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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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.

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0 replies
jlacosta
Mar 2, 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
Problem 5
blug   0
5 minutes ago
Source: Polish Junior Math Olympiad Finals 2025
Each square on a 5×5 board contains an arrow pointing up, down, left, or right. Show that it is possible to remove exactly 20 arrows from this board so that no two of the remaining five arrows point to the same square.
0 replies
blug
5 minutes ago
0 replies
Diophantine equation with large moduli
Assassino9931   2
N 6 minutes ago by Assassino9931
Source: Bulgaria, Concours Generale Minko Balkanski 2024
Solve in positive integers $2^x - 23^y = 9$.
2 replies
Assassino9931
3 hours ago
Assassino9931
6 minutes ago
Problem 4
blug   0
6 minutes ago
Source: Polish Junior Math Olympiad Finals 2025
In a rhombus $ABCD$, angle $\angle ABC=100^{\circ}$. Point $P$ lies on $CD$ such that $\angle PBC=20^{\circ}$. Line parallel to $AD$ passing trough $P$ intersects $AC$ at $Q$. Prove that $BP=AQ$.
0 replies
blug
6 minutes ago
0 replies
Problem 3
blug   0
9 minutes ago
Source: Polish Junior Math Olympiad Finals 2025
Find all primes $(p, q, r)$ such that
$$pq+4=r^4.$$
0 replies
blug
9 minutes ago
0 replies
rows are DERANGED and a SOCOURGE to usajmo .
GrantStar   26
N Today at 6:00 AM by joshualiu315
Source: USAJMO 2024/4
Let $n \geq 3$ be an integer. Rowan and Colin play a game on an $n \times n$ grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is orderly if: [list] [*]no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and [*]no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. [/list] In terms of $n$, how many orderly colorings are there?

Proposed by Alec Sun
26 replies
GrantStar
Mar 21, 2024
joshualiu315
Today at 6:00 AM
Geo equals ABsurdly proBEMatic
ihatemath123   73
N Today at 5:38 AM by joshualiu315
Source: 2024 USAMO Problem 5, JMO Problem 6
Point $D$ is selected inside acute $\triangle ABC$ so that $\angle DAC = \angle ACB$ and $\angle BDC = 90^{\circ} + \angle BAC$. Point $E$ is chosen on ray $BD$ so that $AE = EC$. Let $M$ be the midpoint of $BC$.

Show that line $AB$ is tangent to the circumcircle of triangle $BEM$.

Proposed by Anton Trygub
73 replies
ihatemath123
Mar 21, 2024
joshualiu315
Today at 5:38 AM
average FE
KevinYang2.71   74
N Today at 4:55 AM by joshualiu315
Source: USAJMO 2024/5
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy
\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]for all $x,y\in\mathbb{R}$.

Proposed by Carl Schildkraut
74 replies
1 viewing
KevinYang2.71
Mar 21, 2024
joshualiu315
Today at 4:55 AM
Rip Red/Blue, Long live Amber/Bronze
AwesomeYRY   50
N Today at 3:35 AM by MathLuis
Source: USAMO 2022/1, JMO 2022/2
Let $a$ and $b$ be positive integers. The cells of an $(a+b+1)\times (a+b+1)$ grid are colored amber and bronze such that there are at least $a^2+ab-b$ amber cells and at least $b^2+ab-a$ bronze cells. Prove that it is possible to choose $a$ amber cells and $b$ bronze cells such that no two of the $a+b$ chosen cells lie in the same row or column.
50 replies
AwesomeYRY
Mar 24, 2022
MathLuis
Today at 3:35 AM
Erecting Rectangles
franchester   101
N Today at 3:12 AM by Ilikeminecraft
Source: 2021 USAMO Problem 1/2021 USAJMO Problem 2
Rectangles $BCC_1B_2,$ $CAA_1C_2,$ and $ABB_1A_2$ are erected outside an acute triangle $ABC.$ Suppose that \[\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.\]Prove that lines $B_1C_2,$ $C_1A_2,$ and $A_1B_2$ are concurrent.
101 replies
franchester
Apr 15, 2021
Ilikeminecraft
Today at 3:12 AM
2025 AMC 8 Problem
Kexinshi   8
N Today at 3:00 AM by CJB19
Source: 2025 AMC 8 Problem #15
Kei draws a $6$-by-$6$ grid. He colors $13$ of the unit squares silver and the remaining squares gold. Kei then folds the grid in half vertically, forming pairs of overlapping unit squares. Let $m$ and $M$ equal equal the least and greatest possible number of gold-on-gold pairs, respectively. What is the value of $m+M$?

$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 16\qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 20$

Had fun doing this one!
8 replies
Kexinshi
Jan 31, 2025
CJB19
Today at 3:00 AM
How to get better at AMC 10
Dream9   2
N Today at 2:01 AM by hashbrown2009
I'm nearly in high school now but only average like 75 on AMC 10 sadly. I want to get better so I'm doing like the first 11 questions of previous AMC 10's almost every day because I also did previous years for AMC 8. Is there any specific way to get better scores and understand more difficult problems past AMC 8? I have almost no trouble with AMC 8 problem given enough time (like 23-24 right with enough time).
2 replies
Dream9
Today at 1:17 AM
hashbrown2009
Today at 2:01 AM
have you done DCX-Russian?
GoodMorning   80
N Today at 1:23 AM by bjump
Source: 2023 USAJMO Problem 3
Consider an $n$-by-$n$ board of unit squares for some odd positive integer $n$. We say that a collection $C$ of identical dominoes is a maximal grid-aligned configuration on the board if $C$ consists of $(n^2-1)/2$ dominoes where each domino covers exactly two neighboring squares and the dominoes don't overlap: $C$ then covers all but one square on the board. We are allowed to slide (but not rotate) a domino on the board to cover the uncovered square, resulting in a new maximal grid-aligned configuration with another square uncovered. Let $k(C)$ be the number of distinct maximal grid-aligned configurations obtainable from $C$ by repeatedly sliding dominoes. Find the maximum value of $k(C)$ as a function of $n$.

