AN, BC$ are concurrent.

Proof. By Poncelet's Porism on degenerate $\triangle XX_1X_1$, we get that one common external tangent of $\omega, \Omega_A$ passes through $X_1$. Similarly, the other common external tangent passes through $Y_1$. Therefore, by a well-known lemma, we get that $X_1Y_1$ passes through $BI\cap AC$ and $CI\cap AB$. The conclusion then follows through harmonic. $\blacksquare$
Claim: $K,D,N$ are colinear.

Proof. First, we prove that $KK, XY, BC$ are concurrent. Applying DDIT on the degenerate quadrilateral $XXYY$ and point $A$, we get an involution swapping $(AX, AY)$, $(AT, AT)$, and $(AB, AC)$. Projecting this involution onto $\omega$ gives the conclusion.

Now, let the concurrency point be $U$. Notice that $UB\cdot UC = UK^2 = UD^2$. Thus, $D$ lies on $K$-Appollonius circle of $\triangle BKC$, meaning that $KD$ bisects $\angle BKC$, implying the conclusion. $\blacksquare$
Claim: Let $U = XX_1\cap BC$. Then, $X_1D$ and $AU$ meet at $L\in\omega$.

Proof. Use DDIT again but on point $X_1$ and quadrilateral $ABDC$. We get an involution swapping $(X_1B, X_1C)$, $(X_1A, X_1D)$, and $(X_1X_1, X_1X)$. Projecting this involution on to $\omega$, we get an involution swapping $(B,c)$ ,$(A,X_1D\cap\omega)$, and $(X_1,X)$. This gives the conclusion. $\blacksquare$
Finally, the requested lemma follows from Pascal on $ALX_1Y_1KN$.
Reduction to the lemma:

Let $PQ$ meet $\odot(APR)$ and $\odot(AQR)$ again at $P_1$ and $Q_1$. Now, let $AT$ meet $\odot(KUV)$ again at $K_1$ and let $K_1'$ be the isogonal conjugate of $K_1$ w.r.t. $\triangle TUV$. A straightforward angle chasing gives that
$$\measuredangle K_1'VT = \measuredangle UVK_1 = \measuredangle UKK_1 = \measuredangle Y_1KA = \measuredangle QYA = \measuredangle RQA = \angle RQ_1A,$$and $\angle K_1'UT = \angle AP_1R$. Moreover, since
$$\measuredangle K_1'TV = \measuredangle UTK_1 = \measuredangle PTA = \measuredangle PQA = \measuredangle Q_1QA = \measuredangle Q_1RA,$$so $\triangle K_1'TV \stackrel{-}{\sim} \triangle ARQ_1$. Similarly, $\triangle K_1'TU\stackrel{-}{\sim} \triangle ARP_1$. Therefore, quadrilaterals $K_1'TUV$ and $ARP_1Q_1$ are similar. Hence,
$$\measuredangle RAP = \measuredangle RP_1P = \measuredangle RP_1Q_1 = -\measuredangle TUV$$As $AP\perp UT$, the conclusion follows.
This post has been edited 1 time. Last edited by MarkBcc168, Jul 13, 2022, 3:05 PM
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JAnatolGT_00
559 posts
#10 • 1 Y
Y by GuvercinciHoca
Denote by $S$ intersection of perpendicular to $BC$ through $A$ with tangent $\ell$ to $\Omega_A,$ parallel to $BC.$ Let $XP$ meet $\omega$ again at $X_1$ - by Poncelet theorem we easily conclude that different from $XX_1$ tangent to $\Omega_A$ through $X_1$ also tangent to $\omega.$ Let this common tangent intersect $RX$ at $P_1,$ so we get $$\measuredangle APP_1=\measuredangle AXX_1=\measuredangle AX_1P_1\implies A\in \odot (PP_1X_1)\implies AP_1\perp X_1P_1.$$Let $X_1P_1\cap \ell =X_2,$ $XX_1\cap \ell =X_3,$ $\tau$ denotes reflection over bisector of angle $BAC.$

Claim. $\tau (AX)=AX_2.$
Proof. Let $AX_2$ meet $\omega$ again at $Z,$ and $\phi$ denotes involution which swaps $(B;C),(XX_1\cap BC;AX_2\cap BC).$ DIT on $AXZX_1$ gives $\phi (AX_1\cap BC) =XZ\cap BC.$ By DDIT on $BC,XX_1,X_1X_2,\ell$ there exist involution which swaps $$(AB;AC),(\overline{A(XX_1\cap BC)},AX_2),(A\infty_{BC};AX_1)$$so projecting on $BC$ we get $\phi (AX_1\cap BC)= \infty_{BC}$ and in fact $XZ\parallel BC,$ thus we are done $\Box$

By DDIT on $XX_1X_2X_3$ we get $\tau (AX_1)=AX_3\implies \measuredangle X_2AX_3=\measuredangle X_1AX=\measuredangle X_2X_1X_3\implies A\in \odot (X_1X_2X_3)$.
By Simson theorem $S\in \overline{RPP_1}.$ Analogously $S\in RQ,$ therefore $S=R,$ and we are done.
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timon92
224 posts
#11 • 4 Y
Y by MarkBcc168, megarnie, SatisfiedMagma, LoloChen
This problem was proposed by Burii.
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CyclicISLscelesTrapezoid
372 posts
#12 • 9 Y
Y by ETS1331, samrocksnature, bluelinfish, crazyeyemoody907, CoolCarsOnTheRun, Mango247, Mango247, Mango247, math_comb01
Wow. Did this with geogebra.

