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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:
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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
D1010 : How it is possible ?
Dattier   8
N 17 minutes ago by Dattier
Source: les dattes à Dattier
Is it true that$$\forall n \in \mathbb N^*, (24^n \times B \mod A) \mod 2 = 0 $$?

A=1728400904217815186787639216753921417860004366580219212750904
024377969478249664644267971025952530803647043121025959018172048
336953969062151534282052863307398281681465366665810775710867856
720572225880311472925624694183944650261079955759251769111321319
421445397848518597584590900951222557860592579005088853698315463
815905425095325508106272375728975

B=2275643401548081847207782760491442295266487354750527085289354
965376765188468052271190172787064418854789322484305145310707614
546573398182642923893780527037224143380886260467760991228567577
953725945090125797351518670892779468968705801340068681556238850
340398780828104506916965606659768601942798676554332768254089685
307970609932846902
8 replies
Dattier
Mar 10, 2025
Dattier
17 minutes ago
GOTEEM #5: Circumcircle passes through fixed point
tworigami   21
N 41 minutes ago 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
41 minutes ago
Strike the inequality
giangtruong13   1
N 44 minutes ago by arqady
Source: Idk
Let $a,b,c \geq 0$ satisfy that $a+b+c=3$. Prove that $$\sum a\sqrt{b^3+1} \leq 5$$
1 reply
giangtruong13
5 hours ago
arqady
44 minutes ago
Calculus rather than inequalities
darij grinberg   12
N an hour ago by asdf334
Source: German TST, IMO ShortList 2003, algebra problem 3
Consider pairs of the sequences of positive real numbers \[a_1\geq a_2\geq a_3\geq\cdots,\qquad b_1\geq b_2\geq b_3\geq\cdots\]and the sums \[A_n = a_1 + \cdots + a_n,\quad B_n = b_1 + \cdots + b_n;\qquad n = 1,2,\ldots.\]For any pair define $c_n = \min\{a_i,b_i\}$ and $C_n = c_1 + \cdots + c_n$, $n=1,2,\ldots$.


(1) Does there exist a pair $(a_i)_{i\geq 1}$, $(b_i)_{i\geq 1}$ such that the sequences $(A_n)_{n\geq 1}$ and $(B_n)_{n\geq 1}$ are unbounded while the sequence $(C_n)_{n\geq 1}$ is bounded?

(2) Does the answer to question (1) change by assuming additionally that $b_i = 1/i$, $i=1,2,\ldots$?

Justify your answer.
12 replies
1 viewing
darij grinberg
Jul 15, 2004
asdf334
an hour ago
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
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
Concurrency
Dadgarnia   26
N Today at 12:38 AM by joshualiu315
Source: Iranian TST 2020, second exam day 2, problem 4
Let $ABC$ be an isosceles triangle ($AB=AC$) with incenter $I$. Circle $\omega$ passes through $C$ and $I$ and is tangent to $AI$. $\omega$ intersects $AC$ and circumcircle of $ABC$ at $Q$ and $D$, respectively. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of $CQ$. Prove that $AD$, $MN$ and $BC$ are concurrent.

Proposed by Alireza Dadgarnia
26 replies
Dadgarnia
Mar 12, 2020
joshualiu315
Today at 12:38 AM
Concurrency
G H J
Source: Iranian TST 2020, second exam day 2, problem 4
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Dadgarnia
164 posts
#1 • 7 Y
Y by Purple_Planet, itslumi, jhu08, mathematicsy, Mango247, Mango247, ItsBesi
Let $ABC$ be an isosceles triangle ($AB=AC$) with incenter $I$. Circle $\omega$ passes through $C$ and $I$ and is tangent to $AI$. $\omega$ intersects $AC$ and circumcircle of $ABC$ at $Q$ and $D$, respectively. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of $CQ$. Prove that $AD$, $MN$ and $BC$ are concurrent.

