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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
Inequality => square
Rushil   12
N 17 minutes ago by ohiorizzler1434
Source: INMO 1998 Problem 4
Suppose $ABCD$ is a cyclic quadrilateral inscribed in a circle of radius one unit. If $AB \cdot BC \cdot CD \cdot DA \geq 4$, prove that $ABCD$ is a square.
12 replies
+2 w
Rushil
Oct 7, 2005
ohiorizzler1434
17 minutes ago
p + q + r + s = 9 and p^2 + q^2 + r^2 + s^2 = 21
who   28
N 33 minutes ago by asdf334
Source: IMO Shortlist 2005 problem A3
Four real numbers $ p$, $ q$, $ r$, $ s$ satisfy $ p+q+r+s = 9$ and $ p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $ \left(a,b,c,d\right)$ of $ \left(p,q,r,s\right)$ such that $ ab-cd \geq 2$.
28 replies
who
Jul 8, 2006
asdf334
33 minutes ago
H not needed
dchenmathcounts   44
N an hour ago by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
dchenmathcounts
May 23, 2020
Ilikeminecraft
an hour ago
AMC- IMO preparation
asyaela.   9
N an hour ago by Schintalpati
I'm a ninth grader, and I recently attempted the AMC 12, getting 18 questions correct and leaving 7 empty. I started working on Olympiad math in November and currently dedicate about two hours per day to preparation. I'm feeling a bit demotivated, but if it's possible for me to reach IMO level, I'd be willing to put in more time. How realistic is it for me to get there, and how much study would it typically take?
9 replies
asyaela.
4 hours ago
Schintalpati
an hour ago
Tennessee Math Tournament (TMT) Online 2025
TennesseeMathTournament   29
N an hour ago by NashvilleSC
Hello everyone! We are excited to announce a new competition, the Tennessee Math Tournament, created by the Tennessee Math Coalition! Anyone can participate in the virtual competition for free.

The testing window is from March 22nd to April 5th, 2025. Virtual competitors may participate in the competition at any time during that window.

The virtual competition consists of three rounds: Individual, Bullet, and Team. The Individual Round is 60 minutes long and consists of 30 questions (AMC 10 level). The Bullet Round is 20 minutes long and consists of 80 questions (Mathcounts Chapter level). The Team Round is 30 minutes long and consists of 16 questions (AMC 12 level). Virtual competitors may compete in teams of four, or choose to not participate in the team round.

To register and see more information, click here!

If you have any questions, please email connect@tnmathcoalition.org or reply to this thread!
29 replies
TennesseeMathTournament
Mar 9, 2025
NashvilleSC
an hour ago
IZHO 2017 Functional equations
user01   51
N an hour ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
an hour ago
AIME score for college apps
Happyllamaalways   75
N an hour ago by hashbrown2009
What good colleges do I have a chance of getting into with an 11 on AIME? (Any chances for Princeton)

Also idk if this has weight but I had the highest AIME score in my school.
75 replies
+1 w
Happyllamaalways
Mar 13, 2025
hashbrown2009
an hour ago
AMC 8 discussion
Jaxman8   42
N 2 hours ago by mpcnotnpc
Discuss the AMC 8 below!
42 replies
Jaxman8
Jan 29, 2025
mpcnotnpc
2 hours ago
Segment has Length Equal to Circumradius
djmathman   72
N 2 hours ago by Zhaom
Source: 2014 USAMO Problem 5
Let $ABC$ be a triangle with orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $AHC$ with the internal bisector of the angle $\angle BAC$. Let $X$ be the circumcenter of triangle $APB$ and $Y$ the orthocenter of triangle $APC$. Prove that the length of segment $XY$ is equal to the circumradius of triangle $ABC$.
72 replies
djmathman
Apr 30, 2014
Zhaom
2 hours ago
[Registration Open] Mustang Math Tournament 2025
MustangMathTournament   22
N 2 hours ago by RainbowSquirrel53B
Mustang Math is excited to announce that registration for our annual tournament, MMT 2025, is open! This year, we are bringing our tournament to 9 in-person locations, as well as online!

Locations include: Colorado, Norcal, Socal, Georgia, Illinois, Massachusetts, New Jersey, Nevada, Washington, and online. For registration and more information, check out https://mustangmath.com/competitions/mmt-2025.

MMT 2025 is a math tournament run by a group of 150+ mathematically experienced high school and college students who are dedicated to providing a high-quality and enjoyable contest for middle school students. Our tournament centers around teamwork and collaboration, incentivizing students to work with their teams not only to navigate the challenging and interesting problems of the tournament but also to develop strategies to master the unique rounds. This includes a logic puzzle round, a strategy-filled hexes round, a race-like gallop round, and our trademark ‘Mystery Mare’ round!

Awards:
[list]
[*] Medals for the top teams
[*] Shirts, pins, stickers and certificates for all participants
[*] Additional awards provided by our wonderful sponsors!
[/list]

We are also holding a free MMT prep seminar from 3/15-3/16 to help students prepare for the upcoming tournament. Join the Google Classroom! https://classroom.google.com/c/NzQ5NDUyNDY2NjM1?cjc=7sogth4
22 replies
MustangMathTournament
Mar 8, 2025
RainbowSquirrel53B
2 hours ago
d_k-eja Vu
ihatemath123   46
N 4 hours ago by Ilikeminecraft
Source: 2024 USAMO Problem 1
Find all integers $n \geq 3$ such that the following property holds: if we list the divisors of $n!$ in increasing order as $1 = d_1 < d_2 < \dots < d_k = n!$, then we have
\[ d_2 - d_1 \leq d_3 - d_2 \leq \dots \leq d_k - d_{k-1}. \]
Proposed by Luke Robitaille.
46 replies
ihatemath123
Mar 20, 2024
Ilikeminecraft
4 hours ago
average FE
KevinYang2.71   75
N 5 hours ago by Marcus_Zhang
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
75 replies
KevinYang2.71
Mar 21, 2024
Marcus_Zhang
5 hours ago
apparently circles have two intersections :'(
itised   76
N 5 hours ago by Ilikeminecraft
Source: 2020 USOJMO Problem 2
Let $\omega$ be the incircle of a fixed equilateral triangle $ABC$. Let $\ell$ be a variable line that is tangent to $\omega$ and meets the interior of segments $BC$ and $CA$ at points $P$ and $Q$, respectively. A point $R$ is chosen such that $PR = PA$ and $QR = QB$. Find all possible locations of the point $R$, over all choices of $\ell$.

Proposed by Titu Andreescu and Waldemar Pompe
76 replies
itised
Jun 21, 2020
Ilikeminecraft
5 hours ago
Too Bad I'm Lactose Intolerant
hwl0304   216
N Today at 5:20 PM by AshAuktober
Source: 2018 USAMO Problem 1/USAJMO Problem 2
Let \(a,b,c\) be positive real numbers such that \(a+b+c=4\sqrt[3]{abc}\). Prove that \[2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.\]
216 replies
hwl0304
Apr 18, 2018
AshAuktober
Today at 5:20 PM
GOTEEM #5: Circumcircle passes through fixed point
tworigami   21
N Yesterday at 8:45 PM by Ilikeminecraft
Source: GOTEEM: Mock Geometry Contest
Let $ABC$ be a triangle and let $B_1$ and $C_1$ be variable points on sides $\overline{BA}$ and $\overline{CA}$, respectively, such that $BB_1 = CC_1$. Let $B_2 \neq B_1$ denote the point on $\odot(ACB_1)$ such that $BC_1$ is parallel to $B_1B_2$, and let $C_2 \neq C_1$ denote the point on $\odot(ABC_1)$ such that $CB_1$ is parallel to $C_1C_2$. Prove that as $B_1, C_1$ vary, the circumcircle of $\triangle AB_2C_2$ passes through a fixed point, other than $A$.

Proposed by tworigami
21 replies
tworigami
Jan 2, 2020
Ilikeminecraft
Yesterday at 8:45 PM
GOTEEM #5: Circumcircle passes through fixed point
G H J
G H BBookmark kLocked kLocked NReply
Source: GOTEEM: Mock Geometry Contest
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tworigami
844 posts
#1 • 9 Y
Y by rocketscience, NJOY, amar_04, GeoMetrix, mijail, CyclicISLscelesTrapezoid, Adventure10, Mango247, Rounak_iitr
Let $ABC$ be a triangle and let $B_1$ and $C_1$ be variable points on sides $\overline{BA}$ and $\overline{CA}$, respectively, such that $BB_1 = CC_1$. Let $B_2 \neq B_1$ denote the point on $\odot(ACB_1)$ such that $BC_1$ is parallel to $B_1B_2$, and let $C_2 \neq C_1$ denote the point on $\odot(ABC_1)$ such that $CB_1$ is parallel to $C_1C_2$. Prove that as $B_1, C_1$ vary, the circumcircle of $\triangle AB_2C_2$ passes through a fixed point, other than $A$.

Proposed by tworigami
Z K Y
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rocketscience
466 posts
#2 • 3 Y
Y by tworigami, mijail, Adventure10
My submission, with typos corrected (I think). Also my favorite problem of the test.

The following solution will make use of the spiral similarity lemma: if $P: XY \mapsto WZ$ is a spiral similarity centered at $P$ mapping $\overline{XY}$ to $\overline{WZ}$, then $(PXW)$ and $(PYZ)$ meet a second time at $XY \cap WZ$.

Let $M$ be the midpoint of arc $\widehat{BAC}$ on $(ABC)$. It's easy to see that $M$ is the center of spiral congruence taking $BB_1 \mapsto CC_1$, so $MB_1 = MC_1.$

Define $P = (ABC_1) \cap (AC_1B)$ to be the second intersection of those circles; note that $P$ is the center of the spiral congruence $BB_1 \mapsto C_1C$. Thus $\mathrm{dist}(P, AB)=\mathrm{dist}(P, AC)$, which implies that $P$ lies on the internal $A$-angle bisector. Next we claim that $P = C_1B_2 \cap B_1C_2$. Indeed, the given parallel lines imply $\triangle AC_1C_2 \sim \triangle AB_1B_2$, i.e. $A: C_2B_1 \mapsto C_1B_2$, meaning $P = (ABC_1) \cap (AC_1B)$ is the intersection of $C_1B_2$ and $B_1C_2$, as desired. This implies that $B_1C_1B_2C_2$ is cyclic: $\angle B_1C_2C_1 = \angle BAP = \angle CAP = \angle C_1B_2B_1 = \frac 12 \angle A$.

