Subfield: Geometry


Question: Let $A B C C_{1} B_{1} A_{1}$ be a convex hexagon such that $A B=B C$, and suppose that the line segments $A A_{1}, B B_{1}$, and $C C_{1}$ have the same perpendicular bisector. Let the diagonals $A C_{1}$ and $A_{1} C$ meet at $D$, and denote by $\omega$ the circle $A B C$. Let $\omega$ intersect the circle $A_{1} B C_{1}$ again at $E \neq B$. Prove that the lines $B B_{1}$ and $D E$ intersect on $\omega$.

Solution:
If $A A_{1}=C C_{1}$, then the hexagon is symmetric about the line $B B_{1}$; in particular the circles $A B C$ 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 $A E B C$ and $A_{1} E B C_{1}$, and the circumcircle of the symmetric trapezoid $A C C_{1} A_{1}$; that is the common point of the pairwise radical axes $A C, A_{1} C_{1}$, and $B E$. By the symmetry of $A C$ 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 $\angle A D C$.

Let $F$ be the second intersection of the line $D R$ and the circle $A C D$. From the power of $R$ with respect to the circles $\omega$ and $A C F D$ we have $R B \cdot R E=R A \cdot R C=R D \cdot D F$, so the points $B, E, D$ and $F$ are concyclic.

The line $R D F$ is the external bisector of $\angle A D C$, so the point $F$ bisects the arc $\overparen{C D A}$. By $A B=B C$, on circle $\omega$, the point $B$ is the midpoint of arc $\overparen{A E C}$; let $M$ be the point diametrically opposite to $B$, that is the midpoint of the opposite $\operatorname{arc} \overparen{C A}$ of $\omega$. Notice that the points $B, F$ and $M$ lie on the perpendicular bisector of $A C$, so they are collinear.

<img_3936>

Finally, let $X$ be the second intersection point of $\omega$ and the line $D E$. Since $B M$ is a diameter in $\omega$, we have $\angle B X M=90^{\circ}$. Moreover,

$$
\angle E X M=180^{\circ}-\angle M B E=180^{\circ}-\angle F B E=\angle E D F,
$$

so $M X$ and $F D$ are parallel. Since $B X$ is perpendicular to $M X$ and $B B_{1}$ is perpendicular to $F D$, this shows that $X$ lies on line $B B_{1}$.
Define point $M$ as the point opposite to $B$ on circle $\omega$, and point $R$ as the intersection of lines $A C, A_{1} C_{1}$ and $B E$, and show that $R$ lies on the external bisector of $\angle A D C$, like in the first solution.

Since $B$ is the midpoint of the arc $\overparen{A E C}$, the line $B E R$ is the external bisector of $\angle C E A$. Now we show that the internal angle bisectors of $\angle A D C$ and $\angle C E A$ meet on the segment $A C$. Let the angle bisector of $\angle A D C$ meet $A C$ at $S$, and let the angle bisector of $\angle C E A$, which is line $E M$, meet $A C$ at $S^{\prime}$. By applying the angle bisector theorem to both internal and external bisectors of $\angle A D C$ and $\angle C E A$,

$$
A S: C S=A D: C D=A R: C R=A E: C E=A S^{\prime}: C S^{\prime}
$$

so indeed $S=S^{\prime}$.

By $\angle R D S=\angle S E R=90^{\circ}$ the points $R, S, D$ and $E$ are concyclic.

<img_3964>

Now let the lines $B B_{1}$ and $D E$ meet at point $X$. Notice that $\angle E X B=\angle E D S$ because both $B B_{1}$ and $D S$ are perpendicular to the line $D R$, we have that $\angle E D S=\angle E R S$ in circle $S R D E$, and $\angle E R S=\angle E M B$ because $S R \perp B M$ and $E R \perp M E$. Therefore, $\angle E X B=\angle E M B$, so indeed, the point $X$ lies on $\omega$.