Proposed by Alireza Dadgarnia
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GeoMetrix
924 posts
#5 • 8 Y
Y by mueller.25, amar_04, Mathasocean, Purple_Planet, agwwtl03, IFA, jhu08, sabkx
Nice and easy.
Define $\omega=\odot(CIQ)$ and let $H=\omega \cap \overline{BC}$. Observe that trivially we have that $N \in \overline{AC}$ (just consider the line through $I$ parallel to $BC$). Now since $$\angle NCH=\angle NHC=\angle ABC$$$\implies$ $\overline{NH} \| \overline{AB}$ . Hence by the converse of ceva's theorem we have that $\overline{AH},\overline{BN},\overline{CM}$ are concurrent. Now define $K= \overline{AD} \cap \overline{BC}$ . Observe that to show that $K \in \overline{MN}$ we need to show that $(B,C;H,K)=-1$. But if $\overline{DH} \cap \omega=J$ then we have that $(B,C;H,K)\overset{D}{=}(A,J;B,C)$. Hence we just need to show that $(AJBC)$ is harmonic. For this notice that by reims theorem we have that $\overline{AJ} \| \overline{QH}$. But as $\angle QHC=90^\circ$ we have that $\overline{AJ} \perp \overline{BC}$ by which we have that $J$ is the midpoint of minor $\widehat{BC}$. Done $\blacksquare$.

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Mathematicsislovely
245 posts
#7 • 3 Y
Y by Purple_Planet, jhu08, PRMOisTheHardestExam
CLAIM: $N$ is the centre of $\omega=\odot(CIQ)$.
Proof: If $N'$ is the centre of $\omega$ then $N'C=N'I$ and $\angle N'IA=90^\circ$ implies $N'I$ and $BC$ are parallel .Let $IN'$ cut $AC$ at $N''$.Then $$\angle N''IC=\angle ICB=\angle ICN''$$.It means $N''C=N''I$.So $N''\equiv N'$.Now the circle centre at $N''$ and with radius $N''C=N''I$ is the unique circle tangent to $AI$ at $I$.So $N''=N$..
$\square$

Now let $\omega=\odot(CIQ)$ cut $BC$ at $X$ then $NX=NC$ implies $\angle NXH=\angle NXC=\angle ABC$ so $NX$ and $AB$ are parallel.So the line joining $C$ and the intersection of $BN$ and $AX$ passes through midpoint of $AB$.In other words $AX$,$BN$,$CM$ are concurrent at a point.So let $F$ is the intersection of $MN$ and $BC$.So $(B,C;X,F)=-1$ and let M be the midpoint of $BC$.Then $FC.FB=FX.FM$ [since $F$ is the inverse image of $X$ w.r.t circle with diameter $BC$].

Now assume $Y= XN \cap AD$.We claim that $XYDC$ is cyclic.Indeed, as $XN$ and $AB$ are parallels so $\angle XYD=\angle BAD=180^\circ-\angle XCD$.
Now as $XY$ passes through center $N$ of $omega=\odot(CIQ)$ so $\angle ADX=\angle XDY=90^\circ$ and together with $\angle AMXY=90^\circ$ We get $AMXD$ is cyclic.
Let $F'= BC\cap AD$.So $F'D.F'A=F'C.F'B$ and $F'D.F'A=F'X.F'M$. So we have $F'C.F'B=F'X.F'M$.But we have previously proved that $FC.FB=FX.FM$ so $F \equiv F'$. So $BC,AD,MN$ concur at a point $F$.
This post has been edited 1 time. Last edited by Mathematicsislovely, Mar 26, 2020, 9:58 PM
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AlastorMoody
2125 posts
#8 • 7 Y
Y by GeoMetrix, Purple_Planet, SenatorPauline, Aryan-23, Muaaz.SY, jhu08, PRMOisTheHardestExam
Solution
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Aryan-23
558 posts
#9 • 5 Y
Y by AlastorMoody, Siddharth03, Muaaz.SY, jhu08, PRMOisTheHardestExam
Solution
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AwesomeYRY
579 posts
#10 • 2 Y
Y by jhu08, PRMOisTheHardestExam
Claim: N is the circumcenter of (CIQ)
Proof:
Note that $IN\perp, AI, BC\perp AI \Longrightarrow NI\parallel BC$. Thus, \[\angle NIC = \angle BCI = \angle ICA = \angle ICN\]Thus, $NI=NC$, so the circle with center $N$ and radius $NI$ is both tangent to $AI$ at $I$ and passes through $C$, so we have sufficiently redefined $\omega$. $\square$

