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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.

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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
A number theory problem
super1978   2
N 13 minutes ago by Tintarn
Source: Somewhere
Let $a,b,n$ be positive integers such that $\sqrt[n]{a}+\sqrt[n]{b}$ is an integer. Prove that $a,b$ are both the $n$th power of $2$ positive integers.
2 replies
super1978
May 11, 2025
Tintarn
13 minutes ago
Mega angle chase
kjhgyuio   2
N an hour ago by Jupiterballs
Source: https://mrdrapermaths.wordpress.com/2021/01/30/filtering-with-basic-angle-facts/
........
2 replies
kjhgyuio
3 hours ago
Jupiterballs
an hour ago
power of a point
BekzodMarupov   0
an hour ago
Source: lemmas in olympiad geometry
Epsilon 1.3. Let ABC be a triangle and let D, E, F be the feet of the altitudes, with D on BC, E on CA, and F on AB. Let the parallel through D to EF meet AB at X and AC at Y. Let T be the intersection of EF with BC and let M be the midpoint of side BC. Prove that the points T, M, X, Y are concyclic.
0 replies
BekzodMarupov
an hour ago
0 replies
Interesting inequalities
sqing   1
N 2 hours ago by sqing
Source: Own
Let $a,b,c \geq 0 $ and $ab+bc+ca- abc =3.$ Show that
$$a+k(b+c)\geq 2\sqrt{3 k}$$Where $ k\geq 1. $
Let $a,b,c \geq 0 $ and $2(ab+bc+ca)- abc =31.$ Show that
$$a+k(b+c)\geq \sqrt{62k}$$Where $ k\geq 1. $
1 reply
sqing
2 hours ago
sqing
2 hours ago
euler function
mathsearcher   0
2 hours ago
Prove that there exists infinitely many positive integers n such that
ϕ(n) | n+1
0 replies
mathsearcher
2 hours ago
0 replies
Simple but hard
Lukariman   1
N 3 hours ago by Giant_PT
Given triangle ABC. Outside the triangle, construct rectangles ACDE and BCFG with equal areas. Let M be the midpoint of DF. Prove that CM passes through the center of the circle circumscribing triangle ABC.
1 reply
Lukariman
4 hours ago
Giant_PT
3 hours ago
Floor function and coprime
mofumofu   13
N 3 hours ago by Thapakazi
Source: 2018 China TST 2 Day 2 Q4
Let $k, M$ be positive integers such that $k-1$ is not squarefree. Prove that there exist a positive real $\alpha$, such that $\lfloor \alpha\cdot k^n \rfloor$ and $M$ are coprime for any positive integer $n$.
13 replies
mofumofu
Jan 9, 2018
Thapakazi
3 hours ago
Old problem
kwin   0
3 hours ago
Let $ a, b, c > 0$ . Prove that:
$$(a^2+b^2)(b^2+c^2)(c^2+a^2)(ab+bc+ca)^2 \ge 8(abc)^2(a^2+b^2+c^2)^2$$
0 replies
kwin
3 hours ago
0 replies
Interesting inequalities
sqing   2
N 3 hours ago by sqing
Source: Own
Let $a,b,c \geq 0 $ and $ abc+2(ab+bc+ca) =32.$ Show that
$$ka+b+c\geq 8\sqrt k-2k$$Where $0<k\leq 4. $
$$ka+b+c\geq 8 $$Where $ k\geq 4. $
$$a+b+c\geq 6$$$$2a+b+c\geq 8\sqrt 2-4$$
2 replies
sqing
Yesterday at 2:51 PM
sqing
3 hours ago
RMM 2013 Problem 3
dr_Civot   79
N 3 hours ago by Ilikeminecraft
Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$. The lines $AB$ and $CD$ meet at $P$, the lines $AD$ and $BC$ meet at $Q$, and the diagonals $AC$ and $BD$ meet at $R$. Let $M$ be the midpoint of the segment $PQ$, and let $K$ be the common point of the segment $MR$ and the circle $\omega$. Prove that the circumcircle of the triangle $KPQ$ and $\omega$ are tangent to one another.
79 replies
dr_Civot
Mar 2, 2013
Ilikeminecraft
3 hours ago
Inspired by KhuongTrang
sqing   7
N 3 hours ago by TNKT
Source: Own
Let $a,b,c\ge 0 $ and $ a+b+c=3.$ Prove that
$$\frac{1}{ab+2}+\frac{1}{bc+2}+\frac{1}{ca+2}\ge \frac{6}{abc+5}$$$$\frac{1}{ab+2}+\frac{1}{bc+2}+\frac{1}{ca+2}\ge \frac{7}{abc+6}$$$$\frac{1}{ab+2}+\frac{1}{bc+2}+\frac{1}{ca+2}\ge \frac{21}{17(abc+1)}$$$$\frac{1}{ab+2}+\frac{1}{bc+2}+\frac{1}{ca+2}\ge \frac{42}{17(abc+2)}$$
7 replies
sqing
Jan 21, 2024
TNKT
3 hours ago
Cycle in a graph with a minimal number of chords
GeorgeRP   5
N 3 hours ago by Photaesthesia
Source: Bulgaria IMO TST 2025 P3
In King Arthur's court every knight is friends with at least $d>2$ other knights where friendship is mutual. Prove that King Arthur can place some of his knights around a round table in such a way that every knight is friends with the $2$ people adjacent to him and between them there are at least $\frac{d^2}{10}$ friendships of knights that are not adjacent to each other.
5 replies
GeorgeRP
Wednesday at 7:51 AM
Photaesthesia
3 hours ago
Sum of bad integers to the power of 2019
mofumofu   8
N 4 hours ago by Orthocenter.THOMAS
Source: China TST 2019 Test 2 Day 2 Q6
Given coprime positive integers $p,q>1$, call all positive integers that cannot be written as $px+qy$(where $x,y$ are non-negative integers) bad, and define $S(p,q)$ to be the sum of all bad numbers raised to the power of $2019$. Prove that there exists a positive integer $n$, such that for any $p,q$ as described, $(p-1)(q-1)$ divides $nS(p,q)$.
8 replies
mofumofu
Mar 11, 2019
Orthocenter.THOMAS
4 hours ago
Collinearity with orthocenter
liberator   181
N 4 hours ago by Giant_PT
Source: IMO 2013 Problem 4
Let $ABC$ be an acute triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\omega_1$ is the circumcircle of $BWN$, and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$. Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$, and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$. Prove that $X,Y$ and $H$ are collinear.