Proposed by Holden Mui
80 replies
GoodMorning
Mar 23, 2023
bjump
Today at 1:23 AM
Stanford Math Tournament (SMT) Online 2025
stanford-math-tournament   5
N Today at 12:28 AM by stanford-math-tournament
[center]Register for Stanford Math Tournament (SMT) Online 2025[/center]


[center] :surf: Stanford Math Tournament (SMT) Online is happening on April 13, 2025! :surf:[/center]

[center]IMAGE[/center]

Register and learn more here:
https://www.stanfordmathtournament.com/competitions/smt-2025-online

When? The contest will take place April 13, 2025. The pre-contest puzzle hunt will take place on April 12, 2025 (optional, but highly encouraged!).

What? The competition features a Power, Team, Guts, General, and Subject (choose two of Algebra, Calculus, Discrete, Geometry) rounds.

Who? You!!!!! Students in high school or below, from anywhere in the world. Register in a team of 6-8 or as an individual.

Where? Online - compete from anywhere!

Check out our Instagram: https://www.instagram.com/stanfordmathtournament/

Register and learn more here:
https://www.stanfordmathtournament.com/competitions/smt-2025-online


[center]IMAGE[/center]


[center] :surf: :surf: :surf: :surf: :surf: [/center]
5 replies
stanford-math-tournament
Mar 9, 2025
stanford-math-tournament
Today at 12:28 AM
AIME Math History
hashbrown2009   82
N Yesterday at 11:35 PM by stjwyl
Idk why but I wanted to see how good ppl are
Post all your AIME scores ever (if you qualified for USA(J)MO, you may put that score, too)

(Note: Please do not post fake scores. I legit want to see how good ppl are and see how good I am)
I'll start:

5th grade: AIME : 2 lol
6th grade: AIME : 5
7th grade: AIME : 8
8th grade : AIME : 13 USAJMO: 18
9th grade (rn): AIME: 11 (sold)
82 replies
hashbrown2009
Feb 20, 2025
stjwyl
Yesterday at 11:35 PM
An I for an I
Eyed   65
N Today at 12:51 AM by joshualiu315
Source: 2020 ISL G8
Let $ABC$ be a triangle with incenter $I$ and circumcircle $\Gamma$. Circles $\omega_{B}$ passing through $B$ and $\omega_{C}$ passing through $C$ are tangent at $I$. Let $\omega_{B}$ meet minor arc $AB$ of $\Gamma$ at $P$ and $AB$ at $M\neq B$, and let $\omega_{C}$ meet minor arc $AC$ of $\Gamma$ at $Q$ and $AC$ at $N\neq C$. Rays $PM$ and $QN$ meet at $X$. Let $Y$ be a point such that $YB$ is tangent to $\omega_{B}$ and $YC$ is tangent to $\omega_{C}$.

Show that $A,X,Y$ are collinear.
65 replies
Eyed
Jul 20, 2021
joshualiu315
Today at 12:51 AM
An I for an I
G H J
Source: 2020 ISL G8
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
dolphinday
1310 posts
#57
Y by
Let $K$ be the intersection point of the tangent at $I$ and $\Gamma$. Then $\angle BIC = \angle BIK + \angle CIK = \angle IMB + \angle CNI = \angle YCI + \angle YBI$. Note that $\angle BIC = 90^{\circ} + \frac{\angle A}{2}$ so $\angle BYC = 360^{\circ} - (180^{\circ} - \angle A) = 180^{\circ} - \angle A \implies ABYC$ is cyclic.
Notice that $MNC_1B_1$ is an isosceles trapezoid due to the fact that the perpendicular bisectors of $MB_1$ and $NC_1$ coincide. Note that $PBCQ$ is cyclic.
We can then prove that $\triangle XMN \sim \triangle XQP$. $\angle XNM = 180^{\circ} - (\angle MNC_1 + \angle C_1NQ) = 180^{\circ} - (\angle MBB_1 + 180^{\circ} - \angle C_1CQ)$ $= \angle BPM + \angle C_1CQ - 180^{\circ} = \angle BPM - \angle BPQ = \angle XPQ$.
By using similar triangle ratios, we find that $XM \cdot XP = XN \cdot XQ$, so $AX$ is the radical axis of $\omega_B$ and $\omega_C$. Similarly, we have $XN \cdot XQ = XM \cdot XP \implies AX$ is the radical axis of $(AMP)$ and $(ANP)$. All that is left is to show that $AY$ is the radical axis of $(AMP)$ and $(ANP)$ which is equivalent to $AY$ tangent to both circle. Notice that $PAYB$ is cyclic. Then $\angle PMB = 180^{\circ} - PBY = \angle PAY$, which proves the tangency, so we are done.
This post has been edited 1 time. Last edited by dolphinday, Feb 6, 2024, 7:04 PM
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jp62
53 posts
#58
Y by
Angle chase :sleeping:

We use directed angles mod $180^\circ$ throughout this solution.

ABCY is cyclic
I is the incenter of BCNM
(AMP) and (ANQ) are tangent at A
PMNQ is cyclic
AXY collinear

:wallbash: mfw g8 is just a massive chase
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awesomeming327.
1664 posts
#59
Y by
https://cdn.aops.com/images/a/6/b/a6b5883bce870c8c62780cb1b6b03ad25f549101.png
Take an inversion about I.