Now, define $X=\omega \cap BC$ and $Y=\omega \cap AD$

Claim 1: $NX\parallel AB$
This clearly follows from
\[\angle CNX = \angle NCX = \angle ACB=\angle ABC\]$\square$

Claim 2: $XY\parallel AB$
We angle chase (basically a rederivation of Reim's)
\[\angle DAB = \angle DCB = \angle DCX = \angle DYX\]$\square$

Combining these, we have that $N,X,Y$ are collinear. Since, $X,Y\in \omega$ and $N$ is the center of $\omega$, we have that $N$ is the midpoint of $XY$. Since $M$ is the midpoint of $AB$, combined with $YX\parallel AB$, we have that $AY,MN,BX$ are concurrent which finishes $\blacksquare$
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SatisfiedMagma
449 posts
#11 • 2 Y
Y by jhu08, PRMOisTheHardestExam
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[/asy]
Remarks
Attachments:
This post has been edited 1 time. Last edited by SatisfiedMagma, Aug 13, 2021, 3:06 PM
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MrOreoJuice
594 posts
#12 • 4 Y
Y by jhu08, SatisfiedMagma, PRMOisTheHardestExam, vrondoS
$\angle QIA = \angle QCI = \angle ICB = 90^\circ - \angle (IC , \text{angle bisector}) \implies \angle QIC = 90^\circ \implies N$ is the circumcenter of $\omega$.
Let $E=BC \cap \omega$ and $F=AD \cap \omega$.
$$\angle FEC = 180^\circ - \angle FDC = \angle ABC = \angle NCE = \angle NEC$$Thus $\overline{E-N-F}$ are collinear also $EF \parallel AB$ so by homothety we are done.
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srisainandan6
2811 posts
#13 • 2 Y
Y by PRMOisTheHardestExam, jhu08
To begin, note that it is easy to see that $N$ is the circumcenter of $\omega$ as $\triangle NIC$ is isosceles.

Denote $X= BC \cap AD$. We desire to show that $M,N,X$ are collinear. Denote $E = \omega \cap AD$ and $F = \omega \cap BC$. By Reim's Theorem, we get that $EF \parallel AB$. To finish, take a homotethy centered at $X$ that maps $\triangle XEF$ to $\triangle XAB$ which yields $X,N,M$ collinear. $\blacksquare$
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RM1729
63 posts
#14 • 2 Y
Y by PRMOisTheHardestExam, jhu08
Consider $P$ as $AI \cap BC$.

Claim
$N$ is the centre of the circle $\omega$


Proof
We angle chase.

Note that since $ \angle IQC = \angle PIC = 90^{\circ} - \angle ICP = 90^{\circ} - \angle C/2 $ and $\angle QCI = \angle C/2$

We have that $\angle QIC = 90^{\circ}$ and thus $QC$ is a diameter

But since $N$ is the midpoint of $QC$ it must be the centre of the circle



We now define $X$ as $AD\cap \omega (\neq D)$ and $Y$ as $BC\cap \omega (\neq C)$

Note that $\angle NXC = \angle NCX$ but $\angle NCX = \angle ABC$ since the triangle is isosceles. Thus $\angle NXC = \angle ABC$ and so $NX||AB$

However we also have that $XY||AB$ by Reim's Theorem. Combining these we have that $ABXY$ is a trapezium and $M$ and $N$ are the midpoints of its parallel sides. A simple homothety argument proves that $AY, BX,MN$ that is $AD,BC,MN$ are concurrent.
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mohamad021
1 post
#15 • 1 Y
Y by jhu08
srisainandan6 wrote:
To begin, note that it is easy to see that $N$ is the circumcenter of $\omega$ as $\triangle NIC$ is isosceles.