I'm pretty sure the rest is angle-chaseable, but I succumb to my projective/inversive tendencies. Plus, this is cool. Let $O$ be the circumcenter of $\Gamma = B_1C_1B_2C_2$. Radical center on $\Gamma, (ABC_1), (ACB_1)$ yields a point $Q = C_1C_2 \cap B_1B_2 \cap AP$. Consider the inversion at $P$ which fixes $\Gamma$. It swaps $(ABC_1) \leftrightarrow B_1B_2$ and $(ACB_1) \leftrightarrow C_1C_2$, and thus $A \leftrightarrow Q$. Brokard implies $Q$ lies on the polar of $P$ wrt $\Gamma$, and since the image of the polar of $P$ under our inversion is $(PO)$, we have $A \in (PO)$, i.e. $\angle PAO = 90^\circ$.

We now have two characterizations that uniquely determine $O$: first, it lies on the perpendicular bisector of $\overline{B_1C_1}$, and second it lies on the line through $A$ perpendicular to $AP$. Surprise, we claim $O = M$. Indeed, we verified earlier that $MB_1 = MC_1$, and also $\angle MAP = 90^\circ$ since $AP$ is the internal angle bisector.

Armed with this, we can finally slay the problem. The required fixed point is $M$; observe that $AP$ is the internal angle bisector of $\angle B_2AC_2$ by some earlier angle work, so $AM$ is its external angle bisector; also, $M$ is the center of $\Gamma$ so $MB_2 = MC_2$, and together these imply that $AMB_2C_2$ is cyclic, as desired.
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Sugiyem
115 posts
#3 • 2 Y
Y by Adventure10, Mango247
Let $D$ be the midpoint of arc $\widehat{BAC}$ of $(ABC)$, $E$ the second intersection of $(ABC_{1})$ and $(ACB_{1})$, $F$ the intersection of $C_{1}C_{2}$ and $B_{1}B_{2}$.

It's easy to check by spiral similarity that $D$ is the second intersection of $(AB_{1}C_{1})$ and $(ABC)$.

By spiral similarity also we can get that $\triangle EBB_{1}\thicksim \triangle EC_{1}C$.
Thus $\frac{dist (E,AB)}{dist (E,AC)}=\frac{BB_{1}}{CC_{1}}=1$.
So $E$ lies on the internal bisector of $\angle BAC.$

$\textbf{Claim 1}$: $E$ is the intersection of $B_{1}C_{2}$ and $C_{1}B_{2}$
$\textbf{Proof}$:
Notice that $\angle AB_{1}B_{2}=\angle ABC_{1}=\angle AC_{2}C_{1}$. Moreover $\angle AC_{1}C_{2}=\angle ACB_{1}=\angle AB_{2}B_{1}$
So $\triangle AC_{2}C_{1}\thicksim \triangle AB_{1}B_{2}$
Which is equivalent to $\triangle AB_{1}C_{2}\thicksim \triangle AB_{2}C_{1}$.
$\angle AB_{1}C_{2} +\angle AB_{1}E=\angle AB_{2}C_{1} +\angle AB_{1}E=180^{\circ}$ because $AB_{1}EB_{2}$ is cyclic.
So $E,B_{1},C_{2}$ are collinear. Analagously we can prove that $E,C_{1},B_{2}$ are collinear. So the claim 1 is proven.

Now let's move on to the second claim.

$\textbf{Claim 2}$: $B_{1}C_{1}B_{2}C_{2}$ lies on a circle centered at $D$.
$\textbf{Proof}$:
Note that $\angle B_{1}C_{2}C_{1}=\angle EC_{2}C_{1}=\angle EAC_{1}=\frac{\angle B_{1}AC_{1}}{2}=\frac{\angle B_{1}DC_{1}}{2}$.
With the same way, we can get that $\angle B_{1}B_{2}C_{1}=\frac{\angle B_{1}DC_{1}}{2}$
So the claim 2 is done.

$\textbf{Claim 3}$: $F$ lies on the internal bisector of $\angle BAC$
$\textbf{Proof}$: This is equivalent to show that $A,F,E$ are collinear.
Because $A$ is the center of spiral similarity which sends $B_{1}C_{2}$ to $B_{2}C_{1}$, we have that $AFB_{1}C_{2}$ and $AFC_{1}B_{2}$ are both cyclic.
Thus by radical center at $(B_{1}C_{1}B_{2}C_{2})$,$(AFB_{1}C_{2})$ and $(AFC_{1}B_{2})$, we have $A,F,E$ collinear.
Therefore the claim 3 is done too.

Now we'll prove that point $D$ is the fixed point that the problem wants,
Define $X$ to be the intersection of $B_{1}C_{1}$ and $B_{2}C_{2}$
By Brokard's at $(B_{1}C_{1}B_{2}C_{2})$, we have $EF$ is the polar of $X$ WRT $(B_{1}C_{1}B_{2}C_{2})$.
Hence, $DX\perp EF$, but $AD$ also perpendicular to $EF$ because $AD$ is the external bisector of $\angle BAC$ and $EF$ is the internal bisector of $\angle BAC$.
So, $D,A,X$ collinear.
Therefore, because $ADB_{1}C_{1}$ and $B_{1}C_{1}B_{2}C_{2}$ are both cyclis, we have:
$XA\times XD=XC_{1}\times XB_{1}=XB_{2}\times XC_{2}$.
So, $D$ lies on $(AB_{2}C_{2})$ as we want.
This post has been edited 2 times. Last edited by Sugiyem, Jan 3, 2020, 7:36 AM
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Th3Numb3rThr33
1247 posts
#4 • 2 Y
Y by Adventure10, Mango247
The inversion is probably unnecessary, but I don't like circles so...

The fixed point is the midpoint $T$ of major arc $\widehat{BAC}$.

Let $T'$ denote the Miquel point of $BB_1C_1C$, which is on $(ABC)$ and sends $\overline{BB_1}$ to $\overline{CC_1}$. But $BB_1 = CC_1$ by definition, so $T'B = T'C$ and thus $T = T'$. So $T$ lies on $(AB_1C_1)$.

Now take a $\sqrt{bc}$-inversion at $A$ and reflect about the angle bisector of $\angle BAC$, yielding the following problem (with stars dropped).
Quote:
Let $ABC$ be a triangle, and $T$ be the foot of the $A$-external bisector onto $\overline{BC}$. A variable line $\ell$ through $T$ meets $\overline{AB}$ and $\overline{AC}$ at $B_1$ and $C_1$, respectively. Let $B_2 \neq B_1$ denote the point on $CB_1$ such that $\measuredangle AB_2B_1 = \measuredangle AC_1B$, and let $C_2 \neq C_1$ denote the point on $BC_1$ such that $\measuredangle AC_2C_1 = \measuredangle AB_1C$. Prove that $T,B_2,C_2$ are collinear.

[asy]
defaultpen(fontsize(11pt));
size(10cm);
real t = 0.5;
pair A = dir(160); pair B = dir(225); pair C = dir(540-225);
pair X = t*A + (1-t)*dir(270);
pair C1 = extension(B,X,A,C); pair B1 = extension(C,X,A,B);
path circ1 = circumcircle(A,B1,X); path circ2 = circumcircle(A,C1,X);
pair B2 = intersectionpoints(C--2B1-C, circ2)[1];
pair C2 = intersectionpoints(B--2C1-B, circ1)[1];
pair T = extension(B,C,B1,C1);
pair O = circumcenter(B1,C1,B2);

filldraw(A--B--C--cycle,pink+white+white);

draw(circ1,orange); draw(circ2,orange);
draw(B--C2,red); draw(C--B2,red);
draw(T--C2,gray(0.5)); draw(T--C1,gray(0.5)); draw(T--B);


dot("$A$", A, dir(90));
dot("$B$", B, dir(270));
dot("$C$", C, dir(270));
dot("$X$", X, dir(270));
dot("$B_1$", B1, dir(10));
dot("$C_1$", C1, dir(180));
dot("$B_2$", B2, dir(180));
dot("$C_2$", C2, dir(0));
dot("$T$", T, dir(270));
draw(arc(O,abs(O-B1),225,360),dashed);

[/asy]

For this new problem, let $X = \overline{BC_1} \cap \overline{CB_1}$. Then by definition $A,B_1,C_2,X$ are concyclic, and similarly $A,C_1,B_2,X$ are concyclic. Moreover, by Ceva-Menelaus $X$ is on the $A$-internal bisector. So
$$\measuredangle B_1B_2C_1 = \measuredangle XAC_1 = \measuredangle B_1AX = \measuredangle B_1C_2C_1,$$and thus the self-intersecting quadrilateral $B_1B_2C_1C_2$ is cyclic and evidently has Miquel point $A$. Let $T_0 = \overline{B_1C_1} \cap \overline{B_2C_2}$; then it is well known that $\angle XAT_0 = 90^\circ$. Thus, $T=T_0$, as desired.
This post has been edited 1 time. Last edited by Th3Numb3rThr33, Jan 6, 2020, 1:06 AM
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pinetree1
1206 posts
#5 • 3 Y
Y by tworigami, Adventure10, Rounak_iitr
Here's my solution with a different finish (which might actually be essentially the same, but phrased differently).

Let $L$ be the midpoint of arc $BAC$ and $R=\overline{AT}\cap\overline{B_1C_1}$, and denote by $\omega_B$ and $\omega_C$ the circumcircles of $\triangle ABC_1$ and $\triangle ACB_1$. Finally, let $T = \omega_B\cap \omega_C\neq A$. We will show that $L$ is the desired fixed point.
[asy]
defaultpen(fontsize(10pt));
size(350);
pair A, B, C, B1, C1, B2, C2, L, R, S, T;
A = dir(120);
B = dir(210);
C = dir(330);
real r = 0.7;
B1 = intersectionpoint(Circle(B, r), A--B);
C1 = intersectionpoint(Circle(C, r), A--C);
B2 = intersectionpoints(Line(B1, B1+C1-B, 20), circumcircle(A, C, B1))[1];
C2 = intersectionpoints(Line(C1, C1+B1-C, 20), circumcircle(A, B, C1))[1];
L = dir(90);
S = extension(B1, C1, B2, C2);
T = extension(B1, C2, C1, B2);
R = extension(A, T, B1, C1);
draw(A--B--C--cycle, orange);
draw(circumcircle(A, B, C), red);
draw(circumcircle(A, C, B1), lightblue);
draw(circumcircle(A, B, C1), lightblue);
draw(arc(circumcenter(A, B2, C2), circumradius(A, B2, C2), 55, 155), magenta+dotted);
draw(circumcircle(A, B1, C1), heavygreen+dashed);
draw(arc(L, circumradius(B1, C1, B2), 190, 365), heavycyan+dashed);
draw(B--C1^^C--B1, purple+dotted);
draw(B1--B2^^C1--C2, purple+dotted);
draw(B2--T--C2, heavycyan);
draw(L--S^^B2--S^^C1--S, deepgreen);
draw(A--T, purple+dotted);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$B_2$", B2, dir(20));
dot("$C_2$", C2, dir(150));
dot("$B_1$", B1, dir(210));
dot("$C_1$", C1, dir(0));
dot("$L$", L, dir(90));
dot("$T$", T, dir(270));
dot("$S$", S, dir(180));
dot("$R$", R, dir(50));
dot(extension(B1, B2, C1, C2));
label("$\omega_B$", (-1.21, -0.1));
label("$\omega_C$", (1.33, 0.1));
[/asy]

Claim: We have $T = \overline{B_1C_2}\cap \overline{C_1B_2}$.

