Given a unit square $ABCD$, $O$ is its center. Suppose point $P$ is on side $CD$, $AP$ intersects $OD$ at point $T$, $BP$ intersects $OC$ at point $R$, and $OP$ intersects $TR$ at point $K$.
(1) Try to find the maximum value of $S_{\triangle MPT}$ and the maximum value of $S_{\text{quadrilateral } RRP T}$;
(2) When point $P$ moves along side $CD$, try to find the locus of point $K$ (please specify its type and range).


Solution: (1) As shown in Figure 5, let $D P = x$, then $P C = 1 - x$.

Since $\angle A D T = \angle P D T = 45^{\circ}$, we have
$$
\frac{A T}{T P} = \frac{A D}{D P} = \frac{1}{x},
$$

which means $\frac{A P}{T P} = \frac{1 + x}{x}$.
Thus, we have
$$
\begin{array}{l}
S_{\triangle A P T} = \frac{T P}{A P} S_{\triangle N P-1} \\
= \frac{x}{1 + x} \cdot \frac{1}{2} S_{\triangle A C P} \text{ (since } O \text{ is the midpoint of } A C) \\
= \frac{1}{4} \cdot \frac{x(1 - x)}{1 + x}.
\end{array}
$$

Substituting $1 - x$ for $x$ in equation (1), we get
$$
\begin{array}{l}
= \frac{1}{4} x(1 - x)\left(\frac{1}{1 + x} + \frac{1}{2 - x}\right) \\
= \frac{3}{4} \cdot \frac{x(1 - x)}{2 + x - x^{2}} = \frac{3}{4}\left(1 - \frac{2}{-x^{2} + x + 2}\right) \\
= \frac{3}{4}\left[1 - \frac{2}{\frac{9}{4} - \left(x - \frac{1}{2}\right)^{2}}\right] \text{.} \\
\end{array}
$$

Since $0 \leqslant x \leqslant 1$, it is easy to see from equation (3) that when $x = \frac{1}{2}$,
$S_{\text{quadrilateral IRPT}}$ achieves its maximum value $\frac{1}{12}$.
Notice that
$$
\begin{array}{l}
S_{\triangle \text{IIT}} = \frac{1}{4}\left(1 - \frac{x^{2} + 1}{x + 1}\right) \\
= \frac{1}{4} - \frac{1}{4} \cdot \frac{(x + 1)^{2} - 2(x + 1) + 2}{x + 1} \\
= \frac{1}{4} - \frac{1}{4}\left(x + 1 - 2 + \frac{2}{x + 1}\right) \\
= \frac{3}{4} - \frac{1}{4}\left(x + 1 + \frac{2}{x + 1}\right) \\
= \frac{3}{4} - \frac{1}{4}\left[\left(\sqrt{x + 1} - \sqrt{\frac{2}{x + 1}}\right)^{2} + 2 \sqrt{2}\right] \\
= \frac{3}{4} - \frac{\sqrt{2}}{2} - \frac{1}{4}\left(\sqrt{x + 1} - \sqrt{\frac{2}{x + 1}}\right)^{2}.
\end{array}
$$

From equation (4), it is easy to see that when $\sqrt{x + 1} = \sqrt{\frac{2}{x + 1}}$, i.e., $x = \sqrt{2} - 1$, $S_{\triangle M p r}$ achieves its maximum value $\frac{3 - 2 \sqrt{2}}{4}$.
(2) From $\triangle A B R \sim \triangle C P R$ and $\triangle A B T \sim \triangle P D T$, we get $\frac{B R}{R P} = \frac{A B}{C P} = \frac{1}{1 - x}$, $\frac{T D}{B T} = \frac{P D}{A B} = \frac{x}{1} = x$.
Thus, $\frac{B D}{B T} = x + 1$, i.e.,
$B T = \frac{\sqrt{2}}{x + 1}$.
Therefore, $O T = \frac{\sqrt{2}}{x + 1} - \frac{\sqrt{2}}{2}$.
In $\triangle P O B$ and the transversal $T K R$, by Menelaus' theorem, we have $\frac{B R}{R P} \cdot \frac{P K}{K O} \cdot \frac{O T}{T B} = 1$.
Thus, $\frac{P K}{K O} = \frac{R P}{B R} \cdot \frac{T B}{O T} = (1 - x) \cdot \frac{\frac{\sqrt{2}}{x + 1}}{\frac{\sqrt{2}}{x + 1} - \frac{\sqrt{2}}{2}}$
$= (1 - x) \cdot \frac{1}{1 - \frac{1}{2}(x + 1)}$
$= (1 - x) \cdot \frac{2}{1 - x} = 2$ (a constant).
Now, take points $T_{1}$ and $R_{1}$ on $O D$ and $O C$ respectively, such that $\frac{O T_{1}}{T_{1} D} = \frac{O R_{1}}{R_{1} C} = \frac{1}{2}$, i.e.,
$$
O T_{1} = O R_{1} = \frac{1}{3} \times \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{6}.
$$

Connect $T_{1} R_{1}$. From equation (5), it is easy to see that point $K$ must lie on the line segment $T_{1} R_{1}$.

Therefore, as point $P$ moves along side $C D$, the trajectory of point $K$ is the line segment $T_{1} R_{1}$, whose length is $\frac{1}{3}$. The maximum value is $\frac{1}{12}$, and the trajectory of point $K$ is the line segment $T_{1} R_{1}$.