Let $\angle APX=\alpha$. So, $$\angle O_1CA=\angle O_1BA=90^\circ-\alpha=\angle BAO_1=\angle BCO_1.$$Hence, $$\angle PXA=\angle BXA=\angle BCA=\angle BCO_1+\angle O_1CA=(90^\circ-\alpha) +(90^\circ-\alpha) =180^\circ-2\alpha.$$So, in cyclic quadrilateral $CYAX$ we have $$\angle XCY=\angle XAP=180^\circ-(\angle PXA+\angle APX) =180^\circ-(180^\circ-2\alpha+\alpha)=\alpha.$$Since, $CO_2$ is diameter of $\omega_2$, $\angle O_1BC=90^\circ\implies \angle ABC=90^\circ-\angle O_1BA=\alpha$. So, in cyclic quadrilateral $ABCY$, $\angle CYA=180^\circ-\angle ABC=180^\circ-\alpha $. So, $$\angle CYA+\angle XCY=180^\circ\implies PY\parallel CX$$Since, $\angle YPX=\angle APX=\alpha=\angle XCY$ implies $CY\parallel PX$. Hence, $XPYC$ is a parallelogram. $\blacksquare$

Dear Mathlinkers, just an application of the Reim's theorem in a particular case whih tangent... Sincerely Jean-Louis

(1) <YCO1=<O1AP=<O1PA , <XCO1=<O1BP=<O1BP --> <YCX=<YPX (2) <O1YA=<O1XB , <O1YC=<O1XC=90 --> <PYC=<PXC [END]

Like jayme said it's nothing but Reim. Since $AC$ is tangent to $\omega_1$ , by Reim $AP \parallel CX$. Similarly $BP \parallel CY$. $\blacksquare$

Easy! Claim: $O_1$ is the orthocenter of $\triangle PXY$ Proof: By spiral similarity $\angle AO_1O_2=\angle APX=x$ Also $\angle O_1XA=\frac{\angle O_1O_2A}{2}=90-x$. Therefore $XO\perp YP$.Similarly $YO\perp XP$.Therefore $O_1$ is the orthocenter. Also observe that $O_1$ and $C$ are antipodes of each other Therefore $O_1A\perp CA$ Similar argument for $CB$ Hence $PXCY$ is parallelogram. An interesting fact:Length of $XY$ is invariant $\blacksquare$

EASY- let $\angle AO_1B$ = $\ 2$ $\angle APB$ and since $\ O_1A = O_1B$ then $\angle BCO_1=ACO_1 = 90 - BO_1A/2$ but we also have that $\angle O_1CA = O_1YA$ and since $\angle CYO_1=90$ so we have that $\angle CYP=CYO_1+O_1YA=180-APB$ so we have that $\ YC||PB$ and similarly we have that $\angle CXB= CXO_1+O_1XB= 180-APB$ which implies that $\ YP||CX $ and our result follows:

IGO 2019 Intermediate P1 wrote: Two circles $\omega_1$ and $\omega_2$ with centers $O_1$ and $O_2$ respectively intersect each other at points $A$ and $B$, and point $O_1$ lies on $\omega_2$. Let $P$ be an arbitrary point lying on $\omega_1$. Lines $BP, AP$ and $O_1O_2$ cut $\omega_2$ for the second time at points $X$, $Y$ and $C$, respectively. Prove that quadrilateral $XPYC$ is a parallelogram. Proposed by Iman Maghsoudi Nice and easy! I'll present two solutions by angle chasing and Reim's Theorem. Solution. 1 Since $O_1A=O_1B \implies \angle O_1AB=\angle O_1BA$, we know that $\angle O_1CB=\angle O_1AB=\angle O_1BA=\angle O_1YA=\angle O_1CA=\frac{180^{\circ}-\angle BO_1A}{2}=90^{\circ}-\frac{\angle BO_1A}{2}=90^{\circ}-\angle XPA.$ $\angle PYC=\angle O_1YA+\angle O_1YC=180-\angle XPA \implies \angle XYP+\angle XPY=180^{\circ} \implies XP\parallel CY, PY\parallel XC.\blacksquare$ Solution. 2 $\angle CAO_1= 90^{\circ} \implies AC$ is tangent to $\omega_1$, now by Reim's we have $AP\parallel CX,$ similarly $BP \parallel CY.\blacksquare$

