We first prove that $AZ\parallel BC$. We use Cartesian coordinates. Let $B=(-1,0)$, $C=(1,0)$, $A=(-a,b)$, $D=(-d,0)$. Let $P$ be the circumcenter of $\triangle AXY$. We require that the $OP\perp BC$, or that $P$ lies on the $y$-axis. Line $AB$ passes through $(-1,0)$ and $(-a,b)$, so it satisfies the equation \[ y=\frac b{1-a}(x+1).\]Since $X$ lies on $x=-d$, we have \[X=\left(-d,\frac{b(1-d)}{1-a}\right).\]Similarly, \[Y=\left(-d,\frac{b(1+d)}{1+a}\right).\]Since $P$ lies on the perpendicular bisector of $XY$ and the $y$-axis by assumption, its coordinates are \[\left(0,\frac b2\left(\frac{1-d}{1-a}+\frac{1+d}{1+a}\right)\right)=\left(0,\frac{b(1-ad)}{1-a^2}\right).\]This point must lie on the perpendicular bisector of $AX$. Since $AX$ has slope $\tfrac b{1-a}$, its perpendicular bisector has slope $\tfrac {a-1}{b}$. Furthermore, it passes through the midpoint of $AX$ \[ \left(-\frac{a+d}2, \frac{b(2-a-d)}{2(1-a)}\right).\]Therefore the equation of the line is \[ y-\frac{b(2-a-d)}{2(1-a)}=\frac{a-1}{b}\left(x+\frac{a+d}2\right).\]Plugging in $x=0$ and equating with $P$, we have \begin{align*} \frac{b(1-ad)}{1-a^2}-\frac{b(2-a-d)}{2(1-a)}&=\left(\frac{a-1}{b}\right)\left(\frac{a+d}{2}\right)\\ 2b^2(1-ad)-b^2(1+a)(2-a-d)&=(a-1)(a+d)(1-a^2)\\ b^2(a^2-ad-a+d)&=(a-1)(a+d)(1-a^2)\\ b^2(a-d)&=(a+d)(1-a^2)\\ b^2a-b^2d&=a-a^3+d-a^2d\\ \Aboxed{a^3+a^2d-b^2d+b^2a-a-d&=0}. \end{align*}This is the condition that is necessary for $AZ$ to be parallel to $BC$. We now show it's equivalent to $OD=OX$. Obviously $O$ lies on the $y$-axis, so let its $y$-coordinate be $y$. From $OA=OB$ we get \begin{align*} y^2+1&=a^2+(b-y)^2\\ 2by&=a^2+b^2-1\\ y&=\frac{a^2+b^2-1}{2b}. \end{align*}So the slope of $OD$ is \[\frac{a^2+b^2-1}{2bd},\]while that of $OA$ (which is the negative of $OD$ assuming $OD=OX$) is \[ \frac{a^2-b^2-1}{2ab}.\]So \begin{align*} \frac{a^2+b^2-1}{2bd}&=\frac{-a^2+b^2+1}{2ab}\\ a^3+b^2a-a&=-a^2d+b^2d+d\\ \Aboxed{a^3+a^2d-b^2d+b^2a-a-d&=0}, \end{align*}which is equivalent, proving $AZ\parallel BC$. Now note that $OD$ and $OZ$ are reflections of $OA$ over lines parallel and perpendicular to $BC$, respectively, and are therefore the same line. Since $\angle OAC=90-\angle B=\angle BXD$, $OA$ is tangent to $(AXY)$, and since $OA=OZ$, $OZ$, and thus $DZ$, is also tangent, as desired.

Let $DO$ intersect the minor arc $BC$ at $V$ and major arc $BC$ at $Z'$. Claim: $AO$ is tangent to $(AXY)$. Proof: Since $\frac{OW}{OA} = \frac{OD}{OV}, WD//AV$. This implies $AV \perp BC$. By angle chase, $\angle AYX = \angle BAV = \frac{1}{2}(180^{\circ}-2\angle ABC)=\frac{1}{2}(180^{\circ}-\angle AOC)=\angle OAC$. Claim: $AZ'//BC$. Proof: Since $VZ'$ is the diameter of $(ABC)$, $AZ' \perp AV$, which implies the result. Claim: $DXZ'C$ is cyclic. Proof: This is true since $\angle XDZ' = \angle AVZ' = \angle ACZ'$. Claim: $AYZ'X$ is cyclic. Proof: Note $\angle XZ'C = 90^{\circ}$. It follows that $\angle AZ'X = \angle AZ'C - \angle XZ'C = 180^{\circ} - \angle ABC - 90^{\circ}=90^{\circ}-\angle ABC = \angle BAV = \angle AYX$, thus $AYZ'X$ is cyclic This shows that $Z' = Z$ and it is easy to see that we are done! (by symmetry)

Lovely, despite being easy for G4!

Let $AO\cap BC = T, \angle ACB=\gamma, \angle ABC=\beta$ Claim 1: $OA$ is tangent to $(AYX)$ Proof: $\angle OAC= \angle BYD = 90^{\circ}-\beta$. Hence, $OA$ is tangent to $(AYX)$ Now from Claim 1 and from $OA=OZ$ it follows that $OZ$ is also tangent to $(AYX)$. Hence, $DZ$ is tangent to $(AYX)$ $\Leftrightarrow D,O,Z$ are collinear. Claim 2: $ZYBD$ and $DCZX$ are cyclic Proof: Angle chasing gives $\angle ZYX =\angle ZAX = \angle ZAC = \angle ZBC$ so $ZYBD$ is cyclic. Similarly, $\angle DCZ = \angle BCZ = \angle ZAY = \angle ZXY$ so $DCZX$ is cyclic as well. Claim 3: $BD=CT$ Proof: It is enough to show that $OD=OT$. However, $\triangle WDT$ is a right triangle and $O\in S_{WD}$ $\Rightarrow DO$ is a median of $\triangle WDT \Rightarrow OD=OT=OW$ Claim 4: $\frac{AY}{AX} =\frac{\sin(90^{\circ}-\gamma)}{\sin(90^{\circ}-\beta)}$ Proof: Notice that $\angle AXY=\angle DXC = 90^{\circ}-\gamma$ so applying Law of Sines in $\triangle AYX$ gives $\frac{AY}{AX} =\frac{\sin(90^{\circ}-\gamma)}{\sin(90^{\circ}-\beta)}$. Now $\triangle ZXC \sim \triangle ZYB (\angle XZC=\angle YZB = 90^{\circ}$ and $\angle BYZ=\angle CXZ)$. Therefore, $\frac{XC}{YB}=\frac{ZX}{ZY}$. Applying Menelaus in $\triangle BAC$ and line $D-X-Y$ yields that $\frac{BD}{DC}\cdot \frac{CX}{XA}\cdot\frac{AY}{YB} = 1$. However, from Claim 3: $\frac{BD}{DC} = \frac{CT}{TB} = \frac{CA}{AB}\cdot\frac{\sin( 90^{\circ}-\beta)}{\sin( 90^{\circ}-\gamma)}$. Here we used Ratio Lemma in $\triangle ABC$. Hence, $\frac{CX}{YB} = \frac{AX}{AY}\cdot\frac{DC}{DB} = \frac{BA}{AC}$. Now we have that $\triangle ZXY \sim BAC$ because $\frac{ZX}{ZY}=\frac{XC}{YB} = \frac{BA}{AC}$ and $\angle XZY=\angle BAC$. Therefore, $\angle ZAX = \angle ZYX = \angle BCA$ so $AZ||BC$. We also have that $\angle YBZ=\beta-\gamma = \angle YDZ$ and $\angle WDO = \angle AWX = \beta-\gamma$. Hence, $D,O,Z$ are collinear and we are done.