Proposed by Warut Suksompong and Potcharapol Suteparuk, Thailand
181 replies
liberator
Jan 4, 2016
Giant_PT
4 hours ago
Prove that A, X, O, Y are concyclic
v_Enhance   26
N Dec 24, 2024 by MathLuis
Source: Taiwan 2014 TST2 Quiz 2, P1
Let $ABC$ be a triangle with incenter $I$ and circumcenter $O$. A straight line $L$ is parallel to $BC$ and tangent to the incircle. Suppose $L$ intersects $IO$ at $X$, and select $Y$ on $L$ such that $YI$ is perpendicular to $IO$. Prove that $A$, $X$, $O$, $Y$ are cyclic.

Proposed by Telv Cohl
26 replies
v_Enhance
Jul 18, 2014
MathLuis
Dec 24, 2024
Prove that A, X, O, Y are concyclic
G H J
Source: Taiwan 2014 TST2 Quiz 2, P1
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v_Enhance
6877 posts
#1 • 10 Y
Y by buratinogigle, doxuanlong15052000, anantmudgal09, Davi-8191, Gaussian_cyber, HamstPan38825, LoloChen, Adventure10, Mango247, GeoKing
Let $ABC$ be a triangle with incenter $I$ and circumcenter $O$. A straight line $L$ is parallel to $BC$ and tangent to the incircle. Suppose $L$ intersects $IO$ at $X$, and select $Y$ on $L$ such that $YI$ is perpendicular to $IO$. Prove that $A$, $X$, $O$, $Y$ are cyclic.

Proposed by Telv Cohl
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kaszubki
30 posts
#2 • 3 Y
Y by Ya_pank, Adventure10, Mango247
I wish there were such nice problems on polish TST (unfortunately, we don't have TST :D )

Sketch of proof:
Click to reveal hidden text
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nawaites
204 posts
#3 • 2 Y
Y by Adventure10, Mango247
Oooooh any sokuyion not with complex???
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mathuz
1525 posts
#4 • 1 Y
Y by Adventure10
is it hard problem? :maybe:
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mathuz
1525 posts
#5 • 3 Y
Y by buratinogigle, Adventure10, Mango247
see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=598779&p=3553812#p3553812
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XmL
552 posts
#6 • 10 Y
Y by nima1376, 61plus, Panoz93, buratinogigle, livetolove212, anantmudgal09, Phie11, Adventure10, and 2 other users
My solution:

Let $BI,CI\cap (O)=E,F, IY\cap BC=Y', L\cap (I)=D$, $M$ is the midpoint of $BC$. Since $OM\perp BC$, therefore $I,O,M,Y'$ are concyclic$\Rightarrow \angle YOI=\angle IOY'=\angle IMY'$($Y,Y'$ are reflexive about $IO$. Since it's well known that $IM\parallel AD$, therefore $A,X,Y,O$ are concyclic $\iff \angle XAY=180-\angle YOI=$ $180-\angle IMY'=\angle ADY$ $\iff AY^2=DY*YX=YI\iff AY=YI$.

Note that If $IY\cap FE=Y''$, then by butterfly theorem $IY'=IY''\Rightarrow Y=Y''$, hence $Y$ lies on $EF$ or the perpendicular bisector of $AI$ so $AY=YI$ and we are done.

BTW mathuz's way also works
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TelvCohl
2312 posts
#7 • 12 Y
Y by buratinogigle, v_Enhance, anantmudgal09, Davi-8191, amar_04, AmirKhusrau, Siddharth03, LoloChen, Adventure10, GeoKing, and 2 other users
Thanks for all of you like this problem and here is my solution :)

Lemma :

Given a $ \triangle ABC $ with incenter $ I $ and circmcenter $ O $ .
Let $ L $ be a line passing through $ I $ and perpendicular to $ IO $ .
Let $ L $ intersects the external bisector of $ \angle BAC , BC $ at $ X, Y $, respectively .

Then $ XI=2YI $

Proof of the lemma :

Let $ A', B', C', Y', O' $ be the reflection of $ I $ with respect to $ A, B, C, Y, O $ , respectively.
Let $ I_a, I_b, I_c $ be the $ A $ -excenter, $ B $ -excenter, $ C $ -excenter of $ \triangle ABC $, respectively.

Since $ O' $ is the circumcenter of $ \triangle I_aI_bI_c $ and the circumcenter of $ \triangle A'B'C' $ ,
so $ I_a, I_b, I_c, A', B', C' $ lie on a circle with center $ O' $ and radius $ 2R $ ( $ R $ is the radius of $ \odot (ABC) $ ).
From Butterfly theorem (for quadrilateral $ I_cI_bB'C' $ ) we get $ IX=IY'=2IY $ .
____________________________________________________________
Back to the main problem :

Let $ Q $ be the intersection of $ IY $ and $ BC $ .
Let $ S, R $ be the intersection of $ AI $ and $ BC, L $ , respectively.
Let $ P $ be the intersection of the external bisector of $ \angle BAC $ and $ IY $ .

