In an isosceles right $\triangle ABC$, $\angle C=90^{\circ}, AC$ $=4, P$ is a point inside the triangle. If the inradii of $\triangle PBC$, $\triangle PAB$, and $\triangle PAC$ are all $r$, find the value of $r$.

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First, we can prove: point $P$ is on the bisector of $\angle A C B$. Let the incenters of $\triangle P A B$ and $\triangle P B C$ be $M$ and $N$, respectively. As shown in Figure 2, draw $M D \perp A B$ at point $D$ and $N E \perp B C$ at point $E$.

Given $A C=4$, we know $A B=4 \sqrt{2}$.
Therefore, $B D=2 \sqrt{2}$.
Let $\angle D B M=\alpha, \angle N B E=\beta, \angle N C E=\gamma=\frac{\pi}{8}$.
Then $\alpha+\beta=\frac{\pi}{8}$.
Since $C E=r \cot \frac{\pi}{8}$, we have
$$
B E=4-r \cot \frac{\pi}{8}.
$$

Thus, $\tan \beta=\frac{r}{B E}=\frac{r}{4-r \cot \frac{\pi}{8}}$.
Also, $\tan \alpha=\frac{r}{2 \sqrt{2}}$,
$$
\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}
$$
$$
=\frac{\frac{\sqrt{2} r}{4}+\frac{r}{4-r \cot \frac{\pi}{8}}}{1-\frac{\sqrt{2} r}{4} \times \frac{r}{4-r \cot \frac{\pi}{8}}}
$$
$$
=\frac{\frac{\sqrt{2} r}{4}\left(4-r \cot \frac{\pi}{8}\right)+r}{4-r \cot \frac{\pi}{8}-\frac{\sqrt{2} r^{2}}{4}}.
$$

Substituting $\tan \frac{\pi}{8}=\sqrt{2}-1, \cot \frac{\pi}{8}=\sqrt{2}+1$ into the above equation and simplifying, we get
$$
r^{2}-2(\sqrt{2}+1) r+8-4 \sqrt{2}=0.
$$

Solving this, we get $r=\sqrt{2}+1-\sqrt{6 \sqrt{2}-5}$, or $r=\sqrt{2}+1+\sqrt{6 \sqrt{2}-5}$ (discard the latter).
(Xu Shenglin, Editorial Department of "Mathematical Communications", Central China Normal University, 430079)