SAUDI ARABIAN MATHEMATICAL COMPETITIONS

1) We use inversion to solve this problem. Suppose that $AI$, $A_{1}I$ intersect $(O)$ at $A_0$, $A_1$. Because $AI$ is the bisector then $A_0$ is the midpoint of the minor $\mathrm{arc}BC$. Based on the property of the Mixtilinear circle, we also have $A_1$ as the midpoint of the major arc $BC$. Consider the inversion of center $I$ and ratio equal to the power of $I$ to $(O)$ as the function $f$. We have $f(A)=A_0$, $f(A_1)=A_2$ then $f(A{A}_1)=(I{A}_0{A}_2)$. Define $B_0$, $B_2$, $C_0$, $C_2$ similarly then $f(B{B}_1)=(I{B}_0{B}_2)$, $f(C{C}_1)=(I{C}_0{C}_2)$. It is easy to see that three circles $(I{A}_0{A}_2)$, $(I{B}_0{B}_2)$ and $(I{C}_0{C}_2)$ share the common point $I$. On the other hand, the power of $O$ to the three circles is also equal to $-R^2$ where $R$ is the radius of the circumcircle. Hence, the three circles have two common points and one of them is $I$ which is the center of inversion. Then $A{A}_1$, $B{B}_1$, $C{C}_1$ are concurrent at a point $M$ and $M$, $I$, $O$ are collinear.

2) From the property of the Mixtilinear circle, we have $B_a{B}_c$ and $C_a{C}_b$ have the common midpoint $I$ then $B_a{C}_a{B}_c{C}_b$ is a parallelogram, which implies that $B_a{C}_a$ is parallel to $BC$ and the distance from $I$ to $B_a{C}_a$ and $BC$ are the same. Hence $B_a{C}_a$ is tangent to $(I)$. Similarly, we also have $A_b{C}_b$ and $B_c{A}_c$ are also tangent to $(I)$. It is easy to see that $C_b{B}_c$ coincides with $BC$ then it is tangent to $(I)$. Similarly, we also have $C_a{A}_c$ and $A_b{B}_a$ are tangent to $(I)$. Hence, the 6 sides of the hexagon $C_b{B}_c{A}_c{C}_a{B}_a{A}_b$ are tangent to the circle $(I)$, which finishes the solution.