From the lemma we get $ IP=2IQ $ ,
so $ Y $ is the midpoint of $ IP $ (Since $ I $ is the midpoint of $ YQ $ ),
hence we get $ YA=YP=YI $ . i.e. $ \triangle YAI $ is a isosceles triangle

From $$ \angle OAY=\angle IAY - \angle IAO =\angle YIA - \angle IAO =\angle ACB + \tfrac{1}{2} \angle BAC - \angle AIX =\angle ASB - \angle AIX =\angle IRY - \angle AIX =\angle OXY $$we conclude that$ A, X, O, Y $ are concyclic.

Q.E.D
____________________________________________________________
Generalization : Geometry Marathon Problem 29 , Generalization of 2014 Taiwan TST2 Quiz2 P1
This post has been edited 3 times. Last edited by TelvCohl, Apr 3, 2020, 10:26 PM
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jayme
9795 posts
#8 • 3 Y
Y by I_m_vanu1996, Adventure10, Mango247
Dear XmL and Mathlinkers,
your proof is very nice because it is based on this interesting idea that IM // AD.
Sincerely
Jean-Louis
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I_m_vanu1996
166 posts
#9 • 1 Y
Y by Adventure10
@jayme,,,agree n thnx for ur post!!!!! :D :D
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toto1234567890
889 posts
#10 • 2 Y
Y by Adventure10, Mango247
TelvCohl wrote:
Thanks for all of you like this problem

My solution:

Lemma:

Given a $ \triangle ABC $ with incenter $ I $ and circmcenter $ O $ .
$ L $ is a line pass through $ I $ and perpendicular to $ IO $ .
$ L $ intersects the external bisector of $ \angle BAC , BC $ at $ X, Y $ .

Then $ XI=2YI $

==============================
Proof of the lemma:

Let $ A', B', C', Y', O' $ be the reflection of $ I $ with respect to $ A, B, C, Y, O $ , respectively.
Let $ Ia, Ib, Ic $ be the $ A $ -excenter, $ B $ -excenter, $ C $ -excenter of $ \triangle ABC $, respectively.

Since $ O' $ is the circumcenter of $ \triangle IaIbIc $ and the circumcenter of $ \triangle A'B'C' $ ,
so $ Ia, Ib, Ic, A', B', C' $ lie on a circle with center $ O' $ and radius $ 2R $ ( $ R $ is the radius of $ (ABC) $ ).
From Butterfly theorem (for quadrilateral $ IcIbB'C' $ ) we get $ IX=IY'=2IY $ , so we are done.

==============================
Back to the main problem

Let $ Q $ be the intersection of $ IY $ and $ BC $ .
Let $ S, R $ be the intersection of $ AI $ and $ BC, L $ , respectively.
Let $ P $ be the intersection of the external bisector of $ \angle BAC $ and $ IY $ .

From the lemma we get $ IP=2IQ $ ,
so we get $ Y $ is the midpoint of $ IP $ (Since $ I $ is the midpoint of $ YQ $ ),
hence we get $ YA=YP=YI $ . ie. $ \triangle YAI $ is a isosceles triangle,
so
$ \angle OAY=\angle IAY - \angle IAO =\angle YIA - \angle IAO =\angle ACB + ( \frac { \angle BAC}{2}) - \angle AIX =\angle ASB - \angle AIX =\angle IRY - \angle AIX =\angle OXY $
hence we get $ A, X, O, Y $ are concyclic.

Q.E.D

Same with mine :D And can you post the other versions of this problem that you made, too ?
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buratinogigle
2377 posts
#11 • 2 Y
Y by Adventure10, Mango247
I have seen general problem

Let $ABC$ be a triangle and $P$ is a point on bisector of $\angle BAC$. $D,E,F$ are projections of $P$ on $BC,CA,AB$. $PD$ cuts $(DEF)$ again at $G$. $d$ is the line passing through $G$ and parallel to $BC$. $O$ is circumcenter of $ABC$. $Q$ is isogonal conjugate of $P$. $OQ$ cuts $d$ at $X$ and $Y$ lies on $d$ such that $PY\perp OQ$. Prove that $A,X,Y,O$ lie on a circle.
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buratinogigle
2377 posts
#12 • 2 Y
Y by Adventure10, Mango247
A property is from this configuration

Let $(\omega_a)$ be the circle pass through $A,X,Y,O$. $IY$ cuts $(\omega_a)$ again at $D$. Define cyclically $E,F$. Prove that $AD,BE,CF$ are concurrent on $OI$.
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TelvCohl
2312 posts
#13 • 5 Y
Y by buratinogigle, amar_04, Adventure10, Mango247, and 1 other user
buratinogigle wrote:
I have seen general problem

Let $ABC$ be a triangle and $P$ is a point on bisector of $\angle BAC$. $D,E,F$ are projections of $P$ on $BC,CA,AB$. $PD$ cuts $(DEF)$ again at $G$. $d$ is the line passing through $G$ and parallel to $BC$. $O$ is circumcenter of $ABC$. $Q$ is isogonal conjugate of $P$. $OQ$ cuts $d$ at $X$ and $Y$ lies on $d$ such that $PY\perp OQ$. Prove that $A,X,Y,O$ lie on a circle.
Dear buratinogigle, thank you for your interest and nice generalization :) .

