[Example 3.2.1] Given a right angle $Q O P$, find points $A, B, C$ on $O P$, $O Q$, and inside the angle, respectively, such that $B C + C A = l$ (a fixed length), and maximize the area of quadrilateral $A C B O$.

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[Example 3.2.1] Given a right angle $Q O P$, find points $A, B, C$ on $O P$, $O Q$, and inside the angle, respectively, such that $B C + C A = l$ (a fixed length), and maximize the area of quadrilateral $A C B O$.
In this problem, $O A, O B, A B, A C, B C$ are all variables, with only $B C + C A = l$ being a constant. The variations of these variables all affect the area of quadrilateral $A C B D$.

We first fix $A B$ and let $O A, O B, A C, B C$ vary, thus transforming the problem into two issues related to the area:

Problem 1: In $\triangle A B C$, if $A B$ is a fixed value and $A C + B C = l$ is a fixed value, when is the area of $\triangle A B C$ maximized?

Problem 2: In $\triangle A B O$, if $A B$ is a fixed value and $\angle A O B = 90^{\circ}$ is a fixed angle, when is the area of $\triangle A B O$ maximized?

Problem 1 is an isoperimetric problem, and the conclusion is that the area of $\triangle A B C$ is maximized when $\triangle A B C$ is an isosceles triangle with $A B$ as the base (the two legs are each $\frac{l}{2}$). This is not difficult to prove.
Assume $\triangle A B C$ is an isosceles triangle, and $\triangle A B C^{\prime}$ is not an isosceles triangle. The common base $A B$ is of fixed length, and
$$
A C + C B = A C^{\prime} + C^{\prime} B.
$$

We need to prove that $S_{\triangle A B C} > S_{\triangle A B C^{\prime}}$.
Without loss of generality, assume $C$ and $C^{\prime}$ are on the same side of $A B$. Draw a line $l$ parallel to $A B$ through $C$.
We only need to prove that $C^{\prime}$ lies between the line $l$ and $A B$.
Otherwise, if $A C^{\prime}$ intersects $l$ at $M$, and draw the symmetric point $B^{\prime}$ of $B$ with respect to $l$, since $\triangle A B C$ is an isosceles triangle, then $A, C, B^{\prime}$ are collinear, and $C B^{\prime} = C B = A C$.
Connect $B M$ and $B^{\prime} M$, then we have
$$
\begin{aligned}
A C^{\prime} + C^{\prime} B & \geqslant A M + B M = A M + M B^{\prime} > A B^{\prime} \\
& = A C + C B^{\prime} = A C + C B,
\end{aligned}
$$

which contradicts $A C + C B = A C^{\prime} + C^{\prime} B$.
This solves Problem 1.
For Problem 2.
Since $S_{\triangle A O B} = \frac{1}{2} O A \cdot O B \leqslant \frac{1}{4} (O A^2 + O B^2) = \frac{1}{4} A B^2$,
the equality holds if and only if $O A = O B$, i.e., $\triangle A O B$ is an isosceles triangle, and the area is maximized.
Combining these two problems, we have:
When $O A = O B$ and $A C = C B = \frac{1}{2} l$, the area of quadrilateral $A C B O$ is maximized.
At this point, point $C$ lies on the angle bisector of $\angle P O Q$.
Next, we unfreeze $A B$, i.e., let $A B$ vary, and study the maximum value of the area of $A C B O$.
Since
$$
S_{A C B O} = 2 S_{\triangle A O C},
$$

the area of quadrilateral $A C B O$ is maximized when the area of $\triangle A O C$ is maximized.
The problem of maximizing the area of $\triangle A O C$ returns to Problem 2, because at this point $A C = \frac{1}{2} l$ is a fixed value, and $\angle A O C = 45^{\circ}$ is a fixed angle. By Problem 2, the area of $\triangle A O C$ is maximized if and only if $A O = O C$, thus the area of quadrilateral $A C B O$ is maximized.
From the above, we just need to construct the angle bisector of $\angle P O Q$, and then construct $\angle O A C = 67.5^{\circ}$, and we can also use the sine rule to find the length of $O A$
$$
\begin{array}{l}
\frac{O A}{\sin 67.5^{\circ}} = \frac{A C}{\sin 450^{\circ}}, \\
O A = \frac{\sqrt{2 + 2 \sqrt{2}}}{4}.
\end{array}
$$

At this point, quadrilateral $A C B O$ is one-fourth of a regular octagon inscribed in a circle with radius $O A$.
$$
A C = \frac{1}{2} l, \angle O B C = 67.5^{\circ}, \text{ and } C B = \frac{1}{2} l,
$$

we can obtain the quadrilateral $A C B O$.