1. Let the circumcenter, incenter, and orthocenter of non-isosceles $\triangle ABC$ be $O$, $I$, and $H$, respectively, with the circumradius being $1$ and $\angle A=60^{\circ}$. Then the circumradius of $\triangle OIH$ is $\qquad$.
If $\triangle A B C$ is an acute triangle, since
$$
\angle B O C=\angle B I C=\angle B H C=120^{\circ} \text {, }
$$

then, $O$, $I$, $H$, $B$, and $C$ are concyclic;
if $\triangle A B C$ is an obtuse triangle, since
$$
\angle B O C=\angle B I C=120^{\circ}, \angle B H C=60^{\circ} \text {, }
$$

and $H$ is on the opposite side of $B C$ from $O$, $I$, and $A$, then, $O$, $I$, $H$, $B$, and $C$ are concyclic.

Therefore, the circumradius of $\triangle O I H$ is the same as the circumradius of $\triangle B H C$, which equals the circumradius of $\triangle A B C$, which is 1.