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

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

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

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

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0 replies
jlacosta
May 1, 2025
0 replies
k i Peer-to-Peer Programs Forum
jwelsh   157
N Dec 11, 2023 by cw357
Many of our AoPS Community members share their knowledge with their peers in a variety of ways, ranging from creating mock contests to creating real contests to writing handouts to hosting sessions as part of our partnership with schoolhouse.world.

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

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

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

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

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

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

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

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

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

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

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

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

Advertising paid program or service: never allowed

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

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

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

Original 2013 post
50 replies
v_Enhance
Feb 15, 2013
blawho12
Oct 16, 2017
Tangent circles
Sadigly   0
5 minutes ago
Source: Azerbaijan Junior MO 2025 P6
Let $T$ be a point outside circle $\omega$ centered at $O$. Tangents from $T$ to $\omega$ touch $\omega$ at $A;B$. Line $TO$ intersects bigger $AB$ arc at $C$.The line drawn from $T$ parallel to $AC$ intersects $CB$ at $E$. Ray $TE$ intersects small $BC$ arc at $F$. Prove that the circumcircle of $OEF$ is tangent to $\omega$.
0 replies
Sadigly
5 minutes ago
0 replies
Functional Inequality Implies Uniform Sign
peace09   32
N 10 minutes ago by lelouchvigeo
Source: 2023 ISL A2
Let $\mathbb{R}$ be the set of real numbers. Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function such that \[f(x+y)f(x-y)\geqslant f(x)^2-f(y)^2\]for every $x,y\in\mathbb{R}$. Assume that the inequality is strict for some $x_0,y_0\in\mathbb{R}$.

Prove that either $f(x)\geqslant 0$ for every $x\in\mathbb{R}$ or $f(x)\leqslant 0$ for every $x\in\mathbb{R}$.
32 replies
peace09
Jul 17, 2024
lelouchvigeo
10 minutes ago
Inequality
Sadigly   0
12 minutes ago
Source: Azerbaijan Junior MO 2025 P5
For positive real numbers $x;y;z$ satisfying $0<x,y,z<2$, find the biggest value the following equation could acquire:


$$(2x-yz)(2y-zx)(2z-xy)$$
0 replies
Sadigly
12 minutes ago
0 replies
Calculating sum of the numbers
Sadigly   0
15 minutes ago
Source: Azerbaijan Junior MO 2025 P4
A $3\times3$ square is filled with numbers $1;2;3...;9$.The numbers inside four $2\times2$ squares is calculated,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?

$\text{a)}$ $24,24,25,25$

$\text{b)}$ $20,23,26,29$
0 replies
Sadigly
15 minutes ago
0 replies
9 ARML Location
deduck   38
N 4 hours ago by shawnzeng
UNR -> Nevada
St Anselm -> New Hampshire
PSU -> Pennsylvania
WCU -> North Carolina


Put your USERNAME in the list ONLY IF YOU WANT TO!!!! !!!!!

I'm going to UNR if anyone wants to meetup!!! :D

Current List:
Iowa
UNR
PSU
St Anselm
WCU
38 replies
deduck
May 6, 2025
shawnzeng
4 hours ago
USAMO Medals
YauYauFilter   54
N 5 hours ago by ohiorizzler1434
YauYauFilter
Apr 24, 2025
ohiorizzler1434
5 hours ago
Will I fail again
hashbrown2009   16
N 5 hours ago by ohiorizzler1434
so this year I got 34 on JMO 772 774 and got docked 1 point from top honors + mop

I just got info that I pretty much cannot do math for the rest of summer due to family reasons, and the only time I have is winter break

do you guys think it's enough time to practice/grind to qualify mop through USAMO, or should I tell my parents to reschedule the stuff because I really want to make mop

(Note: I'm aiming for like 25+ on USAMO so at least silver but I'm not sure that's realistic given the circumstances i'm in)
16 replies
hashbrown2009
Yesterday at 1:54 PM
ohiorizzler1434
5 hours ago
Wizard101
El_Ectric   65
N 5 hours ago by HamstPan38825
Source: USAMO 2016, Problem 6
Integers $n$ and $k$ are given, with $n\ge k\ge2$. You play the following game against an evil wizard.

The wizard has $2n$ cards; for each $i=1,\ldots,n$, there are two cards labeled $i$. Initially, the wizard places all cards face down in a row, in unknown order.

You may repeatedly make moves of the following form: you point to any $k$ of the cards. The wizard then turns those cards face up. If any two of the cards match, the game is over and you win. Otherwise, you must look away, while the wizard arbitrarily permutes the $k$ chosen cards and then turns them back face-down. Then, it is your turn again.

We say this game is winnable if there exist some positive integer $m$ and some strategy that is guaranteed to win in at most $m$ moves, no matter how the wizard responds.

For which values of $n$ and $k$ is the game winnable?
65 replies
El_Ectric
Apr 20, 2016
HamstPan38825
5 hours ago
SUMaC Residential vs. Ross
AwesomeDude10   9
N 5 hours ago by Rong0625
Hi! I got into the SUMaC residential i program, and I also recently got off the Ross waitlist. I was wondering if anyone had any insight on which program is
1) More useful in furthering my mathematical knowledge (which has a better curriculum)
2) Since I'm a junior, which is more useful for college apps? (I know this is a little cringe)
Thanks!
9 replies
AwesomeDude10
May 6, 2025
Rong0625
5 hours ago
HCSSiM results
SurvivingInEnglish   63
N Yesterday at 11:52 PM by KevinChen_Yay
Anyone already got results for HCSSiM? Are there any point in sending additional work if I applied on March 19?
63 replies
SurvivingInEnglish
Apr 5, 2024
KevinChen_Yay
Yesterday at 11:52 PM
Sad Algebra
tastymath75025   46
N Yesterday at 11:40 PM by Ilikeminecraft
Source: 2019 USAMO 6, by Titu Andreescu and Gabriel Dospinescu
Find all polynomials $P$ with real coefficients such that $$\frac{P(x)}{yz}+\frac{P(y)}{zx}+\frac{P(z)}{xy}=P(x-y)+P(y-z)+P(z-x)$$holds for all nonzero real numbers $x,y,z$ satisfying $2xyz=x+y+z$.

Proposed by Titu Andreescu and Gabriel Dospinescu
46 replies
tastymath75025
Apr 18, 2019
Ilikeminecraft
Yesterday at 11:40 PM
Aime ll 2022 problem 5
Rook567   1
N Yesterday at 9:39 PM by clarkculus
I don’t understand the solution. I got 220 as answer. Why does it insist, for example two primes must add to the third, when you can take 2,19,19 or 2,7,11 which for drawing purposes is equivalent to 1,1,2 and 2,7,9?
1 reply
Rook567
Yesterday at 9:08 PM
clarkculus
Yesterday at 9:39 PM
MathILy 2025 Decisions Thread
mysterynotfound   41
N Yesterday at 9:11 PM by cappucher
Discuss your decisions here!
also share any relevant details about your decisions if you want
41 replies
mysterynotfound
Apr 21, 2025
cappucher
Yesterday at 9:11 PM
The answer of 2022 AIME II #5 is incorrect
minz32   5
N Yesterday at 8:14 PM by Rook567
Source: 2022 AIME II problem 5
The answer posted on the AOPS page for the 2022 AIME II problem 5 has some flaw in it and the final answer is incorrect.

The answer is trying to use the symmetric property for the a, b, c. However the numbers are not exactly cyclically symmetric in the problem.

For example, the solution works well with a = 20, it derives 4 different pairs of b and c for p2 is in (3, 5, 11, 17) as stated. However, for a = 19, only 3 p2 in the set works, since if the p2 = 17, we will have c = 0, which is not a possible vertex label. So only three pairs of b and c work for a= 19, which are (19, 17, 14), (19, 17, 12), (19, 17, 6).

Same for a = 18.

However, for a = 17, one more possibility joined. c can be bigger than a. So the total triple satisfying the problem is back to 4.

There are two more cases that the total number of working pairs of b, c is 3, when a = 5, and a = 4.

So the final answer for this problem is not 072. It should be 068.