Nice Let $A$'s altitude meet $ABC$ at $E$. Claim $: O,D,E$ are collinear. Proof $:$ Note that $AE || DW$ and $OW=OD$ and $OE=OA$. Claim $: \angle CZX = \angle 90$. Proof $:$ Note that $\angle CZX = \angle 180 - \angle B - \angle AZX = \angle 180 - \angle B - \angle AYX = \angle 180 - \angle B - (\angle 90-\angle B) = \angle 90$. Claim $: E,D,Z$ are collinear. Proof $:$ Note that $\angle EZC = \angle 90-\angle C = \angle DXC = \angle DZC$ Now since $E,D,O$ and $E,D,Z$ both are collinear we have that $D,Z,O$ are collinear. Claim $: AO$ is tangent to $AXY$. Proof $:$ $\angle AYX = \angle 90-\angle B=\angle CAO = \angle XAO$ Now since $AO$ is tangent and $AO = ZO$ we have that $ZO$ is also tangent and since $ZO$ passes through $D$ then $DZ$ is tangent to $AXY$ as desired. we're Done.

Kinda silly (but still cute :3) As a first observation, note that \[ \providecommand{\dang}{\measuredangle} \dang OAX = 90^{\circ} - \dang CBA = \dang AYX, \]so $OA$ is tangent to $(AXY)$. Hence the two circles are orthogonal. Now we phantom point the entire problem. Let $Z'$ be the reflection of $A$ over the perpendicular bisector of $BC$, and let $\gamma$ be the circle orthogonal to $(ABC)$ passing through $A$ and $Z$. Define by $P$ and $Q$ the intersection not equal to $Z'$ of $BZ'$ and $CZ'$ with $\gamma$ respectively. It is now sufficient to prove that $PQ$ meets $AO$ on $BC$ and that $PQ$ is perpendicular to $BC$. As for the second claim, one can observe that $AZ'CB$ is an isosceles trapezoid, and that $BA$ and $BZ'$ intersect $\gamma$ at antipodal points. The claim then follows. As per the second claim, let $PQ$ intersect $BC$ at $D'$. Then $Z'$ is the Miquel point of complete quadrilateral $PD'CZ'QB$, and hence $APD'B$ is cyclic. Now \[ \providecommand{\dang}{\measuredangle} \dang D'AP = \dang D'BP = \dang CBZ' = \dang ACB = \dang AZ'B = \dang AZ'P, \]so $D'A$ is tangent to $\gamma$. Therefore $D'$ lies on $AO$, and we're done.

In the version of the official packet I received, this problem did not include the $W \ne D$ assertion. Without it, the statement is false, since then you could have $W = D$. I'm glad we noticed the issue in time to patch the problem for the USA training camp... it would've been really embarrassing if students discovered they were spending time on a broken problem while taking a practice exam.

Since $\angle OAC=90^\circ-\angle ABC=\angle AYX$ line $AO$ is tangent to $\odot (AXY),$ and because of $|OA|=|OZ|$ line $OZ$ is tangent too. By Miquel point $\measuredangle ZCB=\measuredangle ZXY=\measuredangle ZAY=\measuredangle CBA,$ so $AO$ is symmetric to $AZ$ wrt perpendicular bisector of $BC.$ But equality $|OD|=|OW|$ yields that the same symmetry maps line $AO$ onto $OW$, i.e. $D\in OZ,$ done.

Not bad for a G4 Claim: $AO$ is tangent to $(AXY)$. Proof: Notice that $\angle BYD=90-\angle B$ but also $\angle CAO=90-\angle B$ so this claim is proven. Claim: $D, O, Z$ are collinear Proof: First, by reim, $BC’ \parallel DW$ where $C’=ZX\cap(ABC)$ and so $ZXDC$ is cyclic. Then, suppose $OD$ hit $(ABC)$ at $D’$. We have that $\angle AOD’=2\angle ODW$ from the length condition, and $\angle AOZ=2\angle ACZ=2\angle WDZ$. Thus $\angle D’OZ=2\angle ODZ=2\angle D’DZ$ which means that if $O$ didn’t lie on $DZ,$ then $D$ is on the circumcircle which is absurd. Thus the claim is proven. Now, the first claim gives $OZ$ is a tangent so by the second claim, $DZ$ is a tangent

Let the tangent to $(ABC)$ at $A$ meet $\overline{BC}$ at $K$, and let $\overline{AO}$ meet $\overline{BC}$ at $A'$. Redefine $Z$ to be the point on $(ABC)$ such that $\overline{AZ} \parallel \overline{BC}$. We first claim that $\triangle XDZ \sim \triangle BKA$. Since $AKDW$ is cyclic, we have \[ \angle XDZ = \angle WDO = \angle OWD = \angle BKA. \]Moreover, \[ \frac{XD}{DZ} = \frac{XD}{AA'} = \frac{XD}{\sin C} \div \frac{AA'}{\sin C} = \frac{DC}{\cos C} \div \frac{A'C}{\sin \angle OAC} = \frac{BA'}{A'C} \cdot \frac{\sin \angle OAC}{\sin \angle OAB} = \frac{AB}{AC} = \frac{BK}{KA}, \]and so the desired similarity is proved. Therefore, $\angle ZXY = \angle ABC = \angle ZCA = \angle ZAY,$ so $Z$ is the same as the one in the problem. Now $\angle AYX = 90^\circ - \angle B = \angle OAX,$ so $\overline{OA}$ is tangent to $(AXY)$. Since $OA = OZ$, it follows that $\overline{DOZ}$ is also tangent to $(AXY)$ as desired.

