69th Belarusian Mathematical Olympiad

Since $\angle B{B}_{1}C=\angle C{C}_{1}B=90^{\circ }$, the points $B,C_1,B_1$ and $C$ lie on the circle $\omega$ with the diameter $BC$. Hence the bisectors of angles $\angle B{B}_{1}C$ and $\angle C{C}_{1}B$ pass through the midpoint of the arc $BC$ of $\omega$, so this midpoint is $X$.



First we prove that the quadrilateral $E{C}_1{B}_{1}D$ is cyclic. The angle $\angle BD{B}_1$ is an external angle of the triangle $C{B}_{1}D$, hence $\angle BD{B}_1=\angle ACB+45^{\circ }$. Since $\angle B{B}_{1}C$ is right, $\angle B_{1}BC=90^{\circ }-\angle ACB$. Thus, $$\angle B_1{C}_{1}E=\angle B_1{C}_{1}C+45^{\circ }=135^{\circ }-\angle ACB.$$ $$\angle B_1{C}_{1}E+\angle ED{B}_1=180^{\circ }.$$ Now we prove that the intersection points of the circumcircles of the triangles $A{B}_{1}D$ and $A{C}_{1}E$ lie on the line $AX$. Let the circumcircle of the triangle $A{B}_{1}D$ intersect the line $AX$ at points $A$ and $P$, then $XP\cdot XA=XD\cdot X{B}_1$. Since points $E,C_1,B_1$ and $D$ are concyclic, $XE\cdot X{C}_1=XD\cdot A{B}_1$. Hence $XE\cdot X{C}_1=XP\cdot XA$, and therefore the circumcircle of the triangle $A{C}_{1}E$ passes through $P$.

Finally, we will prove that the circumcircles of the triangles $BEX$ and $CDX$ passes through points $X$ and $P$. Since point $X$ is the midpoint of the arc $BC$, $XB=XC$ and $\angle BXC=90^{\circ }$. Therefore, $\angle BCX=\angle CBX=45^{\circ }$. Since the quadrilateral $A{B}_{1}DP$ is cyclic, $\angle APD=180^{\circ }-\angle D{B}_{1}A=45^{\circ }$. Hence $\angle DPX+\angle DCX=180^{\circ }$, i.e. points $P,D,C$ and $X$ are concyclic. Similarly, the quadrilateral $BEPX$ is cyclic. If point $P$ is distinct from point $X$, the problem is solved. If points $P$ and $X$ coincide, the equalities $\angle DCX=\angle APD$ and $\angle EBX=\angle APE$ imply that the circumcircles of the triangles $BEX$ and $CDX$ are tangent to the line $AX$.

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Alternative solution.

Note that $B{C}_1{B}_{1}C$ is concyclic, and let $\omega ,{\omega }_B$ and ${\omega }_C$ be the circumcircles of $B{C}_1{B}_{1}C$, $BEX$, and $CDX$, respectively. Since $B_{1}X$ and $C_{1}X$ are the bisectors of $\angle B{B}_{1}C$ and $\angle C{C}_{1}B$, they both pass through the midpoint of the arc $BC$ of $\omega$ not containing $B_1$ and $C_1$, thus this midpoint is $X$. Let $AX$ intersect $\omega ,{\omega }_B$ and ${\omega }_C$ the second time at points $T,F_B$ and $F_C$, respectively, and denote $AX\cap BC=P$.

As in the first solution, note that $E{C}_1{B}_{1}D$ is cyclic. Let $\gamma ,{\gamma }_B$, and ${\gamma }_C$ be the circumcircles of $E{C}_1{B}_{1}D$, $BE{C}_1$, and $CD{B}_1$, respectively. Since $C_{1}E$ and $B_{1}D$ are radical axes of ${\gamma }_B,\gamma$ and ${\gamma }_B,\gamma$ and $C_{1}E\cap B_{1}D=X$, the radical axis of ${\gamma }_B,{\gamma }_C$ passes through $X$. On the other hand, it passes through $A$ since ${\text{Pow}}_{{\gamma }_B}A=A{C}_1\cdot AB=A{B}_1\cdot AC={\text{Pow}}_{{\gamma }_C}A$. Thus, $AX$ is the radical axis of ${\gamma }_B,{\gamma }_C$, and it follows from $P\in AX$ that $$P{F}_A\cdot PX=PD\cdot PC=PE\cdot PB=P{F}_B\cdot PX.$$ Consequently, $F_A=F_B$ and the problem is solved.