Let $\measuredangle$ denote directed angles $\!\!\!\mod 180^\circ$. Let $\ell$ be the line parallel to $\overline{BC}$ tangent to $\Omega_A$. Redefine $R$ as the foot of the altitude from $A$ to $\ell$. It's sufficient to show that $\overline{PR}$ is tangent to the circumcircle of $APX$.

Step 1: Removing $P$ and $R$.

Since $\angle APZ=\angle ARZ=90^\circ$, $APRZ$ is cyclic. We claim that proving that the circumcircle of $AXZ$ is tangent to $\overline{RZ}$ solves the problem. Indeed, if that was true, then we have $\measuredangle APR=\measuredangle AZR=\measuredangle AXZ=\measuredangle AXP$, as desired. Thus, we can ignore $P$ and $R$ in the diagram.

https://cdn.discordapp.com/attachments/931295365062348881/997928685829115914/unknown.png

Step 2: Reduction to $\overline{BC} \parallel \ell$.

Let $\overline{XZ}$ intersect $\omega$ at $U \ne X$.

Claim: $U$ is on a common tangent of $\omega$ and $\Omega_A$.
Proof: By Poncelet's porism on $ABC$ and degenerate triangle $UUX$, the tangent at $U$ to $\omega$ is tangent to $\Omega_A$.

Consider the following problem:
Converse of ISL 2021 G8 wrote:
Let $UZV$ be a triangle with incircle $\Omega_A$ and $V$-intouch point $X$. Let the circle through $U$ and $X$ tangent to $\overline{UV}$ intersect the circumcircle of $UZV$ at $A$. Let the tangents from $A$ to $\Omega_A$ intersect the circumcircle of $AUX$ at $B \ne A$ and $C \ne A$. Prove that $\overline{BC} \parallel \overline{ZV}$.

We claim that this problem implies the original problem. Ignore the condition $\ell \parallel \overline{BC}$, and consider all positions of $\ell$ that make $\Omega_A$ the incircle of $UZV$. Since $\angle AVZ$ is monotonic and ranges from $0^\circ$ to $180^\circ$, there is exactly one position of $\ell$ that makes $AUVZ$ cyclic. Choose this position of $\ell$. By Miquel's theorem on $UZV$, the circumcircle of $AXZ$ is tangent to $\overline{ZV}$, so we want to prove that $\ell \parallel \overline{BC}$.

https://cdn.discordapp.com/attachments/882923149186977813/998004285772152862/unknown.png

Step 3: Poncelet's porism and DDIT finish.

We will now solve the modified problem in step 2. Let $\overline{AB}$ and $\overline{AC}$ intersect the circumcircle of $AUVZ$ at $D \ne A$ and $E \ne A$. By Poncelet's porism on $UZV$ and $ADE$, $\overline{DE}$ is tangent to $\Omega_A$. Let $I_A$ be the center of $\Omega_A$, and let $UV$ intersect $\Omega_A$ at $X'$. Since $\measuredangle BUC=\measuredangle BAC=\measuredangle DAE=\measuredangle DUE$, the angle bisectors of $\angle BUE$ and $\angle CUD$ coincide. Consider a line $\ell_1$ perpendicular to the angle bisector of $\angle BUE$, and let $UB$, $UC$, $UD$, $UE$, $UX$, and $UX'$ intersect $\ell_1$ at $B^*$, $C^*$, $D^*$, $E^*$, $X^*$, and $X'^*$, respectively. By symmetry, $B^*C^*=D^*E^*$ in directed lengths, so the midpoints of $B^*E^*$ and $C^*D^*$ coincide at a point we call $M$. By DDIT, there exists a pencil involution at $U$ sending $\overline{UB}$ to $\overline{UE}$, $\overline{UC}$ to $\overline{UD}$, and $\overline{UX}$ to $\overline{UX'}$. Therefore there exists an involution sending $B^*$ to $E^*$, $C^*$ to $D^*$, and $X^*$ to $X'^*$. The involution must be a reflection over $M$, so the angle bisector of $\angle BUE$ also coincides with the angle bisector of $\angle XUX'$. Thus, Let $\overline{UB}$ intersect the circumcircle of $AUVZ$ at $B' \ne U$. Then, $\measuredangle B'UZ=\measuredangle VUE$, so $\overline{B'E} \parallel \ell$. By Reim's theorem, we also have $\overline{B'E} \parallel \overline{BC}$, so $\ell \parallel \overline{BC}$, as desired.

https://cdn.discordapp.com/attachments/882923149186977813/998004783216599070/unknown.png

