Romanian Master of Mathematics

Let $X^{\prime },Y^{\prime },P^{\prime }$, and $Q^{\prime }$ be the reflections across $BC$ of $X,Y,P$, and $Q$, respectively. Then ${\Omega }_1$ and ${\Omega }_2$ are just the circles $PX{X}^{\prime }{P}^{\prime }$ and $QY{Y}^{\prime }{Q}^{\prime }$, respectively. Denote $\alpha =\angle BAC=\angle CBX=\angle YCB$. Let $XY$ cross $BC$ at $W$; the case $XY\parallel BC$ may be treated as a limit case. The symmetry yields that $W$ also lies on the line $X^{\prime }{Y}^{\prime }$. The same symmetry, along with tangency of $PB$ and $QC$ to the circle $ABC$, yields $$\alpha =\angle X^{\prime }BC=\angle PBW=\angle WCQ=\angle BC{Y}^{\prime }.(\ast )$$ This yields that each of the triples $(P,B,X^{\prime })$, $(Q,C,Y^{\prime })$, $(P^{\prime },B,X)$, and $(Q^{\prime },C,Y)$ is collinear, and, moreover, that $PB{X}^{\prime }\parallel Q^{\prime }CY$ and $P^{\prime }BX\parallel QC{Y}^{\prime }$. It follows now that quadrilaterals $PX{X}^{\prime }{P}^{\prime }$ and $YQ{Q}^{\prime }{Y}^{\prime }$ are homothetic at $W$. Therefore, so are ${\Omega }_1$ and ${\Omega }_2$.

Let now ${\Omega }_1$ and ${\Omega }_2$ cross at $D$ and $D^{\prime }$. Let $WD$ and $W{D}^{\prime }$ meet ${\Omega }_1$ again at $E$ and $E^{\prime }$. Since $W=PX\cap P^{\prime }{X}^{\prime }$ and $B=P{X}^{\prime }\cap P^{\prime }X$, the point $B$ lies on the polar of $W$ with respect to ${\Omega }_1$. In other words, $W$ and $B$ are inverse with respect to that circle. This yields that the lines $D{E}^{\prime }$ and $D^{\prime }E$ also cross at $B$.

Now, we have $\angle BDX=\angle E^{\prime }DX=\angle X^{\prime }{D}^{\prime }E=\angle X^{\prime }PE=\angle BPE=\angle CYD$ (the last equality holds by means of homothety). Similarly, we have $\angle DXB=\angle DX{P}^{\prime }=\angle P{X}^{\prime }{D}^{\prime }=\angle PE{D}^{\prime }=\angle PEB=\angle YDC$. Therefore, the triangles $BDX$ and $CYD$ are similar. Firstly, this yields that $\angle DBC=\angle DBX+\angle XBC=\angle YCD+\angle BCY=\angle BCD$, whence $BD=CD$. Secondly, this also implies that $BD/BX=CY/CD$, or $BX\cdot CY=BD\cdot CD=B{D}^2$. But the triangles $BXC$ and $CBY$ are also similar (as both are similar to $ABC$), so $BX/BC=BC/CY$, or $BX\cdot CY=B{C}^2$. Thus, $BC=BD=CD$, and the triangle $BCD$ is equilateral. This finishes the solution.

Solution 2: All angles in the solution are directed. All segment lengths on lines $BX$ and $CY$ (and parallel to them) are also oriented; we assume that the directions $\overrightarrow{BX}$ and $\overrightarrow{CY}$ are positive. As in the solution above, we prove that $BP\parallel CY$. Assume that $ABC$ is oriented anti-clockwise. Let $D$ and $D^{\prime }$ be the points such that the triangles $DBC$ and $D^{\prime }CB$ are equilateral, and oriented anti-clockwise. We will show that $D$ and $D^{\prime }$ lie on the circle ${\Omega }_1$; similarly, they lie on ${\Omega }_2$. Notice that $\alpha =\angle BAC=\angle CBX=\angle YCB=\pi -\angle CBP$; moreover, each of the triangles $XBC$ and $BCY$ is similar to $BAC$ and oriented differently than $BAC$; hence those two triangles are equi-oriented. Let $\Omega$ denote the circle $(D{D}^{\prime }X)$; clearly, its centre lies on the perpendicular bisector of $D{D}^{\prime }$, i.e., on $BC$. We aim to prove that $\Omega$ passes through $P$; that will yield that $\Omega ={\Omega }_1$, which establishes what we are aimed to prove. Denote $Z=XB\cap YC$. Since $\angle CBZ=\angle BAC=\angle ZCB$, we have $ZB=ZC$, and hence $Z$ lies on the perpendicular bisector $D{D}^{\prime }$ of $BC$. By similarity, we get $BX/BC=BC/CY$, or $B{C}^2=BX\cdot CY=BX\cdot (ZY+CZ)$. Since $CY\parallel BP$, the triangles $XZY$ and $XBP$ are similar, so $BX\cdot ZY=ZX\cdot BP$. Therefore, $B{D}^2=B{C}^2=BX\cdot ZY+BX\cdot CZ=ZX\cdot BP+(BZ+ZX)\cdot BZ=ZX\cdot (BP+BZ)+B{Z}^2$. On the other hand, let $M$ be the midpoint of $BC$, and let $XB$ cross $\Omega$ again at $P^{\prime }$. Write the power of point $Z$ with respect to $\Omega$ as $XZ\cdot (P^{\prime }B+BZ)=XZ\cdot P^{\prime }Z=ZD\cdot Z{D}^{\prime }=M{Z}^2-D{M}^2=B{Z}^2-M{B}^2-D{M}^2=B{Z}^2-B{D}^2$. The two obtained relations yield $$ZX\cdot (BP+BZ)=B{D}^2-B{Z}^2=ZX\cdot (P^{\prime }B+BZ),$$ so $BP=P^{\prime }B$, and so $P$ and $P^{\prime }$ are reflections of one another in the line $BC$. Thus, $P$ lies on $\Omega$, as desired.