Let $A',D'$ be the reflections of $A,D$ respectively over the perpendicular bisector of $A.$ First, the length condition tells us that $A,O,D'$ are collinear, so we have $\angle BA'D=\angle D'AC=90-\angle B=\angle BYD,$ so $BYA'C$ is cyclic. Trivially we see $A'ABC$ is cyclic as well, so $A'$ is the Miquel point of quadrilateral $ABDX.$ From the definition of $Z,$ we see that $Z$ is also the Miquel point of $ABDX,$ so $A'=Z$ and we get $AZ||BC$ and $\angle ZAC=\angle ACB.$ We have $\angle XZD=\angle XCD=\angle ACB=\angle ZAC=\angle ZAX,$ so $DZ$ must be tangent to circle $AXY.$

Fakesolved this initially which is why I thought why it turned out to be way too easy than usual lol. Also first diagram in the entire thread... [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 (otherwise throws error due to the missing xmin) are done using bubu-asy.py. This adds back the dps, real xmin coordinates, changes the linewidth, the fontsize and adds the directions to the labellings. */ pair A = (-70,31); pair B = (-100,-60); pair C = (40,-60); pair O = (-30,-32.63186); pair Hp = (-70,-96.26373); pair D = (-47.20404,-60); pair Y = (-47.20404,100.14774); pair Z = (10,31); pair X = (-47.20404,12.14152); pair W = (-47.20404,-5.26373); pair O_1 = (-30,56.14463); pair T = (-47.20404,31); import graph; size(12.1cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); pen ffxfqq = rgb(1,0.49803,0); draw(arc(A,21.93868,-57.84588,-39.60002)--A--cycle, linewidth(0.75) + blue); draw(arc(Y,21.93868,-108.24585,-90)--Y--cycle, linewidth(0.75) + blue); draw((2.11983,35.95359)--(-2.83375,28.07343)--(5.04640,23.11983)--Z--cycle, linewidth(0.75) + blue); draw((-37.89624,-60)--(-37.89624,-50.69220)--(-47.20404,-50.69220)--D--cycle, linewidth(0.75) + blue); draw(A--B, linewidth(0.5)); draw(B--C, linewidth(0.5) + red); draw((-27.69643,-60)--(-30,-62.96172), linewidth(0.5) + red); draw((-27.69643,-60)--(-30,-57.03827), linewidth(0.5) + red); draw(C--A, linewidth(0.5)); draw(circle(O, 75.15992), linewidth(0.5)); draw(A--Hp, linewidth(0.5)); draw(A--O, linewidth(0.5) + blue); draw(A--Y, linewidth(0.5)); draw(Y--D, linewidth(0.5)); draw(circle(O_1, 47.24672), linewidth(0.5) + blue); draw(Hp--O, linewidth(0.5) + linetype("4 4")); draw(O--Z, linewidth(0.5) + blue); draw(A--Z, linewidth(0.5) + red); draw((-27.69643,31)--(-30,28.03827), linewidth(0.5) + red); draw((-27.69643,31)--(-30,33.96172), linewidth(0.5) + red); draw(circle((-3.60202,-23.92923), 56.58830), linewidth(0.5) + linetype("4 4") + ffxfqq); draw(O_1--Z, linewidth(0.5) + blue); dot("$A$", A, 1.5*dir(170)); dot("$B$", B, dir(180)); dot("$C$", C, dir(0)); dot("$O$", O, dir(180)); dot("$H'$", Hp, SW); dot("$D$", D, 1.5*dir(270)); dot("$Y$", Y, N); dot("$Z$", Z, 1.5*dir(10)); dot("$X$", X, dir(180)); dot("$W$", W, dir(180)); dot("$O_1$", O_1, NW); dot("$T$", T, NW); [/asy][/asy] Let $H'$ denote the intersection of $A$-altitude with $\odot(ABC)$. Let $O_1$ denote the center of $\odot(AXYZ)$ and also let $T=DX\cap AZ$. Now take a homothety centered at $O$ that maps $W\mapsto A$. Let $D'$ denote the image of $D$ under this homothety. Firstly, we get that as $OW=OD$, then $OD'=OA$, that is $D'$ lies on $\odot(ABC)$. Now moreover as $WD\perp BC$, we get that $AD'\perp BC$. This finally gives us that $D'\equiv H'$. This gives us that $\overline{O-D-H'}$ are collinear. Now we have that \[\measuredangle OAX=\measuredangle OAC=90^\circ-\measuredangle CBA=90^\circ-\measuredangle DBY=\measuredangle BDY+\measuredangle YBD=\measuredangle BYD=\measuredangle AYX,\]which gives us that $OA$ is tangent to $\odot(AXY)$. This means that $\odot(ABC)$ and $\odot(AXY)$ are orthogonal. Now note that we have $A=BY\cap XC$ and that $Z=\odot(YAX)\cap\odot(ABC)$. This gives us that $Z$ is the center of the spiral similarity mapping $\overline{YX}\mapsto \overline{BC}$, and thus this maps $X\mapsto C$. Now also note that this spiral similarity maps $\odot(ZYX)\mapsto\odot(ZBC)$, so this means that their centers also get mapped, that is $O_1\mapsto O$. Now moreover, note that we have the fact that $\odot(ABC)$ and $\odot(AXY)$ are orthogonal, so this gives us that $\measuredangle O_1ZO=90^\circ$, which means that the angle of rotation of the spiral similarity is $=90^\circ$. Now as we already had that $X\mapsto C$, this gives us that $\measuredangle XZC=90^\circ$. Finally note that $\measuredangle XDC=90^\circ=XZC\implies ZXDC$ is cyclic. Now we have $\measuredangle ZDX=\measuredangle ZCX=\measuredangle ZCA=\measuredangle ZH'A$ and that $\measuredangle DTZ=\measuredangle H'AZ=90^\circ$. This gives us that $\triangle ZTD$ and $\triangle ZAH'$ are homothetic which further implies that $\overline{Z-D-H'}$ are collinear. Now as we already had that $\overline{O-D-H'}$ are collinear, this gives us that $DZ\equiv OZ$ is tangent to $\odot(AXY)$ and we are done.

Claim: $D, O$, and $Z$ are collinear. Proof: Let $Z'$ be the intersection of ray $\overrightarrow{DO}$ with $(ABC)$, and $D'$ be the intersection of ray $\overrightarrow{OD}$ with $(ABC)$. Since $\tfrac{OW}{OD} = \tfrac{OA}{OD'} = 1$, we have $WD \parallel AD'$. Therefore, angle chasing, we get $$\measuredangle Z'DX = \measuredangle Z'D'A = \measuredangle Z'CA = \measuredangle Z'CX,$$so quadrilateral $Z'XDC$ is cyclic. But since $Z'$ lies on $(ABC)$ as well, this means that $Z'$ must be the Miquel point of quadrilateral $AXDB$, and hence it lies on $(AXY)$ too. Thus $Z' \equiv Z$, proving the claim. Then, since $D'Z$ is a diameter of $(ABC)$, we have $AZ \perp AD'$. But since $AD' \parallel WD$ and $WD \perp BC$, we must also have $AZ \parallel BC$. Also, as noted in the proof of the claim, $Z$ is the Miquel point of $AXDB$, so $ZXDC$ is cyclic. Hence, we may angle chase to find $$\measuredangle DZX = \measuredangle DCX = \measuredangle BCA = \measuredangle ZAC = \measuredangle ZAX = \measuredangle ZYX,$$and $DZ$ is indeed tangent to $(AXY)$, as desired.