Remark: The claim in step 2 is a lemma in https://artofproblemsolving.com/community/c6h2313676. The construction of $V$ is motivated by reverse engineering Miquel's theorem and trying to make $\Omega_A$ the incircle of something. The construction of $D$ and $E$ is motivated by a solution to Taiwan TST 2014/3/3.
This post has been edited 1 time. Last edited by CyclicISLscelesTrapezoid, Jul 17, 2022, 12:17 AM
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nobodyknowswhoIam
94 posts
#13
Y by
Here is more synthetic solution I wrote during TST.
Let us redefine point R as intersection of tangent at $D'$ and $A$-altitude and prove $PR$ is tangent to $\omega_{XPA}$. Let us denote some points:
$D'D'\cap AB = B'$,$D'D'\cap AC = C'$,$D'D'\cap XX = B'$,$XX\cap AB = X'$ and $CX\cap AP' = M$. Now let's write Brianchon's theorem on hexagon $DD'XXB"C"$:
Number 1 $DD\cap D'D' = \infty$
Number 2 $D'D'\cap XX = P'$
Number 3 $XX\cap XX = X$
Number 4 $XX\cap B"B" = X'$
Number 5 $B"B"\cap C"C" = A$
Number 6 $C"C"\cap DD = C'$ and lines $14,25,36$ must be concurrent. Since $AP'\cap CX = M$ we can imply that $MX'||BC$
By angle chase $\angle{MX'A} = \angle{P'B'A}=180 - \angle{AB'C'}=180- \angle{ABC}=180- \angle{AXC} = \angle{MXA}$ so $MXX'A$ is concyclic.
Again by angle chase $\angle{MAX}=\angle{MX'P'}=\angle{X'P'B'}=\angle{PAR}$ (since $P'APR$ is concyclic) thus
$\angle{XAP}=\angle{P'AR} = \angle{P'PR}$ and we are done.
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crazyeyemoody907
450 posts
#14 • 6 Y
Y by v4913, strong_boy, Mogmog8, Dukejukem, GuvercinciHoca, Rounak_iitr
i3435 wrote:
('cause every night I lie in bed...) spamming ddit fills my head... (geometry is keeping me awake...)
[asy] 
//21SLG8
//setup 
size(13cm); 
pen blu,grn,blu1,blu2,lightpurple; blu=RGB(102,153,255); grn=RGB(0,204,0); blu1=RGB(233,242,255); blu2=RGB(212,227,255); lightpurple=RGB(234,218,255);// blu1 lighter
//defn 
pair A,B1,C1,Ia,D,D1; A=(14.92,11.07); B1=(0,0); C1=(14,0); Ia=incenter(A,B1,C1); D=foot(Ia,B1,C1); D1=2*Ia-D; 
pair B,C; B=extension(A,B1,D1,D1+(1,0)); C=extension(A,C1,D1,D1+(1,0)); path W,wa; W=circumcircle(A,B,C); wa=incircle(A,B1,C1); 
pair X,R,V,P; X=intersectionpoints(W,wa)[1]; R=foot(A,B1,C1); 
V= extension(X,X+rotate(90)*(X-Ia),B1,C1); P=foot(A,X,V); 
pair U,X1,V1; U=extension(X,V,B,C); X1=extension(A,X,B,C); V1=extension(A,V,B,C); 
//draw 
filldraw(A--B1--C1--cycle,blu1,blu); draw(C1--V^^B--V1,blu); draw(wa,blu); draw(A--R,dashed+blu); 
draw(W,purple); draw(circumcircle(A,P,X),magenta+dashdotted); draw(circumcircle(A,X,V),red+dotted); 
draw(X--A--V,red); draw(R--P--V,magenta); 
clip((-2,-2)--(-2,15)--(16,15)--(22,-2)--cycle);
//label 
label("$A$",A,dir(90)); label("$B$",B,dir(90)); label("$C$",C,-dir(40)); label("$B'$",B1,-dir(90)); label("$C'$",C1,-dir(90)); label("$X$",X,dir(-70)); label("$D$",D,dir(-90)); label("$R$",R,dir(-90)); label("$V$",V,dir(-90)); label("$P$",P,dir(50)); label("$U$",U,dir(60));
label("$V'$",V1,dir(10));label("$X'$",X1,dir(110)); 
label("$\omega$",(10.5,11.2),purple); 
label("$\omega_a$",(8.31,2.73),blu); 
[/asy]
Let the antipode of the $A$-extouch point be $D$; let the tangent to $\omega_a$ at $D$ intersect $\overline{AB},\overline{AC}$ at $B',C'$ respectively. Let line $x$ be tangent to $\omega_a$ at $X$, $U=x\cap\overline{BC}$, and $V=x\cap\overline{B'C'}$. Finally, let $X'=\overline{AX}\cap\overline{BC}$, $V'=\overline{AV}\cap\overline{BC}$.
Claim 1: $AXUV'$ cyclic.
Proof. Apply DDIT to $A$, $UXV\infty_{BC}$ with inconic $\omega_a$, and project onto $\overline{BC}$, to obtain the involution $(BC;UV';\infty_{BC}X')$-- or equivalently, $X'B\cdot X'C=X'U\cdot X'V'$. By power of a point, $X'B\cdot X'C=X'A\cdot X'X$, so the claim follows from power of a point converse on $X'U\cdot X'V=X'A\cdot X'X$. $\qquad\qquad\square$

