International Mathematical Olympiad

Solution 1. If $A{A}_1=C{C}_1$, then the hexagon is symmetric about the line $B{B}_1$; in particular the circles $ABC$ and $A_{1}B{C}_1$ are tangent to each other. So $A{A}_1$ and $C{C}_1$ must be different. Since the points $A$ and $A_1$ can be interchanged with $C$ and $C_1$, respectively, we may assume $A{A}_1<C{C}_1$. Let $R$ be the radical center of the circles $AEBC$ and $A_{1}EB{C}_1$, and the circumcircle of the symmetric trapezoid $AC{C}_1{A}_1$; that is the common point of the pairwise radical axes $AC$, $A_1{C}_1$, and $BE$. By the symmetry of $AC$ and $A_1{C}_1$, the point $R$ lies on the common perpendicular bisector of $A{A}_1$ and $C{C}_1$, which is the external bisector of $\angle ADC$. Let $F$ be the second intersection of the line $DR$ and the circle $ACD$. From the power of $R$ with respect to the circles $\omega$ and $ACFD$ we have $RB\cdot RE=RA\cdot RC=RD\cdot DF$, so the points $B,E,D$ and $F$ are concyclic. The line $RDF$ is the external bisector of $\angle ADC$, so the point $F$ bisects the $\mathrm{arc}\overline{CDA}$. By $AB=BC$, on circle $\omega$, the point $B$ is the midpoint of $\mathrm{arc}\overline{AEC}$; let $M$ be the point diametrically opposite to $B$, that is the midpoint of the opposite $\mathrm{arc}\widetilde{CA}$ of $\omega$. Notice that the points $B,F$ and $M$ lie on the perpendicular bisector of $AC$, so they are collinear. Finally, let $X$ be the second intersection point of $\omega$ and the line $DE$. Since $BM$ is a diameter in $\omega$, we have $\angle BXM=90^{\circ }$. Moreover, $$\angle EXM=180^{\circ }-\angle MBE=180^{\circ }-\angle FBE=\angle EDF,$$ so $MX$ and $FD$ are parallel. Since $BX$ is perpendicular to $MX$ and $B{B}_1$ is perpendicular to $FD$, this shows that $X$ lies on line $B{B}_1$.

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Alternative solution.

Solution 2. Define point $M$ as the point opposite to $B$ on circle $\omega$, and point $R$ as the intersection of lines $AC$, $A_1{C}_1$ and $BE$, and show that $R$ lies on the external bisector of $\angle ADC$, like in the first solution. Since $B$ is the midpoint of the $\mathrm{arc}\overline{AEC}$, the line $BER$ is the external bisector of $\angle CEA$. Now we show that the internal angle bisectors of $\angle ADC$ and $\angle CEA$ meet on the segment $AC$. Let the angle bisector of $\angle ADC$ meet $AC$ at $S$, and let the angle bisector of $\angle CEA$, which is line $EM$, meet $AC$ at $S^{\prime }$. By applying the angle bisector theorem to both internal and external bisectors of $\angle ADC$ and $\angle CEA$, $$AS:CS=AD:CD=AR:CR=AE:CE=A{S}^{\prime }:C{S}^{\prime },$$ so indeed $S=S^{\prime }$. By $\angle RDS=\angle SER=90^{\circ }$ the points $R,S,D$ and $E$ are concyclic. Now let the lines $B{B}_1$ and $DE$ meet at point $X$. Notice that $\angle EXB=\angle EDS$ because both $B{B}_1$ and $DS$ are perpendicular to the line $DR$, we have that $\angle EDS=\angle ERS$ in circle $SRDE$, and $\angle ERS=\angle EMB$ because $SR\perp BM$ and $ER\perp ME$. Therefore, $\angle EXB=\angle EMB$, so indeed, the point $X$ lies on $\omega$.