My solution:

Lemma:

Let $ P $ be a point on the bisector of $ \angle BAC $ .
Let $ Q $ be the isogonal conjugate of $ P $ WRT $ \triangle ABC $ .
Let $ R $ be the second intersection of $ AQ $ and $ \odot (QBC) $ .
Let $ \ell $ be a line passing through $ R $ and perpendicular to $ OP $ .

Then the intersection $ K $ of $ BC $ and the perpendicular bisector of $ AR $ lie on $ \ell $ .
__________________________________________________
Proof of the lemma:

Let $ X=\odot(ABC) \cap \odot (P, PA) $ .
Let $ \Psi $ be the composition of Inversion $ \mathcal{I}(A, \sqrt{AB \cdot AC}) $ and reflection $ \mathcal{R}(AP) $ .

Easy to see $ \odot (ABC) $ is the image of the line $ BC $ under $ \Psi $ .

From $ \angle PBA=\angle CBQ=\angle CRA \Longrightarrow \triangle ABP \sim \triangle ARC $ ,
so we get $ AP \cdot AR=AB \cdot AC \Longrightarrow P $ is the image of $ R $ under $ \Psi $ ,
hence $ \odot(P, PA) $ is the image of the perpendicular bisector of $ AR $ under $ \Psi $ ,
so $ X\equiv \odot(ABC) \cap \odot (P, PA) $ is the image of $ K $ under $ \Psi $ .

Since $ OP $ is the perpendicular bisector of $ AX $ ,
so $ OP $ pass through the center of $ \odot (APX) $ ,
hence the tangent of $ \odot (APX) $ at $ P $ is perpendicular to $ OP $ ,
so from $ \angle (PA, OP)=90^{\circ}-\angle (\ell, PA ) $ we get $ \ell $ is the image of $ \odot (APX) $ under $ \Psi $ .
i.e. $ K \in \ell $
____________________________________________________________
Back to the main problem:

Let $ R=AP \cap \odot (PBC), B'=XY \cap AB, C'=XY \cap AC $ .
Let $ D', E', F' $ be the projection of $ Q $ on $ BC, CA, AB $, respectively .

Since $ \angle CRB=180^{\circ}-\angle BPC=180^{\circ}-\angle BPD-\angle DPC $
$=180^{\circ}- \angle F'FD-\angle DEE'=180^{\circ}-\angle D'F'E'-\angle D'E'F' $
$ =180^{\circ}-\angle PEQ-\angle PFQ =180^{\circ}-\angle PC'B'-\angle PB'C' =\angle C'PB' $ ,
so we get $ \triangle RBC $ and $ \triangle PB'C' $ are homothetic with center $ A $ .

From lemma we get the perpendicular bisector of $ AR $ and the perpendicular from $ R $ to $ OQ $ are concurrent on $ BC $ ,
so after doing honothety with center $ A $ which send $ R \mapsto P $ and $ BC \mapsto B'C' $ we get $ YA=YP $ ,
hence $ \angle YAO=\angle YAP+\angle PAO=\angle APY+\angle PAO =90^{\circ}-\angle OQA+\angle PAO $
$ =90^{\circ}-\angle OQA+\tfrac{1}{2} \angle BAC+\angle CBA-90^{\circ}=\angle(BC, PA)-\angle OQA=\angle YXO $ .
i.e. $ A, X, O, Y $ are concyclic

Q.E.D
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buratinogigle
2377 posts
#14 • 2 Y
Y by Adventure10, Mango247
You are welcome dear Telv. Thanks for nice solution. Actually, the original problem is very nice. I try to find some more properties of circle $(\omega_a)=(AXYO)$. But It seems this circle is quite specially.
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TelvCohl
2312 posts
#15 • 5 Y
Y by THVSH, buratinogigle, amar_04, enhanced, Adventure10
buratinogigle wrote:
A property is from this configuration

Let $(\omega_a)$ be the circle pass through $A,X,Y,O$. $IY$ cuts $(\omega_a)$ again at $D$. Define cyclically $E,F$. Prove that $AD,BE,CF$ are concurrent on $OI$.

My solution :

Let $ A' $ be the reflection of $ A $ in $ L $ and $ T=L \cap \odot (I), H=L \cap AA' $ .

From the solution in post #6 or post #7 we get $ YI=YA \Longrightarrow Y $ is the center of $ \odot (AA'I) $ .

From $ OI= AI \cdot \frac{\sin \angle IAO}{\sin \angle AOI}  \Longrightarrow \sin \angle AOI=\frac{AI \cdot \sin \angle IAO}{OI}= \frac{AI \cdot \sin \angle A'AI}{OI}=\frac{TH}{OI} $ ,

so combine $ \triangle AA'I \sim \triangle AIO $ $ \Longrightarrow \frac{AH}{YI}=\frac{AH}{YA}=\cos \angle YAH=\sin \angle AIA'=\sin \angle AOI=\frac{TH}{OI}  $ ,

hence $ \frac{AH}{TH}=\frac{YI}{OI} \Longrightarrow \triangle AHT \sim \triangle YIO \Longrightarrow \angle HAT=\angle IYO=\angle DAO \Longrightarrow \angle BAD=\angle TAC $ ,

so $ AD $ is the isogonal conjugate of A-Nagel line WRT $ \angle A \Longrightarrow AD $ pass through the exsimilicenter $ Z $ of $ \odot (I) \sim \odot (O) $ .

Similarly, we can prove $ Z \in BE , Z \in CF $ $ \Longrightarrow AD, BE, CF, OI $ are concurrent at the exsimilicenter of $ \odot (I) \sim \odot (O) $ .