In the inverted image, we have $I$ is the orthocenter of $\triangle ABC$, $B,P,M$ collinear and $C,Q,N$ collinear. Note that \[\angle APM=\angle ACB=180^\circ-\angle AIB=\angle AMP\]so $AP=AM$ and similarly $AQ=AN$. This implies that $PMNQ$ is an isosceles trapezoid and $(MAP)$ is tangent to $(NAQ)$. In our uninverted image, note that since $PMNQ$ is cyclic, $X$ is on the radical axis of $\omega_B$ and $\omega_C$. Furthermore, it is on the radical axis of $(MAP)$ and $(NAQ)$. It is also on the common tangents from $I$ to $\omega_B$ and $\omega_C$ and from $A$ to $(MAP)$ and $(NAQ)$.
Note that
\[\angle BIC+\angle MIN=180^\circ-\angle IBC-\angle ICB+\angle MIX+\angle XIN=180^\circ-\angle IBC-\angle ICB+\angle IBM+\angle ICN=180^\circ\]so $\angle BIM+\angle CIN = 180^\circ\implies \angle BPM+\angle CQN=180^\circ\implies \angle ABY+\angle ACY=180^\circ$, so $Y$ is on $(ABC)$.
Furthermore, \[\angle AQN=\angle AQC-\angle NQC=180^\circ-\angle AYC-\angle ACY=\angle YAC\]so $YA$ is tangent to $(NAQ)$. Similarly, it is tangent to $(MAP)$ and so $Y$ is also on the common tangent from $A$ to $(MAP)$ and $(NAQ)$. We are done.
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awesomehuman
496 posts
#60 • 2 Y
Y by OronSH, dolphinday
Invert about $I$ and rename $I$ to $H$ to get the following equivalent statement:

Let $\triangle ABC$ have orthocenter $H$.
Let $\ell_B$, $\ell_C$ be paralell lines through $B$ and $C$, respectively.
Let $P$ and $Q$ be the intersections of
$\ell_B$ and $\ell_C$ with $(ABC)$, respectively.
Let $M$ and $N$ be the intersections of
$\ell_B$ and $\ell_C$ with $(AHB)$ and $(AHC)$, respectively.
Let $X$ be the intersection of $(HPM)$ and $(HQN)$.
Let $Y$ be the second intersection of the circle through $H$ tangent to $\ell_B$ at $B$ and the circle through $H$ tangent to $\ell_C$ at $C$.
Then, $AHYX$ is cyclic.

We will prove this statement instead of the original.
Let $H'$ be the reflection of $H$ over the angle bisector of $BC$ and $PQ$.
Then, $\triangle BHC \cong PH'Q$.

Claim:
$X$ is the intersection of $HH'$ with $(H'PQ)$.

Proof:
We have
\[\angle PXQ = \angle PXH + \angle HXQ = \angle BMH + \angle HNC\]\[= \angle BAH + \angle HAC = \angle BAC = \angle CHB = \angle PH'Q.\]Thus, $X$ lies on $(PH'Q)$.
We have
\[\angle PXH = \angle PMH = \angle BMH = \angle BAH = \angle HCB = \angle PQH' = \angle PXH'.\]Thus, $X$ is on $HH'$.

Claim:
$A$ is the orthocenter of $\triangle PXQ$.

Proof:
Let $D$ be the intersection of $PQ$ and $HH'$.
We have
\[\angle APQ = \angle ACQ = \angle BCQ + \angle ACB\]\[\angle PQX = \angle DXQ + \angle QDX = \angle HCB + \angle QCB.\]Thus,
\[\angle APQ + \angle PQX = \angle BCQ + \angle QCB + \angle ACB + \angle HCB = 90.\]Repeating the same logic on the other side, $A$ is the orthocenter of $\triangle PXQ$.

Claim:
$AHYX$ is cyclic

Proof:
We have
\[\angle AXH = \angle AXP + \angle PXH' = \angle PQA + \angle PQH'\]\[= \angle PBA + \angle HCB = \angle PBC + \angle CBA + \angle HCB = \angle PBC + 90\]\[= \angle PBC + \angle CBH + \angle ACB = \angle PBH + \angle ACB = \angle BYH + \angle AYB\]\[= \angle AYH.\]
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kiemsibongtoi
25 posts
#62
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Eyed wrote:
Let $ABC$ be a triangle with incenter $I$ and circumcircle $\Gamma$. Circles $\omega_{B}$ passing through $B$ and $\omega_{C}$ passing through $C$ are tangent at $I$. Let $\omega_{B}$ meet minor arc $AB$ of $\Gamma$ at $P$ and $AB$ at $M\neq B$, and let $\omega_{C}$ meet minor arc $AC$ of $\Gamma$ at $Q$ and $AC$ at $N\neq C$. Rays $PM$ and $QN$ meet at $X$. Let $Y$ be a point such that $YB$ is tangent to $\omega_{B}$ and $YC$ is tangent to $\omega_{C}$.

Show that $A,X,Y$ are collinear.