Proof. Redefine $B_2 = \overline{TC_1}\cap \omega_C$ and $C_2 = \overline{TB_1}\cap \omega_B$; we will show that $\overline{B_1B_2}\parallel \overline{BC_1}$ and $\overline{C_1C_2}\parallel \overline{CB_1}$.

Indeed, we have
$$\angle AB_1B_2 = \angle ATB_2 = \angle ATC_1 = \angle ABC_1,$$which implies $\overline{B_1B_2}\parallel \overline{BC_1}$. The other pair of parallel lines follows similarly. $\blacksquare$

Now observe that:
  • Since $LB = LC$, $BB_1 = CC_1$, and $\angle B_1BL = \angle C_1CL$, we have $\triangle LBB_1\cong \triangle LCC_1$. Hence $L$ is the center of the spiral similarity (congruence) sending $\overline{BB_1}\to \overline{CC_1}$; in particular, $LAB_1C_1$ is cyclic.
  • By definition, $T$ is the center of the spiral similarity sending $\overline{BB_1}\to \overline{C_1C}$, and since $BB_1 = CC_1$, we actually have $\triangle TBB_1\cong \triangle TC_1C$.
This analysis also shows that $\overline{ART}$ is $\angle A$-bisector (since $BT = TC_1$ and $B_1T = TC$).

Claim: $B_1C_1B_2C_2$ is cyclic.

Proof. We have $\angle BTC_1 = \angle B_1TC = 180^\circ - \angle A$, so
$$\angle B_1C_2C_1 = \angle TB_1C = \angle A/2 = \angle TC_1B = \angle C_1B_2B_1,$$which is enough. $\blacksquare$

Claim: $\overline{AL}$, $\overline{B_1C_1}$, $\overline{B_2C_2}$ concur at a point $S$.

Proof. Let $S = \overline{B_1C_1}\cap \overline{B_2C_2}$; we'll show that $S$ lies on $\overline{AL}$.

By Radical axis on $\omega_B$, $\omega_C$, $(B_1C_1B_2C_2)$, we know that $\overline{AT}$, $\overline{B_1B_2}$, $\overline{C_1C_2}$ concur, which forces $$(S, R; B_1, C_1) = -1.$$On the other hand, $\overline{AT}$ and $\overline{AL}$ are the internal/external angle bisectors of $\angle B_1AC_1$, so
$$(\overline{AL}\cap \overline{B_1C_1}, R; B_1, C_1) = -1,$$which is sufficient. $\blacksquare$

Note that $S$ is the radical center of $(LAB_1C_1)$, $(B_1B_2C_1C_2)$, and $(AB_2C_2)$. Thus $\overline{SAL}$ is the radical axis of $(AB_2C_2)$ and $(LAB_1C_1)$, which forces $L$ to lie on $(AB_2C_2)$, as desired.
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yayups
1614 posts
#6 • 3 Y
Y by amar_04, sameer_chahar12, Adventure10
By spiral similarity, the circle $(AB_1C_1)$ passes through the arc midpoint of $BAC$. We'll now $\sqrt{bc}$-invert the entire problem. Stated below is the inverted problem, which we'll solve.
Inverted Problem wrote:
Let $ABC$ be a triangle, and let $S$ be the intersection of the external angle bisector of $\angle BAC$ with $BC$. Let $B_1\in AC$ and $C_1\in AB$ such that $S,B_1,C_1$ collinear. Let $B_2\in BB_1$ such that $(AB_1B_2)$ is tangent to $(AC_1C)$, and define $C_2$ similarly. Show that $B_2C_2$ passes through a fixed point on $BC$.
[asy]
unitsize(2.5inches);
pair A=dir(125);
pair B=dir(220);
pair C=dir(-40);
pair I=incenter(A,B,C);
pair D=extension(A,I,B,C);
pair E=extension(B,I,A,C);
pair F=extension(C,I,A,B);
pair S=extension(E,F,B,C);
pair P=0.8*D+0.2*A;
pair B1=extension(B,P,A,C);
pair C1=extension(C,P,A,B);
pair Y=extension(B,C1+B1-B,A,C);
pair Z=extension(C,B1+C1-C,A,B);
pair B2=2*foot(circumcenter(A,Z,B1),B,B1)-B1;
pair C2=2*foot(circumcenter(A,Y,C1),C,C1)-C1;
pair Q=extension(A,D,B2,C1);

draw(circumcircle(A,B,C));
draw(A--B--C--cycle);
draw(A--D);
draw(C--C1);
draw(B--B1);
draw(B1--Z,dotted);
draw(C1--Y,dotted);
draw(B2--C1,dotted);
draw(C2--B1,dotted);

dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$B_1$",B1,dir(B1));
dot("$C_1$",C1,dir(C1));
dot("$Y$",Y,dir(Y));
dot("$Z$",Z,dir(Z));
dot("$B_2$",B2,dir(B2));
dot("$C_2$",C2,dir(C2));
dot("$Q$",Q,dir(Q));
dot("$P$",P,dir(P));
dot("$D$",D,dir(D));
[/asy]

As expected, we'll show that the fixed point is $S$. Let $Y\in AC$ and $Z\in AB$ such that $C_1Y\parallel BB_1$ and $B_1Z\parallel CC_1$. Then, we have that $B_2\in(AZB_1)$ and $C_2\in(AYC_1)$. Let $P=BB_1\cap CC_1$, and let $D$ be the foot of the angle bisector from $A$ to $BC$. Since $B_1C_1$ passes through $S$, we have by Ceva-Menalaus that $P\in AD$.

We want to show that $B_1C_1$, $B_2C_2$, and $BC$ are concurrent. By the so called prism lemma, it suffices to show that $(BP;B_2B_1)=(CP;C_2C_1)$, or that
\[(AB,AD;AB_2,AC) = (AC,AD;AC_2,AB).\]By reflecting in $AD$, we see then that it suffices to show $AB_2$ and $AC_2$ are isogonal.

Note that
\[AY\cdot AB = \frac{AC_1}{AB}\cdot AB_1\cdot AB = AB_1\cdot AC_1,\]and similarly $AZ\cdot AC=AB_1\cdot AC_1$. Thus, $Y\mapsto B$ and $Z\mapsto C$ under $\sqrt{bc}$ inversion in triangle $AB_1C_1$. Thus, $(AB_1Z)$ is sent to $C_1C$ and $(AC_1Y)$ is sent to $B_1B$. Since $C_1C$ and $B_1B$ concur at $P$, which is on $AD$, we see that $(AB_1Z)$ and $(AC_1Y)$ concur at $Q$, which is also on $AD$. Then, we have (all angles directed mod $\pi$)
\[\angle AQC_1=\angle AYC_1 = \angle AB_1B = \angle AQB_2,\]so $C_1,Q,B_2$ collinear. Similarly, $C_2,Q,B_1$ collinear.

Now, by DDIT on $C_1C_2B_2B_1$ with pencil point $A$, we see that there is an involution swapping the pairs $\{AC_1,AB_1\}$, $\{AC_2,AB_2\}$, and $\{AQ,AP\}$. The first and third imply that the involution is isogonality, so $AB_2$ and $AC_2$ isogonal, as desired.
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Physicsknight
635 posts
#7 • 1 Y
Y by Adventure10
Let $D $ be the midpoint of arc $\widehat {BAC} $ of $\triangle ABC $. Let $\odot(AB_1C) $ cuts the internal angle bisector of $\widehat {BAC} $ at $H\neq A $, and $\odot (ABH) $ cuts $AC $ at $C_3\neq A $. Note, $HB=HC_3$, $HC=HB_1$, $\widehat {BHC_3}=180^{\circ}-\widehat {BAC}=\widehat {B_1HC}\implies\triangle BHB_1=\triangle CHC_3\implies CC_3=BB_1=CC_1\implies C_3\equiv C_1$.
But, $\widehat {HB_1C}=\widehat {HAC}=\widehat {HC_2C_1}\implies\overline {H,B_1,C_2}$.
Similarly, $\overline {H,C_1,B_2} $. Thus $\widehat {HC_2C_1}=\widehat {HAC}=\widehat {HAB}=\widehat {HB_2B_1}\implies B_1,B_2,C_1,C_2$ are concyclic.
$\triangle DBB_1=\triangle DCC_1\implies DB_1=DC_1$, and $\widehat {B_1DC_1}=\widehat {BAC}=\widehat {BDC}=2\widehat {BAH}=2\widehat {B_1C_2C_1}=\widehat {B_2B_1C}+2\widehat {BAH}+\widehat{C_2C_1B}=\widehat{B_2AC}+\widehat {CAB}+\widehat {BAC_2}=\widehat {B_2AC_2} $.
It is proved that $\odot (AB_2C_2) $ passes through $D $, which is fixed point.
$\blacksquare $
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TheUltimate123
1739 posts
#8 • 2 Y
Y by CyclicISLscelesTrapezoid, Adventure10
[asy]
        size(9cm); defaultpen(fontsize(10pt));

        pair A,B,C,I,D,B1,C1,B2,C2,K,X,Y;
        A=dir(125);
        B=dir(210);
        C=dir(330);
        I=incenter(A,B,C);
        real t=0.7;
        D=(t+1)*I-t*A;
        B1=2*foot(circumcenter(A,D,C),A,B)-A;
        C1=2*foot(circumcenter(A,D,B),A,C)-A;
        B2=2*foot(circumcenter(A,D,C),B1,B1+C1-B)-B1;
        C2=2*foot(circumcenter(A,D,B),C1,C1+B1-C)-C1;
        K=dir(90);
        X=extension(B1,B2,C1,C2);
        Y=extension(B1,C1,B2,C2);

        //draw(B1--B2,Dotted); draw(C1--C2,Dotted);
        draw(B--C1,Dotted); draw(C--B1,Dotted);
        draw(A--D,dashed);
        draw(arc(circumcenter(B1,C1,B2),circumradius(B1,C1,B2),170,10,CCW),gray);
        draw(B2--D--C2,gray);
        draw(circumcircle(A,B1,C1),dashed);
        draw(circumcircle(A,D,C));
        draw(circumcircle(A,D,B));
        draw(A--B--C--A);
        draw(K--Y--C1);
        draw(Y--B2);

        dot("$A$",A,NW);
        dot("$B$",B,SW);
        dot("$C$",C,SE);
        dot("$D$",D,S);
        dot("$B_1$",B1,dir(215));
        dot("$C_1$",C1,E);
        dot("$B_2$",B2,E);
        dot("$C_2$",C2,dir(140));
        dot("$K$",K,N);
        dot("$X$",Y,W);
    [/asy]


Let $K$ be the midpoint of arc $BAC$ on the circumcircle of $\triangle ABC$, and let $(ABC_1)$ and $(ACB_1)$ intersect again at $D$. Denote $X=\overline{B_1B_2}\cap\overline{C_1C_2}$ and $Y=\overline{B_1C_1}\cap\overline{B_2C_2}$. I claim that $K$ is the fixed point.

