161 Given that $P Q R S$ is a square, there is a point $O$ on the line $R Q$ such that the distance from $O$ to point $P$ is 1, then the maximum possible value of the distance from $O$ to point $S$ $=$ $\qquad$
$161 \frac{1+\sqrt{5}}{2}$. Let the side length of the square be $x$. By the Pythagorean theorem, we have
$$
\begin{array}{c}
O Q=\sqrt{1-x^{2}}, O R=x+\sqrt{1-x^{2}}, \\
O S=\sqrt{x^{2}+\left(x+\sqrt{1-x^{2}}\right)^{2}}=\sqrt{1+\left(x^{2}+2 x \cdot \sqrt{1-x^{2}}\right)} .
\end{array}
$$

Let $y=x^{2}+2 x \cdot \sqrt{1-x^{2}}$, we only need to find the maximum value of $y$. Also,
$$
\begin{aligned}
y & =\cos ^{2} \theta+2 \cos \theta \cdot \sin \theta=\sin 2 \theta+\frac{1+\cos 2 \theta}{2} \\
& =\sin 2 \theta+\frac{1}{2} \cdot \cos 2 \theta+\frac{1}{2} \leqslant \sqrt{1+\left(\frac{1}{2}\right)^{2}}+\frac{1}{2} \\
& =\frac{1+\sqrt{5}}{2},
\end{aligned}
$$

the equality holds if and only if $\operatorname{tg} \theta=\frac{1}{2}$ (at this time $\left.x=\sqrt{\frac{5+\sqrt{5}}{10}}\right)$, so the maximum distance from $S$ to $O$ is
$$
\sqrt{\frac{1+\sqrt{5}}{2}+1}=\frac{1+\sqrt{5}}{2} .
$$