Denote $X= BC \cap AD$. We desire to show that $M,N,X$ are collinear. Denote $E = \omega \cap AD$ and $F = \omega \cap BC$. By Reim's Theorem, we get that $EF \parallel AB$. To finish, take a homotethy centered at $X$ that maps $\triangle XEF$ to $\triangle XAB$ which yields $X,N,M$ collinear. $\blacksquare$

Can you explain Reim's theorem please?
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DottedCaculator
7303 posts
#16
Y by
Solution
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guptaamitu1
656 posts
#17 • 2 Y
Y by PRMOisTheHardestExam, Lantien.C
Here's a solution with radical axes ($\overline{MN}$ will become radical axes of two circles)
Let $\Omega = \odot(ABC)$ and $A',C'$ be antipode of $A,C$ wrt $\Omega$. $\angle QIC = \angle AIC - \angle ACI = 90^\circ$. Thus $\overline{QC}$ is a diameter of $\omega$, consequently $N$ is its center. So $\overline{CO},\overline{QD}$ intersect at $C'$. Let $T = \overline{AD} \cap \overline{BC}$, $\Gamma = \odot(BIC)$ and $X$ be a point of $\overline{AC}$ such that $\overline{QD}$ bisects $\angle ADX$, let $\gamma = \odot(AXD)$. hand drawn figure
[asy]
size(250);
pair A=dir(90),B=dir(-130),C=dir(-50),O=(0,0),I=incenter(A,B,C),N=extension(A,C,I,I+C-B),Q=2*N-C,Cp=-C,D=foot(C,Cp,Q),T=extension(A,D,B,C),M=1/2*(A+B),Ap=-A,X=foot(I,A,C);
draw(unitcircle^^circumcircle(C,I,Q),cyan);
draw(CP(circumcenter(A,X,D),X,-210,-60 ),purple);
draw(CP(Ap,I,-10,190 ),purple);
dot("$A$",A,dir(110));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$O$",O,dir(0));
dot("$I$",I,dir(130));
dot("$N$",N,dir(-120));
dot("$Q$",Q,dir(50));
dot("$C'$",Cp,dir(Cp));
dot("$D$",D,dir(D));
dot("$M$",M,dir(M));
dot("$T$",T,dir(T));
dot("$A'$",Ap,dir(Ap));
dot("$X$",X,dir(50));
draw(C--A--B--T^^A--Ap,red);
draw(M--T,purple);
draw(T--A^^Cp--D--X,green);
draw(C--Cp,green);
[/asy]
Claim: Points $M,N,T$ lie on radical axes $\ell$ of $\Gamma,\gamma$.
Proof: $T \in \ell$ is direct. For $M$: Note $\Gamma$ is tangent to $\overline{BC}$ as it has center $A'$. Note $$\angle OAC = \angle OCA = \angle C'CA = \angle C'DA = \angle QDA$$As $\angle BAC = 2 \angle OAC = 2 \angle QDA = \angle XDA$, so $\gamma$ is tangent to $\overline{AB}$. Hence power of $M$ wrt both $\Gamma,\gamma$ equals $MB^2 = MA^2$. For $N$: We will mostly focus of $\triangle ADX$. Observe
$$ \angle NDX = \angle NDQ - \angle XDQ = \angle NQD - \angle ADQ = \angle QAD = \angle NAD $$So $\overline{ND}$ is tangent to $\gamma$ (experts may also directly note that $N$ is the center of $D$-appolonius circle wrt $\triangle ADX$). Hence $NX \cdot NA = ND^2 = NC^2$. Laslty, recall $\overline{NC}$ is tangent to $\Gamma$ (as $A'$ is its center). This proves our Claim. $\square$