5 replies
minz32
Nov 20, 2022
Rook567
Yesterday at 8:14 PM
Guessing Point is Hard
MarkBcc168   31
N Apr 20, 2025 by wu2481632
Source: IMO Shortlist 2023 G5
Let $ABC$ be an acute-angled triangle with circumcircle $\omega$ and circumcentre $O$. Points $D\neq B$ and $E\neq C$ lie on $\omega$ such that $BD\perp AC$ and $CE\perp AB$. Let $CO$ meet $AB$ at $X$, and $BO$ meet $AC$ at $Y$.

Prove that the circumcircles of triangles $BXD$ and $CYE$ have an intersection lie on line $AO$.

Ivan Chan Kai Chin, Malaysia
31 replies
MarkBcc168
Jul 17, 2024
wu2481632
Apr 20, 2025
Guessing Point is Hard
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO Shortlist 2023 G5
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MarkBcc168
1595 posts
#1 • 9 Y
Y by OronSH, peace09, Rounak_iitr, buratinogigle, crazyeyemoody907, Funcshun840, ehuseyinyigit, GeoKing, wizixez
Let $ABC$ be an acute-angled triangle with circumcircle $\omega$ and circumcentre $O$. Points $D\neq B$ and $E\neq C$ lie on $\omega$ such that $BD\perp AC$ and $CE\perp AB$. Let $CO$ meet $AB$ at $X$, and $BO$ meet $AC$ at $Y$.

Prove that the circumcircles of triangles $BXD$ and $CYE$ have an intersection lie on line $AO$.

Ivan Chan Kai Chin, Malaysia
This post has been edited 1 time. Last edited by MarkBcc168, Jul 19, 2024, 2:11 PM
Z K Y
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Math-48
44 posts
#2 • 5 Y
Y by BorivojeGuzic123, kinnikuma, Funcshun840, wizixez, Sadigly
Just bash it :yup:

Set $(ABC)$ as the unit circle then we find:
$$d=-\frac{ac}b ~,~x=\frac{c^2(a+b)}{c^2+ab}$$Now define the point $T$ such that:

$t=\frac{a^2+bc}{b+c}~,~$ clearly $T\in AO$

So if we prove that $B,D,X,T$ are concyclic we will be done by symmetry
and this is indeed true because:
$$\frac{x-b}{t-b}\div\frac{x-d}{t-d}=\frac{a(c^2-b^2)}{c(a^2-b^2)}\in\mathbb{R}~~\blacksquare$$remark
This post has been edited 1 time. Last edited by Math-48, Jul 17, 2024, 4:39 PM
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MarkBcc168
1595 posts
#3 • 8 Y
Y by peace09, OronSH, Math-48, ehuseyinyigit, Rounak_iitr, Funcshun840, Sedro, Kingsbane2139
Synthetic is also equally short :D

[asy]
size(7cm);
defaultpen(fontsize(10pt));
import olympiad;
import geometry;
pair A = (1,3.2);
pair B = (0,0);
pair C = (4,0);
pair O = circumcenter(A,B,C);
pair D = 2*foot(O,B,foot(B,A,C)) - B;
pair X = extension(C,O,A,B);
pair T = 2*foot(circumcenter(B,O,C),A,O)-O;
pair P = extension(X,X+B-D,A,O);
draw(A--B--C--cycle, linewidth(1));
draw(circle(B,X,D), linewidth(0.7)+dashed);
draw(circle(B,O,C), black);
draw(A--T, linewidth(0.7));
draw(C--O, linewidth(0.7));
draw(O--X, linewidth(0.7), StickIntervalMarker(1, 1, p=linewidth(0.7), size=0.2cm));
draw(O--P, linewidth(0.7), StickIntervalMarker(1, 1, p=linewidth(0.7), size=0.2cm));
draw(circle(A,B,C), black);
draw(O--B, linewidth(0.7), StickIntervalMarker(1, 2, p=linewidth(0.7), size=0.2cm));
draw(O--D, linewidth(0.7), StickIntervalMarker(1, 2, p=linewidth(0.7), size=0.2cm));
dot("$A$", A, dir(115));
dot("$B$", B, dir(179));
dot("$C$", C, dir(0));
dot("$O$", O, 1.5*dir(-108));
dot("$D$", D, dir(70));
dot("$X$", X, dir(154));
dot("$T$", T, dir(-56));
dot("$P$", P, dir(165));
[/asy]
Let $AO$ intersect $\odot(BOC)$ again at $T$. We claim that $T$ is the concurrency point.

Note that since angle bisector of $\angle AOC$ is perpendicular to $AC$, lines $AO$, $OX$, and $BD$ form an isosceles triangle. Combining with $OB=OD$, it follows that there exists point $P$ on $AO$ such that $BXPD$ is isosceles trapezoid. Clearly, $P\in\odot(BXD)$. Now, notice that
$$\measuredangle BTP = \measuredangle BTO = \measuredangle BCO = \measuredangle OBC = \measuredangle ABD = \measuredangle XBD = \measuredangle BDP,$$so $T\in\odot(BXDP)$. Analogously, $T\in\odot(CYE)$, so we are done.
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bin_sherlo
720 posts
#4
Y by
Let $AO\cap (BOC)=\{O,K\}$ and $H$ be the orthocenter. We will show that $K$ lies on both $(BXD)$ and $(CYE)$.
Take the inversion centered at $A$ with radius $\sqrt{AB.AC}$ and reflect according to the angle bisector of $A$. We have $K\leftrightarrow H, \ B\leftrightarrow C,\ Y\leftrightarrow Y^*, \ E\leftrightarrow E^*$
Since $O^*$ is the reflection of $A$ with respect to $BC$, we get $Y^*C\parallel BO$ and $Y^\in AB$. Also $\angle E^*AC=\angle BAE=90-\angle B$
Let's show that $Y^*,B,H,E^*$ are cyclic and the other is similar.
Since $ACE^*\sim AHB,$ we have
\[\frac{CE^*}{HB}=\frac{AC}{AH}=\frac{\sin B}{\cos A}=\frac{Y^*C}{Y^*B}\implies \frac{Y^*B}{BH}=\frac{Y^*C}{CE^*}\]Also $\angle HBY^*=90+\angle A=\angle E^*CY$ hence $Y^*BH\sim Y^*CE$. Thus, $\angle Y^*HB=\angle Y^*E^*C=\angle Y^*E^*B$ as desired.$\blacksquare$
This post has been edited 1 time. Last edited by bin_sherlo, Jul 17, 2024, 12:06 PM
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pokmui9909
185 posts
#5
Y by
Let $N = AO \cap (BOC)$. We show that $N$ is the desired intersection point.
[asy]
/*
Converted from GeoGebra by User:Azjps using Evan's magic cleaner
https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
*/
pair O = (-2.74,1.36);
pair A = (-0.22,4.82);
pair B = (-6.49312,-0.69816);
pair C = (0.94790,-0.81287);
pair X = (-3.58643,1.85871);
pair D = (1.52203,0.96367);
pair N = (-6.90682,-4.36111);
pair P = (-6.42790,3.53287);
pair Q = (-2.16161,2.15412);

import graph;
size(10cm);
draw(circle(O, 4.28042), linewidth(1));
draw(A--B, linewidth(1));
draw(B--C, linewidth(1));
draw(C--A, linewidth(1));
draw(circle((-2.80673,-2.96935), 4.32986), linewidth(1));
draw(A--N, linewidth(1));
draw(P--C, linewidth(1));
draw(D--B, linewidth(1));
draw(circle((-1.81924,-3.08088), 5.24619), linewidth(1) + linetype("4 4"));

dot("$O$", O, dir((3, -10)));
dot("$A$", A, dir((8.000, 20.000)));
dot("$B$", B, dir((-10, 0)));
dot("$C$", C, dir((10, 0)));
dot("$X$", X, dir((0, 20)));
dot("$D$", D, dir((15, 15)));
dot("$N$", N, dir((-49.318, -3.889)));
dot("$P$", P, dir((-10, 10)));
dot("$Q$", Q, dir((2, -5)));[/asy]
Let $P = CO \cap \omega$, and $\ell$ be the perpendicular bisector of $BD$. Notice $A, P$ are symmetric wrt $\ell$. Let $Q$ be a point symmetric with $X$ wrt $\ell$, which clearly lies on $AO$. Since
$$\measuredangle QDB = \measuredangle DBX = \measuredangle CBO = \measuredangle ONB,$$we get that $B, X, Q, D, N$ are concyclic. From this, we conclude $N = (BXD) \cap (CYE)$. $\ \ \blacksquare$
This post has been edited 2 times. Last edited by pokmui9909, Jul 17, 2024, 12:44 PM
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AlephG_64
52 posts
#6 • 1 Y
Y by Funcshun840
To solve the problem, we need to solve this subproblem:
“Let $ABC$ be a triangle.
$AO$ meets $(BOC)$ again at $T$ and $OC$ meets $AB$ at $X$.
$(XBT)$ meets $(ABC)$ again at $D$.
Prove that $AC \perp BD$ cyclic.”