Claim 2: $\overline{DV}$ is tangent to $(AXV)$.
Proof.Angle chase using previous claim, and the fact that $\overline{BC}\parallel \overline{B'C'}$: \[\measuredangle XAV \overset{\text{claim }1}= \measuredangle XUV'=\measuredangle XVD.\qquad\qquad\square\]Redefine $R$ as the foot from $A$ to $\overline{B'C'}$. It remains to show,
Claim 3: $\overline{PR}$ touches $(APX')$.
Proof.Since $\angle VPA=\angle VRA=90^\circ$, $APRV$ cyclic, so we may anglechase as follows:
\[\measuredangle APR=\measuredangle AVR \overset{\text{claim }2} =\measuredangle AXV=\measuredangle AXP.\qquad\qquad\square\]
This post has been edited 9 times. Last edited by crazyeyemoody907, Aug 16, 2024, 5:54 AM
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Dukejukem
695 posts
#17 • 1 Y
Y by crazyeyemoody907
Let $\ell \ne BC$ be the line parallel to $BC$ and tangent to $\Omega_A$. The tangent line to $\Omega_A$ at $X$ meets $\ell$ at $S$ and meets $\omega$ for a second time at $K$. By Poncelet's porism, there is a common tangent line to $\omega, \Omega_A$ through $K$. Let this common tangent meet $\ell$ at $T$.

[asy]
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filldraw((11.070163074164075,13.579393494621243)--(9.85222486650305,9.002512170668734)--(15.403417278955729,9.002512170668734)--cycle, evevev, linewidth(1.2)); 
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draw((9.85222486650305,9.002512170668734)--(15.403417278955729,9.002512170668734), linewidth(1.2)); 
draw(circle((13.411126821667693,4.372710352455849), 4.629801818212886), linewidth(1.2)); 
draw(circle((12.62782107272939,10.714399117792942), 3.261056764264046), linewidth(1.2)); 
draw((15.403417278955729,9.002512170668734)--(15.159988564460608,8.659497900319954), linewidth(1.2)); 
draw((11.070163074164075,13.579393494621243)--(11.070163074164139,-0.25709146575662056), linewidth(1.2) + dotted); 
draw((11.070163074164075,13.579393494621243)--(13.541896000323954,10.211634652335071), linewidth(1.2) + dotted); 
draw((11.070163074164075,13.579393494621243)--(14.633491267851694,14.834954496727654), linewidth(1.2) + dotted); 
draw((11.070163074164139,-0.25709146575662056)--(14.633491267851694,14.834954496727654), linewidth(1.2) + qqzzqq); 
draw((7.1695628133431075,9.002512170668735)--(9.85222486650305,9.002512170668734), linewidth(1.2)); 
draw((15.403417278955729,9.002512170668734)--(16.68858884337307,9.002512170668735), linewidth(1.2)); 
draw((11.070163074164075,13.579393494621243)--(-0.7218396074105807,-0.25709146575704056), linewidth(1.2)); 
draw((14.633491267851694,14.834954496727654)--(19.951268167004887,-0.2570914657570312), linewidth(1.2)); 
draw((-0.7218396074105807,-0.25709146575704056)--(19.951268167004887,-0.2570914657570312), linewidth(1.2)); 
draw((11.070163074164075,13.579393494621243)--(8.93702727489717,5.563297653545161), linewidth(1.2) + blue); 
draw((11.070163074164075,13.579393494621243)--(16.773148093569667,7.555771433114837), linewidth(1.2) + blue); 
draw((-0.7218396074105807,-0.25709146575704056)--(15.703530280369414,11.798144516763893), linewidth(1.2) + red); 
 /* dots and labels */
dot((11.070163074164075,13.579393494621243),linewidth(4.pt) + dotstyle); 
label("$A$", (10.829860095768108,13.798104606630401), N * labelscalefactor); 
dot((9.85222486650305,9.002512170668734),linewidth(4.pt) + dotstyle); 
label("$B$", (9.314865562254608,8.395576930516235), N * labelscalefactor); 
dot((15.403417278955729,9.002512170668734),linewidth(4.pt) + dotstyle); 
label("$C$", (15.10333488119387,9.367460216166403), NE * labelscalefactor); 
label("$\Omega_A$", (8.544480158746174,2.10972890532501), N * labelscalefactor); 
label("$\omega$", (9.029392007598816,12.140654148690581), NE * labelscalefactor); 
dot((10.671757998305733,8.105125565896055),linewidth(4.pt) + dotstyle); 
label("$X$", (10.429672860500391,7.423693644866068), NE * labelscalefactor); 
dot((13.541896000323954,10.211634652335071),linewidth(4.pt) + dotstyle); 
label("$P$", (13.859849162795108,9.681893043876752), N * labelscalefactor); 
dot((11.070163074164139,-0.25709146575662056),linewidth(4.pt) + dotstyle); 
label("$R$", (10.829860095768108,-0.8944838881986001), NE * labelscalefactor); 
dot((-0.7218396074105807,-0.25709146575704056),linewidth(4.pt) + dotstyle); 
label("$S$", (-1.004248147148664,-0.54881459493244466), SE * labelscalefactor); 
dot((15.703530280369414,11.798144516763893),linewidth(4.pt) + dotstyle); 
label("$K$", (15.717861326538078,12.083016455483047), E * labelscalefactor); 
dot((19.951268167004887,-0.2570914657570312),linewidth(4.pt) + dotstyle); 
label("$T$", (20.119920914482964,-1.0088230982750905), NE * labelscalefactor); 
dot((7.1695628133431075,9.002512170668735),linewidth(4.pt) + dotstyle); 
label("$S'$", (6.799402940571816,9.138781796013422), NE * labelscalefactor); 
dot((16.68858884337307,9.002512170668735),linewidth(4.pt) + dotstyle); 
label("$T'$", (16.804083822264737,9.167366598532544), E * labelscalefactor); 
dot((14.633491267851694,14.834954496727654),linewidth(4.pt) + dotstyle); 
label("$H$", (14.745978040887907,14.99866631243355), NE * labelscalefactor); 
label("$\ell$", (9.572128784926711,-0.8658990856794776), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]

Claim 1: Quadrilateral $AKST$ is cyclic.
Proof. Let lines $AS, KT$ meet line $BC$ at $S', T'$. Since $\ell \parallel BC$, it suffices to show that quadrilateral $AKS'T'$ is cyclic.

