Initially 161 points $P$, $Q$, $R$ are on the sides $BC$, $CA$, $AB$ of $\triangle ABC$ respectively, and they trisect the perimeter. Let the lengths of the three sides of $\triangle ABC$ be $a$, $b$, $c$. Find the maximum value of $S=AQ \cdot BP + BR \cdot CQ + CP \cdot AR$.
Solution: As shown in Figure 2, let $A Q=$
$$
\begin{array}{l}
x, B R=y, C P=z \text {. Let } t \\
=\frac{1}{3}(a+b+c) \text {, then } \\
S=x(t-y)+ \\
y(t-z)+ \\
z(t-x) \\
=(x+y+z) t-(x y+y z+z x) \text {. } \\
\end{array}
$$

Since $b=x+(t-z), c=y+(t-x)$, then
$$
y=x+c-t, z=x+t-b \text {. }
$$

Thus, $S=-3 x^{2}+(a+3 b-c) x+\left(t^{2}-2 b t+b c\right)$.
Since $S$ is a quadratic function of $x$, and it is easy to verify that
$$
\begin{array}{l}
0<\frac{1}{6}(a+3 b-c)<\min \{t, b\}, \\
0<\frac{1}{6}(b+3 c-a)<\min \{t, c\}, \\
0<\frac{1}{6}(c+3 a-b)<\min \{t, a\} .
\end{array}
$$

Therefore, $S$ reaches its maximum value when and only when $x=\frac{1}{6}(a+3 b-c)$. At this time, $y=\frac{1}{6}(b+3 c-a), z=\frac{1}{6}(c+3 a-b)$.
Thus, $S_{\max }=\frac{1}{6} \sum a^{2}+\frac{1}{36}\left(\sum a\right)^{2}$.
He Bin, Gucheng County No. 3 High School, Hubei Province, 441700.