https://i.imgur.com/zH9EHA5.png

Let $I$ be the circumcenter of $(TBXD)$.
We must show $OI \parallel AC$

Claim 1: $XOIT$ is cyclic.
Proof:
By angle-chase and incenter-excenter lemma,
$\angle XIT = 2\angle XDT = 2(180 - \angle XBT) = 2\angle ABC = \angle AOC = \angle XOT$
as desired. $\square$

Claim 2: $OI \parallel AC$
Proof:
By angle-chase,
$\angle COI = \angle XTI = 90 - \angle XIC/2 = 90 - \angle XDT = 90 - \angle ABC = \angle OCA$,
as desired. $\square$.

Now, $BO=OD$ and $BI=ID \implies OI \perp BD$
By claim 2 and since $OI \perp BD$, we have $AC \perp BD$ and we are done. $\square$
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HaO-R-Zhe
23 posts
#7
Y by
I claim that the intersection point is the second intersection, $T$, of the line $AO$ with the circle $(BOC)$ (other than $O$). We will now employ complex numbers with $(ABC)$ as the unit circle.

We know that $T \in AO$, so $t = \lambda a$, where $\lambda \in \mathbb R$. Moreover, from $T \in (BOC)$, we know that \[\frac{t-b}{t-c} \div \frac{o-b}{o-c} \in \mathbb R \Longleftrightarrow \frac{c(\lambda a - b)}{b(\lambda a - c)} \in \mathbb R \Longleftrightarrow \frac{c(\lambda a - b)}{b(\lambda a - c)} = \frac{\lambda b - a}{\lambda c-a}\]Now, we solve for $\lambda$, which is given by \begin{align*}
    c(\lambda a-b)(\lambda c - a) &= b(\lambda a-c)(\lambda b - a) \\
    ac^2\lambda^2 - (a^2c+bc^2)\lambda + abc &= ab^2\lambda^2 - (a^2b+b^2c) \lambda + abc \\
    a(c-b)(c+b)\lambda &= a^2(c-b) + bc(c-b) \\
    \lambda &= \frac{a^2+bc}{a(b+c)}
\end{align*}Thus, we see that $t = \frac{a^2+bc}{b+c}$. Moreover, we know that $e = -ab\bar c$. We compute for the point $Y$, it is the intersection of $AC$ and the line passing through $B$ and its antipodal point. So we see that \[y = \frac{ac(b+(-b))+b^2(a+c)}{ac+b^2} = \frac{b^2(a+b)}{ac+b^2}\]
Finally, we show that $T, Y, E, C$ are concyclic: \[\frac{t-e}{t-c} \div \frac{y-e}{y-c}\in \mathbb R\]After some tedious computation you'll check that this is true.
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navi_09220114
479 posts
#8 • 12 Y
Y by khina, peace09, Seicchi28, CyclicISLscelesTrapezoid, AlephG_64, Assassino9931, sami1618, Kosiu, SerdarBozdag, pingupignu, Sedro, Funcshun840
My proposal :)
This post has been edited 1 time. Last edited by navi_09220114, Jul 17, 2024, 1:32 PM
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pinetree1
1207 posts
#9 • 1 Y
Y by Funcshun840
Here is a solution with lengths that sheds some light on how one might guess the correct point.

Part I: Setup. First note that $D$ and $E$ are reflections over $\overline{AO}$. Let $S$ and $R$ be the reflections of $B$ and $C$ over $\overline{AO}$; we claim that the desired point of intersection is $P = \overline{BR}\cap \overline{CS}$, which clearly lies on $\overline{AO}$. We will show that $P\in (CYE)$, and by symmetry we will also have $P\in (BXD)$.

Observe that $\triangle BAD\sim \triangle BYC$ and $\triangle CAX\sim \triangle CEB$, so we can write
\[ \frac{CY}{AD} = \frac{BC}{BD}, \quad \frac{BX}{AE} = \frac{BC}{CE}
\implies \frac{CY}{CE} = \frac{BX}{BD}. \]Since $\angle XBD = \angle YCD = 90^\circ - \angle A$, this implies $\triangle BXD \sim \triangle CYE$.
[asy]
defaultpen(fontsize(10pt));
size(300);
pair A, B, C, H, O, D, E, X, Y, R, S, P;
A = dir(115);
B = dir(210);
C = dir(330);
H = orthocenter(A, B, C);
O = (0,0);
D = 2*foot(H, A, C) - H;
E = 2*foot(H, A, B) - H;
X = extension(C, O, A, B);
Y = extension(B, O, A, C);
R = 2*foot(C, A, O) - C;
S = 2*foot(B, A, O) - B;
P = extension(B, R, C, S);
draw(A--B--C--cycle, orange);
draw(circumcircle(A, B, C), red);
draw(B--Y^^C--X^^A--P, grey+dotted);
draw(B--X--D--cycle, lightblue+dotted);
draw(B--P--S, magenta+dashed);
draw(C--Y--E--cycle, lightblue+linewidth(0.9));
draw(B--X--D--cycle, lightblue+linewidth(0.9));
draw(circumcircle(C, Y, E), heavycyan+dashed);
draw(circumcircle(B, X, D), heavycyan+dashed);
draw(B--P--S, magenta+dashed);
dot("$A$", A, dir(A));
dot("$B$", B, dir(200));
dot("$C$", C, dir(0));
dot("$D$", D, dir(70));
dot("$E$", E, dir(150));
dot("$X$", X, dir(150));
dot("$Y$", Y, dir(60));
dot("$R$", R, dir(R));
dot("$S$", S, dir(S));
dot("$P$", P, dir(P));
dot("$O$", O, dir(250));
[/asy]

Remark: [Motivation for $P$] Here's how one might guess the correct construction of $P$ after showing $\triangle BXD\sim \triangle CYD$. The line $AO$ is the perpendicular bisector of $\overline{DE}$, so we want the intersection $P = (CYE)\cap (BXD)$ to satisfy $PD = PE$. The similarity gives a nice way to compute $PD$ and $PE$: we have
\[ \frac{PD}{PE} = \frac{R_{(BXD)}\sin \angle DBP}{R_{(CYE)}\sin \angle ECP} = \frac{BD}{CE} \cdot \frac{\sin \angle DBP}{\sin \angle ECP}, \]so we want $P$ to satisfy $\sin \angle DBP / \sin \angle ECP = CE/BD$. This motivates the construction of $R$ and $S$ since $CE = DR$ and $BD = ES$ give this ratio automatically.

Part II: Length Calculation. The key claim is the following:

Claim: We have the similarity $\triangle XAD\sim \triangle PCE$.