It follows points $M, N,T$ are collinear, solving our problem. $\blacksquare$

Motivation
This post has been edited 1 time. Last edited by guptaamitu1, Feb 14, 2022, 4:19 PM
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BarisKoyuncu
577 posts
#18 • 1 Y
Y by teomihai
Simple angle chasing gives us that $N$ is the center of $(IDC)$.
Let $AD\cap BC=S$. We have
$$\angle ASC=\angle ACB-\angle CAD=\angle ABC-\angle CAD=\angle ACD$$Hence, $NC$ is tangent to $(DCS)$. Since $|NC|=|ND|$, we know that $ND$ is tangent to $(DCS)$ as well. Then, $SN$ is symmedian in $DSC$. Since $DC$ and $AB$ are antiparallels wrt $DSC$, we find that $SN$ bisects $AB$, done.
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BarisKoyuncu
577 posts
#19
Y by
A kind of generalization
This post has been edited 1 time. Last edited by BarisKoyuncu, Mar 7, 2022, 8:36 AM
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JAnatolGT_00
559 posts
#20 • 1 Y
Y by MrOreoJuice
BarisKoyuncu wrote:
A kind of generalization

Proof. All angles are oriented. From $\angle ADB=\angle ACB=\angle CBA=\angle SBA$ we obtain $$\angle CDN=\angle NCD=\angle ABD=\angle CBS.$$Therefore $SN$ is a symmedian in $CDS$ and so bisects $AB.$
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dusicheng20080513
36 posts
#21
Y by
GeoMetrix wrote:
Nice and easy.
Define $\omega=\odot(CIQ)$ and let $H=\omega \cap \overline{BC}$. Observe that trivially we have that $N \in \overline{AC}$ (just consider the line through $I$ parallel to $BC$). Now since $$\angle NCH=\angle NHC=\angle ABC$$$\implies$ $\overline{NH} \| \overline{AB}$ . Hence by the converse of ceva's theorem we have that $\overline{AH},\overline{BN},\overline{CM}$ are concurrent. Now define $K= \overline{AD} \cap \overline{BC}$ . Observe that to show that $K \in \overline{MN}$ we need to show that $(B,C;H,K)=-1$. But if $\overline{DH} \cap \omega=J$ then we have that $(B,C;H,K)\overset{D}{=}(A,J;B,C)$. Hence we just need to show that $(AJBC)$ is harmonic. For this notice that by reims theorem we have that $\overline{AJ} \| \overline{QH}$. But as $\angle QHC=90^\circ$ we have that $\overline{AJ} \perp \overline{BC}$ by which we have that $J$ is the midpoint of minor $\widehat{BC}$. Done $\blacksquare$.

[asy]
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 /* end of picture */
[/asy]

When you use Geogebra-Asymptote conversion, how did you doit. Why did mine have an error?
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guptaamitu1
656 posts
#22
Y by
Firstly you have convert it from Geogebra Classic and not Geogebra Geometry. If you find easier to draw in Geogebra Geomtery (as I also do), then save it and reopen it in Geogebra Classic. After the conversion, you have to do some slight edits. Like changing the size and defaultpen.
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IAmTheHazard
4999 posts
#23 • 1 Y
Y by Lantien.C
We begin with the following key claim.

Claim: $N$ is the center of $\omega$.
Proof: Let $N'$ be the intersection of the perpendicular to $\overline{AI}$ at $I$ with $\overline{AC}$. Then we have
$$\angle N'IC=\angle ICB=\angle ICA=\angle ICN',$$so $N'I=N'C$ and thus $N'$ is the center of $\omega$. Then $Q$ is the reflection of $C$ over $N'$, hence $N'$ is the midpoint of $\overline{CQ}$ and we have $N=N'$, which implies the desired result.