Q.E.D
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tranquanghuy7198
253 posts
#16 • 2 Y
Y by Adventure10, Mango247
My solution for the problem in post #11:
Lemma.
Let $ABC$ be a triangle, $P, Q$ are isogonal conjugates on the bisector of $\angle{BAC}$ and $S$ is the midpoint of arc $BC$ not containing $A$. We have: $SP.SQ = SB^{2} = SC^{2}$

Back to our main problem:
We’ll prove that $YA = YP$. Indeed:
$I$ is the midpoint of $PQ$
Let the points $K, D, M, S, J, G, R$ be constructed as in the figure.
$\angle{PJQ}$ = 90 $\Rightarrow$ $IP = IQ = IJ$
Now:
$YA = YP$ $\Longleftrightarrow$ $PJ.PY = PI.PA$ $\Longleftrightarrow$ $PR.PG = PI.PA$ $\Longleftrightarrow$ $\frac{PR}{SO}.QK.SO = PI.PA$ $\Longleftrightarrow$ $\frac{QP}{QS}.\frac{QK}{SM}.SM.SO = \frac{PQ}{2}.PA$ $\Longleftrightarrow$ $\frac{1}{QS}.\frac{TQ}{TS}.(2.SM.SO) = PA$ $\Longleftrightarrow$ $\frac{1}{QS}.\frac{TQ}{TS}.(SP.SQ) = PA$ (because $2.SM.SO = SM.SZ = SP.SQ$) $\Longleftrightarrow$ $\frac{TQ}{TS} = \frac{PA}{PS}$ $\Longleftrightarrow$ $\frac{SQ}{ST} = \frac{SA}{SP}$ $\Longleftrightarrow$ $SP.SQ = SA.ST$ (this is right)
Now we have: $\angle{OAY} = \angle{PAY}-\angle{PAO} = \angle{APY}-\angle{APG} = \angle{YPG} = \angle{OXY}$ and the conclusion follows.
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THVSH
158 posts
#17 • 3 Y
Y by buratinogigle, Adventure10, Mango247
TelvCohl wrote:
My solution:

Lemma:

Let $ P $ be a point on the bisector of $ \angle BAC $ .
Let $ Q $ be the isogonal conjugate of $ P $ WRT $ \triangle ABC $ .
Let $ R $ be the second intersection of $ AQ $ and $ \odot (QBC) $ .
Let $ \ell $ be a line passing through $ R $ and perpendicular to $ OP $ .

Then the intersection $ K $ of $ BC $ and the perpendicular bisector of $ AR $ lie on $ \ell $ .

(This lemma was mentioned by buratinogigle at here (post # 8).
I did not read Xml's solution but I think my proof is similar to Xml's proof :) .)
__________________________________________________
Proof of the lemma:

Let $ X=\odot(ABC) \cap \odot (P, PA) $ .
Let $ \Psi $ be the composition of Inversion $ \mathcal{I}(A, \sqrt{AB \cdot AC}) $ and reflection $ \mathcal{R}(AP) $ .

Easy to see $ \odot (ABC) $ is the image of the line $ BC $ under $ \Psi $ .

From $ \angle PBA=\angle CBQ=\angle CRA \Longrightarrow \triangle ABP \sim \triangle ARC $ ,
so we get $ AP \cdot AR=AB \cdot AC \Longrightarrow P $ is the image of $ R $ under $ \Psi $ ,
hence $ \odot(P, PA) $ is the image of the perpendicular bisector of $ AR $ under $ \Psi $ ,
so $ X\equiv \odot(ABC) \cap \odot (P, PA) $ is the image of $ K $ under $ \Psi $ .

Since $ OP $ is the perpendicular bisector of $ AX $ ,
so $ OP $ pass through the center of $ \odot (APX) $ ,
hence the tangent of $ \odot (APX) $ at $ P $ is perpendicular to $ OP $ ,
so from $ \angle (PA, OP)=90^{\circ}-\angle (\ell, PA ) $ we get $ \ell $ is the image of $ \odot (APX) $ under $ \Psi $ .
i.e. $ K \in \ell $

See the proof without inversion in this link i don't think it's... :)
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buratinogigle
2377 posts
#18 • 2 Y
Y by baopbc, Adventure10
Here is a consequence of problem in #11, I get a solution base on solution of Telv Cohl in #13.

Problem. Let $ABC$ be a triangle with circumcenter $O$ and $P$ is a point on bisector of $\angle BAC$. $Q$ is isogonal conjugate of $P$. $QB,QC$ cut $CA,AB$ at $E,F$. The line passes through $Q$ and is perpendicular to $OP$ cut $BC,EF$ at $M,N$. Prove that $\frac{QM}{QN}=1+\frac{d(Q,BC)}{d(P,BC)}$.
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doxuanlong15052000
269 posts
#19 • 1 Y
Y by Adventure10
buratinogigle wrote:
I have seen general problem

Let $ABC$ be a triangle and $P$ is a point on bisector of $\angle BAC$. $D,E,F$ are projections of $P$ on $BC,CA,AB$. $PD$ cuts $(DEF)$ again at $G$. $d$ is the line passing through $G$ and parallel to $BC$. $O$ is circumcenter of $ABC$. $Q$ is isogonal conjugate of $P$. $OQ$ cuts $d$ at $X$ and $Y$ lies on $d$ such that $PY\perp OQ$. Prove that $A,X,Y,O$ lie on a circle.