This problem was easier than I expected, is this a property of last year's Geometry shortlist or what? Good for me Let $AO \cap BC=D'$, $OC \cap (ABC)=C'$ and $CY \cap (ABC)=Z'$ $$\angle D'DW=90^{\circ} \implies OD=OW=OD'$$so $D$ is the reflection of $D'$ about the perp bisector of $BC$ and $$\angle AZY = \angle AZC' = \angle ACC' = \angle AXY \implies Z'=(AXY) \cap (ABC)=Z$$ Now let $ZD \cap (ABC)=X$, we get that $YDCZ$ is cyclic from reims theorem which gives $$\angle ACB=\angle ACD=\angle YZD=\angle C'ZX$$which means $AC'BX$ is a cyclic isosceles trapezoid $\implies AX \perp BC \implies Z$ is just the reflection of $A$ about the perp bisector of $BC$ (as the lines $DX$ and $AD'$ are just reflections). So we have $$\angle ZAY=\angle ACB =\angle C'AX = \angle YZD \implies DZ \text{ is tangent to } (AXY) \blacksquare$$

[asy][asy] size(15cm); pair A, B, C, D, W, X, Y, Z, O, B1, C1; A=(2,3sqrt(5)); B=(0,0); C=(8,0); O=circumcenter(A,B,C); B1=2*O-B; C1=2*O-C; W=extension(A,O,B1,C1); D=foot(W,B,C); X=extension(A,C,D,W); Y=extension(A,B,D,W); Z=extension(D,O,A,A+C-B); draw(A--B--C--A--B--B1--C1--C--D--Y--A--X--Z--C1--Z--D); draw(circumcircle(A,B,C)); draw(circumcircle(A,X,Y)); label("$A$", A, WNW); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, S); label("$W$", W, SW); label("$X$", X, NNE); label("$Y$", Y, N); label("$Z$", Z, ENE); label("$O$", O, SSE); label("$B_1$", B1, NE); label("$C_1$", C1, NW); [/asy][/asy] Let $B'$ and $C'$ be the reflections of $B$ and $C$ oveer $O$, respectively. Then, $W$ lies on $B'C'$. We have \begin{align*} \angle AZ&=\angle AYX\\ &=90^{\circ}-\angle DBY\\ &=\angle ACO\\ &=\angle AZC', \end{align*}so $C'$, $X$, and $Z$ are collinear. Similarly, $B'$, $Y$, and $Z$ are collinear. Since $\angle XWC'=90^{\circ}=\angle C'AX$, we get $AXWC'$ is cyclic, so $\angle ZC'B'=\angle WAX=90^{\circ}-\angle ABC$. Therefore, $\angle ZBB'=\angle ACC'$ implies $ACB'Z$ is an isosceles trapezoid. Now, we have \begin{align*} \angle AOZ&=2\angle AC'Z\\ &=2\angle AWX\\ &=\angle ODW+\angle DWO, \end{align*}so $D$, $O$, and $Z$ are collinear. Therefore, since $$\angle OAX=90^{\circ}-\angle ABC=\angle AYX,$$we get $OA$ is tangent to the circumcircle of $AXY$, and since $\angle OZA=\angle OAZ=\angle AYZ$, $OZ$ is also tangent to the circumcircle of $AXY$, so $DZ$ is tangent to the circumcircle of $AXY$.

Wow fun coordbash We begin by proving that $AZ || BC$, using coordinates and phantom points. Let O be $(0,0)$, $A = (-a,b), B = (-c,-d), C = (c,-d), Z' = (a,b)$. Note that $AZ' || BC$ and therefore we need to show that $A,X,Y,$ and $Z'$ are concyclic. In order for $OW = OD$, we must have $D = BC \cap A'O$, where $A' = (-a,-b)$ is the reflection of $A$ over the x-axis. Now, we can easily compute the coordinates of D as $\left(\frac{-ad}{b}, -d\right)$ and $W = \left(\frac{-ad}{b},d\right)$ as it is the reflection of D over the x-axis. Therefore, the equation of line DW is $x = \frac{-ad}{b}$. Now we can intersect DW with $AC = -\frac{b+d}{c+a}x + \frac{bc-ad}{c+a}$ and $AB = \frac{b+d}{c-a}x + \frac{bc+ad}{c-a}$ to get $X = \left(\frac{-ad}{b}, \frac{ad^2+b^2c}{b(a+c)}\right)$ and $Y = \left(\frac{-ad}{b}, \frac{bc^2-ad^2}{b(c-a)}\right)$. Denote V as the intersection of $XY$ and $AZ'$. It is easy to see that $V = \left(\frac{-ad}{b},b\right)$ as it is the intersection of $x = \frac{-ad}{b}$ and $y = b$. Now, we aim to prove that $VX \cdot VY = VA \cdot VZ'$. After a bit of calculation, we get that $VX \cdot VY = \frac{a^2(b^2-d^2)^2}{b^2(c^2-a^2)}$ and $VA \cdot VZ' = \frac{a^2(b^2-d^2)}{b^2}$. However, A and B lie on the same circle centered at the origin and so $a^2+b^2 = c^2+d^2$, or $b^2-d^2 = c^2-a^2$ and so we have now shown that $AZ || BC$. Now, to show the desired tangency, we require $\angle DZX = \angle XAZ$. So, it remains to show $\angle DZX = \angle C$. From quadrilateral $WXZO$, we get that $\angle WXZ + \angle XZO + \angle ZOW+\angle OWX = (180 - \angle B) + (180 - \angle B + \angle C) + (2\angle B- 2\angle C) + \angle DZX = 360 \Rightarrow \angle DZX = \angle C$ and we are done.

Canadian TST Problem 2. Personally thought it was too easy for its spot. Let $D'$ be the intersection point of $AO$ and $BC$, and let $Z'\neq A$ be the point on $(ABC)$ such that $AZ'\parallel (ABC)$. Since $OD=OW$ and $\angle WDD'=90^\circ$, we have $OD=OD'$, so $D$ and $D'$ are reflections across $BC$'s perpendicular bisector. Indeed, $D$, $O$, $Z'$ are collinear. $~$ Note that \[\angle DZ'C=\angle D'AB=90^\circ-\angle C=\angle DXC\]so $DXZ'C$ is cyclic. Therefore, $\angle YAZ'=\angle ABC=\angle Z'CB=\angle YXZ'$ so $Z'=Z$. Thus, it suffices to show that $OZ$ is tangent to $(AXY)$. In fact, $OZ=OA$ so it suffices to show $OA$ is tangent. This is obvious: \[\angle OAX=90^\circ-\angle B=\angle AYX\]We are done.

