In Rt $\triangle ABC$, $\angle ACB=90^{\circ}$, $BC=a$, $CA=b$. Construct $CD$ perpendicular to $AB$, with the foot of the perpendicular at $D$. Let $G$ be the centroid of $\triangle ABC$, and draw a line through point $G$ that intersects $AC$, $CD$, and $BC$ at points $P$, $T$, and $Q$ respectively. If $PG=TQ$, find the length of $CT$ (expressed in terms of $a$ and $b$), and determine the range of $\frac{b}{a}$.

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Solution: As shown in Figure 4, let $M$ be the midpoint of $AB$, then
$$
\begin{array}{l}
\angle BCD = \angle CAM \\
= \angle ACM = \alpha.
\end{array}
$$

Let $\angle CPQ = \theta$, then
$$
\angle CQP = 90^\circ - \theta.
$$

Obviously, $CG = \frac{2}{3} CM = \frac{2}{3} \times \frac{1}{2} AB = \frac{1}{3} \sqrt{a^2 + b^2}$,
$$
\begin{array}{l}
\angle CGT = \theta + \alpha, \\
\angle CTG = \alpha + 90^\circ - \theta = 90^\circ - (\theta - \alpha).
\end{array}
$$

In $\triangle PCG$, $\triangle QCT$, and $\triangle CGT$, by the Law of Sines, we have
$$
\begin{array}{l}
\frac{PG}{\sin \alpha} = \frac{CG}{\sin \theta}, \\
\frac{TQ}{\sin \alpha} = \frac{CT}{\sin (90^\circ - \theta)} = \frac{CT}{\cos \theta}, \\
\frac{GC}{CT} = \frac{\sin (\alpha + 90^\circ - \theta)}{\sin (\theta + \alpha)} = \frac{\cos (\theta - \alpha)}{\sin (\theta + \alpha)}.
\end{array}
$$

From (1) $\div$ (2) and (3), we get
$$
1 = \frac{PG}{TQ} = \frac{GC}{CT} \cdot \frac{\cos \theta}{\sin \theta} = \frac{\cos (\theta - \alpha) \cdot \cos \theta}{\sin (\theta + \alpha) \cdot \sin \theta},
$$

which means $\cos (\theta - \alpha) \cdot \cos \theta = \sin (\theta + \alpha) \cdot \sin \theta$.
Thus, $\frac{1}{2}[\cos (2\theta - \alpha) + \cos (-\alpha)]$
$$
= -\frac{1}{2}[\cos (2\theta + \alpha) - \cos \alpha],
$$

which implies $\cos (2\theta - \alpha) = -\cos (2\theta + \alpha)$.
Therefore, $\cos 2\theta \cdot \cos \alpha = -\cos 2\theta \cdot \cos \alpha$.
Hence, $\cos 2\theta = -\cos 2\theta$, which means $\cos 2\theta = 0$.
Thus, $2\theta = 90^\circ$, $\theta = 45^\circ$.
From equations (3), (4), and (5), we get
$$
\begin{array}{l}
CT = \frac{\sin (\theta + \alpha)}{\cos (\theta - \alpha)} \cdot GC = \frac{\cos \theta}{\sin \theta} \cdot GC \\
= GC = \frac{\sqrt{a^2 + b^2}}{3}.
\end{array}
$$

In $\triangle QCT$, by the Law of Sines, we have
$$
\frac{CQ}{CT} = \frac{\sin (\alpha + 90^\circ - \theta)}{\sin (90^\circ - \theta)} = \frac{\cos (\theta - \alpha)}{\cos \theta}.
$$

Therefore, $CQ = CT \cdot \frac{\cos (\theta - \alpha)}{\cos \theta}$
$$
\begin{array}{l}
= CT \cdot \frac{\cos \theta \cdot \cos \alpha + \sin \theta \cdot \sin \alpha}{\cos \theta} \\
= CT (\cos \alpha + \sin \alpha) \\
= \frac{1}{3} \sqrt{a^2 + b^2} \left( \frac{b}{\sqrt{a^2 + b^2}} + \frac{a}{\sqrt{a^2 + b^2}} \right) \\
= \frac{a + b}{3}.
\end{array}
$$

Similarly, $CP = \frac{a + b}{3}$.
Let $\left\{\begin{array}{l}\frac{a + b}{3} \leq a, \\ \frac{a + b}{3} \leq b.\end{array}\right.$ Solving this, we get $\frac{1}{2} \leq \frac{b}{a} \leq 2$.
In summary, $CT = \frac{\sqrt{a^2 + b^2}}{3}$, and the range of $\frac{b}{a}$ is $\left[\frac{1}{2}, 2\right]$.
(Wu Weizhao, School of Mathematics and Information Science, Guangzhou University, 510006)