My solution:
Let $PY$ cut $OQ$ at $L$, $PD$ cut $d$ at $G$,$OQ$ cut $PD$ at $R$, $U$ and $V$ be the midpoint of arc $BC$ and $BAC$, $M$ is the midpoint of $BC$, $K$ be the midpoint of $PQ, AP$ cut $BC$ at $T$, $QR\perp BC$, $OQ$ cut $PD$ at $H$.
We have $\frac {PG}{UM}=\frac {QR}{UM}=\frac {QT}{TU}$(well-know)$=\frac {PA}{PU}$$ \Longrightarrow$ $\triangle APG\sim \triangle PUM$. Since $UP.UQ=UM.UV$$ \Longrightarrow$$\triangle UQV\sim \triangle PGA$$ \Longrightarrow$$\frac {PQ}{PH}=\frac {UQ}{2UO}=\frac {PG}{PA}$$ \Longrightarrow$$PH.PK=PG.PA$. Let $O_1$ be the midpoint of $AP$$ \Longrightarrow$ $PO_1.PQ=PG.PH=PY.PL$$ \Longrightarrow$$YO_1\perp AP$$ \Longrightarrow$ $YA=YP$. We have $\angle YXO=\angle HPQ+\angle QPL=\angle YPA+\angle OUA=\angle YAO$$ \Longrightarrow$$A,X,Y,O$ lie on a circle
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anantmudgal09
1980 posts
#20 • 3 Y
Y by amar_04, Adventure10, Mango247
Beautiful Problem! Congratulations to Telv Cohl :)
v_Enhance wrote:
Let $ABC$ be a triangle with incenter $I$ and circumcenter $O$. A straight line $L$ is parallel to $BC$ and tangent to the incircle. Suppose $L$ intersects $IO$ at $X$, and select $Y$ on $L$ such that $YI$ is perpendicular to $IO$. Prove that $A$, $X$, $O$, $Y$ are cyclic.

Proposed by Telv Cohl


Let $Y'$ be the reflection of $I$ wrt $Y$. Time Travelling to Sharygin Olympiad 2016, we get $\angle IAY'=90^{\circ}$ so $YA=YI$.

Suppose line $L$ touches the incircle at $D$ and let the $A$-excircle touch $BC$ at point $E$. It is well-known that $A, D, E$ are collinear. Let $K$ be the midpoint of $DE$. $K$ lies on the perpendicular bisector of $BC$.

Observe that $\angle IKO=\angle IDY=90^{\circ}$ and $$\angle YIO=\angle KID=90^{\circ} \Longrightarrow \angle YID=\angle OIK.$$So, $\triangle IDY \sim \triangle IKO$. By spiral similarity, $\triangle YIO \sim \triangle DIK$. Thus, $\angle ADY=\angle IKD=\angle IOY$.

Since $$YA^2=YI^2=YD\cdot YX \Longrightarrow \angle YOI=\angle ADY=\angle XAY,$$points $A, X, O, Y$ lie on a circle, as desired. $\square$
This post has been edited 1 time. Last edited by anantmudgal09, Dec 13, 2016, 9:43 PM
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Leooooo
157 posts
#23 • 1 Y
Y by Adventure10
buratinogigle wrote:
A property is from this configuration

Let $(\omega_a)$ be the circle pass through $A,X,Y,O$. $IY$ cuts $(\omega_a)$ again at $D$. Define cyclically $E,F$. Prove that $AD,BE,CF$ are concurrent on $OI$.
Note that by d'Alembert theorem it's suffice to prove:
$\dfrac {Theorem}{}$: let $A'$ be the tangent point of $A-$mixtilinear circle and $(ABC)$, then $A, A', D$ are collinear.

$\dfrac {Proof}{}$: let $D=AA'\cap IY$, then it's suffice to prove $A, X, O, D$ are concyclic.

$\dfrac {Lemma}{}$: Given a triangle $\Delta ABC$, let $I$, $O$ be the incenter and circumcenter of $\Delta ABC$, respectively. Let $A'$ be the tangent point of $A-$mixtilinear circle and $(ABC)$. $P$ is the midpoint of $\widehat {BAC}$. Then $OI$, the line perpendicular to $AA'$ at $A$, the line tangent to $(ABC)$ at $P$ is concurrent.

$\dfrac {Proof of Lemma}{}$: let the antipode of $A'$ in $(ABC)$ be $A''$, $A_1$ be the midpoint of $\widehat {BC}$(not contain $A$), since it's well-known $A', I, P$ are collinear, then use Pascal's theorem to $PPA_1AA''A'$ complete the proof.

$\dfrac {Back to the main problem}{}$: let $OI\cap AA'=K$, $AA'\cap \odot I=S$, by d'Alembert theorem $K$ is the external homothetic center of $(ABC)$ and $\odot I$, so use the lemma we have $L$, $OI$, the line perpendicular to $AA'$ at $S$ are concurrent $\Rightarrow $ $D, S, I, X$ are concyclic. Since $SI\parallel AO$ $\Rightarrow $ $A, O, D, X$ are concyclic. $\Box $
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EulerMacaroni
851 posts
#24 • 3 Y
Y by anantmudgal09, Adventure10, Mango247
Let $U\equiv L\cap \odot(I)$, $Y'\equiv YI\cap BC$, and suppose $V$ is the $A$-mixtilinear touchpoint on $\odot(ABC)$. By angle chasing and using the fact that $\{AU,AV\}$ are isogonal, we establish that $O$ is the spiral center sending $YY'$ to $AV$; consequently, $Y'V=AY$. On the other hand, we know that $IY'=Y'V$, so $AY=Y'V=Y'I=YI$. Let $M$ be the midpoint of $\overline{BC}$; then $AY^2=YI^2=YU\cdot YX$, so $\angle XAY=\angle AUY=\pi-\angle IMY'=\pi-\angle IOY$ as desired.
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Asuboptimal
25 posts
#25
Y by
[asy]
    size(7cm); defaultpen(fontsize(10pt));
    pair O=(0,0);
    pair A=dir(121);
    pair B=dir(195);
    pair C=dir(345);
    pair I=incenter(A,B,C);
    pair D=foot(I,B,C);
    pair Dp=2I-D;
    pair X=extension(I,O,Dp,Dp+B-C);
    pair Y=extension(X,Dp,I,I+rotate(90)*(O-I));
    pair Xp=2I-X;
    pair Yp=2I-Y;

