In $\triangle A B C$, $A B=A C, M 、 N$ are the midpoints of sides $A B 、 A C$ respectively, and $P$ is an intersection point of line $M N$ with the circumcircle of $\triangle A B C$. Find the range of $w=\frac{P B}{P C}+\frac{P C}{P B}$.
Solution: As shown in Figure 2, let $Q$ be the other intersection point of line $MN$ with the circumcircle of $\triangle ABC$, and connect $CQ$.

Since $AB = AC$, and $M, N$ are the midpoints of sides $AB, AC$ respectively, it follows that quadrilateral $PBCQ$ is an isosceles trapezoid, hence
$$
QC = PB, \quad NQ = PM.
$$

Let $AB = 2b, BC = 2a$, then $MN = a$.
Suppose $PM = x$, then $NQ = x$.
By the intersecting chords theorem, we have $MP \cdot MQ = MA \cdot MB$, which gives $x(x + a) = b^2$.
It is easy to see that $CA$ bisects $\angle PCQ$, thus
$$
\frac{PB}{PC} = \frac{QC}{PC} = \frac{NQ}{PN} = \frac{x}{x + a}.
$$

Therefore, $w = \frac{PB}{PC} + \frac{PC}{PB} = \frac{x}{x + a} + \frac{x + a}{x}$
$$
\begin{array}{l}
= \frac{x^2 + (x + a)^2}{x(x + a)} = \frac{2x(x + a) + a^2}{x(x + a)} \\
= \frac{2b^2 + a^2}{b^2}.
\end{array}
$$

When $\angle BAC < 60^\circ$, i.e., $a > b$, we have $3 < w < 6$.
(Tian Yonghai, Shuozhou Education Institute, Heilongjiang Province, 152054)