Baltic Way 2024 · Problem 12

Triangles $X F B$ and $X E C$ are isosceles (Fig. 12). Note that $X$ lies on the shorter arc $A B$ of $\omega$ since otherwise the perpendicular bisector of $B X$ could not intersect side $A B$. Hence $\angle X B F=\angle X B A=\angle X C A=\angle X C E$ which implies that triangles $X F B$ and $X E C$ are similar. Therefore $\angle X F A=\angle X E A$. Hence $A E F X$ is a cyclic quadrilateral. Let $X F$ intersect $\omega$ again at $X^{\prime} \neq X$ (Fig. 13). By $\angle B X X^{\prime}=\angle B X F=\angle X B F=\angle X B A$, arcs $B X^{\prime}$ and $A X$ of $\omega$ are equal, i.e. $B X^{\prime}=A X$. Since $A X=A M$, it follows that $B X^{\prime}=A M$. So $A B X^{\prime} M$ is an isosceles trapezoid which implies also $A X^{\prime}=B M$, i.e. arcs $A X^{\prime}$ and $B M$ of $\omega$ are equal. Therefore

$$\angle A E F=180^{\circ}-\angle A X F=180^{\circ}-\angle A X X^{\prime}=180^{\circ}-\angle M A B=\angle M C B .$$

By considering the second intersection point of $X E$ with $\omega$, we can analgously prove the equality $\angle E F A=\angle C B M$. Since arcs $M B$ and $M C$ of $\omega$ are equal, $\angle M C B=\angle C B M$, so $\angle A E F=\angle E F A$. Hence $A E=A F$, as desired.

Triangles $X F B, X E C$ and $X A M$ are isosceles (Fig. 144. In addition, note that $\measuredangle F B X=\measuredangle A B X=\measuredangle A M X$ and $\measuredangle F B X=\measuredangle A B X=\measuredangle A C X=\measuredangle E C X$, from which it follows that triangles $X F B, X E C$ and $X A M$ are directly similar (i.e., similar with the same orientation). Denote $\frac{X F}{X B}=\frac{X E}{X C}=\frac{X A}{X M}=k$ and $\measuredangle B X F=\measuredangle C X E=\measuredangle M X A=\alpha$. Then the rotation with center $X$ by angle $\alpha$ along with scaling by ratio $k$ maps points $B, C, M$ to points $F, E, A$, respectively. Thus triangle $A F E$ is similar to triangle $M B C$ by spiral similarity. But $M B=M C$ as $M$ is the midpoint of arc $B C$. Hence also $A E=A F$. Remark: The problem can be approached using complex numbers, taking $\omega$ as a unit circle.

Figure 12

Figure 13

Figure 14