∠APB = 180 - ∠AO1B/2 = ∠AXC = ∠BYC. so we have ∠PBY = ∠CYA = ∠CXP = ∠XAY so CXAY and CXBY are isosceles trapezoids. ∠AXC + ∠CYA = 180 ---> ∠XCY = ∠XPY. now we have ∠XCY = ∠XPY and ∠CXP = ∠CYP so PXCY is parallelogram. we're Done.

jayme wrote: Dear Mathlinkers, just an application of the Reim's theorem in a particular case whih tangent... Sincerely Jean-Louis Can you tell the statement of Reim's theorem?

Titusir wrote: jayme wrote: Dear Mathlinkers, just an application of the Reim's theorem in a particular case whih tangent... Sincerely Jean-Louis Can you tell the statement of Reim's theorem? Let $\omega_1, \omega_2$ be two circles intersecting at$ M,N$. Let line $\ell_M$ through M intersect $\omega_1, \omega_2$ at $A_1, A_2. Let B_1, B_2$ be points on $\omega_1, \omega_2$ respectively, Then $A_1B_1\parallel A_2B_2$ if , and only if, $B_1,N,B_2$ are collinear on a line $\ell_N.$

Note that $AC$ and $BC$ is tangent to $\omega_1$. Therefore, by Reim's we are done (since $CA$ intersects $\omega_1$ at $A$ and $A$, and $XB$ intersects at $B$ and $P$, so $AP$ is parallel to $CX$ and similarly for $YC$ and $XP$)