We'll prove this problem follows these claims :
$\textbf{1/}$ Let $B'$ be the intersection of line $BI$ with $\omega_c$ ($B' \neq I$); $C'$ be the intersection of line $CI$ with $\omega_b$ ($C' \neq I$)
$\hspace{1cm}$$M'$ be the intersection of lines $C'M$ and $AC$; $N'$ be the intersection of lines $B'N$ and $AB$
Then $N'MM'N$ is a isosceles trapezoid and $I$ lie on circumcircle of $N'MM'N$
$\textit{Prove :}$
$\, \,$Cuz $M$, $I$, $B$, $C$ are concyclic and $I$ is incenter of $\triangle ABC$, we see that :
$$ \angle AM'M = \angle MC'C + \angle ACC' = \angle IBC + \angle ICB = \angle C'IB = \angle C'MB $$$\, \,$So $\triangle AMM'$ is a isosceles triangle at $A$. Similar, $\triangle ANN'$ is a isosceles triangle at $A$
$\, \,$Therefore, $N'MM'N$ is a isosceles trapezoid
$\, \,$Next, let $\ell$ be the line which tangent both $\omega_b$, $\omega_c$ at $I$, we see that :
$$ \angle MIN = \angle MI\ell + \angle NI\ell = \angle MBI + \angle NCI = \angle IBC + \angle ICB = \angle AM'M $$$\, \,$Therefore, $M'$ lies on circle $(IMN)$. Similar, $N'$ lies on circle $(IMN)$.
$\textbf{2/}$ $M$, $N$, $P$, $Q$ are concyclic
$\textit{Prove :}$
$\, \,$Let $\omega_b$, $\omega_c$ meet line $BC$ at $T$, $L$ respectively ($T \neq B$, $L \neq C$)
$\hspace{1cm}$$O_b$, $O_c$ be center of $\omega_b$, $\omega_c$ respectively
$\, \,$Cuz $CI$ is the internal bisector of $\angle NCL$ so $O_cI \perp NL$. Similar, $O_bI \perp MT$
$\, \,$But $O_b$, $I$, $O_c$ are colibnear, so $NL$ $\|$ $MT$
$\, \,$Combine with claim $\textbf{1/}$, we see that :
$$\angle N'IN = 2\angle NIA = 2(\angle AIC - \angle NIC) = 180^\circ + \angle BAC - 2\angle NIC = \angle BAC + \angle NQC - \angle NLC = \angle NQC - \angle BMT$$$\, \,$So
$$ \angle PMN = \angle PMA + \angle AMN = \angle PC'B + \angle N'IN = \angle PBT + \angle NQC = 180^\circ - (\angle PQC - \angle NQC) = 180^\circ - \angle PQN$$$\, \,$Therefore, $M$, $N$, $P$, $Q$ are concyclic
$\textbf{3/}$ Let $V$ be the midpoint of arc $BAC$ of $\Gamma$
Then $V$ is the intersection of lines $B'Q$, $C'P$
$\textit{Prove :}$
$\, \,$Cuz $Q$ is the second intersection of $\omega_c$ and $\Gamma$ so $\angle B'QC = \angle BIC = 90^\circ + \frac12 \angle BAC = \angle VQC$
$\, \,$Lead to $V$, $B'$, $Q$ are colinear. Similar, $P$, $C'$, $P$ are colinear
$\textbf{4/}$$\triangle AXI$ is a isosceles triangle at $X$
$\textit{Prove :}$
$\, \,$Let $V_b$, $V_c$ be the intersections of lines $BI$, $CI$ with $\Gamma$ respectively ($V_b \neq B$, $V_c \neq C$)
$\hspace{0.7cm}$$U_b$ be the intersection of lines $V_bQ$ and $B'N$
$\, \,$We ez to see that $V_bV_c$ is the perpendicular bisector of $AI$, so we just need to prove that $V_b$, $X$, $V_c$ are colinear
$\, \,$Cuz $M$, $N$, $P$, $Q$ are concyclic from $\textbf{2/}$, so $X$ lie on radical axis of $\omega_b$ and $\omega_c$
$\, \,$Which means $XI$ tangent with both $\omega_b$, $\omega_c$ at $I$
$\, \,$Lead to $\triangle XNI \sim \triangle XIQ$. As a result : $\dfrac{XN}{XQ} = \dfrac{XN}{XI} .\dfrac{XI}{XQ} = \left( \dfrac{IN}{IQ} \right)^2 $
$\, \,$In the order hand :
$\hspace{0.5cm}$From claim $\textbf{3/}$ and $NB' \perp AI$ at claim $\textbf{1/}$, we see that :
$$\angle V_bQB' = \angle V_bQV = 90^\circ - \frac12(\angle BAC + \angle ABC) = 90^\circ - \angle AIB' = \angle NB'I = \angle V_bB'U_b$$$\hspace{0.5cm}$Lead to $\triangle V_bU_bB' \sim \triangle V_bB'Q$. As a result : $\dfrac{V_bU_b}{V_bQ} = \dfrac{V_bU_b}{V_bB'} .\dfrac{V_bB'}{V_bQ} = \left( \dfrac{B'U_b}{B'Q} \right)^2 $
$\, \,$Therefore
$$ \dfrac{XN}{XQ} = \left( \dfrac{IN}{IQ} \right)^2 = \left( \dfrac{\sin \angle NQI}{\sin \angle INQ} \right)^2 = \left( \dfrac{\sin \angle V_bB'U_b}{\sin \angle V_bB'Q} \right)^2 = \left( \dfrac{V_bU_b}{B'V_b} .\dfrac{V_bB'}{V_bQ} \right)^2 =  \dfrac{V_bU_b}{V_bQ} $$$\, \,$Hence, follow Thales theorem, we see that $XV_b$ $\|$ $NB'$. Similar, $XV_c$ $\|$ $MC'$
$\, \,$But from claim $\textbf{1/}$, we already have that $B'N$ $\|$ $C'M$, so $X$, $V_b$, $V_c$ are colinear
$\textbf{Finally, from these claims, we have : }$
$\, \,$ Cuz $\triangle XMI \sim \triangle XIP$ and $AX = XI$, so $\triangle XMA \sim \triangle XAP$
$\, \,$Lead to
$$\angle MAX = \angle APM = \angle APB - \angle MPB = 180^\circ - \angle ACB - \angle MTC = \angle LNC = \angle BCY = \angle BAY $$$\, \,$Similar, $\angle XAN = \angle CAY$. Therefore $A$, $X$, $Y$ are colinear, done :)
Attachments:
imosl g8 2020.pdf (74kb)
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bin_sherlo
661 posts
#63 • 2 Y
Y by ehuseyinyigit, swynca
Claim: $Y$ lies on $(ABC)$.
Proof:
\[\measuredangle BYC= 180-\measuredangle CBY-\measuredangle YCB=180-(\measuredangle IMB-\frac{\measuredangle B}{2})-(\measuredangle CNI-\frac{\measuredangle C}{2})=270-\frac{\measuredangle A}{2}-\measuredangle CIB=180-\measuredangle A\]Which gives that $Y\in (ABC)$.$\square$