Shameless plug time! Instakilled by the lemma on my blog. Notice that the construction for $Z$ is exactly the construction for the lemma, so we have $\frac{BZ}{ZC} = \frac{BY}{CX} = \frac{BD}{DC} \cdot \frac{\cos C}{\cos B}$. Notice that if $K = AO \cap BC$ and $S$ is the foot of $A$ on $BC$, then $D$ is the reflection of $K$ over the perpendicular bisector of $BC$ and by ratio lemma we have $\frac{BD}{DC} \cdot \frac{BK}{KC} = \frac{AB^2}{AC^2}$ so $\frac{BK}{KC} = \frac{AB \cos C}{AC \cos B}$ so $\frac{BD}{DC} = \frac{AC \cos B}{AB \cos C}$ so $\frac{BZ}{ZC} = \frac{AC}{AB}$ so we must have $AZ \parallel BC$. Now, the rest is easy. Notice that $\angle XAO = \angle XAK = \angle BAS = \angle BYX$ since $O, H$ are isogonal conjugates, so it follows that $AO$ is tangent to $(AXYZ)$. By symmetry over the perpendicular bisector of $BC$, it follows that $ZD$ is tangent to the same circle, finishing.

Seems on the easy end for a G4. Let $M$ be the midpoint of $BC$ and let $H$ be the orthocenter of $ABC$. Note that $OD = OW$ implies $WD = 2OM = AH$, so as $AH \parallel WD$, $AHWD$ is a parallelogram. Thus $HD \parallel AO$. Let $O’$ be the reflection of $O$ in $BC$. Now, as $OO’ \parallel AH$ and $OO’ = AH$, $AHO’O$ is also a parallelogram, and $H, D, O’$ are collinear. Letting $H_A$ be the reflection of $H$ in $A$, we thus obtain that $O, D, H_A$ are collinear. Now note that $\angle OAX = \pi / 2 - \angle B = \angle AYX$, so $OA$ is tangent to $(AXY)$. Thus as $OA = OZ$ and $Z$ lies on $(AYX)$ we obtain $OZ$ tangent to $(AYX)$. Thus we just need to show $O, D, Z$ collinear. To do this we use phantom points. Let $A’$ exist such that $AA’CB$ is an isosceles trapezoid. Note that $A’, O, D, H$ are collinear. Clearly $\angle XCA’ = \angle AH_AA’ = \angle XDA’$ so $X, D, C, A’$ are concyclic. Similarly $A’, Y, B, D$ are concyclic, and we obtain that $A’$ is the Miquel point of the complete quadrilateral $\{AB, BD, DX, XA\}$. Thus $YAXA’$ is cyclic and $A’ = Z$, done.

First, it´s clear that $Z$ is Miquel's point of complete quadrilateral $AXDB$. Let $R = WO \cap BC $ Let $Z' = DO \cap \odot(ABC)$ such that $O$ is between $D$ and $Z$. Since $OW = OD$ and $\angle WDR = 90^{\circ}$ then $O$ is center of $\odot WDR$. It's clear that $AZ \parallel DR$ and $AZRD$ is a isosceles trapezium. Notice that \[\angle DZC = \angle OZC = 90^{\circ} - \angle ZBC = \stackrel{AZ \parallel BC}{=} 90^{\circ} - \angle ACB = \angle DXC \]therefore $DXZC$ is cyclic and by Miquel's theorem $Z' \equiv Z$. As $AYZX$ is cyclic, \[\angle XAZ = \angle CAZ = \angle ACD = \angle XCD = \angle XZD \]so $DZ$ is tangent to $\odot (AYZX)$ $\blacksquare$

REVENGE For context, at MOP I somehow managed to notice $D,O,Z$ was collinear in my diagram and convince myself that this was false by using some weird (wrong) argument. Obviously this kind of leads to getting nowhere on the problem. First, note that $\overline{AO}$ is tangent to $(AXY)$ since $\measuredangle CAO=\measuredangle HAD=\measuredangle DYB=\measuredangle XAY$ where $H$ is the orthocenter of $\triangle ABC$. Since $D$ is clearly the reflection of $\overline{AO} \cap \overline{BC}$ over the midpoint of $\overline{BC}$, it thus suffices to show that $ABCZ$ is an isosceles trapezoid, whence by symmetry $D,O,Z$ are collinear (!!) and the tangency follows. By phantom points, it suffices to prove the following: let $Z$ be the point such that $ABCZ$ is an isosceles trapezoid. Let $D$ be a point on $\overline{BC}$ and define $X,Y$ as before. Then if $AXYZ$ is cyclic, we have $D,O,Z$ collinear. Letting $Z'$ be the $Z$-antipode with respect to $(ABCZ)$, this is equivalent to $D,Z,Z'$ collinear. This would imply the original, because exactly one such point $D$ exists, and if it's $\overline{BC} \cap \overline{OZ}$ then it's the same as $D$ in the original problem. By spiral similarity it is evident that $\triangle ZCB \sim \triangle ZXY$. Since $\overline{XY} \perp \overline{BC}$ it follows that $\angle CZX=\angle BZY=90^\circ$. We now use phantom points again: define $D$ as the point $\overline{BC} \cap \overline{ZZ'}$, and let $X$ lie on $\overline{AC}$ such that $\overline{XD} \perp \overline{BC}$. Then I claim that $\angle CZX=90^\circ$. This would imply that the point $X$ in both versions of the problem are the same, and thus the $D$'s are as well. But this is easy: note that $\overline{AZ'} \parallel \overline{DX}$, so $\measuredangle ZCX=\measuredangle ZCA=\measuredangle ZZ'A=\measuredangle ZDX$. Thus $CDXZ$ is cyclic, so $\angle CZX=\angle CDX=90^\circ$. $\blacksquare$

Firstly, $$\measuredangle XYA = 90 - \measuredangle ABC = \measuredangle CAO$$so $OA$ is tangent to $(AXY)$ and thus $OD$ is tangent to $(AXY).$ Thus, it remains to show $D-O-Z$ are collinear. Drop the Altitude from $A$ and extend it until it hits $(ABC)$ at the point $E \neq A$. Note that $$\measuredangle OEA = \measuredangle EAO = \measuredangle DWO$$by parallel lines, and thus $O-D-E$ are collinear. Finally, it remains to show $Z$ is the $E$ antipode. We find $$\angle EZA = \frac{1}{2}\angle EOA = 90 - \angle ODW = 90-\angle OEA \implies \angle ZAE = 90. \blacksquare$$

Take $D'$ to be the intersection of the $A$-altitude with $(ABC)$. Now the condition of $OW=OD$ implies that $O-D-D'$. Now notice how $$\angle OAX=\angle BAD'=\angle AYX,$$so $OA$ is tangent to $(AXY)$, which also implies that $OZ$ is tangent to $(AXY)$. So the problem is equivalent to showing that $Z-O-D'$. For this let $X'=ZX\cap (ABC)$. Notice how $\angle AZ'X=\angle OAC$ implies that $X'B\perp BC$, so by applying Pascal at $ZD'ACBX'$ we get that $\infty_{XD}$, $X$ and $ZD'\cap BC$ are collinear, impying that $D=ZD'\cap BC$, which ends the problem.

