Question 179, As shown in the figure, in the Cartesian coordinate system, the ellipse $\Gamma: \frac{x^{2}}{2}+y^{2}=1$ has left and right foci $F_{1}$ and $F_{2}$, respectively. Let $P$ be a point on $\Gamma$ in the first quadrant, and the extensions of $P F_{1}$ and $P F_{2}$ intersect $\Gamma$ at points $Q_{1}$ and $Q_{2}$, respectively. Let $r_{1}$ and $r_{2}$ be the inradii of $\triangle P F_{1} Q_{2}$ and $\triangle P F_{2} Q_{1}$, respectively. Find the maximum value of $r_{1}-r_{2}$.
Question 179, Solution: According to the conditions, $F_{1}(-1,0), F_{2}(1,0)$, and the perimeters of $\triangle P F_{1} Q_{2}$ and $\triangle P F_{2} Q_{1}$ are both $\mathrm{l}=4 \mathrm{a}=4 \sqrt{2}$. Therefore, we know
$$
r_{1}-r_{2}=\frac{2 S_{\Delta P F_{1} Q_{2}}}{1}-\frac{2 S_{\Delta P P_{2} Q_{1}}}{1}=\frac{2 S_{\Delta F_{1} F_{2} Q_{2}}-2 S_{\Delta F_{1} F_{2} Q_{1}}}{1}=\frac{\left(\left|y_{Q_{2}}\right|-\left|y_{Q_{1}}\right|\right)\left|F_{1} F_{2}\right|}{1}=\frac{y_{Q_{1}}-y_{Q_{2}}}{2 \sqrt{2}}
$$

Let $P(\sqrt{2} \cos \alpha, \sin \alpha), \alpha \in\left(0, \frac{\pi}{2}\right)$. Then $P F_{1}: y=\frac{\sin \alpha}{\sqrt{2} \cos \alpha+1} \cdot(x+1), \mathrm{PF}_{2}: y=$ $\frac{\sin \alpha}{\sqrt{2} \cos \alpha-1} \cdot(x-1)$.
Solving the system $\left\{\begin{array}{c}y=\frac{\sin \alpha}{\sqrt{2} \cos \alpha+1} \cdot(x+1) \\ \frac{x^{2}}{2}+y^{2}=1\end{array}\right.$, we get
$$
\frac{\left(\frac{\sqrt{2} \cos \alpha+1}{\sin \alpha} y-1\right)^{2}}{2}+y^{2}=1 \Rightarrow\left(\frac{\left(\frac{\sqrt{2} \cos \alpha+1}{\sin \alpha}\right)^{2}}{2}+1\right) y^{2}-\frac{\sqrt{2} \cos \alpha+1}{\sin \alpha} \cdot y-\frac{1}{2}=0
$$
$\frac{1}{2}$
One root of this equation is $\mathrm{y}_{\mathrm{P}}=\sin \alpha$, and the other root is $\mathrm{y}_{Q_{1}}$. By Vieta's formulas, we have $\mathrm{y}_{\mathrm{Q}_{1}}=\frac{\frac{\left(\frac{\sqrt{2} \cos \alpha+1}{\sin \alpha}\right)_{+1}}{2}}{\sin \alpha}=$ $\frac{-\sin \alpha}{3+2 \sqrt{2} \cos \alpha}{ }^{\circ}$
Similarly, solving the system $\left\{\begin{array}{c}y=\frac{\sin \alpha}{\sqrt{2} \cos \alpha-1} \cdot(x-1) \\ \frac{x^{2}}{2}+y^{2}=1\end{array}\right.$, we get
$$
\frac{\left(\frac{\sqrt{2} \cos \alpha-1}{\sin \alpha} y+1\right)^{2}}{2}+y^{2}=1 \Rightarrow\left(\frac{\left(\frac{\sqrt{2} \cos \alpha-1}{\sin \alpha}\right)^{2}}{2}+1\right) y^{2}+\frac{\sqrt{2} \cos \alpha-1}{\sin \alpha} \cdot y-\frac{1}{2}=0
$$

One root of this equation is $\mathrm{y}_{\mathrm{P}}=\sin \alpha$, and the other root is $\mathrm{y}_{\mathrm{Q}_{2}}$. By Vieta's formulas, we have $\mathrm{y}_{\mathrm{Q}_{2}}=\frac{\frac{\left(\frac{\sqrt{2} \cos \alpha-1}{\sin )^{2}}\right.}{2}+1}{\sin \alpha}=$ $\frac{-\sin \alpha}{3-2 \sqrt{2} \cos \alpha}$.
Thus, we have
$$
\begin{array}{l}
\mathrm{y}_{\mathrm{Q}_{1}}-\mathrm{y}_{\mathrm{Q}_{2}}=\frac{-\sin \alpha}{3+2 \sqrt{2} \cos \alpha}-\frac{-\sin \alpha}{3-2 \sqrt{2} \cos \alpha}=\frac{4 \sqrt{2} \sin \alpha \cos \alpha}{9-8 \cos ^{2} \alpha}=\frac{2 \sqrt{2} \sin 2 \alpha}{5-4 \cos 2 \alpha} \\
=\frac{2 \sqrt{2} \sin 2 \alpha}{\sqrt{\left(4^{2}+3^{2}\right)\left(\cos ^{2} 2 \alpha+\sin ^{2} 2 \alpha\right)}-4 \cos 2 \alpha} \leq \frac{2 \sqrt{2} \sin 2 \alpha}{(4 \cos 2 \alpha+3 \sin 2 \alpha)-4 \cos 2 \alpha}=\frac{2 \sqrt{2}}{3}
\end{array}
$$

When $\cos 2 \alpha=\frac{4}{5}, \sin 2 \alpha=\frac{3}{5}$, the equality holds.
Thus, the maximum value of $r_{1}-r_{2}=\frac{y_{Q_{1}}-y_{Q_{2}}}{2 \sqrt{2}}$ is $\frac{\frac{2 \sqrt{2}}{3}}{2 \sqrt{2}}=\frac{1}{3}$.