IGO 2019 I1. Two circles $\omega_1$ and $\omega_2$ with centers $O_1$ and $O_2$ respectively intersect each other at points $A$ and $B$, and point $O_1$ lies on $\omega_2$. Let $P$ be an arbitrary point lying on $\omega_1$. $BP$, $AP$ and $O_1O_2$ cut $\omega_2$ for the second time at points $X$, $Y$ and $C$ respectively. Prove that quadrilateral $XPYC$ is a parallelogram. Solution 1. Since $O_1C$ is a diameter, $\angle O_1AC=90^{\circ}$. Hence $CA$ is tangent to $\omega_1$ at $A$. Then \[\angle PXY=\angle BAY=90^{\circ}-\angle O_1AB-\angle YAC=90^{\circ}-\angle ABO_1-\angle PBA=\angle CBX=\angle CYX\]implies $PX\parallel YC$. Also, \begin{align*}\angle XCY+\angle CYA&=(\angle XCA+\angle ACO_1+\angle O_1CB+\angle BCY)+(\angle CYX+\angle XYA)\\&=(\angle XBA+\angle ABO_1+\angle O_1AB+\angle BAY)+(\angle CBX+\angle XBA)\\&=(\angle PAC+\angle ABO_1+\angle O_1AB+\angle BAY)+(\angle CBX+\angle XBA)\\&=(\angle PAC+\angle O_1AB+\angle BAY)+(\angle ABO_1+\angle CBX+\angle XBA)\\&=90^{\circ}+90^{\circ}=180^{\circ}\end{align*}implies $XC\parallel AY$. Therefore $XPYC$ is a parallelogram. Solution 2. Note that \[\angle PAX+\angle AXP=\angle APB=\frac{360^{\circ}-\angle BO_1A}2=180^{\circ}-\angle CO_1A=180^{\circ}-\angle CBA=\frac{360^{\circ}-\angle CO_2A}2=\angle AXC\implies\widehat{YCX}+\widehat{AO_1B}=\widehat{ABC}\implies\widehat{BY}=\widehat{CX}\implies\angle BXY=\angle CYX\implies BX\parallel YC.\]Similarly, $AY\parallel XC$. Therefore $XPYC$ is a parallelogram. [asy][asy] import graph; size(10cm); 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 = -4.808145471086625, xmax = 46.1979007177304, ymin = -16.576673808330693, ymax = 9.601469832607396; /* image dimensions */ /* draw figures */ draw(circle((6.919650033265665,-4.889780225786552), 5.561862764193502), linewidth(0.8)); draw(circle((1.3578592896709818,-4.918084504379911), 3.656544386039834), linewidth(0.8)); draw((2.5773817079974948,-8.36526914090266)--(4.436201825663398,0.08684537876511722), linewidth(0.8)); draw((1.3578592896709818,-4.918084504379911)--(12.48144077686035,-4.861475947193194), linewidth(0.8)); draw((2.542233601920173,-1.4586662967091928)--(11.95637872528588,-7.248949887066233), linewidth(0.8)); draw((1.3578592896709818,-4.918084504379911)--(2.542233601920173,-1.4586662967091928), linewidth(0.8)); draw((2.542233601920173,-1.4586662967091928)--(4.436201825663398,0.08684537876511722), linewidth(0.8)); draw((4.436201825663398,0.08684537876511722)--(12.48144077686035,-4.861475947193194), linewidth(0.8)); draw((12.48144077686035,-4.861475947193194)--(11.95637872528588,-7.248949887066233), linewidth(0.8)); draw((11.95637872528588,-7.248949887066233)--(2.5773817079974948,-8.36526914090266), linewidth(0.8)); draw((2.5773817079974948,-8.36526914090266)--(1.3578592896709818,-4.918084504379911), linewidth(0.8)); draw((2.542233601920173,-1.4586662967091928)--(2.5773817079974948,-8.36526914090266), linewidth(0.8)); draw((4.436201825663398,0.08684537876511722)--(11.95637872528588,-7.248949887066233), linewidth(0.8)); draw((2.542233601920173,-1.4586662967091928)--(12.48144077686035,-4.861475947193194), linewidth(0.8)); draw((12.48144077686035,-4.861475947193194)--(2.5773817079974948,-8.36526914090266), linewidth(0.8)); /* dots and labels */ dot((6.919650033265665,-4.889780225786552),linewidth(2pt) + dotstyle); label("$O_2$", (7.040909019022137,-4.7827311995712165), NE * labelscalefactor); dot((2.542233601920173,-1.4586662967091928),linewidth(2pt) + dotstyle); label("$A$", (2.659514451749362,-1.3382386152372576), NE * labelscalefactor); dot((1.3578592896709818,-4.918084504379911),linewidth(2pt) + dotstyle); label("$O_1$", (1.4746090027384857,-4.810287140245888), NE * labelscalefactor); dot((2.5773817079974948,-8.36526914090266),linewidth(2pt) + dotstyle); label("$B$", (2.6870703924240336,-8.254779724579848), NE * labelscalefactor); dot((3.911139774088934,-2.3006285611079225),linewidth(2pt) + dotstyle); label("$P$", (4.009755544808268,-2.1924727761520795), NE * labelscalefactor); dot((11.95637872528588,-7.248949887066233),linewidth(2pt) + dotstyle); label("$Y$", (12.056090221812358,-7.124986156918308), NE * labelscalefactor); dot((4.436201825663398,0.08684537876511722),linewidth(2pt) + dotstyle); label("$X$", (4.533318417627027,0.2048940625443558), NE * labelscalefactor); dot((12.48144077686035,-4.861475947193194),linewidth(2pt) + dotstyle); label("$C$", (12.579653094631118,-4.755175258896545), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

$\color{magenta}\boxed{\textbf{SOLUTION}}$ $$\angle PXC=\angle BXC=\angle CXO_1 + \angle BXO_1= 90°+ \frac{\angle BO_2O_1}{2}$$ $$\angle PYC=\angle AYC=\angle CYO_1+\angle AYO_1=90°+\frac{\angle AO_2O_1}{2}$$ But, $\triangle AO_2O_1$ and $\triangle BO_2O_1$ are congruent by $\textbf{SSS}$ congruency So $$\angle BO_2O_1=\angle AO_2O_1 \implies \angle PXC=\angle PYC \implies PXCY$$is a parallelogram $\blacksquare$