Let $\angle BAC=\alpha, \angle ABC=\beta, \angle ACB=\gamma, (AXY)=\omega, (ABC)=\Gamma, DY \cap AZ=T$

Proof:

Now from $\square XDZC$ we have: $\angle XZD=\angle XCD=\gamma \implies \angle XZD=\gamma$ We know from $\Gamma$ that $AZ$ is $antiparallel$ with $BC$ so: $ \angle ATX=\angle ATZ= \angle ADB=90 \implies \angle ATX=90$ From quadrilateral $\square AXBD$ we have: $\angle ABD+\angle BDT+\angle ATD+\angle BAT=360$ $\beta+90+90+\angle BAC+\angle CAT=360 \implies \angle CAT=\gamma$ Now again from $\omega$ we have: $\angle CAT=\angle XAZ=\angle XYZ=\gamma$ So $\angle DZX=\angle XYZ=\gamma \implies \angle DZX=\angle XYZ \implies$ $DZ$ is tangent to the circle $AXY$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.1) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -23.212236464938904, xmax = 4.851247293661304, ymin = -3.94771840789012, ymax = 14.009938895715438; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((-12.441358747202935,9.853453424452061)--(-16.35110992940682,0.3788498233889507)--(-3.5303514564127685,0.40220275321225496)--cycle, linewidth(2) + zzttqq); /* draw figures */ draw((-12.441358747202935,9.853453424452061)--(-16.35110992940682,0.3788498233889507), linewidth(2) + zzttqq); draw((-16.35110992940682,0.3788498233889507)--(-3.5303514564127685,0.40220275321225496), linewidth(2) + zzttqq); draw((-3.5303514564127685,0.40220275321225496)--(-12.441358747202935,9.853453424452061), linewidth(2) + zzttqq); draw(circle((-9.945993365208215,3.2797333801319697), 7.031404099520478), linewidth(2)); draw((-12.441358747202935,9.853453424452061)--(-9.945993365208215,3.2797333801319697), linewidth(2)); draw((-9.945993365208215,3.2797333801319697)--(-11.031448634041308,0.38853955252440703), linewidth(2)); draw((-16.35110992940682,0.3788498233889507)--(-11.054809246071999,13.213515557373334), linewidth(2)); draw((-11.054809246071999,13.213515557373334)--(-11.031448634041308,0.38853955252440703), linewidth(2)); draw(circle((-9.959683268630409,10.795490358915945), 2.654457904515024), linewidth(2)); draw((-11.045993173451837,8.373491688906162)--(-7.474592440448108,9.862500357615547), linewidth(2)); draw((-7.474592440448108,9.862500357615547)--(-12.441358747202935,9.853453424452061), linewidth(2)); draw((-11.054809246071999,13.213515557373334)--(-7.474592440448108,9.862500357615547), linewidth(2)); draw((-7.474592440448108,9.862500357615547)--(-11.031448634041308,0.38853955252440703), linewidth(2)); draw((-7.474592440448108,9.862500357615547)--(-3.5303514564127685,0.40220275321225496), linewidth(2)); draw(circle((-7.288172314932304,4.387847221059207), 5.477826159052833), linewidth(2)); /* dots and labels */ dot((-12.441358747202935,9.853453424452061),dotstyle); label("$A$", (-12.338565352824613,10.096408062653813), NE * labelscalefactor); dot((-16.35110992940682,0.3788498233889507),dotstyle); label("$B$", (-16.25209618588625,0.6345803523402637), NE * labelscalefactor); dot((-3.5303514564127685,0.40220275321225496),dotstyle); label("$C$", (-3.42165965724114,0.6593495348279955), NE * labelscalefactor); dot((-9.945993365208215,3.2797333801319697),linewidth(4pt) + dotstyle); label("$O$", (-9.836877921563694,3.483036338429421), NE * labelscalefactor); dot((-11.04197397863814,6.166953736187139),linewidth(4pt) + dotstyle); label("$W$", (-10.95149113351163,6.356261507006311), NE * labelscalefactor); dot((-11.031448634041308,0.38853955252440703),linewidth(4pt) + dotstyle); label("$D$", (-10.926721951023897,0.5850419873648001), NE * labelscalefactor); dot((-11.045993173451837,8.373491688906162),linewidth(4pt) + dotstyle); label("$X$", (-10.95149113351163,8.560718748414441), NE * labelscalefactor); dot((-11.054809246071999,13.213515557373334),linewidth(4pt) + dotstyle); label("$Y$", (-10.95149113351163,13.415478516009875), NE * labelscalefactor); dot((-7.474592440448108,9.862500357615547),linewidth(4pt) + dotstyle); label("$Z$", (-7.384728855278239,10.07163888016608), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Denote $Z'$ as a point on $(ABC)$ such that $AZ'\parallel BC$. Note $AZ'\perp DY$. Claim: $Z',O,D$ are collinear Proof: Let $D'=Z'O\cap BC$, and $W'$ be the intersection of $AO$ and the line through $D'$ perpendicular to $BC$. We can say: \[\angle OD'W'=90-\angle OD'C=90-\angle AZ'O=\frac{\angle AOZ'}{2}=90-\frac{\angle D'OW'}{2}=\angle OW'D',\]so $OD'=OW'$, so $D=D'$, $W=W'$, as desired $\square$ Claim: $OA$ is tangent to $(AXY)$ Proof: We can say: \begin{align*} \angle AYX &=90-\angle YAZ'\\ &=-90+\angle BAZ'\\ &=-90+\angle BAO+\angle OAX+\angle CAZ'\\ &=-90+90-\angle BCA+\angle BCA+\angle OAX\\ &=\angle OAX\text{ }\square\\ \end{align*}Claim: $Z=Z'$ Proof: We can say: \[\angle DZ'C=\angle OZ'C=90-\angle Z'AC=90-\angle BA=\angle DXC,\]so $D,C,Z',X$ are cyclic. This means that $\angle XZ'C=90$. As a result: \[\angle AYX=\angle OAX=\angle BAZ'-90=\angle AZ'C-90=\angle AZ'X,\]so $Z'$ lies on $(AYX)$, implying the desired result $\square$ Finally, as $OA=OZ$, $Z,O,D$ are collinear, and $OA$ is tangent to $(AXY)$, we are done $\blacksquare$