    draw(Xp--X); draw(Y--Yp);
    draw(circumcircle(A,X,Y));
    draw(incircle(A,B,C));
    draw(A--B--C--A);
    draw(X--Y);

    dot("$O$",O,dir(210));
    dot("$A$",A,N);
    dot("$B$",B,dir(120));
    dot("$C$",C,SE);
    dot("$I$",I,S);
    dot("$D$",D,N);
    dot("$D'$",Dp,S);
    dot("$X$",X,NW);
    dot("$Y$",Y,NE);
    dot("$X'$",Xp,S);
    dot("$Y'$",Yp,S);
[/asy]


Use complex numbers with circumcircle as the unit circle. Let $x'=2i-x$ be $\overline{BC}\cap\overline{OI}$. Then
\begin{align*}
    x'&=\frac{(\overline bc-b\overline c)(i-o)-(\overline io-i\overline o)(b-c)}{(\overline b-\overline c)(i-o)-(\overline i-\overline o)(b-c)}\\
    &=\frac{i\left(\frac cb-\frac bc\right)}{i\left(\frac1b-\frac1c\right)-\overline i(b-c)}=\frac{i\cdot\frac{c^2-b^2}{bc}}{i\cdot\frac{c-b}{bc}+\overline i(c-b)}\\
    &=\frac{i(c+b)}{i+\overline ibc}.
\end{align*}Hence we compute
\begin{align*}
    x&=2i-x'=i\left(2-\frac{b+c}{i+\overline ibc}\right)\\
    &=i\left(\frac{2i+2\overline ibc-b-c}{i+\overline ibc}\right).
\end{align*}
Now let $y'=2i-y$ lie on $\overline{BC}$. The intersections $R$, $S$ of $\overline{IY}$ with circumcircle are the roots of the quadratic in $t$:
\begin{align*}
    \frac{t-i}i\in\mathbb I&\iff\frac{t-i}i+\frac{\frac1t-\overline i}{\overline i}=0\\
    &\iff t^2-ti+i/\overline i-ti=0\\
    &\iff t^2-2i\cdot t+i/\overline i=0.
\end{align*}Hence $r+s=2i$, $rs=i/\overline i$, and
\begin{align*}
    y'&=\frac{bc(r+s)-rs(b+c)}{bc-rs}\\
    &=\frac{2bci-\frac i{\overline i}(b+c)}{bc-\frac i{\overline i}}
    =\frac{2bci\overline i-i(b+c)}{bc\overline i-i}.
\end{align*}Hence we compute
\begin{align*}
    y&=2i-y'=2i-\frac{2bci\overline i-i(b+c)}{bc\overline i-i}\\
    &=\frac{-2i^2+i(b+c)}{bc\overline i-i}=\frac{i(b+c-2i)}{bc\overline i-i}.
\end{align*}Let $P=\frac{a-y}a$, $Q=\frac{x-y}x$, $F=\frac PQ$. It suffices to show $F\in\mathbb R$. Let $a=u^2$, $b=v^2$, $c=w^2$, so that $i=-(uv+vw+wu)$.

