66th Belarusian Mathematical Olympiad

(Solution by A. Goloubitskaya.) Let $\Gamma$ be the circumcircle of the given regular hexagon. It is evident that $B$ is the center of $\Gamma$. Since all sides of the regular hexagon are equal, we have $2\alpha =\text{arc }{A}_1{A}_2=\text{arc }{A}_2{A}_3=\cdots =\text{arc }{A}_9{A}_{10}=\text{arc }{A}_{10}{A}_1=36^{\circ }$. Since $A_1{A}_2=A_9{A}_{10}$, $\angle A_{10}{A}_2{A}_1=\angle A_1{A}_9{A}_{10}=\alpha$ and $\angle A_2{A}_1{A}_9=\angle A_2{A}_{10}{A}_9=7\alpha$, it follows that the triangles $A_1{A}_{2}C$ and $A_{10}{A}_{9}C$ are equal, so $C{A}_1=C{A}_{10}$. Hence the line $BC$ is the bisector of the segment $A_1{A}_{10}$. Then the line $BC$ contains the altitude, the median, and the bisectrix of the isosceles triangle $A_{10}B{A}_1$ ($B{A}_{10}=B{A}_1$). Hence, $\angle A_{1}BC=0.5\text{ arc }{A}_{10}{A}_1=\alpha =18^{\circ }$. Similarly, $\angle A_{2}BA=0.5\text{ arc }{A}_{2}B{A}_4=0.5\text{ arc }{A}_2{A}_4=2\alpha =36^{\circ }$. Since $\angle A_{2}B{A}_1=2\alpha =72^{\circ }$, we have $$\angle ABC=\angle AB{A}_2+\angle A_{2}B{A}_1+\angle A_{1}BC=36^{\circ }+36^{\circ }+18^{\circ }=90^{\circ }.$$

Since $\angle A{A}_{2}C=\angle A_5{A}_2{A}_{10}=5\alpha =90^{\circ }$, we have $\angle ABC+\angle A{A}_{2}C=180^{\circ }$, so the points $A$, $B$, $C$, $A_2$ are concyclic. Then $\angle BCA=\angle B{A}_{2}A=2\alpha =36^{\circ }$ and $\angle CAB=180^{\circ }-\angle ABC-\angle BCA=54^{\circ }$.