Claim: $I$ is the $A-$excenter on $\triangle AMN$.
Proof: Let $O$ be the circumcenter of $(IMN)$. Since $\measuredangle MIN=\measuredangle MBI+\measuredangle ICN=90-\frac{\measuredangle A}{2},$ we have $\measuredangle MON=180-\measuredangle MAN$ hence $O$ lies on $(AMN)$. If $AI$ intersects $(IMN)$ at $J$ for second time, then
\[\measuredangle IMB=\measuredangle MIJ+\frac{\measuredangle A}{2}=\measuredangle MNJ+\measuredangle ONM=\measuredangle ONJ=\measuredangle NJO=\measuredangle NMI\]Similarily we get $\measuredangle CNI=\measuredangle INM$ so $I$ is the $A-$excenter on $\triangle AMN$.$\square$

Claim: $M,N,P,Q$ are concyclic.
Proof: Let $MN$ and incircle of $ABC$ be tangent to each other at $S$. Take the inversion centered at $I$ with radius $IS$. Let $D,E,F$ be the tangency points of the incircle with $BC,CA,AB$ respectively. $A^*,B^*,C^*,M^*,N^*$ are the midpoints of $EF,FD,DE,SF,SE$. $P^*,Q^*$ are the intersections of $B^*M^*,C^*N^*$ with $(A^*B^*C^*)$. Note that $M^*B^*\parallel SD\parallel N^*C^*$ and $M^*N^*\parallel EF\parallel B^*C ^*$.
\[\measuredangle M^*P^*Q^*=180-\measuredangle Q^*P^*B^*=\measuredangle B^*C^*N^*=\measuredangle N^*M^*P^*\]Thus, $N^*M^*P^*Q^*$ is an isosceles trapezoid which yields $M,N,P,Q$ lie on a circle.$\square$

Claim: $A,X,Y$ are collinear.
Proof: Let $PM$ and $QN$ intersect $(ABC)$ at $K,L$ for second time.
\[\measuredangle YCA=\measuredangle YCN=\measuredangle CQN=\measuredangle CQL \]\[\measuredangle ABY=\measuredangle MBY=\measuredangle MPB=\measuredangle KPB\]These give that $AY=CL$ and $AY=BK$. Hence $YK\parallel AB$ and $YL\parallel AC$. Also
\[\measuredangle KLN=\measuredangle KLQ=\measuredangle KPQ=\measuredangle MPQ=\measuredangle MNL\]Thus, $MN\parallel KL$. By Desargues on $\triangle AMN$ and $\triangle YKL$, since $AM\cap YK=AB_{\infty},MN\cap KL=MN_{\infty},NA\cap LY=AC_{\infty}$ are collinear, they are perspective. $AY,MK,NL$ are concurrent. $MK$ and $NL$ intersect at $X$ so $A,X,Y$ are collinear.$\blacksquare$
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GeorgeRP
130 posts
#64
Y by
Why did I solve a G8 synthetically???

Let's denote with $t$ the common tangent of $w_b$ and $w_c$.
Also let $U,V$ be the intersections of $BC$ with $w_b$ and $w_c$

Claim 1: $Y$ lies on $(ABC)$
Proof. $\angle BIU=\angle BCY=\phi$ and $\angle CIV=\angle CBY=\psi, \angle UIt=\angle IBC=\frac{\beta}{2}, \angle VIt=\angle ICB=\frac{\gamma}{2}$
Now:
$$90+\frac{\alpha}{2}=\angle BIC=\angle BIU+\angle UIV+\angle CIV= \phi+\psi+90-\frac{\alpha}{2}$$So $\angle BYC=180-\phi-\psi=180- \alpha=\angle BAC \text{. }\square$

Claim 2: $MU$ || $NV$ || $t$
Proof. $\angle MUI=\angle MBI= \angle UBI = \angle UIt \Rightarrow $ $MU$ || $t$ and similarly $NV$ || $t$. $\square$

Claim 3: $MUVN$ is an isosceles trapezoid
Proof. As $I$ is the midpoint of arcs $MU$ and $NV$ in $w_b,w_c$, this means that $MI=UI, IN=IV \Rightarrow \triangle MIN$ and $\triangle UIV $ are congruent $\Rightarrow MN=UV$. $\square$

Let $Q'$ and $P'$ be the intersections of $QX$ and $PX$ with $(ABC)$.

Claim 4: $BQ'$ || $QC$ || $t$
Proof. Reims theorem on $(ABC), (NQCV)$ and lines $QX,CV$ gives that $BQ'$ || $NV$ and similarly $CP'$ || $UM$. $\square$

Let $K;G$ be the intersections of $AB;BC$ and $NV;MU$

Claim 5: $PMAG$ and $QNAK$ are cyclic
Proof. $\angle KNQ=\angle VCQ=180-\angle BQ'Q=180-\angle BAQ=\angle KAQ$ and similarly $\angle PMG=\angle PAG$. $\square$

Claim 6:
$(PMAG)$ and $(NAKQ)$ are tangent at $A$
Proof. As $GMNK$ is a trapezoid we have that the circumcenters of $GMA$ and $KAN$ lie on a line with $A$. $\square$

Claim 7: $PMNQ$ is cyclic
Proof. $\angle MPQ= \angle BPQ-\angle BPM=(180-\angle BCQ)-\angle MUV=$
$(180-\angle KNQ)-\angle UMN=\angle XNK-\angle MNK=\angle MNX$. $\square$