Proof. Using the reflections over $\overline{AO}$, we have
\[ \angle PCE = 180^\circ - \angle SCE = 180^\circ - \angle BCD = \angle XAD, \]so it suffices to prove that
\[ \frac{PC}{AX} = \frac{CE}{AD} \iff PC\cdot AD = AX\cdot CE \iff PC\cdot AD = AC\cdot BE, \]where we've used $\triangle CAX\sim \triangle CEB$ in the last step. But we have $AD = AE$ and $\triangle AEB\sim \triangle ACP$ (using $\angle AEB = \angle ACP = 180^\circ - \angle C$ and $\angle BAE = \angle PAC = 90^\circ - \angle B$), which gives
\[ \frac{AE}{AC} = \frac{BE}{PC} \implies PC\cdot AE = AC\cdot BE \implies PC\cdot AD = AC\cdot BE,\]as needed. $\blacksquare$

Now $\angle CPE + \angle CYE = \angle AXD + \angle BXD = 180^\circ$, so $P\in (CYE)$. This completes the proof.
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foolish07
24 posts
#10
Y by
Let $ \Psi $ be the inversion centered at $A$ with radius $\sqrt{\frac{AB \cdot AC}{2}}$ and reflect according to the angle bisector of $A$.
And Let $M,N$ is midpoint of $AB, AC$ respectively and $P$ is foot of the altitude from $A$, $H$ is Orthocenter of $\Delta ABC$
Then $\Psi (B)=N, \Psi(C)=M, \Psi(Y)=AB \cap \odot(ANP)$ and $\Psi(E)$ is on $MN$ and $\measuredangle \Psi(E)AP=\measuredangle CAB$
And let $T$ is midpoint of $AH$, $X = MN\cap \odot (ANP)$, $F = X\Psi(Y) \cap AH$, $K=AP \cap MN$, $\Psi(\odot (CEY))=\odot (M\Psi(E)\Psi(Y))$
Since nine-point circle and Reim, we have $AX \parallel NP$
By angle chasing, we have $X,F,A,\Psi(E)$ is cyclic
Also, we have $M,\Psi(Y),F,\Psi(E)$ is cyclic
And by Reim, we have $F,\Psi(E),T,M$ is cyclic
So, $\Psi(E), \Psi(Y), M, T$ is cyclic
Similarly $\Psi(D), \Psi(X), N, T$ is cyclic
Since $\Psi(AO)=AH$, we have $\Psi(T)(=\odot (BXD) \cap \odot (CYE)$ is on $AO$
This post has been edited 2 times. Last edited by foolish07, Jul 20, 2024, 7:53 AM
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VicKmath7
1389 posts
#11
Y by
Solution
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sami1618
907 posts
#12 • 1 Y
Y by Funcshun840
Very Beautiful! I think same point as ISL 2022 G6 :):

Let the reflections of $BC$ over $AB$ and $AC$ meet at $Z$ ($A$ is the $Z$-excenter). Let $I$ be the incenter of $ZBC$ thus $Z$, $I$, $O$, and $A$ are collinear. We show that $Z$ is the desired intersection point. It is sufficient to show that $Z$ lies on $(BXD)$.

Let the parallel line to $CI$ through $Z$ meet $(CIZ)$ again at $P$. As $CIZP$ and $CIBD$ are both isosceles trapezoid, $ZPDB$ is also an isosceles trapezoid (thus cyclic). Notice we also have that $P$, $C$, and $X$ are collinear as $$\angle PCI+\angle ICB+\angle BCO=\angle CIZ+180^{\circ}-\angle A-\angle C=180^{\circ}$$Also $PZBX$ is cyclic as $$\angle XPZ=\angle XCI=\angle B=180^{\circ}-\angle XBZ$$
Attachments:
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KST2003
173 posts
#13
Y by
Let $AO$ intersect $(BOC)$ again at $S$. We will show that both $(BXD)$ and $(CYE)$ passes through $S$.

Let $S'$ be the reflection of $S$ over $\ell$. Then $BSS'D$ is an isosceles trapezoid, and also $S'$ lies on line $OX$ as $\ell$ is the external angle bisector of $\angle AOC$. By symmetry, we see that $\triangle BOS \cong \triangle DOS'$. Therefore,
\[ \measuredangle XBD = \measuredangle OBC = \measuredangle BCO = \measuredangle BSO = \measuredangle OS'D = \measuredangle XS'D, \]which shows that $XBS'D$ is cyclic. Therefore, combining the two, we see that $BXDS$ is cyclic. Similarly, $CYES$ is also cyclic, so we are done.

Some remarks: Personally, this is the most enjoyable problem in the G shortlist this year, barring G8 which I couldn't solve last year. Somehow it reminds me a lot of 2013 ISL G2.This is also the fastest I've ever solved a G5 (<10 mins) because I managed to guess the point $S$ right away.
This post has been edited 1 time. Last edited by KST2003, Jul 22, 2024, 1:40 AM
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SerdarBozdag
892 posts
#14 • 2 Y
Y by ehuseyinyigit, Amy_Chen
Similarity :w00tb: :wow: :w00t:

Let $Z = AO \cap (OBC)$, $BD \cap AZ = P$ and $CE \cap AZ = Q$.

$(1)$ $BZP \sim XCB \sim ACE$ because $\angle ACE = \angle OCB = \angle OZB$ and $\angle AEC = \angle ABC = \angle OBC + \angle OBA = \angle PBA + \angle BAP = \angle BPZ$.

$(2)$ $ABC \sim APD$ because $\angle OAD = 90 -\angle ABD = \angle BAC$ and $\angle ADB = \angle ACB$.

$(3)$ $AE = AD$ because $E$ and $D$ are reflections of $H$ over $AB$ and $AC$.

Claim: $ZBX \sim ZPD$
Proof. $\angle ZBX = \angle ABC + \angle ZBC = \angle B + \angle ZOC = 180 - \angle B = 180 - \angle BPZ = \angle ZPD$. Lastly,
$$\frac{BX}{PD} \overset{3} = \frac{BX}{EA} \cdot \frac{AD}{PD} \overset{1,2} = \frac{CX}{CA} \cdot \frac{AC}{BC} = \frac{CX}{BC} \overset{1} = \frac{ZB}{ZP}$$which is sufficient. $\square$

The claim gives $\angle BXZ = \angle BDZ$ which shows $Z \in (BXD)$. By symmetry, $Z \in (CYE)$ so we are done.
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Hasin_Ahmad
14 posts
#16 • 1 Y
Y by sami1618
Angle Chase:

Let $T = AO \cap (BOC)$. We will show that $T$ is our desired point. It suffices to prove that $BXDT$ is cyclic as the other part can be proved similarly.

First, let $C^\prime$ be the antipode of $C$ in $\omega$, and $Q = DC^\prime \cap AO$.

Claim: $AQXC^\prime$ is cyclic.
Proof: $\angle XC^\prime Q = \angle CC^\prime D = \angle CBD = \angle ABO = \angle BAO = \angle XAQ$. $\blacksquare$

Claim: $BXQD$ is cyclic.
Proof: Since, $AQXC^\prime$ is cyclic

\begin{align*}
\angle AXQ &= AC^\prime Q = \angle AC^\prime D = \angle ABD = \angle CBO\\
&= \angle OCB = \angle C^\prime CB = \angle C^\prime DB\\
&= \angle QDB. \blacksquare
\end{align*}
To finish the problem, observe that: $\angle BTQ = \angle BTO = \angle BCO = \angle BCC^\prime = \angle BDC^\prime = \angle BDQ$. Therefore, making $BXQDT$ cyclic. Hence, we are done.

Nice Problem :)
Attachments:
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ddami
10 posts
#17
Y by
I could not figure out a nice way to construct the point we wanted, so I came up with something rather different.

Let $H$ be the orthocenter of $ABC$ and let $AH$ meet $\omega$ at $P$. Let $O_1$ and $O_2$ be the circumcenters of $CEY$ and $BDX$ respectively. Note that $O_1E = O_1Y$, $OE = OA$ and
\[
\angle EO_1Y = 2\angle ECY = 2\angle EOA.
\]Thus $E$ is the center of spiral similarity that sends $O_1Y$ to $OA$. Hence it also sends $O_1O$ to $YA$. Similarly $D$ sends $O_2O$ to $XA$. Now notice that
\[
\frac{O_1O}{YA} = \frac{EO}{EA} = \frac{DO}{DA} = \frac{OO_2}{AX}.
\]Furthermore $\angle O_1OO_2 = \angle YAX$, thus $OO_1O_2 \sim AYX$. Moreover, note that line $AO$ divides $\angle O_1OO_2$ in the same way line $AP$ divides $\angle YAX$. Thus we might consider $T$ in $AO$, with $O$ between $A$ and $T$, such that quadrilaterals $OO_1TO_2$ and $AYPX$ are similar. Now note that
\[
\frac{XP}{O_2T} = \frac{AX}{OO_2} = \frac{XD}{OO_2}.
\]Since $XP = XD$ we thus obtain $O_2T = OO_2$ which means $T \in (BDX)$. Similarly we get that $T \in (CEY)$.
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ACGNmath
327 posts
#18
Y by
We claim that the desired point of concurrency is the second intersection of $AO$ with $(BOC)$. Call this point $P$. We shall show that $PYCE$ is cyclic, from which the result follows by symmetry.

