BMO 2017

Lemma. If $BCD$ is the isosceles triangle which is outside the triangle $ABC$ and has $$\angle CBD=\angle BCD=90^{\circ }-\alpha :=\beta ,$$ then $AD\perp B{A}_C$.

Proof of the lemma. Construct an isosceles triangle $ABE$ outside the triangle $ABC$, so that $\angle ABE=\angle AEB=\beta$. Then $AE=AB=A{B}_A$ and $\angle EA{B}_A=\alpha$, so a rotation of center $A$ and angle $\alpha$ sends $C_A$ to $C$ and $B_A$ to $E$, hence $\angle (\overrightarrow{B_A{C}_A},\overrightarrow{EC})=\alpha$ (the angle between vectors is considered oriented). Also triangles $EBA$ and $BCD$ are similar, so a rotation of center $B$ and angle $\beta$, followed by a dilation of ratio $\frac{EB}{AB}=\frac{BC}{BD}$ sends $E$ to $A$ and $C$ to $D$, hence $\angle (\overrightarrow{EC},\overrightarrow{AD})=\beta$ (also oriented angle). This shows that $$\angle (\overrightarrow{B_A{C}_A},\overrightarrow{AD})=\angle (\overrightarrow{B_A{C}_A},\overrightarrow{EC})+\angle (\overrightarrow{EC},\overrightarrow{AD})=\alpha +\beta =90^{\circ }.■$$

Returning to the solution of the problem, denote $A^{\prime }$ the intersection of $BC$ with the perpendicular from $A$ to $B_A{C}_A$. Then $A^{\prime }$ belongs to the segment $BC$ and $$\frac{A^{\prime }B}{A^{\prime }C}=\frac{AB\sin (B+\beta )}{AC\sin (C+\beta )}.$$ Since similar relations are true for the intersections $B^{\prime },C^{\prime }$ of the other two perpendiculars with the opposite sides, this yields $$\frac{A^{\prime }B}{A^{\prime }C}\cdot \frac{B^{\prime }C}{B^{\prime }A}\cdot \frac{C^{\prime }A}{C^{\prime }B}=\frac{AB\sin (B+\beta )}{AC\sin (C+\beta )}\cdot \frac{BC\sin (C+\beta )}{BA\sin (A+\beta )}\cdot \frac{CA\sin (A+\beta )}{CB\sin (B+\beta )}=1,$$ whence the conclusion.