Since $KB=KC$, $BB_1=CC_1$, and $\measuredangle KBB_1=\measuredangle KCC_1$, $\triangle KBB_1\cong\triangle KCC_1$, so $K$ is the center of spiral similarity sending $\overline{BB_1}$ to $\overline{CC_1}$ and $K$ lies on $(AB_1C_1)$. Moreover, $D$ is the center of spiral similarity sending $\overline{BB_1}$ to $\overline{C_1C}$. Since $BB_1=CC_1$, we have $DB=DC_1$, so $\overline{AD}$ bisects $\angle BAC$.

Notice that \[\measuredangle ADB_1=\measuredangle ACB_1=\measuredangle AC_1C_2=\measuredangle ADC_2\]and $\measuredangle ADC_1=\measuredangle ADB_2$, so $D=\overline{B_1C_2}\cap\overline{B_2C_1}$. Furthermore \[\measuredangle B_1B_2C_1=\measuredangle B_1AD=\measuredangle DAC_1=\measuredangle B_1DC,\]so $B_1C_1B_2C_2$ is cyclic.

Finally $A$ is the Miquel point of $B_1B_2C_1C_2$, so by a well-known property of complete quadrilaterals, $A$ is the foot from $X$ to $\overline{AD}$, id est $\angle XAD=90^\circ$. It follows that $K\in\overline{AX}$, so \[XB_2\cdot XC_2=XB_1\cdot XC_1=XA\cdot XK,\]and we are done.
This post has been edited 1 time. Last edited by TheUltimate123, Jan 20, 2020, 8:28 AM
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v_Enhance
6857 posts
#9 • 7 Y
Y by Aryan-23, GeoMetrix, amar_04, sameer_chahar12, mijail, v4913, Mango247
Solution with Aditya Khurmi, Anant Mudgal, Anushka Aggarwal, Arindam, Dhrubajyoti Ghosh, Paul Hamrick, Pranjal Srivastava, Rishabh, Robin Son, Rohan Goyal, William Hu, Zifan Wang:

Let $D = \overline{B_1B_2} \cap \overline{C_1C_2}$ and $E = \overline{BC_1} \cap \overline{B_1C}$. Moreover let $X$ be the arc midpoint of $\widehat{BAC}$. We will show that $X$ is the desired fixed point.



[asy] size(15cm);  real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */  pen dotstyle = black; /* point style */  real xmin = -15.08, xmax = 18.82, ymin = -6.34, ymax = 10.22;  /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen zzffff = rgb(0.6,1.,1.); pen ttffqq = rgb(0.2,1.,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen ffdxqq = rgb(1.,0.8431372549019608,0.);

draw((-1.,8.)--(-5.,-2.)--(7.,-2.)--cycle, linewidth(1.) + zzttqq);  /* draw figures */ draw((-1.,8.)--(-5.,-2.), linewidth(1.) + zzttqq);  draw((-5.,-2.)--(7.,-2.), linewidth(1.) + zzttqq);  draw((7.,-2.)--(-1.,8.), linewidth(1.) + zzttqq);  draw(circle((3.,3.), 6.4031), linewidth(1.) + zzffff);  draw(circle((-1.3780,2.3512), 5.6614), linewidth(1.) + zzffff);  draw((-3.34482,2.1379)--(8.3309,6.5469), linewidth(1.));  draw((4.2159,1.4800)--(-6.0296,5.5783), linewidth(1.));  draw(circle((1.8537,-4.3609), 12.686), linewidth(1.) + linetype("2 2"));  draw((-5.,-2.)--(4.2159,1.4800), linewidth(1.));  draw((7.,-2.)--(-3.3448,2.1379), linewidth(1.));  draw(circle((1.,1.4), 6.8963), linewidth(1.) + red);  draw(circle((1.,8.2963), 7.5368), linewidth(1.) + ttffqq);  draw(circle((0.6219,3.9512), 4.3615), linewidth(1.) + red);  draw((0.6219,-2.9451)--(-6.0296,5.5783), linewidth(1.));  draw((0.6219,-2.9451)--(8.3309,6.5469), linewidth(1.));  draw((-1.,8.)--(1.,-5.4963), linewidth(1.));  draw(circle((-2.8092,5.3236), 3.2304), linewidth(1.) + ffdxqq);  draw(circle((3.5076,6.2597), 4.8318), linewidth(1.) + ffdxqq);   /* dots and labels */ dot("$A$", (-1.,8.), dir(90)); dot("$B$", (-5.,-2.), dir(225)); dot("$C$", (7.,-2.), dir(-45)); dot("$B_1$", (-3.3448,2.1379), dir(190)); dot("$C_1$", (4.2159,1.4800), dir(0)); dot("$B_2$", (8.3309,6.5469), dir(30)); dot("$C_2$", (-6.0296,5.5783), dir(160)); dot((1.,-5.4963)); dot("$X$", (1.,8.2963), dir(90)); dot("$D$", (-0.3016,3.2871), dir(75)); dot("$E$", (1.1645,0.3278), dir(-90)); dot("$Y$", (0.6219,-2.9451), dir(-45)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);   /* end of picture */ [/asy]


To begin, note that $B_1DC_1E$ is a parallelogram. Define $\theta = \measuredangle EB_1D = \measuredangle DC_1E$; this angle appears in several places. We now proceed in eight steps.
  1. Since \[ \measuredangle C_2AB = \measuredangle C_2 C_1 B = \theta = \measuredangle CB_1 B_2 = \measuredangle CA_2 B_2. \]it follows $\overline{AC_2}$ and $\overline{AB_2}$ are isogonal with respect to $\angle A$.
  2. Let $D = \overline{B_1B_2} \cap \overline{C_1C_2}$. Since $\measuredangle C_2AB_1 = \theta = \measuredangle C_2DB_1$, we have $C_2ADB_1$ is cyclic; similarly $B_2ADC_2$ are cyclic.
  3. Note that $A$ is Miquel point of self-intersecting quadrilateral $C_1C_2B_1B_2$. It follows that $Y \overset{\text{def}}{=} \overline{C_1B_2} \cap \overline{C_2B_1}$ lies on both $(C_2AC_1)$ and $(B_2AB_1)$.
  4. The isogonal lemma on $\triangle AB_1C_1$ now gives $AD$, $AY$ isogonal with respect to $\angle BAC$. (This is equivalent to DDIT on $B_1B_2C_1C_2DY$ from $A$.)
  5. Since $Y$ is now the spiral center mapping congruent segments $BB_1$ and $C_1C$, it follows $YB_1 = YC$, $YB = YC_1$, so $\overline{AY}$ is an angle bisector. Since $\overline{AD}$ and $\overline{AY}$ were isogonal, we conclude $ADY$ are collinear.
  6. By power of a point from $Y$, we find $B_1B_2C_1C_2$ is cyclic (with Miquel point $A$).
  7. Then it follows by Miquel theory, and from $XB_1 = XC_1$, that $X$ is the circumcenter of $B_1B_2C_1C_2$.
  8. By Miquel theory, $XAB_2C_2$ concyclic as needed.
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NJOY
495 posts
#11 • 2 Y
Y by char2539, Purple_Planet
My $256\text{th}$ post. :)

Common remarks/Constructions. Let $T$ be the mid-point of the arc $\widehat{BAC}$ of $\odot(ABC)$. Again, let $P$ be the point of intersection of $\odot(ABC_1)$ and $\odot(ACB_1)$.

We prove that the desired fixed point is the point $T$.

[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.5) + fontsize(8); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -4.767963430995203, xmax = 11.435801392656167, ymin = -3.0611011252211666, ymax = 8.06902586020495;  /* image dimensions */
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ffffww = rgb(1,1,0.4); pen ffzzcc = rgb(1,0.6,0.8); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); 

draw((0,6)--(-2.34,0.64)--(5.58,0.04)--cycle, linewidth(1) + rvwvcq); 
draw(circle((2.07585201821125,6.357246640388502), 5.306669925109475), linewidth(1) + orange); 
draw(circle((1.7492097630383627,2.0455688721063865), 4.3240328791953795), linewidth(1) + ffzzcc); 
draw(circle((1.9115843695020698,1.1020592167954548), 5.257754151489301), linewidth(1) + sexdts); 
 /* draw figures */
draw((0,6)--(-2.34,0.64), linewidth(1) + rvwvcq); 
draw((-2.34,0.64)--(5.58,0.04), linewidth(1) + rvwvcq); 
draw((5.58,0.04)--(0,6), linewidth(1) + rvwvcq); 
draw((-1.539797717153027,2.472941981222126)--(5.58,0.04), linewidth(1)); 
draw((4.213095871870886,1.4999907891845023)--(-2.34,0.64), linewidth(1)); 
draw(circle((0.7513492591259361,2.4812020025457664), 3.5981196556086097), linewidth(1)); 
draw(circle((2.5440342362231454,2.7897166171350927), 4.096099314393395), linewidth(1)); 
draw((-2.5869508073840293,3.8236690308248895)--(4.213095871870886,1.4999907891845023), linewidth(1)); 
draw((-1.539797717153027,2.472941981222126)--(6.571316432489792,3.5373981703432285), linewidth(1)); 
draw((-2.5869508073840293,3.8236690308248895)--(0,6), linewidth(1)); 
draw((0,6)--(6.571316432489792,3.5373981703432285), linewidth(1)); 
draw((-2.5869508073840293,3.8236690308248895)--(6.571316432489792,3.5373981703432285), linewidth(1)); 
draw((2.075852018211249,6.357246640388503)--(6.571316432489792,3.5373981703432285), linewidth(1) + linetype("4 4") + sexdts); 
draw((0,6)--(2.075852018211249,6.357246640388503), linewidth(1)); 
draw((0,6)--(1.219531477137834,-1.0863280207076442), linewidth(1) + linetype("4 4")); 
draw((-2.34,0.64)--(2.075852018211249,6.357246640388503), linewidth(1) + dtsfsf); 
draw((2.075852018211249,6.357246640388503)--(5.58,0.04), linewidth(1) + dtsfsf); 
draw((-1.539797717153027,2.472941981222126)--(2.075852018211249,6.357246640388503), linewidth(1) + linetype("4 4") + sexdts); 
draw((2.075852018211249,6.357246640388503)--(4.213095871870886,1.4999907891845023), linewidth(1) + linetype("4 4") + sexdts); 
draw((-2.5869508073840293,3.8236690308248895)--(1.219531477137834,-1.0863280207076442), linewidth(1) + wvvxds); 
draw((1.219531477137834,-1.0863280207076442)--(6.571316432489792,3.5373981703432285), linewidth(1) + wvvxds); 
draw((-2.34,0.64)--(1.219531477137834,-1.0863280207076442), linewidth(1) + wvvxds); 
draw((1.219531477137834,-1.0863280207076442)--(5.58,0.04), linewidth(1) + wvvxds); 
draw((-2.5869508073840293,3.8236690308248895)--(2.075852018211249,6.357246640388503), linewidth(1) + linetype("4 4") + sexdts); 
draw((-1.539797717153027,2.472941981222126)--(4.213095871870886,1.4999907891845023), linewidth(1)); 
 /* dots and labels */
dot((0,6),linewidth(3pt) + dotstyle); 
label("$A$", (-0.11,6.12988679114503), NE * labelscalefactor); 
dot((-2.34,0.64),linewidth(3pt) + dotstyle); 
label("$B$", (-2.7,0.40), NE * labelscalefactor); 
dot((5.58,0.04),linewidth(3pt) + dotstyle); 
label("$C$", (5.65,-0.19223784496813395), NE * labelscalefactor); 
dot((-1.539797717153027,2.472941981222126),linewidth(3pt) + dotstyle); 
label("$B_1$", (-2.1,2.33), NE * labelscalefactor); 
dot((4.213095871870886,1.4999907891845023),linewidth(3pt) + dotstyle); 
label("$C_1$", (4.36989738758852,1.3617297651689253), NE * labelscalefactor); 
dot((-2.5869508073840293,3.8236690308248895),linewidth(3pt) + dotstyle); 
label("$C_2$", (-3.02,3.8852669098359445), NE * labelscalefactor); 
dot((6.571316432489792,3.5373981703432285),linewidth(3pt) + dotstyle); 
label("$B_2$", (6.7074896900169145,3.473531902021852), NE * labelscalefactor); 
dot((2.075852018211249,6.357246640388503),linewidth(3pt) + dotstyle); 
label("$T$", (2.1252775062794367,6.435367603394195), NE * labelscalefactor); 
dot((1.219531477137834,-1.0863280207076442),linewidth(3pt) + dotstyle); 
label("$P$", (1.1,-1.43), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]

