SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Let $H$ be the orthocenter of $△ABC$. Since $AH\perp X_1{X}_2$ and $A{X}_1=A{X}_2$, then $H{X}_1=H{X}_2$. Similarly, $H{Y}_1=H{Y}_2$ and $H{Z}_1=H{Z}_2$.



Denote $R$ as the radius of the three circles $({\omega }_A)$, $({\omega }_B)$, $({\omega }_C)$. As $HB\perp Y_1{C}^{\prime }$, $HC\perp Z_1{B}^{\prime }$, we have $$\begin{array}{rl}& H{Y}_1^2-R^2=H{Y}_1^2-B{Y}_1^2=H{C}^2-B{C}^2\\ & H{Z}_1^2-R^2=H{Z}_1^2-C{Z}_1^2=H{B}^{\prime 2}-C{B}^{\prime 2}\end{array}$$ In addition, $AH\perp B^{\prime }{C}^{\prime }$, so $$H{C}^{\prime 2}-B{C}^{\prime 2}=H{C}^{\prime 2}-A{C}^{\prime 2}=H{B}^{\prime 2}-A{B}^{\prime 2}=H{B}^{\prime 2}-C{B}^{\prime 2}$$ From that $H{Y}_1=H{Z}_1$. Similarly, $H{X}_1=H{Y}_1=H{Z}_1$ and $H{X}_2=H{Y}_2=H{Z}_2$.

From (1) and (2), the orthocenter $H$ is indeed the center of the circle which goes through the six points $X_1$, $X_2$, $Y_1$, $Y_2$, $Z_1$, $Z_2$.