Since there are many circles passing through $C$, let us invert with radius $\sqrt{CA\cdot CB}$ and reflect about the $\angle C$-angle bisector. Then we have: $A$ and $B$ swap places, $E$ is sent to the intersection of $CO$ with $AB$, $O$ is sent to the reflection of $C$ across $AB$, and $Y$ is the intersection of $(B^*O^*C)$ with $A^*C^*$. The task is therefore to show that $B^* O^*$, $Y^* E^*$ and $(CA^*O^*)$ are concurrent. Relabelling points for convenience, we have the following equivalent problem:
ISL 2023/G5, inverted wrote:
Let $\triangle ABC$ have circumcenter $O$, and suppose $A'$ is the reflection of $A$ across $BC$. Let $AB$ intersect $(ACA')$ at $D$ and $A'C$ intersect $(ABA')$ at $E$. Show that $DE$, $BC$ and $AO$ are concurrent.
[asy]
size(8cm);
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair AA = 2*foot(A,B,C)-A;
path ABAA = circumcircle(A,B,AA);
path ACAA = circumcircle(A,C,AA);
pair D = intersectionpoints(A--B+10*(B-A),ACAA)[1];
pair E = intersectionpoints(AA--C+10*(C-AA),ABAA)[1];
pair O = circumcenter(A,B,C);
draw(A--B--C--A--cycle);
draw(B--AA--C);
draw(B--D--E--C);
draw(ABAA);
draw(ACAA);
pair P = extension(A,O,B,C);
draw(A--P);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$A'$",AA,dir(AA));
dot("$D$",D,dir(D));
dot("$E$",E,dir(E));
dot("$P$",P,dir(P));
dot("$O$",O,dir(45));
[/asy]
It would be great if someone could provide a synthetic solution, it doesn't seem too hard to find but I can't seem to do it.

Anyway, here's a bary bash. Set $\triangle ABC$ to be the reference triangle, and denote $S_A = \frac{-a^2+b^2+c^2}{2} = bc\cos A$ and likewise $S_B$ and $S_C$ for convenience. Note that $S_A+S_B=c^2$, as well as cyclic variants. Let $AO$ intersect $BC$ at $P$.

First, we have $O=(a^2 S_A:b^2 S_B: c^2 S_C)$ so $P = (0:b^2 S_B : c^2 S_C)$. Next, the foot from $A$ to $BC$ is $(0:b\cos C:c\cos B) = (0:S_C : S_B) = (0, \frac{S_C}{a^2},\frac{S_B}{a^2})$. Therefore,
\[A' = 2\left(0,\frac{S_C}{a^2},\frac{S_B}{a^2}\right)-(1,0,0) = \left(-1,\frac{2S_C}{a^2},\frac{2S_B}{a^2}\right) = (-a^2: 2S_C : 2S_B).\]Now we find the equation for $(ABA')$. The generic equation is $-a^2 yz - b^2 zx - c^2 xy + (x+y+z)(ux+vy+wz)=0$. Substituting in $A=(1,0,0)$, we have $u=0$. Similarly, $B=(0,1,0)$ gives $v=0$. To find $w$, we substitute $A'$ to get
\[-a^2\cdot 4S_B S_C + b^2 \cdot 2a^2 S_B + c^2 \cdot 2a^2 S_C + a^2 \cdot w \cdot 2S_B = 0.\]Cancelling $a^2$ and simplifying, we have
\[w\cdot S_B = 2S_B S_C - b^2 S_B - c^2 S_C = 2S_B S_C - (S_A+S_C) S_B - (S_A+S_B)S_C = -S_A(S_B+S_C) = -a^2 S_A.\]Therefore,
\[w = -\frac{a^2 S_A}{S_B}.\]Similarly, the equation for $(ACA')$ is
\[-a^2 yz - b^2 zx - c^2 xy + (x+y+z)\left(\frac{a^2 S_A}{S_C}\right)y = 0.\]
Let $D = (k,1-k,0)$. Then
\[-c^2 k(1-k) - (1-k)\left(\frac{a^2 S_A}{S_C}\right) = 0 \quad\Rightarrow\quad k = -\frac{a^2 S_A}{c^2 S_C}.\]Therefore, $D = (-a^2 S_A: a^2S_A + c^2 S_C : 0)$.

The strategy now is to intersect $DP$ and $A'C$ and show that this intersection lies on $(ABA')$.

The line $DP$ is given by
\[0  =\begin{vmatrix}x & y & z \\ -a^2 S_A & a^2 S_A + c^2 S_C & 0 \\ 0 & b^2 S_B & c^2 S_C \end{vmatrix} = c^2 S_C(a^2 S_A+c^2 S_C)x + a^2 c^2 S_A S_C y - a^2 b^2 S_A S_B z.\]The line $A'C$ is given by
\[0 = \begin{vmatrix}x& y & z \\ -a^2 & 2S_C & 2S_B \\ 0& 0 & 1 \end{vmatrix} = 2S_C x + a^2 y.\]The intersection of these two lines is
\begin{align*}
&\left(a^4 b^2 S_A S_B : -2a^2 b^2 S_A S_B S_C : a^2 c^2 (a^2 S_A S_C + c^2 S_C^2 - 2S_A S_C^2)\right)
\\&= \left(a^2 b^2 S_A S_B : -2b^2 S_A S_B S_C : c^2 S_C(a^2 S_A + c^2 S_C - 2S_A S_C)\right)
\\&= \left(a^2 b^2 S_A : -2b^2 S_A S_C : c^2 S_C(S_A+S_C)\right)
\\&= \left(a^2 S_A : -2S_A S_C: c^2 S_C\right).
\end{align*}Now we verify that this point lies on $(ABA')$, which has equation $-a^2 yz - b^2 zx - c^2 xy - \frac{a^2 S_A}{S_B}(x+y+z)z = 0$.
\begin{align*}
&2a^2 c^2 S_A S_C^2 - a^2 b^2 c^2 S_A S_C + 2a^2 c^2 S_A^2 S_C - \frac{a^2 S_A}{S_B}\cdot S_B(S_A+S_C)\cdot c^2 S_C 
\\&= 2a^2 c^2 S_A S_C(S_A+S_C) - a^2 c^2 S_A S_C (S_A+S_C) - a^2 c^2 S_A S_C (S_A+S_C) = 0
\end{align*}by using the fact that $S_A+S_C=b^2$.
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Kosiu
1 post
#19
Y by
I'll show that the desired point of concurrency is the second intersection of $AO$ with circle $BOC$. Let's call this point $P'$. I'll show that $P'CYE$ is cyclic.

It's equivalent to $\angle P'YE = \angle CYO$ which is the same as $\angle CYP' = \angle BYE$.

Now let's define a bunch of useful points:

$R$ is the second intersection of $BO$ with circle $ABC$
$G$ is orthogonal projection of $P'$ onto $AC$
$T$ is orthogonal projection of $A$ onto $BC$
$S$ is the second intersection of $AP'$ with circle $GYP'$
$K$ is point on segment $AB$ such that $KY||AR$


See that $\triangle AGP'$~$\triangle ATB$ ~ $\triangle REB$

Now I will show that point $S$ in $\triangle AGP'$ is the same point as $Y$ in $\triangle REB$, by proving: $RY/YB = AS/SP'$.

See that $RY/YB = AK/KB$ (because $KY||AR$)

It's left to show that $KS||BP'$, but fortunately it's easy by simple angle chasing, because $\angle YKA = 90$ and $AYSK$ is cyclic.

$\angle AP'B = \angle OCB = 90 - \angle BAC$ and $\angle ASK = \angle AYK = 90 - \angle BAC$

Then $\angle GSP' = \angle BYE$ but $\angle GSP' = \angle GYP'$

That finishes the proof.

