4. Given that the circumcenter, incenter, and orthocenter of a non-isosceles acute $\triangle ABC$ are $O, I, H$ respectively, and $\angle A=60^{\circ}$. If the altitudes of $\triangle ABC$ are $AD, BE, CF$, then the ratio of the circumradius of $\triangle OIH$ to the circumradius of $\triangle DEF$ is $\qquad$ .
4. 2 .

From $\angle B O C=\angle B I C=\angle B H C=120^{\circ}$, we know that $O, I, H, B, C$ are concyclic.

Let the circumradii of $\triangle A B C$ and $\triangle O I H$ be $R$ and $r$, respectively.
By the Law of Sines, we have
$2 R \sin A=B C=2 r \sin \angle B O C$.
Thus, $r=R$.
Let the circumradius of $\triangle D E F$ be $r^{\prime}$.
Noting that $A, E, H, F$ are concyclic, and $A H$ is the diameter of this circle. By the Law of Sines, we get
$$
E F=A H \sin A=2 R \cos A \cdot \sin A=\frac{\sqrt{3}}{2} R .
$$

Since $A, F, D, C$ and $A, E, D, B$ are concyclic, respectively, we have
$\angle B D F=\angle C D E=\angle B A C=60^{\circ}$.
Thus, $\angle F D E=60^{\circ}$.
By the Law of Sines, we get
$$
E F=2 r^{\prime} \sin \angle F D E=\sqrt{3} r^{\prime} \text {. }
$$

Therefore, $r^{\prime}=\frac{R}{2}$.
Hence, $\frac{r}{r^{\prime}}=2$.
[Note] The circumcircle of $\triangle D E F$ is actually the nine-point circle of $\triangle A B C$, so $r^{\prime}=\frac{R}{2}$.