Nice problem. It seems $Z $ lies on the first circle or $\angle XZY = \frac{5\pi}{6}$

This problem was really awesome. Here is my horrendously long solution. Firstly, we denote the two circles as $\omega_1$ and $\omega_2.$ We claim that $\omega_2$ is the $C$-excentre of $CA'B'.$ Note that, using the Poncelet's Porism, if this fact is true for a single point $C$ on $\omega_1,$ then it's true for all points. Let $C\equiv O_1O_2\cap \omega_1.$ It can be easily verified for this $C$ that $A'B'$ is tangent to $\omega_2,$ therefore it follows that $O_2$ is the $C$-excentre of $CA'B'.$ Denote by $D,$ the point where $A'B'$ touches $\omega_2.$ Using Pascal's theorem in $AABBDD,$ we note that $AA\cap BD, AB\cap DD, BB\cap DA$ are collinear. By the converse of Desargues theorem, it implies that $CA'B'$ and $DAB$ are perspective triangles. Therefore, $CD$ passes through $AA'\cap BB',$ in other words, $CD$ passes through $Z.$ [asy][asy] import graph; size(17cm); 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 = -1.97, xmax = 1.95, ymin = -1.31, ymax = 1.25; /* image dimensions */ pen ffqqqq = rgb(1,0,0); pen ffqqqq = rgb(1,0,0); pen qqzzqq = rgb(0,0.6,0); pen qqzzqq = rgb(0,0.6,0); pen ffzzcc = rgb(1,0.6,0.8); pen ffzzcc = rgb(1,0.6,0.8); pen qqttcc = rgb(0,0.2,0.8); pen ffqqqq = rgb(1,0,0); pen ffqqqq = rgb(1,0,0); pen ccwwff = rgb(0.8,0.4,1); pen yqqqyq = rgb(0.5,0,0.5); pen ccwwqq = rgb(0.8,0.4,0); pen ccwwqq = rgb(0.8,0.4,0); pen qqttcc = rgb(0,0.2,0.8); pen qqttcc = rgb(0,0.2,0.8); pen qqttcc = rgb(0,0.2,0.8); pen qqqqqq = rgb(0,0,0); pen ffffff = rgb(1,1,1); /* draw figures */ draw(circle((-0.87,0), 1), ffqqqq); draw(circle((0.87,0), 1), ffqqqq); draw((xmin, -0.83*xmin-0.58)--(xmax, -0.83*xmax-0.58), qqzzqq); /* line */ draw((xmin, 0.12*xmin + 0.9)--(xmax, 0.12*xmax + 0.9), qqzzqq); /* line */ draw((-0.03,-0.55)--(0.75,0.99), ffzzcc); draw((-0.36,0.86)--(0.23,-0.77), ffzzcc); draw((-0.36,0.86)--(-0.03,-0.55), qqttcc); draw((-0.87,0)--(0.87,0), ffqqqq); draw((xmin, 0*xmin + 0)--(xmax, 0*xmax + 0), ffqqqq); /* line */ draw((-0.11,-0.22)--(0.87,0), ccwwff); draw((-0.87,0)--(-1.57,0.71), yqqqyq); draw((-1.57,0.71)--(0.87,0), ccwwqq); draw((-0.11,-0.22)--(-0.87,0), ccwwqq); draw((-0.36,0.86)--(xmax, -4.4*xmax-0.7), qqttcc); /* ray */ draw((0.75,0.99)--(xmin, 3.41*xmin-1.55), qqttcc); /* ray */ draw((xmin, -2.33*xmin-0.93)--(xmax, -2.33*xmax-0.93), qqttcc); /* line */ draw((xmin, -0.64*xmin-0.29)--(xmax, -0.64*xmax-0.29), qqqqqq); /* line */ /* dots and labels */ dot((0.87,0),dotstyle); label("$O_2$", (0.82,-0.09), S * labelscalefactor); dot((-0.87,0),dotstyle); label("$O_1$", (-0.92,-0.08), S * labelscalefactor); dot((-1.57,0.71),dotstyle); label("$C$", (-1.6,0.76), NE * labelscalefactor); dot((0.75,0.99),dotstyle); label("$A$", (0.76,1.02), NE * labelscalefactor); dot((0.23,-0.77),dotstyle); label("$B$", (0.26,-0.78), NE * labelscalefactor); dot((-0.03,-0.55),dotstyle); label("$A'$", (-0.12,-0.50), S * labelscalefactor); dot((-0.36,0.86),dotstyle); label("$B'$", (-0.34,0.89), NE * labelscalefactor); dot((-0.11,-0.22),dotstyle); label("$D$", (-0.17,-0.25), SW * labelscalefactor); dot((0.11,-1.18),dotstyle); label("$F$", (0.2,-1.21), E * labelscalefactor); dot((0.07,-0.34),dotstyle); label("$Z$", (0.11,-0.35), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); shipout(bbox(ffffff,Fill)); [/asy][/asy] Let $I$ be the incentre of $CA'B'$. Now, note that from simple angle-chasing(for the configuration in the diagram), we get \[\begin{aligned}\angle O_1CO_2&=\frac{\pi}{2}-\angle CA'B'-\frac 12\angle A'CB'\\&=\frac12\left(\angle A'B'C-\angle B'A'C\right)\\&=\angle B'AD-\frac 12\angle B'A'C\\&=\angle B'O_2D-\angle IA'B'\\&=\angle B'O_2D-\angle IO_2B\\&=\angle DO_2C.\end{aligned}\] Since $O_1C=O_2D,$ therefore it follows that $[O_1CO_2]=[DCO_2],$ so that $O_1O_2CD$ is an isosceles trapezoid. Thus $O_1D\parallel CO_2.$ Also, $CO_2\perp AB,$ so that $O_1D\perp AB.$ Now, denote $E=CD\cap \omega_1.$ Let $A'E\cap CA=A_1,$ and $B'E\cap CB=B_1.$ It is well-known that $A_1B_1$ is the polar of $E$ wrt $\omega_1.$ Therefore we get $O_1D\perp AB\ \land \ O_1D\perp A_1B_1\implies AB\parallel A_1B_1,$ On the other hand, observe that $A'B_1\cap B'B_1, AA'\cap BB', AA_1\cap BB_1$ are collinear and lie on $CD,$ so that $AA_1A'$ and $BB_1B'$ are perspective triangles. All of this immplies that $AB, A'B', A_1B_1$ concur. But, since $A_1B_1\parallel AB,$ so it is forced that $A\equiv A_1$ and $B\equiv B_1,$ implying that $Z\equiv E.$ Finally, $Z\in \omega_1,$ so that $\boxed{\angle XZY=150^{\circ}}.$ We are done.$\Box$ Edit: I wrote $\angle XZY=120^{\circ}$ in the last step, and submitted it like that. I donno how much I will get for this problem. In any case, thanks to yunxiu for telling me about this typo.

