If we invert about $P$, the congruent circle condition gives that $P$ is the $Q'$-excenter of triangle $Q'X'Y'$. Now $A'PB'Q'$ is cyclic, so its center lies on $PQ'$. It is well known that $X'Y'P$ has its center at the midpoint of arc $X'Y'$ on circle $(Q'X'Y')$, so this center lies on the angle bisector of $Q'$, which is $Q'P$. Thus $P$ is collinear with the centers of the circles $(PX'Y')$ and $(PA'B')$, so they are tangent at $P$, and $AB\parallel XY$.

Boo inversion Let $O_1$, $O_2$ be the centers of the circles $\alpha$, $\beta$, and let $\gamma$ be the third circle. Note that since $\alpha$ and $\gamma$ have the same radius, $\triangle PXO_1$ is equilateral. Similarly, $\triangle PYO_2$ is also equilateral. Therefore $\angle XPO_1=\angle YPO_2$, so $\widehat{XO_1}=\widehat{YO_2}$ and $XY\parallel O_1O_2$. Now observe that the homothety centered at $P$ with scale factor $2$ takes $O_1O_2$ to $AB$, so $O_1O_2\parallel AB$, implying the desired.

Dear Mathlinkers, 1. AX and BY intersect on the third circle at the antipole Q of P 2. PQ gors through the second intersection of the first two circles 3. According to Reim's theorem (two times), we are done... Sincerely Jean-Louis

Let the second intersection of $ \alpha $ and $ \beta $ be $ Z $. Then, by Johnson's three circles theorem, $ P $ is the orthocenter of $ \triangle XYZ $, i.e $ ZP \perp XY $. On the other hand, $ \angle PZA=\angle PZB=90^{\circ} $, so $ A,Z,B $ are collinear and $ ZP \perp AB $. This gives $ XY \parallel AB $ as desired.

Remark: We call this Three circles theorem ´Titeica´s five lei coin problem´. Happy New Year to everybody! sunken rock

Dera Stan and Mathlinkers, this nice result "The Triquetra Theorem" was known from Johnson in 1916 and also by Titeica who lives at the same period... Do you have a reference for Titeica? Very sincerely Jean-Louis

Let $O_1,O_2,O_3$ be the centers of $\alpha,\beta$ and the other .$O_1XO_3P$ and $O_2YO_3P$ are rhombuses .$\Rightarrow O_3XY \equiv PO_1O_2 \Rightarrow O_1XYO_2$ a parallellogram .Done!! Dear Mathlinkers and all , Wish you a happy new year.

djmathman wrote: Let $O_1$, $O_2$ be the centers of the circles $\alpha$, $\beta$, and let $\gamma$ be the third circle. Note that since $\alpha$ and $\gamma$ have the same radius, $\triangle PXO_1$ is equilateral. Similarly, $\triangle PYO_2$ is also equilateral Hello, djmathman i don't think that $\triangle PXO_1 \text{and} \triangle PYO_2$ are equilateral. Because, the third circle does not fixed, so the length $PX$ and $PY$ varies.

Well the third circle is not fixed, of course, since as you reorient the circles $\alpha$ and $\beta$ its position will change. However, it is given that it has the same radius as those of the other two circles, and as such it is guaranteed to pass through both $O_1$ and $O_2$, making the aforementioned triangles equilateral.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Quickies%209.pdf p. 11-12. Sincerely Jean-Louis

Let $O_1,O_2$ be the centers of $\alpha,\beta$, respectively. Let $O$ be the center of $(PXY)$. Note that $O_1O_2$ is the midline of $\triangle ABP$. We claim that $O_1O_2XY$ is a parallelogram, from which the desired follows. Obviously, $O_1X=O_2Y$. As the radius of $(PXY)$ is equal to the radius of $\alpha,\beta$, we have that $OXO_1P$ and $OPO_2Y$ are parallelograms. This implies that $XO_1\parallel OP\parallel YO_2$, we are done. [asy][asy] import olympiad; size(9cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair O1,O2,P,A,B,X,O,Y; O1=(0,0);O2=(1.6,0);path w=circle(O1,1);path g=circle(O2,1);P=intersectionpoints(w,g)[0];A=2O1-P;B=2O2-P;X=dir(70);O=X+P-O1;path h=circle(O,abs(O-P));Y=intersectionpoints(h,g)[1]; draw(g,deep);draw(w,deep);draw(h,deep);draw(A--P--B,deep);draw(O1--O2,org);draw(X--Y,org);draw(A--B,org);draw(O1--X,light);draw(O2--Y,light); dot("$O_1$",O1,S); dot("$O_2$",O2,S); dot("$P$",P,dir(P)); dot("$X$",X,dir(X)); dot("$O$",O,dir(O)); dot("$Y$",Y,dir(Y)); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); [/asy][/asy]

jayme wrote: Dear Mathlinkers, 1. AX and BY intersect on the third circle at the antipole Q of P 2. PQ gors through the second intersection of the first two circles Actually I don't think this is true at all... (I mean 2.) In fact here is a solution which is a bit different from all the others so far: Let $C$ be the point diametrically opposite to $P$ on the third circle. Then by Thales $X,Y$ are the projections from $P$ on $AC$ resp. $BC$. Since $P$ is the circumcenter of $ABC$, this means that $X,Y$ are the midpoints of $AC$ resp. $BC$ and hence $XY$ is just the midline in triangle $ABC$ and so of course parallel to $AB$.