SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Let $H$ be the orthocenter of triangle $ABC$.



We can see that $HA\cdot HX=HB\cdot HY=HC\cdot HZ=k$. We consider the inversion with center $H$ and ratio $-k$ as the function $f$.

It is easy to see that $$f(A)=X,f(B)=Y,f(C)=Z$$ so $f((O))=(O^{\prime })$ with $(O^{\prime })$ being the 9-point circle (which also passes through $X$, $Y$, $Z$).

On the other hand, because $\angle HEA=\angle HFA=90^{\circ }$, hence $H\in (O_1)$. After the inversion, the circles $(O_1)$ become a line passing through the images of $X$, $Y$; indeed, this line is $BC$ or $f((O_1))=BC$.

Similarly, we also have $f((O_2))=CA$ and $f((O_3))=AB$. So the circle $K$ that is tangent to $(O_1)$, $(O_2)$, $(O_3)$ will become the incircle $(I)$ of triangle $ABC$. But based on Feuerbach's theorem, the two circles $(O^{\prime })$, $(I)$ are tangent to each other.

Therefore, the circles $(K)$ and $(O)$ are also tangent to each other. $\square$