Given circles $\odot O_{1}$ and $\odot O_{2}$ with radii $r$ and $R$ respectively, they are externally tangent at point $P$. A square $ABCD$ is inscribed in $\odot O_{1}$, and point $P$ lies on the minor arc $\overparen{AB}$. Tangents $AA_{1}$, $BB_{1}$, $CC_{1}$, and $DD_{1}$ are drawn from points $A$, $B$, $C$, and $D$ to $\odot O_{2}$, with the points of tangency being $A_{1}$, $B_{1}$, $C_{1}$, and $D_{1}$ respectively. Find the value of $\frac{AA_{1} + BB_{1}}{CC_{1} + DD_{1}}$.
Connect $A P$ and extend it to intersect $\odot \mathrm{O}_{2}$ at point $G$, connect $O_{1} O_{2}$ and extend it in both directions to intersect $\odot O_{1}$ and $\odot O_{2}$ at points $E$ and $F$ respectively. Then point $P$ is on the line segment $E F$, connect $A E$ and $G F$.
Since $E P$ and $F P$ are diameters of $\odot O_{1}$ and $\odot O_{2}$ respectively,
$\Rightarrow \angle E A P = \angle F G P = 90^{\circ}$
$\Rightarrow A E \parallel G F \Rightarrow \frac{G P}{A P} = \frac{P F}{P E}$
$\Rightarrow \frac{A G}{A P} = \frac{E F}{P E} \Rightarrow \frac{A G}{A P} = \frac{r + R}{r}$.
By the secant-secant theorem, we have $A A_{1}^{2} = A P \cdot A G$.
Therefore, $\frac{A A_{1}^{2}}{A P^{2}} = \frac{A G}{A P} = \frac{r + R}{r}$.
Let $t = \sqrt{\frac{r}{r + R}}$. Then $A P = A A_{1} t$.
Similarly, connect $B P$, $C P$, and $D P$ to get $B P = B B_{1} t$, $C P = C C_{1} t$, and $D P = D D_{1} t$.
Let $\angle A D P = \alpha$. Then
$$
\begin{array}{l}
\angle B D P = 45^{\circ} - \alpha, \angle C D P = 90^{\circ} - \alpha, \\
\angle P A D = 135^{\circ} - \alpha .
\end{array}
$$

By the Law of Sines, we have
$$
\begin{array}{l}
\frac{A P + B P}{C P + D P} = \frac{\sin \angle A D P + \sin \angle B D P}{\sin \angle C D P + \sin \angle P A D} \\
= \frac{\sin \alpha + \sin \left(45^{\circ} - \alpha\right)}{\sin \left(90^{\circ} - \alpha\right) + \sin \left(135^{\circ} - \alpha\right)} \\
= \frac{\sqrt{2} \sin \alpha + (\cos \alpha - \sin \alpha)}{\sqrt{2} \cos \alpha + (\sin \alpha + \cos \alpha)} \\
= \sqrt{2} - 1 .
\end{array}
$$

Therefore, $\frac{A A_{1} t + B B_{1} t}{C C_{1} t + D D_{1} t} = \sqrt{2} - 1$, which means
$$
\frac{A A_{1} + B B_{1}}{C C_{1} + D D_{1}} = \sqrt{2} - 1 \text{. }
$$
(Wang Mingli, Beijing Chemical University Affiliated High School, 100029)