This problem is a test to see how well you can use phantom points lol We will instead redefine $Z$ to be the point on $(ABC)$ with $\overline{AZ} \parallel \overline{BC}$, $\omega$ a circle through $A$ and $Z$ such that $\overline{OA}$ and $\overline{OZ}$ are tangent to $\omega$, and $X$ and $Y$ the intersections of $\omega$ with $\overline{AC} \cap \overline{AB}$. It will suffice to show that $\overline{XY} \perp \overline{BC}$ at a point $D$ that lies on $\overline{OZ}$. The first part is just angle chasing. As $\angle YXZ = \angle YAZ = \angle B$ and similar, $\triangle XYZ \sim \triangle BAC$. Then the tangency condition implies that $90^\circ - B + C = \angle AZO = \angle AYZ$, or $\angle AYX = 90^\circ - B$. It follows that $\overline{XY} \perp \overline{BC}$. Now, note that $XZCD$ is cyclic as $\angle XZC = \angle YXZ + \angle AYX = 90^\circ$. So $\angle XZD = \angle C = \angle XZO$, implying the result.

let $P$ be the point on $BC$ so that $AP \perp BC$, and extend $AP$ to meet the circle at point $Q$ we can easily see that $OQ$ intersects $BC$ at $D$ reflect $A,Y,X,W,P,D$ and $Q$ over an axis through $O$ perpencidular to $BC$ to get points $A_1,Y_1,X_1,W_1,P_1,D_1$ and $Q_1$ let $AQ_1$ and $BA_1$ intersect at $E$, and similarly define $E_1$ it is clear $EE_1$ is parallel to $BC$, so easy angle chasing yields that $AEE_1A_1$ is cyclic we can also prove that $AXX_1A_1$ is cyclic fairly easily by spiral similarity, $AEE_1$ and $AX_1A_1$ are similar, and we easily get that $QA_1$ is tangent to the circumcircle of $A_1X_1Y_1$ then, more angle chasing yields that $AXX_1A_1Y_1Y$ are all cyclic, so $A_1=Z$, and $(AXX_1A_1Y_1Y)=(AXY)$, so $DZ=DA_1=QA_1$ and is tangent to $(AXY)$

Let $Z'$ be the reflection of $A$ about the perpendicular bisector of $BC$. Claim: $DOZ'$ collinear $\angle Z'OA=\angle B-\angle C=180^{\circ}-\angle WOD$ Claim: $Z'XDC$ is concylic $\angle XDZ'=\angle WDO=\angle B-\angle C=\angle XCZ'$ Claim: $Z'YDB$ is concyclic $\angle YDZ'=\angle WDO=\angle B-\angle C=\angle YBZ'$ Claim: $Z'XYA$ is concylic $\angle AYZ'=\angle ZDC=\angle ZXC$ Corollary: $Z'=Z$ Claim: $OA$ is tangent to $AXYZ$ $\angle OAX=90^{\circ}-\angle B=\angle AYX$ Claim: $DOZ$ is tangent to $(AXYZ)$ Circles $(ABC)$ and $(AXYZ)$ are orthogonal.

First of all, we notice that $OA$ and $OZ$ are tangent to $(YAXZ)$ as $OA = OZ$ and, \[\measuredangle OAX = \measuredangle OAC = 90 - \measuredangle ABC = \measuredangle BYD = \measuredangle AYX.\] Now, let ray $OD$ intersect $(ABC)$ at $F$. By Thales we have $AF \parallel WD.$ Furthermore as \[\measuredangle OCA = \measuredangle OAC = \measuredangle AZX\] we see that line $OC$ and $XZ$ meet at $(ABC).$ Let this point be $K.$ Now, we redefine points $X'$ and $Z'$ as follows. Let $Z'$ be the other intersection of line $FO$ with the circumcircle and $X'$ be the intersection of $AC$ and $KZ'.$ Then, note that $\measuredangle FAZ = 90$ so $AZ' \parallel BC.$ Addtionally, \[\measuredangle XZD = \measuredangle KZO = \measuredangle ZKO = \measuredangle ZKC = \measuredangle ZAC = \measuredangle ACD = \measuredangle X'CD\] So, $X'Z'CD$ cyclic which implies $XD' \perp BC$ which is enough to show $X' \equiv X$ and $Z' \equiv Z.$ This gives, $D,O,Z$ collinear. Thus $DZ$ is tangent to $(AXY)$ is needed.