Now let $P \neq D$ be the intersection of $\omega$ with $\overline{AD}$, $R \neq C$ be the intersection of $\omega$ with $\overline{BC}$, and $X=\overline{AD} \cap \overline{BC}$. By Reim's, we have $\overline{AB} \parallel \overline{PR}$, so
$$\angle QPR=\angle QCR=\angle ABC=\angle PRC,$$so we have $\overline{PQ} \parallel \overline{CR}$ as well. Then we have $90^\circ=\angle QPC=\angle PCR$, so $CPQR$ is a rectangle, and $N$ is the midpoint of $\overline{PR}$. But triangles $\triangle XAB$ and $\triangle XPR$ are homothetic, which implies that $\overline{MN}$ passes through $X$ as well, hence $\overline{AD}$, $\overline{MN}$, and $\overline{BC}$ concur. $\blacksquare$
This post has been edited 1 time. Last edited by IAmTheHazard, Apr 14, 2022, 4:02 PM
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Lantien.C
7 posts
#24
Y by
mohamad021 wrote:
srisainandan6 wrote:
To begin, note that it is easy to see that $N$ is the circumcenter of $\omega$ as $\triangle NIC$ is isosceles.

Denote $X= BC \cap AD$. We desire to show that $M,N,X$ are collinear. Denote $E = \omega \cap AD$ and $F = \omega \cap BC$. By Reim's Theorem, we get that $EF \parallel AB$. To finish, take a homotethy centered at $X$ that maps $\triangle XEF$ to $\triangle XAB$ which yields $X,N,M$ collinear. $\blacksquare$

Can you explain Reim's theorem please?
JI//FH,as the picture says
Attachments:
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ihatemath123
3426 posts
#25
Y by
By angle chasing, $\angle QIC = 90^{\circ}$, hence $\angle QDC = 90^{\circ}$, hence $\angle ADQ = 90 - \angle B$, hence $(ADQ)$ is tangent to $\overline{AI}$.

If we extend $AB$ to $B'$ and $AC$ to $C'$ such that $BB' = CC' = AQ$, then $M$ and $N$ have equal powers WRT $(AQD)$ and $(BCC'B')$, hence $\overline{MN}$ is their radical axis. Then, obviously $\overline{BC}$ is the radical axis of $(ABC)$ and $(BCC'B')$, and $\overline{AD}$ is the radical axis of $(AQD)$ and $(ABC)$, hence they all concur by the radical axis theorem.
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trk08
614 posts
#26
Y by
Claim:
$N$ is the center of $(CIQ)$
Proof:
By angle-chasing, we can see that:
\[\angle CQI=\angle CIF=90^{\circ}-\angle ICF=90^{\circ}-\angle ICQ.\]Therefore, $\angle QIC=90^{\circ}$, or $N$ is the center $\blacksquare$

Claim:
The line parallel to $AB$ that passes through $N$ goes through $E$ and $F$.
Proof:

Denote $E=(CQD)\cap BC$ and $F=(CQD)\cap AD$. By Reim's theorem, $EF$ is parallel to $AB$.

Also, $\triangle{NEC}$ is isosceles, so it is similar to $\triangle{ABC}$. As a result, $NE\parallel AB$. Therefore, $N,E,F$ are all collinear and parallel to $AB$ $\blacksquare$

As a result, we can take a homothety at $T=AD\cap BC$, which sends $N$ to $M$, implying the desired result $\square$
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asdf334
7575 posts
#27
Y by
Note that $N$ is the center of $\omega$. Let $E$ be the other intersection of $\omega$ with $BC$ and let $E'$ be its antipode wrt $\omega$. Let $F$ be the midpoint of $BC$. It suffices to show that $AFED$ is cyclic as this implies $A,E',D$ collinear.

Here's the interesting part.

Claim: Let $ABCD$ be a cyclic quadrilateral. Let $F$ and $E$ be points where $B,F,E,C$ are on $BC$ in that order. If $\angle BAF=\angle CDE$ then $AFED$ is cyclic.
Proof: Angle chase.

