1 Let $\angle X O Y=90^{\circ}, P$ be a point inside $\angle X O Y$, and $O P=1$. $\angle X O P=30^{\circ}$. Draw any line through point $P$ that intersects rays $O X$ and $O Y$ at points $M$ and $N$. Find the maximum value of $(M O + O N - M N)$. (Wang Jianwei)
Draw a circle $O_{1}$ passing through point $P$, tangent to the rays $O X$ and $O Y$ at points $A$ and $B$, respectively, such that point $P$ lies on the major arc $\overparen{A B}$.

Establish a rectangular coordinate system as shown in the figure. Given the conditions, the coordinates of point $P$ are $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$, and the coordinates of $O_{1}$ are $(a, a)$. Here, $a$ satisfies the equation
$$
\left(\frac{\sqrt{3}}{2}-a\right)^{2}+\left(\frac{1}{2}-a\right)^{2}=a^{2},
$$

which simplifies to $a=\frac{\sqrt{3}+1-\sqrt[1]{12}}{2}$ (taking the smaller root).

Since the angle is $30^{\circ}$, the line $l_{0}$ passing through point $P$ and tangent to circle $O_{1}$ (this is a tangent line) intersects both $OX$ and $OY$. Let the intersection points of $l_{0}$ with $OX$ and $OY$ be $M_{0}$ and $N_{1}$, respectively. It is easy to see that $()M_{0}+()N_{n}-M_{0}N_{0}=2OA$.

If line $l$ does not touch circle $O_{1}$ (but intersects it), draw a tangent line $l_{1}$ (with the tangent point $Q$) such that $l_{1} \parallel l$, and place point $Q$ on either side of line $l$. Let the intersection points of $l_{1}$ with $OX$ and $OY$ be $M_{1}$ and $N_{1}$, respectively. The length of the broken line $N_{1}NM_{1}$ is greater than the length of $M_{1}N_{1}$. By the tangent segment theorem, we have
$$
\begin{aligned}
()M+()N-MN & < (OM_{1} + (ON_{1} - M_{1}N_{1}) \\
& = ((M) + BN_{1}) + (OA + AM_{1}) - (N_{1}Q + QM_{1}) \\
& = 2OA.
\end{aligned}
$$

Therefore, when the line $l$ passing through point $P$ is tangent to circle $O_{1}$, $OM + ON - MN$ achieves its maximum value. The maximum value is $2OA = 2a = \sqrt{3} + 1 - \sqrt[1]{12}$.