172 Acute triangle $ABC$ is inscribed in the unit circle $O, P$ and $Q$ are internal points on sides $AB, AC$ respectively, such that $\frac{AP}{PB}=2, \frac{AQ}{QC}=1$, and points $P, O, Q$ are collinear, then the range of the length of $AB$ is
$172(\sqrt{3}, 2)$. As shown in Figure 17, on the one hand, $Q$ is the midpoint of $A C$, so $O Q \perp A C . \angle A O P$ is an obtuse angle. Let $A B=2 x$, then the perpendicular distance from the center to the chord $A B$ $=\sqrt{1-x^{2}}$.
$$
O P^{2}=\left(1-x^{2}\right)+\left(x-\frac{2 x}{3}\right)^{2}=1-\frac{8}{9} x^{2} .
$$

Since $\angle A O P$ is an obtuse angle, $A P^{2}>O A^{2}+O P^{2}$, that is
$$
\left(\frac{4 x}{3}\right)^{2}>1+\left(1-\frac{8}{9} x^{2}\right) .
$$

Solving this, we get $2 x>\sqrt{3}$, which means $A B>\sqrt{3}$. Also, $A B$ should be less than the diameter of the unit circle, i.e., $A B<2$. Since $\angle A O P>90^{\circ}$, $O$ is between $P$ and $Q$, and $O$ is inside $\triangle A B C$, making $\triangle A B C$ an acute triangle. Therefore, the range of $A B$ is $(\sqrt{3}, 2)$.