First, we prove the following claim.

Claim. The lines $B_1C_2$ and $C_1B_2$ meet at point $P$.
Proof. To prove this, let us redefine the point $B_2$ as the intersection of lines $PC_1$ and $\odot(ACB_1)$. Redefine the point $C_2$ similarly. Then, we have $$\angle AB_1B_2 = \angle APB_2 = \angle APC_1 = \angle ABC_1,$$which implies that the lines $B_1B_2$ and $BC_1$ are parallel. Similarly, $C_1C_2\parallel CB_1$. Thus, we obtain that $B_2$, $C_2$ are the points as defined earlier and hence, the lines $B_1C_2$ and $C_1B_2$ meet at point $P$. $\square$

By the problem, we have that $P$ is the center of spiral similarity that maps $BB_1 \mapsto CC_1$, and thus, we conclude that $AP$ is the internal angle bisector of $\angle BAC$. It follows that $AT\perp AP$. This further implies that the quadrilateral $AB_1C_1T$ is cyclic, which yields $\angle B_1AC_1=\angle B_1TC_1$.

Next, we prove this crucial claim.

Claim. The quadrilateral $B_1B_2C_1C_2$ is cyclic with center $T$.
Proof. Note that $$\angle B_1C_2C_1=\angle PC_2C_1=\angle PAC_1=\frac{\angle B_1AC_1}2=\frac{\angle B_1TC_1}2.$$In the same way, we can prove that $$\angle B_1B_2C_1=\frac{\angle B_1TC_1}2,$$which thus yields that the quadrilateral $B_1C_1B_2C_2$ is cyclic with center $T$. $\square$

The above claim also yields that $$TB_1=TC_1=TB_2=TC_2.$$
Finally, observe that $$\angle BAC_2 = \angle BC_1C_2 = \angle CB_1B_2 = \angle CAB_2,$$which thus implies that the lines $AB_2$ and $AC_2$ are isogonal with respect to $\angle BAC$. It follows that $AP$ is also the angle bisector of $\angle B_2AC_2$. Again, as $AT\perp AP$, we obtain that $AT$ is the external angle bisector of $\angle B_2AC_2$. Now, as $TB_2=TC_2$, we conclude that $T$ is the mid-point of arc $\widehat{B_2AC_2}$, which finally implies that the circumcircle of $\odot(AB_2C_2)$ pass through the point $T$, as desired. $\blacksquare$
This post has been edited 2 times. Last edited by NJOY, Feb 24, 2020, 8:45 AM
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amar_04
1915 posts
#13 • 7 Y
Y by mueller.25, GeoMetrix, lilavati_2005, Hexagrammum16, AmirKhusrau, Purple_Planet, Bumblebee60
Great Problem! Congrats to Tworigami. :) Here's a bashy and a different solution which I got and also this is the first time I carried out a Menelaus and Trig Bash properly. :D
tworigami wrote:
Let $ABC$ be a triangle and let $B_1$ and $C_1$ be variable points on sides $\overline{BA}$ and $\overline{CA}$, respectively, such that $BB_1 = CC_1$. Let $B_2 \neq B_1$ denote the point on $\odot(ACB_1)$ such that $BC_1$ is parallel to $B_1B_2$, and let $C_2 \neq C_1$ denote the point on $\odot(ABC_1)$ such that $CB_1$ is parallel to $C_1C_2$. Prove that as $B_1, C_1$ vary, the circumcircle of $\triangle AB_2C_2$ passes through a fixed point, other than $A$.

Proposed by tworigami

Let $\odot(AB_1C_1)\cap\odot(ABC)=T$. So there $\exists$ a unique Spiral Similarity $(\sigma)$ centered at $T$ such that $\sigma:\triangle TB_1B\mapsto \triangle TC_1C\implies TB=TC$ as $BB_1=CC_1$. So, $T$ is the midpoint of $\widehat{BAC}$. Now we will Invert around $A$ with radius $\sqrt{AB\cdot AC}$ followed by a reflection along the angle bisector of $\angle BAC$. The Problem becomes equivalent to this one. (Diagram taken from #4)
Inverted Problem wrote:
$ABC$ be a triangle and the External Bisector of $\angle BAC$ intersects $BC$ at $T$. A ray passing through $T$ intersects $\{AB,AC\}$ at $\{B_1,C_1\}$ respectively. Let $BC_1\cap CB_1=X$. Let $\dot(AB_1X)\cap BC_1=C_2$ and $\odot(AC_1X)\cap CB_1=B_2$. Then $\overline{T-B_2-C_2}.$

[asy]
defaultpen(fontsize(11pt));
size(10cm);
real t = 0.5;
pair A = dir(160); pair B = dir(225); pair C = dir(540-225);
pair X = t*A + (1-t)*dir(270);
pair C1 = extension(B,X,A,C); pair B1 = extension(C,X,A,B);
path circ1 = circumcircle(A,B1,X); path circ2 = circumcircle(A,C1,X);
pair B2 = intersectionpoints(C--2B1-C, circ2)[1];
pair C2 = intersectionpoints(B--2C1-B, circ1)[1];
pair T = extension(B,C,B1,C1);
pair O = circumcenter(B1,C1,B2);

filldraw(A--B--C--cycle,pink+white+white);

draw(circ1,orange); draw(circ2,orange);
draw(B--C2,red); draw(C--B2,red);
draw(T--C2,gray(0.5)); draw(T--C1,gray(0.5)); draw(T--B);


dot("$A$", A, dir(90));
dot("$B$", B, dir(270));
dot("$C$", C, dir(270));
dot("$X$", X, dir(270));
dot("$B_1$", B1, dir(10));
dot("$C_1$", C1, dir(180));
dot("$B_2$", B2, dir(180));
dot("$C_2$", C2, dir(0));
dot("$T$", T, dir(270));
draw(arc(O,abs(O-B1),225,360),dashed);

[/asy]

Notice that if $AX\cap BC=Y$, then $(TY;BC)$ is harmonic $\implies AX$ is the bisector of $\angle A$.

Claim:- $B_1B_2C_1C_2$ is a cyclic quadrilateral.
Proof:- $\angle C_1B_2B_1=\angle C_1B_2X=\angle C_1AX=\angle XAB_1=\angle B_1C_2X=\angle B_1C_2C_1$. So, $B_1B_2C_1C_2$ is a cyclic quadrilateral.


Now we will use Menelaus Theorem in order to prove that $\overline{T-B_2-C_2}$.


So let's compute $\frac{BC_2}{XC_2}$ and $\frac{XB_2}{CB_2}$ first. We will use PoP to compute them. $$\begin{cases} \frac{BC_2\cdot BX}{XC_2\cdot XC_1}=\frac{BA\cdot BB_1}{XB_2\cdot XB_1}\implies\frac{BC_2}{XC_2}=\frac{BA\cdot BB_1\cdot XC_1}{XB_1\cdot XB_2\cdot XB} \\  \\ \frac{XB_2\cdot XB_1}{CB_2\cdot CX}=\frac{XB_1\cdot XB_2}{CC_1\cdot CA}\implies \frac{XB_2}{CB_2}=\frac{XB_1\cdot XB_2\cdot XC}{CC_1\cdot CA\cdot XB_1} \end{cases}$$Now to prove that $\overline{T-B_2-C_2}$ it suffices to prove this $\longrightarrow$

\begin{align*}
\iff\frac{CT}{TB}\cdot\frac{BC_2}{XC_2}\cdot\frac{XB_2}{CB_2}=1 \\
&\iff \frac{CA}{AB}\cdot\frac{BA\cdot BB_1\cdot XC_1}{XB_1\cdot XB_2\cdot XB}\cdot\frac{XB_1\cdot XB_2\cdot XC}{CC_1\cdot CA\cdot XB_1}=1\\
&\iff \frac{BB_1\cdot XC_1\cdot XC}{XB\cdot CC_1\cdot XB_1}=1\cdots\cdots\cdots (\star)
\end{align*}

Now we will prove the following Lemma.

Lemma:- In a $\triangle ABC$, let $G$ be an arbitary point inside the plane of $\triangle ABC$ and $\triangle XEF$ be the $G-$ Cevian Triangle WRT $\triangle ABC$ such that $AX$ is the bisector of $\angle BAC$. Then $\frac{GE\cdot GC}{GF\cdot GB}=\frac{CE}{BF}$.