Claim 8: $A,X,Y$ are collinear
Proof. By radical axises on $(PMNQ), (PMAG), (NAKQ)$ we get that $AX$ is the tangent of $(PMAG)$ and $(NAKQ)$.
$\angle YAC=\angle YBC= \angle BMU=\angle Q'BM=\angle Q'QA=\angle XAC$. $\square$
Attachments:
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L13832
245 posts
#65 • 2 Y
Y by radian_51, S_14159
Amazing problem! also my first G8 :showoff:
We show $AY$ is tangent to $\odot(AMP), \odot(ANQ)$
$$\angle YAM+\angle MAP = 180^{\circ} - \angle YBP=180^{\circ}-\angle PBI-\angle BPI = \angle BIP =  180^{\circ} - \angle PMA$$Note that $Y\in{\odot(ABC)}$as
\begin{align*}
& \angle BIC=\angle IMB+\angle INC=\angle IBY+\angle ICY=90^{\circ}+\frac{\angle A}{2}\\ 
& \angle BYC=360-\angle YBI-\angle BIC-\angle ICY=360^{\circ}-2\left(90^{\circ}+\frac{\angle A}{2}\right)=180^{\circ}-\angle A
\end{align*}By another angle chase we show $\odot(I)$ is the incircle of $MNCB$. $$\angle MIN = \angle MBI + \angle NCI = \angle IBC + \angle ICB = 180^{\circ}-\angle BIC=90-\frac{\angle A}{2}$$If we show $\odot(MPNQ)$ we get $X$ as the radical center of $(MNQP)$, $(AMP)$, and $(ANQ)$, which will clearly lie on radax of $(AMP)$, and $(ANQ)$ which is clearly $\overline{AY}$ implying $\overline{A-X-Y}$ is collinear.
So finally we go on to another angle chase to prove $\odot(MPNQ)$ which took me a lotta time to figure out :stretcher:
\begin{align*}
\angle PQN=\angle CQN-\angle CQP&= 180^{\circ}-\angle CIN-\angle CQP\\&=180^{\circ}-(180^{\circ}-\angle INC-\angle ICN)-\angle CQP\\&=180^{\circ}-(180^{\circ}-\angle ICY-\angle ICN)-\angle CQP\\&=\angle ICY+\angle ICN-\angle CQP\\&=\angle ACY-\angle CQP\\&=(180^{\circ}-\angle ABY)-(180^{\circ}-\angle CBP)\\&=\angle CBP-\angle ABY\\&=\angle CBA+\angle ABP-\angle CBA-\angle YBC\\&=\angle ABP-\angle YBC\\
\angle PMN=\angle PMB+\angle BMN&=\angle PIB+2\angle BMI\\&=180^{\circ}-(\angle IPB+\angle PBI)+2\angle BMI\\&=180^{\circ}-(\angle IBY+\angle PBI)+2\angle BMI\\&=180^{\circ}-\angle PBY+2\angle BMI\\&=180^{\circ}-\angle PBY+2\angle CBY+\angle CBA\\&=180^{\circ}-\angle PBA+\angle CBY
\end{align*}and we are finally done! :stretcher: :wacko:
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Saucepan_man02
1294 posts
#66
Y by
Here's my solution (angle-chase+invert+radical-axis+trig) :

Claim-1: $Y \in (ABC)$

Notice that: $$\angle ABY+\angle ACY = 360^\circ - (\angle MIB + \angle NIC) =  \angle MIC + \angle BIC = \angle MIX+\angle XIN + 90^\circ + \angle A/2$$$$ = \angle MBI + \angle NCI + 90^ + \angle A/2 = 180^\circ.$$
Claim-2: $MNQP$ is cyclic + $(AMP), (ANQ)$ are tangent.

Invert around incircle. Let $X'$ denote the inverse of point $X$ in this inversion. Notice that:

-$I$ is the orthocenter of $\triangle A'B'C'$.
-$M'A'B'I$ and $N'A'C'I$ are cyclic ($A, M, B$ and $A, N, C$ are collinear).
-$M', P', B'$ and $N', Q', C'$ are collinear ($MPIB$ and $NIAC$ are cyclic).
-$M'P'B' \parallel N'Q'C'$.
-$A',P',B',C',Q'$ lie on same circle.

We will prove $M'N'Q'P'$ is cyclic. It suffice to show $M'N'=P'Q'$. Note that: $\angle M'IB'+\angle N'IC' = 360^\circ - (\angle B'M'I + \angle C'N'I) - (\angle M'B'I + \angle N'C'I) = 180^\circ.$
Now, we have: $$M'B' = \frac{B'I \sin \angle M'IB'}{\sin \angle M'B'I}$$Note that: $$\frac{B'I}{\sin \angle M'B'I} = 2R  = \frac{C'I}{\sin \angle N'C'I}.$$Thus, we have: $\frac{B'I \sin \angle M'IB'}{\sin \angle M'B'I} = \frac{C'I \sin \angle N'IC'}{\sin \angle N'C'I} = N'C'$ which implies $M'B'C'N'$ is a parallelogram.

Note that, from the following sub-claim, we will have: $P'Q' \parallel M'N' \parallel BC$ and $(A'M'P'), (A'N'Q')$ to be tangent at $A'$ which finishes the main-claim.

Sub-Claim: $\triangle A'M'P'$ and $\triangle A'N'Q'$ are isosceles.
Note that: $\angle A'M'B' =  180^\circ - \angle A'IB' = \angle A'C'B'$ and $\angle A'P'M' = \angle A'C'B'$ which shows that $A'M' = A'P'$. Similarly, $A'N'=A'P'$ and we are done.
Hence, Claim-2 is true.

Thus, from Claim-2, $AX$ is the radical axis of $(APM), (ANQ)$ along with $AX$ being tangent to both circles $(APM), (ANQ)$.

Note that, the following claim finishes the problem:

Claim-3: $AY$ is tangent to both circles $(APM), (ANQ)$.

Let $R$ be a point of line $AP$ such that $A, R$ are on different sides with-respect to $P$.
We will show it with an one-liner angle-chase:
$$\angle BAY = 180^\circ  - \angle AYB - \angle ABY = 180^\circ - \angle RPB - \angle BPM  = \angle APM.$$Thus, $AY$ is tangent to $(APM)$. Similarly, $AY$ is tangent to $(ANQ)$ and we are done.