Now we're done; just apply the claim here.
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bjump
962 posts
#28
Y by
Claim:$N$ is the center of $\omega$.
Proof Let $O$ denote the center of $\omega$. $\angle ICQ= \angle ICB = \tfrac{1}{2} \angle QOI$. Since $OI \perp AI$, then $OI \parallel BC$. Since $\angle QOI = \angle  ICQ + \angle ICB= \angle BCA$, $O$ lies on $AC$. Since $OQ=OC$, then $O$ is the midpoint of $CQ$. So $O= N$. $\square$

By Reims on $\omega$ and $(ABC)$ with lines $AD$ and $BC$ we get that the line through $D$ parallel to $AB$ intersects $BC$ on $\omega$ at a point we will call $P$ then let $DP\cap MN= Q$ then by homothety $BP$, $AD$, $MQ$ are concurrent so $MN$, $BC$, and $AD$ are concurrent.
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LuciferMichelson
18 posts
#29
Y by
Define $R$ as midpoint of $BC$
Easy to see $N$ is center of $DQIC$.
After that define $S=(DIQ) \cap BC$
Angle chasing shows $SN//AB$ so $AS,BN,CM$ concurrent.
Now we should show that $(B,C;S,AD \cap BC)=-1$
Let $AD \cap BC= K'$ so $(B,C;S,AD \cap BC)=-1$ is equal to $KD.KA=KS.KR$ and from angle chasing it is easy that show $ADSR$ is cyclic.
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shendrew7
787 posts
#30
Y by
If $N'$ is the intersection of $AC$ with the line through $I$ parallel $BC$, we know $N'I \perp AI$ and
\[\angle ACI = \angle ICB = \angle NIC,\]
so $N'$ is the center of $\omega$. Therefore $N'$ is the midpoint of $CQ$, so $N' = N$.

Next we define the intersections of $AC$ and $BC$ with $\omega$ as $K$ and $L$. Looking at $CD$, Reim's tells us that $AB \parallel KL$, which then gives
\[\angle LKQ = \angle LCQ = \angle ABC = \angle KLC,\]
so $KQ \parallel BC$ as well. As a result, $\angle KCL = \angle QKC = \angle QLC = 90$, so $KL$ is a diameter of $\omega$, and hence passes through $M$. We finish by noting the homothety which maps $KL$ to $AB$ also maps corresponding midpoints $N$ to $M$.
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joshualiu315
2513 posts
#31
Y by
We begin with a simple, but important claim.


Claim: $N$ is the center of $\omega$.

Proof: Let $\angle IAC = \alpha$. Note that $\angle ICA = 90-\tfrac{\alpha}{2}$, and we also have $\angle ICA = \angle AIQ$ from the tangency condition. Therefore,

\[\angle CQI = \angle IAQ+\angle AIQ = \alpha + \left(90-\frac{\alpha}{2} \right) = 90+\frac{\alpha}{2},\]
which means $\angle CQI + \angle ICQ = 90^\circ$, or $\angle CIQ = 90^\circ$. This means that $\overline{CQ}$ is a diameter of $\omega$, and the midpoint of $\overline{CQ}$ is the center of $\omega$, which is $N$. $\square$


Let $C' \neq C =  \overline{AC} \cap \omega$ and $D' \neq D = \overline{AD} \cap \omega$. Note that lines $\overline{AB}$ and $\overline{CD}$ are antiparallel and $\overline{CD}$ and $\overline{C'D'}$ are also antiparallel. By Reim's Theorem, we have $\overline{AB} \parallel \overline{C'D'}$.

It suffices to show that $N$ is the midpoint of $\overline{C'D'}$; if so, the three lines will concur at the center of the homothety that maps $\overline{AB}$ to $\overline{D'C'}$. However, we simply angle chase to find

\[\angle NC'C = \angle NCC' = \angle ACB = \angle ABC,\]
so $\overline{NC'} \parallel \overline{AB}$. This means $N$ lies on $\overline{C'D'}$, and $N$ is the center of $\omega$, which implies $N$ bisects $\overline{C'D'}$. $\blacksquare$
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