Proof:- We will make use of Sine Rule extensively. $$\begin{cases}\frac{GE}{GF}=\frac{\sin\angle GFE}{\sin\angle GEF} \\ \frac{GC}{GB}=\frac{\sin\angle GBC}{\sin\angle GCB}\end{cases}$$Now by Trig Form of Ceva's Theorem we get that $$\frac{\sin\angle BAX}{\sin\angle CAX}\cdot\frac{\sin\angle ACF}{\sin\angle BCF}\cdot\frac{\sin\angle EBC}{\sin\angle EBA}=1\implies\frac{\sin\angle GBC}{\sin\angle GCB}=\frac{\sin\angle EBF}{\sin\angle ECF}$$
Hence putting the values back we get that,
$$\frac{GE}{GF}\cdot\frac{GC}{GB}=\frac{\sin\angle GFE\cdot\sin\angle GBC}{\sin\angle GEF\cdot\sin\angle GCB}=\frac{\sin\angle CFE}{\sin\angle BEF}\cdot\frac{\sin\angle EBF}{\sin\angle ECF}=\frac{CE}{EF}\cdot\frac{EF}{BF}=\frac{CE}{BF}$$

Back to the Inverted Problem:-

From (Lemma) indeed we get that $(\star)=1$. So, by Converse of Menelaus Theorem we get that $\overline{T-B_2-C_2}$. So, Inverting back we get that $\odot(AB_2C_2)$ passes through the midpoint of $\widehat{BAC}$ which is $T$. $\blacksquare$
This post has been edited 1 time. Last edited by amar_04, Feb 29, 2020, 11:26 AM
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GeoMetrix
924 posts
#14 • 6 Y
Y by AmirKhusrau, mueller.25, amar_04, Purple_Planet, Mango247, Mango247
Really nice problem. Although my solution is similiar to the ones above but since this is too nice i cant resist to post.

[asy]
size(10cm);
pair A=(4.089884929030667,7.097525729001119);
pair B=(-0.12839547437282756,-4.869227188455621);
pair C=(16.954144315296645,-4.629892130106486);
pair B1=(1.9358694038884572,1.0243986233918232);
pair C1=(12.317027559782165,-0.4714454912902692);
pair B2=(17.552481961169484,6.559021847715566);
pair C2=(-0.5173149441901711,4.345172557986069);
pair D=(8.218414685553237,8.054865962397658);
pair F=(5.316477103069982,2.1612401505502135);
pair G=(7.560243275093117,-6.843741419835983);
filldraw(A--B--C--cycle,orange+white+white+white,orange);
draw(circumcircle(A,B,C),red);
draw(C--B1,green+dotted);
draw(B--C1,green+dotted);
draw(circumcircle(A,B,C1),purple+dotted);
draw(circumcircle(A,C,B1),purple+dotted);
draw(C1--C2,green+dotted);
draw(B1--B2,green+dotted);
draw(A--G,red+dotted);
draw(C2--B,green);
draw(C--B2,green);
draw(G--B2,cyan);
draw(G--C2,cyan);
draw(circumcircle(A,B1,C1),cyan+dashed);
draw(arc(circumcenter(A, B2, C2), circumradius(A, B2, C2), 55, 150),pink);
draw(arc(D,circumradius(B1,B2,C1),180,360),lightblue);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(D));
dot("$G$",G,dir(G));
dot("$B_1$",B1,dir(B1));
dot("$C_1$",C1,dir(C1));
dot("$B_2$",B2,dir(B2));
dot("$C_2$",C2,dir(C2));
dot("$F$",F,dir(F));
[/asy]
We proceed with several claims.

Claim 1: If $\odot(C_2BC_1) \cap (B_2CB_1)$ and $G \neq A$ then $\overline{GC_2}$ bisects $\angle BC_2C_1$ and $\overline{GB_2}$ bisects $\angle B_1B_2C$

Proof: Notice that there exists a spiral similiarity at $G$ such that $G: \overline{BB_1} \mapsto \overline{CC_1}$. This implies that $\overline{GB}=\overline{GC_1}$ and $\overline{GB_1}=\overline{GC}$ and this proves the claim $\qquad \square$

Claim 2: $B_1\in \overline{GC_2}$ and also $C_1 \in \overline{GB_2}$

Proof: Notice that $$\angle C_1C_2G=\angle C_1AG=\angle CAG=\angle CB_1G$$but since $\overline{CB_1}\parallel \overline{C_1C_2}$ hence we have that $B_1\in \overline{GC_2}$. A similiar arguement shows that $C_1 \in \overline{GB_2}$ and we're done $\qquad \square$

Claim 3: $B_1C_1B_2C_2$ is cyclic

Proof: Notice that $$\angle B_1B_2C_1=\angle BC_1G=\angle BC_2G=\angle GC_2C_1=\angle B_1C_2C_1$$and this proves the claim $\qquad \square$

Claim 4: If $F=\overline{B_1B_2} \cap \overline{C_1C_2}$ then $F \in \overline{AG}$ . Also $AG$ is the angle bisector OF $\angle BAC$

Proof: The first part is just radical axis on $\odot(ABC_1),\odot(ACB_1),\odot(B_1B_2C_1C_2)$. For the second just notice that $$\angle BAG=\angle BC_2G=\angle GC_2C_1=\angle GAC_1$$which proves the second part $\qquad \square$

Claim 5: If $D$ is the midpoint of $\widehat{BAC}$ then $D$ is the circumcenter of $\odot(B_1B_2C_1C_2)$

Proof: Notice that simple congruency stuff implies that $\triangle{DB_1B} \cong \triangle{DC_1C}$ so $\overline{DB_1}=\overline{DC_1}$. Now just notice that this implies that $D\in \odot(AB_1C_1)$. Hence we have that $$\angle B_1DC_1=\angle B_1AC_1=\angle BAC_1=\angle BC_2C_1=2\angle B_1C_2C_1$$and this proves the claim $\qquad \square$

Now back to the main problem. We will claim that $D$ is infact the required fixed point. Notice that $$\angle C_2DB_2=2\angle C_2C_1G=2\angle C_2AG$$But since $$\angle C_2AB=\angle C_2GB=\angle B_2GC=\angle B_2AC$$hence $$2\angle C_2AG=\angle C_2AG+\angle B_2AG=\angle C_2AB_2$$and with this we are done $\qquad \blacksquare$
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mathlogician
1051 posts
#16 • 3 Y
Y by Mango247, Mango247, Mango247
There was a small hole in the previously posted solution, which has now been fixed.

Let $M$ be the second intersection of circles $(AB_1C_1)$ and $(ABC)$.

We begin with some easy observations. Note that $M$ is the Miquel point of $B_1BCC_1$, so $\triangle MB_1B \sim \triangle MC_1C$. But since $B_1B = C_1C$, this similarity is in fact a congruence, so $MB_1= MC_1$ and $MB=MC$. Therefore, $M$ is invariant of $B_1$ and $C_1$. I claim that $M$ is the desired fixed point.

Claim: $A$ is the Miquel Point of $C_1C_2B_1B_2$.

Proof: Let $T = C_1C_2 \cap B_1B_2$. Notice that $$\measuredangle B_2TC_1 = \measuredangle B_2B_1C = \measuredangle B_2AC = \measuredangle B_2AC_1$$whence $B_2ATC_1$ is cyclic. Similarly, $C_2ATB_1$ is cyclic, proving the desired claim.

Let $K = B_1C_2 \cap C_1B_2$. Note that by properties of Miquel point that $AC_1KC_2$ and $AB_1KB_2$ are cyclic.

Claim: $\angle BAK = \angle CAK$.

Proof: Note that there exists a spiral similarity taking $BB_1$ to $C_1C$. Since $B_1B= C_1C$ this similarity is a congruence, so $BK = C_1K$ and by Fact 5 we have our desired angle equality.


Claim: $B_1C_1B_2C_2$ is cyclic.

Proof:
$$\measuredangle C_1C_2B_1 = \measuredangle CB_1K = \measuredangle CAK = \measuredangle KAB = \measuredangle KC_1B = \measuredangle C_1B_2B_1.$$
Claim: $M$ is the circumcenter of $B_1C_1B_2C_2$.

Proof: Let $M'$ be the circumcenter of $B_1C_1B_2C_2$. Note that by Miquel, $M'AB_1C_1$ is cyclic, and $M'B_1 = M'C_1$. But there is only one unique point that satisfies both conditions, namely $M$, so $M$ is the circumcenter.

This last claim finishes the problem by a well-known property of Miquel points.[/hide]

Motivational Remarks
This post has been edited 2 times. Last edited by mathlogician, Jul 12, 2020, 8:42 PM
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MP8148
888 posts
#17
Y by
Here is a slightly different inversion solution:

First, like everyone else, note that by spiral sim $(AB_1C_1)$ passes through the midpoint of $\widehat{BAC}$. Now $\sqrt{bc}$ inversion gives the following (primes are omitted):
inverted problem wrote:
In $\triangle ABC$, points $B_1$, $C_1$ lie on sides $\overline{AB}$, $\overline{AC}$ such that $K = \overline{B_1C_1} \cap \overline{BC}$ lies on the external $A$-bisector. Let $B_2$ be the point on $\overline{CB_1}$ such that $(AB_1B_2)$ is tangent to $(ABC_1)$, and similar for $C_2$. Prove that $B_2$, $C_2$, $K$ are collinear.
[asy]
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pair A = dir(160), B = dir(235), C = dir(305), K = extension(A,dir(90),B,C), B1 = A+dir(A--B)*abs(A-B)*0.6, C1 = extension(A,C,K,B1), S = extension(B,C1,C,B1), B3 = intersectionpoint(circumcircle(A,B,C1),B+dir(C--B1)*0.01--B+dir(C--B1)*100), B2 = extension(A,B3,C,B1), C2 = extension(B,C1,K,B2), C3 = extension(A,C2,C,C+dir(B--C1));

draw(A--B--C--A);
draw(circumcircle(A,B,C1)^^circumcircle(A,C,B1));
draw(A--K--B^^C1--K);
draw(C2--K, red+dashed);
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draw(B1--C2^^C1--B2, purple);
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dot("$A$", A, dir(135));
dot("$B$", B, dir(270));
dot("$C$", C, dir(270));
dot("$C_1$", C1, dir(5));
dot("$B_1$", B1, dir(250));
dot("$B_2$", B2, dir(135));
dot("$B_3$", B3, dir(225));
dot("$C_2$", C2, dir(95));
dot("$C_3$", C3, dir(10));
dot("$K$", K, dir(210));
dot("$S$", S, dir(285));
[/asy]
Let $B_3$ be the point on $(ABC_1)$ such that $\overline{BB_3} \parallel \overline{CB_1}$; by homothety we can redefine $B_2 = \overline{CB_1} \cap \overline{AB_3}$. Do the same thing for $C_3$ and $C_2$. Then $$\measuredangle B_3AC_1 = \measuredangle(\overline{B_3B}, \overline{BC_1}) = \measuredangle(\overline{B_1C}, \overline{CC_3}) = \measuredangle B_1AC_3,$$which implies $\overline{AB_2}$ and $\overline{AC_2}$ are isogonal wrt $\angle BAC$.