Remarks
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Likeminded2017
391 posts
#67
Y by
We prove that $Y$ lies on $(ABC)$ through angle chasing - $\angle ABY+\angle ACY=360-\angle MIB-\angle NIC=180^\circ.$ Next, $YA$ is tangent to both $(ANQ)$ and $(AMP)$ and is thus the radical axis, as $\angle APM=\angle BPA-\angle BPM=(180-\angle BYA)-\angle YBA=\angle YAB=\angle YAM$ and the same works for $(ANQ).$ Finally, we show $X$ lies on the radical axis by showing $MNQP$ are concyclic to finish. Let $\omega_b$ and $\omega_c$ intersect $BC$ at $R,S$ respectively. Observe by Fact 5 $\triangle IRS \cong \triangle IMN$ so $\angle INM=\angle ISR=\angle INC.$
\begin{align*}
    \angle QPM &= \angle MPB-\angle QPB \\
    &= \angle NIC+\angle QCB-180 \\
    &= \angle NIC+\angle BCI-\angle QNI \\
    &=\angle NIC+\angle NCI-\angle QNI \\
    &=180-\angle INC-\angle QNI \\
    &=180-\angle INM-\angle QNI \\
    &=180-\angle QNM
\end{align*}so we are done.
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cj13609517288
1857 posts
#68
Y by
Part 1 - The spiral similarity
First, note that
\[\angle BIC=\angle BMI+\angle CNI=\angle YBI+\angle YCI.\]Therefore,
\[\angle BYC=360^{\circ}-\angle BIC-\angle BIC=180^{\circ}-\angle A.\]Therefore, $Y\in (ABC)$.

Now note that
\[\angle PBY=180^{\circ}-\angle PMB=\angle PMA\]\[\angle PYB=\angle PAB=\angle PAM\]so $\triangle PBY\sim\triangle PMA$.

So there exists a spiral similarity centered at $P$ sending $AM$ to $YB$.

Part 2 - Redefining $X$
Redefine $X$ to be the intersection of the perpendicular bisector of $AI$ with $AY$. Then $\angle IXY=2\angle IAX$, so $IX$ is parallel to the isogonal of $AY$ wrt angle $BAC$. In particular,
\[\angle(BA,IX)=\angle YAC=\angle YBC=\angle(YB,CB)\Longrightarrow
\angle(BI,IX)=\angle(YB,IB)\Longrightarrow \angle BIX=180^{\circ}-\angle YBI=180^{\circ}-\angle BPI.\]Therefore, $XI$ is tangent to $(BPI)$.

Part 3 - The finish
By our spiral similarity from earlier,
\[\angle YAM=\angle YAB=\angle YPB=\angle APM.\]Therefore, $AY$ is tangent to $(APM)$. Then the powers from $X$ to $(APM)$ and $(BPM)$ are $AX^2$ and $IX^2$, respectively, which are equal, so $X$ lies on their radical axis $PM$. Similarly, $X$ lies on $QN$, as desired. $\blacksquare$
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swynca
10 posts
#69 • 1 Y
Y by bin_sherlo
A different approach,
After a simple angle chasing, it appears that $K=AC \cap YP$ and $L=AB \cap YQ$ is on $(MNPQ)$. Using pascal with points $P,K,N,Q,L,M$ gives the desired result.
This post has been edited 1 time. Last edited by swynca, Feb 13, 2025, 8:23 PM
Reason: typo
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Iveela
115 posts
#70
Y by
The ultimate angle chasing.

Let $O_B$ and $O_C$ be the circumcenters of $\omega_B$ and $\omega_C$ respectively. Recall that $\angle BIC = 90^{\circ} + \frac{\angle A}{2}$ which implies $\angle O_BIB + \angle O_CIC = 90^{\circ} - \frac{\angle A}{2}$ and hence
\[\angle IBY + \angle ICY = \angle IPB + \angle IQC = 180 - \angle O_BIB - \angle O_CIC = 90^{\circ} + \frac{\angle A}{2}.\]Consequently, $\angle ABY + \angle ACY = 180^{\circ}$ or $ABCY$ cyclic.

Let $T = BP \cap CQ$. Since $TP \cdot TB = TQ \cdot TC$, it must lie on the radical axis of $\omega_B$ and $\omega_C$. In other words, $TI$ is tangent to both circles. We will now show that the quadrilateral $TPQX$ is cyclic. Indeed, we have
\[\angle XQC = \angle NQC = \angle ACY = 180^{\circ} - \angle ABY = 180^{\circ} - \angle XPB = \angle TPX\]as desired. Now, observe that
\[\angle XTQ = \angle QPX = \angle QPB - \angle ABY = \angle BCY - \angle ACQ = \angle ICY - \angle ICQ = \angle ACQ \]which implies that $T$, $X$, and $I$ are collinear. In particular, $XI$ is tangent to $\omega_B$ and $\omega_C$. This implies $XM \cdot XP = XI^2 = XN \cdot XQ$ or $MNPQ$ cyclic.

Now, we are well equipped to prove that $XA = XI$. We once again, angle chase. Recall that $XI$ is tangent to $\omega_B$ and $\omega_C$. Indeed, we have
\[\angle XIA = \angle AIB - \angle XIB = (90^{\circ} + \frac{\angle C}{2}) + \angle IBY - 180^{\circ} = \angle YAC - \frac{\angle A}{2} = \angle YAI.\]Recall that $XN \cdot XQ = XI^2 = XA^2$. Therefore, $(ANQ)$ is tangent to $XA$. Angle chasing one final time, we receive
\[\angle XAC = \angle AQN = \angle AQC - \angle NQC = \angle BAC + \angle ACB - \angle ACY = \angle CAY\]which implies the result.
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EeEeRUT
48 posts
#71
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Denote $\ell$ a line tangent to $\omega_B$ and $\omega_C$ at $I$.