I enjoyed this problem:)
Attachments:
This post has been edited 5 times. Last edited by Kosiu, Aug 4, 2024, 11:44 PM
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crazyeyemoody907
450 posts
#20 • 1 Y
Y by GeoKing
[asy]
//23slg5 init
//setup
defaultpen(fontsize(10pt));
size(6cm);
pen blu,grn,blu1,blu2,lightpurple; blu=RGB(102,153,255); grn=RGB(0,204,0);
blu1=RGB(233,242,255); blu2=RGB(212,227,255); lightpurple=RGB(234,218,255);// blu1 lighter
//define
int i=0;
pair inverse(pair P) {return P/(P.x*P.x+P.y*P.y); }
pair O,A,B,C; O=(0,0); A=(-.6,.8); B=(-.8,-.6); C=(.8,-.6);
pair X,Y,Z,X1,Y1,Z1; X=extension(A,O,B,C); Y=extension(B,O,A,C); Z=extension(C,O,A,B); X1=inverse(X); Y1=inverse(Y); Z1=inverse(Z);
//draw
filldraw(A--B--C--cycle,blu1,blu); draw(A--X1^^B--Y,blu); draw(circle(O,1),blu);
draw(circumcircle(B,O,C),purple); draw(circumcircle(C,X1,Y),magenta); draw(C--2*foot(C,A,B)-A-B-C,magenta+dashed);
clip((1.2,1.2)--(1.2,-1.2)--(-.4,-1.2)--(-1.2,-.4)--(-1.2,1.2)--cycle);
//label
void pt(string s,pair P,pair v, pen a){filldraw(circle(P,.015),a,linewidth(.1)); label(s,P,v);}
string labels[]={"$O$","$A$","$B$","$C$","","$Y$",""}; //7
pair points[]={O,A,B,C,X1,Y,2*foot(C,A,B)-A-B-C};
real dirs[]={-100,120,-160,-20,0,60,0};
pen colors[]={blu,blu,blu,blu,purple,magenta,magenta};
for (i=0; i<7; ++i) {pt(labels[i],points[i],dir(dirs[i]),colors[i]);}
[/asy]

Desired intersection is stated in earlier posts. Do refactoring of points as follows:
refactoring wrote:
Triangle $ABC$ has circumcircle $\Omega$ with center $O$. Define $X=\overline{AO}\cap\overline{BC}$ and $Y,Z$ similarly, and $X'=\overline{AO}\cap(BOC)$ to be the inverse of $X$ in $\Omega$ (and $Y',Z'$ similarly). Then $A,Y,Z'$, and the intersection of the $A$-altitude with $(ABC)$ are concyclic.
[asy]
//23slg5 final
//setup
defaultpen(fontsize(10pt));
size(9cm);
pen blu,grn,blu1,blu2,lightpurple; blu=RGB(102,153,255); grn=RGB(0,204,0);
blu1=RGB(233,242,255); blu2=RGB(212,227,255); lightpurple=RGB(234,218,255);// blu1 lighter
//define
int i=0;
pair inverse(pair P) {return P/(P.x*P.x+P.y*P.y); }
pair O,A,B,C; O=(0,0); A=dir(110); B=dir(-160); C=dir(-20);
pair X,Y,Z,X1,Y1,Z1; X=extension(A,O,B,C); Y=extension(B,O,A,C); Z=extension(C,O,A,B); X1=inverse(X); Y1=inverse(Y); Z1=inverse(Z);
pair K,T; K=extension(Y,Z1,Z,Y1); T=2*foot(A,B,C)-A-B-C;
pair[] xook=intersectionpoints(circumcircle(Y,Z,Y1),circle(O,1));
//draw
filldraw(A--B--C--cycle,blu1,blu); draw(circle(O,1)^^circumcircle(A,O,B)^^circumcircle(A,O,C),blu); draw(A--X1^^B--Y1^^C--Z1,blu);
draw(Y--Z1^^Z--Y1,purple); draw(circumcircle(Y,Z,Y1),purple); draw(A--T,magenta); draw(circumcircle(A,Y,Z1),magenta+dotted);
draw(1.2*xook[1]-.2*xook[0]--1.2*xook[0]-.2*xook[1],red);
clip( box((-1.5,-2.6),(2.2,1.5)) );
//label
void pt(string s,pair P,pair v, pen a){filldraw(circle(P,.015),a,linewidth(.1)); label(s,P,v);}
string labels[]={"$O$","$A$","$B$","$C$","$X$","$Y$","$Z$","$X'$","$Y'$","$Z'$","$K$","$T$"}; //12
pair points[]={O,A,B,C,X,Y,Z,X1,Y1,Z1,K,T};
real dirs[]={-110,100,-130,-50, -110,-100,-80,0,-40,100,130,-90};
pen colors[]={blu,blu,blu,blu,blu,blu,blu,blu,blu,blu,purple,magenta};
for (i=0; i<12; ++i) {pt(labels[i],points[i],dir(dirs[i]),colors[i]);}
[/asy]
Define $K=\overline{Y'Z}\cap\overline{YZ'}$; observe that there is a spiral similarity at $A$ sending $Z',B\to C,Y'$ so $Y',Z'$ are $\sqrt{bc}$ inverses. As a consequence $\overline{AY'}$ and $\overline{AZ'}$ are isogonal in $\angle BAC$.

Claim 1: $K$ lies on the $A$-altitude.
Proof. Apply DDIT to $A$ and $YY'ZZ'$, to obtain the involution $A(BC;Y'Z';OK)$, which, based on the first two pairs, is isogonality at $\angle BAC$, so $\overline{AK}$ and $\overline{AO}$ are isogonal as well.$\qquad\qquad\square$

Claim 2: $\overline{Y'Z}\cap\overline{YZ'}$ lies on the radical axis of $\Omega$ and $(YY'ZZ')$.
Proof. Said radical axis is also the polar of $O$ wrt $(YY'ZZ')$ because $OA^2=OY\cdot OY'=OZ\cdot OZ'$ by design.
The claim then reduces to Brocard on that quadrilateral.$\qquad\qquad\square$
By PoP at the concurrency point in claim 1, if $T$ is the intersection of the $A$-altitude with $(ABC)$ $KY\cdot KZ'=\text{Pow}(K,\Omega)=KA\cdot KT$, and we win.

remark: guessing the concurrency point
This post has been edited 3 times. Last edited by crazyeyemoody907, Aug 16, 2024, 5:50 AM
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SerdarBozdag
892 posts
#21 • 1 Y
Y by Amy_Chen
I will give a second solution using Ptolemy-Sinus Lemma. Solved with my bros from ENKA.

Let $Z = AO \cap (OBC)$. I will prove that $Z \in (BXD)$ which is sufficient from symmetry.

$$Z \in (BXD) \iff $$$$BD \sin B = BX \sin (270 - 2B - C)+BZ \sin (90-A)$$$$\iff $$$$BD \sin B = BX \cos (B-A)+BZ \cos A$$$$\overset{\text{Sinus Theorem in } \triangle BCX} \iff $$$$BD \sin B = BC \cos A+BZ \cos A $$$$\overset{\text{Sinus Theorem in } \triangle BOZ \text{ and } \triangle ABC} \iff $$$$2R\cos (C-A)\sin B = R\sin 2A + R\sin 2C$$$$\iff $$$$2 \sin(C+A) \cos(C-A) = \sin 2A + \sin 2C$$which is true.
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SHZhang
109 posts
#22
Y by
Let $K$ be the second intersection of $(BOC)$ with $AO$. It now suffices to show that $K \in (BXD) \cap (CYE)$. We will show $K \in (CYE)$; the proof for $K \in (BXD)$ is similar. Let $P$ be the intersection of $AO$ with the line through $Y$ parallel to $CE$.

Claim: $EPYC$ is an isosceles trapezoid, and thus cyclic.
Proof: Let $Q$ be the point diametrically opposite $A$ on $\omega$. Then $PY \parallel CE \parallel BQ$. Since $OB = OQ$ and $\triangle OBQ \sim \triangle OYP$, $OY = OP$, so a line through $O$ perpendicular to $PY$ is the perpendicular bisector of both $PY$ and $CE$. $\square$

Claim: $PYCK$ is cyclic.
Proof: We have $\angle PYC = 180^\circ - \angle YCE = \angle BAC + 90^\circ$, and $\angle PKC = \angle OKC = \angle OBC = 90^\circ - \angle BAC$, giving $\angle PYC + \angle PKC = 180^\circ$.