Remarks. (1) It is also possible to solve the problem via the moving points method. Introduce the points $D$ and $D^{\prime }$ as in Solution 2, and introduce the reflections $X^{\prime }$, $Y^{\prime }$, $P^{\prime }$, and $Q^{\prime }$ of $X$, $Y$, $P$, and $Q$ in the line $BC$, respectively, as in Solution 1, to read ${\Omega }_1$ and ${\Omega }_2$ as the circles $PX{X}^{\prime }{P}^{\prime }$ and $QY{Y}^{\prime }{Q}^{\prime }$, respectively. We need to show that $D$ lies on ${\Omega }_2$ (the other incidences are similar). To this end, it suffices to check that $\angle YDQ=\angle Y{Y}^{\prime }Q=90^{\circ }-\angle Y^{\prime }CB=90^{\circ }-\angle BAC$. Fix $B$, $C$, and the circle $ABC$. As $A$ varies on that circle, the lines $BX$, $CY$, $BP$, and $CQ$ remain constant, and $X$ and $Y$ depend projectively on $A$. Choosing $Q_1$ on $CQ$ such that $\angle YD{Q}_1=90^{\circ }-\angle BAC$, we need to show that $Q_1=Q$, or that $X$, $Y$, and $Q_1$ are collinear. The point $Q_1$ also depends projectively on $A$, so it suffices to check that the points $Q_1$, $X$, and $Y$ are collinear for four specific positions of $A$.

(2) Although inversion, homothety, and the moving point method are essentially the same thing, i.e., a Möbius transformation, we briefly sketch yet another approach below: Let $BP\cap CQ=D$, $BX\cap CY=Z$, and $DZ\cap XY=T$. Let $R_1$ and $R_2$ be distinct points such that triangles $R_{1}BC$ and $R_{2}BC$ are equilateral. We first note that $BZ\parallel CD$ and $CZ\parallel BD$, so the triangles $DPQ$ and $ZYX$ are homothetic from $T$. Hence, $TP\cdot TX=TQ\cdot TY$, so $T$ lies on the radical axis of ${\Omega }_1$ and ${\Omega }_2$. Since their centres both lie on $BC$, their radical axis is the perpendicular from $T$ to $BC$, i.e. the perpendicular bisector $DZ$ of $BC$. As $CY\parallel DP$ and $BX\parallel DQ$, it follows that $BP=YZ\cdot \frac{BX}{XZ}$ and $CQ=XZ\cdot \frac{CY}{YZ}$. As triangles $BXC$ and $CYB$ are similar, this means that $BP\cdot CQ=BX\cdot CY=B{C}^2$. Perform an inversion about $\Omega$, the circle of radius $BC$ centred at $B$. Let $X^{\prime }$, $P^{\prime }$, and ${\Omega }_1^{\prime }$ denote the images of $X$, $P$ and ${\Omega }_1$. As $BX\cdot CY=BP\cdot CQ=B{C}^2$, we get $B{P}^{\prime }=CQ$ and $B{X}^{\prime }=CY$. Since lines $BD$, $CD$ and $BZ$, $CZ$ are symmetric about $DZ$, it follows that $X^{\prime }$, $P^{\prime }$ and $Y$, $Q$ are symmetric about $DZ$. Thus, ${\Omega }_1^{\prime }$ and ${\Omega }_2$ are reflections about $DZ$. Since ${\Omega }_1\cap {\Omega }_2$ and ${\Omega }_1^{\prime }\cap {\Omega }_2$ all lie on $DZ$, it follows that $\Omega$, ${\Omega }_1$, ${\Omega }_1^{\prime }$, ${\Omega }_2$ all meet on $DZ$. This implies that ${\Omega }_1$ and ${\Omega }_2$ both pass through $R_1$ and $R_2$, the two required fixed points.