Example II Construct $\triangle A P B, \triangle B Q C, \triangle C R A$ outside an equilateral $\triangle A B C$ with unit area, such that $\angle A P B = \angle B Q C = \angle C R A = 60^{\circ}$.
(1) Find the maximum area of $\triangle P Q R$;
(2) Find the maximum area of the triangle whose vertices are the incenters of $\triangle A P B, \triangle B Q C, \triangle C R A$.
(2004 Australian Mathematical Olympiad)
Analyzing the problem of the largest area can be transformed into the problem of an equilateral triangle with vertices on a given circle.
Proof (1) Since $\angle A P B=60^{\circ}$.
Therefore, point $P$ lies on the arc symmetrical to $A C B$ about $A B$, and this arc is within a circle that is concentric with the circumcircle of $\triangle A B C$ and has a radius twice that of the circumcircle of $\triangle A B C$ (including the boundary).
Similarly, points $Q$ and $R$ apply to similar conclusions.
Thus, the area of $\triangle P Q R$ is no greater than the area of an equilateral triangle with all three vertices on this larger circle.
In fact, we can choose the point $X$, which is the reflection of point $C$ about $A B$, as $P$; the point $Y$, which is the reflection of point $A$ about $B C$, as $Q$; and the point $Z$, which is the reflection of point $B$ about $C A$, as $R$. At this time, this largest area triangle can be divided into four triangles congruent to $\triangle A B C$. Therefore, the maximum area of $\triangle P Q R$ is 4.
(2) Let $M$ be the incenter of $\triangle A P B$.
Since $\angle A P B=60^{\circ}$,
Therefore, $\angle B A P+\angle A B P=120^{\circ}$.
So $\angle A M B=180^{\circ}-(\angle B A M+\angle A B M)$
$$
\begin{array}{l}
=180^{\circ}-\frac{1}{2}\left(\angle B A P+\angle A^{\prime} B P\right) \\
=180^{\circ}-\frac{1}{2} \times 120^{\circ}=120^{\circ} .
\end{array}
$$

Since $\angle A C B=60^{\circ}$,
Therefore, point $M$ lies on the circumcircle of $\triangle A B C$.
Similarly, the incenters of $\triangle B Q C$ and $\triangle C R A$ also lie on the circumcircle of $\triangle A B C$.
Since the triangle with the largest area and vertices on a given circle is an inscribed equilateral triangle,
the triangle with the largest area and vertices being the incenters of $\triangle A P B, \triangle B Q C, \triangle C R A$ is congruent to $\triangle A B C$, and thus the maximum area of the triangle with vertices being the incenters of $\triangle A P B, \triangle B Q C, \triangle C R A$ is 1.