Let $AO$ intersect $BC$ at $E$. Clearly $D$ and $W$ lie on the circle centered at $O$ and radius $OE$. Also define $Z$ as the point on $(ABC)$ such that $AZ\parallel BC$. We will prove that $AXZY$ is cyclic. This is equivalent to $\angle AYX=\angle AZX$. But notice that $\angle AYX=\angle EAC=90^\circ-B$ and $\angle ZAX=\angle ACE$. So we need to prove that $\triangle AZX\sim \triangle CAE$ or equivalently, that $\frac{AZ}{BC}=\frac{AX}{EC}$. Now notice that $D$ is the reflection of $E$ across the perpendicular bisector of $BC$ so $BD=EC$. Also note that $\angle ABZ=180^\circ-A-2C$, hence by LOS we have $$\frac{AZ}{BC}=\frac{\sin(B-C)}{\sin B}$$Now we compute $AX=b-XC$. We have that $XC=\frac{DC}{\cos C}=\frac{BE}{\cos C}$ and by LOS, we have $BE=\frac{c\cdot \cos C}{\cos(B-C)}$, therefore $XC=\frac{c}{\cos(B-C)}$ so $AX=b-\frac{c}{\cos(B-C)}$ By LOS we have $EC=\frac{b\cdot \cos B}{\cos(B-C)}$ so $$\frac{AX}{EC}=\frac{b\cdot \cos (B-C)-c}{b\cdot \cos B}=\frac{\cos(B-C)-\frac{\sin C}{\sin B}}{\cos B}$$ And we wish to prove that this is equal to $\frac{\sin(B-C)}{\sin B}$, which is equivalent to $$\sin B\cos(B-C)-\sin C=\cos B\sin(B-C)\iff \sin^2B\sin C-\sin C=-\cos^2B\sin C$$ And this is clearly true. Finally, we conclude that $AYZX$ is cyclic so it remains to prove that $DZ$ is tangent to this circle. But notice that since $AE$ is tangent to the circle, the conclusion follows by reflecting across the perpendicular bisector of $BC$. $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 1.3972062084257204, xmax = 42.97592017738355, ymin = -16.210022172949007, ymax = 5.696851441241685; /* image dimensions */ pen qqzzcc = rgb(0.,0.6,0.8); draw((17.,0.)--(14.,-11.)--(26.,-11.)--cycle, linewidth(0.8) + qqzzcc); /* draw figures */ draw((17.,0.)--(14.,-11.), linewidth(0.8) + qqzzcc); draw((14.,-11.)--(26.,-11.), linewidth(0.8) + qqzzcc); draw((26.,-11.)--(17.,0.), linewidth(0.8) + qqzzcc); draw(circle((20.,-6.7272727272727275), 7.365880690528964), linewidth(0.8)); draw((17.,0.)--(21.905405405405403,-11.), linewidth(0.8)); draw((17.,0.)--(18.094594594594593,4.013513513513506), linewidth(0.8)); draw((18.094594594594593,4.013513513513506)--(18.094594594594593,-11.), linewidth(0.8)); draw((18.09459459459459,-1.3378378378378355)--(23.,0.), linewidth(0.8)); draw((23.,0.)--(17.,0.), linewidth(0.8)); /* dots and labels */ dot((17.,0.),dotstyle); label("$A$", (16.477915742793774,0.18022172949002063), NE * labelscalefactor); dot((14.,-11.),dotstyle); label("$B$", (13.262394678492227,-11.21756097560976), NE * labelscalefactor); dot((26.,-11.),dotstyle); label("$C$", (26.371263858093104,-11.21756097560976), NE * labelscalefactor); dot((20.,-6.7272727272727275),linewidth(4.pt) + dotstyle); label("$O$", (20.075698447893554,-6.578093126385812), NE * labelscalefactor); dot((21.905405405405403,-11.),linewidth(4.pt) + dotstyle); label("$E$", (21.97370288248335,-10.853037694013308), NE * labelscalefactor); dot((18.094594594594593,-11.),linewidth(4.pt) + dotstyle); label("$D$", (18.159955654101978,-10.853037694013308), NE * labelscalefactor); dot((18.094594594594597,-2.454545454545456),linewidth(4.pt) + dotstyle); label("$W$", (18.159955654101978,-2.320886917960091), NE * labelscalefactor); dot((18.09459459459459,-1.3378378378378355),linewidth(4.pt) + dotstyle); label("$X$", (18.159955654101978,-1.2033702882483388), NE * labelscalefactor); dot((18.094594594594593,4.013513513513506),linewidth(4.pt) + dotstyle); label("$Y$", (18.159955654101978,4.15361419068736), NE * labelscalefactor); dot((23.,0.),linewidth(4.pt) + dotstyle); label("$Z$", (23.073481152993327,0.14474501108647295), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Clearly $Z$ is miquel point of $AXDB$ therefore $YZDB, ZXDC$ are cyclic. Now let $H_A$ be a point in $(ABC)$ such that $AH_A \perp BC$, then note that we have $DW \parallel AH_A$, we have $AO=OH_A$ and $DO=OW$ , therefore $D$ must be $OH_A \cap BC$ by homothety, now by angle chase: \[\angle AYX=\angle BAH_A=\angle OAX \]Therefore $OA$ is tangent to $(AXY)$ , but by tangent lenght this means that $OZ$ is tangent to $(AXY$) as well so if we prove that $Z,O,D,H_A$ are colinear we are done so... \[\angle H_AZC=\angle H_AAC=\angle DXC=\angle DZC \]Which implies that $Z,D,H_A$ are colinear so by the other colinearity we are done.

Let $DO \cap (ABC)$ at $Z$ ($Z$ and $A$ are on same side of $BC$). Let $\angle OWD=\angle ODW=\alpha$ and $\angle OAC=\angle OCA=\beta$ ,thus, $\angle DYC=\angle OWD+\angle YAW=\alpha+\beta$ $\implies$ $\angle OCB=90-\alpha-2\beta$. $\angle BXD=90-\angle ABC=\beta$ ,thus , $OA$ tangents to $(AXY)$. $\angle AOZ=2\alpha$ $\implies$ $\angle OAZ=90-\alpha$ which means $\angle CAZ=90-\alpha-\beta$. So, $\angle ZBC=\angle CAZ=90-\alpha$. We know that $\angle ZBC=\angle OBC+\angle ZBO$ $\implies$ $\angle ZBO=\beta$.Thus, $\angle XAZ=\alpha$. $\angle XDZ=\angle XAZ$ implies $XBDZ$ cyclic . So, $\angle ZBD=\angle DXZ=\angle YAZ$. Which means $AXZY$ cyclic.Since, $AO=ZO$ and $AO$ tangents to $AXYZ$ that means $OZ$ tangents to $(AXYZ)$.

First off, note that the $OW=OD$ condition implies $D$ is the reflection of $AO \cap BC$ across the midpoint of $BC$ and that $AO,BC, (OD)$ are concurrent at the point we call $W'$ as in the reflection of $W$ across $O$. Now if we reflect $DO$ across the perpendicular bisector of $BC$ then we get the line $AO$ so $\overrightarrow{DO} \cap (ABC)$ is the reflection of $A$ over perpendicular bisector of $BC$. Now it is easy to see by some angle chasing that $OA$ is tangent to $(AXY)$ and since $O$ lies on perpendicular bisector of $AZ$, $OZ$ is also tangent to $(AXY)$. Also notice that $XZ \perp CZ$ so $XDCZ$ is cyclic. Now take $Z'$ as $\overrightarrow {DO} \cap (ABC)$ and see that it satisfies the above properties of $Z$ which can be proven by angle chasing, so $ Z \equiv Z'$ and now some more angle chasing finishes.

Let $H$ be the orthocentre of $\Delta ABC$, $H'_A$ be the reflection of $H$ across $BC$. Note that $OA,OD$ are reflections across the line through $O$ parallel to $BC$. It follows that ray $OD$ intersects $(ABC)$ at $H'_A$. Let $Z'$ be the reflection of $A$ across the perpendicular bisector of $BC$, so $H'_A-D-O-Z'$ collinear. Let $\omega$ be the circle through $A,Z'$ such that $OA,OZ'$ are tangent to $\omega$. It suffices to show that $X,Y$ lie on $\omega$. First note that $\measuredangle DXC = \measuredangle H'_AAC = \measuredangle DZ'C$, so $(XDCZ')$ concyclic. Next, $\measuredangle OZX = \measuredangle DCA = \measuredangle Z'AX$, so $OZ'$ is tangent to $(AXZ')$. It follows that $X$ lies on $\omega$. By symmetry, $Y$ also lies on $\omega$ so we are done. $\square$