Furthermore, note that $S = \overline{BC_1} \cap \overline{CB_1}$ lies on the internal $A$-bisector by Ceva-Menelaus, so by isogonal lemma $\overline{C_1B_2} \cap \overline{B_1C_2}$ lies on $\overline{AS}$. Now note that $\overline{B_2C_2}$ meets both $\overline{AK}$ and $\overline{B_1C_1}$ at the harmonic conjugate of $\overline{AS} \cap \overline{B_2C_2}$: the former by a well-known lemma (right angle and angle bisector), the the latter by Ceva-Menelaus. This implies the lines are concurrent, as desired. $\blacksquare$
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rcorreaa
238 posts
#18
Y by
We claim that the desired point is the arc midpoint $X$ of $\widehat{BAC}$. Let $Y= (ABC_1) \cap (ACB_1)$, with $A \neq Y$, $Z= B_1B_2 \cap C_1C_2$ and $W= B_1C \cap BC_1$.

Observe that $Y$ is the center of the spiral similarity mapping $BB_1 \mapsto C_1C$, so $\angle B_1YB= \angle B_1WB= \angle CB_1B_2$ (since $B_1B_2 \parallel BC_1$), and also $\angle CB_1B_2= \angle C_2C_1B$, because $ZB_1WC_1$ is a parallelogram. Thus, $\angle CB_1B_2= \angle C_2C_1B= \angle C_2YB \implies \angle B_1YB= \angle C_2YB$, so $C_2,B_1,Y$ are collinear. Similarly, $B_2,C_1,Y$ are collinear.

Now, notice that $\angle B_1C_2C_1= \angle YC_2C_1= \angle YBC_1$, and since $Y$ is the center of the spiral similarity mapping $BB_1 \mapsto C_1C$, we have that $YB_1B ~ YCC_1$, then since $BB_1=CC_1 \implies YB=YC_1,YC=YB_1$, hence $\angle YBC_1= \angle YC_1B= \angle YCB_1= \angle YB_2B_1= \angle C_1B_2B_1$. Therefore, $\angle B_1C_2C_1= \angle C_1B_2B_1$, so $B_1C_1B_2C_2$ is cyclic. Furthermore, observe that $\angle BAC= 180º- \angle BYC_1= 2 \angle YBC_1= 2 \angle B_1C_2C_1$, so $A$ is the circumcenter of $B_1C_1B_2C_2$. Let $\Omega= (B_1C_1B_2C_2)$.

Consider the inversion $\Phi$ about $\Omega$ centered at $A$. Clearly, since $A \in (C_1C_2B)$, $\Phi(B)=B' \in C_1C_2$. Similarly, $\Phi(C)=C' \in B_1B_2$. Our goal is to prove that $B_1C_1,B'C', B_2C_2$ are concurrent (it's well-known that $X=(AB_1C_1) \cap (ABC)$). Now, notice that since $B_1C_1B_2C_2$ is cyclic $ZB_1.ZB_2=ZC_1.ZC_2$, so $Z$ lies on the radical axis of $(ABC_1),(AB_1C)$, which is $AY$. Thus, since $A,Y,Z$ are collinear, by Desargues' Theorem, we have that $B_1B'C_2,C_1C'B_2$ are in perspective, so $B_1C_1,B'C', B_2C_2$ are concurrent, as desired.
$\blacksquare$
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jeteagle
480 posts
#19
Y by
Very fun problem.

I claim this point is the point $M$ that is the midpoint of arc $BAC$. It is well known that $M$ is the Miquel Point of $BB_1C_1C$ and it lies on the perpendicular bisector of $B_1C_1$. Let $N \ne A$ and be on both $(AC_1B)$ and $(AB_1C)$ and $F = B_1C \cap BC_1$. This is the Miquel Point of complete quadrilateral $B_1CC_1B$ so $BB_1FN$ and $CC_1FN$ are cyclic quads. We also have $NBB_1 \sim NC_1C$ and $BB_1 = C_1C$ so $NB = NC_1$ and $NB_1 = NC$.

Let $D$ and $E$ be the second points of intersection of $B_1C$ and $BC_1$ with $(ABC)$ respectively. Additionally we have $$\measuredangle{BDC} = \measuredangle{BAC} = \measuredangle{BAC_1} = \measuredangle{BC_2C_1}$$so $B, D, C_2$ are collinear. Similarly $C, E, B_2$ are collinear. We also have $$\measuredangle{NC_2C_1} = \measuredangle{NBC_1} = \measuredangle{NBF} = \measuredangle{NB_1F}$$so $N, B_1, C_2$ are collinear and so is $N, C_1, B_2$. Now because $AC_1NC_2$ and $AB_1NB_2$ are cyclic quads and also $B_1C_2 \cap C_1B_2 = N$, this means $A$ is the Miquel Point of $B_1B_2C_1C_2$. Let $G = B_1B_2 \cap C_1C_2$. Notice that $$\measuredangle{B_1AG} = \measuredangle{BC_2G} = \measuredangle{NC_2C_1} = \measuredangle{NBC_1} = \measuredangle{BC_1N} = \measuredangle{BAN} = \measuredangle{B_1AN}$$so $A, G, N$ are collinear. This means $B_1B_2C_1C_2$ is a cyclic quad and let $M'$ be its center. It is well-known that $AB_1B_2M'$ and $M'$ lies on the perpendicular bisector of $B_1B_2$. There are two points that satisfy this with one of them being $M$. Clearly the other point doesn't work I'm too lazy to rigorously prove so $M = M'$. Finally, it is again well-known that $(AC_1C_2)$ contains the center of this circle so it always passes through $M$ which is fixed. $\blacksquare$
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cursed_tangent1434
548 posts
#20 • 1 Y
Y by GeoKing
Very beautiful problem. I actually found this pretty straightforward. Doing certain other problems such as 2014 EGMO 2 with the given mysterious length condition helped. We let $M$ denote the arc midpoint of major arc $BC$ in $(ABC)$ (containing $A$). We know that the given condition translates to the following claim.

Claim : Points $A$ , $B_1$ , $C_1$ and $M$ are concyclic.
Proof : It is not hard to see that since $MC=MB$ , $CC_1=BB_1$ and
\[\measuredangle C_1CM = \measuredangle ACM = \measuredangle  ABM = \measuredangle  B_1BM\]it follows that $\triangle MBB_1 \cong \triangle MCC_1$. In particular, this implies that $MB_1=MC_1$. Further, we can also note that there must exist a spiral similarity centered at $M$ mapping $BB_1 \mapsto CC_1$ and in turn, there must also exist a spiral similarity centered at $M$ mapping $B_1C_1 \mapsto BC$. This gives us that $\triangle MB_1C_1 \sim \triangle MBC$ and thus,
\[\measuredangle  C_1MB_1 = \measuredangle  CMB = \measuredangle  CAB = \measuredangle  C_1AB_1\]which most clearly implies the claim.

Let $\omega$ denote the circle centered at $M$ with radius $MB_1=MC_1$. We can proceed with proving the following key claim.

Claim : Quadrilateral $B_1C_1B_2C_2$ is cyclic (with circumcenter $M$).
Proof : Let $R = (ABC_1) \cap (AB_1C)$. Then, $BB_1=CC_1$ , $\measuredangle RB_1B = \measuredangle  RCA$ and $\measuredangle  RBA = \measuredangle  CC_1R$. Thus, $\triangle RB_1B \cong \triangle RCC_1$. Now, this means there exists a spiral similarity centered at $R$ mapping $BB_1 \mapsto CC_1$. Thus, there must also exist a spiral similarity centered at $R$ mapping $BC_1 \mapsto CB_1$. This allows us to conclude that $\triangle RBC_1 \sim \triangle RCB_1$. Let $C_2 '  = \overline{RB_1} \cap (AC_1B)$ and $B_2' = \overline{RC_1} \cap (AB_1C)$. Then,
\[\measuredangle CB_1R = \measuredangle  C_1BR = \measuredangle  C_1C_2'R\]and thus, $\overline{C_1C_2'} \parallel \overline{CB_1}$ which implies that $C_2'=C_2$. Similarly we conclude that $B_2'=B_2$. Thus, lines $\overline{B_1C_2}$ and $\overline{B_2C_1}$ intersect at $R$. Clearly, $R$ is the arc midpoint of minor arc $BC_1$ (since $RB=RC_1$ due to the previously establish congruency). So, $\overline{C_2B_1}$ is the $\angle C_1C_2B$-bisector. Now, we can note that,
\[2\measuredangle C_1C_2B_1 = \measuredangle  C_1CB = \measuredangle  C_1AB_1 = \measuredangle  C_1MB_1 \]so it follows that $C_2$ lies on $\omega$. A similar argument implies that $B_2$ also lies on $\omega$ which finishes the proof of the claim.

Now, the finish is clear. Simply note that,
\begin{align*}
\measuredangle B_2MC_2 &= \measuredangle  B_2MC_1 + \measuredangle  C_1MB_1 + \measuredangle  B_1MC_2\\
&= 2\measuredangle  B_2B_1C_1 + \measuredangle  C_1AB + 2\measuredangle  B_1C_1C_2\\
&= 2(\measuredangle  B_2B_1C_1 + \measuredangle  C_1B_1C) + \measuredangle  C_1AB\\
&= 2\measuredangle  B_2B_1C + \measuredangle  C_1AB_1\\
&= \measuredangle  B_2B_1C + \measuredangle  BC_1C_2 + \measuredangle  C_1AB_1\\
&= \measuredangle  B_2AC + \measuredangle  BAC_2 + \measuredangle  CAB\\
&= \measuredangle  B_2AC_2
\end{align*}and thus, $M$ lies on $(AB_2C_2)$. Thus, as $B_2$ and $C_2$ vary along $\overline{AB}$ and $\overline{AC}$, the circle $(AB_2C_2)$ passes through the fixed point $M$, which is the midpoint of arc $BC$ not containing $A$ of $(ABC)$ which is what we wished to show.
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bronzetruck2016
87 posts
#21
Y by
This is a very interesting problem. It took me a while but it was very fun.

Define $O$ as the circumcenter of $ABC$, $O_1$ as the circumcenter of $AB_1C$, $O_2$ as the circumcenter of $ABC_1$, and $O_3$ as the circumcenter of $AB_2C_2$. Let $D$ and $E$ be the midpoints of sides $AB$ and $AC$, respectively.

Lemma 1: O_2O=O_1O
Proof of Lemma 1: Let $O'$ be circumcenter of $AB_1C_1$. Then, the sides of quadrilateral $O'O_1OO_2$ are the perpendicular bisectors of segments $AC_1$, $AC$, $AB_1$, and $AB$, which are 2 pairs of parallel lines, with each pair separated by the same distance $\frac{1}{2}C_1C=\frac{1}{2}B_1B$. Therefore, $O'O_1OO_2$ is a rhombus, so we are done.