By the dual of Desargues' involution theorem for quadrilateral $SKT'\infty_{BC}$ with inconic $\Omega_A$, there is an involution swapping $(AB, AC), (AS, AT'), (AK, A\infty_{BC})$. Projecting onto line $BC$, we obtain an involution swapping $(B, C), (S', T'), (Z, \infty_{BC})$, where $Z := AK \cap BC$. Since $Z$ is swapped with the point at infinity, this involution is an inversion with pole $Z$. We conclude using power of a point: \[ ZS' \cdot ZT' = ZB \cdot ZC = ZA \cdot ZK. \quad \blacksquare \]
Claim 2: The Simson line of $A$ w.r.t. $\triangle KST$ is tangent to $\odot(APX)$ at $P$.
Proof. Let $H$ be the projection of $A$ onto line $KT$ (so the Simson line is $PH$). Since $APKH$ is cyclic (with diameter $\overline{AK}$) and line $KH$ is tangent to $\omega$, we obtain \[ \measuredangle HPA = \measuredangle HKA = \measuredangle KXA = \measuredangle PXA. \quad \blacksquare \]
We conclude by observing that the projection $R'$ of $A$ onto $\ell$ lies on the Simson line of $A$ w.r.t. $\triangle KST$. By claim 2, line $R'P$ is tangent to $\odot(APX)$. Similarly, line $R'Q$ is tangent to $\odot(AQY)$. Therefore, $R' = R$, and $AR \perp BC$.
This post has been edited 2 times. Last edited by Dukejukem, Dec 19, 2023, 3:30 PM
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MathLuis
1556 posts
#18 • 1 Y
Y by crazyeyemoody907
Let $\Omega_A$ hit $BC$ at $Z$. $ZZ'$ the diameter in $\Omega_A$, the tangent from $Z'$, hit $AB, AC$ at $B_1, C_1$ respectivily, let the tangent from $X$ to $\Omega_A$ hit $BC, \omega, B_1C_1$ at $K, X', G_X$ respectivily, let $AG_X$ hit $BC$ at $G$, redefine $R$ as a point in $B_1C_1$ such that $AR \perp BC$, by poncelet porism $X'$ lies in one of the common tangents of $\omega, \Omega_A$ so let this mentioned line hit $\Omega_A$ at $T_X$ and $B_1C_1$ at $P_X$, let $AB, AC$ hit $\Omega_A$ at $E,F$.
Now by DDIT over $G_XXK \infty_{BC}$ we get that $(AG_X, AK), (A \infty_{BC}, AX), (AE, AF)$ are pairs of involution, projecting onto $BC$ we get $(G, K), (AX \cap BC, \infty_{BC}), (B, C)$ are pairs of involution so by PoP over the center of (negative) inversion which turns out to be $AX \cap BC$ we get $AGXK$ cyclic but by reims over $(AGXK), (AXG_X)$ we get that $(AXG_X)$ is tangent to $B_1C_1$ at $G_X$ so now there is a spiral sim centered at $A$ sending $XX' \to G_XP_X$ so $AX'P_XG_X$ is cyclic, now let $(APX') \cap X'P_X=P'$ by angle chase:
$$\angle APP'=\angle AX'P'=\angle AG_XP_X=\angle AXK \implies PP' \; \text{tangent to} \; (APX)$$But by simson line over $\triangle X'P_XG_X$ we get that $P',P,R$ are colinear so $PR$ is tangent to $(APX)$, and in a symetric way u can also get $QR$ tangent to $(AQY)$ hence $AR \perp BC$ (in original problem) as desired!.
This post has been edited 1 time. Last edited by MathLuis, May 4, 2023, 10:36 PM
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alinazarboland
172 posts
#19
Y by
Fix $\Omega_A , \omega$ and let triang;e $\triangle ABC$ move along $\omega$. By poncelet's porism $\Omega_A$ remains unchanged.It's well-known that the tangent lines to $\Omega_A$ at $X$ intersects $\omega$ at the common tangents of $\omega$,$\Omega_A$ (or we can say this by Poncelet's again).Now , for an arbitrary point $A$ on $\omega$ , let the perpendiculars through $A$ to those common tangents intersect them at $D',E'$ and tangent line at $P$ to the circumcircle of the triangle $APX$ intersects it at $D$ and define $E$ for $Q$ similarly. by simple angle chasing $(APD)$ passes through the intersection of common tangents of $\omega , \Omega_A$ and $\omega$. So by the definition of $P$ , $D'=D$ and $E'=E$ similarly. So $R=PD \cap QE$. Now we'll use degree bounds . $A$ moves along $\omega$ with degree 2 but since $A=D$ for some case $D$ has degree on. similar for $P$. So $PD$ has degree 2 and $R$ has degree 4 and perpendicular from $A$ to $BC$ has degree 2 since both moves with degree at most two 2 becouse the foot of perpendicular coinsides at some cases. So it's enough to check 5 cases for triangle $ABC$ and there's in fact 6 degenerated cases for the triangle and we're done.
This post has been edited 1 time. Last edited by alinazarboland, May 8, 2023, 12:45 PM
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Marinchoo
407 posts
#20 • 1 Y
Y by VicKmath7
One of the hybrid solutions of all time.