Combining the two claims gives $EPYCK$ cyclic, so $K \in (CYE)$.
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Entrepreneur
1173 posts
#23
Y by
MarkBcc168 wrote:
Synthetic is also equally short :D

[asy]
size(7cm);
defaultpen(fontsize(10pt));
import olympiad;
import geometry;
pair A = (1,3.2);
pair B = (0,0);
pair C = (4,0);
pair O = circumcenter(A,B,C);
pair D = 2*foot(O,B,foot(B,A,C)) - B;
pair X = extension(C,O,A,B);
pair T = 2*foot(circumcenter(B,O,C),A,O)-O;
pair P = extension(X,X+B-D,A,O);
draw(A--B--C--cycle, linewidth(1));
draw(circle(B,X,D), linewidth(0.7)+dashed);
draw(circle(B,O,C), black);
draw(A--T, linewidth(0.7));
draw(C--O, linewidth(0.7));
draw(O--X, linewidth(0.7), StickIntervalMarker(1, 1, p=linewidth(0.7), size=0.2cm));
draw(O--P, linewidth(0.7), StickIntervalMarker(1, 1, p=linewidth(0.7), size=0.2cm));
draw(circle(A,B,C), black);
draw(O--B, linewidth(0.7), StickIntervalMarker(1, 2, p=linewidth(0.7), size=0.2cm));
draw(O--D, linewidth(0.7), StickIntervalMarker(1, 2, p=linewidth(0.7), size=0.2cm));
dot("$A$", A, dir(115));
dot("$B$", B, dir(179));
dot("$C$", C, dir(0));
dot("$O$", O, 1.5*dir(-108));
dot("$D$", D, dir(70));
dot("$X$", X, dir(154));
dot("$T$", T, dir(-56));
dot("$P$", P, dir(165));
[/asy]
Let $AO$ intersect $\odot(BOC)$ again at $T$. We claim that $T$ is the concurrency point.

Note that since angle bisector of $\angle AOC$ is perpendicular to $AC$, lines $AO$, $OX$, and $BD$ form an isosceles triangle. Combining with $OB=OD$, it follows that there exists point $P$ on $AO$ such that $BXPD$ is isosceles trapezoid. Clearly, $P\in\odot(BXD)$. Now, notice that
$$\measuredangle BTP = \measuredangle BTO = \measuredangle BCO = \measuredangle OBC = \measuredangle ABD = \measuredangle XBD = \measuredangle BDP,$$so $T\in\odot(BXDP)$. Analogously, $T\in\odot(CYE)$, so we are done.

Did you mean perpendicular bisector?
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kamatadu
480 posts
#24 • 4 Y
Y by SilverBlaze_SY, S.Ragnork1729, HACK_IN_MATHS, GeoKing
Solved with SilverBlaze_SY, HACK_IN_MATHS and S.Ragnork1729.

[asy]
/* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */ pair A = (10.69888,16.57077); pair B = (1.51632,-12.50733); pair C = (34.50552,-13.10250); pair O = (18.20964,-1.78998); pair D = (32.29102,12.18297); pair X = (7.29373,5.78779); pair T = (13.17850,10.50910); pair P = (30.63286,-32.15967);
import graph; size(12cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5;
draw(A--B, linewidth(0.6)); draw(B--C, linewidth(0.6)); draw(C--A, linewidth(0.6)); draw(circle(O, 19.83755), linewidth(0.6)); draw(circle((24.58069,-9.73103), 23.23086), linewidth(0.6) + blue); draw(circle((17.88747,-19.64757), 17.86050), linewidth(0.6) + linetype("4 4") + red); draw(X--T, linewidth(0.6)); draw(A--P, linewidth(0.6)); draw(B--D, linewidth(0.6)); draw(C--X, linewidth(0.6));
dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$O$", O, NE); dot("$D$", D, NE); dot("$X$", X, NW); dot("$T$", T, W); dot("$P$", P, SE);  [/asy]

Let the line through $X$ parallel to $BD$ intersect $AO$ at $T$.

Claim: $T$ lies on $\odot(BXD)$.
Proof. Firstly, we have, \begin{align*} \measuredangle OTX &= \measuredangle ATX\\ &= \measuredangle (AO,BD)\\ &= \measuredangle OAB+\measuredangle ABD\\ &= (90^{\circ}-\measuredangle BCA) + (90^{\circ}-\measuredangle CAB)\\ &= \measuredangle ACB + \measuredangle BAC\\ &= \measuredangle ABC .\end{align*}Also, \begin{align*} \measuredangle OXT &= \measuredangle (OX,BD)\\ &= \measuredangle DBX + \measuredangle BXO\\ &= (90^{\circ}-\measuredangle CAB)+ (\measuredangle ABC+\measuredangle BCO)\\ &= (90^{\circ}-\measuredangle BCO)+\measuredangle ABC+ \measuredangle BAC\\ &= \measuredangle CAB + \measuredangle ABC + \measuredangle BAC\\ &= \measuredangle ABC .\end{align*}This gives us that $\measuredangle OTX=\measuredangle ABC= \measuredangle OXT$. This further implies that $OX=OT$. We also have that $OB=OD$. Combining these with the fact that $XT\parallel BD$, we get that $BXTD$ is an isosceles trapezium and thus is also cyclic. $\blacksquare$

Let $P=AO \cap \odot(BXD)$. Then note that, \[ \measuredangle OPB=\measuredangle TPB=\measuredangle TDB =\measuredangle DBX =90^{\circ}-\measuredangle BAC =\measuredangle OCB .\]
So $OBPC$ is cyclic. Similarly, if we define $Q=AO\cap \odot(CYE)$, then $OBQC$ is also cyclic. Now note that $P=AO\cap \odot(OBC)=Q$ which gives $P\equiv Q$. We are done.
This post has been edited 2 times. Last edited by kamatadu, Oct 1, 2024, 4:34 PM
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EthanWYX2009
863 posts
#25
Y by
Bashing is nice!
take $\odot (ABC)$ as unit circle, then $d=-ac/b,$ $x=tc$ where $t\in\mathbb R.$ By $x+ab\overline x=a+b,$ $t=\frac{(a+b)c}{ab+c^2},$ $x=\frac{(a+b)c^2}{ab+c^2}.$ Some observation makes me guess that the intersection is on $\odot (BOC),$ say $\lambda a$ where $\lambda\in\mathbb R.$ We want $\frac{\lambda a-b}{\lambda a-c}/\frac bc\in\mathbb R.$ Here $\lambda =\frac{a^2+bc}{a(b+c)},$ $\lambda a=\frac{a^2+bc}{b+c}.$ Now we only need $b,-ac/b,\frac{(a+b)c^2}{ab+c^2},\frac{a^2+bc}{b+c}$ are concyclic. $$\iff \dfrac{\frac{(a+b)c^2}{ab+c^2}-b}{\frac{(a+b)c^2}{ab+c^2}+\frac {ac}b}\div\dfrac{\frac{a^2+bc}{b+c}-b}{\frac{a^2+bc}{b+c}+\frac {ac}b}=\frac{a(c+b)(c-b)}{c(a+b)(a-b)}\in\mathbb R$$which is obvious.$\Box$
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Saucepan_man02
1338 posts
#26
Y by
Heres my solution:

Define point $T = AO \cap (BOC)$ with $T \neq O$.

Note that: $\angle BTO = \angle BCO = 90^\circ - A$. The following claim finishes the problem: Let $P = (BXD) \cap AO$.

Claim : $BXPD$ is a cyclic trapeziod.
Proof: Notice that: $BD$ and the angle bisector of $\angle AOC$ are parallel which implies that $BD$ and the angle bisector of $\angle XOA$ are perpendicular. Let $P'$ be a point on $AO$ such that $XP' \parallel BD$. Therefore, $XP'$ and $BD$ have same perpendicular bisector which implies $BCP'D$ is an isosceles trapezoid (and thus cyclic) and therefore $P=P'$.

Due to the above claim: $\angle PDB =\angle XBD = 90^\circ-A = \angle BTP$ and therefore $BXDT$ is cyclic.

