Saudi Arabian Mathematical Competitions

1) Consider the inversion with center $J$ and ratio equal to the power of $J$ to the three circles $(O_{1})$, $(O_{2})$, $(O_{3})$ as a function $f$.
It is easy to see that
$$f\left((O_{1})\right) = (O_{1}), \quad f\left((O_{2})\right) = (O_{2}), \quad f\left((O_{3})\right) = (O_{3}) .$$
On the other hand, $f(X) = A_1$, $f(Y) = B_1$, $f(Z) = C_1$, thus
$$f((XYZ)) = (A_1B_1C_1) = (O) .$$
Because $(O)$ is tangent to all three circles $(O_{1})$, $(O_{2})$, $(O_{3})$, then $(XYZ)$ is also tangent to $(O_{1})$, $(O_{2})$, $(O_{3})$, based on the property of inversion.

2) We will prove that $AA'$, $BB'$, $CC'$ are concurrent at the homothety center of $(O)$ and $(I)$. Indeed,
Suppose that $AA'$ intersects $IO$ at $M$. It is easy to see that $A$, $I$, $O_1$ are collinear and $O$, $O_1$, $A'$ are also collinear. Denote $r_A$ as the radius of $(O_1)$.

From the ratio of radii, we can see that:
$$\frac{\overline{AI}}{\overline{AO_1}} = \frac{r}{r_A} \text{ and } \frac{\overline{A'O}}{\overline{A'O_1}} = -\frac{R}{r_A} .$$
By applying Menelaus' theorem, we have
$$\frac{\overline{AI}}{\overline{AO_1}} \cdot \frac{\overline{A'O_1}}{\overline{A'O}} \cdot \frac{\overline{MO}}{\overline{MI}} = 1 \Leftrightarrow \frac{\overline{MO}}{\overline{MI}} = \frac{\overline{AO_1}}{\overline{AI}} \cdot \frac{\overline{A'O}}{\overline{A'O_1}} = \frac{r_A}{r} \left(-\frac{R}{r_A}\right) = -\frac{R}{r} .$$
Hence the line $AA'$ passes through the point $M$ that internally divides the segment $OI$ with ratio $\frac{R}{r}$. Similarly for $BB'$, $CC'$. Therefore, $AA'$, $BB'$, $CC'$ are concurrent at a point $M$ belonging to the segment $IO$. $\square$