We have
\begin{align*}
    P&=\frac{a-y}a=1-\frac ya
    =1-\frac{i(b+c-2i)}{a(bc\overline i-i)}=\frac{abc\overline i-i(a+b+c-2i)}{a(bc\overline i-i)}\\
    &=\frac{-u^2v^2w^2\left(\frac1{uv}+\frac1{vw}+\frac1{wu}\right)+(uv+vw+wu)(u+v+w)^2}{u^2\left(-v^2w^2\left(\frac1{uv}+\frac1{vw}+\frac1{wu}\right)+uv+vw+wu\right)}\\
    &=\frac{(u+v+w)\left[-uvw+(uv+vw+wu)(u+v+w)\right]}{u\left(u^2-vw\right)(v+w)}\\
    &=\frac{u+v+w}{u\left(u^2-vw\right)(v+w)}\left[2uvw+\sum_\mathrm{sym}u^2v\right]\\
    &=\frac{(u+v+w)(u+v)(v+w)(w+u)}{u\left(u^2-vw\right)(v+w)}\\
    &=\frac{(u+v+w)(u+v)(u+w)}{u\left(u^2-vw\right)}.
\end{align*}The main computation:
\begin{align*}
    Q&=\frac{x-y}x=1-\frac yx=1-\left(\frac{i(b+c-2i)}{bc\overline i-i}\bigg/\frac{i(2i+2\overline ibc-b-c)}{i+\overline ibc}\right)\\
    &=\frac{(bc\overline i-i)(2i+2\overline ibc-b-c)-(b+c-2i)(i+\overline ibc)}{(bc\overline i-i)(2i+2\overline i-b-c)}\\
    &=\frac{2(i+\overline ibc)(bc\overline i-i)-(b+c)(bc\overline i-i)-(b+c)(bc\overline i+i)+2i(i+\overline ibc)}{(bc\overline i-i)(2i+2\overline ibc-b-c)}\\
    &=\frac{2(i+\overline ibc)bc\overline i-2(b+c)bc\overline i}{(bc\overline i-i)(2i+2\overline ibc-b-c)}\\
    &=\frac{2bc\overline i(i+\overline ibc-b-c)}{(bc\overline i-i)(2i+2\overline ibc-b-c)}\\
    &=\frac{2v^2w^2\left(\frac1{uv}+\frac1{vw}+\frac1{wu}\right)\left(v^2+w^2+uv+vw+wu+v^2w^2\left(\frac1{uv}+\frac1{vw}+\frac1{wu}\right)\right)}{\left(v^2w^2\left(\frac1{uv}+\frac1{vw}+\frac1{wu}\right)-uv-vw-wu\right)\left(v^2+w^2+2(uv+vw+wu)+2v^2w^2\left(\frac1{uv}+\frac1{vw}+\frac1{wu}\right)\right)}\\
    &=\frac{\left[2\frac{vw}u(u+v+w)\right]\cdot\left[\frac1u\left(2uvw+\sum_\mathrm{sym}u^2v\right)\right]}{\left[\frac1u\left(u^2-vw\right)(v+w)\right]\cdot\left[\frac1u\left(2uvw(u+v+w)+u(u+v+w)^2-u^3\right)\right]}\\
    &=\frac{2vw(u+v+w)(u+v)(u+w)}{\left(u^2-vw\right)\big(2vw(u+v+w)+u(v+w)(2u+v+w)\big)}.
\end{align*}
From here, \[F=\frac PQ=\frac{2uvw(u+v+w)+u(v+w)(2u+v+w)}{2uvw},\]and we can check
\begin{align*}
    \overline F&=\frac{\frac2{vw}\left(\frac1u+\frac1v+\frac1w\right)+\frac1u\left(\frac1v+\frac1w\right)\left(\frac2u+\frac1v+\frac1w\right)}{\frac2{uvw}}\\
    &=\frac{2u(uv+vw+wu)+(v+w)(2vw+wu+uv)}{2uvw}\\
    &=\frac{2vw(u+v+w)+2\left(u^2v+u^2w\right)+(v+w)(uw+uv)}{2uvw}\\
    &=\frac{2vw(u+v+w)+u(v+w)(2u+v+w)}{2uvw}=F.
\end{align*}This implies $F\in\mathbb R$, and we are done.
This post has been edited 1 time. Last edited by Asuboptimal, Apr 3, 2020, 2:10 AM
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FISHMJ25
293 posts
#27 • 1 Y
Y by Mango247
For anyone who is bashing with $I=0$ we get : $\bar x=-\frac{2 \sum (xy)}{(y+z)(x^2+yz)+2xyz}$ and $\bar y=\frac {2\sum (xy)}{(y+z)(yz-x^2)}$
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jayme
9795 posts
#28
Y by
Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/24.%203.%20Cocyclite.pdf p. 6…

Sincerely
Jean-Louis
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Ru83n05
170 posts
#29 • 1 Y
Y by GeoKing
It is a well known lemma that the locus of points such that the sum of their oriented distances to the sides of $\triangle ABC$ is constant, is a line perpendicular to $OI$. In particular, letting $r$ denote the inradius of $\triangle ABC$, and using $YI\perp OI$ this implies that
$$3r=d(I, BC)+d(I, AC)+d(I, AB)=d(Y, BC)+d(Y, AB)+d(Y, AC)=2r+d(Y, AB)+d(Y, AC),$$so we learn $d(Y, AB)+d(Y, AC)=r$. Now, I claim this implies the equality $YA=YI$. Let $K\in AC$ such that $IK=IA$, whence $IK\parallel AB$. By the above we find that $d(Y, IK)=d(Y, KA)$, so $Y$ lies on the interior angle bisector of $\angle AKI$, and the claim follows.

Finally, we can conclude by a long but straightforward angle chase (directing angles modulo 180). First we compute
$$\angle AYX=\angle (AY, BC)=\angle ACB+\angle IAC-\angle IAY=\angle ACB+\angle IAC-\angle YIA.$$Now, by definition $\angle YIA=\angle OIA-90$, so upon substituting
$$\angle AYX=90+\angle ACB+\angle IAC-\angle OIA.$$On the other hand
$$\angle AOX=\angle AOI=180-\angle OIA-\angle IAO=180-\angle OIA-(90-\angle ACB-\angle BAI)=90+\angle ACB+\angle BIA-\angle OIA,$$so indeed $\angle AOX=\angle AYX$, so $AYXO$ is cyclic, as desired.
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MathLuis
1535 posts
#30
Y by
Very nice config, let $IY \cap BC=Z$ and also $M$ midpoint of $BC$ and also $OY \cap (OIMZ)=J$ (where that circle exists due to $\angle OIZ=90=\angle OMZ$), also let $D'$ touchpoint of incircle with $L$, from I-E Lemma we have that perpendicular bisector of $AI$ is the line connecting midpoints of minor arcs $AB,AC$ in $(ABC)$ and from a little angle chase we notice that $AO,ID'$ are symetric on the perpendicular bisector of $AI$, so now also note that $IY=IZ$ and $\angle YJZ=90$ which means that $IJ=IZ$ which means from arcs that $\angle XAY=\angle IZJ=\angle IJZ=\angle IMB=\angle AD'Y$ (from $IM \parallel AD'$ by homothety)m which means that $YA^2=YD' \cdot YX=YI^2$ so $YA=YI$ which means $Y$ lies on the perpendicular bisector of $AI$ and therefore $\angle YAO=\angle YID'=\angle YXO$ therefore $AXOY$ is cyclic as desired, thus we are done :cool:.
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