I wish there were such nice problems on polish TST (unfortunately, we don't have TST ) Sketch of proof:

Oooooh any sokuyion not with complex???

is it hard problem?

see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=598779&p=3553812#p3553812

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

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

Dear XmL and Mathlinkers, your proof is very nice because it is based on this interesting idea that IM // AD. Sincerely Jean-Louis

@jayme,,,agree n thnx for ur post!!!!!

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 And can you post the other versions of this problem that you made, too ?

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.

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 $OI$.

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

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.

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

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.

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...

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 $\frac{QM}{QN}=1+\frac{d(Q,BC)}{d(P,BC)}$.

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

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. $\square$

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 $

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.

[asy][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][/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.

For anyone who is bashing with $I=0$ we get : $\bar x=-\frac{2 \sum (xy)}{(y+z)(x^2+yz)+2xyz}$ and $\bar y=\frac {2\sum (xy)}{(y+z)(yz-x^2)}$

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/24.%203.%20Cocyclite.pdf p. 6… Sincerely Jean-Louis

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.

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 .