Now, define $M$ and $N$ as the midpoints of $AB_2$ and $AC_2$. I claim that $\measuredangle{OO_2O_3}=\measuredangle{O_3O_1O}$.
Time to angle chase!
$$\measuredangle{OO_2O_3}=\measuredangle{DO_2N}=\measuredangle{DO_2A}-\measuredangle{NO_2A}=\measuredangle{BC_1A}-\measuredangle{C_2C_1A}=\measuredangle{BC_1C_2}$$$$=\measuredangle{B_2B_1C}=\measuredangle{CB_1A}+\measuredangle{AB_1B_2}=\measuredangle{AO_1D}+\measuredangle{MO_1A}=\measuredangle{MO_1D}=\measuredangle{O_3O_1O}$$
Combining this result with Lemma 1 gets that triangle $O_1O_2O_3$ is isosceles and $O_1O_3O$ is isosceles. This means that $O_3O$ is the perpendicular bisector of $O_1O_2$, which we will call $l$. Since $O_1$ and $O_2$ always lie on the perpendicular bisectors of $AC$ and $AB$, by Lemma 1, $l$ does not change as $O_1$ and $O_2$ vary. Therefore, $O_3$ always lies on a fixed line through $O$, so our desired fixed point is simply the reflection of $A$ across $l$.
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Saucepan_man02
1294 posts
#22
Y by
Let's G(e)o:

Let $D = C_1C_2 \cap B_1 B_2$ and $F = B_1 C_2 \cap C_1 B_2$.

Claim: $D, F$ lie on the angle bisector on angle $\angle BAC$.
Note that: $A$ is the Miquel point of $B_1 B_2 C_1 C_2$. Thus: $AC_2 B_1 D, A B_2 C_1 D, A C_2 F C_2, A B_2 F B_1$ are cyclic. Thus; $C_2, B_2$ and $C_1, B_1$ are isogonal wrt $\angle BAC$. Therefore, due to isogonal lemma, $D, F$ are isogonal wrt $\angle BAC$.
Note that, $F$ is the spiral center sending $BB_1$ to $C C_1$. Since $BB_1 = C C_1$, then we have: $BF=C_1F$ and $B_1 F = CF$.
Therefore; $F$ lies on the angle bisector of $\angle BAC$.

Let $X$ be the midpoint of arc $BAC$ in $(ABC)$.
Now, we finish the problem with Miquel Properties:
Notice that: $\triangle XBB_1 \cong \triangle XCC_1$ which implies $XB_1 = XC_1$. Thus: $X$ maps $BB_1$ to $CC_1$ (spiral similarity) and by Miquel theorem on $BB_1 C_1 C$, we have: $X$ to be its Miquel center with $AXB_1 C_1$ to be cyclic. Due to Miquel theorem on $C_1 B_1 C_1 B_2$, we have $X$ to be its circumcenter.

By Miquel theorem on $C_1 B_1 C_1 B_2$, we must have $AXB_2C_2$ to be cyclic. Therefore, $X$ is the claimed fixed point and we are done,
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HamstPan38825
8846 posts
#23
Y by
I thought this was fairly straightforward if one knows Miquel theory well. I claim the fixed point is the Miquel point $X = (AB_1C_1) \cap (ABC)$ of $B_1BCC_1$.

Claim: $M = (ABC_1) \cap (AB_1C)$ is the arc midpoint of $\widehat{BC_1}$ and $\widehat{B_1C}$, and it lies on $\overline{B_1C_2}$ and $\overline{C_1B_2}$.

Proof: A spiral similarity at $M$ takes $\overline{BB_1} \to \overline{CC_1}$, so in fact it must be a spiral congruence. Thus $MB = MC_1$ and $MB_1 = MC$, and the first part follows.

For the second part, define $C_2' = \overline{MB_1} \cap (ABC_1)$ and $B_2'$ similarly; I will show $\overline{C_1C_2'} \parallel \overline{B_1C}$. This is true because $\measuredangle C_1C_2M = \measuredangle C_1AM = \measuredangle CB_1M$, as needed. $\blacksquare$

Claim: $X$ is the circumcenter of $C_2B_1C_1B_2$.

Proof: We will show that $X$ is the circumcenter of both triangles $C_2B_1C_1$ and $B_2C_1B_1$. In fact, as $XB_1 = XC_1$ by the same Miquel point properties, it suffices to show that \[\measuredangle B_1XC_1 = \measuredangle B_1AC_1 = 2\measuredangle MAC_1 = 2\measuredangle B_1C_2C_1. \ \blacksquare\]
Now note that $XB=XC$, so $\overline{AX}$ bisects the external $\angle BAC$. But $\measuredangle C_2AB = \measuredangle C_2C_1B = \measuredangle CB_1B_2 = \measuredangle CAB_2$ by parallelogram properties, so the external bisectors of $\angle BAC$ and $\angle C_2AB_2$ coincide; so $\overline{AX}$ bisects the external $\angle C_2AB_2$ and as $XC_2 = XB_2$, $X$ lies on $(AB_2C_2)$ by Fact 5.
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Ywgh1
136 posts
#24
Y by
[asy]

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draw(circle((-1.282348432280575,-1.3081121458651874), 1.636734468293493), linewidth(1.) + cqcqcq); 
draw((0.047202520399816095,-2.2626769597603187)--(-1.1557880179790645,0.32372184729459846), linewidth(1.) + wrwrwr); 
draw((-2.7431815848701495,-2.046262686063586)--(-1.1557880179790645,0.32372184729459846), linewidth(1.) + wrwrwr); 
draw((1.4167121369496614,-0.908709886091022)--(-2.1418963035901135,0.08475438637227817), linewidth(1.) + wrwrwr); 
draw((-2.1418963035901135,0.08475438637227817)--(-3.5523384591232348,-1.223272337444277), linewidth(1.) + wrwrwr); 
 /* dots and labels */
dot((-3.0426529553799324,-3.1076201200981206),dotstyle); 
label("$B$", (-3.006207676845374,-3.0114890444703097), NE * labelscalefactor); 
dot((0.7993241017788306,-3.069197778685229),dotstyle); 
label("$C$", (0.841079865294838,-2.973954531864064), NE * labelscalefactor); 
dot((-2.1418963035901135,0.08475438637227817),dotstyle); 
label("$A$", (-2.1053793742954707,0.17894452706060046), NE * labelscalefactor); 
dot((-2.7431815848701495,-2.046262686063586),dotstyle); 
label("$B_2$", (-2.705931575995406,-1.9511390633438603), NE * labelscalefactor); 
dot((0.047202520399816095,-2.2626769597603187),linewidth(4.pt) + dotstyle); 
label("$C_1$", (0.0810059850183572,-2.185729767132898), NE * labelscalefactor); 
dot((1.4167121369496614,-0.908709886091022),linewidth(4.pt) + dotstyle); 
label("$B_2$", (1.451015695146335,-0.8344873133080416), NE * labelscalefactor); 
dot((-3.5523384591232348,-1.223272337444277),linewidth(4.pt) + dotstyle); 

label("$D$", (-1.120098418381514,0.39476797454651497), NE * labelscalefactor); 
dot((-1.259878807867243,-3.55492424279874),linewidth(4.pt) + dotstyle); 
label("$Y$", (-1.2233183280486903,-3.480670452048385), NE * labelscalefactor); 
dot((-1.694943236930499,-1.7596143850546555),linewidth(4.pt) + dotstyle); 
label("$J$", (-1.654965223020519,-1.6883974751001383), NE * labelscalefactor); 
dot((-1.001035827539835,-2.549325260769249),linewidth(4.pt) + dotstyle); 

label("$C_2$", (-3.5129235970296944,-1.1441470423095712), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */


[/asy]
Here is a Sketch

Let $J$ be the intersection of $B_2B_1$ and $C_2C_1$, and let $(AB_1C)$ intersect $(ABC_1)$ again at $Y$.

We claim that the fixed point is the mid point of arc $BAC$. And let it be $D$.

Firstly we show that $D$ lies on $AB_1C_1$ and then we show that $AY$ is the angle bisector of $\angle BAC$, then we show $Y-C_1-B_2$ and $Y-B_1-C_2$ are collinear.

After that we get that $A$ is the Miquel point of $YB_1C_1D$ and we get that $J$ lies on $AY$. Finally we show that $D$ is the Center of $(B_1B_2C_1C_2)$. After that it’s easy to see that $D$ lies on circle $(AB_2C_2)$ hence we are done.
This post has been edited 4 times. Last edited by Ywgh1, Jan 20, 2025, 2:58 PM
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Ilikeminecraft
283 posts
#25
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Let $P$ be the second intersection of $(ABC_1), (AB_1C).$
Note that $\angle AC_2C_1 = \angle ABC_1  = \angle AB_1B_2,$ with implies $AB_1B_2\sim AC_2C_1.$
By Spiral similarly, we get that $P = B_1C_2\cap B_2C_1.$
Furthermore, there is spiral similarity centered at $P$ mapping $BB_1$ to $CC_1$, and so $P$ lies on the angle bisector of $\angle BAC.$
Take $\angle C_1C_2P=\angle PAC_1 = \angle BAP = \angle B_11B_2P,$ so $B_1C_2B_2C_1$ is cyclic.

Let $M$ be Miquel point of quad $BB_1C_1C.$
By the given length condition, we have $M$ is the midpoint of arc $BAC.$

Take a $\sqrt{bc}$-inversion and we get:
Let $ABC$ be a triangle
Let $M$ be the intersection of the $A$-external bisector and $BC$
Let $B_1$ be on $AB,$ and $C_1$ be the intersection of $MB_1, AC.$
Let $Q$ be a point on the $A$ angle bisector
Let $B_2$ be intersections of $(AB_1Q)$ with $CB_1.$
Define $C_2$ similarly.
Show that $B_2C_2$ passes through $M$.

Note that from inversion, we also have that $B_1C_2B_2C_2$ are cyclic. Recall that $P$ is the second intersection $(AB_1C_2), (AB_2C_1).$
Observe that the Miquel point of $B_1B_2C_1C_2$ is $A.$ If we define $T$ to be the intersection of $B_1C_1, B_2C_2,$ by an application of Miquel’s master theorem, we get $A, T_0$ are inverses with respect to $(B_1C_2B_2C_1).$
By Ceva-Menelaus, note that $(M, B_1C_1\cap AP; B_1C_1) = -1$ , so $M$ lies on the polar of $AP\cap B_1C_1$. By Brokard's on $B_1C_2B_2C_1,$ we get $M$ lies on the polar of $P.$ By La Hire's, $AP$ is the polar of $M.$ However, $AP\perp AM,$ which implies that $M$ and $A$ are inverses.
Thus, $M = T_0.$
This post has been edited 1 time. Last edited by Ilikeminecraft, Yesterday at 9:04 PM
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