一、(50 points) As shown in Figure 4, in quadrilateral $ABCD$,
$AC, BD$
intersect at point
$O, AO=$
$OC, \quad BO$
$>DO$. Let
points $M, N$
be the midpoints of
$DO, BO$ respectively, draw $OP \parallel BC$, intersecting $AD$ at point $P$, and connect $PM, PN$. Try to determine: when $\angle ACB$ is what value,
$$
\angle MPO \leqslant \angle NPO ?
$$
One, as shown in Figure 8, take the midpoints $S, T$ of $AO, OC$ respectively, and connect $SM, NT$.
Then $SM \parallel AD, NT \parallel BC$.
Also, since $M, N$ are the midpoints of $DO, BO$ respectively, $S, T$ are the midpoints of $AO, CO$ respectively.

Let the intersection of lines $SM, NT$ be point $Q$, and $SQ$ intersects $OP$ at point $E$. Since $S$ is the midpoint of $AO$, $E$ is the midpoint of $OP$. Since $TQ \parallel OP$, and $O$ is the midpoint of $ST$, $E$ is the midpoint of $SQ$. Therefore, connecting $PQ$, we have $PQ \parallel AC$.
Construct $OF \parallel NP$, intersecting $MP$ at $F$, and connect $EF$, then $\angle OPN = \angle EOF$.
Since $OF \parallel NP$; and $PO \parallel NQ$, we have
$$
\begin{array}{l}
\frac{MF}{FP} = \frac{MO}{ON} = \frac{ME}{EQ} \Rightarrow EF \parallel PQ \\
\Rightarrow \angle OEF = \angle EOC = \angle OCB \\
\Rightarrow \angle EPF \leqslant \angle EOF \xlongequal{\text { in } \triangle OP \text { }} OF \leqslant PF .
\end{array}
$$

Since $OE = EP$, and $EF$ is a common side, it is equivalent to $\angle OEF \leqslant \angle PEF$.

Since $\angle OEF + \angle PEF = 180^{\circ}$, it is equivalent to $\angle OEF \leqslant 90^{\circ}$, i.e., $\angle OCB \leqslant 90^{\circ}$.
Therefore, when and only when $\angle ACB \leqslant 90^{\circ}$,
$\angle MPO \leqslant \angle NPQ$.
Two, first prove: if $i^{2} = -1$, then
$$
\begin{array}{l}
\prod_{k=1}^{n-1}\left(x - \cot \frac{k \pi}{n}\right) \\
= \frac{i}{2n}\left[(x - i)^{n} - (x + i)^{n}\right].
\end{array}
$$
$$
\text{Let } P(x) = \frac{i}{2n}\left[(x - i)^{n} - (x + i)^{n}\right].
$$

Then $P(x)$ is an $(n-1)$-degree polynomial, with the leading coefficient
$$
\frac{i}{2n}\left[\mathrm{C}_{n}^{1}(-i) - \mathrm{C}_{n}^{1} i\right] = 1.
$$

When $x = \cot \frac{k \pi}{n} (1 \leqslant k \leqslant n-1)$,
$$
\begin{array}{l}
(x + i)^{n} = \left(\cot \frac{k \pi}{n} + i\right)^{n} \\
= \left(\csc \frac{k \pi}{n}\right)^{n}\left(\cos \frac{k \pi}{n} + i \sin \frac{k \pi}{n}\right)^{n} \\
= \left(\csc \frac{k \pi}{n}\right)^{n}\left[\left(\cos \frac{\pi}{n} + i \sin \frac{\pi}{n}\right)^{n}\right]^{k} \\
= \left(\csc \frac{k \pi}{n}\right)^{n}(-1)^{k} \\
= \left(\csc \frac{k \pi}{n}\right)^{n}\left(\cos \frac{k \pi}{n} - i \sin \frac{k \pi}{n}\right)^{n} \\
= (x - i)^{n}.
\end{array}
$$

Therefore, $P\left(\cot \frac{k \pi}{n}\right) = 0$.
By the factor theorem,
$$
P(x) = \prod_{k=1}^{n-1}\left(x - \cot \frac{k \pi}{n}\right).
$$

In equation (1), let $x = 2 \cot \frac{\pi}{n} + i$. Then
$$
\begin{array}{l}
\prod_{k=1}^{n-1}\left(2 \cot \frac{\pi}{n} - \cot \frac{k \pi}{n} + i\right) \\
= \frac{i}{2n}\left[\left(2 \cot \frac{\pi}{n}\right)^{n} - \left(2 \cot \frac{\pi}{n} + 2i\right)^{n}\right] \\
= \frac{2^{n-1} i}{n}\left[\left(\cot \frac{\pi}{n}\right)^{n} - \left(\cot \frac{\pi}{n} + i\right)^{n}\right] \\
= \frac{2^{n-1} i}{n}\left(\csc \frac{\pi}{n}\right)^{n} \left[\left(\cos \frac{\pi}{n}\right)^{n} - \left(\cos \frac{\pi}{n} + i \sin \frac{\pi}{n}\right)^{n}\right] \\
= \frac{2^{n-1} i}{n}\left(\csc \frac{\pi}{n}\right)^{n} \left[\left(\cos \frac{\pi}{n}\right)^{n} + 1\right] \\
= \frac{2^{n-1} i}{n}\left[\left(\cot \frac{\pi}{n}\right)^{n} + \left(\csc \frac{\pi}{n}\right)^{n}\right].
\end{array}
$$