Claim : $\angle MIB + \angle NIC = 180^{\circ}$
Proof : Consider \begin{align*}
\angle MIB + \angle NIC &= 360^{\circ} - \angle MIN - \angle BIC\\ &= 360^{\circ} - (\angle IBM + \angle ICN) - (180^{\circ} - \angle IBC - \angle ICB)\\ &= 360^{\circ} - (\angle IBC + \angle ICB) - (180^{\circ} - \angle IBC - \angle ICB) = 180^{\circ}
\end{align*}Let $BP$ intersect $CQ$ at $T$. By radical axis, $T$ lies on $\ell$. Note that $$\angle TPX + \angle TQX = 360^{\circ} - (\angle MIB + \angle NIC) = 180^{\circ}$$Thus, $TPXQ$ is cyclic.

Claim : $X$ lies on $\ell$.
Proof : Let $\ell$ intersect $BC$ at $S$.
Consider $$\angle BMX = 180^{\circ} - \angle PIB = 180^{\circ} - \angle ISB$$Hence, $PXSB = \Omega_B$ is cyclic. Similarly, $QXSC = \Omega_C$ is cyclic.
That is the radical axis of these $2$ circles is $XS$
By radical center of $\Omega_B, \Omega_C$ and $(BCPQ)$ is $T$. Thus, $T,X,S$ are collinear, which is the line $\ell$.

Claim : $XA = IA$.
Proof : Consider \begin{align*}
\angle MPA + \angle NQA &= \angle APB - \angle BPM + \angle AQC - \angle CQN\\
&= -\angle ABC - \angle ACB + \angle MIB + \angle NIC\\
&= \angle CAB
\end{align*}Thus, $(PMA)$ is tangent to $(QNA)$
By power of point, the segment $XA = XI$ is tangent to $(PMA), (QNA)$.

Claim : $A,B,C,Y$ cyclic.
Proof : $$\angle ABY + \angle YCA = 360^{\circ} - \angle MIB - \angle NIC = 180^{\circ}$$Thus, $A,B,C,Y$ are concyclic.

Let $\angle XAI = \gamma$
Now, we are left to consider the following angle chasing \begin{align*} 
\angle XAC &= \angle CAB - \angle BAX\\ 
&= \angle CAB -(360^{\circ} - \angle AMI - \angle PIX - \angle IXA)\\
&= \angle CAB + 180^{\circ} - \angle IAM - \gamma - \angle IBC\\
&= \angle CAB - \angle IMB - \gamma\\
&= \angle CAB - \angle IBY\\
&= \angle CAB - \angle YAB\\
&= \angle YAC
\end{align*}
Hence, $ A,X,Y$ are collinear as desired.

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This post has been edited 1 time. Last edited by EeEeRUT, Mar 4, 2025, 9:44 AM
Reason: Diagram
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joshualiu315
2513 posts
#72 • 1 Y
Y by OronSH
first g8 yippee! :coolspeak:


Denote the centers of $\omega_B$ and $\omega_C$ as $O_B$ and $O_C$, respectively.


Claim: $Y$ lies on $\Gamma$.

Proof: Notice that

\begin{align*}
\angle ABY + \angle ACY &= \angle BPM + \angle CQN \\
&= (180^\circ - \angle BIM) + (180^\circ - \angle CIN) \\
&= \angle BIC + \angle MIN \\
&= \angle BIC + \angle MBI + \angle NCI \\
&= \angle BIC + \angle IBC + \angle ICB = 180^\circ,
\end{align*}
so $ABYC$ is cyclic. $\square$


Let $D \neq B = \omega_B \cap \overline{BC}$ and $E \neq C = \omega_C \cap \overline{BC}$.


Claim: $MDEN$ is an isosceles trapezoid.

Proof: Since $\overline{BI}$ and $\overline{CI}$ are angle bisectors, $I$ is the midpoint of $\widehat{MD}$ in $\omega_B$ and the midpoint of $\widehat{NE}$ in $\omega_C$. Thus, $\overline{O_BO_C}$ is the perpendicular bisector of both $\overline{MD}$ and $\overline{NE}$. $\square$


Claim: Points $M$, $N$, $P$, and $Q$ are concyclic.

Proof: It suffices to show that $\triangle XMN \sim \triangle XQP$, or that $\angle XMN = \angle XQP$. We proceed with angle chasing:

\begin{align*}
\angle XMN &= 180^\circ - \angle PMD - \angle DMN \\
&= \angle PBD - \angle MDE \\
&= \angle PBD - \angle NEC \\
&= (180^\circ - \angle PQC) - (180^\circ - \angle NQC) \\
&= \angle NQC - \angle PQC = \angle NQP,
\end{align*}
as desired. $\square$


Then, consider the radical axis of $(APM)$ and $(AQN)$, denoted as $\ell$.


Claim: Points $A$, $X$, and $Y$ all lie on $\ell$

Proof: Since $A$ lies on both circles, it obviously lies on the radical axis. Also, we have $XM \cdot XP = XN \cdot XQ$ from Power of a Point with respect to $(MNQP)$. This implies $X$ has equal power with respect to $(APM)$ and $(AQN)$, meaning $X$ lies on $\ell$.

Finally, it suffices to show that $Y$ lies on $\ell$. We prove the stronger statment that $\overline{AY}$ is the common tangent of $(APM)$ and $(AQN)$. Notice that

\begin{align*}
\angle YAP &= 180^\circ - \angle YBP \\
&= 180^\circ - (\angle PBD + \angle YBD) \\
&= 180^\circ - (180^\circ - \angle PMD + \angle BMD) \\
&= \angle PMD - \angle BMD \\
&= \angle PMB = 180^\circ - \angle AMP.
\end{align*}
Thus, $\overline{AY}$ is tangent to $(APM)$, and similarly we can find that $\overline{AY}$ is tangent to $(AQN)$. This means $Y$ lies on $\ell$ as well. $\square$


Since $A$, $X$, and $Y$ all lie on $\ell$, they are obviously collinear. $\blacksquare$
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