Let $\ell\neq\overline{BC}$ be the line parallel to $\overline{BC}$ and tangent to $\Omega_{A}$. We'll show that $R$ is the foot of the perpendicular from $A$ onto $\ell$ (this can be noticed from a large, well-drawn diagram). In order to do this, redefine $R$ as said foot and we'll show that $\overline{RP}$ is tangent to the circumcircle of $\triangle APX$ (the other tangency would follow analogously).

Let $S = \overline{PX} \cap \overline{AB}$, $T = \overline{PX} \cap \ell$, $V = \overline{AT} \cap \overline{CX}$. Then $\angle ART = 90^{\circ} = \angle APT$, so $TRPA$ is cyclic. Hence $\overline{RP}$ being tangent to the circumcircle of $\triangle APX$ is equivalent to:
\[\angle PAX = \angle RPX = \angle RPT = \angle RAT \iff \triangle TAR \sim \triangle XAP\]The last is equivalent to $\angle RTA = \angle PXA$. However, this holds if $\overline{VS} \parallel \ell$. Indeed, if $\overline{VS} \parallel \ell$, then
\[\angle SAX = \angle BAX = \angle BCX = \angle XVS\]which implies that $AXSV$ is cyclic and moreover, $\angle PXA = \angle SVA = \angle RTA$ as desired. Furthermore, $\overline{VS} \parallel \ell$ is equivalent to $\frac{AV}{VT} = \frac{AS}{SW}$. However, point $V$ is not particularly nice to work with. Let $\overline{TS} \cap \overline{AC} = L$, $\overline{AS} \cap \ell = W$, and let $\Omega_{A}$ touch $\overline{BC}$, $\ell$, $\overline{AB}$, and $\overline{AC}$ at points $D, E, F, G$ respectively. By Menelaus's theorem applied on $\triangle ATL$ and line $V, X, C$, we get that:
\[\frac{AV}{VT} = \frac{XL}{TX} \cdot \frac{CA}{LC}\]Now in order to prove $\frac{AS}{SW} = \frac{XL}{TX} \cdot \frac{CA}{LC}$ and be done, we employ complex numbers with $\Omega_{A}$ as the unit circle. Let $D = 1$, $E = -1$, and $b, c, x$ denote the complex numbers of points $F, G, X$, respectively. Then we get:
\[A = \frac{2bc}{b+c}, \quad L = \frac{2cx}{c+x}, \quad C = \frac{2c}{c+1}, \quad S = \frac{2bx}{b+x}, \quad T = \frac{2x}{1-x}, \quad W = \frac{2b}{1-b}\]so all the points we need are easily computed and now we just have:
\[|A-S| = \left|\frac{2bc}{b+c} - \frac{2bx}{b+x} \right| = \frac{2|c-x|}{|b+c|\cdot |b+x|}\]\[|S-W| = \left|\frac{2bx}{b+x} - \frac{2b}{1-b}\right| = \frac{2|x+1|}{|b+x| \cdot |1-b|}\]\[|X-L| = \left| x - \frac{2cx}{c+x}\right| = \frac{|x-c|}{|c+x|}\]\[|T-X| = \left|\frac{2x}{1-x} - x\right| = \frac{|x+1|}{|1-x|}\]\[|C-A| = \left|\frac{2bc}{b+c} - \frac{2c}{c+1}\right| = \frac{2|b-1|}{|b+c| \cdot |c+1|}\]\[|L-C| = \left|\frac{2cx}{c+x} - \frac{2c}{c+1}\right| = \frac{2|x-1|}{|c+x|\cdot |c+1|}\]Combining all of the above, we get that:
\[\frac{|A-S|}{|S-W|} = \frac{|x-c|\cdot |b-1|}{|b+c| \cdot |x+1|} = \frac{|X-L|}{|T-X|}\cdot \frac{|C-A|}{|L-C|} = \frac{|A-V|}{|V-T|}\]therefore $\overline{VS} \parallel \ell$, as desired.
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pacelaaa
1 post
#21
Y by
Complex bash
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Leo.Euler
577 posts
#22
Y by
We begin with some line redefinition. Let $M$ and $N$ be the touchpoint of the external tangents of $\omega$ and $\Omega_A$ closer to $C$ and $B$, respectively. Then redefine the ``tangent lines to $\Omega_A$ at $X$ and $Y$" as the lines $\overline{XM}$ and $\overline{YN}$, respectively. This works by Poncelet's porism.

Let $B_0$ be on $\overline{AB}$ and $C_0$ on $\overline{AC}$ such that $B_0C_0$ is parallel to $BC$ and $\Omega_A$ is the incircle of $BCC_0B_0$. Let $S$ be the intersection of lines $\overline{XM}$ and $\overline{BC}$.