Remark
This post has been edited 2 times. Last edited by Saucepan_man02, Nov 1, 2024, 3:26 AM
Reason: EDIT
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GeoKing
518 posts
#27
Y by
Sol:- Let $(BOC)$ meet $AO$ again at $V$ .Let $l$ be the line through $O$ parallel to $AB$. Since $AO=OB$ ,the reflection $F$ of $Y$ across $l$ lies on $AO$.Since $l$ is also the perpendicular bisector of $CE$ ,$CYFE$ is a cyclic isosceles trapezoid.
$\measuredangle CVF=\measuredangle CVO=\measuredangle CBO=\measuredangle OCB=\measuredangle YCE=\measuredangle CEF \implies V \in (CYEF)$. Similarly $V\in (BXD)$.
https://cdn.discordapp.com/attachments/1247512024687181896/1316092227696594944/image.png?ex=6759c9e6&is=67587866&hm=abc2f48fb0ccc7d6b6d8baa84f813f2805ba6a447199a6e0f60e37f2de7a73d4&
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wizixez
168 posts
#28
Y by
Easy G5
Solution With Complex Numbers
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Autistic_Turk
9 posts
#29
Y by
MarkBcc168 wrote:
Let $ABC$ be an acute-angled triangle with circumcircle $\omega$ and circumcentre $O$. Points $D\neq B$ and $E\neq C$ lie on $\omega$ such that $BD\perp AC$ and $CE\perp AB$. Let $CO$ meet $AB$ at $X$, and $BO$ meet $AC$ at $Y$.

Prove that the circumcircles of triangles $BXD$ and $CYE$ have an intersection lie on line $AO$.

Ivan Chan Kai Chin, Malaysia
Define X': intersection of line through X and parallel to BD with AO and we define Y' similarly
Define X": second intersection of AO and BXD define Y" similarly
Claim 1: BXX'D is cyclic
Proof:
notice XX' and BD are parallel thus it's enough to show angels XBD and X'DB are equal with is equivalence to showing tan(A)=cot(pi/2-A)=cot(X'DB) now by similarity we have XB/AX =X'T/AX' and by law of sin in triangles CXB and CXA we have X'T/AX'=XB/AX=(sin(A)*cos(A))/((sin(B)*cos(B)) thus AX'/X'T=cos(B)*sin(B)/sin(A)*cos(A)
now by law of sin in triangles DX'A and DX'T we have (sin(C-X'DB)/sin(X'DB))*(sin(B)/(Sin(A))=AX'/X'T
thus we have sin(C)*cot(X'DB)-Cos(C)=Cos(B)/Cos(A)
thus it's enough to prove sin(C)*tan(A)-Cos(C)=Cos(B)/Cos(A) ifoif sin(A)*sin(C)-cos(A)*cos(C)=cos(B) ifoif
-Cos(A+C)=Cos(B) with is obvious
Claim 2: AY"=AX"= (sin(B)*sin(C)*2R)/Cos(A)
proof:by power of point we have AX*AB=AX'*AX" thus we have AX"=(AX/AX')*AB
by law of sin and famous identity sin(x)=sin (pi-x) we have AX/AX'=sin(B)/Cos(A) thus we have AX"= sin(B)*sin(C)*2R/cos(A) we could have similar argument for AY" thus claim is proved now for notice X", and Y" are at same side of ray AO because triangle is acute thus X"=Y" and problem is solved
This post has been edited 1 time. Last edited by Autistic_Turk, Feb 16, 2025, 2:38 PM
Reason: Typo
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ihategeo_1969
235 posts
#30 • 1 Y
Y by babarazamtruefan152-0
How is this a G5 bro

Let $T=\overline{AO} \cap (BOC)$ and we will prove this is the concurrency point.

We will do a sketch with complex numbers with $(ABC)$ as unit circle. See that \[d=-\frac{ac}b, \text{ } t=\frac{a^2+bc}{b+c},\text{ }x=\frac{c^2(a+b)}{c^2+ab}\]And done just check \[\frac{x-d}{x-b}:\frac{t-d}{t-b}=
\frac{c(a^2b+abc+ac^2+b^2c}{ab(c-b)(c+b)}:\frac{a^2b+abc+ac^2+b^2c}{b(a-b)(a+b)}=
\frac{c(a-b)(a+b)}{a(c-b)(c+b)} \in \mathbb{R}\]And done.
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Ilikeminecraft
623 posts
#31
Y by
triv???
Let $T$ be the second intersection of $(OBC)$ and $AO.$ I claim $T$ is the desired concurrency point.
Let $P$ be the intersection inside of $ABC$ of $AO$ and $(BXD).$
Claim: $OX = OE$ and $BXED$ is isosceles trapezoid
Proof: Observe that the angle bisector of $EOC$ is perpendicular to the angle bisector of $AOX.$
Also note that we have $\angle(BH, OC) = 180 - \angle HBC - \angle OCB = 180 -(90 - C) - (90 - A) = 180 - \angle OAB - \angle HBA = \angle(BH, AO)$. Thus, the angle bisector of $AOX$ is perpendicular to $EX$ and $BD,$ which finishes.

Note that $\angle XET = \angle OEX = \angle(BH, OC) = B = 90 - C + 90 - A = \angle BAO + \angle OTB = \angle XBT,$ which finishes.
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Circumcircle
69 posts
#32
Y by
Let the intersections of $BXD$ and $CYE$ be points $T$ and $S$ with $T$ being on the opposite side of $BC$ wrt $A$.
Let $L$ be the intersection of the $A$ altitude with $(ABC)$.

Claim 1: $H$ lies on $ST$.
Proof: Since $BH\cdot HD=CH\cdot HE$ it means that $H$ has equal powers to circles $BXD$ and $CYE$ so it lies on $ST$.

Claim 2: $ASLT$ is cyclic
Proof: $SH\cdot HT=BH\cdot HD=AH\cdot HL$ which implies that $ASLT$ is cyclic.

Let $OH$ intersect $(AOL)$ again at $J$.

Claim 3: $JSOT$ is cyclic
Proof: $JH\cdot HO=AH\cdot HL=SH\cdot HT$ so $JSOT$ is cyclic.

Claim 4: $JSHL$ is cyclic
Proof: $\angle JSH+\angle JLH=\angle JOT+\angle JOA=180^\circ$ so $JSHL$ is cyclic.

Now by all the cyclics we found, we can chase angles

$\angle STO=\angle SJO=\angle SLO=\angle STA$ which implies that $A$, $O$, and $T$ are collinear.
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wu2481632
4239 posts
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cute!

Let $H$ be the orthocenter of $ABC$. Perform a $\sqrt{bc}$ inversion with the standard reflection over the $A$-angle bisector. Noting that $\angle CAD = 90 - \angle C$, it follows that $D$ maps to a point $D'$ on ray $CB$ with $\angle D'AB = 90 - \angle C$; similarly, $E$ maps to $E'$ on ray $BC$ with $\angle E'AC = 90 - \angle B$. As $H$ and $O$ are isogonal conjugates, and as $(ABC)$ maps to $BC$, we note that $O$ maps to the reflection of $A$ over $BC$, point $O'$.

Thus point $Y$ maps to $(ACO') \cap AB$. We claim that $Y'BHE'$ is cyclic. This will, by symmetry, give us that $X'CHD'$ is cyclic, and we will be done, as $(Y'BHE')$ would be the image of $(CYE)$.

Observe that $\angle Y'CE' = \angle Y'CO' + \angle E'CO' = 90 - \angle B + 180 - \angle C = 90 + \angle A$, and that $\angle Y'BH = 180 - \angle ABH = 90 + \angle A$. Next, we'll show that $\frac{Y'B}{Y'C} = \frac{BH}{CE'}$, which will yield similar triangles $Y'BH$ and $Y'CE'$, from which we easily obtain $Y'BHE'$ cyclic.

Note that by the Law of Sines, $\frac{Y'B}{Y'C} = \frac{\cos \angle A}{\sin \angle B}$. Moreover, we have $BH = 2 R \cos \angle B$ and $CE' = \frac{b \cos \angle B}{\cos \angle A}$, again by the Law of Sines. Putting this together, we see that $\frac{BH}{CE'} = \frac{2R \cos \angle A}{b}$, so as $\frac{b}{\sin \angle B} = 2R$, we are done.
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