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 $$f\left((O_1)\right)=(O_1),f\left((O_2)\right)=(O_2),f\left((O_3)\right)=(O_3).$$ On the other hand, $f(X)=A_1$, $f(Y)=B_1$, $f(Z)=C_1$, thus $$f((XYZ))=(A_1{B}_1{C}_1)=(O).$$ 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 $A{A}^{\prime }$, $B{B}^{\prime }$, $C{C}^{\prime }$ are concurrent at the homothety center of $(O)$ and $(I)$. Indeed, Suppose that $A{A}^{\prime }$ intersects $IO$ at $M$. It is easy to see that $A$, $I$, $O_1$ are collinear and $O$, $O_1$, $A^{\prime }$ are also collinear. Denote $r_A$ as the radius of $(O_1)$.



From the ratio of radii, we can see that: $$\frac{\overline{AI}}{\overline{A{O}_1}}=\frac{r}{r_A}\text{ and }\frac{\overline{A^{\prime }O}}{\overline{A^{\prime }{O}_1}}=-\frac{R}{r_A}.$$ By applying Menelaus' theorem, we have $$\frac{\overline{AI}}{\overline{A{O}_1}}\cdot \frac{\overline{A^{\prime }{O}_1}}{\overline{A^{\prime }O}}\cdot \frac{\overline{MO}}{\overline{MI}}=1\iff \frac{\overline{MO}}{\overline{MI}}=\frac{\overline{A{O}_1}}{\overline{AI}}\cdot \frac{\overline{A^{\prime }O}}{\overline{A^{\prime }{O}_1}}=\frac{r_A}{r}\left(-\frac{R}{r_A}\right)=-\frac{R}{r}.$$ Hence the line $A{A}^{\prime }$ passes through the point $M$ that internally divides the segment $OI$ with ratio $\frac{R}{r}$. Similarly for $B{B}^{\prime }$, $C{C}^{\prime }$. Therefore, $A{A}^{\prime }$, $B{B}^{\prime }$, $C{C}^{\prime }$ are concurrent at a point $M$ belonging to the segment $IO$. $\square$