Potla wrote: It can be easily verified for this $C$ that $A'B'$ is tangent to $\omega_2,$ therefore it follows that $O_2$ is the $C$-excentre of $CA'B'.$ Can you explain why this is correct?

ts0_9 wrote: Potla wrote: It can be easily verified for this $C$ that $A'B'$ is tangent to $\omega_2,$ therefore it follows that $O_2$ is the $C$-excentre of $CA'B'.$ Can you explain why this is correct? This is known as Poncelet's Porism. http://mathworld.wolfram.com/PonceletsPorism.html

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Indeed, a very nice problem. Let $w_1,w_2$ be the two circles, $O_1,O_2$ their centers. Let $w$ be the circle with diameter $XY$, (radius $1/2$), center $O$. Then it is easy to see that $w,w_2$ are inverses with respect to $w_1$ and a dilation with factor 2, center $O_1$ takes $w$ into $w_2$. Similar relations hold for $w,w_1$. Let $B'L$ be the other tangent from $B'$ to $w_2$ and assume that $B'L$ intersects $BC$ at $A_1$. Then $w_2$ is the excircle of $CB'A_1$, and its points of tangency with those triangular sides form a new triangle whose midpoints are the inverses of $CB'A_1$ in $w_2$. So the inverse of the circumcircle of $CB'A_1$ has radius $1/2$ as well. But that circle shares two points with $w$ (the inverses of $C,B'$), so it must be in fact $w$. (The other circular possibility is ruled out because its inverse in $w_2$ is actually $AC$). So the circumcircle of $CB'A_1$ is $w_1$ and $A_1=A'$. Thus $w$ is the nine point circle of $ABL$. Then if $R$ is the foot of the altitude from $L$ to $AB$, $R$ lies on $w$ as well. Since $O$ is the midpoint of $O_1O_2$, $O_1$ must the orthocenter of $ABL$, which means that $O_1,L,R$ are collinear and that $O_1R \perp AB$. But $w,w_2$ are inverses in $w_1$, so $L,R$ are inverses, which means that $AB$ is the pole of $L$ in $w_1$. Now extend $CL$ to intersect $w_1$ at $Z$, and assume that $A'Z,B'Z$ intersect $CA,CB$ at $A_2,B_2$ respectively. Since quadrilateral $CB'ZA'$ is inscribed in $w_1$, $A_2B_2$ must be the pole of $L$ in $w_1$, which of course means that $A=A_2,B=B_2$.

Attachments:
sharygin 2012.pdf (394kb)

Hi! Sorry for reviving such an old topic... but I can't understand something.. Potla wrote: We claim that $\omega_2$ is the $C$-excentre of $CA'B'$.Note that, using the Poncelet's Porism, if this fact is true for a single point $C$ on $\omega_1,$ then it's true for all points. As I red Poncelet porism states that given a circle inside another, if there is a triangle/quadrilateral/n-gon inscribed in the outter circle and circumscribed to the small circle, then there are an infinity. Didn't I understand correctly? I can't see how Poncelet porism can be applied here...

Lumpa wrote: As I red Poncelet porism states that given a circle inside another, if there is a triangle/quadrilateral/n-gon inscribed in the outter circle and circumscribed to the small circle, then there are an infinity. Didn't I understand correctly? It holds for any two conic sections that are given, in general. The examples given are only a special case. Here I'm using the fact that if the property holds for one point on $\omega_1,$ it will hold for all points, and hence for $C$ as well. Choose a convenient point to check the tangency, and then we are done. ^_^