Baltic Way shortlist

Let $B^{\prime }$ be the reflection of $A$ through $B$. Since the perimeter of $△ABC$ equals the length of the segment $AX$, $AB$ is less than half of $AX$ and, therefore, $B^{\prime }$ lies on the segment $AX$. Let the point $C^{\prime }$ lie on $AY$ such that $B^{\prime }{C}^{\prime }$ touches $\omega$ in the point $Z$. Let $C^{\prime \prime }$ be the midpoint of $A{C}^{\prime }$. Since $B$ and $C^{\prime \prime }$ are midpoints of the sides $A{B}^{\prime }$ and $A{C}^{\prime }$, respectively, we have that the perimeter of $△A{B}^{\prime }{C}^{\prime }$ is double that of $△AB{C}^{\prime \prime }$. We also have that the perimeter of $△A{B}^{\prime }{C}^{\prime }$ equals $$\begin{array}{rl}|A{B}^{\prime }|+|B^{\prime }{C}^{\prime }|+|C^{\prime }A|& =|A{B}^{\prime }|+|B^{\prime }Z|+|Z{C}^{\prime }|+|C^{\prime }A|\\ & =|A{B}^{\prime }|+|B^{\prime }X|+|Y{C}^{\prime }|+|C^{\prime }A|\\ & =|AX|+|AY|\\ & =2|AX|.\end{array}$$ Therefore, the perimeter of triangles $△ABC$ and $△AB{C}^{\prime \prime }$ is the same, namely $|AX|$.



If we assume that $C^{\prime \prime }$ lies between $A$ and $C$ we have that $|B{C}^{\prime \prime }|+|A{C}^{\prime \prime }|=|BC|+|AC|$, so $$|B{C}^{\prime \prime }|=|BC|+|C{C}^{\prime \prime }|.$$ Which contradicts the triangle inequality, so $C^{\prime \prime }$ does not lie between $A$ and $C$. Similarly, $C^{\prime \prime }$ cannot lie between $C$ and $Y$, and must therefore lie on $C$. Hence, $C$ and $C^{\prime \prime }$ are the same point so $C$ is the midpoint of $A{C}^{\prime }$. Now, $\omega$ is tangent to the extensions of the sides $A{B}^{\prime }$ and $A{C}^{\prime }$ of $△A{B}^{\prime }{C}^{\prime }$ as well as being tangent to the side $B^{\prime }{C}^{\prime }$ and is therefore an excircle of the triangle. Also, $B$ and $C$ are the midpoints of sides $A{B}^{\prime }$ and $A{C}^{\prime }$, respectively. We have that the point $D$ lies on the line $B^{\prime }{C}^{\prime }$, for $D$, $B^{\prime }$ and $C^{\prime }$ are reflections of $A$ through points on $BC$. Also, since $BC\parallel B^{\prime }{C}^{\prime }$, and $AD\perp BC$, we have that $AD\perp B^{\prime }{C}^{\prime }$. So $D$ is the foot of the altitude from $A$ in $△A{B}^{\prime }{C}^{\prime }$. Thus, the circle through $B,D,C$ is the nine-point circle of $△A{B}^{\prime }{C}^{\prime }$. According to Feuerbach's theorem, it touches the excircle $\omega$, as required.