Just bash it 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:

Synthetic is also equally short [asy][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][/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.

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.$\blacksquare$

Let $N = AO \cap (BOC)$. We show that $N$ is the desired intersection point. [asy][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][/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 $N = (BXD) \cap (CYE)$. $\ \ \blacksquare$

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.” 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. $\square$

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.

My proposal

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][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][/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.

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 $AO$

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 $$\angle XPZ=\angle XCI=\angle B=180^{\circ}-\angle XBZ$$

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.

Similarity 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.

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

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 $T \in (CEY)$.

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][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][/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 $S_A+S_C=b^2$.

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:)

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.

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.

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 $K \in (CYE)$.

MarkBcc168 wrote: Synthetic is also equally short [asy][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][/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?

Solved with SilverBlaze_SY, HACK_IN_MATHS and S.Ragnork1729. [asy][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][/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.

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.$\Box$

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.

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 $V\in (BXD)$.

Easy G5