Claim: It suffices to prove that $(ASX)$ is tangent to $\overline{B_0C_0}$.
Proof. Since $(APR'S)$ is cyclic, $\angle ASX= \angle ASR = \angle AR'P$. Realize that $(AXP)$ and $(ASR')$ intersect at $S=\overline{XP} \cap \overline{SR'}$, so $A$ is the center of the spiral similarity mapping $XP$ to $SR'$. Thus, \[ \angle SAX = \angle XSD' = \angle PSR' = \angle R'AP, \]which implies that $(ASX)$ is tangent to $\overline{B_0C_0}$. Since all steps above are reversible, the claim is true.
:yoda:

Let $S' = \overline{AS} \cap \overline{BC}$ and $S_0' = \overline{XM} \cap \overline{BC}$.

Claim: $A$, $X$, $S'$, and $S_0'$ are concyclic.
Proof. Let $X' = \overline{AX} \cap \overline{BC}$. Apply DDIT on $A$ with respect to $SXS_0'\infty_{BC}$, so that $(AB, AC)$, $(AS', AS_0')$, and $(AX', A\infty_{BC})$ are pairs of an involution. By Desargues assistant theorem, this involution must be an inversion (negative or positive) centered at $X'$. Since $B$ and $C$ are swapped by this inversion, it is a negative inversion with radius $-\sqrt{BX' \cdot CX'} = -\sqrt{XX' \cdot AX'}$. Since $S'$ and $S_0'$ are swapped, the desired concyclicity follows by power of a point.
:yoda:

Since $\overline{B_0C_0} \parallel \overline{BC}$, it follows by Reim's theorem that $\overline{B_0C_0}$ is tangent to $(ASX)$, as desired.
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math_comb01
662 posts
#23
Y by
Very Instructive Problem, so I will post my solution although same as others. Apparently, I am not used to to using Poncelet Porism again and again lol that's why this took lot of time.
Let The line parallel to $BC$ and tangent to excircle intersect $AB,AC$ at $B_1,C_1$, Let the tangent at $X$ intersect $(ABC)$ at $V$ and $B_1C_1$ at $S$. Redefine $R$ to be $AR \perp B_1C_1$ s.t. $R \in B_1C_1$. Note that it suffices to prove that $AXS$ is tangent to $B_1C_1$. Now before proceeding further we prove the following lemma:
Lemma: $V$ is the point of tangency of tangents between circumcircle and excircle, $V \in (ABC)$.
Proof
Now let the other tangent from $V$ to exircle meet $B_1C_1$ at $T$. Notice that we can restate the converse of problem as follows:
Converse of ISL 2021 G8 wrote:
Let $SVT$ be a triangle with incircle $\Omega_A$ and $T$-intouch point $X$. Let the circle through $V$ and $X$ tangent to $\overline{ST}$ intersect the circumcircle of $SVT$ at $A$. Let the tangents from $A$ to $\Omega_A$ intersect the circumcircle of $AVX$ at $B \ne A$ and $C \ne A$. Prove that $\overline{BC} \parallel \overline{ST}$.
It is easy to see by miquel there is a unique position for $B_1C_1$, hence this implies the orignal problem.
Now, the idea is to force poncelet type configuration and apply DDIT.
Let $AB,AC$ meet $(SVT)$ at $D,E$ then by Poncelet's porism $(ADE)$ has incircle as $\omega_A$ Notice that the angle bisectors of $BVE$ and $CVD$ coincide. Consider a line $\ell_1$ perpendicular to the angle bisector of $\angle BUE$, and let $VB$, $VC$, $VD$, $VE$, $VX$, and $VX'$ intersect $\ell_1$ at $B^*$, $C^*$, $D^*$, $E^*$, $X^*$, and $X'^*$, respectively. By symmetry, $B^*C^*=D^*E^*$ in directed lengths, so the midpoints of $B^*E^*$ and $C^*D^*$ coincide at a point we call $M$. By DDIT,there exists an involution sending $B^*$ to $E^*$, $C^*$ to $D^*$, and $X^*$ to $X'^*$. THis must be reflection over mispoint of $B*E*$, Now let $VB$ intersect $(SVT)$ at $B'$, so $B'E  \parallel \ell$. By reims we're done.
Remark: This is a very projective type of problem where we need to extend lines to get poncelet type config, this often leads us to getting symmetry which makes DDIT a good option.
Sidenote
This post has been edited 3 times. Last edited by math_comb01, Jan 27, 2024, 2:59 PM
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BlizzardWizard
108 posts
#24 • 1 Y
Y by EthanWYX2009
The complex bash 3 posts above is neat, but it contains synthetic steps (including guessing the characterization of $R$). Here's a bash with no synthetic steps whatsoever, which requires no insights except for a burning desire for terms to factor or cancel. The finish is a touch too tedious for my liking, though, and I would appreciate help finding a cleaner method!

Complex bash
This post has been edited 2 times. Last edited by BlizzardWizard, Mar 28, 2024, 10:56 PM
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BlizzardWizard
108 posts
#25 • 1 Y
Y by ihatemath123
I found a cleaner way to finish.

