THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

Since $b\cos C=A_{1}C$ and $c\cos B=B{A}_1$, we deduce that $\frac{C_{1}X}{X{B}_1}=\frac{C{A}_1}{A_{1}B}$ or $\frac{C_{1}X}{C_1{B}_1}=\frac{C{A}_1}{CB}$ or, equivalently, $\frac{C_{1}X}{C{A}_1}=\frac{C_1{B}_1}{CB}$. (1)

Triangles $A{C}_1{B}_1$ and $ACB$ being similar, we obtain $\frac{C_1{B}_1}{CB}=\frac{A{C}_1}{AC}$, hence $\frac{C_{1}X}{C{A}_1}=\frac{A{C}_1}{AC}$. Also, $\angle A{C}_{1}X=\angle AC{A}_1$, so it follows that triangles $A{C}_{1}X$ and $AC{A}_1$ are similar, and we deduce that $AX\perp B_1{C}_1$.

On the other hand, the tangent at $A$ to the circumcircle of the triangle $ABC$, centered at $O$, is parallel to $B_1{C}_1$, and hence $O$ lies on $AX$.

Similarly, $O$ lies on $BY$ and $CZ$ as well, therefore the lines $AX$, $BY$ and $CZ$ are concurrent.