Complex bash
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dolphinday
1329 posts
#26
Y by
Let $B'C'$ be the points so that $B'C' \parallel BC$ and $\Omega_A$ is the incircle of $\triangle AB'C'$. It suffices to show that $PR$ is tangent to $(APX)$.
Then let $XP \cap RB = K$. Since $\angle APK = \angle ARK = 90^\circ$ we have $APRK$ cyclic.
Then let $X'$, $P'$, $K'$ be points $AX \cap BC$, $XP \cap BC$ and $AK \cap BC$. By DDIT on $P'XK\infty_{BC}$ from point $A$ gives that $(AP', AK)$, $(AX, A\infty_{BC)}$ and since $P'XK\infty_BC$ has incircle $\Omega_A$ we have that $(AB', AC')$ are involutive pairs as well. Projecting onto $BC$ gives us that $(P', K')$, $(X', \infty_BC)$, $(B, C)$ are involutive pairs. Then since the involution is a negative inversion, and since $(X, \infty_BC)$ get swapped we know that the inversion is centered at $X'$ which gives us that $X'P' \cdot X'K' = X'B \cdot X'C$. By PoP we have $X'B \cdot X'C = X'X \cdot X'A$ so $P'XK'A$ is cyclic. Note that $\measuredangle{APR} = \measuredangle{AKR} = \measuredangle{AK'B} = \measuredangle{AXP}$ which proves $RP$ tangent to $(AXP)$.
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This post has been edited 2 times. Last edited by dolphinday, Jul 27, 2024, 4:08 PM
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ohiorizzler1434
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#27
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dDIT RIZZ

We reverse reconstruct by starting with $R$ as the foot onto $B'C'$. Define the points obviously when not said. $P$ is the foot onto the tangent from $A$, $S$ the meeting with $B'C'$. It is enough by uniqueness to prove that $RP$ is a tangent. Then $APRS$ cyclic. Draw $T$ as well. We also observe that proving $(AXV)$ tangent to $B'C'$ finishes the problem after a quick angle chase.

Now take dDIT on the triangle with sides $ST$, $TC$, and $SD$ so the invoution gives pairs $A(\infty, X), (T, S), (B,C)$, which when projected onto line $BC$ gives that it is an inversion at $X'$, or that $AS'XT$ is cyclic. This implies that $\angle XAS' = \angle XTS' = \angle XSR$ which gives the tangency, and we are done.
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TestX01
341 posts
#28
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ok unc

Use the definitions in ohiorizzler's diagram. We claim that if $R$ is redefined as the foot of the perpendicular from $D$ to the altitude at $A$, then $RP$ is tangent to $(APX)$. Indeed, note that $APRS$ is cyclic from right angles. Now, by Alternate segment theorem, clearly we just wish to show that $\angle RAP=\angle SAX$.

Let $D'$ be the extouch with $BC$. Let $E,F$ be the extouch with $AC$ and $AB$ respectively. Brianchon on $DD'EFXX$ means that if $PX\cap AB$ is defined as $G$, then the parallel to $BC$ through $G$, $AS$, and $CX$ concur. Let this point be $H$.

Note that $AHGX$ is cyclic as
\[180^\circ-\angle HGA=\angle B=\angle AXC=180^\circ-\angle HXA\]By Bowtie and corresponding angles.

This means that $\angle SAX=180^\circ- \angle XGH$. Yet this is equivalent to showing that the angle formed by $BC$ and $XP$ same as angle formed by $AP$, $AR$. Easy to see by the right angles.
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bin_sherlo
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#31
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Let $J$ be $A-$excenter. Let $D,D'$ be the tangency point of $A-$excircle with $BC$ and its antipode, respectively. Let $D'D'\cap AB=B',D'D'\cap AC=C',XP\cap B'C'=U,UX\cap BC=S,UX\cap (ABC)=L,AU\cap (ABC)=K$
$AS\cap (ABC)=V,AD\cap (ABC)=E,AU\cap BC=U',JX\cap (ABC)=F$. Let $W\in B'C'$ and $AW\perp BC$.

Claim: $KL\parallel BC$.
Proof: By DDIT, since $(\overline{UX},\overline{UD'}),(\overline{UA},\overline{UD}),(\overline{UB},\overline{UC})$ is an involution thus, $(S,BC_{\infty}),(U',D),(B,C)$ is an involution which implies $SU'.SD=SB.SC$ Also $(\overline{AS},\overline{AS})(\overline{AX},\overline{AD}),(\overline{AB},\overline{AC})$ is an involution hence $VV,XE,BC$ concur. This yields $A,U',V,D$ are concyclic. Pascal at $VVLXEA$ gives $VV\cap XE,VL\cap EA,S$ are collinear so $V,D,L$ are collinear.
\[\measuredangle AXS=\measuredangle AXL=\measuredangle AVD=\measuredangle AU'D=\measuredangle AU'S\]Hence $A,U',X,S$ are concyclic. $\measuredangle XLK=\measuredangle XAK=\measuredangle XSU'$ which implies $KL\parallel BC$.

Note that $\measuredangle FKL=\measuredangle FXL=90$ thus, $FK\perp BC$. Since $A,P,W,U$ lie on the circle with diameter $AU$,
\[\measuredangle WPX=\measuredangle WAU=\measuredangle WAK=\measuredangle FKA=\measuredangle FXA=\measuredangle PAX\]Hence $(APX)$ and $AW$ are tangent to each other. Similarily $(AQY)$ and $AW$ are tangent so $R=W$ and $AW\perp BC$ as desired.$\blacksquare$
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